2/4 Spectral Geometric Unification
Chamseddine, Ali
Classification of finite spaces and basis for geometric unification.
 Let's get right into the song. So last week I introduced the idea that non-committal geometry can be taken as a basis to unify all fundamental interactions. And so what I did is that I defined what we mean by non-committal space. I gave the definitions. And so with certain properties, and just as I remind you, just set the spectral triple and I defined these five quantities and the relation among them. And then the question, of course, is if a non-committal space is the basis to unify the interactions, then the question is what kind of non-committal space should we take, you know? And as a starter, I made an assumption, and the assumption I made is that space time r is a non-committal space which is formed as a product of continuous four-dimensional space times some finite space. Now four-dimensional spaces are well understood, and then I went to see essentially what one should take for the finite space, you know, what kind of space should one take. And so for that, started by classifying spaces which fit certain physical criteria. And the physical criteria that we need, and the following, that, you know, the Hilbert space, which would be space of, you know, fermions essentially for us, would include all the fermions, and in this case, actually, the fermions just satisfy certain properties. And essentially we said we have fermions, but because actually of the presence of what we call the reality operator, one really requires essentially what we would like to require essentially, that the fermions are real, which in physical language means Majorana. And in this respect, of course, actually the presence also of the gamma operator, which is a current operator, would allow us always to impose this condition. So what we really have, we know, we say we have current spinors. Now then, of course, actually if you have to on J on psi, you are really getting to get something like the conjugate spinner. And the question is that is this an independent set of fields? If it's an independent set of fields, then you really immediately face the problem that all your particles will be doubled. You are really going to get, for every particle we know, a mirror particle. And you know, sometimes it's called the mirror fermions problem or the mirror doubling problem. And usually it's not easy to solve this problem. Now we are really working in Euclidean space. And the issue, of course, is that if you really go to Euclidean space and you said I would like, sorry, but the physical space, of course, is Minkowski. And if you would like to impose both conditions, the J psi is psi, then this actually would impose certain conditions on the properties of the space. As I have shown last time, that would impose a condition what we know as the k-o dimension. So in Minkowski space, you really need to impose both conditions. The J psi is psi and gamma psi is psi, and this would imply actually that the k-o dimension of this space is equal to zero. Which implies actually that J squared, which is epsilon, should be one. J d is, I think, epsilon prime d J. And I need one more sign. J gamma is epsilon dot the prime gamma J. And d gamma is minus gamma J. So these are the properties that we need. And essentially, it really forces, now we know actually that in Minkowski, you have four dimensional space with signatures, say, minus plus plus plus or the opposite. And these would somehow cancel each other. And then you are left with k-o dimension two, which means that the internal space or the finite dimension space should be dimension six, k-o dimension six. So that's actually forced the finite space to have certain properties and k-o dimension six. Then you need certain sets of epsilon, for example, epsilon is really minus one. I forgot actually, I think, epsilon prime is equal plus one and epsilon prime is minus one. Something like this. So then actually one goes ahead and try to understand these finite spaces. And what one discovers, and I would go through, not through the proof, but through the realization of the proof, is that one discovers that the algebras of the finite space are essentially what one finds, that the complexification of the algebra. So essentially here, one assumes that the finite space is characterized by A f, H f, d f, gamma f, and g f. Everything is for finite. And one discovers that the complex extension of this algebra, if you take the center of it, there are only two possibilities which are either C or C plus C. And this actually is inconsistent with k-o dimension six, which leaves us only one possibility that A C is C plus C, and really corresponds to an algebra which is a variant of, you know, let me call it M N of C plus M N of C. In addition, when we imposed certain isometry on the first algebra, and this is a point actually we don't really, you know, understand, you know, we cannot give it a mathematical characterization why it should be like that, but it is like that. And the other possibilities are not relevant. And this really reduces this instead of matrices on the complex numbers. You really take matrices over the catorials for the first, and in which actually N is 2A in this case. So N is even. And also, because one would like to impose the chirality operator on this algebra and elements of the algebra, for example, gamma A equal A gamma. So in this case, what happened is that this M of 2A of H would split into M A of H plus M A of H, for example, plus M N of C or 2A of C. So the upshot of this is that really the first non-trivial example you can do is the following, that A should be one in this case. Sorry, A is, yeah, this is a matrix, you know, M A of H. This should be 4A or something. N is, let me see, M is. N is 2A, yeah. So essentially, yeah, so essentially this, yeah. So essentially, really the first non-trivial example is M1 of H or H plus H. Okay, let me write it like this, and this actually is like 4A. And in this case, the first example would be H plus H for the first algebra plus M4 of C for the second. Of course, other algebras are possible. But Ali, what is important to say here is that you have the gamma, which is the diagonal matrix. So, on the first, right? For M2 of H. Right. So it takes the even part. It's the even part of M2 of H, which is 8 plus H. Yeah, okay, so I have to say, okay, this I would say M2 of H even, and then it really, at least for the first non-trivial case, it belongs to H plus H. Now it's up to this point, you know, I went last time. And the only thing we're, you know, the assumptions I made are there. So at one point, one did impose on isometry. If one didn't impose on isometry, then you are really going to get, instead of, you know, H plus H, you're going to get M2C plus M2C. And one really can show that these cases would not be physical, but one cannot rule them immediately, unfortunately. What it is, is that if you have a real algebra, then you have three cases. The symplectic case, the pteric case, and the also. Exactly, there are three. And you're okay, we choose, you know, the symplectic, you know, case. So this is a choice. This is something which is not imposed by the... But you could have chosen once. We could have chosen, yeah. Symplectic, symplectic, or... We could have chosen this section. So this is the really... the other is actually not relevant because they would really give you real algebra. For instead of H plus H, you can take M2C plus M2C. What would it really give essentially? It gives you, you know... It gives you an extra U1, actually, you know, which is anomalous. So one really can work it out. It gives... it's a little variation. Everything almost the same except this little variation. It gives you an extra vector. An extra vector which is anomalous, you know. Maybe there is an extra vector lurking in the spectrum, but, you know, people are talking about, you know... Recently, there was some papers in which they said they observed something which deviates from the standard model. And, you know, some people said maybe it's an extra vector. So who knows you? Anyway, but this is actually the most symmetric possibility. The only assumption I made also that left action... This is the left action. And then this is, say, the right action. And then we said this is J B star J inverse. And we said... And this actually is defined to be the opposite. So this is one of the axioms that left action and right action commute with each other. So this actually has been used. This property really has been used up to this point. Now, from here, actually, one needs further assumption. And, you know, I will discuss both possibilities. In that... And this is called order one condition. And now, actually, we understand this condition. It's really a condition on the linearity of the connection. If one drops this condition, then what happened is that the connection would really get quadratic terms and that really can work out the possibility when this is not so, when this is not satisfied. But this I would leave to the last lecture to allow, because actually in non-convenient geometry, there are examples where this condition is not satisfied. And you know, this is what the quantum spheres. Okay? Anyway, so here it will have a really big advantage that the connections are linear. Everything is very nice. Everything, you know, falls into place. Okay? What happens if this is not satisfied? You know, this would lead us to grand unified theories and Patti-Sellem models and things like that. But this will come much later. Okay. So what are the consequences of this condition that's satisfied? Okay. So I will assume, actually, that this is satisfied. And on top of it, actually, now there is one more condition. Is the question that if we take the commutator of the Dirac operator with the center of the algebra, is it zero or not zero, actually? This is a question. Well, what we have shown is that there are two possibilities. If it is zero, then this symmetry is really intact. And if this symmetry is intact, one can show, actually, that you are really going to get, when one work out a model, which I'm really going to do a little bit later, is that you really get SU4 color, which really corresponds to this M4C, as a symmetry of the quantum. So essentially, in this case, you really get a higher color symmetry, which is not observed in nature. And on top, actually, you'll find out that the neutrinos are all massless. They are exactly massless, which now we know that it's not true. So from the physics point of view, we know, actually, that the three must be different than zero. Because if it is zero, then you are already in contradiction with the experiment that neutrinos will become all massless, which would have been OK until 1995 or something like that, because people didn't know, were not sure, actually, whether neutrinos are massive or not. Because every this tiny mass, there are only limits on the masses of the neutrino. Now we know, actually, that they are really massive, but the mass is really extremely small. But that they are massive is well established. Now what happened, actually, if this difference is zero? Now the beauty of it is that if this is different than zero, and for this condition to be satisfied, the Dirac operator must have the following form. And so in this case, actually, instead of telling you in description what really happens, I would start being more concrete. I will tell you exactly what are the representations, what goes on, so that you have an idea. That this actually, as I would say, field is not simply talk, you know, you really can go and compute everything unambiguously, not much ambiguity, and with all the details. So, all right, so essentially let me examine this in a minute. But I really can tell you that the fact that this is non-zero really gives a unique possibility that you really can give the neutrinos a Majoranum mass, and it will explain, actually, why the masses of the neutrinos are very small. It really, it's an example of what was known as the CISO mechanism. So here it really comes out naturally, and it solves the problem in a really very nice way. So, let's see how things would look like if I wanted to go ahead and do some computations, you know, what should I do? So this is really my starting point. I'm really going to take H, the algebra A, to be H plus H plus of course. And this, remember, almost uniquely we were led to this. It didn't come out of the sky, you know, out of classification. This was the first non-trivial result that we have. Okay, so essentially if I would like actually to represent an element of A, how would I write that? Then I am going to write it in matrix form, you know. I'm going to show you that everything that we do is really matrices, essentially. We really can use tensorial notation if one would like to simplify his life. And you know, recently, you know, everything that we do can be put on the computer, so in the end one doesn't do much calculation. You just write it on Mathematica and press a button and it gives you the answer. So, you know, it's, all right. So I'm really going to use, you know, index notation. So essentially here what I'm really going to say, all right. So in order to tell you what my notation, I really have to tell you what are my spinners. Okay, so the spinner actually, as you see, it belongs to this guy. So what I'm really going to write, I'm going to write as psi alpha i, where this alpha is, would see the H plus H and the i would see the 4. So obviously i is 1 to 4 and the alpha is also 1 to 4, but it's a different form. It's H plus H. Now, and then, of course, actually I'm going to write this as, you know, a dot a, where this has two indices, like 1 and 2, and this is 1 dot 2 dot. So this actually tells me that it's a doublet under the catorians, you know, doublet left, doublet right. And what we write actually, we write as H right plus H left, because we have already graded the algebra. And in this respect, my spinners would be psi a dot i and psi i. Okay. So how many spinners I have? I have 16 spinners. And of course, we have also the conjugates, but the conjugates, we're going to see that they're not the independent spinners, because the thermion doublet problem is solved through, you know, the Majorana condition. We are not really going to say Geppsi equal to psi. The problem in Euclidean space is really solved in a very clever way by, you know, in the path integral formalism that only chiral thermions get integrated and they give square root of determinant instead of determinant. So it's really solved in really very subtle way. You know, this is a nice discovery. You get the fulfillment. You know, I'll talk. Anyway, so you see actually now, you know, without redoing any work, I have already obtained the correct classifier. First, actually, we already predicted that you have 16 fermions, which is exactly per family of course, which is exactly what we have. We will also actually find, let me, you know, let me write this one of this. The i. The i eventually, you know, when this condition, the d with z of a different than 0, will also be broken into 1 and i, where i is 1, 2, 3. So that actually will be the color index and this is the lepton index. So this is lepton and this is the quark. And then you can see that the lepton is the fourth color. It comes out of the fourth color. And in this respect, you're going to get psi 1.i, psi 2. Sorry, psi 1.1, psi 2.1, psi a, sorry, 1.2. And then I have psi i, which I'm going to leave it as psi a1. Then I have psi 1.i, psi 2.i, and then I have psi a.1, psi a. What's in this? You know, remember, this is a dot means, when you see dot, dot means right. So this would be what I'm going to call neutrino, right? Why give it the name? Because if it comes out like that, the quantum numbers tell me that this will be a neutral guy. But this comes out. You know, I name it according to the usual naming and it would agree. And then, of course, this is a, right? And this, of course, is upright. This is down right. Similarly, psi a1 is the doublet, which is a new a, a new e. It's a doublet. You know, a is 1, 2, left. And this is up, down, left. OK? So you see, actually, without much work, we obtain the correct representations of the particle spectrum of all the fermions. We know which representations they are. So one really can write that the 16 in this case is a 4, 4, 4. And this 4 is decomposed into 2 plus 2. And this 4 is decomposed into 1 plus 3. And then this is 2 right plus 2 left. And this is everything that we see belong to that representation. Everything falls into place. OK. So now, so these are my 16 spinners. But remember, actually, you know, I have the j psi. And so my spinners will be psi alpha i and then psi alpha i conjugate, which I'm really going to go psi alpha prime i prime. They're not independent, but they are related through the j. OK? What does it mean, actually? It means if I would like now to write the element of the algebra, you know, I can write it as follows. I can write it as this one, you know. Remember this belongs to the second guy to M4 of C. So it is something like this. And this is delta alpha beta. And this is like yij. So you see, actually, how an element of the algebra is satisfied. And then you can form the b opposite, which is jb star or j inverse. And you discover, actually, what happened is that here you are going to get the y times 1. You are going to get yij, essentially, transposed into 1. And here you are going to get the x alpha beta into delta ij. And if you look at, you know, this element and this element, obviously, they commute with each other y because then this is don't talk to each other. This is in the first algebra. Well, this is now in the second. And they don't talk to each other. So essentially, jb star. Yeah, absolutely. Right. And this is actually, I wrote it as transposed, but, you know, a star is, OK. So essentially, we can write it in this form. So the first a b opposite is satisfied automatically. Now where do I go from here? Then I have to satisfy this condition. So this actually is a condition on the d and the algebra, both, you know. Because what happened in the following? You try to satisfy this condition, you work out the matrix representation, and you obtain actually. I mean, you mean the commutation with the center, right? So you are right here. This actually, you are right. Yeah, exactly. Yeah, it's exactly. Yeah. This is the difference here. This is the difference here. OK. So this actually, we have to assume it's different than zero. And then actually, this really would require the form of the direct operator to be as follows. You can show that you are going to get something like this. Let me call it, I don't know, let me call it dab or something. And here we are going to get the bar. Here we are going to get something. And then one can show through a little algebra. Actually, I learned it through abstract proofs, which you know, beyond me. But through a little algebra, one can show actually that the only operator that satisfies in the following, that here you can have only one non-zero entry. Only one. Anything else would not satisfy this condition. OK. Of course, actually, this, together with that, we have also to make sure that dA would be opposite is equal to zero, putting the two firsts actually. First we know that here we must have a non-zero element. The question is that the whole thing could be non-zero. But because of this condition, the first condition, you discover actually that this is possible if and only if you have only one non-zero entry. What does this mean? Actually, this one non-zero entry. Let me write actually the expression for you. And let me see how does it look like. You know, I have, remember, when I write Dirac equation, what do I write? I write psi star deep psi, or I can write it this way, you know, psi deep psi, OK, in a product. So this is my, say, Fermi reaction. And this Fermi reaction, if I expand, and then I just look at this term, what does this mean? This actually, if I look at that classification, it really tells me that you really have an element which is this one. If I have more, it would mean that, for example, the up, the down, everybody will start getting Majorana mass, but you know, you're getting a Majorana mass would break charge, would break the charge conservation, because in this case they would not have zero charge. So the only term that you're in physics wise that you can write is a term which is reneutral. It must be neutral, otherwise you destroy charge conservation. And it's not an accident, actually, then that you can have only one nonzero term, and this one nonzero term would really correspond to the mass, Majorana mass of the right-handed neutrino. Now, I'll explain later, actually, that why this is really extremely important in order to solve, it to explain, essentially, why the neutrino masses are so small, OK? So this actually is a result. It tells us that, you know, in order to agree with the observation, the Dirac operator should take this form. It has only one nonzero entry, and the rest are zero. This d and this d bar are correlated. They are, you know, one in the complex conjugate of each other. And the only thing I have to find is that what's the form of the Dirac operator in the finite space, OK? So what shape does it take? You know, the whole idea, the things really are really constrained, you know, because in a way, we have to explain why the standard model. This actually, you know, why the standard model. So if the hypothesis that everything comes from the geometry, we must really arrive at this answer in an almost unique way. Maybe one day one would have fine and say, OK, here it comes, this is without any ambiguity. But you know, there's always a possibility that you have to put some physics, because if you don't put any physics, in principle, you can get anything, right? Because, you know, if I don't want to explain nature, then, you know, I can start with the smaller space or bigger space, OK? You can start with too small, because if you already assume that it's four dimensional continuum times finite space, OK, and the finite space has k dimension six, and then you make a very minor hypothesis on that space, then you are already led to the sixteenth, so it cannot be too small, you know? This is the first example, but then you cannot rule out why it's not bigger. Oh, no, of course you cannot swap. You know, the question is that can I say uniquely, you know, this is the only thing I would ever get, you know? But this is the first possibility that one gets. OK, so now actually, where do we go from here? And what we do is the following. So we would like actually to see what type of Dirac operator can we have, this DAB. Now remember, actually, the D satisfies many properties. I have written them up, so you really have to show folks, you have to verify that what the area that G da is, Gd, still I'm talking about the finite space, remember, is equal to dj and gamma j equal to minus j gamma and d gamma equal to minus gamma g. So these properties, I have to satisfy, and so if you do that, you'll immediately discover actually that the Dirac operator of the finite space really takes simple form. Remember, you know, let me write it in terms of new right, E right, because I don't want to write one dot and two dot, because then it's more physical. And in your right, E right, and then I have, say, new left, E left is a doublet. And then up, right, down, right, and up left, down left. Okay? This is really a doublet, but these are really singlets. Okay? Why is it singlet? Because actually, this sigma really breaks the algebra from, you know, for this condition and the other condition to be satisfied. And then you put the sigma. What you discover in the following, that this algebra, h plus h plus m4 of c is really broken. And obviously, it must be broken, because you put the sigma, remember. This is like one, and this is the i, and this really breaks the color index. It breaks the color index into one and the three. And in this respect, the m4 of c is really broken into c plus m3 of c. However, because that condition dA, b opposite is equal to zero, it really, what happened, and then it breaks, actually. This is also broken, you know, because this is one index. We'll talk this one index. And then, essentially, this is a c plus c. It breaks, it breaks, and the three are not independent. What one discover, actually, that the element of the algebra A takes this form, in this way. So you are going to get, you know, x, x bar. This is my first h. These are my first ketonion. In other words, actually, remember, a ketonion takes the form like alpha beta alpha bar minus beta bar. It tells me that in the first h, it must be of this form. Now here, what do I have? I have a ketonion. So this is my first four, and this is my bottom four. And in the bottom four, what do we really have? We discover, actually, that this is x, and of course, this is zero, and this is m, where m is the element of m3 of c. In other words, actually, what we discovered is that the three c, the c here, the c here, the c here are not independent, and they are essentially one c. And the finite algebra of the finite space really breaks down to c plus h plus m3 of c, which is really a symmetry of a central model. It really comes out of the requirement that there is a Majorana mass for the neutrino. No, this really has drastic consequences. OK, so with this action, the element of the algebra, I go with the d, and then we can really classify the d, satisfying all these properties. And I want to cut the story short. What happened is that the only non-vanishing elements of the d. There is one more requirement that we have to put in order to really completely single out the d, and that the d commutes with, no, actually, yeah. d commutes with the element, which is lambda lambda bar lambda lambda bar in here. If you do that, then you discover actually that only the non-ucala coupling of the d-operator are non-zero, and really we have the d-operator completely singled out. If I want to write it down, I can write it down. Let me write it down, actually. So how does it look like? When you zeroes, I think here you are going to get k nu, and here you are going to get ke, which is actually the yocawa coupling of the neutrino and the yocawa coupling of the e. And similarly here, you are going to get, this is new left, e left, and this is k nu right, ke right, like that. And similarly for the quarts and the only off diagonal elements, you are going to get. And here also. So this is k up, k down, and diagonal form. So in other words, actually, after some matrix algebra, one would know exactly what the d-operator is. So what do we do with this now? What we do with this is the following. Having determined that. Exactly. So you know, many, you'll get many, essentially a point that you get many zeros. By, you know, gamma d minus d gamma, I tell you all diagonal elements are zero, you only get non-diagonal elements, and so on. So it's little matrix algebra, nothing much to it. And however, once one is finished, one can proceed as follows. Note that we have not really talked about dynamics or anything, we just talked about classifications and representations, and satisfying the properties of the uncompetitive space. And you are led almost uniquely, actually. What do I call that of the standard model? Now, remember now, I said that the Dirac operator of the full non-competitive space is equal to the Dirac operator of the manifold cross one plus gamma five cross df. We have been, what I was talking about up to now is the df. We know now exactly how the df looks like. dm, of course, we know because it is the Dirac operator of the usual, say, remaining manifold. And similarly, we can say a equal a of the manifold cross af and j. Everything is a product, actually. Don't have to write it down. All right? Now, remember, actually, here, when all starts with what's non-fluxuating d. So we start with a non-fluxuating d. Say, let's start with a d which corresponds to, say, some flat space. Starting from a d from a flat space, for example, you can really start with gamma mu d mu like that, actually, for the d. And we know, actually, that you really can generate the metric of the dm out of fluctuations. These are the outer fluctuations. And this is easy, actually. What we have to do is that you say this is e mu a gamma a. You make this as a function of x. You discover that you have to add a spin connection in order to make the thing, in order to make the thing coherent. And so that, actually, is easy to do. So here, we have seen last time that the d always go into dA, where dA is a d plus a plus j. But that's a good for the global. Yeah? What you have written there is for the global. Here is the global. Here is the n-arithomorphism, no? No, but I mean, if you want to do the fluctuations, you do that for everybody, of course. I do for you, exactly. You know, we get both n-arithomorphism. I'm trying to say that if you'd like to get the gravity, you have to do the outer. And you want to get the other, the a. Yeah, but these are enough fluctuations by the physicians, you know. Exactly. So they are one form in this language. And it is adb. This looks, actually, very simple. But remember, actually, the a we have written is what? The a we have written was a 32 by 32 matrix. The d was a 32 by 32 matrix. Tensor do the Clifford-Dellscherprin. So you're talking about. Now, in addition, actually, we said everything we know is really for one family. And I have written here, for example, k and k, I have written as a number. But in reality, this k is a 3 by 3 matrix in generation space. So you mean it's a 96 by 36? So it's 396. So 32 times 32. And tensor do the Clifford-Dellscherprin. So it's 3, 384 by 384 matrix. This is actually why it's much easier instead of writing matrices because they look huge to put in tensor notation. Because in tensor notation, you just put an index and you don't care how much it runs because you know where does it sum. So it becomes really efficient to do it. Of course, we didn't do it like that in the beginning because for years, you have to struggle with all these matrices to see what acts. And it took some time, actually. This was progressing developments. All right. So this actually here. You say, all right, I would like to compute this guy. How does it look like? Remember? So I will give you an example. The reason I want to show you that things come out, that it's not that you say, oh, the gauge fields come out, the Higgs field come out. And people think that we are cheating because we know the answer. We put it in and then it's not true. Everything comes out in a very well-defined way. We really cannot cheat in any way whatsoever. So I can say, OK, AB, and then you can put indices like AA prime, and then you can do DB, like AC, CB, and so on. So then I can look, actually, at all the matrix elements AB. And how do they look like? So I say, OK, how do they look like? It's the following. I go back to my representation. And then I'm really going to get an element here, like 1 dot, 1, 1 dot, 1. Remember, I said A alpha I beta J, and this alpha could be 1 dot or 2 dot on the I. It could be 1 on the I. And you see, now I have to take all possibilities into account. If I think about this, what is this, this would be the gauge field acting on the right-handed neutrino because we defined the right-hand neutrino to be on this corner. And in this respect, you'll discover, actually, that this guy would be 0 because the neutrino, the right-hand neutrino, is a neutral guy. So then, of course, actually, the next guy would be A 2 dot 1, 2 dot 1, and so on. And this, actually, we discover, is a vector. And then it is given by minus I over 2 G, I don't know, B mu. So it's a vector. And with a gamma mu. Everything tends to be the cliff or the algebra. So you get that this guy is a vector. In addition, you start, actually, to get something like a guy like A 1 dot 1, 1 dot A. And then it's a doublet index. A, remember, is an S2 index. And you discover, actually, and then you put A is 1, 2. And what do you discover that this is really proportional to gamma 5 HA? And this is where the H is entered. So the H is entered as inner fluctuation. And now I say, what is this thing? Why are we going this way? This is an off-diagonal element. And off-diagonal element, it means that it's really trying to connect two elements of the left and right algebras with each other. I mean, in fact, it's very invisible when you write D, because D F, that's what, there's going to be 5 turns D F. I mean, you know, D F, that's what will generate the gauge, both of them. Right, yeah. And the other one will generate the H. Oh, yeah. One really can see it. But of course, you really cannot see, for example, here it's really easy if you see that it's a doublet. What's not so easy, actually, is to show, for example, the other guy, the A 2 dot 2 dot, for example, 1A. In this case, you say it's gamma 5. And what happened, actually, it is A BH bar B. You really cannot get, guess ahead of time, that you are only going to get 1 Higgs and not 2 Higgs doublets. The fact that you get only 1 Higgs is really related to the fact that these guys are related. Remember, I told you I have C plus C. And then, because we are talking about the same C, it tells you that, OK, you know, you really cannot have different Higgs. You are really going to get the same Higgs. So the upshot of this is that one really can do this calculation, is a matrix calculation, no complications, and after some algebra, one finds the Dirac operator, including fluctuations. What does it mean? It gives me the connections. It gives me the connections on the space. And the connections are nothing but the gauge fields and the hex fields. Nothing else. So up to here, you know, we have, we can achieve that. And before doing anything, I can tell you that the Dirac action for the fermions, OK, you know, one really has to write the gypsy. I'll tell you actually now, maybe I can tell you why we can write the gypsy. The reason we write the gypsy, because in euclidean space, I really cannot say gypsy equal to psi. Yeah. So if in order to get, once you write gypsy, what really happened is that when you integrate on the gypsy and the path integral, then you really, this is an anti-symmetry product, and you don't get determinant of Da. You get the pfafian of Da and the pfafian of the square root, which means that it's really a chiral integration, and you have not doubled the degrees of Frito. OK? No. Anyway, if I expand this, what do I do? What do I really get? I really get all fermionic kinetic terms, for example, nu bar D nu plus e bar D e plus left and right at everybody, plus vector Fermi interactions plus tibs Fermi. And you get exactly the correct hypercharges and everything. Actually, why one would get the correct hypercharges exactly because of this, actually, if you really work out the jA star j inverse, they really come and they bring, actually, the u1s in really completely funny way in order to have the matching. And here, actually, one of the lots actually makes an assumption that one takes SU of the algebra, not the algebra. In other words, one has to make a requirement that the trace of the A should be equal to zero. This is another condition that we put in. OK. So, here, actually, it's a first check. Everything fermionic-wise works, and it works perfectly. So what's next, actually? What's next is that you need to make, so fermions are dynamical, but the bosons are not. So how do they get the dynamics? For all, we say that we have a care space. The care space should have dynamics. And we should be able to describe the full interactions of the system. See up to now I have been speaking classically. Everything here is classical. Now, so here, actually, comes the idea of the spectrum action. And the basic idea is that the Dirac operator, which was really the building block of the non-committed space, in principle, one can study the spectrum of this Dirac operator. So you study the eigenvalue problem, and you look at the eigenvalues. And we know, actually, from studying, say, the remaining geometry or the Dirac operators on remaining manifolds, that the eigenvalues are geometric invariants. In other words, all the geometric invariants are included, actually, or lambda is a function of all the geometric invariants. So now the question is that how can we extract this dynamic? How can we extract the information about the lambda? Of course, it's not easy to simply work out lambda of every space, so we need something more general. And so the basic idea is that since the spectrum knows about the geometry, we took as the dynamics of our system is conjectured to be given by the trace of the function of a d, where f is a positive function. I think you have to put a scale, so why is it that it makes sense? Yeah. Exactly. Yeah, I was going to talk about the scale in a minute, because, yeah, d has dimension on operator, remember it is, say, gamma mu, say, d over dx mu, and this actually has dimension one, therefore you really have to divide it by something of dimension one. Now, of course, actually, you may ask what is lambda, and I will talk a little bit about, you know, dealing with the scale later, because this is a very sensitive issue. What you should do with the scale? Can you get rid of the scale, or what are we talking about? Of course, I want to add something which is that this expression, trace of a function of d, is the only expression that you can write which is additive, which is joined spaces. See, the action should have at least the properties that if you take two disjoint spaces, it should have. Right. This is the only expression you can write. So you want to say F, but d1 and d2 should be disjoint in this case, all right? Yeah, sure. If you take d1 and d2 to be disjoint, it takes the fact that the full direction, f of d1 plus d2, okay, that it certainly proves that, you know, the trace of that will be the sum of the... Okay, I got it.... function. And there is a proverse to that, which can be that it is a functional, which is spec-wise. And full field is this additivity properties, then it has to be learned this way. But what about the positivity? Positive because... Positivity, no, no, no. It's... But it's important for Euclidean actions, no? No, no, of course. I mean, if you... Z, like Y, on top of that, it's positivity. But something that I've been, you know, toying with the idea that maybe it's self-positive and things like that. Anyway, so now essentially, how can we deal with this space? Obviously, we really, for an arbitrary function, it's not that easy to work out, but what we can really work out is asymptotic expansion. This way, one would know how to do. And the idea, the following, we use methods of heat cannon expansion. You know, it's not the only methods, but they actually... At least for remaining many forces and for product almost committed very force, they're really very, very effective. So what's the basic idea here? You say, okay, you know, suppose that I have a function of some operator P, you make some, I don't know, I don't know, a normal expansion. You say, say, SPS. And then... And then you are... You have something like trace of P minus S. The new identity trace of P minus S is given by, you know, you use the Mellon transform formula. And with the Mellon transform formula, you obtain that, you know, it's the opposite of the definition of the gamma function essentially. T S minus 1, trace E minus T P will be the real part of S is larger than or equal to 0. And then the next step would be to use formulas of trace E minus T P using heat cannon expansions. And then, of course, actually, this is given by A n of the... Sorry, T n minus, actually, in this case, minus, well, n over 2 minus 2. Okay. At least in four dimensions. And where n is larger than or equal to 0. And this is A n of XB1 dv. And these actually are called the serial discussions. And they're defined in terms of connections of the dirac operator. Yeah? It's an asymptotic series. It's an asymptotic series. Okay. I would say why it's useful in physics because exactly at a point that this is, you know, the fact of d over lambda. And then it will be clear, actually, when I write the expansion, it will be an expansion of 1 over lambda. So the first term will be 1 over lambda to the 4. Then you are going to get lambda 4, sorry. Then lambda squared, then 1, then 1 over lambda squared, that's all. Now, assuming, actually, the scale is extremely large, then these terms will become, you know, less relevant. Of course, actually, the expansion will break down once you approach the scale. And you know, the scale, you know, you can take the scale to be, say, unification scale or Planck scale. In this respect, actually, the expansion would definitely break down. And in that case, one has to learn different ways of dealing with such expansions. But you know, we did some calculations of this expansion for, you know, Robertson-Walker metrics and things like that. And really, it falls almost up to the Planck scale. Up to the Planck scale, it's only when you are orders of, you know, like twice or three times the Planck scale, then you start to see that it breaks down. But not before, you know, which is really amazing that this action is so accurate that it goes all the way up almost, you know, to the Planck scale, but not beyond, you know. It's just there. All right. So what do we do with this expansion? It doesn't give us anything sensible, because obviously, this looks far-fetched, I would say. That you take an arbitrary function, you say, okay, this arbitrary function. The fermic part is straightforward, say, psi, d psi, okay? And you get, you know, the usual expression. You can say it's by construction. But for the bosonic part, it's not really clear that this idea should work, you know, that it should give you anything accurate, essentially, after all you're using some expansion. All right. So how to proceed, actually, here, you know, it's known, as I said, here we assume that the continuous manifold is four-dimensional. And in this respect, in this respect, the expansion, yeah, what happens in the following that only an equal to zero, if n is off, for manifolds, without boundary, okay? What happens for manifolds with boundary? We have already addressed this question. It's more, it's actually very interesting. But I will talk about it later. But for manifolds with boundary, n equals zero, and z, it means, actually, the only non-zero ones are the a0, the a2, a4, and things like that. And these formulas were worked out long ago by Gilkey. He has some, you know, standard formulas. And because of that, actually, what really becomes relevant, and this is the d squared of the d, because, huh? Disabase squared. And what the trick of the following, you write this as g mu d mu plus a remainder. And this d mu will be mu plus omega mu, which is the connection. And this omega mu, okay, this also is given by, actually, there's minus g mu d mu d mu plus a mu d mu plus b. So what Gilkey did, actually, you know, he gave an expression. This actually is, there are two important terms in this expansion. The e, which is an invariant, and the connection, the omega mu, and, you know, curvatures, which are constructed out of this connection. And this omega mu, one reads from the following. You write this expression into this form, into this form, and then you can know, actually, that a mu is given by, you know, some expression, which is half g mu a mu plus a mu plus comma mu. This is a crystal, a contraction of a crystal. In other words, actually, one really can compute all these expressions in a unique way. And remember, we have already computed our d. So we have this d, huge matrix. You take it, you square it, you get another huge matrix, and you write it in this form, and then you try to find the curvature of the space. Remember, it's re-enlightened curvature now. And you also get the e-inverted of the space, another huge matrix, you know, 384 by 384. So what Gilkey does, I'll look, you know, this a0, the first time in the expansion, is given by 1 over 16 pi root g, trace 1. Trace 1, of course, is 384 because it's 384 by 384 matrix. So this is actually nothing to compute. See, R, which is independent of lambda, obviously. This is, no, no, this is actually lambda 4. This is the re-cosmological constant. And it's huge, you know. The cosmological constant is huge here. So if I write it as 0, that's a different story. Okay? This is not quite high, actually, because what if it's not 1, in the sense that you have the... Things is effective, you know. You want to talk about effective. You want to talk about, yeah, I'm talking about, you know, 0th order cosmological constant. Of course, there are contributions coming from the vacuum expectation value of the higgs, but these are supposed to be much smaller, actually. I'm saying that when you compute this coefficient as 0 and when you take into account the Major Anna-Mass matrix, you get another term. Okay, but... Yeah, okay, you know, at... So you get this time. Yeah. That actually, but... But that comes from the higgs, no? That comes from the Major Anna-Mass matrix. You mean the G-squared term? Yeah, sigma. This is a sigma, actually. Sigma, yeah. Yeah, but remember, yeah, okay. But this takes you after it. Sigma would come later because I see sigma lives here now. Sigma lives here because a priori we don't know. Okay. Anyway, so the interesting part, actually, is this guy. So, you get, actually, R over 6 is the Einstein term, and then you say trace E. Now, you... Let me actually give you an idea of what trace E is in this case, or what E is, actually. You know, E, you have to read by squaring the D, by squaring the D. And so, obviously, let me remind you that you have all this D-slash part. You have here gamma 5h part, gamma 5h dagger. You have some D-slash, okay? When you look at the D-squared, yes, the off diagonal terms are really going to give you what? H-h dagger terms, which is really nothing but the higgs mass, okay? So you discover, actually, essentially, and then you trace it, and you only get trace H-squared. So if you really work it out, actually, in this example, what you really get, you're really going to get the following, actually. You get this Gibbs up to a factor. So you see, this is the guy that would eventually give mass to the neutrinos. And this is the higgs mass. These are essentially your coupling, your square of your coupling. So we look at this expression, and then we notice the following. The curvature term, you mean the whole minus sign? Yeah, there's a whole, overall minus sign, yeah. See, I write minus, okay. So if you'd like, yeah. Okay. So. You have included the one half R into E, or not? Minus 2 over pi squared, if you wanted this, F2 lambda squared. Yes? Root, gd4x. There's a minus sign here. Well, the notation that R is negative for Euclidean spaces. Anyway, so you see, we get a mass term and with a minus sign, this is really extremely important if you, as we'll see for the standard model, because the standard model is really characterized with a potential V, as you see, lambda over 4h2h bar h squared minus m squared over minus mu squared over 2. And then the minus sign that gives the potential the Mexican hat. It's essential that you get a minus sign. Now obviously, you are really going to get a minus sign. Why? Because if you look at it, in the end, you are taking trace of E minus Tp, P has D squared, D squared has S squared. Obviously, you are going to get the first term in the expansion, a minus sign. The next term will be plus sign. It alternates, which is extremely nice, because of course, but you really cannot guarantee that, okay? In principle, before you start. You don't know. So this actually, important points we can point that the curvature, which is the Einstein term and the mass term are really not equal footing. So h bar h is like curvature of, is the curvature of the antenna space. And okay, you know, scale lambda squared. What's lambda squared? Of course, actually, one would like this lambda squared to be related to 8 punch or 16 punch, whatever, you know. Okay? It should be related to the, so this actually is the plank, plank squared. So from here, one would read that if you would like to make sense of the scale lambda that we started with, the natural interpretation of it is that it's almost, of course, you have numbers, you know. You have F2, it's a maliant transform of the function. So you can get factors of 10 or things like that. You have the pi squared. So it is of the order of the plank mass. It's not exactly the plank mass, because you have some factors here. But at least in order, we know that it's of the order of the plank mass. The next time, day four. And according to Gilky, what's the formula? The formula tells me that, what is it? Okay, you have, let me write here. Okay, I'm not going to write the surface terms. So what re-enters is the curvature squared, everything squared actually here. E squared, curvature squared. Now you know, this E squared, the omega mu squared, they really give me invariance in terms of curvatures, square of curvatures. What are the squares of curvature? Of course, R mu nu squared is there. But here we're really going to get what? We're going to get F mu nu squared, which is curvature squared. And we are going to get kinetic energy of the hex term and the sigma term and things like that. So, and, yeah, exactly. Yeah, in mill terms, there's nothing but curvature squared terms. Because what's F mu nu? It's if you take the A squared, as you see, D plus A squared, you are really going to get F. The square of the curvature, the square of the connection is the curvature. And it really lives here, it lives here, and one really can compute it, some long calculation, but in the end, it's well-defined calculation. The important thing is there's no ambiguity at any point, okay? So in this action, we are going to get F0 over 2 pi squared. Then R mu nu squared,code v squared, dou Improved plus. plus conformal factors, kinetic terms for the H, quartic terms for the potential, and similarly everything I write here, I have to write also for the sigma field, and so on. Our sigma score, anyway. It's an important thing that first of all, this time there's no scale. Second, we notice that we have a vile squared term. Now this is really important for, you know, if you are really doing, say, a realization of gravity. You know, this term would make, would improve actually the propagator of the graviton. It makes the theory, of course, non-unitary, but you know, this we don't really worry about because in that case we are cutting, you have to cut the expansion and you know, if you would like to treat unitary, you should not cut the expansion. You have to take the full Fletch theory, but at least this conformal tensor or the vile tensor squared is important for the propagator of the graviton. And now here we see here actually that we really have a kinetic term for the u1, for the electromagnetism, a kinetic term for the SU2 weak, and a kinetic term for the color, SU3. And, exactly, and then we look here and then we immediately find a relation that tells me that g1 squared equals g2 squared, that's 5 over 3, sorry, g3 squared, g2 squared, 5 over 3, g1 squared, which is the famous unification one obtained for SU5 or SO10 gratification theory. We will have actually, we have to talk about this later in the next lecture because this is not exactly satisfied. It's not exactly satisfied because as we are going to see next time, you plot the, you know, okay, the logarithm of the developments of the alphas, which is g squared over 4 pi, alphas inverse of the alphas inverse, then you discover that they don't exactly meet, but, you know, this is a different story, you know. Whether they meet, they don't meet, they almost meet, but in principle, you know, there's so much in between that we don't really know, we don't control, and it's premature to say whether this is a bad point or not a bad point. At least they are the same order. They are, you know, it's almost, you know, okay, within, let me say, within 10% it was, within 10%, but you have extrapolated 15 or 16 orders of magnitude, actually, which is huge, okay. By a variant, the normalization. Yeah, you know, so there is, and, you know, things, we have discovered that things have not really been calculated, you know, to two loops in the model that we have, because this field actually turned out to play an extremely, very important role. The sigma, which a priori we ignored, at one time we even make a prediction that the Higgs mass is given by a certain number, and this turned out not to be correct, because we have ignored actually this field. However, if you take this field into account, then, you know, you're completely in agreement with the experiment, and in addition, actually, the normalization group equations taking the sigma into effect has also only been carried to one loop order, and nobody has carried it to two loop order, so we really don't know. So maybe it will change a little bit, actually, you know. My guess is that it will change, but I don't know whether it will change in the right amount in the right direction, because this is up in the air. So let me summarize again. So we started from a very general idea, the general idea that space-time can be approximated by a product of a continuous time discrete manifold. The real answer, we don't know. You know, it may be just completely discrete, but to a very good start, it's because we know that generativity works very well, so we know the continuous part up to the Planck scale works very well, so we can up to that energy level or a little bit below. We know that the, you know, Raymaine and manifold part works. So to assume that we have an invisible extension, which makes it almost commutative, but in reality, it's not commutative, that we can extend it by a finite space, we entertain this idea, and we discovered that to almost uniquely, we can predict the structure of the model. We know the Fermi-Nichro presentations, we know actually why there are 16 fermions, we know what are the representations of these fermions, we know the gauge group. This actually, yeah, I'm saying that what we can, that things what we cannot do is much more, of course, because what we cannot do, for example, predicts say the electron mass, and you know, oh, the masses, because in principle, you know, if there is more restrictions on the, one should be able to, for this, remember, the finite space contains all the Yukawa couplings, so it contains actually information about, so if you know that, for example, that K is like 10 to the minus 2 or something, then you know that the ratio of the electron mass to the top quark mass is like 10 to the 4 or something, so the minus 4. So in that case, like this information we are putting in by hand, we are not predicting, but if one really is completely successful, everything comes out, but you know, this is really far-fetched for time being, because you need some extra structure which we really cannot put our hands on, not yet, you know. So I think there is another aspect, the criteria and the properties that we verified, do not really still tell us actually why there are three generations, this we don't know, this must come out somewhere, you know, so this aspect, we have no idea. No one has any explanation for that. We need an explanation. And as in the explanation for this, there is no, as even a guess whether there should not be a false generation. Yeah, this actually almost rolled out now actually, you know. No, no, but there are two things to say. There is one, these two Japanese who got the Nobel Prize, and the reason why we need more than two generations. And that's the cap-de-bose Kobayashi-Maskawa mix in the text, because it involves a complex number only if you have more than two. So I mean, three is the first number of generations for which you get a complex number, so for which you get violation of charge conjugation. But then, okay, I think now experimentally it's known that there is no false generation. There are very strong tests which are true. Yeah, so why three? Actually, it's not that we don't know, nobody knows also. It's up in the air, why the masses of the firms are the way they are, you know, the standard model after all has what, 17 or 18 parameters? Sure, but there you can say one thing, which is it's exactly, you know, like if somebody would come and would tell you why is the metric of space times the way it is, okay? You could say, well, Kepler's atom gas, but you know, you could say, well, it's like this, it's because of circumstances. So the issue is, what is the geometry of the finite space? Yes. Is it due to circumstances or does it have a deeper meaning? Right. And this we don't know. This we don't know. One would hope actually, see that, okay, if maybe, well, 20 years ago, you know, okay, you know, before say that, ventals, not geometry, if somebody asks you why the standard model, you say you could have answered the same. But now actually, I think we are very close to knowing that it is, it must be like actually three cross-situ, it must be like 16. So we must have a hex field. So this actually have been answered satisfactorily, but we are far from, you know, the full answer and this is the. It could be both a white and a hex. Not here actually. Here, okay, if later, if, you know, we take off this condition and we really want to push everything to grand unification, then of course you have many more hexes actually. Not only one, we are going to get, but that actually will be relevant at higher energies. And we will not be able to see what really goes on at lower energy except for the six doublet. You know, it's not clear. But that's another story, you know. Anyway, I would stop here, I think, because next time I will continue by analyzing the quantum aspects of the model and maybe, you know, theorizing a little bit about. So what you have so far is the first characterization. Of course, yes. So answer what Thibault has last. It's a very edge of geometry and some of the space set up. Right. But what is quite, one should stress one thing, you know, which is that there are seven or eight sides which have to be correct. And they are all correct. And moreover, you know, it encapsulates the state of most complexity of the Lagrangian into something which is. This thing was done before, you know, this would have been like, and then people come and verify it. It's not like a revolution. But the thing actually lost the luster because people didn't think that, okay, you know, the answer, you are fitting it. And this is the, you know, biggest thing that one has to fight in the misconception that there are billions of possibilities and simply were given an interpretation of these posts of the one possibility, which is correct. And there are many other possibilities, which is not really correct because, you know, according to this specification that I went through, that there are not really many possibilities. It's almost, it comes uniquely, you know. But that one would have to explain why if you want the correction by the final space improves the ordinary four-dimension continuum. And the reason why it improves it, it changes the care dimension. Yeah. It changes the care dimension from four to two. And that's a big improvement because then you can write this profile. But it needs to be finite. Well, it's the most economical to be finite. I mean, it could be that the splitting is actually more complex, but I mean, the simplest way to see it is that you have a four dimensional space. And then you make it two-dimensionality dimension without changing the metric dimension, which is remaining four. But if you have a, you know, if you start finite, then you'll have to fight all this Kaluza Klein modes, as I would say. You know, you have many degrees of freedom, which we don't see. So, it's a new dimension by something finite. Exactly. By something finite, which doesn't have this infinitesimal mode, like in Kaluza Klein. I mean, that's a very basic thing. Okay. Okay.
 Let's get right into the song.  So last week I introduced the idea that non-committal geometry can be taken as a basis to unify all  fundamental interactions.  And so what I did is that I defined what we mean by non-committal space.  I gave the definitions.  And so with certain properties, and just as I remind you, just set the spectral triple  and I defined these five quantities and the relation among them.  And then the question, of course, is if a non-committal space is the basis to unify  the interactions, then the question is what kind of non-committal space should we take,  you know?  And as a starter, I made an assumption, and the assumption I made is that space time r  is a non-committal space which is formed as a product of continuous four-dimensional  space times some finite space.  Now four-dimensional spaces are well understood, and then I went to see essentially what one  should take for the finite space, you know, what kind of space should one take.  And so for that, started by classifying spaces which fit certain physical criteria.  And the physical criteria that we need, and the following, that, you know, the Hilbert  space, which would be space of, you know, fermions essentially for us, would include  all the fermions, and in this case, actually, the fermions just satisfy certain properties.  And essentially we said we have fermions, but because actually of the presence of what  we call the reality operator, one really requires essentially what we would like to  require essentially, that the fermions are real, which in physical language means Majorana.  And in this respect, of course, actually the presence also of the gamma operator, which  is a current operator, would allow us always to impose this condition.  So what we really have, we know, we say we have current spinors.  Now then, of course, actually if you have to on J on psi, you are really getting to get  something like the conjugate spinner.  And the question is that is this an independent set of fields?  If it's an independent set of fields, then you really immediately face the problem that  all your particles will be doubled.  You are really going to get, for every particle we know, a mirror particle.  And you know, sometimes it's called the mirror fermions problem or the mirror doubling problem.  And usually it's not easy to solve this problem.  Now we are really working in Euclidean space.  And the issue, of course, is that if you really go to Euclidean space and you said I would  like, sorry, but the physical space, of course, is Minkowski.  And if you would like to impose both conditions, the J psi is psi, then this actually would  impose certain conditions on the properties of the space.  As I have shown last time, that would impose a condition what we know as the k-o dimension.  So in Minkowski space, you really need to impose both conditions.  The J psi is psi and gamma psi is psi, and this would imply actually that the k-o dimension  of this space is equal to zero.  Which implies actually that J squared, which is epsilon, should be one.  J d is, I think, epsilon prime d J. And I need one more sign.  J gamma is epsilon dot the prime gamma J. And d gamma is minus gamma J.  So these are the properties that we need.  And essentially, it really forces, now we know actually that in Minkowski, you have  four dimensional space with signatures, say, minus plus plus plus or the opposite.  And these would somehow cancel each other.  And then you are left with k-o dimension two, which means that the internal space or the  finite dimension space should be dimension six, k-o dimension six.  So that's actually forced the finite space to have certain properties and k-o dimension  six.  Then you need certain sets of epsilon, for example, epsilon is really minus one.  I forgot actually, I think, epsilon prime is equal plus one and epsilon prime is minus  one.  Something like this.  So then actually one goes ahead and try to understand these finite spaces.  And what one discovers, and I would go through, not through the proof, but through the realization  of the proof, is that one discovers that the algebras of the finite space are essentially  what one finds, that the complexification of the algebra.  So essentially here, one assumes that the finite space is characterized by A f, H f,  d f, gamma f, and g f.  Everything is for finite.  And one discovers that the complex extension of this algebra, if you take the center of  it, there are only two possibilities which are either C or C plus C. And this actually  is inconsistent with k-o dimension six, which leaves us only one possibility that A C is  C plus C, and really corresponds to an algebra which is a variant of, you know, let me call  it M N of C plus M N of C.  In addition, when we imposed certain isometry on the first algebra, and this is a point  actually we don't really, you know, understand, you know, we cannot give it a mathematical  characterization why it should be like that, but it is like that.  And the other possibilities are not relevant.  And this really reduces this instead of matrices on the complex numbers.  You really take matrices over the catorials for the first, and in which actually N is  2A in this case.  So N is even.  And also, because one would like to impose the chirality operator on this algebra and  elements of the algebra, for example, gamma A equal A gamma.  So in this case, what happened is that this M of 2A of H would split into M A of H plus  M A of H, for example, plus M N of C or 2A of C.  So the upshot of this is that really the first non-trivial example you can do is the following,  that A should be one in this case.  Sorry, A is, yeah, this is a matrix, you know, M A of H. This should be 4A or something.  N is, let me see, M is.  N is 2A, yeah.  So essentially, yeah, so essentially this, yeah.  So essentially, really the first non-trivial example is M1 of H or H plus H.  Okay, let me write it like this, and this actually is like 4A.  And in this case, the first example would be H plus H for the first algebra plus M4 of  C for the second.  Of course, other algebras are possible.  But Ali, what is important to say here is that you have the gamma, which is the diagonal  matrix.  So, on the first, right?  For M2 of H.  Right.  So it takes the even part.  It's the even part of M2 of H, which is 8 plus H.  Yeah, okay, so I have to say, okay, this I would say M2 of H even, and then it really,  at least for the first non-trivial case, it belongs to H plus H.  Now it's up to this point, you know, I went last time.  And the only thing we're, you know, the assumptions I made are there.  So at one point, one did impose on isometry.  If one didn't impose on isometry, then you are really going to get, instead of, you know,  H plus H, you're going to get M2C plus M2C.  And one really can show that these cases would not be physical, but one cannot rule them  immediately, unfortunately.  What it is, is that if you have a real algebra, then you have three cases.  The symplectic case, the pteric case, and the also.  Exactly, there are three.  And you're okay, we choose, you know, the symplectic, you know, case.  So this is a choice.  This is something which is not imposed by the...  But you could have chosen once.  We could have chosen, yeah.  Symplectic, symplectic, or...  We could have chosen this section.  So this is the really... the other is actually not relevant because they would really give  you real algebra.  For instead of H plus H, you can take M2C plus M2C.  What would it really give essentially?  It gives you, you know...  It gives you an extra U1, actually, you know, which is anomalous.  So one really can work it out.  It gives... it's a little variation.  Everything almost the same except this little variation.  It gives you an extra vector.  An extra vector which is anomalous, you know.  Maybe there is an extra vector lurking in the spectrum, but, you know, people are talking  about, you know...  Recently, there was some papers in which they said they observed something which deviates  from the standard model.  And, you know, some people said maybe it's an extra vector.  So who knows you?  Anyway, but this is actually the most symmetric possibility.  The only assumption I made also that left action...  This is the left action.  And then this is, say, the right action.  And then we said this is J B star J inverse.  And we said...  And this actually is defined to be the opposite.  So this is one of the axioms that left action and right action commute with each other.  So this actually has been used.  This property really has been used up to this point.  Now, from here, actually, one needs further assumption.  And, you know, I will discuss both possibilities.  In that...  And this is called order one condition.  And now, actually, we understand this condition.  It's really a condition on the linearity of the connection.  If one drops this condition, then what happened is that the connection would really get quadratic  terms and that really can work out the possibility when this is not so, when this is not satisfied.  But this I would leave to the last lecture to allow, because actually in non-convenient  geometry, there are examples where this condition is not satisfied.  And you know, this is what the quantum spheres.  Okay?  Anyway, so here it will have a really big advantage that the connections are linear.  Everything is very nice.  Everything, you know, falls into place.  Okay?  What happens if this is not satisfied?  You know, this would lead us to grand unified theories and Patti-Sellem models and things  like that.  But this will come much later.  Okay.  So what are the consequences of this condition that's satisfied?  Okay.  So I will assume, actually, that this is satisfied.  And on top of it, actually, now there is one more condition.  Is the question that if we take the commutator of the Dirac operator with the center of the  algebra, is it zero or not zero, actually?  This is a question.  Well, what we have shown is that there are two possibilities.  If it is zero, then this symmetry is really intact.  And if this symmetry is intact, one can show, actually, that you are really going to get,  when one work out a model, which I'm really going to do a little bit later, is that you  really get SU4 color, which really corresponds to this M4C, as a symmetry of the quantum.  So essentially, in this case, you really get a higher color symmetry, which is not observed  in nature.  And on top, actually, you'll find out that the neutrinos are all massless.  They are exactly massless, which now we know that it's not true.  So from the physics point of view, we know, actually, that the three must be different  than zero.  Because if it is zero, then you are already in contradiction with the experiment that  neutrinos will become all massless, which would have been OK until 1995 or something  like that, because people didn't know, were not sure, actually, whether neutrinos are  massive or not.  Because every this tiny mass, there are only limits on the masses of the neutrino.  Now we know, actually, that they are really massive, but the mass is really extremely  small.  But that they are massive is well established.  Now what happened, actually, if this difference is zero?  Now the beauty of it is that if this is different than zero, and for this condition to be satisfied,  the Dirac operator must have the following form.  And so in this case, actually, instead of telling you in description what really happens,  I would start being more concrete.  I will tell you exactly what are the representations, what goes on, so that you have an idea.  That this actually, as I would say, field is not simply talk, you know, you really can  go and compute everything unambiguously, not much ambiguity, and with all the details.  So, all right, so essentially let me examine this in a minute.  But I really can tell you that the fact that this is non-zero really gives a unique possibility  that you really can give the neutrinos a Majoranum mass, and it will explain, actually, why the  masses of the neutrinos are very small.  It really, it's an example of what was known as the CISO mechanism.  So here it really comes out naturally, and it solves the problem in a really very nice  way.  So, let's see how things would look like if I wanted to go ahead and do some computations,  you know, what should I do?  So this is really my starting point.  I'm really going to take H, the algebra A, to be H plus H plus of course.  And this, remember, almost uniquely we were led to this.  It didn't come out of the sky, you know, out of classification.  This was the first non-trivial result that we have.  Okay, so essentially if I would like actually to represent an element of A, how would I  write that?  Then I am going to write it in matrix form, you know.  I'm going to show you that everything that we do is really matrices, essentially.  We really can use tensorial notation if one would like to simplify his life.  And you know, recently, you know, everything that we do can be put on the computer, so  in the end one doesn't do much calculation.  You just write it on Mathematica and press a button and it gives you the answer.  So, you know, it's, all right.  So I'm really going to use, you know, index notation.  So essentially here what I'm really going to say, all right.  So in order to tell you what my notation, I really have to tell you what are my spinners.  Okay, so the spinner actually, as you see, it belongs to this guy.  So what I'm really going to write, I'm going to write as psi alpha i, where this alpha  is, would see the H plus H and the i would see the 4.  So obviously i is 1 to 4 and the alpha is also 1 to 4, but it's a different form.  It's H plus H. Now, and then, of course, actually I'm going to write this as, you know, a dot  a, where this has two indices, like 1 and 2, and this is 1 dot 2 dot.  So this actually tells me that it's a doublet under the catorians, you know, doublet left,  doublet right.  And what we write actually, we write as H right plus H left, because we have already  graded the algebra.  And in this respect, my spinners would be psi a dot i and psi i.  Okay.  So how many spinners I have?  I have 16 spinners.  And of course, we have also the conjugates, but the conjugates, we're going to see that  they're not the independent spinners, because the thermion doublet problem is solved through,  you know, the Majorana condition.  We are not really going to say Geppsi equal to psi.  The problem in Euclidean space is really solved in a very clever way by, you know, in the  path integral formalism that only chiral thermions get integrated and they give square root of  determinant instead of determinant.  So it's really solved in really very subtle way.  You know, this is a nice discovery.  You get the fulfillment.  You know, I'll talk.  Anyway, so you see actually now, you know, without redoing any work, I have already obtained  the correct classifier.  First, actually, we already predicted that you have 16 fermions, which is exactly per  family of course, which is exactly what we have.  We will also actually find, let me, you know, let me write this one of this.  The i.  The i eventually, you know, when this condition, the d with z of a different than 0, will also  be broken into 1 and i, where i is 1, 2, 3.  So that actually will be the color index and this is the lepton index.  So this is lepton and this is the quark.  And then you can see that the lepton is the fourth color.  It comes out of the fourth color.  And in this respect, you're going to get psi 1.i, psi 2.  Sorry, psi 1.1, psi 2.1, psi a, sorry, 1.2.  And then I have psi i, which I'm going to leave it as psi a1.  Then I have psi 1.i, psi 2.i, and then I have psi a.1, psi a.  What's in this?  You know, remember, this is a dot means, when you see dot, dot means right.  So this would be what I'm going to call neutrino, right?  Why give it the name?  Because if it comes out like that, the quantum numbers tell me that this will be a neutral  guy.  But this comes out.  You know, I name it according to the usual naming and it would agree.  And then, of course, this is a, right?  And this, of course, is upright.  This is down right.  Similarly, psi a1 is the doublet, which is a new a, a new e.  It's a doublet.  You know, a is 1, 2, left.  And this is up, down, left.  OK?  So you see, actually, without much work, we obtain the correct representations of the  particle spectrum of all the fermions.  We know which representations they are.  So one really can write that the 16 in this case is a 4, 4, 4.  And this 4 is decomposed into 2 plus 2.  And this 4 is decomposed into 1 plus 3.  And then this is 2 right plus 2 left.  And this is everything that we see belong to that representation.  Everything falls into place.  OK.  So now, so these are my 16 spinners.  But remember, actually, you know, I have the j psi.  And so my spinners will be psi alpha i and then psi alpha i conjugate, which I'm really  going to go psi alpha prime i prime.  They're not independent, but they are related through the j.  OK?  What does it mean, actually?  It means if I would like now to write the element of the algebra, you know, I can write  it as follows.  I can write it as this one, you know.  Remember this belongs to the second guy to M4 of C.  So it is something like this.  And this is delta alpha beta.  And this is like yij.  So you see, actually, how an element of the algebra is satisfied.  And then you can form the b opposite, which is jb star or j inverse.  And you discover, actually, what happened is that here you are going to get the y times  1.  You are going to get yij, essentially, transposed into 1.  And here you are going to get the x alpha beta into delta ij.  And if you look at, you know, this element and this element, obviously, they commute  with each other y because then this is don't talk to each other.  This is in the first algebra.  Well, this is now in the second.  And they don't talk to each other.  So essentially, jb star.  Yeah, absolutely.  Right.  And this is actually, I wrote it as transposed, but, you know, a star is, OK.  So essentially, we can write it in this form.  So the first a b opposite is satisfied automatically.  Now where do I go from here?  Then I have to satisfy this condition.  So this actually is a condition on the d and the algebra, both, you know.  Because what happened in the following?  You try to satisfy this condition, you work out the matrix representation, and you obtain  actually.  I mean, you mean the commutation with the center, right?  So you are right here.  This actually, you are right.  Yeah, exactly.  Yeah, it's exactly.  Yeah.  This is the difference here.  This is the difference here.  OK.  So this actually, we have to assume it's different than zero.  And then actually, this really would require the form of the direct operator to be as follows.  You can show that you are going to get something like this.  Let me call it, I don't know, let me call it dab or something.  And here we are going to get the bar.  Here we are going to get something.  And then one can show through a little algebra.  Actually, I learned it through abstract proofs, which you know, beyond me.  But through a little algebra, one can show actually that the only operator that satisfies  in the following, that here you can have only one non-zero entry.  Only one.  Anything else would not satisfy this condition.  OK.  Of course, actually, this, together with that, we have also to make sure that dA would be  opposite is equal to zero, putting the two firsts actually.  First we know that here we must have a non-zero element.  The question is that the whole thing could be non-zero.  But because of this condition, the first condition, you discover actually that this is possible  if and only if you have only one non-zero entry.  What does this mean?  Actually, this one non-zero entry.  Let me write actually the expression for you.  And let me see how does it look like.  You know, I have, remember, when I write Dirac equation, what do I write?  I write psi star deep psi, or I can write it this way, you know, psi deep psi, OK, in  a product.  So this is my, say, Fermi reaction.  And this Fermi reaction, if I expand, and then I just look at this term, what does this  mean?  This actually, if I look at that classification, it really tells me that you really have an  element which is this one.  If I have more, it would mean that, for example, the up, the down, everybody will start getting  Majorana mass, but you know, you're getting a Majorana mass would break charge, would  break the charge conservation, because in this case they would not have zero charge.  So the only term that you're in physics wise that you can write is a term which is reneutral.  It must be neutral, otherwise you destroy charge conservation.  And it's not an accident, actually, then that you can have only one nonzero term, and this  one nonzero term would really correspond to the mass, Majorana mass of the right-handed  neutrino.  Now, I'll explain later, actually, that why this is really extremely important in order  to solve, it to explain, essentially, why the neutrino masses are so small, OK?  So this actually is a result.  It tells us that, you know, in order to agree with the observation, the Dirac operator should  take this form.  It has only one nonzero entry, and the rest are zero.  This d and this d bar are correlated.  They are, you know, one in the complex conjugate of each other.  And the only thing I have to find is that what's the form of the Dirac operator in the  finite space, OK?  So what shape does it take?  You know, the whole idea, the things really are really constrained, you know, because in  a way, we have to explain why the standard model.  This actually, you know, why the standard model.  So if the hypothesis that everything comes from the geometry, we must really arrive at  this answer in an almost unique way.  Maybe one day one would have fine and say, OK, here it comes, this is without any ambiguity.  But you know, there's always a possibility that you have to put some physics, because  if you don't put any physics, in principle, you can get anything, right?  Because, you know, if I don't want to explain nature, then, you know, I can start with the  smaller space or bigger space, OK?  You can start with too small, because if you already assume that it's four dimensional  continuum times finite space, OK, and the finite space has k dimension six, and then  you make a very minor hypothesis on that space, then you are already led to the sixteenth,  so it cannot be too small, you know?  This is the first example, but then you cannot rule out why it's not bigger.  Oh, no, of course you cannot swap.  You know, the question is that can I say uniquely, you know, this is the only thing I would ever  get, you know?  But this is the first possibility that one gets.  OK, so now actually, where do we go from here?  And what we do is the following.  So we would like actually to see what type of Dirac operator can we have, this DAB.  Now remember, actually, the D satisfies many properties.  I have written them up, so you really have to show folks, you have to verify that what  the area that G da is, Gd, still I'm talking about the finite space, remember, is equal  to dj and gamma j equal to minus j gamma and d gamma equal to minus gamma g.  So these properties, I have to satisfy, and so if you do that, you'll immediately discover  actually that the Dirac operator of the finite space really takes simple form.  Remember, you know, let me write it in terms of new right, E right, because I don't want  to write one dot and two dot, because then it's more physical.  And in your right, E right, and then I have, say, new left, E left is a doublet.  And then up, right, down, right, and up left, down left.  Okay?  This is really a doublet, but these are really singlets.  Okay?  Why is it singlet?  Because actually, this sigma really breaks the algebra from, you know, for this condition  and the other condition to be satisfied.  And then you put the sigma.  What you discover in the following, that this algebra, h plus h plus m4 of c is really broken.  And obviously, it must be broken, because you put the sigma, remember.  This is like one, and this is the i, and this really breaks the color index.  It breaks the color index into one and the three.  And in this respect, the m4 of c is really broken into c plus m3 of c.  However, because that condition dA, b opposite is equal to zero, it really, what happened,  and then it breaks, actually.  This is also broken, you know, because this is one index.  We'll talk this one index.  And then, essentially, this is a c plus c.  It breaks, it breaks, and the three are not independent.  What one discover, actually, that the element of the algebra A takes this form, in this  way.  So you are going to get, you know, x, x bar.  This is my first h.  These are my first ketonion.  In other words, actually, remember, a ketonion takes the form like alpha beta alpha bar minus  beta bar.  It tells me that in the first h, it must be of this form.  Now here, what do I have?  I have a ketonion.  So this is my first four, and this is my bottom four.  And in the bottom four, what do we really have?  We discover, actually, that this is x, and of course, this is zero, and this is m, where  m is the element of m3 of c.  In other words, actually, what we discovered is that the three c, the c here, the c here,  the c here are not independent, and they are essentially one c.  And the finite algebra of the finite space really breaks down to c plus h plus m3 of  c, which is really a symmetry of a central model.  It really comes out of the requirement that there is a Majorana mass for the neutrino.  No, this really has drastic consequences.  OK, so with this action, the element of the algebra, I go with the d, and then we can  really classify the d, satisfying all these properties.  And I want to cut the story short.  What happened is that the only non-vanishing elements of the d.  There is one more requirement that we have to put in order to really completely single  out the d, and that the d commutes with, no, actually, yeah.  d commutes with the element, which is lambda lambda bar lambda lambda bar in here.  If you do that, then you discover actually that only the non-ucala coupling of the d-operator  are non-zero, and really we have the d-operator completely singled out.  If I want to write it down, I can write it down.  Let me write it down, actually.  So how does it look like?  When you zeroes, I think here you are going to get k nu, and here you are going to get  ke, which is actually the yocawa coupling of the neutrino and the yocawa coupling of  the e.  And similarly here, you are going to get, this is new left, e left, and this is k nu  right, ke right, like that.  And similarly for the quarts and the only off diagonal elements, you are going to get.  And here also.  So this is k up, k down, and diagonal form.  So in other words, actually, after some matrix algebra, one would know exactly what the  d-operator is.  So what do we do with this now?  What we do with this is the following.  Having determined that.  Exactly.  So you know, many, you'll get many, essentially a point that you get many zeros.  By, you know, gamma d minus d gamma, I tell you all diagonal elements are zero, you only  get non-diagonal elements, and so on.  So it's little matrix algebra, nothing much to it.  And however, once one is finished, one can proceed as follows.  Note that we have not really talked about dynamics or anything, we just talked about  classifications and representations, and satisfying the properties of the uncompetitive space.  And you are led almost uniquely, actually.  What do I call that of the standard model?  Now, remember now, I said that the Dirac operator of the full non-competitive space is equal  to the Dirac operator of the manifold cross one plus gamma five cross df.  We have been, what I was talking about up to now is the df.  We know now exactly how the df looks like.  dm, of course, we know because it is the Dirac operator of the usual, say, remaining  manifold.  And similarly, we can say a equal a of the manifold cross af and j.  Everything is a product, actually.  Don't have to write it down.  All right?  Now, remember, actually, here, when all starts with what's non-fluxuating d.  So we start with a non-fluxuating d.  Say, let's start with a d which corresponds to, say, some flat space.  Starting from a d from a flat space, for example, you can really start with gamma mu  d mu like that, actually, for the d.  And we know, actually, that you really can generate the metric of the dm out of fluctuations.  These are the outer fluctuations.  And this is easy, actually.  What we have to do is that you say this is e mu a gamma a.  You make this as a function of x.  You discover that you have to add a spin connection in order to make the thing, in order to make  the thing coherent.  And so that, actually, is easy to do.  So here, we have seen last time that the d always go into dA, where dA is a d plus a  plus j.  But that's a good for the global.  Yeah?  What you have written there is for the global.  Here is the global.  Here is the n-arithomorphism, no?  No, but I mean, if you want to do the fluctuations, you do that for everybody, of course.  I do for you, exactly.  You know, we get both n-arithomorphism.  I'm trying to say that if you'd like to get the gravity, you have to do the outer.  And you want to get the other, the a.  Yeah, but these are enough fluctuations by the physicians, you know.  Exactly.  So they are one form in this language.  And it is adb.  This looks, actually, very simple.  But remember, actually, the a we have written is what?  The a we have written was a 32 by 32 matrix.  The d was a 32 by 32 matrix.  Tensor do the Clifford-Dellscherprin.  So you're talking about.  Now, in addition, actually, we said everything we know is really for one family.  And I have written here, for example, k and k, I have written as a number.  But in reality, this k is a 3 by 3 matrix in generation space.  So you mean it's a 96 by 36?  So it's 396.  So 32 times 32.  And tensor do the Clifford-Dellscherprin.  So it's 3, 384 by 384 matrix.  This is actually why it's much easier instead of writing matrices because they look huge  to put in tensor notation.  Because in tensor notation, you just put an index and you don't care how much it runs  because you know where does it sum.  So it becomes really efficient to do it.  Of course, we didn't do it like that in the beginning because for years, you have to struggle  with all these matrices to see what acts.  And it took some time, actually.  This was progressing developments.  All right.  So this actually here.  You say, all right, I would like to compute this guy.  How does it look like?  Remember?  So I will give you an example.  The reason I want to show you that things come out, that it's not that you say, oh,  the gauge fields come out, the Higgs field come out.  And people think that we are cheating because we know the answer.  We put it in and then it's not true.  Everything comes out in a very well-defined way.  We really cannot cheat in any way whatsoever.  So I can say, OK, AB, and then you can put indices like AA prime, and then you can do  DB, like AC, CB, and so on.  So then I can look, actually, at all the matrix elements AB.  And how do they look like?  So I say, OK, how do they look like?  It's the following.  I go back to my representation.  And then I'm really going to get an element here, like 1 dot, 1, 1 dot, 1.  Remember, I said A alpha I beta J, and this alpha could be 1 dot or 2 dot on the I.  It could be 1 on the I.  And you see, now I have to take all possibilities into account.  If I think about this, what is this, this would be the gauge field acting on the right-handed  neutrino because we defined the right-hand neutrino to be on this corner.  And in this respect, you'll discover, actually, that this guy would be 0 because the neutrino,  the right-hand neutrino, is a neutral guy.  So then, of course, actually, the next guy would be A 2 dot 1, 2 dot 1, and so on.  And this, actually, we discover, is a vector.  And then it is given by minus I over 2 G, I don't know, B mu.  So it's a vector.  And with a gamma mu.  Everything tends to be the cliff or the algebra.  So you get that this guy is a vector.  In addition, you start, actually, to get something like a guy like A 1 dot 1, 1 dot  A.  And then it's a doublet index.  A, remember, is an S2 index.  And you discover, actually, and then you put A is 1, 2.  And what do you discover that this is really proportional to gamma 5 HA?  And this is where the H is entered.  So the H is entered as inner fluctuation.  And now I say, what is this thing?  Why are we going this way?  This is an off-diagonal element.  And off-diagonal element, it means that it's really trying to connect two elements of the  left and right algebras with each other.  I mean, in fact, it's very invisible when you write D, because D F, that's what, there's  going to be 5 turns D F.  I mean, you know, D F, that's what will generate the gauge, both of them.  Right, yeah.  And the other one will generate the H.  Oh, yeah.  One really can see it.  But of course, you really cannot see, for example, here it's really easy if you see  that it's a doublet.  What's not so easy, actually, is to show, for example, the other guy, the A 2 dot 2 dot,  for example, 1A.  In this case, you say it's gamma 5.  And what happened, actually, it is A BH bar B. You really cannot get, guess ahead of time,  that you are only going to get 1 Higgs and not 2 Higgs doublets.  The fact that you get only 1 Higgs is really related to the fact that these guys are related.  Remember, I told you I have C plus C. And then, because we are talking about the same  C, it tells you that, OK, you know, you really cannot have different Higgs.  You are really going to get the same Higgs.  So the upshot of this is that one really can do this calculation, is a matrix calculation,  no complications, and after some algebra, one finds the Dirac operator, including fluctuations.  What does it mean?  It gives me the connections.  It gives me the connections on the space.  And the connections are nothing but the gauge fields and the hex fields.  Nothing else.  So up to here, you know, we have, we can achieve that.  And before doing anything, I can tell you that the Dirac action for the fermions, OK,  you know, one really has to write the gypsy.  I'll tell you actually now, maybe I can tell you why we can write the gypsy.  The reason we write the gypsy, because in euclidean space, I really cannot say gypsy  equal to psi.  Yeah.  So if in order to get, once you write gypsy, what really happened is that when you integrate  on the gypsy and the path integral, then you really, this is an anti-symmetry product,  and you don't get determinant of Da.  You get the pfafian of Da and the pfafian of the square root, which means that it's really  a chiral integration, and you have not doubled the degrees of Frito.  OK?  No.  Anyway, if I expand this, what do I do?  What do I really get?  I really get all fermionic kinetic terms, for example, nu bar D nu plus e bar D e plus  left and right at everybody, plus vector Fermi interactions plus tibs Fermi.  And you get exactly the correct hypercharges and everything.  Actually, why one would get the correct hypercharges exactly because of this, actually, if you  really work out the jA star j inverse, they really come and they bring, actually, the  u1s in really completely funny way in order to have the matching.  And here, actually, one of the lots actually makes an assumption that one takes SU of the  algebra, not the algebra.  In other words, one has to make a requirement that the trace of the A should be equal to  zero.  This is another condition that we put in.  OK.  So, here, actually, it's a first check.  Everything fermionic-wise works, and it works perfectly.  So what's next, actually?  What's next is that you need to make, so fermions are dynamical, but the bosons are not.  So how do they get the dynamics?  For all, we say that we have a care space.  The care space should have dynamics.  And we should be able to describe the full interactions of the system.  See up to now I have been speaking classically.  Everything here is classical.  Now, so here, actually, comes the idea of the spectrum action.  And the basic idea is that the Dirac operator, which was really the building block of the  non-committed space, in principle, one can study the spectrum of this Dirac operator.  So you study the eigenvalue problem, and you look at the eigenvalues.  And we know, actually, from studying, say, the remaining geometry or the Dirac operators  on remaining manifolds, that the eigenvalues are geometric invariants.  In other words, all the geometric invariants are included, actually, or lambda is a function  of all the geometric invariants.  So now the question is that how can we extract this dynamic?  How can we extract the information about the lambda?  Of course, it's not easy to simply work out lambda of every space, so we need something  more general.  And so the basic idea is that since the spectrum knows about the geometry, we took as the dynamics  of our system is conjectured to be given by the trace of the function of a d, where f  is a positive function.  I think you have to put a scale, so why is it that it makes sense?  Yeah.  Exactly.  Yeah, I was going to talk about the scale in a minute, because, yeah, d has dimension  on operator, remember it is, say, gamma mu, say, d over dx mu, and this actually has dimension  one, therefore you really have to divide it by something of dimension one.  Now, of course, actually, you may ask what is lambda, and I will talk a little bit about,  you know, dealing with the scale later, because this is a very sensitive issue.  What you should do with the scale?  Can you get rid of the scale, or what are we talking about?  Of course, I want to add something which is that this expression, trace of a function  of d, is the only expression that you can write which is additive, which is joined spaces.  See, the action should have at least the properties that if you take two disjoint spaces, it should  have.  Right.  This is the only expression you can write.  So you want to say F, but d1 and d2 should be disjoint in this case, all right?  Yeah, sure.  If you take d1 and d2 to be disjoint, it takes the fact that the full direction, f of d1  plus d2, okay, that it certainly proves that, you know, the trace of that will be the sum  of the...  Okay, I got it. ... function.  And there is a proverse to that, which can be that it is a functional, which is spec-wise.  And full field is this additivity properties, then it has to be learned this way.  But what about the positivity?  Positive because...  Positivity, no, no, no.  It's...  But it's important for Euclidean actions, no?  No, no, of course.  I mean, if you...  Z, like Y, on top of that, it's positivity.  But something that I've been, you know, toying with the idea that maybe it's self-positive  and things like that.  Anyway, so now essentially, how can we deal with this space?  Obviously, we really, for an arbitrary function, it's not that easy to work out, but what we  can really work out is asymptotic expansion.  This way, one would know how to do.  And the idea, the following, we use methods of heat cannon expansion.  You know, it's not the only methods, but they actually...  At least for remaining many forces and for product almost committed very force, they're  really very, very effective.  So what's the basic idea here?  You say, okay, you know, suppose that I have a function of some operator P, you make some,  I don't know, I don't know, a normal expansion.  You say, say, SPS.  And then...  And then you are...  You have something like trace of P minus S. The new identity trace of P minus S is given  by, you know, you use the Mellon transform formula.  And with the Mellon transform formula, you obtain that, you know, it's the opposite of  the definition of the gamma function essentially.  T S minus 1, trace E minus T P will be the real part of S is larger than or equal to  0.  And then the next step would be to use formulas of trace E minus T P using heat cannon expansions.  And then, of course, actually, this is given by A n of the...  Sorry, T n minus, actually, in this case, minus, well, n over 2 minus 2.  Okay.  At least in four dimensions.  And where n is larger than or equal to 0.  And this is A n of XB1 dv.  And these actually are called the serial discussions.  And they're defined in terms of connections of the dirac operator.  Yeah?  It's an asymptotic series.  It's an asymptotic series.  Okay.  I would say why it's useful in physics because exactly at a point that this is, you know,  the fact of d over lambda.  And then it will be clear, actually, when I write the expansion, it will be an expansion  of 1 over lambda.  So the first term will be 1 over lambda to the 4.  Then you are going to get lambda 4, sorry.  Then lambda squared, then 1, then 1 over lambda squared, that's all.  Now, assuming, actually, the scale is extremely large, then these terms will become, you know,  less relevant.  Of course, actually, the expansion will break down once you approach the scale.  And you know, the scale, you know, you can take the scale to be, say, unification scale  or Planck scale.  In this respect, actually, the expansion would definitely break down.  And in that case, one has to learn different ways of dealing with such expansions.  But you know, we did some calculations of this expansion for, you know, Robertson-Walker  metrics and things like that.  And really, it falls almost up to the Planck scale.  Up to the Planck scale, it's only when you are orders of, you know, like twice or three  times the Planck scale, then you start to see that it breaks down.  But not before, you know, which is really amazing that this action is so accurate that  it goes all the way up almost, you know, to the Planck scale, but not beyond, you know.  It's just there.  All right.  So what do we do with this expansion?  It doesn't give us anything sensible, because obviously, this looks far-fetched, I would  say.  That you take an arbitrary function, you say, okay, this arbitrary function.  The fermic part is straightforward, say, psi, d psi, okay?  And you get, you know, the usual expression.  You can say it's by construction.  But for the bosonic part, it's not really clear that this idea should work, you know,  that it should give you anything accurate, essentially, after all you're using some expansion.  All right.  So how to proceed, actually, here, you know, it's known, as I said, here we assume that  the continuous manifold is four-dimensional.  And in this respect, in this respect, the expansion, yeah, what happens in the following  that only an equal to zero, if n is off, for manifolds, without boundary, okay?  What happens for manifolds with boundary?  We have already addressed this question.  It's more, it's actually very interesting.  But I will talk about it later.  But for manifolds with boundary, n equals zero, and z, it means, actually, the only  non-zero ones are the a0, the a2, a4, and things like that.  And these formulas were worked out long ago by Gilkey.  He has some, you know, standard formulas.  And because of that, actually, what really becomes relevant, and this is the d squared  of the d, because, huh?  Disabase squared.  And what the trick of the following, you write this as g mu d mu plus a remainder.  And this d mu will be mu plus omega mu, which is the connection.  And this omega mu, okay, this also is given by, actually, there's minus g mu d mu d mu  plus a mu d mu plus b.  So what Gilkey did, actually, you know, he gave an expression.  This actually is, there are two important terms in this expansion.  The e, which is an invariant, and the connection, the omega mu, and, you know, curvatures, which  are constructed out of this connection.  And this omega mu, one reads from the following.  You write this expression into this form, into this form, and then you can know, actually,  that a mu is given by, you know, some expression, which is half g mu a mu plus a mu plus comma  mu.  This is a crystal, a contraction of a crystal.  In other words, actually, one really can compute all these expressions in a unique way.  And remember, we have already computed our d.  So we have this d, huge matrix.  You take it, you square it, you get another huge matrix, and you write it in this form,  and then you try to find the curvature of the space.  Remember, it's re-enlightened curvature now.  And you also get the e-inverted of the space, another huge matrix, you know, 384 by 384.  So what Gilkey does, I'll look, you know, this a0, the first time in the expansion,  is given by 1 over 16 pi root g, trace 1.  Trace 1, of course, is 384 because it's 384 by 384 matrix.  So this is actually nothing to compute.  See, R, which is independent of lambda, obviously.  This is, no, no, this is actually lambda 4.  This is the re-cosmological constant.  And it's huge, you know.  The cosmological constant is huge here.  So if I write it as 0, that's a different story.  Okay?  This is not quite high, actually, because what if it's not 1, in the sense that you have  the...  Things is effective, you know.  You want to talk about effective.  You want to talk about, yeah, I'm talking about, you know, 0th order cosmological constant.  Of course, there are contributions coming from the vacuum expectation value of the higgs,  but these are supposed to be much smaller, actually.  I'm saying that when you compute this coefficient as 0 and when you take into account the Major  Anna-Mass matrix, you get another term.  Okay, but...  Yeah, okay, you know, at...  So you get this time.  Yeah.  That actually, but...  But that comes from the higgs, no?  That comes from the Major Anna-Mass matrix.  You mean the G-squared term?  Yeah, sigma.  This is a sigma, actually.  Sigma, yeah.  Yeah, but remember, yeah, okay.  But this takes you after it.  Sigma would come later because I see sigma lives here now.  Sigma lives here because a priori we don't know.  Okay.  Anyway, so the interesting part, actually, is this guy.  So, you get, actually, R over 6 is the Einstein term, and then you say trace E. Now, you...  Let me actually give you an idea of what trace E is in this case, or what E is, actually.  You know, E, you have to read by squaring the D, by squaring the D. And so, obviously,  let me remind you that you have all this D-slash part.  You have here gamma 5h part, gamma 5h dagger.  You have some D-slash, okay?  When you look at the D-squared, yes, the off diagonal terms are really going to give you  what?  H-h dagger terms, which is really nothing but the higgs mass, okay?  So you discover, actually, essentially, and then you trace it, and you only get trace H-squared.  So if you really work it out, actually, in this example, what you really get, you're  really going to get the following, actually.  You get this Gibbs up to a factor.  So you see, this is the guy that would eventually give mass to the neutrinos.  And this is the higgs mass.  These are essentially your coupling, your square of your coupling.  So we look at this expression, and then we notice the following.  The curvature term, you mean the whole minus sign?  Yeah, there's a whole, overall minus sign, yeah.  See, I write minus, okay.  So if you'd like, yeah.  Okay.  So.  You have included the one half R into E, or not?  Minus 2 over pi squared, if you wanted this, F2 lambda squared.  Yes?  Root, gd4x.  There's a minus sign here.  Well, the notation that R is negative for Euclidean spaces.  Anyway, so you see, we get a mass term and with a minus sign, this is really extremely  important if you, as we'll see for the standard model, because the standard model is really  characterized with a potential V, as you see, lambda over 4h2h bar h squared minus m squared  over minus mu squared over 2.  And then the minus sign that gives the potential the Mexican hat.  It's essential that you get a minus sign.  Now obviously, you are really going to get a minus sign.  Why?  Because if you look at it, in the end, you are taking trace of E minus Tp, P has D squared,  D squared has S squared.  Obviously, you are going to get the first term in the expansion, a minus sign.  The next term will be plus sign.  It alternates, which is extremely nice, because of course, but you really cannot guarantee  that, okay?  In principle, before you start.  You don't know.  So this actually, important points we can point that the curvature, which is the Einstein  term and the mass term are really not equal footing.  So h bar h is like curvature of, is the curvature of the antenna space.  And okay, you know, scale lambda squared.  What's lambda squared?  Of course, actually, one would like this lambda squared to be related to 8 punch or 16 punch,  whatever, you know.  Okay?  It should be related to the, so this actually is the plank, plank squared.  So from here, one would read that if you would like to make sense of the scale lambda that  we started with, the natural interpretation of it is that it's almost, of course, you  have numbers, you know.  You have F2, it's a maliant transform of the function.  So you can get factors of 10 or things like that.  You have the pi squared.  So it is of the order of the plank mass.  It's not exactly the plank mass, because you have some factors here.  But at least in order, we know that it's of the order of the plank mass.  The next time, day four.  And according to Gilky, what's the formula?  The formula tells me that, what is it?  Okay, you have, let me write here.  Okay, I'm not going to write the surface terms.  So what re-enters is the curvature squared, everything squared actually here.  E squared, curvature squared.  Now you know, this E squared, the omega mu squared, they really give me invariance in  terms of curvatures, square of curvatures.  What are the squares of curvature?  Of course, R mu nu squared is there.  But here we're really going to get what?  We're going to get F mu nu squared, which is curvature squared.  And we are going to get kinetic energy of the hex term and the sigma term and things  like that.  So, and, yeah, exactly.  Yeah, in mill terms, there's nothing but curvature squared terms.  Because what's F mu nu?  It's if you take the A squared, as you see, D plus A squared, you are really going to  get F.  The square of the curvature, the square of the connection is the curvature.  And it really lives here, it lives here, and one really can compute it, some long calculation,  but in the end, it's well-defined calculation.  The important thing is there's no ambiguity at any point, okay?  So in this action, we are going to get F0 over 2 pi squared.  Then R mu nu squared,code v squared, dou Improved  plus.  plus conformal factors, kinetic terms for the H,  quartic terms for the potential, and similarly everything I write here, I have to write also for the sigma field,  and so on.  Our sigma score, anyway.  It's an important thing that first of all, this time there's no scale.  Second, we notice that we have a vile squared term.  Now this is really important for, you know, if you are really doing, say, a realization of gravity.  You know, this term would make, would improve actually the propagator of the graviton.  It makes the theory, of course, non-unitary, but you know, this we don't really worry about because in that case we are cutting,  you have to cut the expansion and you know, if you would like to treat unitary, you should not cut the expansion.  You have to take the full Fletch theory, but at least this conformal tensor or the vile tensor squared is important for the propagator of the graviton.  And now here we see here actually that we really have a kinetic term for the u1, for the electromagnetism,  a kinetic term for the SU2 weak, and a kinetic term for the color, SU3.  And, exactly, and then we look here and then we immediately find a relation that tells me that g1 squared equals g2 squared,  that's 5 over 3, sorry, g3 squared, g2 squared, 5 over 3, g1 squared, which is the famous unification one obtained for SU5 or SO10 gratification theory.  We will have actually, we have to talk about this later in the next lecture because this is not exactly satisfied.  It's not exactly satisfied because as we are going to see next time, you plot the, you know, okay, the logarithm of the developments of the alphas,  which is g squared over 4 pi, alphas inverse of the alphas inverse, then you discover that they don't exactly meet, but, you know, this is a different story, you know.  Whether they meet, they don't meet, they almost meet, but in principle, you know, there's so much in between that we don't really know, we don't control,  and it's premature to say whether this is a bad point or not a bad point.  At least they are the same order. They are, you know, it's almost, you know, okay, within, let me say, within 10% it was, within 10%,  but you have extrapolated 15 or 16 orders of magnitude, actually, which is huge, okay.  By a variant, the normalization.  Yeah, you know, so there is, and, you know, things, we have discovered that things have not really been calculated, you know, to two loops in the model that we have,  because this field actually turned out to play an extremely, very important role.  The sigma, which a priori we ignored, at one time we even make a prediction that the Higgs mass is given by a certain number,  and this turned out not to be correct, because we have ignored actually this field.  However, if you take this field into account, then, you know, you're completely in agreement with the experiment,  and in addition, actually, the normalization group equations taking the sigma into effect has also only been carried to one loop order,  and nobody has carried it to two loop order, so we really don't know.  So maybe it will change a little bit, actually, you know.  My guess is that it will change, but I don't know whether it will change in the right amount in the right direction, because this is up in the air.  So let me summarize again.  So we started from a very general idea, the general idea that space-time can be approximated by a product of a continuous time discrete manifold.  The real answer, we don't know.  You know, it may be just completely discrete, but to a very good start, it's because we know that generativity works very well,  so we know the continuous part up to the Planck scale works very well, so we can up to that energy level or a little bit below.  We know that the, you know, Raymaine and manifold part works.  So to assume that we have an invisible extension, which makes it almost commutative, but in reality, it's not commutative,  that we can extend it by a finite space, we entertain this idea, and we discovered that to almost uniquely, we can predict the structure of the model.  We know the Fermi-Nichro presentations, we know actually why there are 16 fermions, we know what are the representations of these fermions,  we know the gauge group.  This actually, yeah, I'm saying that what we can, that things what we cannot do is much more, of course,  because what we cannot do, for example, predicts say the electron mass, and you know, oh, the masses,  because in principle, you know, if there is more restrictions on the, one should be able to, for this,  remember, the finite space contains all the Yukawa couplings, so it contains actually information about,  so if you know that, for example, that K is like 10 to the minus 2 or something, then you know that the ratio of the electron mass to the top quark mass is like 10 to the 4 or something, so the minus 4.  So in that case, like this information we are putting in by hand, we are not predicting, but if one really is completely successful,  everything comes out, but you know, this is really far-fetched for time being, because you need some extra structure which we really cannot put our hands on, not yet, you know.  So I think there is another aspect, the criteria and the properties that we verified, do not really still tell us actually why there are three generations,  this we don't know, this must come out somewhere, you know, so this aspect, we have no idea.  No one has any explanation for that.  We need an explanation.  And as in the explanation for this, there is no, as even a guess whether there should not be a false generation.  Yeah, this actually almost rolled out now actually, you know.  No, no, but there are two things to say.  There is one, these two Japanese who got the Nobel Prize, and the reason why we need more than two generations.  And that's the cap-de-bose Kobayashi-Maskawa mix in the text, because it involves a complex number only if you have more than two.  So I mean, three is the first number of generations for which you get a complex number, so for which you get violation of charge conjugation.  But then, okay, I think now experimentally it's known that there is no false generation.  There are very strong tests which are true.  Yeah, so why three?  Actually, it's not that we don't know, nobody knows also.  It's up in the air, why the masses of the firms are the way they are, you know, the standard model after all has what, 17 or 18 parameters?  Sure, but there you can say one thing, which is it's exactly, you know, like if somebody would come and would tell you why is the metric of space times the way it is, okay?  You could say, well, Kepler's atom gas, but you know, you could say, well, it's like this, it's because of circumstances.  So the issue is, what is the geometry of the finite space?  Yes.  Is it due to circumstances or does it have a deeper meaning?  Right.  And this we don't know.  This we don't know.  One would hope actually, see that, okay, if maybe, well, 20 years ago, you know, okay, you know, before say that, ventals, not geometry, if somebody asks you why the standard model, you say you could have answered the same.  But now actually, I think we are very close to knowing that it is, it must be like actually three cross-situ, it must be like 16.  So we must have a hex field.  So this actually have been answered satisfactorily, but we are far from, you know, the full answer and this is the.  It could be both a white and a hex.  Not here actually.  Here, okay, if later, if, you know, we take off this condition and we really want to push everything to grand unification, then of course you have many more hexes actually.  Not only one, we are going to get, but that actually will be relevant at higher energies.  And we will not be able to see what really goes on at lower energy except for the six doublet.  You know, it's not clear.  But that's another story, you know.  Anyway, I would stop here, I think, because next time I will continue by analyzing the quantum aspects of the model and maybe, you know, theorizing a little bit about.  So what you have so far is the first characterization.  Of course, yes.  So answer what Thibault has last.  It's a very edge of geometry and some of the space set up.  Right.  But what is quite, one should stress one thing, you know, which is that there are seven or eight sides which have to be correct.  And they are all correct.  And moreover, you know, it encapsulates the state of most complexity of the Lagrangian into something which is.  This thing was done before, you know, this would have been like, and then people come and verify it.  It's not like a revolution.  But the thing actually lost the luster because people didn't think that, okay, you know, the answer, you are fitting it.  And this is the, you know, biggest thing that one has to fight in the misconception that there are billions of possibilities and simply were given an interpretation of these posts of the one possibility, which is correct.  And there are many other possibilities, which is not really correct because, you know, according to this specification that I went through, that there are not really many possibilities.  It's almost, it comes uniquely, you know.  But that one would have to explain why if you want the correction by the final space improves the ordinary four-dimension continuum.  And the reason why it improves it, it changes the care dimension.  Yeah.  It changes the care dimension from four to two. And that's a big improvement because then you can write this profile.  But it needs to be finite.  Well, it's the most economical to be finite.  I mean, it could be that the splitting is actually more complex, but I mean, the simplest way to see it is that you have a four dimensional space.  And then you make it two-dimensionality dimension without changing the metric dimension, which is remaining four.  But if you have a, you know, if you start finite, then you'll have to fight all this Kaluza Klein modes, as I would say.  You know, you have many degrees of freedom, which we don't see.  So, it's a new dimension by something finite.  Exactly. By something finite, which doesn't have this infinitesimal mode, like in Kaluza Klein.  I mean, that's a very basic thing.  Okay.  Okay.
Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_0_59.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_60_119.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_120_491.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_492_1967.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_1968_2747.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_2748_3887.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_3888_5015.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_5016_5567.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_5568_5615.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_5616_5699.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_5700_5963.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_5964_6659.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_6660_6779.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_6780_12011.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_12012_13211.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_13212_15503.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_15504_18299.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_18300_18839.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_18840_19799.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_19800_20147.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_20148_20615.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_20616_20819.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_20820_21671.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_21672_21755.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_21756_21923.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_21924_22187.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_22188_22355.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_22356_22871.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_22872_24527.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_24528_24671.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_24672_25631.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_25632_26759.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_26760_26843.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_26844_28079.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_28080_28883.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_28884_30335.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_30336_30383.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_30384_31799.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_31800_32075.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_32076_32723.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_32724_34127.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_34128_37031.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_37032_37511.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_37512_37571.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_37572_38591.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_38592_39431.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_39432_40499.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_40500_42419.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_42420_42479.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_42480_42647.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_42648_43667.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_43668_44207.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_44208_46019.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_46020_46667.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_46668_47195.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_47196_49367.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_49368_55391.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_55392_55847.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_55848_56123.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_56124_57179.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_57180_57767.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_57768_58067.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_58068_58115.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_58116_58907.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_58908_59303.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_59304_59363.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_59364_59879.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_59880_60767.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_60768_62495.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_62496_64811.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_64812_65207.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_65208_69923.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_69924_73991.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_73992_75719.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_75720_76667.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_76668_78215.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_78216_81335.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_81336_82163.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_82164_84863.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_84864_88919.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_88920_89663.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_89664_89723.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_89724_89843.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_89844_90083.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_90084_91691.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_91692_91955.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_91956_92627.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_92628_93107.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_93108_93167.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_93168_93227.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_93228_94103.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_94104_94151.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_94152_96827.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_96828_97079.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_97080_101483.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_101484_102863.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_102864_103043.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_103044_103475.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_103476_104327.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_104328_105839.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_105840_106043.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_106044_106103.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_106104_109895.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_109896_114767.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_114768_114959.png Ali_Chamseddine_-_2_4_Spectral_Geometric_Unification_114960_126592.png
Let's get right into the song. So last week I introduced the idea that non-committal geometry can be taken as a basis to unify all fundamental interactions. And so what I did is that I defined what we mean by non-committal space. I gave the definitions. And so with certain properties, and just as I remind you, just set the spectral triple and I defined these five quantities and the relation among them. And then the question, of course, is if a non-committal space is the basis to unify the interactions, then the question is what kind of non-committal space should we take, you know? And as a starter, I made an assumption, and the assumption I made is that space time r is a non-committal space which is formed as a product of continuous four-dimensional space times some finite space. Now four-dimensional spaces are well understood, and then I went to see essentially what one should take for the finite space, you know, what kind of space should one take. And so for that, started by classifying spaces which fit certain physical criteria. And the physical criteria that we need, and the following, that, you know, the Hilbert space, which would be space of, you know, fermions essentially for us, would include all the fermions, and in this case, actually, the fermions just satisfy certain properties. And essentially we said we have fermions, but because actually of the presence of what we call the reality operator, one really requires essentially what we would like to require essentially, that the fermions are real, which in physical language means Majorana. And in this respect, of course, actually the presence also of the gamma operator, which is a current operator, would allow us always to impose this condition. So what we really have, we know, we say we have current spinors. Now then, of course, actually if you have to on J on psi, you are really getting to get something like the conjugate spinner. And the question is that is this an independent set of fields? If it's an independent set of fields, then you really immediately face the problem that all your particles will be doubled. You are really going to get, for every particle we know, a mirror particle. And you know, sometimes it's called the mirror fermions problem or the mirror doubling problem. And usually it's not easy to solve this problem. Now we are really working in Euclidean space. And the issue, of course, is that if you really go to Euclidean space and you said I would like, sorry, but the physical space, of course, is Minkowski. And if you would like to impose both conditions, the J psi is psi, then this actually would impose certain conditions on the properties of the space. As I have shown last time, that would impose a condition what we know as the k-o dimension. So in Minkowski space, you really need to impose both conditions. The J psi is psi and gamma psi is psi, and this would imply actually that the k-o dimension of this space is equal to zero. Which implies actually that J squared, which is epsilon, should be one. J d is, I think, epsilon prime d J. And I need one more sign. J gamma is epsilon dot the prime gamma J. And d gamma is minus gamma J. So these are the properties that we need. And essentially, it really forces, now we know actually that in Minkowski, you have four dimensional space with signatures, say, minus plus plus plus or the opposite. And these would somehow cancel each other. And then you are left with k-o dimension two, which means that the internal space or the finite dimension space should be dimension six, k-o dimension six. So that's actually forced the finite space to have certain properties and k-o dimension six. Then you need certain sets of epsilon, for example, epsilon is really minus one. I forgot actually, I think, epsilon prime is equal plus one and epsilon prime is minus one. Something like this. So then actually one goes ahead and try to understand these finite spaces. And what one discovers, and I would go through, not through the proof, but through the realization of the proof, is that one discovers that the algebras of the finite space are essentially what one finds, that the complexification of the algebra. So essentially here, one assumes that the finite space is characterized by A f, H f, d f, gamma f, and g f. Everything is for finite. And one discovers that the complex extension of this algebra, if you take the center of it, there are only two possibilities which are either C or C plus C. And this actually is inconsistent with k-o dimension six, which leaves us only one possibility that A C is C plus C, and really corresponds to an algebra which is a variant of, you know, let me call it M N of C plus M N of C. In addition, when we imposed certain isometry on the first algebra, and this is a point actually we don't really, you know, understand, you know, we cannot give it a mathematical characterization why it should be like that, but it is like that. And the other possibilities are not relevant. And this really reduces this instead of matrices on the complex numbers. You really take matrices over the catorials for the first, and in which actually N is 2A in this case. So N is even. And also, because one would like to impose the chirality operator on this algebra and elements of the algebra, for example, gamma A equal A gamma. So in this case, what happened is that this M of 2A of H would split into M A of H plus M A of H, for example, plus M N of C or 2A of C. So the upshot of this is that really the first non-trivial example you can do is the following, that A should be one in this case. Sorry, A is, yeah, this is a matrix, you know, M A of H. This should be 4A or something. N is, let me see, M is. N is 2A, yeah. So essentially, yeah, so essentially this, yeah. So essentially, really the first non-trivial example is M1 of H or H plus H. Okay, let me write it like this, and this actually is like 4A. And in this case, the first example would be H plus H for the first algebra plus M4 of C for the second. Of course, other algebras are possible. But Ali, what is important to say here is that you have the gamma, which is the diagonal matrix. So, on the first, right? For M2 of H. Right. So it takes the even part. It's the even part of M2 of H, which is 8 plus H. Yeah, okay, so I have to say, okay, this I would say M2 of H even, and then it really, at least for the first non-trivial case, it belongs to H plus H. Now it's up to this point, you know, I went last time. And the only thing we're, you know, the assumptions I made are there. So at one point, one did impose on isometry. If one didn't impose on isometry, then you are really going to get, instead of, you know, H plus H, you're going to get M2C plus M2C. And one really can show that these cases would not be physical, but one cannot rule them immediately, unfortunately. What it is, is that if you have a real algebra, then you have three cases. The symplectic case, the pteric case, and the also. Exactly, there are three. And you're okay, we choose, you know, the symplectic, you know, case. So this is a choice. This is something which is not imposed by the... But you could have chosen once. We could have chosen, yeah. Symplectic, symplectic, or... We could have chosen this section. So this is the really... the other is actually not relevant because they would really give you real algebra. For instead of H plus H, you can take M2C plus M2C. What would it really give essentially? It gives you, you know... It gives you an extra U1, actually, you know, which is anomalous. So one really can work it out. It gives... it's a little variation. Everything almost the same except this little variation. It gives you an extra vector. An extra vector which is anomalous, you know. Maybe there is an extra vector lurking in the spectrum, but, you know, people are talking about, you know... Recently, there was some papers in which they said they observed something which deviates from the standard model. And, you know, some people said maybe it's an extra vector. So who knows you? Anyway, but this is actually the most symmetric possibility. The only assumption I made also that left action... This is the left action. And then this is, say, the right action. And then we said this is J B star J inverse. And we said... And this actually is defined to be the opposite. So this is one of the axioms that left action and right action commute with each other. So this actually has been used. This property really has been used up to this point. Now, from here, actually, one needs further assumption. And, you know, I will discuss both possibilities. In that... And this is called order one condition. And now, actually, we understand this condition. It's really a condition on the linearity of the connection. If one drops this condition, then what happened is that the connection would really get quadratic terms and that really can work out the possibility when this is not so, when this is not satisfied. But this I would leave to the last lecture to allow, because actually in non-convenient geometry, there are examples where this condition is not satisfied. And you know, this is what the quantum spheres. Okay? Anyway, so here it will have a really big advantage that the connections are linear. Everything is very nice. Everything, you know, falls into place. Okay? What happens if this is not satisfied? You know, this would lead us to grand unified theories and Patti-Sellem models and things like that. But this will come much later. Okay. So what are the consequences of this condition that's satisfied? Okay. So I will assume, actually, that this is satisfied. And on top of it, actually, now there is one more condition. Is the question that if we take the commutator of the Dirac operator with the center of the algebra, is it zero or not zero, actually? This is a question. Well, what we have shown is that there are two possibilities. If it is zero, then this symmetry is really intact. And if this symmetry is intact, one can show, actually, that you are really going to get, when one work out a model, which I'm really going to do a little bit later, is that you really get SU4 color, which really corresponds to this M4C, as a symmetry of the quantum. So essentially, in this case, you really get a higher color symmetry, which is not observed in nature. And on top, actually, you'll find out that the neutrinos are all massless. They are exactly massless, which now we know that it's not true. So from the physics point of view, we know, actually, that the three must be different than zero. Because if it is zero, then you are already in contradiction with the experiment that neutrinos will become all massless, which would have been OK until 1995 or something like that, because people didn't know, were not sure, actually, whether neutrinos are massive or not. Because every this tiny mass, there are only limits on the masses of the neutrino. Now we know, actually, that they are really massive, but the mass is really extremely small. But that they are massive is well established. Now what happened, actually, if this difference is zero? Now the beauty of it is that if this is different than zero, and for this condition to be satisfied, the Dirac operator must have the following form. And so in this case, actually, instead of telling you in description what really happens, I would start being more concrete. I will tell you exactly what are the representations, what goes on, so that you have an idea. That this actually, as I would say, field is not simply talk, you know, you really can go and compute everything unambiguously, not much ambiguity, and with all the details. So, all right, so essentially let me examine this in a minute. But I really can tell you that the fact that this is non-zero really gives a unique possibility that you really can give the neutrinos a Majoranum mass, and it will explain, actually, why the masses of the neutrinos are very small. It really, it's an example of what was known as the CISO mechanism. So here it really comes out naturally, and it solves the problem in a really very nice way. So, let's see how things would look like if I wanted to go ahead and do some computations, you know, what should I do? So this is really my starting point. I'm really going to take H, the algebra A, to be H plus H plus of course. And this, remember, almost uniquely we were led to this. It didn't come out of the sky, you know, out of classification. This was the first non-trivial result that we have. Okay, so essentially if I would like actually to represent an element of A, how would I write that? Then I am going to write it in matrix form, you know. I'm going to show you that everything that we do is really matrices, essentially. We really can use tensorial notation if one would like to simplify his life. And you know, recently, you know, everything that we do can be put on the computer, so in the end one doesn't do much calculation. You just write it on Mathematica and press a button and it gives you the answer. So, you know, it's, all right. So I'm really going to use, you know, index notation. So essentially here what I'm really going to say, all right. So in order to tell you what my notation, I really have to tell you what are my spinners. Okay, so the spinner actually, as you see, it belongs to this guy. So what I'm really going to write, I'm going to write as psi alpha i, where this alpha is, would see the H plus H and the i would see the 4. So obviously i is 1 to 4 and the alpha is also 1 to 4, but it's a different form. It's H plus H. Now, and then, of course, actually I'm going to write this as, you know, a dot a, where this has two indices, like 1 and 2, and this is 1 dot 2 dot. So this actually tells me that it's a doublet under the catorians, you know, doublet left, doublet right. And what we write actually, we write as H right plus H left, because we have already graded the algebra. And in this respect, my spinners would be psi a dot i and psi i. Okay. So how many spinners I have? I have 16 spinners. And of course, we have also the conjugates, but the conjugates, we're going to see that they're not the independent spinners, because the thermion doublet problem is solved through, you know, the Majorana condition. We are not really going to say Geppsi equal to psi. The problem in Euclidean space is really solved in a very clever way by, you know, in the path integral formalism that only chiral thermions get integrated and they give square root of determinant instead of determinant. So it's really solved in really very subtle way. You know, this is a nice discovery. You get the fulfillment. You know, I'll talk. Anyway, so you see actually now, you know, without redoing any work, I have already obtained the correct classifier. First, actually, we already predicted that you have 16 fermions, which is exactly per family of course, which is exactly what we have. We will also actually find, let me, you know, let me write this one of this. The i. The i eventually, you know, when this condition, the d with z of a different than 0, will also be broken into 1 and i, where i is 1, 2, 3. So that actually will be the color index and this is the lepton index. So this is lepton and this is the quark. And then you can see that the lepton is the fourth color. It comes out of the fourth color. And in this respect, you're going to get psi 1.i, psi 2. Sorry, psi 1.1, psi 2.1, psi a, sorry, 1.2. And then I have psi i, which I'm going to leave it as psi a1. Then I have psi 1.i, psi 2.i, and then I have psi a.1, psi a. What's in this? You know, remember, this is a dot means, when you see dot, dot means right. So this would be what I'm going to call neutrino, right? Why give it the name? Because if it comes out like that, the quantum numbers tell me that this will be a neutral guy. But this comes out. You know, I name it according to the usual naming and it would agree. And then, of course, this is a, right? And this, of course, is upright. This is down right. Similarly, psi a1 is the doublet, which is a new a, a new e. It's a doublet. You know, a is 1, 2, left. And this is up, down, left. OK? So you see, actually, without much work, we obtain the correct representations of the particle spectrum of all the fermions. We know which representations they are. So one really can write that the 16 in this case is a 4, 4, 4. And this 4 is decomposed into 2 plus 2. And this 4 is decomposed into 1 plus 3. And then this is 2 right plus 2 left. And this is everything that we see belong to that representation. Everything falls into place. OK. So now, so these are my 16 spinners. But remember, actually, you know, I have the j psi. And so my spinners will be psi alpha i and then psi alpha i conjugate, which I'm really going to go psi alpha prime i prime. They're not independent, but they are related through the j. OK? What does it mean, actually? It means if I would like now to write the element of the algebra, you know, I can write it as follows. I can write it as this one, you know. Remember this belongs to the second guy to M4 of C. So it is something like this. And this is delta alpha beta. And this is like yij. So you see, actually, how an element of the algebra is satisfied. And then you can form the b opposite, which is jb star or j inverse. And you discover, actually, what happened is that here you are going to get the y times 1. You are going to get yij, essentially, transposed into 1. And here you are going to get the x alpha beta into delta ij. And if you look at, you know, this element and this element, obviously, they commute with each other y because then this is don't talk to each other. This is in the first algebra. Well, this is now in the second. And they don't talk to each other. So essentially, jb star. Yeah, absolutely. Right. And this is actually, I wrote it as transposed, but, you know, a star is, OK. So essentially, we can write it in this form. So the first a b opposite is satisfied automatically. Now where do I go from here? Then I have to satisfy this condition. So this actually is a condition on the d and the algebra, both, you know. Because what happened in the following? You try to satisfy this condition, you work out the matrix representation, and you obtain actually. I mean, you mean the commutation with the center, right? So you are right here. This actually, you are right. Yeah, exactly. Yeah, it's exactly. Yeah. This is the difference here. This is the difference here. OK. So this actually, we have to assume it's different than zero. And then actually, this really would require the form of the direct operator to be as follows. You can show that you are going to get something like this. Let me call it, I don't know, let me call it dab or something. And here we are going to get the bar. Here we are going to get something. And then one can show through a little algebra. Actually, I learned it through abstract proofs, which you know, beyond me. But through a little algebra, one can show actually that the only operator that satisfies in the following, that here you can have only one non-zero entry. Only one. Anything else would not satisfy this condition. OK. Of course, actually, this, together with that, we have also to make sure that dA would be opposite is equal to zero, putting the two firsts actually. First we know that here we must have a non-zero element. The question is that the whole thing could be non-zero. But because of this condition, the first condition, you discover actually that this is possible if and only if you have only one non-zero entry. What does this mean? Actually, this one non-zero entry. Let me write actually the expression for you. And let me see how does it look like. You know, I have, remember, when I write Dirac equation, what do I write? I write psi star deep psi, or I can write it this way, you know, psi deep psi, OK, in a product. So this is my, say, Fermi reaction. And this Fermi reaction, if I expand, and then I just look at this term, what does this mean? This actually, if I look at that classification, it really tells me that you really have an element which is this one. If I have more, it would mean that, for example, the up, the down, everybody will start getting Majorana mass, but you know, you're getting a Majorana mass would break charge, would break the charge conservation, because in this case they would not have zero charge. So the only term that you're in physics wise that you can write is a term which is reneutral. It must be neutral, otherwise you destroy charge conservation. And it's not an accident, actually, then that you can have only one nonzero term, and this one nonzero term would really correspond to the mass, Majorana mass of the right-handed neutrino. Now, I'll explain later, actually, that why this is really extremely important in order to solve, it to explain, essentially, why the neutrino masses are so small, OK? So this actually is a result. It tells us that, you know, in order to agree with the observation, the Dirac operator should take this form. It has only one nonzero entry, and the rest are zero. This d and this d bar are correlated. They are, you know, one in the complex conjugate of each other. And the only thing I have to find is that what's the form of the Dirac operator in the finite space, OK? So what shape does it take? You know, the whole idea, the things really are really constrained, you know, because in a way, we have to explain why the standard model. This actually, you know, why the standard model. So if the hypothesis that everything comes from the geometry, we must really arrive at this answer in an almost unique way. Maybe one day one would have fine and say, OK, here it comes, this is without any ambiguity. But you know, there's always a possibility that you have to put some physics, because if you don't put any physics, in principle, you can get anything, right? Because, you know, if I don't want to explain nature, then, you know, I can start with the smaller space or bigger space, OK? You can start with too small, because if you already assume that it's four dimensional continuum times finite space, OK, and the finite space has k dimension six, and then you make a very minor hypothesis on that space, then you are already led to the sixteenth, so it cannot be too small, you know? This is the first example, but then you cannot rule out why it's not bigger. Oh, no, of course you cannot swap. You know, the question is that can I say uniquely, you know, this is the only thing I would ever get, you know? But this is the first possibility that one gets. OK, so now actually, where do we go from here? And what we do is the following. So we would like actually to see what type of Dirac operator can we have, this DAB. Now remember, actually, the D satisfies many properties. I have written them up, so you really have to show folks, you have to verify that what the area that G da is, Gd, still I'm talking about the finite space, remember, is equal to dj and gamma j equal to minus j gamma and d gamma equal to minus gamma g. So these properties, I have to satisfy, and so if you do that, you'll immediately discover actually that the Dirac operator of the finite space really takes simple form. Remember, you know, let me write it in terms of new right, E right, because I don't want to write one dot and two dot, because then it's more physical. And in your right, E right, and then I have, say, new left, E left is a doublet. And then up, right, down, right, and up left, down left. Okay? This is really a doublet, but these are really singlets. Okay? Why is it singlet? Because actually, this sigma really breaks the algebra from, you know, for this condition and the other condition to be satisfied. And then you put the sigma. What you discover in the following, that this algebra, h plus h plus m4 of c is really broken. And obviously, it must be broken, because you put the sigma, remember. This is like one, and this is the i, and this really breaks the color index. It breaks the color index into one and the three. And in this respect, the m4 of c is really broken into c plus m3 of c. However, because that condition dA, b opposite is equal to zero, it really, what happened, and then it breaks, actually. This is also broken, you know, because this is one index. We'll talk this one index. And then, essentially, this is a c plus c. It breaks, it breaks, and the three are not independent. What one discover, actually, that the element of the algebra A takes this form, in this way. So you are going to get, you know, x, x bar. This is my first h. These are my first ketonion. In other words, actually, remember, a ketonion takes the form like alpha beta alpha bar minus beta bar. It tells me that in the first h, it must be of this form. Now here, what do I have? I have a ketonion. So this is my first four, and this is my bottom four. And in the bottom four, what do we really have? We discover, actually, that this is x, and of course, this is zero, and this is m, where m is the element of m3 of c. In other words, actually, what we discovered is that the three c, the c here, the c here, the c here are not independent, and they are essentially one c. And the finite algebra of the finite space really breaks down to c plus h plus m3 of c, which is really a symmetry of a central model. It really comes out of the requirement that there is a Majorana mass for the neutrino. No, this really has drastic consequences. OK, so with this action, the element of the algebra, I go with the d, and then we can really classify the d, satisfying all these properties. And I want to cut the story short. What happened is that the only non-vanishing elements of the d. There is one more requirement that we have to put in order to really completely single out the d, and that the d commutes with, no, actually, yeah. d commutes with the element, which is lambda lambda bar lambda lambda bar in here. If you do that, then you discover actually that only the non-ucala coupling of the d-operator are non-zero, and really we have the d-operator completely singled out. If I want to write it down, I can write it down. Let me write it down, actually. So how does it look like? When you zeroes, I think here you are going to get k nu, and here you are going to get ke, which is actually the yocawa coupling of the neutrino and the yocawa coupling of the e. And similarly here, you are going to get, this is new left, e left, and this is k nu right, ke right, like that. And similarly for the quarts and the only off diagonal elements, you are going to get. And here also. So this is k up, k down, and diagonal form. So in other words, actually, after some matrix algebra, one would know exactly what the d-operator is. So what do we do with this now? What we do with this is the following. Having determined that. Exactly. So you know, many, you'll get many, essentially a point that you get many zeros. By, you know, gamma d minus d gamma, I tell you all diagonal elements are zero, you only get non-diagonal elements, and so on. So it's little matrix algebra, nothing much to it. And however, once one is finished, one can proceed as follows. Note that we have not really talked about dynamics or anything, we just talked about classifications and representations, and satisfying the properties of the uncompetitive space. And you are led almost uniquely, actually. What do I call that of the standard model? Now, remember now, I said that the Dirac operator of the full non-competitive space is equal to the Dirac operator of the manifold cross one plus gamma five cross df. We have been, what I was talking about up to now is the df. We know now exactly how the df looks like. dm, of course, we know because it is the Dirac operator of the usual, say, remaining manifold. And similarly, we can say a equal a of the manifold cross af and j. Everything is a product, actually. Don't have to write it down. All right? Now, remember, actually, here, when all starts with what's non-fluxuating d. So we start with a non-fluxuating d. Say, let's start with a d which corresponds to, say, some flat space. Starting from a d from a flat space, for example, you can really start with gamma mu d mu like that, actually, for the d. And we know, actually, that you really can generate the metric of the dm out of fluctuations. These are the outer fluctuations. And this is easy, actually. What we have to do is that you say this is e mu a gamma a. You make this as a function of x. You discover that you have to add a spin connection in order to make the thing, in order to make the thing coherent. And so that, actually, is easy to do. So here, we have seen last time that the d always go into dA, where dA is a d plus a plus j. But that's a good for the global. Yeah? What you have written there is for the global. Here is the global. Here is the n-arithomorphism, no? No, but I mean, if you want to do the fluctuations, you do that for everybody, of course. I do for you, exactly. You know, we get both n-arithomorphism. I'm trying to say that if you'd like to get the gravity, you have to do the outer. And you want to get the other, the a. Yeah, but these are enough fluctuations by the physicians, you know. Exactly. So they are one form in this language. And it is adb. This looks, actually, very simple. But remember, actually, the a we have written is what? The a we have written was a 32 by 32 matrix. The d was a 32 by 32 matrix. Tensor do the Clifford-Dellscherprin. So you're talking about. Now, in addition, actually, we said everything we know is really for one family. And I have written here, for example, k and k, I have written as a number. But in reality, this k is a 3 by 3 matrix in generation space. So you mean it's a 96 by 36? So it's 396. So 32 times 32. And tensor do the Clifford-Dellscherprin. So it's 3, 384 by 384 matrix. This is actually why it's much easier instead of writing matrices because they look huge to put in tensor notation. Because in tensor notation, you just put an index and you don't care how much it runs because you know where does it sum. So it becomes really efficient to do it. Of course, we didn't do it like that in the beginning because for years, you have to struggle with all these matrices to see what acts. And it took some time, actually. This was progressing developments. All right. So this actually here. You say, all right, I would like to compute this guy. How does it look like? Remember? So I will give you an example. The reason I want to show you that things come out, that it's not that you say, oh, the gauge fields come out, the Higgs field come out. And people think that we are cheating because we know the answer. We put it in and then it's not true. Everything comes out in a very well-defined way. We really cannot cheat in any way whatsoever. So I can say, OK, AB, and then you can put indices like AA prime, and then you can do DB, like AC, CB, and so on. So then I can look, actually, at all the matrix elements AB. And how do they look like? So I say, OK, how do they look like? It's the following. I go back to my representation. And then I'm really going to get an element here, like 1 dot, 1, 1 dot, 1. Remember, I said A alpha I beta J, and this alpha could be 1 dot or 2 dot on the I. It could be 1 on the I. And you see, now I have to take all possibilities into account. If I think about this, what is this, this would be the gauge field acting on the right-handed neutrino because we defined the right-hand neutrino to be on this corner. And in this respect, you'll discover, actually, that this guy would be 0 because the neutrino, the right-hand neutrino, is a neutral guy. So then, of course, actually, the next guy would be A 2 dot 1, 2 dot 1, and so on. And this, actually, we discover, is a vector. And then it is given by minus I over 2 G, I don't know, B mu. So it's a vector. And with a gamma mu. Everything tends to be the cliff or the algebra. So you get that this guy is a vector. In addition, you start, actually, to get something like a guy like A 1 dot 1, 1 dot A. And then it's a doublet index. A, remember, is an S2 index. And you discover, actually, and then you put A is 1, 2. And what do you discover that this is really proportional to gamma 5 HA? And this is where the H is entered. So the H is entered as inner fluctuation. And now I say, what is this thing? Why are we going this way? This is an off-diagonal element. And off-diagonal element, it means that it's really trying to connect two elements of the left and right algebras with each other. I mean, in fact, it's very invisible when you write D, because D F, that's what, there's going to be 5 turns D F. I mean, you know, D F, that's what will generate the gauge, both of them. Right, yeah. And the other one will generate the H. Oh, yeah. One really can see it. But of course, you really cannot see, for example, here it's really easy if you see that it's a doublet. What's not so easy, actually, is to show, for example, the other guy, the A 2 dot 2 dot, for example, 1A. In this case, you say it's gamma 5. And what happened, actually, it is A BH bar B. You really cannot get, guess ahead of time, that you are only going to get 1 Higgs and not 2 Higgs doublets. The fact that you get only 1 Higgs is really related to the fact that these guys are related. Remember, I told you I have C plus C. And then, because we are talking about the same C, it tells you that, OK, you know, you really cannot have different Higgs. You are really going to get the same Higgs. So the upshot of this is that one really can do this calculation, is a matrix calculation, no complications, and after some algebra, one finds the Dirac operator, including fluctuations. What does it mean? It gives me the connections. It gives me the connections on the space. And the connections are nothing but the gauge fields and the hex fields. Nothing else. So up to here, you know, we have, we can achieve that. And before doing anything, I can tell you that the Dirac action for the fermions, OK, you know, one really has to write the gypsy. I'll tell you actually now, maybe I can tell you why we can write the gypsy. The reason we write the gypsy, because in euclidean space, I really cannot say gypsy equal to psi. Yeah. So if in order to get, once you write gypsy, what really happened is that when you integrate on the gypsy and the path integral, then you really, this is an anti-symmetry product, and you don't get determinant of Da. You get the pfafian of Da and the pfafian of the square root, which means that it's really a chiral integration, and you have not doubled the degrees of Frito. OK? No. Anyway, if I expand this, what do I do? What do I really get? I really get all fermionic kinetic terms, for example, nu bar D nu plus e bar D e plus left and right at everybody, plus vector Fermi interactions plus tibs Fermi. And you get exactly the correct hypercharges and everything. Actually, why one would get the correct hypercharges exactly because of this, actually, if you really work out the jA star j inverse, they really come and they bring, actually, the u1s in really completely funny way in order to have the matching. And here, actually, one of the lots actually makes an assumption that one takes SU of the algebra, not the algebra. In other words, one has to make a requirement that the trace of the A should be equal to zero. This is another condition that we put in. OK. So, here, actually, it's a first check. Everything fermionic-wise works, and it works perfectly. So what's next, actually? What's next is that you need to make, so fermions are dynamical, but the bosons are not. So how do they get the dynamics? For all, we say that we have a care space. The care space should have dynamics. And we should be able to describe the full interactions of the system. See up to now I have been speaking classically. Everything here is classical. Now, so here, actually, comes the idea of the spectrum action. And the basic idea is that the Dirac operator, which was really the building block of the non-committed space, in principle, one can study the spectrum of this Dirac operator. So you study the eigenvalue problem, and you look at the eigenvalues. And we know, actually, from studying, say, the remaining geometry or the Dirac operators on remaining manifolds, that the eigenvalues are geometric invariants. In other words, all the geometric invariants are included, actually, or lambda is a function of all the geometric invariants. So now the question is that how can we extract this dynamic? How can we extract the information about the lambda? Of course, it's not easy to simply work out lambda of every space, so we need something more general. And so the basic idea is that since the spectrum knows about the geometry, we took as the dynamics of our system is conjectured to be given by the trace of the function of a d, where f is a positive function. I think you have to put a scale, so why is it that it makes sense? Yeah. Exactly. Yeah, I was going to talk about the scale in a minute, because, yeah, d has dimension on operator, remember it is, say, gamma mu, say, d over dx mu, and this actually has dimension one, therefore you really have to divide it by something of dimension one. Now, of course, actually, you may ask what is lambda, and I will talk a little bit about, you know, dealing with the scale later, because this is a very sensitive issue. What you should do with the scale? Can you get rid of the scale, or what are we talking about? Of course, I want to add something which is that this expression, trace of a function of d, is the only expression that you can write which is additive, which is joined spaces. See, the action should have at least the properties that if you take two disjoint spaces, it should have. Right. This is the only expression you can write. So you want to say F, but d1 and d2 should be disjoint in this case, all right? Yeah, sure. If you take d1 and d2 to be disjoint, it takes the fact that the full direction, f of d1 plus d2, okay, that it certainly proves that, you know, the trace of that will be the sum of the... Okay, I got it.... function. And there is a proverse to that, which can be that it is a functional, which is spec-wise. And full field is this additivity properties, then it has to be learned this way. But what about the positivity? Positive because... Positivity, no, no, no. It's... But it's important for Euclidean actions, no? No, no, of course. I mean, if you... Z, like Y, on top of that, it's positivity. But something that I've been, you know, toying with the idea that maybe it's self-positive and things like that. Anyway, so now essentially, how can we deal with this space? Obviously, we really, for an arbitrary function, it's not that easy to work out, but what we can really work out is asymptotic expansion. This way, one would know how to do. And the idea, the following, we use methods of heat cannon expansion. You know, it's not the only methods, but they actually... At least for remaining many forces and for product almost committed very force, they're really very, very effective. So what's the basic idea here? You say, okay, you know, suppose that I have a function of some operator P, you make some, I don't know, I don't know, a normal expansion. You say, say, SPS. And then... And then you are... You have something like trace of P minus S. The new identity trace of P minus S is given by, you know, you use the Mellon transform formula. And with the Mellon transform formula, you obtain that, you know, it's the opposite of the definition of the gamma function essentially. T S minus 1, trace E minus T P will be the real part of S is larger than or equal to 0. And then the next step would be to use formulas of trace E minus T P using heat cannon expansions. And then, of course, actually, this is given by A n of the... Sorry, T n minus, actually, in this case, minus, well, n over 2 minus 2. Okay. At least in four dimensions. And where n is larger than or equal to 0. And this is A n of XB1 dv. And these actually are called the serial discussions. And they're defined in terms of connections of the dirac operator. Yeah? It's an asymptotic series. It's an asymptotic series. Okay. I would say why it's useful in physics because exactly at a point that this is, you know, the fact of d over lambda. And then it will be clear, actually, when I write the expansion, it will be an expansion of 1 over lambda. So the first term will be 1 over lambda to the 4. Then you are going to get lambda 4, sorry. Then lambda squared, then 1, then 1 over lambda squared, that's all. Now, assuming, actually, the scale is extremely large, then these terms will become, you know, less relevant. Of course, actually, the expansion will break down once you approach the scale. And you know, the scale, you know, you can take the scale to be, say, unification scale or Planck scale. In this respect, actually, the expansion would definitely break down. And in that case, one has to learn different ways of dealing with such expansions. But you know, we did some calculations of this expansion for, you know, Robertson-Walker metrics and things like that. And really, it falls almost up to the Planck scale. Up to the Planck scale, it's only when you are orders of, you know, like twice or three times the Planck scale, then you start to see that it breaks down. But not before, you know, which is really amazing that this action is so accurate that it goes all the way up almost, you know, to the Planck scale, but not beyond, you know. It's just there. All right. So what do we do with this expansion? It doesn't give us anything sensible, because obviously, this looks far-fetched, I would say. That you take an arbitrary function, you say, okay, this arbitrary function. The fermic part is straightforward, say, psi, d psi, okay? And you get, you know, the usual expression. You can say it's by construction. But for the bosonic part, it's not really clear that this idea should work, you know, that it should give you anything accurate, essentially, after all you're using some expansion. All right. So how to proceed, actually, here, you know, it's known, as I said, here we assume that the continuous manifold is four-dimensional. And in this respect, in this respect, the expansion, yeah, what happens in the following that only an equal to zero, if n is off, for manifolds, without boundary, okay? What happens for manifolds with boundary? We have already addressed this question. It's more, it's actually very interesting. But I will talk about it later. But for manifolds with boundary, n equals zero, and z, it means, actually, the only non-zero ones are the a0, the a2, a4, and things like that. And these formulas were worked out long ago by Gilkey. He has some, you know, standard formulas. And because of that, actually, what really becomes relevant, and this is the d squared of the d, because, huh? Disabase squared. And what the trick of the following, you write this as g mu d mu plus a remainder. And this d mu will be mu plus omega mu, which is the connection. And this omega mu, okay, this also is given by, actually, there's minus g mu d mu d mu plus a mu d mu plus b. So what Gilkey did, actually, you know, he gave an expression. This actually is, there are two important terms in this expansion. The e, which is an invariant, and the connection, the omega mu, and, you know, curvatures, which are constructed out of this connection. And this omega mu, one reads from the following. You write this expression into this form, into this form, and then you can know, actually, that a mu is given by, you know, some expression, which is half g mu a mu plus a mu plus comma mu. This is a crystal, a contraction of a crystal. In other words, actually, one really can compute all these expressions in a unique way. And remember, we have already computed our d. So we have this d, huge matrix. You take it, you square it, you get another huge matrix, and you write it in this form, and then you try to find the curvature of the space. Remember, it's re-enlightened curvature now. And you also get the e-inverted of the space, another huge matrix, you know, 384 by 384. So what Gilkey does, I'll look, you know, this a0, the first time in the expansion, is given by 1 over 16 pi root g, trace 1. Trace 1, of course, is 384 because it's 384 by 384 matrix. So this is actually nothing to compute. See, R, which is independent of lambda, obviously. This is, no, no, this is actually lambda 4. This is the re-cosmological constant. And it's huge, you know. The cosmological constant is huge here. So if I write it as 0, that's a different story. Okay? This is not quite high, actually, because what if it's not 1, in the sense that you have the... Things is effective, you know. You want to talk about effective. You want to talk about, yeah, I'm talking about, you know, 0th order cosmological constant. Of course, there are contributions coming from the vacuum expectation value of the higgs, but these are supposed to be much smaller, actually. I'm saying that when you compute this coefficient as 0 and when you take into account the Major Anna-Mass matrix, you get another term. Okay, but... Yeah, okay, you know, at... So you get this time. Yeah. That actually, but... But that comes from the higgs, no? That comes from the Major Anna-Mass matrix. You mean the G-squared term? Yeah, sigma. This is a sigma, actually. Sigma, yeah. Yeah, but remember, yeah, okay. But this takes you after it. Sigma would come later because I see sigma lives here now. Sigma lives here because a priori we don't know. Okay. Anyway, so the interesting part, actually, is this guy. So, you get, actually, R over 6 is the Einstein term, and then you say trace E. Now, you... Let me actually give you an idea of what trace E is in this case, or what E is, actually. You know, E, you have to read by squaring the D, by squaring the D. And so, obviously, let me remind you that you have all this D-slash part. You have here gamma 5h part, gamma 5h dagger. You have some D-slash, okay? When you look at the D-squared, yes, the off diagonal terms are really going to give you what? H-h dagger terms, which is really nothing but the higgs mass, okay? So you discover, actually, essentially, and then you trace it, and you only get trace H-squared. So if you really work it out, actually, in this example, what you really get, you're really going to get the following, actually. You get this Gibbs up to a factor. So you see, this is the guy that would eventually give mass to the neutrinos. And this is the higgs mass. These are essentially your coupling, your square of your coupling. So we look at this expression, and then we notice the following. The curvature term, you mean the whole minus sign? Yeah, there's a whole, overall minus sign, yeah. See, I write minus, okay. So if you'd like, yeah. Okay. So. You have included the one half R into E, or not? Minus 2 over pi squared, if you wanted this, F2 lambda squared. Yes? Root, gd4x. There's a minus sign here. Well, the notation that R is negative for Euclidean spaces. Anyway, so you see, we get a mass term and with a minus sign, this is really extremely important if you, as we'll see for the standard model, because the standard model is really characterized with a potential V, as you see, lambda over 4h2h bar h squared minus m squared over minus mu squared over 2. And then the minus sign that gives the potential the Mexican hat. It's essential that you get a minus sign. Now obviously, you are really going to get a minus sign. Why? Because if you look at it, in the end, you are taking trace of E minus Tp, P has D squared, D squared has S squared. Obviously, you are going to get the first term in the expansion, a minus sign. The next term will be plus sign. It alternates, which is extremely nice, because of course, but you really cannot guarantee that, okay? In principle, before you start. You don't know. So this actually, important points we can point that the curvature, which is the Einstein term and the mass term are really not equal footing. So h bar h is like curvature of, is the curvature of the antenna space. And okay, you know, scale lambda squared. What's lambda squared? Of course, actually, one would like this lambda squared to be related to 8 punch or 16 punch, whatever, you know. Okay? It should be related to the, so this actually is the plank, plank squared. So from here, one would read that if you would like to make sense of the scale lambda that we started with, the natural interpretation of it is that it's almost, of course, you have numbers, you know. You have F2, it's a maliant transform of the function. So you can get factors of 10 or things like that. You have the pi squared. So it is of the order of the plank mass. It's not exactly the plank mass, because you have some factors here. But at least in order, we know that it's of the order of the plank mass. The next time, day four. And according to Gilky, what's the formula? The formula tells me that, what is it? Okay, you have, let me write here. Okay, I'm not going to write the surface terms. So what re-enters is the curvature squared, everything squared actually here. E squared, curvature squared. Now you know, this E squared, the omega mu squared, they really give me invariance in terms of curvatures, square of curvatures. What are the squares of curvature? Of course, R mu nu squared is there. But here we're really going to get what? We're going to get F mu nu squared, which is curvature squared. And we are going to get kinetic energy of the hex term and the sigma term and things like that. So, and, yeah, exactly. Yeah, in mill terms, there's nothing but curvature squared terms. Because what's F mu nu? It's if you take the A squared, as you see, D plus A squared, you are really going to get F. The square of the curvature, the square of the connection is the curvature. And it really lives here, it lives here, and one really can compute it, some long calculation, but in the end, it's well-defined calculation. The important thing is there's no ambiguity at any point, okay? So in this action, we are going to get F0 over 2 pi squared. Then R mu nu squared,code v squared, dou Improved plus. plus conformal factors, kinetic terms for the H, quartic terms for the potential, and similarly everything I write here, I have to write also for the sigma field, and so on. Our sigma score, anyway. It's an important thing that first of all, this time there's no scale. Second, we notice that we have a vile squared term. Now this is really important for, you know, if you are really doing, say, a realization of gravity. You know, this term would make, would improve actually the propagator of the graviton. It makes the theory, of course, non-unitary, but you know, this we don't really worry about because in that case we are cutting, you have to cut the expansion and you know, if you would like to treat unitary, you should not cut the expansion. You have to take the full Fletch theory, but at least this conformal tensor or the vile tensor squared is important for the propagator of the graviton. And now here we see here actually that we really have a kinetic term for the u1, for the electromagnetism, a kinetic term for the SU2 weak, and a kinetic term for the color, SU3. And, exactly, and then we look here and then we immediately find a relation that tells me that g1 squared equals g2 squared, that's 5 over 3, sorry, g3 squared, g2 squared, 5 over 3, g1 squared, which is the famous unification one obtained for SU5 or SO10 gratification theory. We will have actually, we have to talk about this later in the next lecture because this is not exactly satisfied. It's not exactly satisfied because as we are going to see next time, you plot the, you know, okay, the logarithm of the developments of the alphas, which is g squared over 4 pi, alphas inverse of the alphas inverse, then you discover that they don't exactly meet, but, you know, this is a different story, you know. Whether they meet, they don't meet, they almost meet, but in principle, you know, there's so much in between that we don't really know, we don't control, and it's premature to say whether this is a bad point or not a bad point. At least they are the same order. They are, you know, it's almost, you know, okay, within, let me say, within 10% it was, within 10%, but you have extrapolated 15 or 16 orders of magnitude, actually, which is huge, okay. By a variant, the normalization. Yeah, you know, so there is, and, you know, things, we have discovered that things have not really been calculated, you know, to two loops in the model that we have, because this field actually turned out to play an extremely, very important role. The sigma, which a priori we ignored, at one time we even make a prediction that the Higgs mass is given by a certain number, and this turned out not to be correct, because we have ignored actually this field. However, if you take this field into account, then, you know, you're completely in agreement with the experiment, and in addition, actually, the normalization group equations taking the sigma into effect has also only been carried to one loop order, and nobody has carried it to two loop order, so we really don't know. So maybe it will change a little bit, actually, you know. My guess is that it will change, but I don't know whether it will change in the right amount in the right direction, because this is up in the air. So let me summarize again. So we started from a very general idea, the general idea that space-time can be approximated by a product of a continuous time discrete manifold. The real answer, we don't know. You know, it may be just completely discrete, but to a very good start, it's because we know that generativity works very well, so we know the continuous part up to the Planck scale works very well, so we can up to that energy level or a little bit below. We know that the, you know, Raymaine and manifold part works. So to assume that we have an invisible extension, which makes it almost commutative, but in reality, it's not commutative, that we can extend it by a finite space, we entertain this idea, and we discovered that to almost uniquely, we can predict the structure of the model. We know the Fermi-Nichro presentations, we know actually why there are 16 fermions, we know what are the representations of these fermions, we know the gauge group. This actually, yeah, I'm saying that what we can, that things what we cannot do is much more, of course, because what we cannot do, for example, predicts say the electron mass, and you know, oh, the masses, because in principle, you know, if there is more restrictions on the, one should be able to, for this, remember, the finite space contains all the Yukawa couplings, so it contains actually information about, so if you know that, for example, that K is like 10 to the minus 2 or something, then you know that the ratio of the electron mass to the top quark mass is like 10 to the 4 or something, so the minus 4. So in that case, like this information we are putting in by hand, we are not predicting, but if one really is completely successful, everything comes out, but you know, this is really far-fetched for time being, because you need some extra structure which we really cannot put our hands on, not yet, you know. So I think there is another aspect, the criteria and the properties that we verified, do not really still tell us actually why there are three generations, this we don't know, this must come out somewhere, you know, so this aspect, we have no idea. No one has any explanation for that. We need an explanation. And as in the explanation for this, there is no, as even a guess whether there should not be a false generation. Yeah, this actually almost rolled out now actually, you know. No, no, but there are two things to say. There is one, these two Japanese who got the Nobel Prize, and the reason why we need more than two generations. And that's the cap-de-bose Kobayashi-Maskawa mix in the text, because it involves a complex number only if you have more than two. So I mean, three is the first number of generations for which you get a complex number, so for which you get violation of charge conjugation. But then, okay, I think now experimentally it's known that there is no false generation. There are very strong tests which are true. Yeah, so why three? Actually, it's not that we don't know, nobody knows also. It's up in the air, why the masses of the firms are the way they are, you know, the standard model after all has what, 17 or 18 parameters? Sure, but there you can say one thing, which is it's exactly, you know, like if somebody would come and would tell you why is the metric of space times the way it is, okay? You could say, well, Kepler's atom gas, but you know, you could say, well, it's like this, it's because of circumstances. So the issue is, what is the geometry of the finite space? Yes. Is it due to circumstances or does it have a deeper meaning? Right. And this we don't know. This we don't know. One would hope actually, see that, okay, if maybe, well, 20 years ago, you know, okay, you know, before say that, ventals, not geometry, if somebody asks you why the standard model, you say you could have answered the same. But now actually, I think we are very close to knowing that it is, it must be like actually three cross-situ, it must be like 16. So we must have a hex field. So this actually have been answered satisfactorily, but we are far from, you know, the full answer and this is the. It could be both a white and a hex. Not here actually. Here, okay, if later, if, you know, we take off this condition and we really want to push everything to grand unification, then of course you have many more hexes actually. Not only one, we are going to get, but that actually will be relevant at higher energies. And we will not be able to see what really goes on at lower energy except for the six doublet. You know, it's not clear. But that's another story, you know. Anyway, I would stop here, I think, because next time I will continue by analyzing the quantum aspects of the model and maybe, you know, theorizing a little bit about. So what you have so far is the first characterization. Of course, yes. So answer what Thibault has last. It's a very edge of geometry and some of the space set up. Right. But what is quite, one should stress one thing, you know, which is that there are seven or eight sides which have to be correct. And they are all correct. And moreover, you know, it encapsulates the state of most complexity of the Lagrangian into something which is. This thing was done before, you know, this would have been like, and then people come and verify it. It's not like a revolution. But the thing actually lost the luster because people didn't think that, okay, you know, the answer, you are fitting it. And this is the, you know, biggest thing that one has to fight in the misconception that there are billions of possibilities and simply were given an interpretation of these posts of the one possibility, which is correct. And there are many other possibilities, which is not really correct because, you know, according to this specification that I went through, that there are not really many possibilities. It's almost, it comes uniquely, you know. But that one would have to explain why if you want the correction by the final space improves the ordinary four-dimension continuum. And the reason why it improves it, it changes the care dimension. Yeah. It changes the care dimension from four to two. And that's a big improvement because then you can write this profile. But it needs to be finite. Well, it's the most economical to be finite. I mean, it could be that the splitting is actually more complex, but I mean, the simplest way to see it is that you have a four dimensional space. And then you make it two-dimensionality dimension without changing the metric dimension, which is remaining four. But if you have a, you know, if you start finite, then you'll have to fight all this Kaluza Klein modes, as I would say. You know, you have many degrees of freedom, which we don't see. So, it's a new dimension by something finite. Exactly. By something finite, which doesn't have this infinitesimal mode, like in Kaluza Klein. I mean, that's a very basic thing. Okay. Okay.
ahead and try to understand these finite spaces. calculations of this expansion for, you know,