Why do mathematicians care about Moment maps? I get why some physics people might, due to its use in a very neat restatement of Noether's theorem and the general idea of momenta in various physics theories, but why is it of interest to a mathematician?

@SayanChattopadhyay Quotienting symplectic manifolds isn't interesting to you?

I'd argue that's exactly why it's relevant in physics as well. Quotienting physical theories are hard; eg this is what gauge fixing is about.

@BalarkaSen I haven't read this quotienting stuff yet, but is the idea that if $\mu$ is regular then $\mu^{-1}(0)$ is a coisotropic submanifold?

And then if $G$ acts freely then you get the orbit space?

@Sayan Why is $\mu^{-1}(0) \subset M$ being coisotropic relevant? I really don't know any symplectic geometry.

Oh no, I see what you are saying. $M/G$ does not literally make sense. It might be odd-dimensional. It will almost never be naturally a symplectic manifold.

So you have to be clever about how you want to define a $"M/G"$

The moment map $\mu : M \to \mathscr{g}^*$ really is defined as follows: the action of $G$ on $M$ gives rise to a map $G \to \text{Diffeo}(M)$, taking the derivative at identity gives $\mathscr{g} \to \text{Vect}(M)$. If the action is Hamiltonian, the image goes inside $\text{Ham}(M) \subset \text{Vect}(M)$, in which case you can come back to $C^\infty(M)$ by inverse of $C^\infty(M) \to \text{Ham}(M)$, the Hamiltonian.

So you get a map $\mathfrak{g} \to C^\infty(M)$, dualizing which gives $M \to \mathfrak{g}^*$

The former is easier to interpret; it means every infinitesimal group element gives rise to a flow on $M$ which has a symplectic potential -- take that potential. That's $\mathfrak{g} \to C^\infty(M)$.

Uncurry: $\mathfrak{g} \times M \to \Bbb R$ takes an infinitisimal group element, and evaluates the symplectic potential at a point $p \in M$.

So $\mu^{-1}(0)$ is all the points on $M$ along which the symplectic potential of all infinitisimal group elements is $0$. They are the "ground energy state" points or whatever you want to call em

Then as you say $\mu^{-1}(0)$ is coisotropic etc if $\mu$ is regular, so $\mu^{-1}(0)/G$ will be a symplectic stratified space.

I cannot tell you the precise reason why this is a smart way to define $"M/G"$ but $\mu$ is sort of like an energy functional or Lagrangian or something (it is literally potential after all) and you're picking out points from the orbit of $G$ on which $\mu = 0$. This is a natural idea in gauge-fixing.

In electromagnetism eg you see a convention for gauge-fixing $B = \text{curl} A$ by saying $\text{grad} A = 0$

I think they call this Coulomb fixing

You should look into Kempf-Ness theorem, which says that if $M$ is a smooth projective variety and $G$ is a compact Lie group acting on $M$ algebraically then the symplectic reduction is the same as the GIT quotient $M//G^{\Bbb C}$, where $G^{\Bbb C}$ is the complexification. This explains why the symplectic reduction has dimension $\dim M - 2\dim G$

I think approximately this must be because if $G$ is equipped with an invariant Riemannian metric, $|\mu|^2 : M \to [0, \infty)$ is a $G$-equivariant Morse function with a critical level set only at $0$. So you get a map $M \to \mu^{-1}(0)/G$ by flowing along the gradient of this guy. And so maybe you can pass to $M//G^{\Bbb C} \to \mu^{-1}(0)/G$

is there an easy way to show that if$ a_0,a_1,a_2$ are integers, and$ c$ is the real cube root of $2$, that $a_0 + a_1c + a_2c^2 = 0$ implies $a_0=a_1=a_2 = 0$?

Yeah $M/G$ might not be necessarily symplectic. Thanks a ton man, you gave me a lot of stuff to look for @Balarka

Show that $X^3-2$ is irreducible in $\Bbb Z[X]$

@SayanChattopadhyay A good person to ask would be @ACuriousMind

Which is the same as proving $2^{1/3}$ is not an integer

Yeah, I have seen him hang around here, I will ask him if I more such questions.

@Astyx sorry im pretty rusty with algebra, so after showing that $X^3 - 2$ is irreducible how do we get to the conclusion?

One should try to understand the reduction in the "abelian gauge" context. Namely, the map $\mu : M \to \mathfrak{g}^*$ writes down the vector of symplectic potentials at every point (fix a basis for the Lie algebra $\mathfrak{g}$, whatever). If $G = S^1$, then the Lie algebra is one-dimensional; so $\mu$ is an honest potential function.

That means that $\Bbb Z[X]/(X^3-2)$ is a 3-dimensional $\Bbb Z$-module

hm, the thing is this was on the first page of the notes for my algebra course, and modules are covered at the very end

Generated by $1, X, X^2$

Try $S^1$ acting on $\Bbb{CP}^2$ by $\theta \cdot [z_1 : z_2 : z_3] = [\theta z_1 : \theta z_2 : \theta^{-1} z_3]$. This is coming from the algebraic action of $(S^1)^{\Bbb C} = \Bbb C^*$ so it is symplectic wrt the Kahler 2-form. Then understand the reduction (which we know is going to be $\Bbb{CP}^1$) extremely explicitly.

more elementary:

The minimal polynomial of $c$ is of order $3$ because it divides $X^3-2$ which is irreducible, so it is $X^3-2$

That means that the familly $1, c, c^2$ is free in $\Bbb R$ considered as a $\Bbb Z$-module

oh of course!

i get the minimal poly argument

thank you

You're welcome

@Astyx and I suppose in general, if $m \in \mathbb{Z}[X]$ is the minimal polynomial of some $\alpha \in \mathbb{C}$, then the same approach should show that $\mathbb{Z}[\alpha]$'s elements are of the form $k(\alpha)$ where $\deg(k) < deg(m)$, and that $k'(\alpha) = k(\alpha)$ iff $k = k'$ as polys in $\mathbb{Z}[X]$?

oh whoops, if $m$ is irreducible and the minimal polynomial

The minimal polynomial is irreducible in an integral domain

But yeah

i see

(because the minimal polynomial $P$ is such that $P(\alpha)=0$, so if it is not irreducible $P=QR$, then $Q(\alpha)=0$ or $R(\alpha)=0$)

ah yes, that makes sense

@BalarkaSen I can certainly explain why the symplectic reduction is interesting to physicists, but I'm afraid I don't know either why mathematicians care about it ;)

Yeah this is part of one of my HW problems. I have to completely understand the $U(n)$ action on $C^{n^2}$.

There is this example in McDuff where she defines a symplectic structure on $\operatorname{Imm}(L,M)$ where $L$ is a compact riemannian manifold of dimension $n$ and $M$ is an exact symplectic manifold. Then she defines a very interesting moment map here. I really want to read that example sometime

Yes, there are lots of natural infinite dimensional symplectic manifolds.

@ACuriousMind Mathematicians kinda suck anyway

Hello, I'm thinking of Riesz representation theorem for C([0,1]) for the following steps, but I do not quite know how to proceed by using this method. Here is a usual approach for proving the same statement:math.stackexchange.com/q/…. I'll post on the community if needed. Thank you so much!

how would one work out for which pairs $a_0,a_1 \in \mathbb{Z}[\frac{1}{2}]$ does $a_0 + a_1\alpha \in \mathbb{Z}[\alpha]$ where $\alpha$ is a root of $x^2 - \frac{1}{2}x + 1$?

i guess the answer would come from figuring out the form of all remainders of polys with integer coefficients , dividing by $x^2 - \frac{1}{2}x + 1$ but this seems tedious

@porridgemathematics might you be studying in IC by any chance?

@LeakyNun yes :)

and might I know you by any chance?

oh probably not

did you go to the first Q&A session?

for algebra 3?

yeah

i did yeah, but i joined halfway through

well I was discussing this with the professor at the end

oh i see - did you metnion it in chat?

what did he say?

well I noticed that $\beta := 2 \alpha$ is an algebraic integer with minimal polynomial $x^2 - x + 4$

oh no, this case is more troublesome

isn't the min poly $x^2 - x + 2$?

oh nvm

so the thing is that $\Bbb Z[\beta]$ is not a UFD

anyway the case for $x^2 - \frac12 x - 1$ is easier so if you don't mind I'll just tell you what I calculated for that one

sure

Z[α] = Ζ[β][1/(3-β)]

that's the min poly of $\alpha$ as a poly in $\mathbb{Q}[X]$ right

so elements of $\Bbb Z[\alpha]$ look like $\frac{e + f \beta}{(3-\beta)^n}$ where $e, f \in \Bbb Z$ and $n \in \Bbb Z_{\ge 0}$

where $\alpha$ is a root of $x^2 - \frac12 - 1 = 0$ and $\beta = 2\alpha$

@porridgemathematics yeah

btw $\frac1{3-\beta} = 1 + \alpha$

oh so they look like $(e+2f\alpha)(1+\alpha)^n$ for $e, f \in \Bbb Z$ and $n \in \Bbb Z_{\ge 0}$

hmm, but is that easily simplifiable to $a_0 + a_1\alpha$?

also how did you compute $Z[α] = Ζ[β][1/(3-β)]$?

@porridgemathematics yeah but not every element of the form $a_0 + a_1 \alpha$ is in $\Bbb Z[\alpha]$

I guess my question is why is that form more desirable, since it still involves a bunch of powers of $\alpha$ that could be arbitrarily large

sure, but we want to figure out which pairs $(a_0,a_1)$ are s.t. thats in $\mathbb{Z}[\alpha]$ right

is it sufficient to say, those that are equal to $(e+2f\alpha)(1+\alpha)^n$ for some $e,f,n$?

I guess you're asking for an algorithm of some sorts

here's what I think you can do

given $\frac{e+f\alpha}{2^n}$

working in $\Bbb Z[\alpha]$ is not very desirable

let's work in $\Bbb Z[\beta]$ instead

ok sure

so you're given $\dfrac{e+f\beta}{2^n} = \dfrac{e+f\beta}{(2+\beta)^n (3-\beta)^n}$

it's in $\Bbb Z[\alpha]$ iff $(2+\beta)^n \mid (e+f\beta)$

note that $2+\beta$ is a prime, so you can just factorize $e+f\beta$ and count the multiplicity of $2+\beta$

$N(2+\beta) = (2+\beta)(3-\beta) = 2$, so $N(e+f\beta)$ must firstly be divisible by $2^n$ (necessary not sufficient)

isn't this a bit above the first few pages of the notes?

yes, that's why the professor refrained from talking about it

i just mean in terms of techniques were using, im guessing even the minimal poly is introduced later in the course

so i wonder why this is asked so early

and just said that it doesn't have a nice description

I think his answer was just that $\Bbb Z[\alpha] = \{ a_0 + a_1 \alpha \mid a_0, a_1 \in \Bbb Z[1/2], a_0 + a_1 \alpha \in \Bbb Z[\alpha] \}$

essentially this is a tautology

I think he said something like it's a subset of $\Bbb Z[1/2] \times \Bbb Z[1/2]$ but we don't know which subset

oh lol

that does look like a tautology

you're welcome to ask him in the next Q&A session :P

the only thing being novel is that $a_0,a_1$ are both in $\mathbb{Z}[\frac{1}{2}]$ there which would technically need to be proved

right

but that isn't too troublesome

yeah true i'll do so

what other modules are you doing this year btw?

number theory (same professor), measure and integration, commutative algebra, group theory

oh i see, were both doing measure and integration then

(and algebra 3)

anyway, thanks for your help!

$\dfrac{e+f\beta}{2^n} = \dfrac{e+f\beta}{(2+\beta)^n (3-\beta)^n} \in \Bbb Z[\alpha] \iff \dfrac{e+f\beta}{(2+\beta)^n} = \dfrac{(e+f\beta)(3-\beta)^n}{2^n} \in \Bbb Z[\beta]$

@porridgemathematics so all you have to do is compute $(e+f\beta)(3-\beta)^n$ and see if $2^n$ divides the coefficients

now that's an algorithm

Determine where the function $f(x,y) = x^2y^2 + x^3 - 3x^2y + i(3x^2y) is complex differentiable. Where is the function $f$ analytic?

when I did the cauchy-reimann equations it fails

so its not analytic?

Hello, I have posted a question in the community about an alternative method for proving Riesz theorem for C([0,1]). Details are included in the post. Can anyone take a look and possibly help? Thank you. math.stackexchange.com/q/3869735/792125

@Mike Are you sure it's not a special case?

what are the places where convex hull is preferred over concave hull and vice-versa . Like some real life examples

@Azmuth What do you mean? Can you be more specific?

@Mike No, In general, I'm saying that 50-60% of the all the alternative proof check in the Math SE are actually don't by taking special case, so, in case you can check if your work relies on such special case (sometimes that happens by mistake), that'd be helpful for those who are willing to help you in the site :-)

How many Mikes are here?

No, I'm just asking the details for the proof. I did not manage to complete this by myself, so I post to the community. Can you go and post an answer, if you know how to complete it and answer my questions? Thanks. @Azmuth

@Mike I checked the proof, I'm afraid, it's not my field. I guess there are more abled and advanced mathematicians here @TedShifrin They may help you better.

That's alright, thank you for the help and recommendation!

@Azmuth

@Mike great :-)

are there any two uncountable sets such that there is no bijection between them?

pow(N) and pow(pow(N)

Ted, where does the theory of formal power series lie? It's algebra, but smells like analysis, too

@JoeShmo: Why does it smell like analysis? There's absolutely no notion of convergence. Formal makes it algebra :P

@TedShifrin what would be a good piece of advice that you would give to your 20s mathematical self?

the advice I would give myself in my 20s....don't be afraid to ask questions, don't memorize, don't be ashamed to be wrong.

@user1993 "don't memorize" can u expand a bit on that? obviously some memorizing is good, maybe "focus less on memorization" might get to the point

like don't memorize everything to get the answer.

like get to know definitions and concepts.

@StupidQuestionsInc ideally you want to get to the point where you can craft a definition based on what it is needed for

haha oh I see on wikipedia.. they're not dealing with issues of convergence. I still don't quite see the sense, though

This first time i took graph theory it was a graduate class that let undergrads in but i forgot what day school started so i missed the first 10 days of school, i downloaded the assignment and did it but i had to make assumptions as to what the definitions actually meant as i didnt have the textbook. i did it that day and took the assignment to the prof to discuss with him. he was pretty mad i hadn't showed up for the first 10 days.

However he read the assignment and said it was good but i made an arbitrary choice in on of my definitions and it was actually the other.

@StupidQuestionsInc like, DON'T memorize ANYTHING

read it 20 times if you have to, but don't do it senselessly.

@StupidQuestionsInc don't be like the young me.

which I memorize and is biting me in the ass.

if your just starting out memorizing is ok but its better to understand why the defintion is the way it is.

same with theorems

The ring of germs of holomorphic functions injects into the formal power series ring, the ring of germs of smooth functions surjects onto the formal power series ring

its better to learn the key steps to prove them than it is to remember them

@JoeShmo

germs are especially dangerous these days, Balarka, wash your hands.

lol

I agree with @Faust

@user1993 do you agree that at the other extreme one shouldn't also "stop revising for too long" (for instance it's easy to forget the key steps in a proof, or the main ideas connection some theorems if one leaves them for too long)

@BalarkaSen, I don't know what germs are, but OK.. are formal power series used for anything more basic, or theyre just an algebraic construction that's waiting for a topological interpretation..?

@StupidQuestionsInc I agree with it but don't turn it into habit.

when it becomes a habit, is difficult to stop it.

@user1993 i don't see how having some revision as a habit is dangerous

when you memorize everyrhing

everything*

are you learning?

@user1993 revising doesn't necessarily mean memorizing

it just means refreshing one's memory to something that you have already learned

you can revise and forget details, that's fine. but make sure you understand.

it's ultimately up to you. if youre doing it senselessly, youre cheating yourself

@JoeShmo yeah i agree 100%

TIL the age difference between the oldest and youngest Nobel prize laureates is 97 - 17 = 80 years!

Hello all! I have a very simple problem that is struggling me with the answer. It's about the area of an annulus

The arm of a windshield wiper is 50 cm; and the brush, 38 cm. If it travels through an angle of 120º, what surface of the windshield does it clean?

hi all, has anyone here worked with hidden markov models?

I proposed the formula $\pi(R^2-r^2)\alpha/360^\circ$, and replacing, $\text{Area}=\pi(50^2-38^2)120^\circ/360^\circ=352\pi\approx1105.84$

I had a tiny conceptual question with regards to marginal probabilities for hmms

But the answer says $2465.95$. What did I do wrong?

depends on where the brush is relative to the wiper.

and where the wiper is attached with respect to the wind shield.

@Jonathan its been awhile whats the question?

@Faust For some particular hidden state of an hmm, to compute the marginal probability,

I want to use the message passing algorithm.

@Jonathan thanks for the fast comment! This is a problem for ~12 years old children, so I don't think they put it hard

So for all the hidden states prior, i compute the probability that the state at time t is the probability of it being that state given the combination of all the states prior.

but this seems like quite a long stirng of joint probabilities to compute. is there a proper mathematical formula that shortens it?

@Jonathan isn't there a whole bunch of message passing algorithms?

Yes, but I'm only concerned about the message passing algorithm that computes all the different paths.

the base algorithm.

ah ok

@manooooh then you simply do the (area of the outer circle - area of inner circle)/3.

@manooooh (50^2pi - 12^2pi)/3 = 2467...

You're doing /3 because 120/360.

@Jonathan i will have a look around for my old textbook but i don't know off the top of my head.

@Faust sounds good. I asked a question on it but no responses. So feel free to add an answer on that post

and i can accept

@Jonathan why did you use $12$ instead of $38$?

Well the outer circle has radius 50. the inner circle has radius 12, because the brush is typically appended at the end of the wiper.

so 50-38.

so the inner cirlce is the portion that is not being wiped cause that's where there is no brush. so you subract that portion.

Oh I drawn this:

@Jonathan I don't understand your point, could you draw please?

The 38 is for the shaded region.

That's why i initially asked where is the brush.

On a wiper, the brush is appended at the end, not the focal point.

@Jonathan that's true

and brush length is 38.

so the radius of the inner sircle is 12, not 38 as you've written.

How would be the graph? I mean, where do we see the $50-38$?

We are dealing with these figures (don't care about Spanish):

@Jonathan Ohhh I understand you, you mean:

Is it correct @Jonathan?

The 38 and 12 should be flipped.

That makes sense, thanks again!!

Can anyone help solve this? $ \int_0^1 g(x)f(x)~dx =\int_0^1f^2(x)~dx$

Solve?

I mean find all the functions $f$ and $g$ that make it true

I found a trivial space of solutions $f=g$

you're asking for functions $f,g$ such that $f\perp g-f$ wrt the standard inner product and the space is infinite-dimensional, so there's not gonna be a nice description

@Thorgott what is that upside down T supposed to mean?

orthogonal

okay

can you suggest a way to make the question more focused?

should I pick a specific function for $f(x)?$

Suppose you had a closed binary operation $a\cdot b$ on a set $A$. Can you always define a function $f:A\to S_A$ (the set of permutations of $A$) such that the binary operation $f(a)(a\cdot b)$ is associative?

Or rather: If you wanted to find the answer to this, would you sooner try to find a counterexample, where it can't be done, or try to show that it can always be done?