A quotient of the ring of symmetricfunctions generalizing quantum cohomology
Grinberg, Darij
One of the many connections between Grassmannians and combinatorics is cohomological: The cohomology ring of a Grassmannian Gr(k,n) is a quotient of the ring S of symmetric polynomials in k variables. More precisely, it is the quotient of S by the ideal generated by the k consecutive complete homogeneous symmetric polynomials hn−k,hn−k+1,…,hn. We deform this quotient, by replacing the ideal by the ideal generated by hn−k−a1,hn−k+1−a2,…,hn−ak for some k fixed elements a1,a2,…,ak of the base ring. This generalizes both the classical and the quantum cohomology rings of Gr(k,n). We find three bases for the new quotient, as well as an S3-symmetry of its structure constants, a “rim hook rule” for straightening arbitrary Schur polynomials, and a fairly complicated Pieri rule. We conjecture that the structure constants are nonnegative in an appropriate sense (treating the ai as signed indeterminate), which suggests a geometric or combinatorial meaning for the quotient.
 Thanks for the invitation. This is something that I've been working on for the last two or three years, some of the time at least, sometimes all of the time. It has quantum comology in its name. It is actually just mostly a combinatorics and algebra project because it is related to quantum comology in a way you will see, but it is completely elementary in the sense all you need is some symmetric function theory. To some extent, this is a weakness. To some extent, this is a strength. If you actually understand quantum comology, unlike me, there's going to be some interesting questions in there for you, I believe, which I so far have not come close to answering. The slides are online. There is some kind of preprint online. There is a survey that I wrote for FIPSAC back before there were not this many results. This story starts with the Schubert calculus, which is from a modern perspective about the following two co-molotry rings, the co-molotry ring of the Grasmanian and that of the flag variety, both over the complex numbers. We're only going to be looking at the first one in the stock, so we're going to be studying things that are in a way generalizing this co-molotry ring. Classically, it is well known since I think Borrell and his time, what this co-molotry ring is as a ring. It is the ring of symmetric polynomials in K variables over the integers. Modulo, the ideal spent by K consecutive complete homogeneous symmetric polynomials. Now somewhere in the 80s and 90s, I think Maxim can talk, can say a lot more about that. Comology has been deformed. The new thing, well at least the simplest version of this deformation is called the small quantum co-molotry ring. And for the Grasmanian, it's been obtained that it is very similar to the co-molotry ring except you're now dividing not by hn, but by a deformed version of hn. And this Q is the polynomial and determinates that you add to the base ring. So instead of working over the integers you're working over that Q. And it has since been shown that this quantum co-molotry still has a lot of the properties of classical co-molotry. In particular, it has a basis of the module over that Q formed by the Schur polynomials, but not all of them, only the ones where the partition fits into a K by n rectangle, sorry, K by n minus K rectangle. The structure constants are the so-called Gromov-Wiednann variants. Well, the simplest ones again, but... And by now, people have written about it in a way that even I can understand. So a lot of this has been studied. There are some pretty interesting properties. Now the goal of this talk is to deform the co-molotry ring instead of just deforming it with a single Q to deform it with K parameters. And this generalizes QH. This new ring, I don't know what it means geometrically, but I have some indications that it does have a geometric meaning, or at least some kind of meaning. Maybe it's K theory of something, maybe it's a representation ring of something very weird. So let me start from scratch and quickly go over standard notions around symmetric polynomials. I see too much of it as known, but it's something Sir Worsh mentioning nevertheless. And then I'm going to introduce this deformed co-molotry ring purely algebraically. So we start with the commutative ring, boldface K, n is the non-negative integers, K is a fixed non-negative integer. This P is going to be the polynomial ring in K variables over the base ring. That's the store, of course, different case. So standard notations, if alpha is a tuple and i is the number, then alpha i is the i's entry of the tuple alpha. Same for infinite sequences. If alpha is a tuple, then x to the alpha means the monomial x1 to the alpha 1, x2 to the alpha 2, and so on. And this means the degree of this monomial, so the sum of the entries of alpha. So what is a symmetric polynomial? It's a polynomial that's invariant under permuting the variables. And I call the set of the symmetric polynomials s. It's the sum ring of P. Now, a classical theorem I think by Artyn and his Galois theory book, I'm not fully sure where it first appears. It's saying all polynomials are a free module over the symmetric polynomials. And the basis is by the basis of what's called sub staircase monomials. It's the monomials of the form x1 to the power of something smaller than 1, x2 to the power of something smaller than 2, and so on xk to the power of something smaller than k. So k factorial monomials in total. This is another way of writing this down. And the only place where I have really found the proof written up easily to understand is the LLPT notes, Luxov, Lascaux, Prague, and T-Torup. But it's been in the air for a lot of for many years. So if k is 3, this basis is just the six monomials. So now what about the ring of symmetric polynomials itself? That ring has several bases, and most of them are indexed by certain sets of integer partitions. So a partition is a weakly decreasing sequence of non-negative integers that has only finitely many nonzero entries. So things like this, or this, or this. This does not qualify because it's not decreasing. This does not qualify because it does not have only finitely many nonzero entries. And I will lazily say k partition for a partition which has at most k nonzero entries, or in simpler terms a weakly decreasing k tuple of non-negative integers. Why are these the same thing? Because if I have a k partition lambda, I can just stuff it up with zeros and I get an actual partition. And conversely, if a partition has at most k nonzero entries, I can remove the zeros and this trail of zeros until I have only a k tuple left. Any k partition has a young diagram. And the important thing for this audience, I guess, is that I'm writing it in English notation. You probably know what it is. So rows are the lengths of a row is lambda i and the number of rows is at most k because it's a k partition. Now here is the usual symmetric functions or rather symmetric polynomials. I'm not going to use infinitely many variables until the very end. So for each integer m, e m is the elementary symmetric polynomial is just the sum of all possible square free monomials of degree m. That's one way to put it. Another way is to say it's just sums of products of m element subsets of this axis. So e 0 is 1, e something negative is 0 in particular. Now what is e nu where nu is a whole tuple of integers. It's just the product of the e nu i's for the i's and for i's going, ranging over the indexes. In particular, if any of the nu i's is negative, then this is 0. So Gauss proved that the symmetric functions as a k algebra are freely generated as a commutative k algebra actually by the elementary symmetric polynomials. Equivalently, you can write it as a statement about the k module s. Namely, that k module has a basis which is namely this family of all elambdas, where lambda is the partition whose all entries are at most k. Note that e m is 0 when m is bigger than k because you won't have anything in the sum. It will be an empty sum. Now here is an analog. Instead of the e i's, I now have the h i's or h m's, the complete homogeneous symmetric functions, or rather polynomials. These are the sums of all monomials of degree m. So again, h 0 is 1, h something negative is 0. And again, I extend this notion to tuples by just multiplying through the entries. And again, the k algebra s is freely generated by this complete homogeneous symmetric functions if you go from the h 1 to the h k. Now, of course, this time, it's not true that all the bigger h's are 0. So if you want free generators, you need to stop at h k even so they keep going on. And equivalently, the h lambda is the partition whose entries are at most k, former basis of s. Now, let me mention this also. The h lambda is the k partition, also former basis of s. And these two bases are actually different. So in finitely many variables, you have two h bases. Of course, the lambdas here are the transposers of the lambdas here or conjugates, you can call them. And now I'm going to briefly introduce the Schur polynomials. So for each k partition lambda, one way to define it is this. It's a fraction. On the top, you have this modified van der Mont determinant. And on the bottom, you have the actual, the original van der Mont determinant. And yes, this is going to be a polynomial. Another way to define it is based on the so-called Jacobi Trude formula, which is another determinant, basically. Now, it's well known that this equality holds, that this is a symmetric polynomial with no negative coefficients. And a third way to define it is via semi standard Young tableaus. This is not something that I'm going to use much in the stock. So I'm just mentioning this as something that exists. If you've seen it, it's, it's maybe the simplest way to define the Schur polynomials. And again, if you restrict lambda to a k partition, you get a basis of the k-module s. So one thing that's neat about the Schur polynomials, it's not only are they themselves polynomials with no negative coefficients, but also if you multiply two of them, you can expand the result in the Schur polynomials again with no negative integer coefficients. They're called C lambda mu nu. They're known as little wood reaches and coefficients. Yeah, so there is actually many theorems of this shape because this coefficients, they count many different things. Now, to preempt, to preempt something I will have to do later on, we have, let me slightly extend the Schur polynomials. We have defined as lambda for any k partition lambda. Now I'm just going to blindly expand this definition to lambda in z to the k. So I allow nonpartitions. I allow k tuples with negative entries and with increases. And I define them in the exact same way. Now it turns out that I don't actually gain anything new this way because if alpha is a k tuple, then this newly defined s alpha is either zero or it's one of the old slumdas for partitions lambda, except possibly with the minus sign. And there is a rule for actually finding this lambda. Basically, you have to raise the entries of alpha by the numbers k minus one, k minus two, and so on, k minus k. If the resulting tuple still has a negative enter, you have zero. If the resulting tuple has two equal entries, you have zero. Otherwise, you have to sort it in decreasing order. You have to watch the sign of the permutations that you do, and then you have to un-raise these numbers again, and then you get the lambda. To be honest, this is probably not worth stating because this is what you get if you just correspondently permute the rows of this matrix. Note that the alternate formula still holds in most cases. It holds if all this lambda i plus k minus i are no negative. Now, the reason I'm mentioning is this is not an exciting result. This is just something that will come up later naturally. I will need those s alphas. Now, so let's define this new quotient. First of all, we pick an integer n greater or equal to k, and we pick k polynomials, not necessarily symmetric at this point, such that the degree of ai is less than n minus k plus i for each i. In particular, they can be constants, and in many applications they are constants. Let j be the ideal of the polynomial ring that's defined as follows. It's generated by these h's, except from each h I subtract the corresponding a. Note that this can count as a deformation because the degree of each a is just smaller than the degree of the corresponding h, so I'm subtracting smaller terms from each of them. And here is the first result that I actually had to work for, that isn't the literature, that the quotient ring as a k module is again free with the same kind of sub-staircase basis, except you have to adapt it a bit. The basis has the form x1 to the power of something smaller than n minus k plus 1, x2 to the something smaller than n minus k plus 2, and so on xn to the something smaller than n. Overline always means projection because it's a quotient ring. And this is how many elements this basis has. So this ring is zero-dimensional as its geometers would call it, or it is a free module of this rank. So from now on, I will assume that this a's are symmetric polynomials, not just arbitrary polynomials. Then I can use the same differences to generate an ideal of s. Previously it was an ideal of p, but now that they lie in s, I can also generate an ideal of s from them, so an ideal of the symmetric polynomials. And they can take that quotient. What is that quotient? Well, I can characterize that as follows. We define omega to be the permutation that has k entries of size n minus k each. So it's a k by n minus k rectangle. And I define pkn to be the set of all k partitions that fit into omega. So this is this classical notation for partitions fitting inside each other. And this is the same thing restated in pedestrian terms. All it has to do is it has at most k entries and the first entries at most n minus k. Then the k module s over i, so symmetric polynomials model this ideal is free with basis the sure polynomials that fit into this rectangle. Next, I'm going to assume something even more restrictive. I'm going to assume that the a's are actually constants, not just symmetric polynomials, but constants. Actually, some of what I'm going to do is does not require this assumption, but I think it's okay to make it here, particularly because the classical cases are particular cases of this. So the classical co-molitary ring, you can just get it when the a's are all zero and k is the integers. Then s module i becomes this classical co-molitary ring. And this s lambda bars, they are the Schubert classes. Quantum co-molitary still fits in this definition, except you have to put the indeterminate in the base ring. The q, so k must be that q. And this time not all of the a's are zero, one of them is actually q with a sign. And then you get the quantum co-molitary ring. So this above theorem allows us to forget about the geometry if you want. So we can just define the quotient and the theorem tells us that the quotient looks like it should look. It has its nice basis. It doesn't, for example, collapse to zero or something else. So a lot of papers on the subject have, in a way, been using geometry sometimes to derive combinatorial consequences. With the theorems, this is not necessary, at least if you only want the consequences. So there is more to be said. So let's look at these bases one more time. Since it's a basis, we can take the dual basis. So for each mu, for each partition mu that fits into this rectangle, we can take the linear form on s modulo i that just sends every element to its s mu coordinate with respect to these bases. Moreover, for every k partition mu, we define the complement of mu. This is going to be the partition that you get. Remember, mu fits into this rectangle. And now if you remove mu from the rectangle and you turn the result around by 180 degrees, you get a new partition. That's the result. So this is how you can define it formally. And finally, for any three k partitions alpha beta gamma, you can look at the coefficient of s gamma check in s alpha s beta in this quotient ring, of course. Not in the actual symmetric polynomials, but the projections of the quotient ring. This is the generalization of little wood Richardson coefficients and also of the three point Gromov Wittman variance. So as a theorem, that is that this things have the same as three symmetry as the Gromov Wittens. So I can basically permute alpha beta and gamma as they want without changing the coefficient. And even more symmetrically, I can rewrite it as a coefficient of the rectangular box in the product of all three-shoot polynomials. And there is a way to restate it. So what about some other bases? And actually, I must say I have read, I have seen surprisingly little about bases of the Cromology ring, even for classical Cromology. Everybody says that the Schubert classes form a basis, but then what else? Nobody, I haven't seen it written down. Well, now here's a couple more. So the H is also for my basis. And the transfer matrix is unit triangular with respect to a reasonable order, but I don't know how to describe it. It's not like I have this kind of analog of cost-conumbers handy. There is actually a formula for expanding a single H in terms of the Schur's, except in the example that if this M is too big, then the Schur's themselves will not fit into the rectangle and then you will have to expand them. And this is going to get messy. But it's interesting that if you're expanding an H, you get Schur's of hook shape. Okay, here is something else. Remember the Peary rule for symmetric polynomials. It tells you how to multiply a Schur polynomial by a complete homogeneous by an H. The result is the sum of Schur polynomials and the sum is ranges over all k partitions such that the new partition over the old partition is a horizontal J strip. So what is a horizontal J strip? So basically, it's every, it's all possible ways to add J boxes to your young diagram in such a way that no two boxes are in the same column and that you get a partition in the end. This is another way to describe it. So basically the entries of the two partitions interleave each other. So this Peary rule works in the ring of symmetric polynomials and by quotienting it also works in classical homology. Quantum homology has a more complicated Peary rule. Now what about this generalized quotient ring? Well, it turns out that it does have a Peary rule too. So this part is familiar. This is just the normal Peary rule. However, the normal Peary rule will often give you some partitions that do no longer fit into the box, into the rectangle. And then you have to reduce the model of the ideal and you get some kind of decay product, so to speak, error terms, maybe. And here is a way to actually explicitly write this error term down. So this is a very common rule. All of these new things fit into the box. And these are little wood Richardson coefficients. I'm wondering if you've ever seen a Peary rule that involves little wood Richardson coefficients. I haven't before. This is kind of new. And that involves the AIs, but it only involves them linearly, which is also a bit weird. And yes, it generalizes the Peary terminology by Bertram Cio, Kant, Fentany and Fulton. However, notice that this little wood Richardson coefficients here can be bigger than two, bigger than one. So this is not a multiplicity free rule. And this is an example for fairly reasonable N and K. Now you might wonder how about multiplying by an E instead of an H in classical symmetric function theory? There is a complete symmetry between the two. In homologies, there is still a symmetry. It's a little bit tricky because you have to swap k with n minus k, but there's still a symmetry. Even in quantum homologies, the symmetry is still there. Posnickov, I think, has shown it. Well, you can forget about symmetry now because at least it's not obvious if it exists. The thing is in this Peary rule for the Hs, you have linear terms. Only one A1 here, only A2 here, only A3 here, but here you get squares, for example. And I have not been able to make any sense of this formula. I don't even have a conjecture for what the general rule would be if there is a Peary rule for E. And here is another example for those who might think it should be easy to multiply at least with a K. It's kind of the extreme case where you're just adding a column. Well, but you're adding a column and then you're reducing models ideal and reducing can give any kind of mess. So even this case is not easy. Now, I've been talking about reducing. In quantum homologies, there is a pretty nice trick for reducing any Schur polynomial to fit it into the box. In a way, you basically just keep removing certain kinds of hooks from it until it fits into the box. And what you get is a Schur polynomial multiplied with some sine and some power of Q. Now, is there such a thing for the generalized quotient? So here is an example where we are reducing 443. Now, 443 isn't a far cry from the rectangle. The rectangle here is 333. 443 just has two extra cells. So if you get such a messy result, it feels kind of hopeless. It feels like it gets messy. That's sort of all you can say. Well, it turns out that this mess can actually be controlled. So here is a reduction rule. More likely, I call this a straightening rule. It's one step of the reduction rule. If you want to fully reduce something, you have to apply it many times. So given a k partition that does not fit into the box, so its first entry is bigger than n minus k, I take all possible k tuples whose first entry is mu1 minus n. So it's my old first entry lowered by n. And then all the other entries are increased by one or remain fixed. So 2 to the k minus 1 many terms, which are usually not partitions, they have a very negative first entry usually. And even the next entry, they might wiggle instead of being decreasing. So it's not a partition at all. Then my Schur polynomial in this quotient equals this sum. Oh, and I don't sum over the entire set. I sum only over a certain slice of it. So only a given size. Also, I guess with this sum, including this sum, it's a whole set. So this is weird. And this is the reason why I introduced Schur polynomials for non partitions. Because as I said, these lambdas are not partitions usually, but you can straighten them out. So this, and again, you see only linear terms in the ace. However, in the general case, these things will not fit into the rectangle as well. They will still not fit into the rectangle. And then you have to apply it many times. And this is how you get all this higher degree terms. So I call it a rim hook algorithm. But frankly, I don't know what the rim hooks here are. I understand this only in terms of the stuples. I don't understand this in terms of young diagrams. I suspect that this old method I think by King with the slinkies can say something about it. Well, unfortunately, I don't quite understand this method. So maybe somebody here can help me with that. Okay, now for the Holy Grail. And the Holy Grail is still well in its place. Do we have non negativity? So are the structure constants non negative? This has to be interpreted correctly because we have all these AIs now. So the structure constants will be polynomials in the AIs. And if you look here, you will hear, for example, you see a lot of negative signs. But since they're polynomials in the AIs, you may think that you could put some signs in front of the AIs and then it will be non negative. Well, apparently that's true. So I re-labeled, I re-indexed the AIs. It's as BIs with the signs in front of them. I picked three partitions, I picked three partitions, slum them, you knew that fit into the rectangle. The claim is that the coefficient of this, what's going on with the sound? So is this always the same sign? Christian, I'm with you. Yes, it is strange, but it is. So we hear you here, Daryng. So the coefficient of another S in the product of two Ss has a predictable sign. I'm not going to say it's positive because the sign is there. So you would have to also change the signs for all the Ss in terms of the degree. So this has been checked using Sage for all N up to 8. I'm not fully sure I should believe it because honestly I would like to see a few more Ns, but it gets harder and harder to check this. And if it's true, it would generalize the positivity of Gramovitian variance. Seeing that it has only recently been proved combinatorially, it would probably not be very easy to prove combinatorially. And this is of course one of the reasons why I think this ring could have geometric meaning. So as I already mentioned, there is another basis. You can take the monomials, symmetric polynomials, and then you take their projections. And that's still a basis. I don't know what the structure constants are, but I'm just saying it is the thing that's, it's not hard to prove using triangularity again. Here is, here is a thing, something that's not a basis. The power sums do not form a basis. And that's not because of positive characteristics things, even in characteristic zeros, this is not a basis. Even if AI is zero, this is not a basis. I'm kind of curious as what's going on there, what kind of subring we get, but I don't know. I haven't actually looked at forgotten symmetric functions, but there is a bunch of other bases that could be checked. So what are reasonable questions? One is whether S-modulo i has a geometric meaning, and if not, why do we have all these nice properties? Because this ideals, there is no guarantee that an ideal like this could would behave nicely in any way. In theory, you could have some weird big Gröbner bases, it would reduction would give you big masses expressions with no patterns. The next question is partly resolved. This is due to something Kristen Krattenthaler has asked in one of his papers, what happens if we replace the edges of the ideal by the power sum symmetric functions? So as a high school student working in the MIT prime's project last year, he has proved that the basis theorem still holds. So the basis that I mentioned, the sure basis is still a basis for this quotient. So I don't know any further properties for so far. I don't think there is a symmetry. I haven't looked at the theory rules. Well, the other catch all question is what other properties of quantum globalize. I have tried post-nicov's curious duality and cyclic him and symmetry. I've tried this duality between k and n minus k. They don't seem to generalize. Maybe somebody with more intuition for the geometry can say how I should probably tweak them to make them generalize. But if you just follow the algebra, it doesn't seem to work out. Is there an analogous construction for the flag manifold? Now I have to admit the quantum cosmology of the flag manifold already has a lot of deformation parameters. I don't quite see how to deform it further. So I'm not actively expecting something to happen here. Is there an quiverient analog? I started working on this. I haven't gotten very far. And what about quotients of quasi-symmetric polynomials? This is a wild card. Sometimes you get something interesting in there. Sometimes you just don't. There is an SK module structure on the quotient. So this time I'm talking about the quotient of all polynomials, not of the symmetric polynomials, because the symmetric polynomials have a trivial SK action. But what if I quotient all polynomials? What is the SK model structure? Well, it turns out, and I've been too lazy to write this up for over a year now, that this is the SK module you would basically expect. It's the power of the regular representation. If you've seen diagonal harmonics or not diagonal, just regular harmonics, you would not be surprised by this. And finally, how much time do I have left? It is up to 20. Oh, okay. I can actually... It's 13 or 10 or 13 minutes, all depends. Okay, I can talk about the proofs then. So let me quickly talk about a question that just feels natural. I don't know how natural it is. So instead of looking at symmetric polynomials and K variables, I can look at symmetric functions in infinitely many variables. This is like the grown-up version, except it's easier. And I'm going to use the same notations for them, except I will put them in boldface because... So I don't get to confuse them with the polynomials. So the symmetric polynomials can be viewed as the symmetric functions, modulo all the high E's that would be zero in the symmetric polynomials. So as modulo i, what's the quotient of a quotient? It's just a quotient by all those things. So you get it from lambda by quotienting out both the E's and the corresponding Hs minus A's. Now this looks a bit asymmetric. I mean, it's asymmetric in two ways. You have infinitely many E's here and only infinitely many H's, but also you are deforming the H's but not deforming the E's. Now I cannot do anything about the first asymmetry. I've tried. It feels like you should be able to kind of have... You should be able to arbitrarily mix H's and E's here, but it's not clear how. If you think about it, you don't want to mix them randomly. That will give you very random rings. But the second asymmetry, you can just try to fix it by also deforming the E's. So here's what comes out. I will... And I'm switching to boldface letters here because they will be elements of lambda. They will be symmetric functions. So A1, A2 up to Ak and B1, B2, B3 and so on should be symmetric functions with appropriate degree conditions. Then this quotient is a free k module with the same basis that we know and love already from comology. So the same Schubert classes. So sure functions that fit into this rectangle. So now you have infinitely many parameters. I do not know much more about this quotient. There is probably some really low hanging fruit here like other bases. I don't think... I wouldn't really expect much to hold for it honestly because this one really does not have any hint of a geometric meaning. But it just feels like a nice thing to have seen. So let me say a few words about the proofs, well except for the SK action. So many ideas. First of all, the quotient of all polynomials, not the symmetric ones, can be done with Gröbner bases. And in hindsight, this has already been done by Konka, Krattenthaler and Watanabe, already sketched. And Gröbner bases actually, so for everybody who is a bit scared of this, so people usually talk about Gröbner bases over a field. Well, the existence theorem requires a field. But if you have a Gröbner basis, so if you just want to use a given Gröbner basis, if you already know that something's Gröbner basis, you only need a commutative ring. It's like this vector spaces. For a vector space to have a basis, you need a field. But if you have a free module with a basis over a ring, you can use it. You don't need that ring to be a field anymore. So in the same way, this boils down to actually finding an explicit Gröbner basis for this ideal J. Using that and Jacoby Trudy, you can show that the symmetric polynomial quotient also has the right basis. And the rest, more or less, is about computing the symmetric functions. So I will probably not talk about Gröbner bases here. It will probably take a long time. This is all just in case slides. But here is the Gröbner basis of this ideal J. So x i dot dot dot k, so x i dot dot dot k is a shorthand for x i, x i plus one, and so on x k. So the Gröbner basis does not use the complete homogeneous polynomials of the whole axis. It uses them only for certain sub sequences of the axis. So if not for the deformation parameters, this would be a Gröbner basis. Well, now you have to deform it. And you deform it in this kind of strange way with a coefficients and with ETs, so elementary symmetric polynomials in the other axis. With respect to the degree of psychographic ordering on the monomials. And so you have a very nice Gröbner basis with really nice leading terms, which is just x i to the n minus k plus i. So, McCullis basis theorem tells you what the basis is for the quotient ring. Now, how do you jump from all polynomials down to symmetric polynomials? Okay, so using Jacobi-Truidi, you can reduce each shorthand polynomial that does not fit into the rectangle. Two shorthand polynomials that fit into the rectangle. It's a bit of a straightening rule. So the shorthand polynomials that fit into the rectangle span the quotient. Now, we need to prove that this is linearly independent. And as usual, this is the harder part. It's like PBW. However, we know that we know already from Artein that this family is span the polynomials as a symmetric polynomials module. And combining these two gives that these products span the quotient of all polynomials. But we already have a basis for this quotient. That's what we did on the previous slide with Grobler basis. So what if you have a k module with the basis and the spanning family that have the same finite size? It's well known or an easy exercise that the spanning family must then also be a basis. So we conclude that this family is the basis of the big quotient. And therefore, it's subfamily is the basis of the small quotient. So it's a bit of a ping-pong argument. It feels natural if you're into symmetric functions to just work with symmetric functions all the time or with symmetric polynomials. But we couldn't, I couldn't prove it dependence with symmetric polynomials. If you actually try to find the Grobler basis of i inside s, I don't think this has a good answer. I think the Grobler basis become very ugly very soon. So you climb up into the ring of all polynomials, you look at the quotient there, and you have, and then you argue using this fact that the spanning set that has the same size as a basis must also be a basis. I think this is a trick that's a good takeaway if you don't know it. I feel like it can be useful in many other situations. So basically, if you don't find a nice Grobler basis, maybe it helps to go into the bigger ring. And the rest of the proofs, as I said, it's about computing with symmetric functions a lot. I'm just going to briefly notice this identity commonly ascribed to Joseph Bernstein, but it appeared first in the book by Zilevinsky, that it's basically saying if you have a Schur function and you want to insert a new entry at the front, well preferably of course it should be bigger than the first entry, so you still have a partition afterwards, you can achieve this by a certain operator which is made out of skewing and multiplying, skewing by e's and multiplying by h's. This operator is now known as the Bernstein operator, or I think Garcia has called it the Schur-Row-Edder for an obvious reason. So this is in a way the driving force, but there is a lot of computations with Peary-Rolls and Jacobi-Truzzi matrices. Okay, so much for this and I hope to actually have two papers out soon enough about it, not just a big preprint with a lot of messy proofs. So thanks are due to Sashar Posniko for the whole project that I started working on back at MIT. Gerard Duchamp, thank you for inviting me again and a lot of people have done some, have contributed some ideas and thank you all. Thank you dear Dari, now time for questions, remarks or comments, maybe? I do have a basic question. Can you, maybe I missed that point from the beginning, but Dari, please, can you lead us again the connection between the little Richard von Coefficient and the Gromov-Wittern? That's the very beginning, I believe. Can you explain us again the link? Because I think if this is happening maybe for the little Richard von, you may have a rule similar with another invariant for, for instance, the Kroniker by changing the law. So I think this connection may be fruitful for doing other stuff, this is what I want to know. So let's see. Maybe just, you know, if you can, yeah, keep up with that. So this one is a coefficient of a Schur polynomial in the product of two Schur polynomials, except you're working in the quotient ring. But if you're, if you're working in the classical cosmology, then the a parameter is zero. So the quotient ring is basically just looking at like a little piece of the symmetric functions ring. It's not, it's in that case, there is no error terms. It's not being in any way contaminated by the a's. So that coefficient, of course, if you choose alpha, alpha, beta and gamma appropriately, you can get any little Richard from coefficient like this is such a thing. But this is only happening for the deformed Schur the deformed Schur polynomial, right? I mean the one with the bars, not with the classic one, as you say, right? So if you pick, if you take the a is to be zero, then the bars don't really change that much. So basically in low, in sufficiently low degrees, everything behaves as it was. Okay. Okay, thank you. Are there questions? Yes. Do you see another way of deforming so that the duality k to n minus k be respected? So this is tricky. In theory, I could well imagine that it is a duality if you properly somehow, I don't know. No, okay, no, this is probably nonsense. But so the thing is that if I have k, I have k parameters. If I have n minus k, I have n minus k parameters. So it's not quite clear. Okay. So I probably should have mentioned that this is all looking kind of similar to what is called factorization algebra and splitting algebras in the numerator geometry or even in constructive Galois theory. This algebra is where you just pick some polynomial and you declare it to be a product of two polynomials. And the indeterminate coefficients of these two polynomials will be my new indeterminates. But I wouldn't, I don't know if this is in any way as a morphic or a particular case or a generalization of that. They just feel alike. Yeah, actually, maybe a comment because sorry, I missed half of your talk, but still, as I wrote you in chat, this is in fact infinitely many variations, different deformations of algebra with the basis and this is proper to be Frobenius algebra. This is quantum, or big quantum homology gives and k actually are natural, we are in classes of tautological k dimension bundles and n minus k will be a class of n minus k that tautological bundle on grass mining. Yeah, so definitely there are natural k parameter families of deformations from quantum commology from bickering. Thank you. Do you think they could be, they could give the same algebra originalization of mine? Why not? Yeah, I would be surprised to have two different different deformations of same algebra with same number of parameters. Of course, it's a question of how to interpret combinatorially. Dari, I have a quick question. So the deformations that you that you consider all come by tweaking the Jacobi Trudy identity or basically perturbing the Hs. Yes. Is there a way of doing deformations so where you exploit say the Jacobi, the sorry, the Giambelli identity instead? So for instance, instead of perturbing the Hs, you perturb hook shapes. Good question. Or I mean, that's another way is can you, can your constructions be realized through a, you know, through a perturbation of Giambelli? Good question. And I'm thinking about this more in the case of sure functions, the infinite case, because all of the Yeah, no, I don't even know this. Do you know this in classical homology? If you don't deform anything, if you just quotient by Giambelli by hooks instead of by Hs. Oh, I think I, I think I remember something by mother and Bible, maybe, but I think it's not quite, is it hooks? You sure finite elements and lumbering that something similar. I need to read, I need to read that one again. Maybe small small question. I probably missed some part of your talk, but when you consider this deformation, you work with a commutative group, their basis. Yes. In spite of this deformation, what, what do you see any place for non commutative group basis there? Maybe you can directly work in the different situation and construct some non commutative group basis to get the basis which you need to work further. Of course, probably it's difficult to answer right away, but it's something probably, which can probably give some simplification. Some other things to do. I don't know. I'm curious as to how non commutativity will simplify things here, but I have to say right away that I don't know what what non commutative sure functions would be. I mean, there are several versions of them, but none of them feels particularly canonical. Yeah, of course, it should be chosen correctly, appropriately. So yeah, it's difficult to say, just suggestion to try. Because if you work straight away in in quotient, you can still do the same as when you work with commutative monomials. And maybe it could be easier, not necessary, but it can. There is one version with practical classes inside the world, inside the free algebra. Oh, okay. Okay. Oh, since like three ribbons. I don't remember the first authors, but I know that Schützenberg and Lascaux worked a lot on this. Other comments or remarks? So we can thank you, Darij.
 Thanks for the invitation. This is something that I've been working on for the last two  or three years, some of the time at least, sometimes all of the time. It has quantum  comology in its name. It is actually just mostly a combinatorics and algebra project  because it is related to quantum comology in a way you will see, but it is completely  elementary in the sense all you need is some symmetric function theory. To some extent,  this is a weakness. To some extent, this is a strength. If you actually understand quantum  comology, unlike me, there's going to be some interesting questions in there for you, I believe,  which I so far have not come close to answering. The slides are online. There is some kind of  preprint online. There is a survey that I wrote for FIPSAC back before there were not this many  results. This story starts with the Schubert calculus, which is from a modern perspective  about the following two co-molotry rings, the co-molotry ring of the Grasmanian and that of  the flag variety, both over the complex numbers. We're only going to be looking at the first one  in the stock, so we're going to be studying things that are in a way generalizing this co-molotry  ring. Classically, it is well known since I think Borrell and his time, what this co-molotry ring  is as a ring. It is the ring of symmetric polynomials in K variables over the integers. Modulo, the  ideal spent by K consecutive complete homogeneous symmetric polynomials. Now somewhere in the 80s  and 90s, I think Maxim can talk, can say a lot more about that. Comology has been deformed. The  new thing, well at least the simplest version of this deformation is called the small quantum  co-molotry ring. And for the Grasmanian, it's been obtained that it is very similar to the  co-molotry ring except you're now dividing not by hn, but by a deformed version of hn. And this  Q is the polynomial and determinates that you add to the base ring. So instead of working over the  integers you're working over that Q. And it has since been shown that this quantum co-molotry  still has a lot of the properties of classical co-molotry. In particular, it has a basis of the  module over that Q formed by the Schur polynomials, but not all of them, only the ones where the  partition fits into a K by n rectangle, sorry, K by n minus K rectangle. The structure constants  are the so-called Gromov-Wiednann variants. Well, the simplest ones again, but... And by now,  people have written about it in a way that even I can understand. So a lot of this has been  studied. There are some pretty interesting properties. Now the goal of this talk is to deform  the co-molotry ring instead of just deforming it with a single Q to deform it with K parameters.  And this generalizes QH. This new ring, I don't know what it means geometrically, but I have some  indications that it does have a geometric meaning, or at least some kind of meaning. Maybe it's K  theory of something, maybe it's a representation ring of something very weird. So let me start  from scratch and quickly go over standard notions around symmetric polynomials. I see  too much of it as known, but it's something Sir Worsh mentioning nevertheless. And then I'm going to  introduce this deformed co-molotry ring purely algebraically. So we start with the commutative  ring, boldface K, n is the non-negative integers, K is a fixed non-negative integer. This P is going  to be the polynomial ring in K variables over the base ring. That's the store, of course, different  case. So standard notations, if alpha is a tuple and i is the number, then alpha i is the  i's entry of the tuple alpha. Same for infinite sequences. If alpha is a tuple, then x to the  alpha means the monomial x1 to the alpha 1, x2 to the alpha 2, and so on. And this means the degree  of this monomial, so the sum of the entries of alpha. So what is a symmetric polynomial? It's a  polynomial that's invariant under permuting the variables. And I call the set of the symmetric  polynomials s. It's the sum ring of P. Now, a classical theorem I think by Artyn and his  Galois theory book, I'm not fully sure where it first appears. It's saying all polynomials are a  free module over the symmetric polynomials. And the basis is by the basis of what's called  sub staircase monomials. It's the monomials of the form x1 to the power of something smaller than  1, x2 to the power of something smaller than 2, and so on xk to the power of something smaller  than k. So k factorial monomials in total. This is another way of writing this down. And the only  place where I have really found the proof written up easily to understand is the LLPT notes,  Luxov, Lascaux, Prague, and T-Torup. But it's been in the air for a lot of for many years.  So if k is 3, this basis is just the six monomials.  So now what about the ring of symmetric polynomials itself? That ring has several bases,  and most of them are indexed by certain sets of integer partitions. So a partition is a  weakly decreasing sequence of non-negative integers that has only finitely many nonzero entries.  So things like this, or this, or this. This does not qualify because it's not decreasing.  This does not qualify because it does not have only finitely many nonzero entries.  And I will lazily say k partition for a partition which has at most k nonzero entries,  or in simpler terms a weakly decreasing k tuple of non-negative integers.  Why are these the same thing? Because if I have a k partition lambda, I can just  stuff it up with zeros and I get an actual partition. And conversely, if a partition has  at most k nonzero entries, I can remove the zeros and this trail of zeros until I have only a k tuple left.  Any k partition has a young diagram. And the important thing for this audience, I guess,  is that I'm writing it in English notation. You probably know what it is.  So rows are the lengths of a row is lambda i and the number of rows is at most k because it's a k partition.  Now here is the usual symmetric functions or rather symmetric polynomials. I'm not going to use  infinitely many variables until the very end. So for each integer m, e m is the elementary  symmetric polynomial is just the sum of all possible square free monomials of degree m.  That's one way to put it. Another way is to say it's just sums of products of m element subsets of  this axis. So e 0 is 1, e something negative is 0 in particular. Now what is e nu where nu is a  whole tuple of integers. It's just the product of the e nu i's for the i's and for i's going,  ranging over the indexes. In particular, if any of the nu i's is negative, then this is 0.  So Gauss proved that the symmetric functions as a k algebra are freely generated as a commutative  k algebra actually by the elementary symmetric polynomials. Equivalently, you can write it as  a statement about the k module s. Namely, that k module has a basis which is namely this family  of all elambdas, where lambda is the partition whose all entries are at most k.  Note that e m is 0 when m is bigger than k because you won't have anything in the sum. It will be an  empty sum. Now here is an analog. Instead of the e i's, I now have the h i's or h m's, the complete  homogeneous symmetric functions, or rather polynomials. These are the sums of all monomials of degree m.  So again, h 0 is 1, h something negative is 0. And again, I extend this notion to tuples by just  multiplying through the entries. And again, the k algebra s is freely generated by this complete  homogeneous symmetric functions if you go from the h 1 to the h k. Now, of course, this time,  it's not true that all the bigger h's are 0. So if you want free generators, you need to stop at h k  even so they keep going on. And equivalently, the h lambda is the partition whose entries are at most  k, former basis of s. Now, let me mention this also. The h lambda is the k partition, also former  basis of s. And these two bases are actually different. So in finitely many variables, you have two h  bases. Of course, the lambdas here are the transposers of the lambdas here or conjugates, you can call them.  And now I'm going to briefly introduce the Schur polynomials. So for each k partition lambda,  one way to define it is this. It's a fraction. On the top, you have this modified van der  Mont determinant. And on the bottom, you have the actual, the original van der Mont determinant.  And yes, this is going to be a polynomial. Another way to define it is based on the so-called Jacobi  Trude formula, which is another determinant, basically. Now, it's well known that this equality holds,  that this is a symmetric polynomial with no negative coefficients. And a third way to define it is  via semi standard Young tableaus. This is not something that I'm going to use much in the  stock. So I'm just mentioning this as something that exists. If you've seen it, it's, it's maybe the  simplest way to define the Schur polynomials. And again, if you restrict lambda to a k partition,  you get a basis of the k-module s. So one thing that's neat about the Schur polynomials, it's not  only are they themselves polynomials with no negative coefficients, but also if you multiply  two of them, you can expand the result in the Schur polynomials again with no negative integer  coefficients. They're called C lambda mu nu. They're known as little wood reaches and coefficients.  Yeah, so there is actually many theorems of this shape because this coefficients,  they count many different things. Now, to preempt, to preempt something I will have to do later on,  we have, let me slightly extend the Schur polynomials. We have defined as lambda for any k  partition lambda. Now I'm just going to blindly expand this definition to lambda in z to the k.  So I allow nonpartitions. I allow k tuples with negative entries and with increases.  And I define them in the exact same way. Now it turns out that I don't actually gain anything  new this way because if alpha is a k tuple, then this newly defined s alpha is either zero or it's  one of the old slumdas for partitions lambda, except possibly with the minus sign. And there is  a rule for actually finding this lambda. Basically, you have to raise the entries of alpha by the  numbers k minus one, k minus two, and so on, k minus k. If the resulting tuple still has a  negative enter, you have zero. If the resulting tuple has two equal entries, you have zero.  Otherwise, you have to sort it in decreasing order. You have to watch the sign of the permutations  that you do, and then you have to un-raise these numbers again, and then you get the lambda.  To be honest, this is probably not worth stating because this is what you get if you just  correspondently permute the rows of this matrix.  Note that the alternate formula still holds in most cases.  It holds if all this lambda i plus k minus i are no negative. Now, the reason I'm mentioning is  this is not an exciting result. This is just something that will come up later naturally.  I will need those s alphas.  Now, so let's define this new quotient. First of all, we pick an integer n greater or equal to k,  and we pick k polynomials, not necessarily symmetric at this point, such that the degree of ai  is less than n minus k plus i for each i. In particular, they can be constants, and in many  applications they are constants. Let j be the ideal of the polynomial ring that's defined as  follows. It's generated by these h's, except from each h I subtract the corresponding a.  Note that this can count as a deformation because the degree of each a is just smaller  than the degree of the corresponding h, so I'm subtracting smaller terms from each of them.  And here is the first result that I actually had to work for, that isn't the literature,  that the quotient ring as a k module is again free with the same kind of sub-staircase basis,  except you have to adapt it a bit. The basis has the form x1 to the power of something smaller  than n minus k plus 1, x2 to the something smaller than n minus k plus 2, and so on xn to the something  smaller than n. Overline always means projection because it's a quotient ring.  And this is how many elements this basis has. So this ring is zero-dimensional as its geometers  would call it, or it is a free module of this rank. So from now on, I will assume that this  a's are symmetric polynomials, not just arbitrary polynomials. Then I can use the same differences  to generate an ideal of s. Previously it was an ideal of p, but now that they lie in s,  I can also generate an ideal of s from them, so an ideal of the symmetric polynomials.  And they can take that quotient. What is that quotient? Well, I can characterize that as follows.  We define omega to be the permutation that has k entries of size n minus k each.  So it's a k by n minus k rectangle. And I define pkn to be the set of all k partitions that fit  into omega. So this is this classical notation for partitions fitting inside each other.  And this is the same thing restated in pedestrian terms. All it has to do is it has at most k  entries and the first entries at most n minus k. Then the k module s over i, so symmetric  polynomials model this ideal is free with basis the sure polynomials that fit into this rectangle.  Next, I'm going to assume something even more restrictive. I'm going to assume that the a's  are actually constants, not just symmetric polynomials, but constants. Actually, some of what I'm going  to do is does not require this assumption, but I think it's okay to make it here, particularly  because the classical cases are particular cases of this. So the classical co-molitary ring, you  can just get it when the a's are all zero and k is the integers. Then s module i becomes this  classical co-molitary ring. And this s lambda bars, they are the Schubert classes. Quantum  co-molitary still fits in this definition, except you have to put the indeterminate in the base  ring. The q, so k must be that q. And this time not all of the a's are zero, one of them is actually  q with a sign. And then you get the quantum co-molitary ring. So this above theorem allows us to  forget about the geometry if you want. So we can just define the quotient and the theorem tells us  that the quotient looks like it should look. It has its nice basis. It doesn't, for example,  collapse to zero or something else. So a lot of papers on the subject have, in a way, been using  geometry sometimes to derive combinatorial consequences. With the theorems, this is not  necessary, at least if you only want the consequences. So there is more to be said.  So let's look at these bases one more time. Since it's a basis, we can take the dual basis. So for  each mu, for each partition mu that fits into this rectangle, we can take the linear form on s  modulo i that just sends every element to its s mu coordinate with respect to these bases.  Moreover, for every k partition mu, we define the complement of mu. This is going to be the  partition that you get. Remember, mu fits into this rectangle. And now if you remove mu from the  rectangle and you turn the result around by 180 degrees, you get a new partition. That's the  result. So this is how you can define it formally. And finally, for any three k partitions alpha  beta gamma, you can look at the coefficient of s gamma check in s alpha s beta in this  quotient ring, of course. Not in the actual symmetric polynomials, but the projections  of the quotient ring. This is the generalization of little wood Richardson coefficients and also  of the three point Gromov Wittman variance. So as a theorem, that is that this things have the same  as three symmetry as the Gromov Wittens. So I can basically permute alpha beta and gamma as they want  without changing the coefficient. And even more symmetrically, I can rewrite it as a coefficient  of the rectangular box in the product of all three-shoot polynomials. And there is a way to restate it.  So what about some other bases? And actually, I must say I have read, I have seen surprisingly  little about bases of the Cromology ring, even for classical Cromology. Everybody says that the  Schubert classes form a basis, but then what else? Nobody, I haven't seen it written down. Well,  now here's a couple more. So the H is also for my basis. And the transfer matrix is unit  triangular with respect to a reasonable order, but I don't know how to describe it. It's not like  I have this kind of analog of cost-conumbers handy. There is actually a formula for expanding a single  H in terms of the Schur's, except in the example that if this M is too big, then the Schur's  themselves will not fit into the rectangle and then you will have to expand them. And this is  going to get messy. But it's interesting that if you're expanding an H, you get Schur's of hook shape.  Okay, here is something else. Remember the Peary rule for symmetric polynomials. It tells you  how to multiply a Schur polynomial by a complete homogeneous by an H. The result is the sum of  Schur polynomials and the sum is ranges over all k partitions such that the new partition  over the old partition is a horizontal J strip. So what is a horizontal J strip? So basically,  it's every, it's all possible ways to add J boxes to your young diagram in such a way that no two  boxes are in the same column and that you get a partition in the end. This is another way to  describe it. So basically the entries of the two partitions interleave each other.  So this Peary rule works in the ring of symmetric polynomials and by quotienting it also works in  classical homology. Quantum homology has a more complicated Peary rule. Now what about this  generalized quotient ring? Well, it turns out that it does have a Peary rule too. So this part  is familiar. This is just the normal Peary rule. However, the normal Peary rule will often give  you some partitions that do no longer fit into the box, into the rectangle. And then you have to  reduce the model of the ideal and you get some kind of decay product, so to speak, error terms,  maybe. And here is a way to actually explicitly write this error term down.  So this is a very common rule. All of these new things fit into the box. And these are little  wood Richardson coefficients. I'm wondering if you've ever seen a Peary rule that involves little  wood Richardson coefficients. I haven't before. This is kind of new. And that involves the AIs,  but it only involves them linearly, which is also a bit weird. And yes, it generalizes the Peary  terminology by Bertram Cio, Kant, Fentany and Fulton. However, notice that this little wood  Richardson coefficients here can be bigger than two, bigger than one. So this is not a multiplicity  free rule. And this is an example for fairly reasonable N and K. Now you might wonder how  about multiplying by an E instead of an H in classical symmetric function theory? There is a  complete symmetry between the two. In homologies, there is still a symmetry. It's a little bit  tricky because you have to swap k with n minus k, but there's still a symmetry. Even in quantum  homologies, the symmetry is still there. Posnickov, I think, has shown it. Well, you can forget about  symmetry now because at least it's not obvious if it exists. The thing is in this Peary rule for  the Hs, you have linear terms. Only one A1 here, only A2 here, only A3 here, but here you get squares,  for example. And I have not been able to make any sense of this formula. I don't even have a  conjecture for what the general rule would be if there is a Peary rule for E. And here is another  example for those who might think it should be easy to multiply at least with a K. It's kind of the  extreme case where you're just adding a column. Well, but you're adding a column and then you're  reducing models ideal and reducing can give any kind of mess. So even this case is not easy.  Now, I've been talking about reducing. In quantum homologies, there is a pretty nice  trick for reducing any Schur polynomial to fit it into the box. In a way, you basically just  keep removing certain kinds of hooks from it until it fits into the box. And what you get is a Schur  polynomial multiplied with some sine and some power of Q. Now, is there such a thing for the  generalized quotient? So here is an example where we are reducing 443. Now, 443 isn't a far cry from  the rectangle. The rectangle here is 333. 443 just has two extra cells. So if you get such a messy  result, it feels kind of hopeless. It feels like it gets messy. That's sort of all you can say.  Well, it turns out that this mess can actually be controlled. So here is a reduction rule. More  likely, I call this a straightening rule. It's one step of the reduction rule. If you want to fully  reduce something, you have to apply it many times. So given a k partition that does not fit into the  box, so its first entry is bigger than n minus k, I take all possible k tuples whose first entry  is mu1 minus n. So it's my old first entry lowered by n. And then all the other entries are increased  by one or remain fixed. So 2 to the k minus 1 many terms, which are usually not partitions,  they have a very negative first entry usually. And even the next entry, they might wiggle instead  of being decreasing. So it's not a partition at all. Then my Schur polynomial in this quotient  equals this sum. Oh, and I don't sum over the entire set. I sum only over a certain slice of it.  So only a given size. Also, I guess with this sum, including this sum, it's a whole set.  So this is weird. And this is the reason why I introduced Schur polynomials for non partitions.  Because as I said, these lambdas are not partitions usually, but you can straighten them out.  So this, and again, you see only linear terms in the ace. However, in the general case,  these things will not fit into the rectangle as well. They will still not fit into the rectangle.  And then you have to apply it many times. And this is how you get all this higher degree terms.  So I call it a rim hook algorithm. But frankly, I don't know what the rim hooks here are. I understand  this only in terms of the stuples. I don't understand this in terms of young diagrams.  I suspect that this old method I think by King with the slinkies can say something about it.  Well, unfortunately, I don't quite understand this method. So maybe somebody here can help me with  that. Okay, now for the Holy Grail. And the Holy Grail is still well in its place. Do we have  non negativity? So are the structure constants non negative? This has to be interpreted correctly  because we have all these AIs now. So the structure constants will be polynomials in the AIs.  And if you look here, you will hear, for example, you see a lot of negative signs.  But since they're polynomials in the AIs, you may think that you could put some signs in front of  the AIs and then it will be non negative. Well, apparently that's true. So I re-labeled, I re-indexed  the AIs. It's as BIs with the signs in front of them. I picked three partitions,  I picked three partitions, slum them, you knew that fit into the rectangle. The claim is that the  coefficient of this, what's going on with the sound?  So is this always the same sign?  Christian, I'm with you.  Yes, it is strange, but it is.  So we hear you here, Daryng.  So the coefficient of another S in the product of two Ss has a predictable sign.  I'm not going to say it's positive because the sign is there. So you would have to also change  the signs for all the Ss in terms of the degree. So this has been checked using Sage for all N up  to 8. I'm not fully sure I should believe it because honestly I would like to see a few more Ns,  but it gets harder and harder to check this. And if it's true, it would generalize the  positivity of Gramovitian variance. Seeing that it has only recently been proved combinatorially,  it would probably not be very easy to prove combinatorially.  And this is of course one of the reasons why I think this ring could have geometric meaning.  So as I already mentioned, there is another basis. You can take the monomials, symmetric  polynomials, and then you take their projections. And that's still a basis.  I don't know what the structure constants are, but I'm just saying it is the thing that's,  it's not hard to prove using triangularity again.  Here is, here is a thing, something that's not a basis. The power sums do not form a basis.  And that's not because of positive characteristics things, even in characteristic zeros, this is  not a basis. Even if AI is zero, this is not a basis. I'm kind of curious as what's going on there,  what kind of subring we get, but I don't know. I haven't actually looked at forgotten symmetric  functions, but there is a bunch of other bases that could be checked. So what are reasonable  questions? One is whether S-modulo i has a geometric meaning, and if not, why do we have  all these nice properties? Because this ideals, there is no guarantee that an ideal like this  could would behave nicely in any way. In theory, you could have some weird big Gröbner bases,  it would reduction would give you big masses expressions with no patterns.  The next question is partly resolved. This is due to something Kristen Krattenthaler has asked  in one of his papers, what happens if we replace the edges of the ideal by the power sum symmetric  functions? So as a high school student working in the MIT prime's project last year, he has  proved that the basis theorem still holds. So the basis that I mentioned, the sure basis is  still a basis for this quotient. So I don't know any further properties for so far. I don't think  there is a symmetry. I haven't looked at the theory rules. Well, the other catch all question is what  other properties of quantum globalize. I have tried post-nicov's curious duality and cyclic  him and symmetry. I've tried this duality between k and n minus k. They don't seem to generalize.  Maybe somebody with more intuition for the geometry can say how I should probably tweak them to make  them generalize. But if you just follow the algebra, it doesn't seem to work out.  Is there an analogous construction for the flag manifold? Now I have to admit the quantum  cosmology of the flag manifold already has a lot of deformation parameters. I don't quite see how  to deform it further. So I'm not actively expecting something to happen here. Is there an  quiverient analog? I started working on this. I haven't gotten very far. And what about quotients  of quasi-symmetric polynomials? This is a wild card. Sometimes you get something interesting in there.  Sometimes you just don't. There is an SK module structure on the quotient. So this time I'm talking  about the quotient of all polynomials, not of the symmetric polynomials, because the symmetric  polynomials have a trivial SK action. But what if I quotient all polynomials? What is the SK  model structure? Well, it turns out, and I've been too lazy to write this up for over a year now,  that this is the SK module you would basically expect. It's the power of the regular representation.  If you've seen diagonal harmonics or not diagonal, just regular harmonics, you would not be surprised  by this. And finally, how much time do I have left?  It is up to 20. Oh, okay. I can actually...  It's 13 or 10 or 13 minutes, all depends.  Okay, I can talk about the proofs then. So let me quickly talk about a question that just feels  natural. I don't know how natural it is. So instead of looking at symmetric polynomials and  K variables, I can look at symmetric functions in infinitely many variables. This is like the  grown-up version, except it's easier. And I'm going to use the same notations for them, except I will  put them in boldface because... So I don't get to confuse them with the polynomials.  So the symmetric polynomials can be viewed as the symmetric functions, modulo all the high  E's that would be zero in the symmetric polynomials. So as modulo i, what's the quotient of a quotient?  It's just a quotient by all those things. So you get it from lambda by quotienting out  both the E's and the corresponding Hs minus A's. Now this looks a bit asymmetric. I mean,  it's asymmetric in two ways. You have infinitely many E's here and only infinitely many H's,  but also you are deforming the H's but not deforming the E's. Now I cannot do anything  about the first asymmetry. I've tried. It feels like you should be able to kind of have...  You should be able to arbitrarily mix H's and E's here, but it's not clear how. If you think about  it, you don't want to mix them randomly. That will give you very random rings. But the second  asymmetry, you can just try to fix it by also deforming the E's. So here's what comes out.  I will... And I'm switching to boldface letters here because they will be elements of lambda.  They will be symmetric functions. So A1, A2 up to Ak and B1, B2, B3 and so on should be  symmetric functions with appropriate degree conditions. Then this quotient is a free k  module with the same basis that we know and love already from comology. So the same Schubert classes.  So sure functions that fit into this rectangle. So now you have infinitely many parameters.  I do not know much more about this quotient. There is probably some really low hanging fruit here  like other bases. I don't think... I wouldn't really expect much to hold for it honestly because  this one really does not have any hint of a geometric meaning. But it just feels like a nice  thing to have seen. So let me say a few words about the proofs, well except for the SK action.  So many ideas. First of all, the quotient of all polynomials, not the symmetric ones,  can be done with Gröbner bases. And in hindsight, this has already been done by Konka,  Krattenthaler and Watanabe, already sketched. And Gröbner bases actually, so for everybody who  is a bit scared of this, so people usually talk about Gröbner bases over a field. Well, the  existence theorem requires a field. But if you have a Gröbner basis, so if you just want to use  a given Gröbner basis, if you already know that something's Gröbner basis, you only need a commutative  ring. It's like this vector spaces. For a vector space to have a basis, you need a field. But if  you have a free module with a basis over a ring, you can use it. You don't need that ring to be a  field anymore. So in the same way, this boils down to actually finding an explicit Gröbner basis for  this ideal J. Using that and Jacoby Trudy, you can show that the symmetric polynomial quotient  also has the right basis. And the rest, more or less, is about computing the symmetric functions.  So I will probably not talk about Gröbner bases here. It will probably take a long time.  This is all just in case slides. But here is the Gröbner basis of this ideal J. So x i dot dot  dot k, so x i dot dot dot k is a shorthand for x i, x i plus one, and so on x k. So the Gröbner  basis does not use the complete homogeneous polynomials of the whole axis. It uses them only  for certain sub sequences of the axis. So if not for the deformation parameters,  this would be a Gröbner basis. Well, now you have to deform it. And you deform it in this kind of  strange way with a coefficients and with ETs, so elementary symmetric polynomials in the other  axis. With respect to the degree of psychographic ordering on the monomials. And so you have a  very nice Gröbner basis with really nice leading terms, which is just x i to the n minus k plus i.  So, McCullis basis theorem tells you what the basis is for the quotient ring.  Now, how do you jump from all polynomials down to symmetric polynomials?  Okay, so using Jacobi-Truidi, you can reduce each shorthand polynomial that does not fit  into the rectangle. Two shorthand polynomials that fit into the rectangle. It's a bit of a  straightening rule. So the shorthand polynomials that fit into the rectangle span the quotient.  Now, we need to prove that this is linearly independent. And as usual, this is the harder  part. It's like PBW. However, we know that we know already from Artein that this family is  span the polynomials as a symmetric polynomials module. And combining these two  gives that these products span the quotient of all polynomials.  But we already have a basis for this quotient. That's what we did on the previous slide  with Grobler basis. So what if you have a k module with the basis and the spanning  family that have the same finite size? It's well known or an easy exercise that the  spanning family must then also be a basis. So we conclude that this family is the basis of the big  quotient. And therefore, it's subfamily is the basis of the small quotient. So it's a bit of a  ping-pong argument. It feels natural if you're into symmetric functions to just work with symmetric  functions all the time or with symmetric polynomials. But we couldn't, I couldn't prove it dependence  with symmetric polynomials. If you actually try to find the Grobler basis of i inside s,  I don't think this has a good answer. I think the Grobler basis become very ugly very soon. So you  climb up into the ring of all polynomials, you look at the quotient there, and you have, and then  you argue using this fact that the spanning set that has the same size as a basis must also be a  basis. I think this is a trick that's a good takeaway if you don't know it. I feel like it can be  useful in many other situations. So basically, if you don't find a nice Grobler basis, maybe it helps  to go into the bigger ring. And the rest of the proofs, as I said, it's about computing with  symmetric functions a lot. I'm just going to briefly notice this identity commonly ascribed to Joseph  Bernstein, but it appeared first in the book by Zilevinsky, that it's basically saying if you have  a Schur function and you want to insert a new entry at the front, well preferably of course it should  be bigger than the first entry, so you still have a partition afterwards, you can achieve this by a  certain operator which is made out of skewing and multiplying, skewing by e's and multiplying by h's.  This operator is now known as the Bernstein operator, or I think Garcia has called it the  Schur-Row-Edder for an obvious reason. So this is in a way the driving force, but there is a lot of  computations with Peary-Rolls and Jacobi-Truzzi matrices. Okay, so much for this and I hope to  actually have two papers out soon enough about it, not just a big preprint with a lot of messy proofs.  So thanks are due to Sashar Posniko for the whole project that I started working on back at MIT.  Gerard Duchamp, thank you for inviting me again and a lot of people have done some,  have contributed some ideas and thank you all.  Thank you dear Dari, now time for questions, remarks or comments, maybe?  I do have a basic question. Can you, maybe I missed that point from the beginning,  but Dari, please, can you lead us again the connection between the little Richard  von Coefficient and the Gromov-Wittern? That's the very beginning, I believe. Can you explain us  again the link? Because I think if this is happening maybe for the little Richard von,  you may have a rule similar with another invariant for, for instance, the Kroniker by changing the  law. So I think this connection may be fruitful for doing other stuff, this is what I want to know.  So let's see. Maybe just, you know, if you can, yeah, keep up with that.  So this one is a coefficient of a Schur polynomial in the product of two Schur polynomials,  except you're working in the quotient ring. But if you're, if you're working in the classical  cosmology, then the a parameter is zero. So the quotient ring is basically just looking at like  a little piece of the symmetric functions ring. It's not, it's in that case, there is no error  terms. It's not being in any way contaminated by the a's. So that coefficient, of course,  if you choose alpha, alpha, beta and gamma appropriately, you can get any little Richard  from coefficient like this is such a thing. But this is only happening for the deformed Schur  the deformed Schur polynomial, right? I mean the one with the bars, not with the classic one,  as you say, right? So if you pick, if you take the a is to be zero, then the bars don't really  change that much. So basically in low, in sufficiently low degrees, everything behaves as it was.  Okay. Okay, thank you.  Are there questions?  Yes. Do you see another way of deforming so that the duality k to n minus k be respected?  So this is tricky. In theory, I could well imagine that it is a duality if you properly somehow,  I don't know. No, okay, no, this is probably nonsense. But so the thing is that if I have k,  I have k parameters. If I have n minus k, I have n minus k parameters. So it's not quite clear.  Okay. So I probably should have mentioned that this is all looking kind of similar to what is  called factorization algebra and splitting algebras in the numerator geometry or even in  constructive Galois theory. This algebra is where you just pick some polynomial and you declare it  to be a product of two polynomials. And the indeterminate coefficients of these two polynomials  will be my new indeterminates. But I wouldn't, I don't know if this is in any way as a morphic  or a particular case or a generalization of that. They just feel alike. Yeah, actually,  maybe a comment because sorry, I missed half of your talk, but still, as I wrote you in chat,  this is in fact infinitely many variations, different deformations of algebra with the basis  and this is proper to be Frobenius algebra. This is quantum, or big quantum homology gives and k  actually are natural, we are in classes of tautological k dimension bundles and n minus k  will be a class of n minus k that tautological bundle on grass mining. Yeah, so definitely there  are natural k parameter families of deformations from quantum commology from bickering. Thank you.  Do you think they could be, they could give the same algebra originalization of mine?  Why not? Yeah, I would be surprised to have two different different  deformations of same algebra with same number of parameters. Of course, it's a question of  how to interpret combinatorially. Dari, I have a quick question. So the deformations that you  that you consider all come by tweaking the Jacobi Trudy identity or basically perturbing  the Hs. Yes. Is there a way of doing deformations so where you exploit say the Jacobi, the sorry,  the Giambelli identity instead? So for instance, instead of perturbing the Hs, you perturb hook  shapes. Good question. Or I mean, that's another way is can you, can your constructions be realized  through a, you know, through a perturbation of Giambelli? Good question.  And I'm thinking about this more in the case of sure functions, the infinite case, because  all of the Yeah, no, I don't even know this. Do you know this in classical  homology? If you don't deform anything, if you just quotient by Giambelli by hooks instead of by Hs.  Oh, I think I, I think I remember something by mother and Bible, maybe, but I think it's  not quite, is it hooks? You sure finite elements and lumbering that something similar. I need to  read, I need to read that one again. Maybe small small question. I probably missed some part of  your talk, but when you consider this deformation, you work with a commutative group, their basis.  Yes. In spite of this deformation, what, what do you see any place for non commutative group  basis there? Maybe you can directly work in the different situation and construct some  non commutative group basis to get the basis which you need to work further.  Of course, probably it's difficult to answer right away, but it's something probably,  which can probably give some simplification. Some other things to do.  I don't know. I'm curious as to how non commutativity will simplify things here,  but I have to say right away that I don't know what what non commutative sure functions would be.  I mean, there are several versions of them, but none of them feels particularly canonical.  Yeah, of course, it should be chosen correctly, appropriately. So yeah, it's difficult to say,  just suggestion to try. Because if you work straight away in in quotient, you can still do the same  as when you work with commutative monomials. And maybe it could be easier, not necessary, but it can.  There is one version with practical classes  inside the world, inside the free algebra.  Oh, okay.  Okay. Oh, since like three ribbons.  I don't remember the first authors, but I know that Schützenberg and Lascaux worked a lot on this.  Other comments or remarks?  So we can thank you, Darij.
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Thanks for the invitation. This is something that I've been working on for the last two or three years, some of the time at least, sometimes all of the time. It has quantum comology in its name. It is actually just mostly a combinatorics and algebra project because it is related to quantum comology in a way you will see, but it is completely elementary in the sense all you need is some symmetric function theory. To some extent, this is a weakness. To some extent, this is a strength. If you actually understand quantum comology, unlike me, there's going to be some interesting questions in there for you, I believe, which I so far have not come close to answering. The slides are online. There is some kind of preprint online. There is a survey that I wrote for FIPSAC back before there were not this many results. This story starts with the Schubert calculus, which is from a modern perspective about the following two co-molotry rings, the co-molotry ring of the Grasmanian and that of the flag variety, both over the complex numbers. We're only going to be looking at the first one in the stock, so we're going to be studying things that are in a way generalizing this co-molotry ring. Classically, it is well known since I think Borrell and his time, what this co-molotry ring is as a ring. It is the ring of symmetric polynomials in K variables over the integers. Modulo, the ideal spent by K consecutive complete homogeneous symmetric polynomials. Now somewhere in the 80s and 90s, I think Maxim can talk, can say a lot more about that. Comology has been deformed. The new thing, well at least the simplest version of this deformation is called the small quantum co-molotry ring. And for the Grasmanian, it's been obtained that it is very similar to the co-molotry ring except you're now dividing not by hn, but by a deformed version of hn. And this Q is the polynomial and determinates that you add to the base ring. So instead of working over the integers you're working over that Q. And it has since been shown that this quantum co-molotry still has a lot of the properties of classical co-molotry. In particular, it has a basis of the module over that Q formed by the Schur polynomials, but not all of them, only the ones where the partition fits into a K by n rectangle, sorry, K by n minus K rectangle. The structure constants are the so-called Gromov-Wiednann variants. Well, the simplest ones again, but... And by now, people have written about it in a way that even I can understand. So a lot of this has been studied. There are some pretty interesting properties. Now the goal of this talk is to deform the co-molotry ring instead of just deforming it with a single Q to deform it with K parameters. And this generalizes QH. This new ring, I don't know what it means geometrically, but I have some indications that it does have a geometric meaning, or at least some kind of meaning. Maybe it's K theory of something, maybe it's a representation ring of something very weird. So let me start from scratch and quickly go over standard notions around symmetric polynomials. I see too much of it as known, but it's something Sir Worsh mentioning nevertheless. And then I'm going to introduce this deformed co-molotry ring purely algebraically. So we start with the commutative ring, boldface K, n is the non-negative integers, K is a fixed non-negative integer. This P is going to be the polynomial ring in K variables over the base ring. That's the store, of course, different case. So standard notations, if alpha is a tuple and i is the number, then alpha i is the i's entry of the tuple alpha. Same for infinite sequences. If alpha is a tuple, then x to the alpha means the monomial x1 to the alpha 1, x2 to the alpha 2, and so on. And this means the degree of this monomial, so the sum of the entries of alpha. So what is a symmetric polynomial? It's a polynomial that's invariant under permuting the variables. And I call the set of the symmetric polynomials s. It's the sum ring of P. Now, a classical theorem I think by Artyn and his Galois theory book, I'm not fully sure where it first appears. It's saying all polynomials are a free module over the symmetric polynomials. And the basis is by the basis of what's called sub staircase monomials. It's the monomials of the form x1 to the power of something smaller than 1, x2 to the power of something smaller than 2, and so on xk to the power of something smaller than k. So k factorial monomials in total. This is another way of writing this down. And the only place where I have really found the proof written up easily to understand is the LLPT notes, Luxov, Lascaux, Prague, and T-Torup. But it's been in the air for a lot of for many years. So if k is 3, this basis is just the six monomials. So now what about the ring of symmetric polynomials itself? That ring has several bases, and most of them are indexed by certain sets of integer partitions. So a partition is a weakly decreasing sequence of non-negative integers that has only finitely many nonzero entries. So things like this, or this, or this. This does not qualify because it's not decreasing. This does not qualify because it does not have only finitely many nonzero entries. And I will lazily say k partition for a partition which has at most k nonzero entries, or in simpler terms a weakly decreasing k tuple of non-negative integers. Why are these the same thing? Because if I have a k partition lambda, I can just stuff it up with zeros and I get an actual partition. And conversely, if a partition has at most k nonzero entries, I can remove the zeros and this trail of zeros until I have only a k tuple left. Any k partition has a young diagram. And the important thing for this audience, I guess, is that I'm writing it in English notation. You probably know what it is. So rows are the lengths of a row is lambda i and the number of rows is at most k because it's a k partition. Now here is the usual symmetric functions or rather symmetric polynomials. I'm not going to use infinitely many variables until the very end. So for each integer m, e m is the elementary symmetric polynomial is just the sum of all possible square free monomials of degree m. That's one way to put it. Another way is to say it's just sums of products of m element subsets of this axis. So e 0 is 1, e something negative is 0 in particular. Now what is e nu where nu is a whole tuple of integers. It's just the product of the e nu i's for the i's and for i's going, ranging over the indexes. In particular, if any of the nu i's is negative, then this is 0. So Gauss proved that the symmetric functions as a k algebra are freely generated as a commutative k algebra actually by the elementary symmetric polynomials. Equivalently, you can write it as a statement about the k module s. Namely, that k module has a basis which is namely this family of all elambdas, where lambda is the partition whose all entries are at most k. Note that e m is 0 when m is bigger than k because you won't have anything in the sum. It will be an empty sum. Now here is an analog. Instead of the e i's, I now have the h i's or h m's, the complete homogeneous symmetric functions, or rather polynomials. These are the sums of all monomials of degree m. So again, h 0 is 1, h something negative is 0. And again, I extend this notion to tuples by just multiplying through the entries. And again, the k algebra s is freely generated by this complete homogeneous symmetric functions if you go from the h 1 to the h k. Now, of course, this time, it's not true that all the bigger h's are 0. So if you want free generators, you need to stop at h k even so they keep going on. And equivalently, the h lambda is the partition whose entries are at most k, former basis of s. Now, let me mention this also. The h lambda is the k partition, also former basis of s. And these two bases are actually different. So in finitely many variables, you have two h bases. Of course, the lambdas here are the transposers of the lambdas here or conjugates, you can call them. And now I'm going to briefly introduce the Schur polynomials. So for each k partition lambda, one way to define it is this. It's a fraction. On the top, you have this modified van der Mont determinant. And on the bottom, you have the actual, the original van der Mont determinant. And yes, this is going to be a polynomial. Another way to define it is based on the so-called Jacobi Trude formula, which is another determinant, basically. Now, it's well known that this equality holds, that this is a symmetric polynomial with no negative coefficients. And a third way to define it is via semi standard Young tableaus. This is not something that I'm going to use much in the stock. So I'm just mentioning this as something that exists. If you've seen it, it's, it's maybe the simplest way to define the Schur polynomials. And again, if you restrict lambda to a k partition, you get a basis of the k-module s. So one thing that's neat about the Schur polynomials, it's not only are they themselves polynomials with no negative coefficients, but also if you multiply two of them, you can expand the result in the Schur polynomials again with no negative integer coefficients. They're called C lambda mu nu. They're known as little wood reaches and coefficients. Yeah, so there is actually many theorems of this shape because this coefficients, they count many different things. Now, to preempt, to preempt something I will have to do later on, we have, let me slightly extend the Schur polynomials. We have defined as lambda for any k partition lambda. Now I'm just going to blindly expand this definition to lambda in z to the k. So I allow nonpartitions. I allow k tuples with negative entries and with increases. And I define them in the exact same way. Now it turns out that I don't actually gain anything new this way because if alpha is a k tuple, then this newly defined s alpha is either zero or it's one of the old slumdas for partitions lambda, except possibly with the minus sign. And there is a rule for actually finding this lambda. Basically, you have to raise the entries of alpha by the numbers k minus one, k minus two, and so on, k minus k. If the resulting tuple still has a negative enter, you have zero. If the resulting tuple has two equal entries, you have zero. Otherwise, you have to sort it in decreasing order. You have to watch the sign of the permutations that you do, and then you have to un-raise these numbers again, and then you get the lambda. To be honest, this is probably not worth stating because this is what you get if you just correspondently permute the rows of this matrix. Note that the alternate formula still holds in most cases. It holds if all this lambda i plus k minus i are no negative. Now, the reason I'm mentioning is this is not an exciting result. This is just something that will come up later naturally. I will need those s alphas. Now, so let's define this new quotient. First of all, we pick an integer n greater or equal to k, and we pick k polynomials, not necessarily symmetric at this point, such that the degree of ai is less than n minus k plus i for each i. In particular, they can be constants, and in many applications they are constants. Let j be the ideal of the polynomial ring that's defined as follows. It's generated by these h's, except from each h I subtract the corresponding a. Note that this can count as a deformation because the degree of each a is just smaller than the degree of the corresponding h, so I'm subtracting smaller terms from each of them. And here is the first result that I actually had to work for, that isn't the literature, that the quotient ring as a k module is again free with the same kind of sub-staircase basis, except you have to adapt it a bit. The basis has the form x1 to the power of something smaller than n minus k plus 1, x2 to the something smaller than n minus k plus 2, and so on xn to the something smaller than n. Overline always means projection because it's a quotient ring. And this is how many elements this basis has. So this ring is zero-dimensional as its geometers would call it, or it is a free module of this rank. So from now on, I will assume that this a's are symmetric polynomials, not just arbitrary polynomials. Then I can use the same differences to generate an ideal of s. Previously it was an ideal of p, but now that they lie in s, I can also generate an ideal of s from them, so an ideal of the symmetric polynomials. And they can take that quotient. What is that quotient? Well, I can characterize that as follows. We define omega to be the permutation that has k entries of size n minus k each. So it's a k by n minus k rectangle. And I define pkn to be the set of all k partitions that fit into omega. So this is this classical notation for partitions fitting inside each other. And this is the same thing restated in pedestrian terms. All it has to do is it has at most k entries and the first entries at most n minus k. Then the k module s over i, so symmetric polynomials model this ideal is free with basis the sure polynomials that fit into this rectangle. Next, I'm going to assume something even more restrictive. I'm going to assume that the a's are actually constants, not just symmetric polynomials, but constants. Actually, some of what I'm going to do is does not require this assumption, but I think it's okay to make it here, particularly because the classical cases are particular cases of this. So the classical co-molitary ring, you can just get it when the a's are all zero and k is the integers. Then s module i becomes this classical co-molitary ring. And this s lambda bars, they are the Schubert classes. Quantum co-molitary still fits in this definition, except you have to put the indeterminate in the base ring. The q, so k must be that q. And this time not all of the a's are zero, one of them is actually q with a sign. And then you get the quantum co-molitary ring. So this above theorem allows us to forget about the geometry if you want. So we can just define the quotient and the theorem tells us that the quotient looks like it should look. It has its nice basis. It doesn't, for example, collapse to zero or something else. So a lot of papers on the subject have, in a way, been using geometry sometimes to derive combinatorial consequences. With the theorems, this is not necessary, at least if you only want the consequences. So there is more to be said. So let's look at these bases one more time. Since it's a basis, we can take the dual basis. So for each mu, for each partition mu that fits into this rectangle, we can take the linear form on s modulo i that just sends every element to its s mu coordinate with respect to these bases. Moreover, for every k partition mu, we define the complement of mu. This is going to be the partition that you get. Remember, mu fits into this rectangle. And now if you remove mu from the rectangle and you turn the result around by 180 degrees, you get a new partition. That's the result. So this is how you can define it formally. And finally, for any three k partitions alpha beta gamma, you can look at the coefficient of s gamma check in s alpha s beta in this quotient ring, of course. Not in the actual symmetric polynomials, but the projections of the quotient ring. This is the generalization of little wood Richardson coefficients and also of the three point Gromov Wittman variance. So as a theorem, that is that this things have the same as three symmetry as the Gromov Wittens. So I can basically permute alpha beta and gamma as they want without changing the coefficient. And even more symmetrically, I can rewrite it as a coefficient of the rectangular box in the product of all three-shoot polynomials. And there is a way to restate it. So what about some other bases? And actually, I must say I have read, I have seen surprisingly little about bases of the Cromology ring, even for classical Cromology. Everybody says that the Schubert classes form a basis, but then what else? Nobody, I haven't seen it written down. Well, now here's a couple more. So the H is also for my basis. And the transfer matrix is unit triangular with respect to a reasonable order, but I don't know how to describe it. It's not like I have this kind of analog of cost-conumbers handy. There is actually a formula for expanding a single H in terms of the Schur's, except in the example that if this M is too big, then the Schur's themselves will not fit into the rectangle and then you will have to expand them. And this is going to get messy. But it's interesting that if you're expanding an H, you get Schur's of hook shape. Okay, here is something else. Remember the Peary rule for symmetric polynomials. It tells you how to multiply a Schur polynomial by a complete homogeneous by an H. The result is the sum of Schur polynomials and the sum is ranges over all k partitions such that the new partition over the old partition is a horizontal J strip. So what is a horizontal J strip? So basically, it's every, it's all possible ways to add J boxes to your young diagram in such a way that no two boxes are in the same column and that you get a partition in the end. This is another way to describe it. So basically the entries of the two partitions interleave each other. So this Peary rule works in the ring of symmetric polynomials and by quotienting it also works in classical homology. Quantum homology has a more complicated Peary rule. Now what about this generalized quotient ring? Well, it turns out that it does have a Peary rule too. So this part is familiar. This is just the normal Peary rule. However, the normal Peary rule will often give you some partitions that do no longer fit into the box, into the rectangle. And then you have to reduce the model of the ideal and you get some kind of decay product, so to speak, error terms, maybe. And here is a way to actually explicitly write this error term down. So this is a very common rule. All of these new things fit into the box. And these are little wood Richardson coefficients. I'm wondering if you've ever seen a Peary rule that involves little wood Richardson coefficients. I haven't before. This is kind of new. And that involves the AIs, but it only involves them linearly, which is also a bit weird. And yes, it generalizes the Peary terminology by Bertram Cio, Kant, Fentany and Fulton. However, notice that this little wood Richardson coefficients here can be bigger than two, bigger than one. So this is not a multiplicity free rule. And this is an example for fairly reasonable N and K. Now you might wonder how about multiplying by an E instead of an H in classical symmetric function theory? There is a complete symmetry between the two. In homologies, there is still a symmetry. It's a little bit tricky because you have to swap k with n minus k, but there's still a symmetry. Even in quantum homologies, the symmetry is still there. Posnickov, I think, has shown it. Well, you can forget about symmetry now because at least it's not obvious if it exists. The thing is in this Peary rule for the Hs, you have linear terms. Only one A1 here, only A2 here, only A3 here, but here you get squares, for example. And I have not been able to make any sense of this formula. I don't even have a conjecture for what the general rule would be if there is a Peary rule for E. And here is another example for those who might think it should be easy to multiply at least with a K. It's kind of the extreme case where you're just adding a column. Well, but you're adding a column and then you're reducing models ideal and reducing can give any kind of mess. So even this case is not easy. Now, I've been talking about reducing. In quantum homologies, there is a pretty nice trick for reducing any Schur polynomial to fit it into the box. In a way, you basically just keep removing certain kinds of hooks from it until it fits into the box. And what you get is a Schur polynomial multiplied with some sine and some power of Q. Now, is there such a thing for the generalized quotient? So here is an example where we are reducing 443. Now, 443 isn't a far cry from the rectangle. The rectangle here is 333. 443 just has two extra cells. So if you get such a messy result, it feels kind of hopeless. It feels like it gets messy. That's sort of all you can say. Well, it turns out that this mess can actually be controlled. So here is a reduction rule. More likely, I call this a straightening rule. It's one step of the reduction rule. If you want to fully reduce something, you have to apply it many times. So given a k partition that does not fit into the box, so its first entry is bigger than n minus k, I take all possible k tuples whose first entry is mu1 minus n. So it's my old first entry lowered by n. And then all the other entries are increased by one or remain fixed. So 2 to the k minus 1 many terms, which are usually not partitions, they have a very negative first entry usually. And even the next entry, they might wiggle instead of being decreasing. So it's not a partition at all. Then my Schur polynomial in this quotient equals this sum. Oh, and I don't sum over the entire set. I sum only over a certain slice of it. So only a given size. Also, I guess with this sum, including this sum, it's a whole set. So this is weird. And this is the reason why I introduced Schur polynomials for non partitions. Because as I said, these lambdas are not partitions usually, but you can straighten them out. So this, and again, you see only linear terms in the ace. However, in the general case, these things will not fit into the rectangle as well. They will still not fit into the rectangle. And then you have to apply it many times. And this is how you get all this higher degree terms. So I call it a rim hook algorithm. But frankly, I don't know what the rim hooks here are. I understand this only in terms of the stuples. I don't understand this in terms of young diagrams. I suspect that this old method I think by King with the slinkies can say something about it. Well, unfortunately, I don't quite understand this method. So maybe somebody here can help me with that. Okay, now for the Holy Grail. And the Holy Grail is still well in its place. Do we have non negativity? So are the structure constants non negative? This has to be interpreted correctly because we have all these AIs now. So the structure constants will be polynomials in the AIs. And if you look here, you will hear, for example, you see a lot of negative signs. But since they're polynomials in the AIs, you may think that you could put some signs in front of the AIs and then it will be non negative. Well, apparently that's true. So I re-labeled, I re-indexed the AIs. It's as BIs with the signs in front of them. I picked three partitions, I picked three partitions, slum them, you knew that fit into the rectangle. The claim is that the coefficient of this, what's going on with the sound? So is this always the same sign? Christian, I'm with you. Yes, it is strange, but it is. So we hear you here, Daryng. So the coefficient of another S in the product of two Ss has a predictable sign. I'm not going to say it's positive because the sign is there. So you would have to also change the signs for all the Ss in terms of the degree. So this has been checked using Sage for all N up to 8. I'm not fully sure I should believe it because honestly I would like to see a few more Ns, but it gets harder and harder to check this. And if it's true, it would generalize the positivity of Gramovitian variance. Seeing that it has only recently been proved combinatorially, it would probably not be very easy to prove combinatorially. And this is of course one of the reasons why I think this ring could have geometric meaning. So as I already mentioned, there is another basis. You can take the monomials, symmetric polynomials, and then you take their projections. And that's still a basis. I don't know what the structure constants are, but I'm just saying it is the thing that's, it's not hard to prove using triangularity again. Here is, here is a thing, something that's not a basis. The power sums do not form a basis. And that's not because of positive characteristics things, even in characteristic zeros, this is not a basis. Even if AI is zero, this is not a basis. I'm kind of curious as what's going on there, what kind of subring we get, but I don't know. I haven't actually looked at forgotten symmetric functions, but there is a bunch of other bases that could be checked. So what are reasonable questions? One is whether S-modulo i has a geometric meaning, and if not, why do we have all these nice properties? Because this ideals, there is no guarantee that an ideal like this could would behave nicely in any way. In theory, you could have some weird big Gröbner bases, it would reduction would give you big masses expressions with no patterns. The next question is partly resolved. This is due to something Kristen Krattenthaler has asked in one of his papers, what happens if we replace the edges of the ideal by the power sum symmetric functions? So as a high school student working in the MIT prime's project last year, he has proved that the basis theorem still holds. So the basis that I mentioned, the sure basis is still a basis for this quotient. So I don't know any further properties for so far. I don't think there is a symmetry. I haven't looked at the theory rules. Well, the other catch all question is what other properties of quantum globalize. I have tried post-nicov's curious duality and cyclic him and symmetry. I've tried this duality between k and n minus k. They don't seem to generalize. Maybe somebody with more intuition for the geometry can say how I should probably tweak them to make them generalize. But if you just follow the algebra, it doesn't seem to work out. Is there an analogous construction for the flag manifold? Now I have to admit the quantum cosmology of the flag manifold already has a lot of deformation parameters. I don't quite see how to deform it further. So I'm not actively expecting something to happen here. Is there an quiverient analog? I started working on this. I haven't gotten very far. And what about quotients of quasi-symmetric polynomials? This is a wild card. Sometimes you get something interesting in there. Sometimes you just don't. There is an SK module structure on the quotient. So this time I'm talking about the quotient of all polynomials, not of the symmetric polynomials, because the symmetric polynomials have a trivial SK action. But what if I quotient all polynomials? What is the SK model structure? Well, it turns out, and I've been too lazy to write this up for over a year now, that this is the SK module you would basically expect. It's the power of the regular representation. If you've seen diagonal harmonics or not diagonal, just regular harmonics, you would not be surprised by this. And finally, how much time do I have left? It is up to 20. Oh, okay. I can actually... It's 13 or 10 or 13 minutes, all depends. Okay, I can talk about the proofs then. So let me quickly talk about a question that just feels natural. I don't know how natural it is. So instead of looking at symmetric polynomials and K variables, I can look at symmetric functions in infinitely many variables. This is like the grown-up version, except it's easier. And I'm going to use the same notations for them, except I will put them in boldface because... So I don't get to confuse them with the polynomials. So the symmetric polynomials can be viewed as the symmetric functions, modulo all the high E's that would be zero in the symmetric polynomials. So as modulo i, what's the quotient of a quotient? It's just a quotient by all those things. So you get it from lambda by quotienting out both the E's and the corresponding Hs minus A's. Now this looks a bit asymmetric. I mean, it's asymmetric in two ways. You have infinitely many E's here and only infinitely many H's, but also you are deforming the H's but not deforming the E's. Now I cannot do anything about the first asymmetry. I've tried. It feels like you should be able to kind of have... You should be able to arbitrarily mix H's and E's here, but it's not clear how. If you think about it, you don't want to mix them randomly. That will give you very random rings. But the second asymmetry, you can just try to fix it by also deforming the E's. So here's what comes out. I will... And I'm switching to boldface letters here because they will be elements of lambda. They will be symmetric functions. So A1, A2 up to Ak and B1, B2, B3 and so on should be symmetric functions with appropriate degree conditions. Then this quotient is a free k module with the same basis that we know and love already from comology. So the same Schubert classes. So sure functions that fit into this rectangle. So now you have infinitely many parameters. I do not know much more about this quotient. There is probably some really low hanging fruit here like other bases. I don't think... I wouldn't really expect much to hold for it honestly because this one really does not have any hint of a geometric meaning. But it just feels like a nice thing to have seen. So let me say a few words about the proofs, well except for the SK action. So many ideas. First of all, the quotient of all polynomials, not the symmetric ones, can be done with Gröbner bases. And in hindsight, this has already been done by Konka, Krattenthaler and Watanabe, already sketched. And Gröbner bases actually, so for everybody who is a bit scared of this, so people usually talk about Gröbner bases over a field. Well, the existence theorem requires a field. But if you have a Gröbner basis, so if you just want to use a given Gröbner basis, if you already know that something's Gröbner basis, you only need a commutative ring. It's like this vector spaces. For a vector space to have a basis, you need a field. But if you have a free module with a basis over a ring, you can use it. You don't need that ring to be a field anymore. So in the same way, this boils down to actually finding an explicit Gröbner basis for this ideal J. Using that and Jacoby Trudy, you can show that the symmetric polynomial quotient also has the right basis. And the rest, more or less, is about computing the symmetric functions. So I will probably not talk about Gröbner bases here. It will probably take a long time. This is all just in case slides. But here is the Gröbner basis of this ideal J. So x i dot dot dot k, so x i dot dot dot k is a shorthand for x i, x i plus one, and so on x k. So the Gröbner basis does not use the complete homogeneous polynomials of the whole axis. It uses them only for certain sub sequences of the axis. So if not for the deformation parameters, this would be a Gröbner basis. Well, now you have to deform it. And you deform it in this kind of strange way with a coefficients and with ETs, so elementary symmetric polynomials in the other axis. With respect to the degree of psychographic ordering on the monomials. And so you have a very nice Gröbner basis with really nice leading terms, which is just x i to the n minus k plus i. So, McCullis basis theorem tells you what the basis is for the quotient ring. Now, how do you jump from all polynomials down to symmetric polynomials? Okay, so using Jacobi-Truidi, you can reduce each shorthand polynomial that does not fit into the rectangle. Two shorthand polynomials that fit into the rectangle. It's a bit of a straightening rule. So the shorthand polynomials that fit into the rectangle span the quotient. Now, we need to prove that this is linearly independent. And as usual, this is the harder part. It's like PBW. However, we know that we know already from Artein that this family is span the polynomials as a symmetric polynomials module. And combining these two gives that these products span the quotient of all polynomials. But we already have a basis for this quotient. That's what we did on the previous slide with Grobler basis. So what if you have a k module with the basis and the spanning family that have the same finite size? It's well known or an easy exercise that the spanning family must then also be a basis. So we conclude that this family is the basis of the big quotient. And therefore, it's subfamily is the basis of the small quotient. So it's a bit of a ping-pong argument. It feels natural if you're into symmetric functions to just work with symmetric functions all the time or with symmetric polynomials. But we couldn't, I couldn't prove it dependence with symmetric polynomials. If you actually try to find the Grobler basis of i inside s, I don't think this has a good answer. I think the Grobler basis become very ugly very soon. So you climb up into the ring of all polynomials, you look at the quotient there, and you have, and then you argue using this fact that the spanning set that has the same size as a basis must also be a basis. I think this is a trick that's a good takeaway if you don't know it. I feel like it can be useful in many other situations. So basically, if you don't find a nice Grobler basis, maybe it helps to go into the bigger ring. And the rest of the proofs, as I said, it's about computing with symmetric functions a lot. I'm just going to briefly notice this identity commonly ascribed to Joseph Bernstein, but it appeared first in the book by Zilevinsky, that it's basically saying if you have a Schur function and you want to insert a new entry at the front, well preferably of course it should be bigger than the first entry, so you still have a partition afterwards, you can achieve this by a certain operator which is made out of skewing and multiplying, skewing by e's and multiplying by h's. This operator is now known as the Bernstein operator, or I think Garcia has called it the Schur-Row-Edder for an obvious reason. So this is in a way the driving force, but there is a lot of computations with Peary-Rolls and Jacobi-Truzzi matrices. Okay, so much for this and I hope to actually have two papers out soon enough about it, not just a big preprint with a lot of messy proofs. So thanks are due to Sashar Posniko for the whole project that I started working on back at MIT. Gerard Duchamp, thank you for inviting me again and a lot of people have done some, have contributed some ideas and thank you all. Thank you dear Dari, now time for questions, remarks or comments, maybe? I do have a basic question. Can you, maybe I missed that point from the beginning, but Dari, please, can you lead us again the connection between the little Richard von Coefficient and the Gromov-Wittern? That's the very beginning, I believe. Can you explain us again the link? Because I think if this is happening maybe for the little Richard von, you may have a rule similar with another invariant for, for instance, the Kroniker by changing the law. So I think this connection may be fruitful for doing other stuff, this is what I want to know. So let's see. Maybe just, you know, if you can, yeah, keep up with that. So this one is a coefficient of a Schur polynomial in the product of two Schur polynomials, except you're working in the quotient ring. But if you're, if you're working in the classical cosmology, then the a parameter is zero. So the quotient ring is basically just looking at like a little piece of the symmetric functions ring. It's not, it's in that case, there is no error terms. It's not being in any way contaminated by the a's. So that coefficient, of course, if you choose alpha, alpha, beta and gamma appropriately, you can get any little Richard from coefficient like this is such a thing. But this is only happening for the deformed Schur the deformed Schur polynomial, right? I mean the one with the bars, not with the classic one, as you say, right? So if you pick, if you take the a is to be zero, then the bars don't really change that much. So basically in low, in sufficiently low degrees, everything behaves as it was. Okay. Okay, thank you. Are there questions? Yes. Do you see another way of deforming so that the duality k to n minus k be respected? So this is tricky. In theory, I could well imagine that it is a duality if you properly somehow, I don't know. No, okay, no, this is probably nonsense. But so the thing is that if I have k, I have k parameters. If I have n minus k, I have n minus k parameters. So it's not quite clear. Okay. So I probably should have mentioned that this is all looking kind of similar to what is called factorization algebra and splitting algebras in the numerator geometry or even in constructive Galois theory. This algebra is where you just pick some polynomial and you declare it to be a product of two polynomials. And the indeterminate coefficients of these two polynomials will be my new indeterminates. But I wouldn't, I don't know if this is in any way as a morphic or a particular case or a generalization of that. They just feel alike. Yeah, actually, maybe a comment because sorry, I missed half of your talk, but still, as I wrote you in chat, this is in fact infinitely many variations, different deformations of algebra with the basis and this is proper to be Frobenius algebra. This is quantum, or big quantum homology gives and k actually are natural, we are in classes of tautological k dimension bundles and n minus k will be a class of n minus k that tautological bundle on grass mining. Yeah, so definitely there are natural k parameter families of deformations from quantum commology from bickering. Thank you. Do you think they could be, they could give the same algebra originalization of mine? Why not? Yeah, I would be surprised to have two different different deformations of same algebra with same number of parameters. Of course, it's a question of how to interpret combinatorially. Dari, I have a quick question. So the deformations that you that you consider all come by tweaking the Jacobi Trudy identity or basically perturbing the Hs. Yes. Is there a way of doing deformations so where you exploit say the Jacobi, the sorry, the Giambelli identity instead? So for instance, instead of perturbing the Hs, you perturb hook shapes. Good question. Or I mean, that's another way is can you, can your constructions be realized through a, you know, through a perturbation of Giambelli? Good question. And I'm thinking about this more in the case of sure functions, the infinite case, because all of the Yeah, no, I don't even know this. Do you know this in classical homology? If you don't deform anything, if you just quotient by Giambelli by hooks instead of by Hs. Oh, I think I, I think I remember something by mother and Bible, maybe, but I think it's not quite, is it hooks? You sure finite elements and lumbering that something similar. I need to read, I need to read that one again. Maybe small small question. I probably missed some part of your talk, but when you consider this deformation, you work with a commutative group, their basis. Yes. In spite of this deformation, what, what do you see any place for non commutative group basis there? Maybe you can directly work in the different situation and construct some non commutative group basis to get the basis which you need to work further. Of course, probably it's difficult to answer right away, but it's something probably, which can probably give some simplification. Some other things to do. I don't know. I'm curious as to how non commutativity will simplify things here, but I have to say right away that I don't know what what non commutative sure functions would be. I mean, there are several versions of them, but none of them feels particularly canonical. Yeah, of course, it should be chosen correctly, appropriately. So yeah, it's difficult to say, just suggestion to try. Because if you work straight away in in quotient, you can still do the same as when you work with commutative monomials. And maybe it could be easier, not necessary, but it can. There is one version with practical classes inside the world, inside the free algebra. Oh, okay. Okay. Oh, since like three ribbons. I don't remember the first authors, but I know that Schützenberg and Lascaux worked a lot on this. Other comments or remarks? So we can thank you, Darij.
co-molotry ring. Classically, it is well known symmetric polynomials in K variables over the consecutive complete homogeneous symmetric co-molotry ring. And for the Grasmanian, it's geometric meaning, or at least some kind of non-negative integers, K is a fixed non-negative is freely generated by this complete homogeneous Schur polynomials. We have defined as lambda for ring. And this is how many elements this basis put the indeterminate in the base ring. The q, so of the quotient ring. This is the generalization k partitions such that the new partition over the in the ring of symmetric polynomials and by an H in classical symmetric function theory? likely, I call this a straightening rule. It's So I call it a rim hook algorithm. But frankly, I non negativity? So are the structure constants edges of the ideal by the power sum symmetric modulo i, what's the quotient of a quotient? It's appropriate degree conditions. Then this quotient have any hint of a geometric meaning. But it just