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5:03 PM
@Ryan Let $X$ be a normed space. $C\subseteq X$ is called a set of uniqueness if for every $x\in X$ there is at most one $c\in C$ with $d(x,C)=d(x,c)$. $C\subseteq X$ is called a set of existence if for every $x\in X$ there is some $c\in C$ with $d(x,c)=d(x,C)$. $C\subseteq X$ is called a Chebyshev set if it is bot a set of uniqueness and a set of existence.
Let $X$ be a normed space, then $X$ is strictly convex iff every nonempty convex set in $X$ is a set of uniqueness iff every nonempty closed convex set in $X$ is a set of uniqueness

Sup there, I invite all of you to math chat with built-in LaTeX to talk about various branch of mathematics, atm we're talkin here about graph theory and doin some proofs
https://hack.chat/?math
It's completely anonymous chat without registration, come and visit

Ciao @Alessandro

@Secret what are your thoughts on consciousness? Do you think matter and mind are equivalent representations of reality and there is some isomorphism that links the two representations of reality

A normed space is strictly convex and reflexive iff every nonempty closed convex set in $X$ is a Chebyshev set
Oh and reflexive iff every closed convex subset is a set of existence of course
See Megginson's "An Introduction to Banach Space Theory", 5.1.17, 5.1.18 and 5.1.19 @RyanUnger
Ciao @Mathei

Hai visto il mio messaggio? Il post di blog è pubblicato

5:08 PM
Sì, ma non lo ho ancora letto

Howdy @Mathein and demonic @Alessandro

Hi @Ted

Hey @Ted
How are you doing?

Doing fine, thanks, just trying to get organized to move in a month. You doing better?

yes, I'm no longer at a hospital

5:13 PM

Where are you moving to?

Oh, just a few miles.

HI TED

But that doesn't make it so much easier. :P
hi @anakhro

Hey Ted!

5:14 PM
ÍgjøgnumMeg praised my prose, which is nice since he's a native speaker

prose in what language?

English, on my blog

hi @F.White

You are probably better at English than I am @MatheinBoulomenos

@TedShifrin how to write the exterior derivative on a sphere in cartesian coordinates $x,y,z$?
Or is that not easy

5:15 PM
Yes, @Mathein, your English is better than most Americans'
@anakhro: I don't know what you're looking for. It's just the restriction of the $\Bbb R^3$ exterior derivative ($d$ commutes with pullback).

Thanks, that makes me less self-conscious about not being anglophone

You just want to take a parametrisation:
$$\Phi(\theta,\phi)=(\cos(\theta)\sin(\phi),\sin(\theta)\sin(\phi),\cos(\phi)),$$
and then you want to compute $\Phi^*(d\omega)$?

@AlessandroCodenotti I’ll take a look
@F.White do you still need help

I wrote it down a slightly more readable form as an answer to my MSE question

@RyanUnger Not with that now I guess. I just experienced that computing the christoffel symbols and then the components of the levi-civita connection, and then the riemann curvature tensor in that order isn't so bad
And I found using the Koszul formula painful

5:20 PM
As I said, @F.White, I don't know what @anakhro is looking for. If we have the $1$-form as the restriction of a $1$-form on $\Bbb R^3$ in the first place, then we're just working in the situation where $x\,dx+y\,dy+z\,dz=0$ and this relation can allow us to simplify.

I have an area form $\Omega$ and a vector field $u$ on $S^2$ (everything written in Cartesian coordinates), and I want to compute $d(\iota_u\Omega)$. The part with $d$ of course doesn't work when I treat it just like the exterior derivative in R^3 (thanks Balarka) and so I need to either convert to spherical coordinates, or figure out a way to calculate the exterior derivative of $\iota_u\Omega$ on $S^2$ without this change.

What do you mean of course doesn't work?

Well...I don't get the right answer. But if I understand your last message correctly, maybe using that relation would help...

How are you given the vector field $u$?

@AlessandroCodenotti cool

5:22 PM
Are you sure you don't have "the right answer"?

Well it disagrees with the textbook.
$u := (xz-y)\frac{\partial}{\partial x} + (yz+x)\frac{\partial}{\partial y} - (x^2+y^2)\frac{\partial}{\partial z}$
$\Omega := x\,dy\wedge dz + y\,dz\wedge dx + z\, dx\wedge dy$

And how do we know $u$ is a vector field on the sphere?

It's just said that it is.

Well, I'm asking you what you would do to check.
At any rate, did you do the interior product correctly?

Not to bombard you with questions Ted (although I am)
If we let $S$ be the surface given by rotating the curve $(f(v),0,g(v))$ for $v_0<v<v_1$ around the $z$-axis, and we assume the parametrisation is an immersion:
$$\Phi:U\subset \Bbb R^2\to S\subset \Bbb R^3,\qquad \Phi(u,v)=(f(v)\cos(u),f(v)\sin(u),g(v)),$$

where $U=\{(u,v)\in\Bbb R^2\mid u_0<u<u_1,\quad v_0<v<v_1\}$

Then one can deduce the pulled back metric on this manifold is:

$$g=f(v)^2du^2+(f'(v)^2+g'(v)^2)dv^2$$

is there a smart way to compute the Gaussian curvature?

5:25 PM
I can do it easily using moving frames, @F.White, but I assume you don't know them.

So far I've got $R(\partial_1,\partial_2)\partial_3=\frac{f(v)f'(v)g'(v)g''(v)+g'(v)^2(-f'(v)^2-f(v)f''(v)+f'(v))}{f^2(v)(f'(v)^2+g'(v)^2)}$

There's an easy formula when the first fundamental form is diagonal (i.e., no $du\,dv$ term). See my undergraduate textbook (free).

Moving frames as in using parallel transport?

No, moving frames as in using differential forms to do everything.

Ahh I'll have to look at that, the only procedure I know at the moment is to compute it as sectional curvature:
$$\frac{\langle R(\partial_1,\partial_2)\partial_2,\partial_1\rangle}{\|\partial_1\wedge\partial_2\|}$$
Which seems super cumbersome

5:28 PM
You want the formula for $R$ in terms of Christoffel symbols when $g_{12}=0$. It's not that bad.
Usually one assumes the curve is arclength parametrized, and that makes things neater.

I just computed by hand $R(\partial_1,\partial_2)\partial_2 = \nabla_1\nabla_2\partial_2 - \nabla_2\nabla_1\partial_2$ in terms of christoffel symbols, so I suppose implicitly I have it in terms of christoffel symbols
It's takes a little bit over a page though, which would suck in a short exam
I guess I probably wouldn't be given such an arbitrary surface though in an exam (hopefully :))

@TedShifrin I'd probably apply the vector field $u$ to $f(x,y,z) = x^2 + y^2 + z^2$. And yes, I am fairly certain that I calculated the interior product correctly.

OK. There's a nice formula, as I said, which you should check once in your life. (I always assigned that mess of algebra to my undergrad class to turn in.) It is $$K=-\frac1{2\sqrt{EG}}\Big((G/\sqrt{EG})_u+(E/\sqrt{EG})_v\Big),$$ where $E=g_{11}$ and $G=g_{22}$.

oof

Easier, @anakhro, check that the vector is orthogonal to the position vector.

5:32 PM
That would definitely be easier.

So I worked out the interior product. It's sorta messy, but taking $d$ won't be that bad. What did you get?

@TedShifrin I remember using this in one of my exams. The professor was shocked that I didn't take any of the two sheets he had given to find the Gaussian Curvature.

@Albas My professor would probably give me no marks for 'memorising a formula that isn't proved' :P

For example, note that $d\big((x^2+y^2)dz\big)=0$ on the sphere, @anakhro. Why, immediately?

@TedShifrin Thanks Ted, I will use this to verify I'm not messing up

5:34 PM
Moving frames wins. If you've never seen it, I have a 5-page section on it in my text/notes.

@TedShifrin I'll look it up and check it out

@F.White I mean later he did think that I had copied it from somewhere and had called me to his office and had me derive that mess to give me credits.

But my notes are completely classical for students who know only multivariable calculus and linear algebra.

Where do I find the notes?
This one?

I gave my students the relevant formulas on their exams/final, because the point is not to memorize ridiculous formulas. If you know differential forms, everything is super easy, but the classical approach is full of formulas.
What's linked in my profile, @F.White, yeah.

5:36 PM
@TedShifrin I do know differential forms, although I'm not sure how you're using them here yet

Well, if you take a gander at section 3.3, you'll see.

Yeah, it is. Sadly we have some people here who still believe in making us memorize. One had asked us to memorize all kinds of wonky probability distributions

@anakhro: If you use the equation of the sphere several times, that form simplifies enormously.

Sadly I deleted my exterior derivative, but for the interior product I got that it is
\begin{align*}\iota_u\Omega&=x((yz+x)\,dz + (x^2+y^2)\,dy)\\&\phantom{{}=1} + y(-(x^2 + y^2)\,dx - (xz-y)\,dz)\\&\phantom{{}=1} + z((xz-y)\,dy - (yz+x)\,dx)\\ &= -(x^2y + y^3 + yz^2 + xz)\,dx\\&\phantom{{}=1} + (x^3 + xy^2 + xz^2-yz)\,dy\\&\phantom{{}=1} + (x^2 +y^2)\,dz,\end{align*}

@Albas: I made my students know definitions of things and basic formulas that one just reasons out, but not the ridiculous complicated ones.
Simplify that before going any further, @anakhro.

5:38 PM
With respect to the relation you gave?

x^2 + y^2 + z^2 = 1 stuff?

Of course.

@anakhro x^2+y^2 =

Great, I will do that. Thanks.
@F.White Yes, Ted gave that hint with his exterior derivative.

5:39 PM
@TedShifrin That's nice. We were asked to remember things like the beta and gamma distributions, zipf and I dont know what else.

Why it vanishes under $d$.
$d(1-z^2) = 0$, si

@Albas Good to memorise the first two antiderivatives of the error function

@anakhro: The answer is that $d$ of it is zero.

Yes.
$-(y + xz)\,dx + (x-yz)\,dy + (1-z^2)\,dz$

OK, good. Now taking $d$ is very quick.
Oh wait. I didn't get $0$. I get $2\,dx\wedge dy$.

5:43 PM
Oh you meant the answer to the entire exterior derivative?

Yes.

Let me see....
$2\,dx\wedge dy + x\,dx\wedge dz + y\,dy\wedge dz$
That's what I got.
Which agrees with the book.
\begin{align*} d(\iota_u\Omega) &= d((-y - xz)\,dx + (x-yz)\,dy + (1-z^2)\,dz,)\\ &= -dy\wedge dx -d(xz)\wedge dx + dx\wedge dy - d(yz)\wedge dy + d(1-z^2)\wedge dz\\ &= 2\,dx\wedge dy + dx\wedge(z\,dx+x\,dz) + dy\wedge(z\,dy + y\,dz)\\ &= 2\,dx\wedge dy + x\,dx\wedge dz + y\,dy\wedge dz \end{align*}

The last two terms disappear.

Why is that?

What is $x\,dx+y\,dy$?

5:55 PM
$-z\,dz$
Si
Ted is sharper than Mr. Geiges.

Of course :)

I better go, thanks for before @TedShifrin bye

bubye

Okay, so Geiges then wants to conclude that the divergence is positive/negative on the upper/lower hemisphere.
He did not eliminate those terms, but he then wedges his result with those additional two terms with $(x\,dx +y\,dy +z\,dz)$.
But that's 0 isn't it? Or what is he even trying to do with that, do you think?
$(\text{div}_\Omega u)\Omega = d(\iota_u\Omega)$

sorry busy for now

6:03 PM
That's fine, thanks for your help, Ted!
I will think about it some more now that I have replicated it.

6:17 PM
Hi guys, Consider a function $f$ ($\mathbb{R}\to \mathbb{R}$) that has an Oblique asymptote in $y=2x-1$. I want to find the limit $\lim_{x\to\infty}\sin\left(\frac{2}{x}\right)f(x)$. How can I do it?
I know from the Oblique asymptote definition that $lim_{x\to\infty} (f(x)-2x)=-1$, but how does it help me?

6:29 PM
What happens if you multiply your equation by $\sin(2/x)$?

clever

@MatheinBoulomenos Mm OK. Interesting, I thought that argument was just me being dumb. Thanks!

Hi a @Balarka

using spectral sequences for group cohomology is not that rare, we used them in our ANT classes as well

Hi @Ted!

6:46 PM
b a l a r k a

Hey @anakhro

Hey
I need some quick complex maths help
exam tomorrow... can someone verify that

I verify what
enjoy

$\frac{2R}{1+iwC2R}+iwL$ comes out to, if wL=3R and 1/wC = 5R

I wish I knew how to verify what

6:52 PM
and i = imaginary unit
So far, I'm thinking its 1+0,86i

what is going on

Hey everyone!

hi Pertur b

Hi @Perturbative

Hey @TedShifrin and @MatheinBoulomenos

6:57 PM
I am heavily confused and need help

Quick question, what's the usually recommended text to learn about spectral sequences? I know of a few, Lecture Notes in Algebraic Topology by Kirk and Davis, Hatcher's incomplete SSAT, A User's Guide to Spectral Sequences and Weibel's Homological Algebra

Hatcher's SSAT is very good and where I learnt (and am learning) these monsters from.
Although really the first time I actually understood them was from Bott-Tu

Oh yeah forgot about Bott and Tu, I've heard they have a nice exposition of spectral sequences
I'll take a second look at Hatcher's SSAT

7:17 PM
@BalarkaSen Were there any computations using spectral sequences that you found particularly cool?

That's the whole point of them :) Sure, lots of them. You can compute nontrivial things about homotopy groups of spheres out of them
I recently realized you can see that $\pi_6 S^3$ has a $\Bbb Z_3$ component using a spectral sequence argument.
Classical examples are that $\pi_m S^n$ for $m > n$ are finite unless $n$ is even and $m = 2n -1$, in which case it's rank 1

Ahh I see, I think I read in Kirk and Davis' notes that you could also use them to give quick proofs of some big theorems, in fact, I think I remember seeing somewhere that you can prove the Thom isomorphism theorem with a certain spectral sequence

Requires more than spectral sequences technically speaking, you need the rationalization construction, but essentially spectral sequence arguments.
You also need Serre's theorem but that's also essentially a spectral sequence argument.
@Perturbative Ya ok but Thom isomorphism theorem is very easy to see by hand
if you're looking for lo-fi computations, a quick corollary of spectral sequences is the Gysin sequence, with which you can prove groups acting freely on spheres have periodic cohomology
They're slowly becoming essential to my toolbox, although I still need a lot of practice to get used to them

7:46 PM
Oh I didn't know Thom iso was easy to see by hand (haven't seen a proof of it yet other than the sketch using spectral sequences)

If $E$ is an orientable rank $k$ vector bundle on $M$ then the map $H^{n+k}_{vc}(E) \to H^n(M)$ from the vertically supported cohomology of $E$ to the base is given by evaluating on the fiberwise orientation class, basically
If you're working deRham, it's integration along the fibers

How's it going, @Balarka?

Hi @AlexWertheim!

Not too much! Things moving along. Writing a paper

Nice

7:57 PM
What's new with you? Seems you're a university student now!

Yeah, finished my first year
Vacations going on

Oh, nice! How are you liking university life?

Meh :P

Lmao

'sprettybad. But whatever, I get to do math. It's something.

8:02 PM
Oh damn, I'm sorry to hear that. I'm glad the math is good though. What are you thinking about mathematically these days?

A disjoint number of things. I was reading about h-principles and stratified spaces but got sidetracked and started learning about localizations of spaces etc

Neat! Still big into geometry, then :)

Haha yeah but mostly doing algebra for the last few days
"weeks" is a stretch
Long spectral sequence calculations, fiddling with group cohomology, etc

Good stuff

8:10 PM
Do you know what groups of multiplicative type are?

Nope, tell me

Haha, sure. Do you know what an algebraic group is?

Yep.

Great. A diagonalizable (affine) group scheme over a field F is one whose representing Hopf algebra F[H], the group algebra over F for some abelian group H. (The comultiplication, inversion and identity maps are what you'd expect they are - I can elaborate if need be)

@BalarkaSen for 3-manifolds, trivial normal bundle <--> orientable manifold?
<-- is true, si
but --> ?

8:14 PM
A (n affine) group of multiplicative type is a group scheme which is diagonalizable when one extends scalars to the separable closure of F

@anakhro trivial normal bundle of what in what?

Sorry, of a circle
knot
si
I thought I said that.

still weirdly parsed. You mean every embedded circle has trivial normal bundle? Yeah that means it's orientable.

A beautiful fact is that there is an equivalence of categories between groups of MT over F and abelian groups with a continuous action (by homomorphisms) of Gal(F_sep/F)

The circle is orientable

8:17 PM
@AlexWertheim Oh, interesting

Given a group G of MT, the corresponding abelian group is the character group G* of G, i.e. the group of group scheme homomorphisms between G and G_m

@BalarkaSen sorry, I am confusing. I want a trivial normal bundle for a knot K. It's enough to assume that K is orientable?
Or are you saying it is enough to assume M is orientable?

@anakhro What does "K is orientable" even mean? A circle is an orientable manifold. I am saying the following is true: Every knot in M has trivial normal bundle $\iff$ M is an orientable manifold (regardless of dimension)
So yes, it is enough to assume M is orientable

On the other hand, given an abelian group H with the corresponding T := Gal(F_sep/F) action, one takes the group scheme represented by the Hopf algebra (F_sep[H])^{T} (that is, take T-fixed points)

Aha

8:20 PM
@BalarkaSen Yes, okay. I just wanted to make sure what you were saying. That is, what "it's" was in your answer here.

Anyway, the goal of the paper is essentially to think about G-torsors (principal G-bundles, if you prefer) for G a group of multiplicative type

I see. I didn't quite catch how the Hopf algebra comes in the story; what's the corresponding Hopf algebra associated to a group scheme?
Oh it's ring of global sections, probably.
If it's an affine group scheme
Because the multiplication becomes the comultiplication there

Precisely. Sorry, I should have specified - all group schemes are affine in my story, haha

All my schemes are affine so no harm done
Anything beyond that is too hard for me

one (af)fine scheme

8:27 PM
fine af

Lol I get that feeling. I've spent a lot of time trying to understand the technical details of general scheme theory, and only now after years am starting to feel like I understand the basics. It is a lot of baggage

love u guyz

It's very surprising to me that people can actually do algebraic geometry, it's too difficult lmao
Working in technical fields is a pain but everything that is not technical has been done by Hilbert and Poincare
Math is hard

What's a good way to explain that you cannot canonically choose the "zero" element in the set of knot framings?
That is, the framing with "zero" twists.
(with language used above, that the framings form a $\mathbb Z$-torsor).

LOL

8:34 PM
i didn't think it was that good of a joke

I liked it
yeah so the (orthogonal) frame bundle of the normal bundle to the knot is a principal SO(2)-bundle I guess
if you had a canonical section it'd canonically trivialize
iunno something like that

But do you always have a canonical section?

Apriori the normal bundle is just a random oriented rank 2 bundle on S^1, there is no canonical trivialization of such a thing
they are always trivial but not canonically so
but justifying that would lead me to setup a functorial meaning of canonical trivialization which i really dont want to

Heh.
I agree with that intuition that it is just a random rank 2 bundle.
So I think I can agreeably say that there is no 0-framing canonically without much more discussion.
It's not really a formal statement anyway, I guess.

Right, because given any oriented rank 2 bundle you can build an oriented manifold in which has a knot with exactly that normal bundle; just take the double of it's disk bundle
@anakhro Yeah I think you can say that just fine

8:45 PM
Neat, thanks.
I need more people to talk to in real life about geometry things.
Balarka, you are one of my only geo friends. geo dudes

By geometry do you mean geometry on etale sites?
:3

I mean anything that doesn't look like dirty algebra

LOL

I love you Alex.
I have only ever taken algebra courses.
Fave one was rep theory of $S_n$ or maybe the superalgebras one.
Love algebra, I am only joking.
pls don't kill me
holds out his "I Love Algebra" pin.

8:50 PM
Lmao

Okay I am getting out of here before an analyst comes.
Bye!

9:44 PM
I could manipulate you

9:58 PM
Hi , how do I show that z^4 -z^3 -4z +1 has no roots for |z|>=2 ? Sorry for no latex
I know about Rouche theorem and all of that stuff
Actually, I need to show that there are exactly three roots in the area { 1 < |z| < 2 }. But it is easy to show there is only 1 root in the { |z| <= 1 } and since there is 4 in total ...

2 hours later…
11:54 PM
@Ultradark I think they probably be representing some notions of reality but I have no reason to believe they are isomorphic (otherwise we would have detected ghosts and such with science already)

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