Mathematics - 2021-02-25

Karim Mansour

2:40 AM

Hi

Karim Mansour

2:59 AM

Hi @TedShifrin

robjohn

Pretty quiet in here.

Karim Mansour

yeah

user 85795

3:17 AM

yup

over 9,000 hours later...

user586228

5:06 AM

Lets say these are the two curves.They have drawn common tangent connecting three points.Firstly are they really the minimia individually?I can find minimum values individually for quite somemore points.Also what does a single point in the common tangent represent?A detailed answer to this from a real anaysis point of view is certainly welcome.

Ted Shifrin

5:40 AM

@user586228 No, not minima. The tangent line needs to be horizontal.

Semiclassical

6:05 AM

@user586228 ooo, Gibbs free energy

Tho I can’t say I remember more than that

Something something phase coexistence

Is the meaning behind that common tangent

robjohn

@user586228 not to mention I see 3 curves; black, blue, and red.

or is the black line a common tangent to the blue and red curves?

Semiclassical

BothI think. But that seems pretty unusual physically

Some good discussion of the thermodynamics here: chemistry.stackexchange.com/questions/75035/…

@user586228 what system is this physically?

Oh, found it: doitpoms.ac.uk/tlplib/phase-diagrams/phasediags2.php

Boy eutectics are weird

@robjohn this context does confirm your query btw: the red and blue curves are in just the right positions that the black line has three tangent points

love_sodam

9:01 AM

If $f:X\to Y$ is a homotopy equivalence and $A\subset X$ be a subspace such that $f|_A A\to Y$ is surjective, then $f|_A$ is a homotopy equivalence?

Leaky Nun

9:42 AM

@love_sodam $X = [0,1]$, $Y = \{\ast\}$, $A = \{0, 1\}$

porridgemathematics

10:02 AM

likely a stupid question but, if I have $v_0,...,v_m$ vectors such that there are $c_0,c_1$ nonzero with $c_0v_0 + c_1v_1 + c_2v_2 + ... + c_mv_m = 0$, why can I be sure that they do not span a subspace of dimension $\geq m$?

sorry, actually it still seems possible, the original question I wanted to ask was why is it impossible for there to be $c_0,...,c_m \geq 0, \sum c_i = 1$ and $c_0',...,c_m' \geq 0 \sum c_i' = 1$ such that $\sum c_iv_i = \sum c_i' v_i$ if we know $\{v_0,...,v_m \} \subset \mathbb{R}^n$ do not lie in an affine subspace of dimension $< m$?

Leonhard Euler

12:54 PM

Is $\mathbb{R}^\omega$ normal in the box topology?

Alessandro Codenotti

I think that's still open

It's known to not be hereditarily normal and it is known to be normal under CH iirc

Balarka Sen

I saw this in Munkres once

@LeakyNun ok

Alessandro Codenotti

Yeah I think Munkres mentions that it is open as a remark somewhere

Leonhard Euler

Okay

It was published in '80s probably

Is it still open?

Alessandro Codenotti

sciencedirect.com/science/article/pii/0016660X72900220 this is the reference for the CH result

Leonhard Euler

1:03 PM

Thanks I would see it when I get time

Alessandro Codenotti

Looks like a very short and readable paper, maybe I should read it too

matwbn.icm.edu.pl/ksiazki/fm/fm88/fm88113.pdf this is the reference for the result that it is not hereditarily normal (he builds a sequence $X_n$ of subspaces of $\Bbb R$ such that $\prod X_n$ is not normal in the box topology, and this embeds as a subspace of $\Bbb R^\omega$)

Balarka Sen

@Alessandro I have a curiosity related to set theory and GGT, do you have some time to listen? It might be nothing though.

Alessandro Codenotti

Sure but there's no guarantee I'll have a good answer

Balarka Sen

I think you will. Anyway, a friend sent me the following axiom of choice puzzle: Suppose there are 100 rooms with a countable family of boxes indexed by $\Bbb N$ in each room, and each box contains a real number. You know that the $n$-th box in each room contains the same real number, regardless of the room.

After some strategizing, 100 mathematicians each go in the 100 rooms and do not communicate once they go in. Each of them are allowed to open countably many boxes in their room to see what real number it contains, but not all of them. Then they are asked to pick an unopened box and guess what real number it contains.

Alessandro Codenotti

I think I've seen some variation of this before

Balarka Sen

1:15 PM

If 99 of them have the right guess, they win. Find a strategy to win.

I have a very simple reason that this is possible. It is because geodesic coalescence in trees.

Astyx

unopened by any of the mathematicians ?

(Do they all have to make a guess for the same box)

Balarka Sen

Each mathematician goes in a unique room

They all operate in their own respective rooms.

Uninfluenced by the rest

schn

Is the convolution of two normalized distributions (not normal, i.e Gaussians) also normalized?

See snst-hu.lzu.edu.cn/zhangyi/ndata/Voigt_profile.html

Alessandro Codenotti

@BalarkaSen what do you mean?

Balarka Sen

Here is my reformulation: Take an unrooted $3$-regular tree $T$ and consider it's set of ends $\partial T$, a Cantor set. Two (not 100) mathematicians are trying to guess a proper (not necessarily non-backtracking) path $\gamma$ in the tree, which is unknown to them. They are allowed to sample a proper subset of the vertices of $\gamma$. They take their separate samples, unbeknownst to each other, and then guess a vertex they have not sampled. If one of them succeeds, they win.

Astyx

1:24 PM

So if $r_{n,i}$ is the number in the n-th box of the i-th room, then we have $r_{n,i} = r_{n,j}=r_n$ ? And at the end we ask each mathematician for $r_{n_i}$ for some $n_i$. Is $n_i$ the same for all mathematicians ?

schn

According to that website, the convolution of a Cauchy-Lorentzian and a Gaussian is. But does it hold in general.

Balarka Sen

@Astyx Yes to the first question. At the end we ask the mathematician to guess $r_{m_i}$ for some $m_i$ such that $m_i$-th box in the $i$-th room is unopened by the $i$-th mathematician.

Alessandro Codenotti

@BalarkaSen hmm I'm not convinced this is the same problem

Balarka Sen

Maybe not, but I think they're the same in spirit.

Astyx

I assume they hear each other's answers?

Balarka Sen

1:27 PM

No.

No interaction. An independent 101-th person collects everyone's answers and then tells them if 99/100 of them guessed the right thing or not.

Alessandro Codenotti

@BalarkaSen maybe, so how do you do your version?

Balarka Sen

@Alessandro By axiom of choice, we can choose a ray $\gamma(e)$ realizing the end $e$ for every end $e \in \partial T$.

Now partition the set of vertices of the unknown path $\gamma$ into two disjoint infinite subsets $A, B$. $V(\gamma) = A \sqcup B$.

$A$ has a limit end, let's call it $e_A$. Similarly, $B$ has a limit end, let's call it $e_B$. The only information the two mathematicians know is that $e_A = e_B$, some common end $e$.

Thorgott

@schn what does normalized mean

Balarka Sen

Consider the three sets of vertices $A, B, \gamma(e)$. Then by geometry of the tree, $\gamma(e)$ and $A$ must coalesce after some time $t_A$ and $\gamma(e)$ and $B$ must coalesce after some time $t_B$.

Alessandro Codenotti

coalesce?

Balarka Sen

1:32 PM

If you have two paths in the tree giving rise to the same end, then they merge after some finite time.

Alessandro Codenotti

sure

Balarka Sen

So the strategy is as follows. The first mathematician samples $A$, second samples $B$. The first computes $t_A$, the second computes $t_B$, because $\gamma(e)$ is known to both of them.

Alessandro Codenotti

Wasn't $\gamma$ to be guessed? How do they know $A$ and $B$?

Balarka Sen

Because they are allowed to sample a proper countable subset of vertices from $\gamma$.

Alessandro Codenotti

Oh ok, I misread

Balarka Sen

1:35 PM

This is similar to allowing the mathematicians in the first problem to open countably many but not all the boxes

Alessandro Codenotti

I thought they were sampling from the tree

Balarka Sen

Ah ok sorry for being confusing. I mean they can do so from the path.

Alessandro Codenotti

ok but how do they guess $\gamma$ before the smallest of $t_A$ and $t_B$?

Balarka Sen

The point is, either $t_A \leq t_B$ or $t_B \leq t_A$. So if the first mathematician picks any vertex from $\gamma(e)$ beyond time $t_A$, and the second mathematician picks any vertex from $\gamma(e)$ beyond time $t_B$, one of them have to be right that the pick is indeed a vertex from $\gamma$

Because once coalescence happens, $\gamma$ and $\gamma(e)$ are really the same path.

Alessandro Codenotti

Ah right they guess a single vertex

Balarka Sen

1:41 PM

Does that make sense?

Yeah

Alessandro Codenotti

Hmm it's not clear to me why $\gamma=\gamma(e)$ after a while if backtracking is allowed. Can't one keep going back and forth?

Balarka Sen

Sorry, yes, I mean, $V(\gamma) \supseteq V(\gamma(e))$ after coalescence still right?

Alessandro Codenotti

yeah

Ok I agree now

Balarka Sen

So my question is to what extent can you push this? The collection of paths representing the ends is sorting of like a geodesic combing but towards the ends? Also, can you say something with delta-hyperbolic background instead of a 0-hyperbolic background?

It doesn't feel like the axiom of choice aspect is interesting

There is geometry happening

This "coalescence time" is the Gromov inner product I think

Astyx

Isn't there a unique path between two vertices? meaning you can find two distinct elements and claim one in between is in $\gamma$ ?

Alessandro Codenotti

1:47 PM

Hm I'm not sure, this feels very choichey to me. This problems usually work that you use AC to pick representatives for some equivalence relations, and then the prisoners are allowed to do some stuff that let's them figure out in which equivalence class they are, and the relation is nice enough that someone has to guess correcrtly if they all guess the representative

Which is kind of what is happening here as well

Balarka Sen

Hm I see

Alessandro Codenotti

I'm not seeing how do 99 correct guesses in the puzzle above though

Eminem

In terms of big O notation is $\sum_{i=0}^{\log n}i=O(\log n)$?

Balarka Sen

@Astyx You're right, so maybe I don't want this exact setup.

One second

Astyx

I thought you'd want the mathematicians to also guess the index of the vertex, but I'm not sure your solution works then

I'm working it out rn

Ah yeah that seems to work

Balarka Sen

1:55 PM

What is your fix?

Oh, ok, the time parameter in the domain of $\gamma$, you mean

Astyx

Yes

Balarka Sen

Yeah OK

Thanks haha

Astyx

But even then, if you look at $\gamma_1$ and $\gamma_3$, you know $\gamma_2$

Balarka Sen

Yeah there's something off about the setup, I'll think about it later

@AlessandroCodenotti It doesn't seem like a problem to me because you have 100 paths with the same end instead of two, and you have the canonical one as well, and you measure the last coalescence time

Anyway, the setup needs to be fixed I'll try it later

robjohn

@Eminem No, it is $O\!\left(\log(n)^2\right)$

Clarinetist

2:11 PM

Let $X_n$ be a sequence of random variables. I've seen sources online use what I'm interpreting as the following inequality (letting $\epsilon > 0$):
$$\mathbb{P}\left(\limsup_{n \to \infty}\left\{X_n \geq \epsilon\right\}\right) \leq \mathbb{P}\left(\limsup_{n \to \infty}\left\{X_n\right\} \geq \epsilon\right)$$
Is this true in general, and also true if I change the inequality? I'm struggling to see *why* this is true if at all.

Astyx

@BalarkaSen If you take $\gamma$ to be a sequence of vertices that stabilizes into a ray I think your solution works and mine doesn't

Balarka Sen

Hm yeah maybe that's what I want, a sequence converging to an end

Quasigeodesic or whatever they call them

@Clarinetist LHS is probability that $X_n \geq \varepsilon$ for all but finitely many $n$. But that definitely implies $\limsup_n X_n \geq \varepsilon$, yes?

So one event seems to be contained in the other

Clarinetist

Great, thanks, and so that also holds if I change the inequality

Balarka Sen

As in? With $\leq \varepsilon$?

Clarinetist

Yeah

Balarka Sen

2:21 PM

Seems right.

Clarinetist

Well, actually, I'll have to think about that

Yeah

It does seem right

Balarka Sen

That's the morally correct inequality I think. For above, you actually get $X_n \geq \varepsilon$ for all but finitely many $n$ implies $\liminf_n X_n \geq \varepsilon$, which is stronger.

Astyx

You need to stabilize into a ray to get $V(\gamma(e))\subset V(\gamma)$

Wait no now I'm convinced your argument doesn't work

Because $t_A$ or $t_B$ can be infinite

Balarka Sen

That is not true in any reasonable setup

Coalescence always happens in finite time in tree

Astyx

Take $\gamma = \gamma(e)$, but every so often you make a small detour to a neighborhood

Balarka Sen

2:31 PM

OK

Not sure what your point is

Astyx

Then A or B is going to capture those detours

So $t_A$ or $t_B$ is going to be infinite

Balarka Sen

What is your A and B?

I do not understand

Astyx

Same as yours, a partition of $V(\gamma) = A\cup B$

user2103480

@Clarinetist One tipp to handle these, since you had problems before:

BigSocks

what do the paths in your tree represent, and how does it match up with the boxes?

Balarka Sen

2:34 PM

Oh, I see your point, but I think this is not an issue upto "quasi-ness"

Astyx

We're trying to figure that out

user2103480

convert, at the least, the set lim sup to FOL

Astyx

@BalarkaSen Right, i think you want to stabilize into a ray

(like a proper ray, without backtracking)

BigSocks

is a ray like a path that is infinite on one side?

Balarka Sen

Hm, probably, yeah. But there should be a nice and natural way to phrase that.

Astyx

2:35 PM

@BigSocks yes, and that never backtracks

BigSocks

what is meant by backtracking?

Astyx

You can't have $v_1\to v_2\to v_1$ as a subpath

BigSocks

oh ok like well founded

to have that it could not be a tree though, bc that's a cycle, right?

so no rays on trees could ever backtrack

Balarka Sen

By definition of a ray

user2103480

@BalarkaSen LHS should be "Probability of the set of omega such that for all n there is an m>=n with X_m(omega) >= eps", so this should mean X_n>= eps infinitely often. IIRC lim inf is what you mean

Balarka Sen

2:40 PM

liminf{A_n} is "all but finitely many A_n's happen" yeah, maybe I messed it up above

user2103480

I literally have to google every time which is which

Clarinetist

@user2103480 What do you mean by FOL?

Balarka Sen

The inequality is still correct, fortunately.

Thorgott

first-order logic

BigSocks

ok so if I understand correctly the end $e$ is the number the mathematicians want to guess and the two disjoint sets of vertices $A$ and $B$ along the ray to the end $e$, $\gamma(e)$, are the other boxes with real numbers they can freely look at?

Astyx

2:47 PM

they know e, they want to guess vertices of $\gamma$

user2103480

@Clarinetist the probability statement is basically just a logic statement. If infinitely many X_n are >= eps, then the lim sup must be >= eps

Astyx

If you only deal in rays, that's like asking for decimals of a real number you know

Clarinetist

Oh, I see. Thanks

Astyx

which is not interesting

BigSocks

ok then my understanding is almost completely off. the vertices of a given ray $\gamma$ give you a real number? if I were to guess how, I would think they'd be each successive digit in a representation of a real number, but you say this is boring, so how do we interpret the vertices in this ray?

anakhro

2:56 PM

Hi all.

BigSocks

and if they both know the end $e$, what could the analogue of this be for the boxes problem? maybe I am still connecting it too much with the first problem, but I thought they were meant to be analogous. I understand if maybe the attention is no longer on that problem so much and that this one is maybe more interesting,

Balarka Sen

@Astyx Not quite because there's no preferred starting point of the ray

Any basepoint on the tree gives an isomorphism of $\partial T$ with a Cantor set

But this is non-canonical

Depends on the basepoint.

But yeah

Leaky Nun

@BalarkaSen Let $C_1, C_2, \cdots$ be distinct closed subsets of $[0,1]$ that cover $[0,1]$. Show that all except one is empty.

Balarka Sen

This is going to be some Baire thing isn't it

Alessandro Codenotti

@LeakyNun I'm confused. $C_1=[0,1/2],C_2=[1/2,1]$?

Leaky Nun

3:07 PM

@AlessandroCodenotti sorry, disjoint not distinct

@BalarkaSen who knows :)

Alessandro Codenotti

Then it's a theorem by Sierpinski

Leaky Nun

aha, why didn't i know that

what does it say?

Alessandro Codenotti

That if you cover a continuum with countably many disjoint closed sets at most one is not empty

Usually people stumble into this trying to decide whether a countable set with the cofinite topology is path connected

Or whether path connected implies uncountable more generally

Balarka Sen

Just delete the interiors

Its a direct Baire corollary

Astyx

Maybe you could take the first vertex of $\gamma$ as the basepoint to work around that ?

BigSocks

3:13 PM

also in the original problem idk how there is any way to make a consistently planned informed choice if all the boxes could be potentially random reals? so why would we think that looking at any of the vertices would help us guess any other vertex?

Balarka Sen

Oh, no, ok I was thinking of the case where $C_i$ are intervals. In that case you can throw away all the endpoints, then you have $[0, 1]$ minus a countable set as union of countably many opens

Alessandro Codenotti

Can I cover it with disjoint $F_\sigma$ sets instead? What's the least $\xi$ such that it can be covered by countably many disjoint $\mathbf{\Pi}^0_\xi$ sets?

Akiva Weinberger

Is there a fixed-point-free involution of the Möbius band?

(continuous)

Astyx

Can you not just do half a circle translation ?

Akiva Weinberger

Doing that twice flips it along the other axis

Alessandro Codenotti

3:24 PM

Ah yeah, $F_\sigma$ sets is trivial

Leaky Nun

@BalarkaSen and what's the contradiction?

Balarka Sen

@LeakyNun The collection of all endpoints is a perfect set

But there are very few things such as a countable perfect set

This probably goes through for general closed sets but I haven't been thinking carefully

Alessandro Codenotti

mathoverflow.net/a/48991/49381 yes it can work in general

Balarka Sen

ah yah this is the other variant of baire that i never remember

Alessandro Codenotti

It's a variant that is often used

I think the usual BCT arguments in functional analysis use it for example

Akiva Weinberger

3:45 PM

0

Q: Is there a fixed-point free involution of the Möbius strip?

Is there a fixed-point free continuous involution of the Möbius strip? The Lefschetz fixed-point theorem is useless here because the Möbius band is homotopic to a circle, which does have fixed-point free involutions. Beyond that, I don't know what tools to use. (You can't just rotate it 180 degre...

This should be low-hanging fruit

'cause I feel like there should be a simple answer but I don't see it

so (probably) easy points^

Balarka Sen

If there was, you'd get an involution of RP^2 with exactly one fixed point, that lifts to a Z/2-equivariant fixed point on S^2 with exactly two fixed points, but then you're screwed because all those are actually the linear ones

Which gives identity on the quotient

@AkivaWeinberger

user2103480

@BalarkaSen when you say the boundary map is literally the boundary, is that just for manifolds or is there a more general statement.

I'm can only think of the following theorem I'm given: Say $M$ is smooth compact orientable $n$-dimensional, so it's boundaries are closed smooth compact orientable n-1-dimensional. Then under the boundary map in the LES, the fundamental class of $H_n(M, \partial M)$ gets mapped to the sum of the fundamental classes of the boundary components

(or, equivalently, gets mapped to a vector with each entry one fundamental class)

Balarka Sen

$\partial[\xi] = [\partial \xi]$

That is how the snake map behaves if you go back to the proof of the long exact sequence

Akiva Weinberger

I don't understand why that screws me

What do you mean the linear ones?

Mike Miller

@AkivaWeinberger He means that every action of a finite group on $S^2$ is conjugate to an action by a finite subgroup of $O(3)$.

Balarka Sen

3:54 PM

An involution of S^2 with exactly two fixed points has to be rotation by $\pi$ about an axis, is what I mean

Mike Miller

This is not obvious, easy, or trivial, but it is true

Akiva Weinberger

Interesting. How do you show that?

I see how that gives us the result

Mike Miller

Bah

Balarka Sen

Orbifold shit man

Geometrization for 2-orbifolds

Mike Miller

First show that every continuous group action is smoothable (this is not too different from the proof that every topological surface has a unique compatible smooth structure up to diffeo)

Balarka Sen

3:56 PM

Mike gonna do equivariant tOpOlGy now

Im gonna leave

Mike Miller

Then as Balarka says generalize the proof of the uniformization theorem to orbifolds. The uniformization theorem says that every surface can be given constant curvature; the uniformization theorem for orbifolds says that this may be done G-invariantly when G is a finite group acting on the surface. In particular for M = S^2/G this justifies that the "universal covering action" is G acting on S^2 by isometries w/r/t the round metric, which is an action by elements of O(3)

Balarka Sen

Oh thank god

Mike Miller

This is what I mean, it's too high-level. It's a good tool for big hard questions but this doesn't need that much machinery.

Balarka Sen

Oh no

Leaky Nun

how does this relate to RP^2 at all?

Balarka Sen

3:57 PM

Equivariant topology incoming

@LeakyNun Cap off the boundary circle

Leaky Nun

done, now what

Akiva Weinberger

Or quotient up the boundary

Leaky Nun

oh, does that give you RP^2?

Balarka Sen

Algebra brain

Learning RP^2 = Mobius/bd for the first time

Leaky Nun

really???

Balarka Sen

3:58 PM

Knows proof of FLT

Akiva Weinberger

@LeakyNun The Möbius band is a cylinder mod antipodes

aka sphere minus poles mod antipodes

Leaky Nun

oh wow that's nice

Akiva Weinberger

(or sphere minus open neighborhood of poles, depends if you want boundary)

Alessandro Codenotti

@Mike when did you start your phd at UCLA?

Mike Miller

@AkivaWeinberger Surfaces with Euler characteristic zero are torus, Klein bottle, cylinder, Mobius strip. Quotient of mobius strip by your antipodal action gives degree 2 cover of Mobius strip to a surface of Euler characteristic zero. This surface must again be a Mobius strip (non-orientable, noncompact and/or has nonempty boundary depending on your defn of mobius strip). But classification of covering spaces says there is one connected degree 2 cover of Mobius strip: the cylinder. Contradiction

user2103480

4:00 PM

@BalarkaSen fckn hell this is so easy and I swear I followed this 5 times in the past without realizing that this is immediate. relative chain in C(X,A) = quotient of (singular simplex) such that quotient of (boundary of singular simplex) is 0 <=> boundary of singular simplex is singular chain in A, and unique up to homology

Mike Miller

@AlessandroCodenotti 2014?

Balarka Sen

@user2103480 yup!

Alessandro Codenotti

I see, you might have met my advisor then, she was a postdoc there in 2013/14 iirc (Aleksandra Kwiatkowska)

Mike Miller

nah

Akiva Weinberger

@MikeMiller Oh yeah that makes sense

If you want points you can copy/paste that into the answer field

Mike Miller

4:03 PM

I don't have an MSE account

You can write that up if you want

Balarka Sen

The answer is of course not

It should not be written

Akiva Weinberger

What about the Möbius band minus an even number of points

I think you can do the Möbius band minus two points

No never mind

Balarka Sen

You know all involutions of RP^2 by my argument

Akiva Weinberger

Right OK yeah

Mike Miller

@BalarkaSen These don't need to compactify to involutions of RP^2.

Akiva Weinberger

4:08 PM

Does the end compactification not do that?

Balarka Sen

Surely you can do some equivariant crap

At the ends

Mike Miller

Meh

Balarka Sen

Lol

Do your topology Mike

It is your time to shine

Mike Miller

One can just google...

covering spaces of surfaces with boundary...

These are $N_{1,k+1}$

Balarka Sen

lol i guess

Akiva Weinberger

4:12 PM

Sorry, what's $N_{a,b}$? $a$ RP2s minus $b$ points?

Balarka Sen

yee

Akiva Weinberger

In any case, the answer to "can I fix it by deleting some points" is "no"

Balarka Sen

Z/2-actions definitely compactify cuz you can just take the homeomorphism given by multiplication by 1 which has a compactification

@AkivaWeinberger A space is k (mod n) if it has a free Z/n action after deleting k points?

You're insane

But number of fixed points of a group action is an interesting question which was studied by Smith and only @MikeMiller among us knows Smith theory

Akiva Weinberger

If you weaken "continuous" to "tame" then I'm pretty sure that notion is exactly the combinatorial Euler characteristic

meaning the Euler characteristic of the cohomology with compact supports

Balarka Sen

How? You only get $\chi(X) - k$ is divisible by $n$

Well, for surfaces

Akiva Weinberger

4:23 PM

The only number such that $\chi(X)=k\pmod n$ for all $n$ is $k$

Balarka Sen

Oh, you're demanding for all n? I agree then

Akiva Weinberger

I'm fine with saying "there's a vague sense in which an open interval has $-1$ points" because that's what $\chi_c$ says anyway

Balarka Sen

Ok, got it

Studying Z/n actions for all n is also not far from studying an S^1 action

But I don't know if that can be said in a precise way

The number of fixed points under an S^1 action is the Euler characteristic

Akiva Weinberger

How does that work for $\Bbb R$

Balarka Sen

Not a clue. Smooth action of R on a smooth manifold is pretty much a vector field, or at least you can linearize it to be a vector field

Noncompact everything breaks down, but compact you have Poincare-Hopf

Akiva Weinberger

4:29 PM

No I mean how does this compute $\chi(\Bbb R)$

Balarka Sen

Oh. Erm

There is not really any good S^1-action on R lol

Akiva Weinberger

Oh I see it only works in the one direction

Balarka Sen

good S^1 action = finite number of fixed points

Yeah

Akiva Weinberger

Also surely "the number of fixed points" can never be negative while $\chi$ can be

Balarka Sen

Poincare-Hopf fixes this by computing the indices of the zeroes of the vector field

schn

4:46 PM

@Thorgott with normalized I meant that the integrals of the PDFs equal one

Thorgott

huh? that's part of the definition of a PDF

schn

You’re right, that was redundant. So the question was, if one takes the convolution of two PDFs is this then also a pdf (in the case of the Voigt profile it is)?

Thorgott

yes

all you have to check is that the total integral of the convolution is 1

schn

5:16 PM

Is this stated in a theorem or is it just very obvious?

Leaky Nun

@BalarkaSen algebra time

direct limit of Z/n is Q/Z, which is dense in S1 = R/Z

Balarka Sen

lol

Thorgott

it's a one-line calculation

Balarka Sen

It's a 0-line calculation. Convolution of f and g is the pdf of X + Y where X and Y are RVs with pdf f and g.

Leaky Nun

that's 1 line

Balarka Sen

5:22 PM

It's 1 line but not 1 line of calculation

Mass at x times Mass at y sum over all pairs x + y = z is new Mass at z

Convolutions

Astyx

quick mass

2

Thorgott

continuous, not discrete

what you're saying is still correct, but it does take 1 line of calculation to justify

Balarka Sen

Mass is mass

Just discretize

Take epsilon covering and collapse these to midpoints and all of mass over the interval to that point

Take limit

Woosh

schn

5:39 PM

Thanks for the replies!

user2103480

5:50 PM

@Thorgott $(\mu \ast \nu)(\Bbb R) = \int_{\Bbb R} \underbrace{\mu(\Bbb R - x)}_{=1} \nu(\mathrm{d}x) = 1$

whats a probability density

Thorgott

woke

user2103480

category theorist brain

abstract away all the difficulties and any possibility to do this practically then be smug

gÖdEls IncOMplETeneEeez iS jUsT a SpeCIaL cEsÉ oF CuRtegery ThearRy

Balarka Sen

Im a lyrical miracle categorical individual

Skibididbidbididbid dap

Mike Miller

Skibadee skibadanger I am the rearranger

Balarka Sen

LMAO

Mike Miller

6:05 PM

not mine, german techno band said it first

Balarka Sen

its fire

Mike Miller

@AkivaWeinberger Indeed, and the only $S^1$-surfaces have nonnegative Euler char

geocalc33

the lap lacian is a differential operator invented by lap lace

geocalc33

6:28 PM

you know that little "1" that's on the browser tab when 1 message has been sent

not showing up for me anymore

Can someone help me prove or disprove the existence of roots to this function? $\Phi(s)=\Gamma\left(1+\frac1s\right)+\sum_{n=0}^\infty\frac{(-1)^n}{n!}\zeta(-ns).$

$\zeta$ is the analytically continued Riemann Zeta function

$\Gamma$ is the gamma function

$\Phi(s)$ is the analytic extension of the series $\sum_{n\ge 1} e^{-n^s}$ to the complex plane

for $\Re(s)<1$ and complex $s\ne 0$

fido9dido

9:18 PM

3

Q: Minimization of Sum of Squares Error Function

Given that $y(x,{\bf w}) = w_0 + w_1x + w_2x^2 + \ldots + w_mx^m = \sum_{j=0}^{m} w_jx^j$ and there exists an error function defined as $E({\bf w})=\frac{1}{2} \sum_{n=1}^{N} \{y(x_n, w)-t_n\}^2$ (where $t_n$ represents the target value). I'm having trouble making sense of a passage in my textbo...

statistics polynomials optimization

I dont understand this guy's answer at all, where do I start
I know derivatives and chain rule , But I dont know how did he come up with y(x,w+μw′) or E(w+h). I am having hard time understanding the bloody math behind Least squares in Bishop Pattern recognition book

loch

9:38 PM

@fido9dido are you familiar with multivariable calculus?

fido9dido

9:53 PM

@Nope maybe I took it under different name long time ago

well I see it contain vector and matrix math which I know

loch

then you might want to review that first

but essentially the answer consists of two parts:

1) computing the first derivative of the function w.r.t. all the w_i's, and set them all the zero, and the answer argues that the zero is unique

2) check that this really does define a minimum, which is the part about E(w+h) (which amounts to checking that the hessian matrix is positive definite)

this should remind you of single variable calculus, where to find the minimum of a function you first 1) compute the first derivative and set it to zero, 2) check that it genuinely defines a (local) minim…

fido9dido

Ah Ok Thanks i will start there

Ted Shifrin

10:45 PM

@Semiclassic: I realize from this post that I have no understanding whatsoever of Euler angles.

Semiclassical

ugh, euler angles

Ted Shifrin

I thought it was all you physicists who adore them.

Semiclassical

mechanical engineers, maybe

Ted Shifrin

Now it makes absolutely no sense to me. Seems very non-unique, to boot.

Semiclassical

that's very likely

it's even a headache in actual applications. c.f. gimbal lock:

Jorge

10:49 PM

hello

Semiclassical

@TedShifrin In formal language, gimbal lock occurs because the map from Euler angles to rotations (topologically, from the 3-torus T^3 to the real projective space RP^3 which is the same as the space of 3d rotations SO(3)) is not a local homeomorphism at every point, and thus at some points the rank (degrees of freedom) must drop below 3, at which point gimbal lock occurs.

Euler angles provide a means for giving a numerical description of any rotation in three-dimensional space using three numbers, but not only is this description not unique, but there are some points where not every chang…

Jorge

can any of you guys endorse for combinatorics in arxiv?

Ted Shifrin

Yes, I was basically observing some of that ad hoc when I tried to understand the guy's question.

Semiclassical

from the Wikipedia article on gimbal lock

Ted Shifrin

Love the topology thrown in :)

Semiclassical

10:51 PM

ya

i really wish I could vote "????" on some questions