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# the  geometry of numbers

The geometry of numbers is q branch of number theory created by Minkowski.

The idea of this theory is to consider the intersection of some set with a lattice in the Euclidean space $\mathbb{R}^n$ to get the existence or non-existence of some particular number.

Start with the basics. Suppose $L$ is a lattice in $\mathbb{R}^n$(i.e. a discret subgroup of the additive group of $\mathbb{R}^n$ without point of accumulation, which viewed as a set of vectors in $\mathbb{R}^n$ generate the whole space). Using the fact that $\mathbb{R}^n$ has a finite set as basis, we deduce easily that $L$ is finitely generated. Using the fact that $\mathbb{Z}$ is a principal domain, we see that $L$ has, in fact, a finite set as $\mathbb{Z}$-basis(what is more, the cardinality of this set is independent of the choice of the basis, it is an invariant under isomorphism, also a property of the abelian group). That is we can assume that $L$ looks like $L=\mathbb{Z}$. What is more, we can for any other basis $\{f_1,f_2,...,f_n\}$ of $L$, the matrix of these $f_i$ expressed with respect to these $e_j$ lies in the group $GL_n(\mathbb{Z})$. After choosing the canonical basis of $\mathbb{R}^n$, we can calculate the determinant of the matrix $(e_1,...,e_n)$. The absolute value of this determinant is, coincidently, the volume of the fundamental domain of the lattice $L$, or in other words, the volume of $\mathbb{R}^n/L$ equipped with the quotient measure(we denote it by $covol(L)$, the co-volume of $L$). This value is, in some sense, the generator of the image of the homomorphism of groups $f:L^n\rightarrow \mathbb{R}, (a_1,...,a_n)\mapsto det(a_1,...,a_n)$.

The fundamental result in the geometry of numbers concerns the intersection of a convex set with the lattice. We say a convex subset $C\subset \mathbb{R}^n$ is symmetric if for all $x\in C$, we have $-x\in C$. So if $x,y\in C$, we have that $latex\frac{x-y}{2}$. This is really a simple fact, but it is also a really important observation. Now we are ready to state the theorem of Minkowski:

Suppose that $C$ a convex symmetric subset of $\mathbb{R}^n$, and $L$ is a lattice of the latter. What is more, $covol(L)<\frac{vol(C)}{2^n}$. Then there is a non-zero element $0\neq x\in L$ which lies in $C$ also. That is to say, $\#C\bigcap L>1$.

The point of the proof is to express the above inequality as a comparison between two volumes. Noting that if we dilate the lattice $L$ by $2$, then we have that $covol(2L)=2^ncovol(L)$. So the inequality turns into $covol(2L). Then we can say that there are two distinct elements $x\neq y\in 2L$ such that the half of their difference $\frac{x-y}{2}\in C$. If it is not the case, then after choosing a measurable representation $X$ of $\mathbb{R}^n/2L$ in $\mathbb{R}^n$, we can express $C=\bigcup_{x\in 2L}(C\bigcap(X+x))$. And so $C'=\bigcup_{x\in 2L}(C\bigcap(X+x)-x)$ has the same volume as $C$. But $C'$ is now a measurable subset of $X$(because $C$ is a measurable subset, see here), so we have $vol(C)\leq covol(2L)$, which is a contradiction to the assumption. So we get a non-zero element $\frac{x-y}{2}\in C\bigcap L$, thus proved the theorem.

Now come to the applications of this fundamental result. As a first example, we consider one theorem of Fermat’s, that is a prime number $p\equiv 1(mod 4)$ if and only if $p$ can be written as a sum of two squares of integers, $p=a^2+b^2(a,b\in\mathbb{Z})$. Note that $p\equiv 1(mod 4)$ is the same as that $-1$ is quadratic residue mod $p$. So we can find some integer $u$ such that $u^2\equiv -1(mod p)$. We will use this $u$ to construct some lattice $L$ in $\mathbb{R}^2$. That is, ze define $L=\{(a,b)\in\mathbb{Z}^2,a\equiv ub(mod p)\}$. Clearly this is a lattice(a routine way of showing this is, $p\mathbb{Z}^2\subset L\subset\mathbb{Z}^2$). What is more, the indice of $L$ in $\mathbb{Z}^2$ is $p$(just consider the group homomorphism $\phi:\mathbb{Z}^2\rightarrow\mathbb{Z}/p\mathbb{Z}, (x,y)\mapsto x-uy$, which is surjective, with kernel $ker(\phi)=L$. So we have $\#\mathbb{Z}/L=p$).And thus $covol(L)=p*covol(\mathbb{Z}^2)=p$. Now consider the disk $D=\{(x,y)\in\mathbb{R}^2, x^2+y^2<2p\}$. This disk is clearly convex and symmetric, with volume $vol(D)=2p\pi$. Now that the assumptions are satisfied just by noting that $p<\frac{2p\pi}{2^2}$, so there is $0\neq x=(a,b)\in L\bigcap D$. Note that since $a=ub(mod p)$, we have $a^2=u^2b^2=-b^2(mod p)$, in other words, $a^2+b^2=np$ for some integer $n$.But recall the definition of $D$, we have that $a^2+b^2<2p$, so the only possibility is that $a^2+b^2=p$.

So by carefully choosing a lattice, we proved the theorem due to Fermat. This is a small success of the geometry of numbers. In fact, the above reasoning can be generalized to any polynomials like $a^2+db^2$ where $d$ is a positive integer.

The following is one such result:

Suppose that integer $d>0$ is a quadratic residue mod $p$, then at least one of the numbers $p,2p,..., hp$($h$ is the largest integer such that $h\leq\frac{4d}{\pi}$) can be expressed in the form $a^2+db^2$.


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# How to manage the absolute value on a differential equation $|T(x)'+A(T,x)+B(T,x)| = f(T,x)$

Hi everyone I need to solve an equation of this type:

$|T(x)'+A(T,x)+B(T,x)| = f(T,x)$

with boundaries conditions.

The absolute value is my problem. Of course without it, the solution of these is well treated in the literature.

After search in the questions I found this: Differential equation with absolute value

So, can I do the same procedure? Or there is another way to solve this? Thanks.

-
Crosspost on Computational Science: scicomp.stackexchange.com/questions/2212 –  user17762 May 14 '12 at 18:35
Cross posting on different sites simultaneously is discouraged. meta.stackexchange.com/questions/64068/… Wait for a couple of days before deciding to cross-post. –  user17762 May 14 '12 at 18:37
ok I deleted the other one. –  Nikko May 14 '12 at 18:45

Your equation does not determine $T'$ in terms of $T$ and $x$, so uniqueness of solutions may be a problem. You can say $T'$ is either $f(T,x) - A(T,x) - B(T,x)$ or $-f(T,x) - A(T,x) - B(T,x)$. Presumably there will be one region where it's the first and one region where it's the second, and (assuming $f,A,B$ are continuous) if you want $T'$ to exist everywhere it'll be impossible to switch between one and the other except when $T' = 0$.
Thank you Robert. But can you explain me why you only take two combinations and not more? I mean why not $f(T,x)+A(T,x)−B(T,x)$ for instance ? –  Nikko May 14 '12 at 18:47
$|t|$ is either $t$ or $-t$ (assuming we're talking about real numbers). –  Robert Israel May 14 '12 at 19:07

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# Launch Center Pro and Day One

I’ve recently discovered Launch Center Pro for iOS that lets you create actions and generally extends the abilities of iOS a bit (basically allowing you to create shortcuts to tasks you do often).

There’s a number of things I do on the iPhone or iPad consistently, so being able to do these from a single app can make it shorter to do. It’s a one stop shop for accessing stuff. For example, I’ve set up a shortcut so I can message both my mum and dad at the same time, without having to open Messages, select new message (or find a previous message I’ve sent to them both) and add them both to the message and then compose. This one lets me just open Launch Center (it’s on my dock), select contacts and then the message parents option. Easier.

Other tasks I do on the iPad regularly, one of which is logging using Day One. I’ll often do this in bed on the iPad after I’ve put my kindle down or stepped away from the Mac.

However, whilst I’ve managed to journal everyday for the past year, I’m always looking to improve.

## Activity Logging

One of the things I would like to add is activity logging. So I cycle fairly regularly and upload most of my rides (not my standard weekday commute though) to Strava and Garmin for tracking (as I do like statistics, even if they’re just to create graphs).

However, there wasn’t really an easy way to add these to Day One without just adding them to the text. I’d like to include them in an easy to read format and perhaps in instances separate to my main journal entries.

Then I stumbled across this post when investigating Launch Center. It has a number of ready made logging shortcuts for Day One that are launched from Launch Center itself and populates it with a table. This was what I’d been looking for!

After some playing, the basic table I came up with was the following:

Daily Activity
Type
Distance (Km)
Time (minutes)
Calories Burnt (KCal)
URL

This would give me a basic table. However, the benefit of Launch Center was that I could also repopulate the table when inserting it into Day One, using prompts for the user. I was therefore then able to add prompts to allow the user to fill it in as they inserted it. This allows the user to select from either Cycling, Running, Walking, Swimming or Other (this covers my own activities) and then lets you specify the other item - the URL field is for linking to online tracks of the data (on Strava or Garmin).

This gives the following code to insert into Launch Center.

<code>dayone://post?entry=%0D%0A%7C%20Daily%20Activity%20%7C%20%7C%0D%0A%7C%20---%20%7C%20---%20%7C%20---%20%7C%0D%0A%7C%20Type%20%7C%20[list:Activity Type|Cycling|Running|Walking|Swimming|Other]%20%7C%0D%0A%7C%20Distance%20%7C%20[prompt-num:How long was it?]Km%20%7C%0D%0A%7C%20Time%20%7C%20[prompt-num:How long did you do it for?]%20minutes%20%7C%0D%0A%7C%20Calories%20Burnt%20%7C%20[prompt-num:Amount of calories burnt?]Kcal%20%7C%0D%0A%7C%20URL%20%7C%20[ [list:URL Location|Strava|Garmin|] ]([prompt:URL Location])%20%7C
</code>

You can also grab the URL from Launch Center’s site here. This allows a laid out entry to inserted into Day One, such as this.

(Ignore the music log, that’s to be sorted slightly and discussed in another post).

The Strava or Garmin URL is easily taken from the Strava/Garmin app and inserted and as the URL doesn’t mess with the clipboard, I can open Strava, copy the URL, open Launch Center, select the action and walk through it and paste the URL in.

## Keyboard Maestro

However, I don’t just log on my iPad, I also log on my iMac as well, depending on my mood or where I am.

How could I do the same on the Mac? Well, I’ve been a keen user of Keyboard Maestro since I discovered it was a lot more powerful than Textexpander (but still allowed me to use it for basic text expansion like Textexpander). Now it’s a program I miss when not at a Mac (the big one for me is highlighting a word and then pressing specific hotkey to replace it with a Markdown link!)

Anyhow, after some playing about, I was able to get the Launch Center Pro URL translated fairly well into Keyboard Maestro.

Here’s how (I have this set to only run in Day One).

Typing activity in Day One will then bring me up the following input dialogue box that lets me repopulate the table for pasting.

This then outputs exactly the same table so I’m able to log regardless of what platform I’m on!

If you want to download this macro, I have it on Dropbox.

## Conclusion

With this basic table template completed, I’m looking at expanding this to a few more items - I’m playing with recording all my Last.fm scribbles into a text file using IFTTT to help me log them. I’ve already got the basic working, with the scribbles being download to my Dropbox in a Markdown formatted table.

Hopefully, I’ll be able to add some more in the near future - I’m using the Book Logging one from the previously mentioned blog post. The next time I run this, I’ll convert it to a Keyboard Maestro macro as well.


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# What does imaginary number maps to physically?

I am taking undergraduate quantum mechanics currently, and the concept of an imaginary number had always troubled me. I always feel that complex numbers are more of a mathematical convenience, but apparently this is not true, it has occurred in way too many of my classes, Circuits, Control Theory and now Quantum Mechanics, and it seems that I always understand the math, but fail to grasp the concept in terms of its physical mapping. Hence my question, what does imaginary number maps to physically?

Any help would be much appreciated

• More on complex numbers and physics: physics.stackexchange.com/q/11396/2451 , physics.stackexchange.com/q/32422/2451 , physics.stackexchange.com/questions/tagged/… – Qmechanic Sep 7 '13 at 22:18
• It seems to me that negative numbers are also a 'mathematical convenience'. – Greg Sep 8 '13 at 3:13
• Eigenvalues distinguish a quantum state. As we can't measure two quantities simultaneously, eigenvalues values are real numbers. And to produce real eigenvalue we need hermitian matrix with complex element. – Self-Made Man Sep 8 '13 at 17:19
• I concur with Greg. In history there was a period where algebra was used without explicit negative numbers. Every problem can be stated in such a way that explicit negative numbers are evaded. Once mathematicians started using negative numbers explicitly many problem could be stated much more efficiently, enabling more efficient mathematical thinking. It's analogous for complex numbers. It's possible to create a set of rules for manipulating pairs of numbers that never mentions the square root of a negative number explicitly, yet mathematically equivalent to manipulating complex numbers. – Cleonis Sep 8 '13 at 22:24

As far as describing nature is concerned mathematics is another language that can describe it.

In ancient times there was no mathematics other than arithmetic, and that was the way nature was described: cycles of the moon, trajectories of planets etc.

Then came geometry and it used mathematical tools to describe nature further, based on arithmetic but using geometrical axioms and theorems.

Algebra came late, from the arabs, during the middle ages where the dogmatism of the church had stopped progress in science. With enlightment, mathematics took off and became a beautiful tool describing/modeling nature by the time of Newton, the co-inventor of calculus. Negative numbers and complex numbers became necessary by using algebra and are now incorporated in all mathematics useful for describing nature, as you have found out.

We can plot negative numbers in an x plot, if that is what you mean by "maps to physically". Complex numbers, one axis real one imaginary are another plot where the solutions for physical systems can be displayed economically. They simplify notations and calculations. A relevance to physics problems comes in that they describe/incorporate trigonometric functions which continually appear in solutions of physics boundary problems due to the harmonic nature of many set ups. They are economical, in the same way that algebra became economical, not needing the convoluted sentences of the classical (BC) times mathematics.

Complex numbers for physics are, as all mathematical tools used, a convenient representation fitting the data.

• Thanks for the explanation, although I am still very confused, for example I always thought that everything in this world can be map with real numbers, energy, distance, time, temperature....etc, because they give meaning such as order and reference to the physical concept, but what does imaginary number give, what is a quantity that has an imaginary value and what does it even mean? How does an imaginary object interact with real object? Those are the questions in my mind that has been unanswered – user2654176 Sep 8 '13 at 4:28
• the use of the word "imaginary" is confusing. It is just another type of coordinate. It does map one to one to the real numbers with an i infront to indicate it is in a different "space". There are even more complicated extensions of this, called "quaternion" en.wikipedia.org/wiki/Quaternion . Complex numbers are useful for calculation in usual physics problems so we know about them:). If you think of mathematics as a tool it is not necessary to angst about a physical identity other than the convenience of calculations. – anna v Sep 8 '13 at 5:32
• Sorry for using the word "imaginary", it is confusing when asking questions, I do realize imaginary numbers are just another mathematical space, just like real number is an extension of rational numbers, complex numbers are the extension of real numbers, but what is intriguing is the fact that we can directly measure integer, rational, and to some degree of precision irrational and real, but we can not directly measure complex, and this really troubles me as it seems that you cannot map to anything physically using complex numbers? – user2654176 Sep 8 '13 at 5:39
• In truth, every type of coordinate is mapped one to one with real numbers, with just a designation to separate it from 0 to infinity. For example negative numbers map on real numbers with just a - in front meaning subtraction. the i in fron of the imaginary axis of complex numbers is infront of a one to one map to real numbers. If we take complicated vector spaces the same is true. we just have to keep track of which direction we are talking about by a symbol that differentiates this axis from that axis. There is nothing more esoteric about it. – anna v Sep 8 '13 at 5:55
• @user2654176 Do you have some examples of mathematical conveniences that are directly measurable? In other words, what do you mean by "directly measurable"? If you mean "measurable without calculation", then almost nothing is directly measurable as it involves calculation in the measuring process. If you mean "all information can be extracted via measurements", well, the complex numbers used in physics can be measures in their real and imaginary parts, and the information can be joined together to get the complex number. – Manishearth Sep 8 '13 at 12:16

Phase and amplitude! While complex numbers are used in the analysis of Classical Harmonic Oscillator or the LCR circuit as a part of Mathematical technique, wherein once the solution is obtained the real part is extracted. However in Quantum mechanics they play a central role, and QM cannot be formulated without them.

We necessarily need both the concept of Phase and Amplitude in the description of quantum mechanics. When 2 event are separated by multiple paths, then the information about later event occurring given first happens is encoded in the Probability Amplitude. Amplitude of a Single path is essentially the quantity $e^{iS}$, where S is the action. To calculate the Probability amplitude sum over the amplitude for each and every possible path in which the final event can be realized if the Initial event had happened. The Probability that the final event happens is Modulus square of the probability amplitude.

The use of Complex numbers(And Probability Amplitudes) in Quantum Mechanics follows very naturally as a generalization of the concept of Probability, And Feynman very nicely Illustrates this in his paper, "Space-Time Approach to Non-Relativistic Quantum Mechanics" or "The Concept of Probability in Quantum Mechanics"

While one can always rewrite all the formulas in Quantum mechanics as 2 real numbers, However then the beauty, and its simplicity is lost.

• +1 This is the correct answer. QM mechanics can not be formulated without complex numbers or something isomorphic to that field, which again, will be complex numbers regardless of how we call them. – Bubble Sep 8 '13 at 15:39
• This answer seems the best so far. The theory of epistemology put forth by Objectivism requires that concepts are grounded to reality via perceptual evidence. Im not satisfied i can explain explain complex numbers to teenager yet but this helps me get started. – Aaronium112 Oct 1 '16 at 14:48

If you really wanted to, you could formulate the laws of physics without using imaginary numbers - see, for example, Can one do the maths of physics without using $\sqrt{-1}$?

Let's say you need to do a fourier decomposition of a function $f$ in order to find how some responds to being driven at $f$.

If you think back, fourier decomposition is normally introduced with only real numbers, and a function $f$ was decomposed into odd and even parts, which can be represented as infinite series of cosines and sines respectively, and you could certainly do that. However, if we view $f$ as the real part of some complex-valued function, the equations become much simpler.

If this makes you wonder whether the complex-valued function is "more real" than the real part, and what is the physical significance, you could see $f$ as $\frac{1}{2}(z + \bar{z})$ where $z$ is the complex function; then, $z$ and $\bar{z}$ both have no physical significance, but they formally satisfy the equations governing the system, hence their sum must also.

@Prathyush's answer is completely correct and provides nice reference, but I just wanted to add that behind his discussion about phases and path integrals is the concept of unitary and conservation of probability, which is a physical concept, and the OP asked about what imaginary numbers "map to physically". Thinking in the Heisenberg picture one can write every operator as a real diagonal matrix and a set of mappings (rotations) between these operators, but in order for everything to conserve probability it is absolutely essential that these rotations contain imaginary numbers in general as any unitary operator may be written as $\exp{i \hat{H}}$, where $\hat{H}$ is a Hermitian operator (with real eigenvalues). It is similar for another very special rotation in Hilbert space, which is time evolution, without which there would be no physics as such. This has a deep connection to the relation between Lie algebras and Lie groups.

Without complex numbers we would have no unitarity and no conservation of probability or would be led to a trivial quantum theory which is not dynamical!

It's just a mathematical tool1, which is especially useful to portray some kinds of vectors and sine waves. It doesn't have a real "meaning" here, and we can do fine whthout it (calculations just become more tedious)

For example, in circuit analysis, we convert the coordinate pair $(V,\phi)$ (which isn't linear; adding two voltages need not add their phases) into the coordinate pair $(V\cos\phi,V\sin\phi)$, which is linear. Now, instead of writing it as a coordinate pair, we write it as a complex number (which is essentially just a pair of numbers) and then notice that we can extend this to the concept of impedance and even be able to divide by these numbers and get the correct answer. Writing voltage as $(V\cos\phi)\hat i +(V\sin\phi)\hat j$ would still give a linear system, but we can't divide by vectors so it's limited in application.

In quantum mechanics, the use is similar -- we are expressing a linear quantity with a phase. Again, this is a nonlinear coordinate pair, but rewriting it as a complex number gives us a linear coordinate pair that plays nice with multiplication and division. When we bring linear algebra into it, one sees that the representation is even more apt.

Again, it's just a tool or a representation. It has no real meaning.

1. As @MBN mentioned below, real numbers are a mathematical tool, too. We can't "directly" measure real numbers, either.

• So does the complex part of the impedance exist? – user2654176 Sep 8 '13 at 5:41
• @user2654176 Impedance is a human constructed quantity. This is like asking "does resistance exist?" -- resistance is just a quantity that we defined. Yes, impedance is measurable. All you have to do is measure the current waveform as you pass a voltage $(V,0)$ through it, and record $(I,\phi)$. Now, $\frac{I}{V}e^{i\phi}$ is your complex impedance. If you're asking "can we directly get the complex part without calculations" -- I find the question meaningless as we can't directly measure much anyway. – Manishearth Sep 8 '13 at 5:46
• "It's just a mathematical tool" can be said about any part of maths, but it is missleading. – MBN Sep 8 '13 at 12:03
• @MBN What do you mean? Basically, there's no real physical significance to it in the sense that you can't "measure" a complex number. Also, we're talking about physics, not maths. Mathematical tools in mathematics have no physical meaning. Representations in physics can have a physical meaning. – Manishearth Sep 8 '13 at 12:06
• Once upon a time complex numbers were treated as a pair of numbers. You can treat it as a pair or a number in the number plane or a vector in a 2D plane but all have the same property. – Self-Made Man Sep 8 '13 at 17:29

Both sides of a complex number are each real numbers. Either the real part or the imaginary part can be used for computing the value of a measurable quantity. And the results are always the same. Then we are left with only 'i', to determine it's Mathematical and then physical meaning. The allegory of 'i' harbors a real Mathematical meaning which when unveiled, discloses the 'reality' of a 'complex' number. In other words, 'complex numbers' are 'real numbers' in disguise.
One can say the above in another way by saying that (-1)^1/2 is completely a real number.

A rotation by a right angle. So complex numbers come up whenever you have periodicity/oscillations/phases etc.

For a simple example, consider the physical intuition for Euler's formula -- if you push something with force $F=kx$, you get exponential motion, but $F=-kx$ is periodic motion. The latter, where you have periodic, trigonometric functions, is where there are complex numbers/imaginary exponents.

Similarly when describing light, the polarisation of a light wave behaves as a phase. The same idea carries over to quantum mechanics.

## protected by Qmechanic♦Apr 1 '14 at 8:37

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).


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# Thread: Using residue calculus when poles are not exactly known

1. ## Using residue calculus when poles are not exactly known

Can the poles of this integrand be calculated exactly? I don't think so but not sure. How might one still use residues to solve (approximately) the integral even if the poles are only approximated? I posted a relevant plot which I'll leave unexplained just for fun for those interested.

$\mathop\oint\limits_{|z|=2}\frac{1}{1.3\sin(1.7z)+ 0.6\sin(4z)}dz$

2. For the function $1/f(z)$, if there is a simple pole at $z_0$ then the residue there is $1/f'(z_0)$. So if you can get a good approximation to the location of the poles then you can also get a reasonable approximation to the values of the residues.

3. Originally Posted by Opalg
For the function $1/f(z)$, if there is a simple pole at $z_0$ then the residue there is $1/f'(z_0)$. So if you can get a good approximation to the location of the poles then you can also get a reasonable approximation to the values of the residues.
Can you provide the error terms?

4. Originally Posted by Opalg
For the function $1/f(z)$, if there is a simple pole at $z_0$ then the residue there is $1/f'(z_0)$.
Oh yea, I forgot about that. So let's say the upper right pole is about $1+0.25i$. Then using Mathematica's $\text{FindRoot}$, I get for the residue there (last line):

Code:
In[7]:= f[z_] = 1.3*Sin[1.7*z] + 0.6*Sin[4*z];
urroot = FindRoot[f[z] == 0, {z, 1 + 0.25*I}]
r1 = 1/D[f[z], z] /. urroot

Out[8]= {z -> 1.2794588067848602 +
0.3743082054455279*I}

Out[9]= 0.05685562855624362 - 0.2787028735656702*I
Suppose could do that 4 more times and get a good approximation to the integral via residues. Just news to me, that's all.


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# Find fundamental group of space $X=(S^1\times S^1)/(S^1\times \{s_0\})$

Find fundamental group of space $X=(S^1\times S^1)/(S^1\times \{s_0\})$

I am not sure how to do it. It seems obvious that $\pi_1(X)=\mathbb{Z}$ because $X$ is torus where one class of nontrivial loops is contracted to the point.

There is also fact I found which may be used here to show that $X$ is homotopically equivalent to $S^2\vee S^1$ which would surely make above result obvious. However I am not sure how to use this fact which says:

For $i=1,2$ let $Z_i=X \sqcup_{h_i}X$ be topological space resulting from gluing $\overline{B^n}$ to $X$ alongside map $h_i:S^{n-1}\to X$. If $h_1$ and $h_0$ are homotopic, then $Z_0$ and $Z_1$ are homotopically equivalent.

• You may either use the fact that $X$ is homotopic to $S^2 \vee S^1$ as you say, which is best proved pictorially (yes most would consider this a perfectly rigorous proof), or one may prove this by using the Seifert-van Kampen theorem using a covering of the space by a small neighbourhood around the contracted circle and its complement. – Dan Rust Jul 2 '18 at 12:02
• Here's one point where the groupoid-version of Van Kampen's theorem is very useful, because there's a very interesting contractible subset in this space – Max Jul 2 '18 at 14:55

Let $X$ be your space, and note that it can be realized as a quotient of $I^2$. In particular, the usual structure of the torus is a quotient map $q:I^2 \to I^2/\sim$ by taking $(x,0) \sim (x,1)$ and $(0,y) \sim (1,y)$.
There is another quotient now $p:I^2 \to I^2/\sim$ by taking $(x_1,0) \sim (x_2,0)$ and $(x_1,1) \sim (x_2,1)$. Your space is done by doing $q$ first and then $p$ (or technically the map induced by $p$.)
However, the situation is clearer if you do $p$ first and then $q$.
Pinching two sides of the square down to a point gives a structure of $2$ zero cells, and two $1$-cells joining them, and filling this with a $2$-cell. The induced bap by $q$ glues these two $1$ cells together giving a sphere, but we still have to identify the $0$-cells, so this is a sphere with two points identified.
There is an excellent illustration that a sphere with two points identified is homotopically just $S^2 \vee S^1$ here.


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Question

# Evaluate the following derivatives. \frac{d}{dx}\int_{x}^{5}\sin w^{6}dw

Derivatives
Evaluate the following derivatives.
$$\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\int_{{{x}}}^{{{5}}}}{{\sin{{w}}}^{{{6}}}{d}}{w}$$

2021-02-16
Step 1
We first switch the limits by changing the sign
$$\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\int_{{{x}}}^{{{5}}}}{{\sin{{w}}}^{{{6}}}{d}}{w}$$
$$\displaystyle={\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[-{\int_{{{5}}}^{{{x}}}}{{\sin{{w}}}^{{{6}}}{d}}{w}\right]}$$
$$\displaystyle=-{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{\int_{{{5}}}^{{{x}}}}{{\sin{{w}}}^{{{6}}}{d}}{w}\right]}$$
Step 2
Then we use the second fundamental theorem of calculus to find the derivative
$$\displaystyle=-{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{\int_{{{5}}}^{{{x}}}}{{\sin{{w}}}^{{{6}}}{d}}{w}\right]}$$
$$\displaystyle=-{\sin{{\left({x}^{{{6}}}\right)}}}$$
Answer: $$\displaystyle-{\sin{{\left({x}^{{{6}}}\right)}}}$$


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# Splitting fields of representations of generalized symmetric groups, 4

First a technical definition.

Let $A=C_\ell^n$. Let $\eta_k(z)=z^k$, for $z\in C_\ell$ and $1\leq k\leq \ell-1$. For $\eta\in C_\ell^*$, let $\mu\otimes \eta =(\mu_1\eta,\mu_2\eta,...,\mu_n\eta)$ where $\mu=(\mu_1,\mu_2,...,\mu_n)$. This defines an action of $C_\ell^*$ on $A^*$ and hence on the set of equivalence classes of $G$, $G^*$. We call two representations $\theta_{\mu,\rho}$, $\theta_{\mu',\rho'}$ $C_\ell^*$-equivalent, and write

$\theta_{\mu,\rho}\sim_\ell \theta_{\mu',\rho'},$

if $\rho=\rho'$ and $\mu'=\mu\otimes \eta$ for some $\eta\in C_\ell^*$. Similarly, we call two characters $\mu$, $\mu'$ of $C_\ell^n$ $C_\ell^*$-equivalent, and write

$\mu\sim_\ell \mu',$

if $\mu'=\mu\otimes \eta$ for some $\eta\in C_\ell^*$.

For example, Let $\ell =9$, $n=3$ and $\mu=(\eta_2,\eta_5,\eta_8)$. Then $\mu\sim \mu\otimes\eta_3$.

Let $\theta_{\mu,\rho}$ be as in the previous post. Note that

$\theta_{\mu\otimes \eta,\rho} = \theta_{\mu,\rho}\otimes \eta ,$

for $\eta\in C_\ell^*$. Therefore, the matrix representations of two $C_\ell^*$-equivalent representations differ only
by a character.

Let $G=C_\ell^n\, >\!\!\lhd \, S_n$.

The results in the above section tells us how to construct all the irreducible representations of $G$. We must

1. write down all the characters (i.e., 1-dimensional representations) of $A=C_\ell^n$,
2. describe the action of $S_n$ on $A^*$,
3. for each $\mu\in [A^*]$, compute the stabilizer $(S_n)_{\mu}$,
4. describe all irreducible representations of each $(S_n)_{\mu}$,
5. write down the formula for the character of $\theta_{\mu,\rho}$.

Write $\mu\in [A^*]$ as $\mu=(\mu_1,...,\mu_n)$, where each component is a character of the cyclic group $C_\ell$, $\mu_j\in C_\ell^*$. Let $\mu'_1,...,\mu'_r$ denote all the distinct characters which occur in $\mu$, so

$\{\mu'_1,...,\mu'_r\}=\{ \mu_1,...,\mu_n\}.$

Let $n_1$ denote the number of $\mu'_1$‘s in $\mu$, $n_2$ denote the number of $\mu'_2$‘s in $\mu$, …, $n_r$ denote the number of $\mu'_r$‘s in $\mu$. Then $n=n_1+...+n_r$. Call this the partition associated to $\mu$.

If two characters $\mu=(\mu_1,...,\mu_n)$, $\mu'=(\mu'_1,...,\mu'_n)$ belong to the same class in $[(C_\ell^n)^*]$, under the $S_n$-equivalence relation, then their associated partitions are equal.

The Frobenius formula for the character of an induced representation gives the following character formula. Let $\chi$ denote the character of $\theta_{\mu,\rho}$. Then

$\chi(\vec{v},p)=\sum_{g\in S_n/(S_n)_\mu} \chi^o_\rho(gpg^{-1})\mu^g(\vec{v}),$

for all $\vec{v}\in C_\ell^n$ and $p\in S_n$. In particular, if $p=1$ then

$\chi(\vec{v},1)=({\rm dim}\ \rho)\sum_{g\in S_n/(S_n)_\mu} \mu^g(\vec{v}).$

# Splitting fields of representations of generalized symmetric groups, 3

The representations of a semi-direct product of a group $H$ by an abelian group $A$, written $G=A\, >\!\!\lhd \, H$ (so $A$ is normal in $G$) can be described explicitly in terms of the representations of $A$ and $H$. The purpose of this post is to explain how this is done.

Again, a good reference for all this is Serre’s well-known book, Linear representations of finite groups.

Let $f$ be a class function on $H$. Extend $f$ to $G$ trivially as follows:

$f^0(g)= \left\{ \begin{array}{cc} f(g),&g\in H,\\ 0, & g\notin H, \end{array} \right.$

for all $g\in G$. This is not a class function on $G$ in general. To remedy this, we “average over $G$” using conjugation: Define the function $f^G=Ind_H^G(f)$ induced by $f$ to be

$Ind_H^G(f)(g)={1\over |H|}\sum_{x\in G} f^0(x^{-1}gx)=\sum_{x\in G/H}f^0(x^{-1}gx).$

This is referred to as the Frobenius formula.

Since $A$ is normal in $G$, $G$ acts on the vector space of formal complex linear combinations of elements of $A^*$ (=the characters of $A$),

$V={\mathbb{C}}[A^*]=span\{\mu\ |\ \mu\in A^*\},$

by

$(g\mu)(a)=\mu(g^{-1}ag),\ \ \ \ \forall g\in G,\ a\in A,\ \mu\in A^*.$

We may restrict this action to $H$, giving us a homomorphism $\phi^*:H\rightarrow S_{A^*}$, where $S_{A^*}$ denotes the symmetric group of all permutations of the set $A^*$. This restricted action is an equivalence relation on $A^*$ which we refer to below as the $H$-equivalence relation}. Let $[A^*]$ denote the set of equivalence classes of this equivalence relation. If $\mu,\mu'$ belong to the same equivalence class then we write

$\mu'\sim \mu$

(or $\mu'\sim_H\mu$ if there is any possible ambiguity). When there is no harm, we identify each element of $[A^*]$ with a character of $A$.

Suppose that $H$ acts on $A$ by means of the automorphism given by a homomorphism $\phi:H\rightarrow S_{A}$, where $S_{A}$ denotes the symmetric group of all permutations of the set $A$. In this case, two characters $\tau,\tau'\in A^*$ are equivalent if there is an element $h\in H$ such that, for all $a\in A$, we have $\tau'(a)=\tau(\phi(h)(a))$.

For each $\mu\in [A^*]$, let

$H_{\mu}=\{h\in H\ |\ h\mu = \mu\}.$

This group is called the stabilizer of $\mu$ in $H$. Let

$G_{\mu}=A\, >\!\!\lhd \, H_{\mu},$

for each $\mu\in [A^*]$. There is a natural projection map

$p_{\mu}:G_{\mu}\rightarrow H_{\mu}$

given by $ah\longmapsto h$, i.e., by $p_\mu(ah)=a$.

Extend each character $\mu\in [A^*]$ from $H_{\mu}$ to $G_{\mu}$ trivially by defining

$\mu(ah)=\mu(a),$

for all $a\in A$ and $h\in H_{\mu}$. This defines a character $\mu\in G^*_{\mu}$. For each $\rho\in H_{\mu}^*$, say $\rho:H_{\mu} \rightarrow Aut(V)$, let $\tilde{\rho}\in G_{\mu}^*$ denote the representation of $G_{\mu}$ obtained by pulling back $\rho$ via the projection $p_\mu:G_{\mu}\rightarrow H_{\mu}$, i.e., define

$\tilde{\rho}=\rho\circ p_{\mu}.$

For each $\mu \in [A^*]$ and $\rho\in H_\mu^*$ as above, let

$\theta_{\mu,\rho}=Ind_{G_\mu}^G(\mu\cdot \tilde{\rho}).$

Finally, we can completely describe all the irreducible representations of $G=A\, >\!\!\lhd \, H$. (This is Proposition 25 in chapter 8 of Serre’s book.)

Theorem:

1. For each $\mu \in [A^*]$ and $\rho\in H_\mu^*$, as above, then $\theta_{\mu,\rho}$ is an irreducible representation of $G$.
2. Suppose $\mu_1,\mu_2 \in [A^*]$, $\rho_1\in H_{\mu_1}^*$, $\rho_2\in H_{\mu_2}^*$. If $\theta_{\mu_1,\rho_1}\cong \theta_{\mu_2,\rho_2}$ then $\mu_1\sim \mu_2$ and $\rho_1\cong \rho_2$.
3. If $\pi$ is an irreducible representation of $G$ then $\pi\cong \theta_{\mu,\rho}$, for some $\mu \in [A^*]$ and $\rho\in H_{\mu}^*$ as above.

In the next post, we will examine the special case $A=C_\ell^n$ and $H=S_n$.

# Splitting fields of representations of generalized symmetric groups, 2

In general, there are three types of (complex) representations of a finite group $G$. (A good reference for all this is Serre’s well-known book, Linear representations of finite groups.)

Let $\rho:H\rightarrow Aut(W)$ be an $n$-dimensional irreducible representation of a finite group $G$ on a complex vector space $W$. Let $\chi$ denote the character of $\rho$.

Exactly one of the following possibilities must hold:

• One of the values of the character $\chi$ is not real. Such representations will be called complex (or type 1 or unitary).
• All the values of $\chi$ are real and $\rho$ is realizable by a representation over a real vector space. Such representations will be called real (or type 2 or orthogonal).
• All the values of $\chi$ are real but $\rho$ is not realizable by a representation over a real vector space. Such representations will be called quaternionic (or type 3 or symplectic).

Proposition (Frobenius-Schur): Let $\rho:H\rightarrow Aut(W)$ be an irreducible representation of a finite group $G$ on a complex vector space $W$ with character $\chi$. Then

${1\over |G|} \sum_{g\in G}\chi(g^2)= \left\{ \begin{array}{cc} 0,&\rho\ {\rm complex},\\ 1,&\rho\ {\rm real},\\ -1,&\rho\ {\rm quaternionic}. \end{array} \right.$

This quantity is sometimes called the Frobenius-Schur indicator of $\rho$.

It can be shown that if $\rho$ $\rho'\cong \rho$ are equivalent representations then $\rho$ and $\rho'$ have the same type.

In the next post we will examine the types that the irreducible representations of semi-direct product $G=C_\ell^n\, >\!\!\lhd \, S_n$ fall into.

# Splitting fields of representations of generalized symmetric groups, 1

The main result of this series of blog posts (originally written 1998), which is expository, is to determine (in a sense made precise later) the splitting field of any irreducible character of a generalized symmetric group. This was basically solved by M. Benard in a 1976 J. of Algebra paper. We use one of his results to make the splitting field explicit.

Notation and definitions

Let $C_\ell$ denote the cyclic group of order $\ell\geq 1$, let $S_n$ denote the symmetric group of degree $n\geq 1$, and let $G$ denote the semi-direct product $G=C_\ell^n\, >\!\!\lhd \, S_n$. We think of this as the set of pairs $(v,p)$, with

• $v=(v_1,...,v_n)$, where each $v_i\in C_\ell=\{0,1,...,\ell-1\}$,
• $p\in S_n$,
• $S_n$ acts on $C_\ell^n$ by $p(v)=(v_{p(1)},v_{p(2)},...,v_{p(n)})$,
• multiplication given by $(v,p)*(v',p')=(v+p(v'),pp'),\ \ \ \ \ (v,p),(v',p')\in G.$

A group of the form $C_\ell^n\, >\!\!\lhd \, S_n$, also written $S_n \ {\rm wr}\ C_\ell$ (here wr denotes the wreath product), will be called a generalized symmetric group.

We may identify each element of $C_\ell^n\, >\!\!\lhd \, S_n$ with an $n\times n$ monomial matrix (a matrix with exactly one non-zero entry in each row and column) with entries in $C_\ell$.

Our main goal will be to investigate the following question: What is the smallest extension of ${\mathbb{Q}}$ required to realize (using matrices) a given irreducible character of a generalized symmetric group?


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# Are large halo orbits around L₁'s and L₂'s preferred over small orbits for reasons other than geometry?

There have been many examples of the placement of satellites in orbits around Lagrange points, most have been sun-earth and earth-moon $$L_1$$ and $$L_2$$ due to their proximity to earth. In each case both bodies and both points fall on a straight line, so to have the right combination of non-blocked line-of-sight to earth for communications, safe angular separation from the sun, and non-blocked line-of-sight to the sun for solar power, the size and shape orbits about the Lagrange points are carefully optimized.

Besides those reasons, are there orbital mechanical or other reasons why placement in orbit around L₁ and L₂ would be preferred over placement near the points themselves? Does it require significantly more delta-v to get there (shortening station keeping lifetime)? Would station keeping be more difficult or costly (delta-v) near the point than it would be in a larger orbit about the point (excluding communications issues)?

• I think the biggest issue would actually be the $\Delta v$ required to stop, which would mean a very massive fuel load, thus a very expensive and risky launch. Further, if you managed to keep the spacecraft "on" the Lagrange point, it would require a lot of fuel to maintain that location, thus more fuel mass and more money required... – honeste_vivere May 9 '16 at 14:43
• @honeste_vivere Can you give any quantitative backup to "very massive fuel load"? Active station keeping absolutely necessary no matter "on" or "near" or "in orbit around" these things. Actually the "on" concept doesn't really mean anything since these are not circular orbits and there are perturbations - CR3BP is an abstraction of reality, back when people used feathers for pens. (of course we still use it as a starting point, but usually we turn to the lumps of silicon for assistance pretty quickly) – uhoh May 9 '16 at 15:13
• Well, the Wind spacecraft orbits about the Earth-sun $L_{1}$ point and needs very small (e.g., few cm/s) $\Delta v$'s once every ~3-4 months. This results in the spacecraft's fuel consumption being extremely low and giving it ~50+ years of station-keeping-only fuel. My hand-wavy statement about the amount of fuel to maintain an exact $L_{j}$ position is multi-faciteted but does have some merit... – honeste_vivere May 9 '16 at 15:52
• Every station keeping (SK) burn has a finite amount of uncertainty due to the inherent tolerances of the thrusters. Thus, no matter how well you plan, you will be constantly performing SK maneuvers just trying to correct for previous "hot" or "cold" burns. The net error for elliptical or Lissajous orbits about $L_{j}$ is a much smaller fraction than an attempt to stay exactly at $L_{j}$... This is my "knee-jerk" thought, based upon all previous work with any "small amplitude oscillations about a mean"-like problems, but perhaps I am overly pessimistic. – honeste_vivere May 9 '16 at 15:56

There are several reasons why spacecraft are sent into pseudo-orbits (they aren't actually "orbits") about the unstable Lagrange points. The least important reason is that only one spacecraft can be at one of those Lagrange points. Wide pseudo-orbits allow multiple spacecraft to simultaneously operate in the vicinity of one of those Lagrange points.

More importantly, those Lagrange points are a bit of a fiction, in a spherical cow sense. The Lagrange points are stationary only in the context of one body in a circular orbit about another more massive body. Add in elliptical orbits and the points move. Add in perturbations from other bodies such as the Moon in the case of the Sun-Earth Lagrange points, the Sun in the case of the Earth-Moon Lagrange points, plus Jupiter, Venus, and the other planets, and the exact location of the Lagrange points becomes a bit fuzzy.

Finally, suppose some organization wanted to have a spacecraft be at exactly one of those fuzzy points. That's not possible, for a number of reasons. The spacecraft's inertial measurement unit cannot sense that it's at a Lagrange point. It would instead have to be calculated. This calculated position (the guidance state) will inherently be in error. To make matters worse, where the spacecraft thinks it is (the navigated state) will also inherently be in error. To make matters worse yet, the propulsion and control systems are also imperfect.

The best that could be done would be to keep the spacecraft within some control volume surrounding the point. The control algorithm would necessarily involve deadbands, regions where the thrusters don't fire, for a couple of reasons. One is the nature of the thrusters on those vehicles, which are either on or off (they are not throttleable). This suggests using what is called bang-bang control. More importantly, the cameras or other instrumentation typically can't be used during thruster firings. Bang-bang control is optimal from both a control theory and operational perspective.

Making the control volume too small will result in infeasibility. (It can't be done.) A small feasible control volume will result in a system that involves frequent thruster firings that consume large amounts of fuel and that make the science part of the mission worthless. A somewhat larger control volume is what's needed.

This however misses the goal of being close to the Lagrange point. This approach also doesn't take advantage of orbital mechanics. A spacecraft in a pseudo-orbit about one of those Lagrange points explicitly does take advantage of orbital mechanics. While these pseudo-orbits are not stable, the stationkeeping requirements for them are vastly reduced compared those for a comparably sized control volume approach.

• @uhoh -- First off, comments are not intended to be used for answers, all across the stackexchange network. The Snickers bars represent fuel (but I'll delete that part of the answer). The stationkeeping costs to stay at the Lagrange points are huge. It doesn't make sense. Robert Farquhar tried to use that argument to the designers of ISEE-3. They didn't understand orbital mechanics, so they didn't understand his arguments. The argument that did win was geometry. There's no way to receive data transmitted by a satellite exactly at the Earth-Sun L1 point. – David Hammen May 12 '16 at 0:11
• OK great. The question does stick to the idea of being "near" and not "at" but you are right it's a common misconception since most things out there are CR3BP simplifications. Is there a way I can see that riding a manifold into a small orbit near sun-earth $L_1$ would require a "huge" amount of propellants (delta-v) compared to riding a manifold into the current large orbit around sun-earth $L_1$? (thinking DSVOVR for example) I'm looking for reasons besides geometry, hopefully something quantitative. Thanks! – uhoh May 12 '16 at 5:09
• I'm going to edit the title of the question a little bit, to make sure that my question is about being "near to" vs "far from" the vicinity of the L point. – uhoh Jun 4 '16 at 14:41
• David, is there shadow in L2 point of sun-earth? How huge umbra, penumbra and antumbra are there? – osgx Jun 4 '16 at 18:07
• @osgx good point! Sun-Eath $L_1$ and $L_2$ are about 1.5 million km from earth, almost exactly 1% of distance from the earth to the sun. Since the earth is about 1% the diameter of the sun, that means there is plenty of penumbra, so power is seriously reduced for small orbits. But that's a line-of-site, or "geometry" issue, I'm asking about issues other than those. – uhoh Jun 5 '16 at 2:21

One thought is traffic collisions. Once you put the first satellite exactly at the L point in question no other satellites can occupy that position. And because that are gravitationally stable points shifting them out again at the end of their life time is relatively expensive. So to be good neighbors to the other missions in the neighborhood you will prefer to orbit the L point which increases the number of missions that can go on simultaneously.

• That's a good point! If two satellites are both near a Lagrange point, then it increases the risk for each satellite that it could get hit by the other, and may make careful station keeping more demanding on ground personnel. I did try to word my question carefully by always saying "near"" and not "*at", and especially not "exactly at". Both the $L_1$ and $L_2$ points, and orbits around those points are all exponentially unstable by the way. – uhoh May 9 '16 at 2:09
• Ahh yes, mistook which pair of Lagrange points you were talking about. – BSteinhurst May 9 '16 at 2:29

Additional useful property of large orbit is easier avoidance of eclipses:

Due to the size of the Lissajous orbit to be limited to less than 15◦ unfavourable conditions with respect to the moon can occur, where the moon is causing a partial eclipse for a prolonged period of time (up to 42 hours). While not problematic with respect to solar power generation it is undesirable with respect to the thermal balance ... this situation could not occur anymore due to changed insertion conditions, which resulted in more favourable conditions with respect to the trajectory of the moon due to larger out-of-plane amplitudes.

Earth Eclipse Avoidance: Any small amplitude Lissajous orbit about SEL2 will have an eclipse following the natural motion. By choosing appropriate initial conditions the time to an eclipse can usually be extended to 6.5 years for orbits with SSCE ∼ 15◦ , if appropriate initial phase angles are selected

JWST paper "LAUNCH WINDOW TRADE ANALYSIS FOR THE JAMES WEBB SPACE TELESCOPE" 2016 lists some constraints on Sun-Earth L2 orbit:

Table 1. JWST Launch Window Constraints.
Constraint. Value. Requirements/Constraint Driver(s)
LPO Box: Maximum RLP Y - 832,000 km - Communications
LPO Box: Maximum RLP Z - 520,000 km - Science (Stray light)
Maximum SCAT finite-burn duration - 11,425 sec - Propulsion
Minimum precision of SCAT finite-burn duration - 0.2 sec - Propulsion
MCC Available ΔV - 58.5 m/s - Mass & Propulsion
Mission Lifetime Goal - 10.5 years - Science
Lunar / Earth Eclipse - None allowed - Power and Thermal

The LPO restrictions define a box centered at the L2 point with three dimensions in the RLP X, Y and Z directions ... The Y and Z constraints are based on communications and stray light requirements. The stray light requirement protects the JWST instruments from solar, lunar and Earth stray light. As seen in Fig. 3, the LPO requirements can be fulfilled by a general LPO category, which includes tori, quasi-halo, and Lissajous.

• These are really interesting/helpful links, thank you! While the question asks about reasons other than geometry, it's good to have a reminder that geometry is quite important. I wonder if this is even more helpful to mention at this question as well, since it asks about solar power specifically? I see your comment there, but maybe posting more of this as an answer there would be good as well? – uhoh Jan 22 '18 at 5:03

This is a supplementary answer because it still involves geometry, even though it's really future geometry.

tl;dr: Large halo orbits are definitely preferred to small halo orbits because the small ones are not halo orbits! They will be criss-crossy Lissajous orbits and sooner or later end up along the axis between the two bodies (geometry reasons) even though they originally clear that area.

The article Chang'e-4 lunar far side satellite named 'magpie bridge' from folklore tale of lovers crossing the Milky Way about Queqiao, which will be (among other things) Chang'e-4's relay satellite, was found in this interesting answer. They both point out that Queqiao will be in a halo orbit around the Earth-Moon L2 point so that it has a line of sight to both the far side of the Moon as well as the Earth at all times.

This led me to re-review Robert W. Farquhar's hundred page tome The Utilization of Halo Orbits in Advanced Lunar Operations, NASA Tech. Note D-6365.

In section II.B.2.b, he points out:

For every value of $A_y$ > 32,871 km, there is a corresponding value of $A_z$ that will produce a nominal path where the fundamental periods of the y-axis and z-axis oscillations are equal. In this case, the nominal path as seen from the earth will never pass behind the moon. The exact relationship between $A_y$, and $A_z$, for this family of nominal paths is given in Figure 5.

To first order, the motion of an object near the L2 equilibrium point is periodic in the x, y plane of the Moon's orbit, and it is also periodic out of plane in the z direction. However, the xy period and the z period are not necessarily equal! The name for the pattern made by harmonic motion in two axes where the periods are not constrained to be identical is called a Lissajous pattern, and likewise these orbits are called Lissajous orbits. The problem is that unlike equal periods resulting in a "nice, halo orbit" these will sooner or later move so close to the Earth-Moon axis that communications with the Earth will be blocked.

This is a well known problem. For example, sooner or later DSCOVR will end up passing so close to the Earth's line of sight with the Sun that communications will be lost for weeks or longer.

Therefore, the reason that large halo orbits are definitely preferred to small halo orbits is that small halo orbits are not halo orbits! While they might be lumped into the "halo category" if they are not large enough they will not be able to have equal in-plane and out-of-plane periods, and really be just Lissajous orbits with unequal periods. While if they are large enough, they can be constructed as Lissajous orbits with equal periods and therefore be proper halo orbits. Though this is not automatic careful adjusting of the ratio of in-plane to out-of-plane amplitudes (the "For every value of $A_y$ > 32,871 km, there is a corresponding value of $A_z$..." block quote from Farquhar.


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# Presidential Election Predictions 2016 (an ASA competition)

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Guest post by Jo professor of mathematics, Pomona College.

ASA’s Prediction Competition

In this election year, the American Statistical Association (ASA) has put together a competition for students to predict the exact percentages for the winner of the 2016 presidential election. They are offering cash prizes for the entry that gets closest to the national vote percentage and that best predicts the winners for each state and the District of Columbia. For more details see:

http://thisisstatistics.org/electionprediction2016/

To get you started, I’ve written an analysis of data scraped from fivethirtyeight.com. The analysis uses weighted means and a formula for the standard error (SE) of a weighted mean. For your analysis, you might consider a similar analysis on the state data (what assumptions would you make for a new weight function?). Or you might try some kind of model – either a generalized linear model or a Bayesian analysis with an informed prior. The world is your oyster!

Getting the Data

Thanks to the Internet, there is a lot of polling data which is publicly accessible. For the competition, you are welcome to get your data from anywhere. However, I’m going to take mine from 538. http://projects.fivethirtyeight.com/2016-election-forecast/national-polls/ (Other good sources of data are http://www.realclearpolitics.com/epolls/latest_polls/ and http://elections.huffingtonpost.com/pollster/2016-general-election-trump-vs-clinton and http://www.gallup.com/products/170987/gallup-analytics.aspx)

Note the date indicated above as to when this R Markdown file was written. That’s the day the data were scraped from 538. If you run the Markdown file on a different day, you are likely to get different results as the polls are constantly being updated.

Because the original data were scraped as a JSON file, it gets pulled into R as a list of lists. The data wrangling used to convert it into a tidy format is available from the source code at https://github.com/hardin47/prediction2016/blob/master/predblog.Rmd.

url = “http://projects.fivethirtyeight.com/2016-election-forecast/national-polls/” doc <- htmlParse(url, useInternalNodes = TRUE)

sc = xpathSApply(doc, “//script[contains(., ‘race.model’)]”,                 function(x) c(xmlValue(x), xmlAttrs(x)[[“href”]])) jsobj = gsub(“.*race.stateData = (.*);race.pathPrefix.*”, \\1″, sc) data = fromJSON(jsobj) allpolls <- data\$polls #unlisting the whole thing indx <- sapply(allpolls, length) pollsdf <- as.data.frame(do.call(rbind, lapply(allpolls, length<-, max(indx))))

A Quick Visualization

Before coming up with a prediction for the vote percentages for the 2016 US Presidential Race, it is worth trying to look at the data. The data are in a tidy form, so ggplot2 will be the right tool for visualizing the data.

ggplot(subset(allpolldata, ((polltypeA == “now”) & (endDate > ymd(“2016-08-01”)))),                         aes(y=adj_pct, x=endDate, color=choice)) +   geom_line() + geom_point(aes(size=wtnow)) +   labs(title = “Vote percentage by date and poll weight\n,     y = “Percent Vote if Election Today”, x = “Poll Date”,     color = “Candidate”, size=“538 Poll\nWeight”)

A Quick Analysis

Let’s try to think about the percentage of votes that each candidate will get based on the now cast polling percentages. We’d like to weight the votes based on what 538 thinks (hey, they’ve been doing this longer than I have!), the sample size, and the number of days since the poll closed.

Using my weight, I’ll calculate a weighted average and a weighted SE for the predicted percent of votes. (The SE of the weighted variance is taken from Cochran (1977) and cited in Gatz and Smith (1995).) The weights can be used to calculate the average or the running average for the now cast polling percentages.

allpolldata2 <- allpolldata %>%   filter(wtnow > 0) %>%   filter(polltypeA == “now”) %>%   mutate(dayssince = as.numeric(today() – endDate)) %>%   mutate(wt = wtnow * sqrt(sampleSize) / dayssince) %>%   mutate(votewt = wt*pct) %>%   group_by(choice) %>%   arrange(choice, -dayssince) %>%   mutate(cum.mean.wt = cumsum(votewt) / cumsum(wt)) %>%   mutate(cum.mean = cummean(pct))

Plotting the Cumulative Mean / Weighted Mean

In tidy format, the data are ready to plot. Note that the cumulative mean is much smoother than the cumulative weighted mean because the weights are much heavier toward the later polls.

ggplot(subset(allpolldata2, ( endDate > ymd(“2016-01-01”))),                         aes(y=cum.mean, x=endDate, color=choice)) +   geom_line() + geom_point(aes(size=wt)) +     labs(title = “Cumulative Mean Vote Percentage\n,     y = “Cumulative Percent Vote if Election Today”, x = “Poll Date”,     color = “Candidate”, size=“Calculated Weight”)

ggplot(subset(allpolldata2, (endDate > ymd(“2016-01-01”))),                         aes(y=cum.mean.wt, x=endDate, color=choice)) +   geom_line() + geom_point(aes(size=wt)) +   labs(title = “Cumulative Weighted Mean Vote Percentage\n,     y = “Cumulative Weighted Percent Vote if Election Today”, x = “Poll Date”,     color = “Candidate”, size=“Calculated Weight”)

Additionally, the weighted average and the SE of the average (given by Cochran (1977)) can be computed for each candidate. Using the formula, we have our prediction of the final percentage of the popular vote for each major candidate!

pollsummary <- allpolldata2 %>%   select(choice, pct, wt, votewt, sampleSize, dayssince) %>%   group_by(choice) %>%   summarise(mean.vote = weighted.mean(pct, wt, na.rm=TRUE),            std.vote = sqrt(weighted.var.se(pct, wt, na.rm=TRUE)))

pollsummary

## # A tibble: 2 x 3 ##    choice mean.vote  std.vote ##               ## 1 Clinton  43.64687 0.5073492 ## 2   Trump  39.32071 1.1792667


text:
# LocalCoordinatesSite¶

class OpenMM::LocalCoordinatesSite

This is a VirtualSite that uses the locations of three other particles to compute a local coordinate system, then places the virtual site at a fixed location in that coordinate system. The origin of the coordinate system and the directions of its x and y axes are each specified as a weighted sum of the locations of the three particles:

origin = w1 r1 + w2 r2 + w3 r3 None

xdir = w1 r1 + w2 r2 + w3 r3 None

ydir = w1 r1 + w2 r2 + w3 r3 None

For the origin, the three weights must add to one. For example if (w1 , w2 , w3 ) = (1.0, 0.0, 0.0), the origin of the local coordinate system is at the location of particle 1. For xdir and ydir, the weights must add to zero. For excample, if (w1 , w2 , w3 ) = (-1.0, 0.5, 0.5), the x axis points from particle 1 toward the midpoint between particles 2 and 3.

The z direction is computed as zdir = xdir x ydir. To ensure the axes are all orthogonal, ydir is then recomputed as ydir = zdir x xdir. All three axis vectors are then normalized, and the virtual site location is set to

origin + x*xdir + y*ydir + z*zdir

Methods

 LocalCoordinatesSite() Create a new LocalCoordinatesSite() virtual site. getOriginWeights() Get the weight factors for the three particles when computing the origin location. getXWeights() Get the weight factors for the three particles when computing xdir. getYWeights() Get the weight factors for the three particles when computing ydir. getLocalPosition() Get the position of the virtual site in the local coordinate system.
LocalCoordinatesSite(int particle1, int particle2, int particle3, const Vec3 &originWeights, const Vec3 &xWeights, const Vec3 &yWeights, const Vec3 &localPosition)

Create a new LocalCoordinatesSite() virtual site.

• particle1 – the index of the first particle
• particle2 – the index of the second particle
• particle3 – the index of the third particle
• originWeights – the weight factors for the three particles when computing the origin location
• xWeights – the weight factors for the three particles when computing xdir
• yWeights – the weight factors for the three particles when computing ydir
• localPosition – the position of the virtual site in the local coordinate system
const Vec3 &getOriginWeights() const

Get the weight factors for the three particles when computing the origin location.

const Vec3 &getXWeights() const

Get the weight factors for the three particles when computing xdir.

const Vec3 &getYWeights() const

Get the weight factors for the three particles when computing ydir.

const Vec3 &getLocalPosition() const

Get the position of the virtual site in the local coordinate system.