@Perturbative so I was very much looking at Notre Dame, in part for topology/geometry (though perhaps leaning more to algebra/number theory)

There's a lot of TQFT stuff going on, recently Stolz has been liking it, also Mnev, and I think Chris Schommer-Pries. Behrens is nice in algebraic topology, Mike already mentioned Nicolaescu/Stolz/Putman (who in hindsight is the person I'd have most likely ended up working with)

Pretty good amount of diffgeo there is well, I remember Gursky and Brain Hall

If I have a closed simple polygon in the plane, then the sum of the angles between the points should be 2*pi.

The generalization of that seems to be: If I have a simple closed curve in the plane, then the integral of the tangent is 2pi.

How do I show this?

This should be the fundamental theorem of calculus in some form, right?

Actually, I want actually to talk about the integral of the normal. But it seems like the answer should be the same.

It seems I was after the winding number.

So, for planar curves this is dealt with nicely by the complex theory (via argument function).

For higher dimensional curves, how does this work?

@moteutsch: Google Hopf Umlaufsatz. I don't know what you mean by the "integral of the tangent." But I know what it's supposed to mean.

this Hopf guy seems to be good at maths. maybe he has some potential

@moteutsch It's not really the winding number that is at play here, the number you are looking for is the turning number

Hi @Ted; thanks for the comment in my answer - I was careless, as always

Hi, does the Erdos-Kac theorem only hold when the distribution of the integer is uniform?

@AminIdelhaj @MikeMiller Thanks for your replies! Stolz and Schommer-Pries are closest to my interests I would say, hence my interest at Notre Dame. It's nice to have Behrens there as well for algebraic topology stuff.

@MikeMiller Sorry I didn't see that and disappeared

That's ok, I do that a lot

I can write down something later though

(remove)

3 hours later...

i someone is interested in elementary question about totient and sigma, ping me, we can discuss abut that question

@nitsua60 I want to talk to you, can I?

psst @LeakyNun, do u have a sec?

?

do you know why this identity holds?

here, $\Phi(A)=A^TA$

I feel like it must be by definition, but I'm a bit confused

It is by definition

then I am confused x)

because I don't see it

how did $\gamma$ get into the definition of the push forward of $\Phi$?

shouldn't we be looking at $d\Phi(B)\gamma$ or something

$\Phi \circ \gamma(t) = \Phi(I_n + hB) = \Phi(I_n) + t d\Phi_{I_n}(B) + o(t^2)$

If you derive this in regards to $t$

You get what you're looking for

@Astyx Hello! How are you?

I'm fine, what about you ?

I’m fine too, just struggling too much with Organic Chemistry

I’m unable to find the products of a given reactants

@ShaVuklia Are you convinced yet ?

yes yes, thanks, I was trying to formulate that I sometimes forget that the differential is just the ordinary derivative

My problem is that I sometimes have a hard time making the translation back to ordinary calculus. It seems that the reason what they're doing works, is because they're making the identification $T_{I_n}GL(n,\mathbb R)=M(n,\mathbb R)$. But something's still odd to me. I'll have to go through the details, but I might ask another question later.

Math Seminars (beta)

@Knight what's up?

@nitsua60 Hello! How are you?

Can you please come here:

?

I read on a pdf that considering SU(2) the spinor $(\xi_1, \xi_2)^T$ transform the same way as $(-\xi_2^*, \xi_1^*)^T$. What does it mean that they transform the same way? I don't know what's the meaning of "two things transform in the same way"

I tried using the chain rule, by writing out the Jacobian of $\beta$ and $\gamma$

however, since we have maps from and intro matrix spaces, I'm not sure if using the Jacobian is insightful?

Can you write out the differential of $\beta$ ?

Well I wrote it as $\begin{bmatrix}D^1_1\beta&D^1_2\beta&D^2_1\beta&D^2_2\beta\end{bmatrix}$

Where $D^i_j$ is the $(i,j)$-th partial derivative

Having made the identification $M_2\mathbb R=R^{2\times 2}$

But in terms of $(\cdot, \cdot)$ ?

I should expres $D^i_j\beta$ in terms of $(\cdot,\cdot)$?

You can compute the maps

For instance, do you know the differential of $(\cdot, \cdot)$ ?

Uh let me see, first of all, $\beta$ is a composition of $(\cdot,\cdot)$ and a linear map which sends $(x,y)$ to $(Ax,Ay)$. And since $(\cdot,\cdot)$ is bilinear, I'm assuming it is equal to its differential?

Not quite

the map that sends (x,y) to (Ax, Ay) is linear so it is equal to its differential

But if you write [x+h, y+h] out (I'm calling $(\cdot, \cdot)$ $[\cdot, \cdot]$ for clarity, if that's ok with you)

ah, one sec

Okay, so the differential of $[\cdot,\cdot]$ at $x,y$ is the linear map (which sends $h$ to) $[x+h,y]+[x,y+h]$, which really makes perfect sense

Not quite

What do you get when you develop [x+h,y+h] ?

but.. [x+h,y+h]-[x,y]=[x,y+h]+[x+h,y], no?

wait wait wait

sry x)

No issue, take your time :)

ah, okay, so then it's equal to the map which sends h to [x,h]+[h,y]

jesus

You're making an assumption when you do this

I left out [h,h] since that is o(h)

is that the assumption?

Nope

You're assuming the bilinear form is symmetric

oh weps, well that holds. let me see where I used that assumption tho

Which is kind of my bad as well, I should have you compute [x+h, y+k]

Greetings, @Sha @Astyx

Heya

How are you ?

elo @Ted

where are we using the symmetry?

because now we just have the map which sends (k,h) to [x,k]+[h,y], no?

+ o(h)

Here you're not using symmetry

you need the symmetry to argue that [h,k] is o(h,k)?

oh no

right I get what you're saying

When you wrote [y,h] you were using symmetry

But I think you've edited your message or I misread you, idk

oh right, ye those were typos

I was editing them indeed

let me see if I can figure it out now x)

Anyways, yes we have $d[\cdot, \cdot]_{(x,y)}(h,k) = [x,k] + [h,y]$

So now you have $\beta\circ \gamma = [\cdot,\cdot]\circ A\circ \gamma$ (where I abusively denote $A:(x,y) \to (Ax,Ay)$)

Now you know the differential of all of these operators, so you can apply the chain rule

right ye

sry, I got confused for a sec, because I thought that (x,y) -> (Ax,Ay) was bilinear, as I was confusing (.,.) with [.,.]

That's why I wanted to use [] instead of ()

ye, I realised that x)

brb

gotta cook

right, have a good meal

and massive thanks

(I had completely forgotten to use the definition of the derivative to compute one x) but this was a good reminder)

Is the involution of a $C^*$-algebra an isometry?

@user193319 Yes

Follows from the C*-property, try to think about it, it's not hard

@TedShifrin so happy to see you here, doing your magic! Hope all is well with you and for you!

Hey hot cats.

Hi

What's up, Alessandro?

waddup boiz

Not much

@EdwardEvans what up with you?

Just got back from a nice cycle, and now doing a quiz on Minkowski stuff again

What Minkowski stuff?

Not physics, if that's the purpose of your question

haha

Minkowski theory within algebraic number theory

Never heard of it. Same Minkowski?

Yeah it's the same guy :)

The only Minkowski things I know occur in geometry and analysis.

What is Minkowski theory about?

One takes the perspective that elements of a number field $K$ are somehow points in an $n$-dimensional $\Bbb R$-vector space $K_\Bbb R := \left[\prod_{\tau \in \operatorname{Hom}_\Bbb Q(K, \Bbb C)} \Bbb C\right]^+$, where the $+$ means that the elements are fixed by complex conjugation

oops

What did Minkowski do to get his name attached to this?

so he put the theory together, and he also proved some theorems in lattice theory that enable one to prove that the class group of a number field is finite

Minkowski's lattice point theorem

Actually it used to just be called the geometry of numbers until arithmetic algebraic geometry got in the way

Well "arithmetic geometry" basically means "geometry of numbers". :P

right, but this isn't algebraic geometry

Oh, I see. I thought that's what you were saying it was subsumed under now.

@EdwardEvans luckily

well I haven't had any experience with algebraic geometry yet so

Pete L. Clark has notes on geometry of numbers.

@AlessandroCodenotti do you like distribution stuff?

@NoName nice, thanks for this

Not really, why?

I know a little of it

Oh, I just was wondering.

I had since then realized my problem was very simple. :P

(it was why the Dirac delta is tempered, but that's easy).

I don't really know what tempered distributions are but I'll trust you :P

Dual space of Schwartz space.

That's all. :P

Well now I should ask what the Schwarts space is, but I don't know if I want to know

Smooth functions which, together with their derivatives, vanish at infinity faster than any power of $|x|$.

Oh ok

Well I guess the Dirac delta is in the dual of pretty much anything anyway

Here it just follows from the topology of Schwartz space. In particular, $\varphi_n\to \varphi \implies \varphi_n\to\varphi$ uniformly.

Conway died today

Thus $\langle\delta,\varphi_n\rangle = \varphi(0)\to \varphi(0) = \langle\delta,\varphi\rangle$

Wait, really?

No way

Which Conway?

JH

twitter.com/CardColm/status/1249038195880341505

ah, sad, but he was probably close to 85-90. We had him down to give some lectures (mostly for undergraduates, but also a colloquium, I believe) about 15-20 years ago.

Pretty playful of a mathematician.

Wrote a bunch of good nice papers.

He was 82

@anakhro I have a question which I feel should be trivial but I'm missing something

That's exactly the kind of question that stumps me.

I can try to help.

I have a sequence $f_n\in L^p(\Bbb R)$ that converges pointwise to $f\in L^p(\Bbb R)$. Does this sequence converge in $L^p(\Bbb R)$?

Certainly not in general.

I know for sure that the converse is false

Think of the standard example from Riemann integration: $\int_0^1 f_n = 1$ for all $n$ but $f_n\to 0$ pointwise.

Right

So I was too optimistic. What I want to prove is that as $t\to 0$, $f(x+t)\to f(x)$ in $L^p(\Bbb R)$ for all $p<\infty$ (and I already have a counterexample showing that this is false in $L^\infty$)

Dominated Convergence?

Uhm, maybe I'm just tired, but I'm not seeing the dominating function

Hmm.

I guess I was thinking continuous.

this seems as an unfortunate designation

somehow misleading

Hey chat

re @Astyx

heya a @Balarka

Hi @Ted!

Sorry about being picky with your answer the other day.

No, thanks! I was careless!

@TedShifrin I had to teach some kids about curvature for an online camp

Kids? Curvature of curves?

Yeah high schoolers. Nah, curvature in general. I gave two talks, first was Gaussian curvature and the second was angle defect and combinatorial Gauss-Bonnet

Do you want to see my slides

Oh, well, curvature of surfaces isn't quite like "in general." :)

You want to email me?

Haha yeah I suppose. Sure, I'll email it to you

Sent, @Ted

Yikes. 71 slides in just the first one. I'll look later.

Must have taken you lots of time drawing all the pictures.

It did take some time to prepare, yeah.

Tried to give them a coherent story

Hi Balarka

@Balarka: You gave a formula for the unit normal but "forgot" to say the surface was a level surface of $f$. :P

That's my definition of a surface

Oh, OK, so then the Möbius strip isn't a surface.

So no fair giving them a homework assignment about a Möbius strip :P

(Even the infinite, closed Möbius strip. :P)

Yeah I just wanted one of them to ask the question "which direction does the normal point towards?" which one of them did

So you didn't show them the pseudosphere?

I did!

Oh, there it is.

So the only thing missing is curvature of curves and what it has to do with principal curvatures :P

But looks like you did a nice job.

Thanks. I referred them to your notes.

LOL, that's cheating.

Told them once they have some working knowledge of multicalc that's where it all is

@Ted If you have a photoluminescent surface in $\Bbb R^3$ you should be able to understand the curvature from seeing the light spots and dark spots on the glowing bright surface, right? Positive curvature points are where there are focal points of multiplicity 0 or 2, and negative curvature points are where there are focal points of multiplicity 1

So elliptic points will look either very bright or very dark, and hyperbolic points will be something in between, maybe?

Mathematica actually does do shading by curvature, if you ask it to. But I'm confused. The focal points are outside the surface, in general, so I'm not quite sure how you're doing this.

So my thought is the light rays are coming out normally to the surface, and if there are principal directions with negative principal curvatures a bunch of those nearby normals will intersect outside the surface at a focal point

So the light rays get interfered, which will create a dark spot

Something something

I don't know physics

So you're not shining light from some point and letting it reflect off the surface.

Ah no I am imagining the surface as a glowing source of light

Ah, OK.

But most surfaces have only a finite number of focal points (singular points of the normal bundle map).

Hm, yeah, so what do I mean.

Brb gonna make a saddle bulb

LOL

I've never worked out what the focal point(s) of a saddle are.

Maybe there is a curve in that case.

Yeah that makes sense

It's not a pleasant computation to do.

Evening

hi Edward

Hey @TedShifrin :)

What's the situation looking like in the US at the moment?

Not very good, tbh

Besides totally incompetent leadership?

Well, that's a given

Hi Captain

Hey, Ted

We're gonna get another update from the government on the 19th or 20th, the university in Heidelberg has provisionally said that lectures etc. can resume from the 20th but I find that hard to believe

They haven't gone online yet?

@Ted So one of the high schoolers asked me a question which is too hard for me

the semester hasn't started yet lol

Might be too hard altogether :P

I mentioned Hilbert's theorem that $\Bbb H^2$ has no smooth isometric embedded in $\Bbb R^3$. He asked me if it embeds in $\Bbb R^n$ for $n > 3$. I know two things (1) If I plug the bounds in Nash's theorem I get there's a $C^\infty$-isometric embedding in $\Bbb R^{51}$ (2) The Klein disk model is a short embedding $\Bbb H^2 \to \Bbb R^3$ so running the Nash h-principle gives me a $C^1$ isometric embedding $\Bbb H^2 \to \Bbb R^3$.

What is the minimal $n$ such that $\Bbb H^2$ has a $C^\infty$-isometric embedding in $\Bbb R^n$? :P

No goddamn clue

Robert Bryant doubtless has a good answer.

Of course, if we allow the Lorentz metric, it's easy.

Yeah

I wonder if I should tell them the hyperboloid model. It's the cleanest description of $\Bbb H^2$ that I have learnt to be honest

But then maybe not

Takes some work to set it up

Is the student's question local or global? Obviously global.

Yeah global. I should emphasize that the pseudosphere is a local answer to him maybe

Thought I made that point in the slides

Yeah, I was gonna say that ...

Nash-Kuiper says there is a $C^1$ embedding into $\Bbb R^3$, though.

But no $C^2$ one.

Yeah, that's my point (2).

Yup

I should carefully work out the details of Hilbert's theorem once again. It uses the curvature to bound the volume of a candidate embedding of H^2 in R^3, and thus arrive at a contradiction, because H^2 has infinite volume, iirc

Aha. Moishe Kohan says it embeds in $\Bbb R^6$ and immerses in $\Bbb R^5$. $\Bbb R^4$ unknown. Here.

Oh wow

The proof in that paper seems very short

I didn't look.

Maybe I'll read it and rant about it here afterwards

It was a joy to talk to the kids. They're also running a student colloquium where some students give 10 minute talks, and I learnt a smart proof of the fact that square roots of distinct squarefree numbers are linearly independent which doesn't use field/Galois theory.

Some infinitely clever quadratic reciprocity trick...

Sounds like a bunch of kids like you were in high school.

A little scarier, because they are Olympiad guys

Neat. Well, still fun for you to talk/mentor a little/be challenged.

Yeah.

OK, I will go and watch another Twin Peaks episode. Maybe study some Riemannian geometry afterwards

See ya, @Ted and everyone

See ya, a @Balarka.

where does one use the notion of surjectiveness in showing that the quotient topology is a topology/

?