Mathematics - 2025-02-25

Thorgott

00:16

it also exists if $G$ is finitely generated

Ben Steffan

I'm happy to assume that $X$ and $G$ are as ugly as possible

(you may also replace $H\mathbb{Z}$ with any other spectrum of your choice)

Thorgott

sure, if you want a counter-example, take $X$ to be an infinite discrete set and $G=\mathbb{Q}$

Ben Steffan

I don't think that's a counterexample(?)

$\mathbb{Q} \otimes \prod_{i = 1}^\infty \mathbb{Z}$ and $\prod_{i = 1}^\infty \mathbb{Q}$ are isomorphic, even if the canonical map is not an isomorphism

if math.stackexchange.com/a/860273/681666 is to be believed

Thorgott

the map in your SES is the canonical one, though?

Ben Steffan

...right, ok

thanks

Derso

02:12

Hi, guys! Hope you all doing well. The curve $(\cos t, \sin t, t)$ is called a "helix". Does the "hyperbolic version" $(\cosh t, \sinh t, t)$ also have a nice name? If not, what name would you guys suggest?

Xander Henderson

02:38

@Derso Helices have a name because they come up a lot, especially outside of mathematics. The things you're talking about don't. So why does it deserve a name?

one potato two potato

03:06

@BenSteffan why is that?

oh it says that the universal line bundle is a tautological line bundle over $CP^\infty$

@Thorgott I was wondering about the correspondence between a single map and a single line bundle

Ben Steffan

@onepotatotwopotato For instance its transition functions wrt. the standard coordinate charts are rational

if you fancy AG, it's $\mathcal{O}(-1)$ (or $\mathcal{O}(1)$, depending on convention)

one potato two potato

Uh, I don't know anything about AG. Is there any written literature on that?

Ben Steffan

on AG? I'm sure at least 500k pages :)

this will be in any standard text

you can disregard it here, but you should be aware that "holomorphic" means "algebro-geometric" in a sense one can make relatively precise

complex geometry is close to AG

@onepotatotwopotato this doesn't really make sense. If you have a holomorphic map, you can pull back the universal bundle along it to get a holomorphic line bundle. If you have two holomorphic maps and you pull the universal bundle along each, when do you get the same bundle? ...who knows!

one potato two potato

03:30

If the pullback of the universal bundle along a certain smooth map turns out to be a holomorphic line bundle, it does not necessarily mean that a certain map is holomorphic right?

Ben Steffan

you can pull back the trivial bundle (which is obviously holomorphic) along any nullhomotopic map, of which most will not be holomorphic

I'm not sure whether given a holomorphic line bundle there is always a holomorphic map such that pullback of the universal bundle along that map gives this exact bundle

...but proving the classification theorem involves producing explicit maps for each vector bundle, so perhaps all it takes is to inspect that construction and test it for holomorphicity :))

one potato two potato

03:54

thank you for the patience and the answers

porridgemathematics

07:40

is it necessarily true that for a sequence of (holomorphic) polynomials $P_n$ of degree $d$ which are all monic and are such that $1$ is a critical value of smallest modulus, that any limit polynomial $P$ which the $P_n$ converge to on compact subsets of the plane, must also have smallest critical value of modulus $1$ and be monic?

I want to say yes, the monic part seems clear, but the smallest critical value part less so

i think I have an argument for the smallest critical value part also necessarily being of modulus 1, it just comes down to drawing small circles around each of the critical points of the limit polynomial $P$, and then using the argument principle plus uniform convergence

just want to check if this argument makes sense

The argument is that if $P$ had no critical value on the unit circle, then by drawing small circles around each of its critical points and applying the argument principle we would find that for large enough $n$ , $P_n$ shares the same property (because the derivatives also converge locally uniformly ), which is not possible by assumption. Therefore $P$ has at least one critical value lying on the unit circle.

By a similar argument one gets that $P$ cannot have a critical value lying inside the unit disk.

porridgemathematics

11:16

Asked more succinctly, is it true that the critical values of a polynomial are continuous functions of its coefficients?

And my answer , written more succinctly is : Yes, because the roots are continuous functions of the coefficients, and the critical points are the roots of the derivative, hence the critical values which are the polynomial evaluated at the respective critical points must also be continuous functions of the coefficients, and thus continuous as functions of polynomials in the compact open topology.

Just want to make sure this checks out…

Derso

12:05

@XanderHenderson Well, I'm working on classifying ruled Weingarten surfaces in Thurston model geometries and this curve came up to be important in the case of the product space $\mathbb{H}^2\times \mathbb{R}$ just as the helix $(\cos t, \sin t, t)$ was important for $\mathbb{S}^2\times \mathbb{R}$. Part of me wants to call it "hyperbolic helix", but looking at the curve trace that seems really outrageous... Can't think of any other name

*Edit: the helix is also important for the Heisenberg group $\mathrm{Nil}_3$ space

Thorgott

12:47

@BenSteffan the answer to that is no, but don't ask me to remember why

@porridgemathematics they are continuous as a function taking values in unordered tuples, yes

Madder

15:09

i am seeking help i understanding this seminal paper.
https://journals.aps.org/pr/abstract/10.1103/PhysRev.65.117

The mathematics discussed here if very complicated, furthermore, even with the massive number of citations, it seems no one was able to produce a detailed explanation of the work discussed here.

I am fairly educated in mathematics, but i am having great difficulities understanding what is done. The notation and compactness of the paper makes it semi -impossible to understand for an undergrad. Having no resources who explain the work in detail (trust me, i have been searching for weeks) , is making this even harder.

Xander Henderson

Offering to pay for help here is not appropriate.

Madder

Oh sorry, i did not know that was a rule.

Xander Henderson

Also, that is a physics paper, not a math paper. I have no idea how much help you are going to get here.

Madder

It is more mathematics than physics. My main issues start with the "Quaternion Algebra" . that is also where the hard mathematics come into play. I have also asked in the physics part of this site, it is really difficult for most people to understand what is happening there, and most known solutions are published later in simpler language and other methodology.

Xander Henderson

I would very much encourage you not to make sweeping statements about what is true for "most people". Most academic papers have an audience of just a small number of experts---I don't expect that anything I've written is terribly accessible to more than a couple of dozen people in the world.

You need to talk to someone who understands "crystal statistics" or whatever that paper is about. Or you need to read a lot of literature which both leads up to that paper (i.e. read all of the cited papers, and then the papers cited by those papers, until you get to a point where you understand what is going on), and that comes after (i.e. read papers which cite that paper).

one potato two potato

15:19

@Thorgott Hmmmm

Xander Henderson

Generally speaking, a masters or phd advisor would help you know what literature you need to learn in order to access that paper.

On the bright side, that series of papers seems to be about 80 years old, so there are likely a good number of people who have learned the material (assuming that it is material that continues to be relevant in modern physics).

Madder

Sadly, this is not the case. Otherwise i have not asked here. There seems to be zero resources explaining this paper. The resources after are different methods of solutions. I have been searching for weeks, is it possible i have not searched hard enough? Yes.
And i am not getting the needed help from the advisor. But thank you for your comments.I will look somewhere else :)

Xander Henderson

Like I said, I have no idea how much help you are going to get here.

I can't even read the paper, as I lack institutional access.

Ben Steffan

15:45

@Thorgott good to know

psie

16:15

I'm fooled by the simplicity of this. Rudin claims for $\beta>0$ that $$\frac{\cos(\alpha+\beta)-\cos\alpha}{\beta}+\sin\alpha=\frac{1}{\beta }\int _{\alpha }^{\alpha +\beta }\left(\sin \alpha -\sin t\right)dt.$$Then he says that since $|\sin\alpha-\sin t|\leq |\alpha-t|$, the right hand side is at most $\beta/2$ in absolute value. I get $$\frac1{\beta}\int_{\alpha}^{\alpha+\beta}|t-\alpha|\,dt\leq \frac1{\beta}\int_{\alpha}^{\alpha+\beta}|\beta+\alpha-\alpha|\,dt=\beta.$$

How does he get $\beta/2$?

Xander Henderson

What have you done?

My first impulse is to expand the left using the angle addition formula for cosine.

Oh, left vs right.

I'm dyslexic.

Still, what have you done?

Ben Steffan

What have I done, my goodness, what have I done?

Xander Henderson

@BenSteffan MURDERER!

Ben Steffan

Become a thief in the night, become a thief on the run?

psie

@XanderHenderson he is saying since $|\sin\alpha-\sin t|\leq |\alpha-t|$, the absolute value of the right-hand side of my first equation is at most $\beta/2$. What I did was I took the absolute value of the right-hand side of the first equation, then used $|\int f|\leq\int|f|$ and then the bound Rudin alluded to. Then I realized that the difference $t-\alpha$ is at its largest when $t=\alpha+\beta$.

Xander Henderson

16:22

I'm confused about where you got that last integral.

$$\int_{\alpha}^{\alpha+\beta} |t-\alpha| \,\mathrm{d}t = \int_{\alpha}^{\alpha+\beta} \alpha - t \,\mathrm{d}t = \left( \alpha t - \frac{1}{2} t^2 \right|_{t=\alpha}^{\alpha+\beta}, $$ no?

(assuming that $\beta > 0$).

Why use such a sloppy upper bound when you can compute the integral directly?

psie

ah yes, makes sense. Since $t\geq\alpha$, indeed, $|t-\alpha|=t-\alpha$.

Xander Henderson

Oh, shoot... I reversed a sign.

Ugh...

psie

Happens to the best of us.

Jakobian

18:09

0

Q: Product and homotopy type of normal spaces

I know that if $X$ and $Y$ are normal, the product $X \times Y$ may be not normal. But if $X$ and $Y$ have the same homotopy type as normal spaces $P$ and $Q$, has the product $X \times Y$ the homotopy type of a normal space $N$ (that might be different from $P \times Q$ if this one is not normal...

if the OP means $T_4$ when they say normal, then I have a counter-example

Jakobian

18:23

it's interesting how there's 2.2k watchers for general topology and 1.5k watchers for algebraic topology, because the actual numbers of people who engage with the subject are way way less

I think it must be the same kind of psychology that leads people to think that if they study together, it will somehow be easier to study

I doubt the people who subscribed to those tags are serious about the subjects

Thorgott

20:03

@Ben what math.stackexchange.com/questions/5039527/quatzalcoatl-lemma

Ben Steffan

my thoughts exactly lol

it's clearly a play on "snake lemma"

Thorgott

ive heard the salamander lemma before, but this one is new to me

but the actual statement is just (homotopy) pullbacks commute with another, no?

Ben Steffan

should be something on that order

porridgemathematics

21:12

@Thorgott thanks. I’m asking because there is something I am working on and a key step in my argument is showing that the set of monic polynomials of degree $d$ with smallest critical value equal to $1$ (I.e of modulus 1), and satisfying $P(0) = 0$ is compact (in the topology of uniform convergence on compact subsets of the plane)

I seem to be able to show that this set is compact, but at the same time the fact that this set is compact is running me into contradictions. So I am wondering if my argument showing this set is compact is what the issue is, and this set isn’t actually compact.

Is it obvious at all whether this set is compact or not?

Oh whoops , I need more normalizations. All of the ones I mentioned AND also I require that the largest critical value is no larger than some prefixed number $A >= 1$

I forgot to mention this last one.

So is it obvious whether or not this set of polynomials, with all the normalisations I mention is compact?

Thorgott

21:30

well, it's closed by what we've argued so far, so it remains to argue it is bounded

I would hope that follows from some basic estimates, but it's not immediately obvious to me

porridgemathematics

Right it’s definitely closed

Not sure if it does, I used an estimate but it’s not an estimate that is known to many I think

And now I’m worried the estimate isn’t true

Basically I tried to show that the capacity of the component of the preimage of the closed unit disk of any of these polynomials is necessarily bounded below by a positive constant

Here capacity is logarithmic capacity

Here the component us the one containing zero

Then using an argument from a paper of eremenko involving showing the greens functions with pole at infinity for the complement of these components , after passing to a subsequence, concerned uniformly on compact subsets outside the disk of radius 4, and majorizing the corresponding polynomials on such compact sets by some factor times the limit, I can show they are locally bounded outside this disk

Hence there is a subsequence of them that converges on some dish in the plane, uniformly

But then that means the coefficients converge, and we get local uniform convergence on the whole plane

The reason I need to bound the capacity of these sets below us because if this shrinks to zero then the greens function of the complement goes to infinity

The reason I need to fix the largest critical value to be bounded below some finite number is basically because if it is allowed to go to infinity, then it turns out that the component of preimage of the non closed unit disk containing zero can shrink to a set of zero capacity, which means I have no lower bound anymore

This is also pretty non obvious, I’m worried my estimate is wrong basically and the capacity of these preimage sets isn’t bounded below

That the largest critical value is bounded by a finite number seems also necessary because the polynomials are monic, which means any limit polynomial must have the same degree, but if we take a sequence of polynomials with these normalizations without the one on the largest critical value, I can move that to infinity

Then there is no hope of convergence because any limit polynomial can’t be of the same degree

Fuck I’m on my phone just noticed autocorrect butchered a lot of things

Concerned should be converges , dish should be disk

But if it is possible to prove this with a more direct approach I would like to know

Thorgott

22:20

half those terms I don't even know, but I don't think this should be that complicated

Thorgott

22:56

Ok, I think I got it. There is a map $\mathbb{C}^{n+1}\rightarrow\mathbb{C}^{n-1}/S_{n-1}$ taking a polynomial $a_1+\dotsc+a_{n-1}z^{n-1}+z^n$ to its unordered set of critical values with multiplicity. I claim this is a proper map. Your set is the preimage under this map of the set of unordered tuples with values between $1$ and $A$, which is clearly compact, so the claim follows.
To prove this, factor the map as $\mathbb{C}^{n-1}\rightarrow\mathbb{C}^{n-1}\times\mathbb{C}^{n-1}\rightarrow\mathbb{C}^{n-1}\times\mathbb{C}^{n-1}/S_{n-1}\rightarrow\mathbb{C}^{n-1}/S_{n-1}$. The first map is th…

@porridge

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