Mathematics - 2024-02-05 (page 2 of 3)

Alessandro Terminiello

19:00

wait

Balarka Sen

Hi @Ted

Ted Shifrin

Heya, Balarka!

Alessandro Terminiello

$$2x - \frac{9x^2}{2} + \frac{73x^3}{6} + o(x^3) \approx 2x$$

~*

2x comes from here

Ted Shifrin

I'm asking what the exact problem you are doing is, without using that.

Alessandro Terminiello

i need to use that

Ted Shifrin

19:02

It is incorrect to use that approximation if you need a Taylor polynomial of degree 2 or more.

Alessandro Terminiello

i need to use this result

Balarka Sen

A strange curiosity: Given a smooth family of smooth embedded curves $C_t \subset \Bbb R^3$, can you choose a smooth family of smooth embedded minimal surfaces $D_t \subset \Bbb R^3$ bounding $C_t$?

Ted Shifrin

No, you do not.

Alessandro Terminiello

ok I guess it can't be done then

Ted Shifrin

@Balarka I actually have no idea.

Alessandro Terminiello

19:04

$$2x - \frac{9x^2}{2} + \frac{73x^3}{6} + o(x^3) \approx 2x$$

Ted Shifrin

Alessandro. Stop.

Alessandro Terminiello

but is correct said that right?

Ted Shifrin

It depends on what error is allowed when you write $\approx$.

Alessandro Terminiello

no I mean is it right to say that 2x is asymptotic to that result?

~*

i dont know how to insert this ~

Ted Shifrin

It is too sloppy a statement. I will not agree with it.

Alessandro Terminiello

19:06

and what could I do?

Ted Shifrin

$2x-\frac{9x^2}2 = 2x + o(x^2)$, but that approximation may be useless.

You refuse to tell me the actual question as it was assigned.

Alessandro Terminiello

was to solve this exercise:

wait

$$\log(1+3x) \cdot e^{x^2} - \sin(x)$$

and i did

I get those numbers I sent earlier

Ted Shifrin

When you're doing the limit of $\frac{f(x)}{\log(1+x)-\tan x}$, WHAT precisely is $f(x)$?

Balarka Sen

@TedShifrin I was curious because it is not too hard to write a discontinuous family of smooth embedded minimal surfaces $D_t$ bounding a smooth family of smooth embedded minimal curves. (Think of, I guess, a pair of coaxial circles -- if they're too far you fill each by a planar disk, if they're too close you fill by a catenoid)

Alessandro Terminiello

f(x) is the third order polynomial of the above function

Balarka Sen

19:09

But all these examples seem to involve passing through a time $t = t_0$ where the curve has some extra symmetries making it bound two distinct minimal surfaces, and for $t < t_0$ and $t > t_0$ patching the "two different branches" of families of surfaces.

Alessandro Terminiello

$$\lim_{{x \to 0^+}} \frac{f(x)}{\log(1+x) - \tan(x)}$$

its also with lim

Ted Shifrin

Yeah, Balarka, I was thinking about my exercise with two minimal (not both area-minimizing) catenoids with certain boundary circles. And as you vary the boundary circles continuous, there is some interesting behavior.

Balarka Sen

Aha.

Ted Shifrin

@Alessandro, then you must use the third-order polynomial, not first-order.

Alessandro Terminiello

okk, so I write the whole result, without saying asymptotic to 2x?

Ted Shifrin

19:13

What is the first nonzero term appearing in the Taylor expansion of the denominator?

Alessandro Terminiello

ok wait

-x^2 / 2

log(1+x) = x - x^2/2 + o(x^2)

tan(x) = x + o(x^2)

Ted Shifrin

So the Taylor polynomial of the denominator starts with an $x^2$. That means you MUST use a Taylor polynomial of degree/order $2$ or more for the numerator.

This is all explained in my handout.

Alessandro Terminiello

but my result is -4/0

that is - infinity

on wolfram I find it correct

maybe it's just a coincidence

Xander Henderson

@TedShifrin Oh? Is he back?

Alessandro Terminiello

I don't understand, I have to develop the numerator up to the second degree, but if I don't do it, why i found the correct result?

Maybe it's something I can't do

Ted Shifrin

19:22

No, Tanner brought him up.

Wolfram is finding it correct because you're giving Wolfram the WRONG question.

You did NOT find the correct result. You are so stubborn.

Xander Henderson

@TedShifrin Ah, okay. Good.

@TedShifrin As I am told VSV would say: "You are not LISTENING!"

Alessandro Terminiello

wolframalpha.com/…

Ted Shifrin

@Alessandro Truly, I am going to stop talking with you. You put in the WRONG question. That is the wrong numerator.

@Xander Who is VSV?

Xander Henderson

@TedShifrin Oh, sorry. en.wikipedia.org/wiki/Veeravalli_S._Varadarajan

Ted Shifrin

Oh, I know him by his actual name, not by his initials.

Alessandro Terminiello

19:25

but I wrote that the result was asymptotic at 2x :(

Xander Henderson

He was the advisor of a very good friend of mine. It sounds like he was a slightly abusive advisor, but also a very dear friend and mentor at the end of the day.

Ted Shifrin

@Alessandro I'm done. Reread everything I've typed. I have no more to say. Done.

@Xander I think Chern was probably the opposite ... too nice. Not that I deserved a nasty adviser.

Alessandro Terminiello

so if I understand, I will write in the numerator all the results I obtained before

I'll proceed

Xander Henderson

@TedShifrin Heh.

From what little I know, Chern seems like he was a real mensch.

Alessandro Terminiello

wolframalpha.com/…

Balarka Sen

19:28

I have heard a lot about Varadarajan but I don't understand his math at all

Alessandro Terminiello

itstill - infinity

Ted Shifrin

@Xander There are some wonderful DVDs about him.

Didn't he do representation theory? That's my recollection.

Balarka Sen

Some mix of that, probability, ...

Quantum probability something something

Xander Henderson

@TedShifrin Yes, that is probably what he is best known for. But he also did some probability theory and mathematical physics.

Ted Shifrin

OK, @Alessandro, the answer is that the limit does not exist but you need to understand that you cannot just do polynomials of whatever degree you want. If the numerator is $2x + o(x)$, then $\dfrac{2x+o(x)}{x^2+o(x^2)} = \dfrac{\frac 2x+ \frac{o(x)}{x^2}}{1+\frac{o(x^2)}{x^2}}$, and this limit becomes $\lim\dfrac{\frac2x}{1}$, which does not exist.

Jakobian

19:34

@TedShifrin oh really? That guy?

Ted Shifrin

Tanner brought him up. Not I.

Jakobian

Didn't realize this is what the paper was about

Ted Shifrin

He claimed he was solving the RH.

Alessandro Terminiello

@TedShifrin thankss! :)

Jakobian

@TedShifrin Well, the title is saying he's "drawing connections" to RH

> employing advanced mathematical tools such as the Taylor series expansion, the argument principle, and the inverse Mellin transform

though listing Taylor series as advanced baffles me a little

Ted Shifrin

19:38

Advanced mathematical tools that most undergraduate engineering students know.

Xander Henderson

@Jakobian He made more extravagant claims in chat.

Ted Shifrin

None of those is advanced.

Undergraduate applied math students learn all that.

Jakobian

Not sure what argument principle is, and Mellin transform is advanced for me at least

Ted Shifrin

Argument principle is from basic complex analysis. In every undergraduate complex analysis course.

Mellin transform is in every engineering math book.

Jakobian

um... what exactly is argument principle?

Soumik Mukherjee

19:40

Argument principle

In complex analysis, the argument principle (or Cauchy's argument principle) is a theorem relating the difference between the number of zeros and poles of a meromorphic function to a contour integral of the function's logarithmic derivative. == Formulation == If f(z) is a meromorphic function inside and on some closed contour C, and f has no zeros or poles on C, then 1 2 π i ∮ C...

Ted Shifrin

Integrating $f'/f$ around a closed curve counts zeroes - poles inside the curve. (Factor of $2\pi i$, as always in complex analysis.)

Xander Henderson

@Jakobian The argument principle is about the zeros and poles of a meromorphic function. It is usually taught in a first course on complex anal. The Mellin transform is likely a little more mysterious to most working mathematicians, but show shows up in physics and engineering a lot. It is also useful in complex anal, and shows up in the analysis of uh... Dirichlet series (is that the term I want?).

Yes... that's what I want: en.wikipedia.org/wiki/Dirichlet_series

I type too slow... :/

Ted Shifrin

The argument principle also generalizes nicely to smooth functions and higher dimensions using degree of a mapping.

Jakobian

now I'm not sure if I knew about it before or if I'm hearing it for the first time

Ted Shifrin

That was one of my favorite topics in teaching differential topology to prove the Hopf Degree Theorem (awesome theorem, Jakobian: two maps $M\to S^n$ — $M$ compact, oriented $n$-manifold — are homotopic iff they have the same degree).

Oh, and I should say (since I discussed this in a post on main recently) @Jakobian, that that holds for continuous maps, not just smooth maps.

Thorgott

19:45

ah yes, a classic calculation of cohomotopy!

Balarka Sen

My favorite application of Yoneda's lemma.

Ted Shifrin

What is, a Balarka?

I never remember what Yoneda's lemma is. I remember that it came up once in algebraic topology, but that was 1974-5.

Balarka Sen

The Hopf degree theorem for continuous maps, that is.

Ted Shifrin

Oh, how is it Yoneda?

Thorgott

I don't really see how Hopf generalizes the argument principle, to be honest

Ted Shifrin

19:48

The proof of Hopf uses the generalization. You can excise little balls around the preimage points of a regular value and get that the global degree is the sum of local degrees.

You need to argue that if the net degree is $0$, for example, then you can cancel things out in pairs.

Thorgott

that's fair

but it's not a one statement implies the other kind of deal

Ted Shifrin

I used to present this proof when I taught diff top.

I never said that.

I said "to prove the Hopf theorem"

Read carefully.

Balarka Sen

@TedShifrin It's a bit cheeky, but here goes: Let $K(\Bbb Z, n)$ denote Eilenberg-MacLane space. One may check $X \mapsto [X, K(\Bbb Z, n)]$ is an assignment which satisfies all the axioms of an ordinary cohomology theory. Therefore, there must be a natural isomorphism $[X, K(\Bbb Z, n)] \to H^n(X; \Bbb Z)$ as there is only one ordinary cohomology theory upto isomorphism (Eilenberg-Steenrod). What Yoneda does for you in these situations is tell you what the isomorphism explicitly is.

Ted Shifrin

I'm confused. A sphere is not a $K(\Bbb Z,n)$.

Balarka Sen

Ah, but $[M, S^n] = [M, K(\Bbb Z, n)]$!

If $M$ is an $n$-dimensional closed manifold.

Ted Shifrin

19:52

Oh, cellular approximation blah blah.

Balarka Sen

Yup

Ted Shifrin

In my retirement, I'm forgetting more than I knew.

Thorgott

this is unironically how I think about this result

shame on me, I know

though I prefer constructing an explicit isomorphism $[X,K(\mathbb{Z},n)]\rightarrow H^n(X;\mathbb{Z})$ to the axiomatic approach

I learned a lot working that one out

Ted Shifrin

Yeah, I hate axiomatic bullshit.

Balarka Sen

As I said, it's very cheeky.

Ted Shifrin

19:56

Well, this was all very educational :) All because of the "advanced" argument principle.

Balarka Sen

Maybe the OP can find use of this jacked up version in their "proof" of Riemann Hypothesis

We sincerely hope, yeah?

Thorgott

well, neither the fact that the Eilenberg-MacLane spectrum represents ordinary cohomology nor the cellular approximation are quite easy

so this "cheeky" method still requires genuine effort

Ted Shifrin

mumble mumble Postnikov towers ...

That, too, goes back to 1975.

leslie townes

everything is "just" the yoneda lemma

Ted Shifrin

Cellular approximation at least is quite geometric.

Balarka Sen

19:59

I love the beautiful PL topology proof of cellular approximation in Hatcher.

Ted Shifrin

@Balarka Do you sorta get the relevance of my remarks about the catenoids?

Balarka Sen

Yeah, I did. I'll have to see what happens with the two catenoids to see if I can make use of that.

I can find this in your book, yes?

Ted Shifrin

This was exercise 16 on p. 43-44.

Balarka Sen

Thanks!

Ted Shifrin

I can send you a Mathematica graph of the answer, if you need.

Balarka Sen

20:02

Maybe I will fiddle with the exercise a bit first before asking :)

Ted Shifrin

I think the fact that there are unstable and non-minimizing minimal surfaces makes these things hard/subtle, but the experts certainly know the answer to your question.

Balarka Sen

I love going back to your diffgeo book/lecture notes, btw. I had to go back and re-learn the Crofton formula a while ago.

Ted Shifrin

I'm actually proud of a number of exercises in there that appear nowhere else.

Balarka Sen

Did I tell you Donaldson and Gromov use the Crofton formula left right and center to prove surprising results in symplectic geometry?

I was amazed

Ted Shifrin

Crofton shows up all over in various guises. Griffiths's paper (with a serious error) on Milnor numbers and singularities was based on versions of Crofton ... and led to my Ph.D. thesis, too.

Balarka Sen

20:04

It's so weird! It shows up in places where it has no business being.

Ted Shifrin

Integral geometry is boss.

Balarka Sen

Haha

Ted Shifrin

It's all about flag manifolds and integration over the fiber :)

Thorgott

@BalarkaSen I've always found that one unreadable

Ted Shifrin

pops popcorn and waits for a fight

Balarka Sen

20:09

How could you?! jk

Thorgott

I think Balarka and I have mellowed out too much to argue over such a small thing

Ted Shifrin

That's a disappointment :D

Balarka Sen

It's true, alas.

Thorgott

still haven't read a proof of Blakers-Massey because they all are too PL for me

Balarka Sen

Oh yeah I love that one too

Thorgott

20:11

perhaps I should learn it from Brown :P

Balarka Sen

Oh no!

Ted Shifrin

This geometric topology stuff is all stuff I know zero about.

But at least I know how to prove Gauss Bonnet by seeing that the Euler class is Poincaré dual to a particularly important Schubert cycle in the oriented Grassmannian :D

That was a proof I never found written down anywhere.

Balarka Sen

I learnt a little bit of Schubert calculus once but never quite got proficient at it.

Ted Shifrin

I'm a fan ... and a fan of doing duality by explicit moving-frames integral computations with curvature forms. Who cares about torsion, anyway.

Those ideas were all in Chern's original papers from the 50s.

Balarka, I hope you're teaching some of your cohorts the beauties of moving frames :P

Balarka Sen

I am not nearly as skilled in it, but I am teaching a bunch of people about pseudoholomorphic curves and almost all computations I do in moving frames.

It is a nightmare to prove Weitzenbock formula without moving frames, say.

Ted Shifrin

20:18

I should have taught you better :)

Yeah, but Weitzenbock is a pointwise thing. That's just choosing a frame so that the connection vanishes at a point.

Balarka Sen

Or rather I should have picked some of the moving frames to characteristic class viewpoint up from your course notes.

Thorgott

I've learned Schubert calculus, but sadly it's not that useful from the AlgTop perspective

Balarka Sen

@TedShifrin Right, it's not really a serious expositional choice.

Ted Shifrin

No, it's a lot more meaningful in algebraic geometry and singularity theory.

It showed up all over in my integral geometry work, too, but that was a long time ago.

Balarka Sen

I got too Gromov-pilled and stopped doing differential geometry computations to try to understand differential geometry better.

He never does computations, maddening how he still understands.

Ted Shifrin

20:22

He's smarter than we are.

Balarka Sen

True

Ted Shifrin

And Chern could do a 3-page moving frames computation in his head (like his original Gauss Bonnet proof). He just knew what would happen.

Balarka Sen

Nuts

Alessandro Terminiello

$$\sum_{{n=1}}^{\infty} (-1)^n \cdot \log(1+\sin(\sqrt{n+1}-\sqrt{n}))$$

Balarka Sen

It's so easy to mess up. Takes skill

Alessandro Terminiello

20:23

How do I understand if I can use Leibniz here?

Balarka Sen

I was reading an old paper of Kronheimer-Mrowka recently and found it riddled with errors

Alessandro Terminiello

the idea is the series has a constant sign

Mr. Feynman

@TedShifrin I can't do a 3-page moving frames computation in 10 pages :P

Alessandro Terminiello

I made the limit of the sin, and it tends to 0, so the log tends to a positive quantity

Thorgott

I kinda miss doing differential geometry computations

Ted Shifrin

20:25

I do moving frames computations quite painlessly. And if someone writes down a page of covariant derivative symbols, I just tune out.

Thorgott

it's the type of thing you start being very bad at if you don't do it very regularly

Alessandro Terminiello

can someone confirm my idea?

Ted Shifrin

@Alessandro You need to know how it tends to $0$. Is it $o(1)$, $o(1/n)$, $o(1/n^2)$ ... ?

Alessandro Terminiello

what is o

Ted Shifrin

Oh, I guess with the $(-1)^n$, you just need to know that the terms decrease in size.

You were writing o in your work earlier.

Mr. Feynman

20:26

@TedShifrin I'm fine with it as long as it's not in components. If I can't do something coordinate-free I have the feeling I cannot do it

And I assure that is painful when you are a physicist :P

Ted Shifrin

Moving frames is not about components at all.

Physicists do everything in coordinates and with tortured tensor notation. I can't deal with it. They hate differential forms.

Alessandro Terminiello

but don't I have to see if Leibniz can be used or not?

Balarka Sen

You don't come across a physicist every day in the wild saying "If I can't do something coordinate-free I have the feeling I cannot do it"

Mr. Feynman

Yes yes, I meant that I'm fine with covariant derivatives as long as they're not written by components

Ted Shifrin

Yes, it is about Leibniz. What do you need for that?

Mr. Feynman

20:28

@TedShifrin It depends on the level, though. More advanced courses use differential forms more afaik

Ted Shifrin

Most physicists hate them. I have it on good authority from close friends who are physicists.

Alessandro Terminiello

the teacher told me that I have to do a check on the logarithm, but it's definitely not that decreasing thing

Balarka Sen

I came across a meme clip a while ago where a guy was doing index computations and ran out of boardspace so he started saying it out loud and it went "mew new mew mew plus new mew new ..."

Sine of the Time

@AlessandroTerminiello you can rationalize and the use Taylor

Thorgott

physics exposition tends to be painfully coordinate-dependent

Balarka Sen

20:30

@Thorgott mew new mew mew plus new mew new new

Thorgott

I remember having to do write down the geodesic equation in coordinates in like a first- or second-semester physics course

Alessandro Terminiello

@SineoftheTime I have to use leibniz

Mr. Feynman

@TedShifrin I see. Are they possibly more into General Relativistic stuff? I don't expect people who deal with gauge theory to hate differential forms, for example

Thorgott

@BalarkaSen lol

Alessandro Terminiello

@SineoftheTime but I should first see if it can be used

I know that i need to set the limit of a_n which tends to +infinity and which must be equal to 0. Then I have to verify that the sequence a_n decreases

right?

Balarka Sen

20:31

@Mr.Feynman I know people in gauge theory who write nothing in differential forms

$F_{\mu \nu} F^{\mu \nu}$ all the way

Mr. Feynman

@Thorgott Imagine the pain I have to bear now doing GR, where I write it as $k^\mu k_{\nu;\mu}=0$

Balarka Sen

And then theres abomination like $A_{\mu;[\nu} B_{\rho}^\nu]$

Alessandro Terminiello

but first I should do log (of the argument) > 0 , but I don't know why

Mr. Feynman

@BalarkaSen ew

Sine of the Time

@AlessandroTerminiello $\log(1+\sin (\sqrt{n+1}-\sqrt n))=\log(1+\sin(\frac1{\sqrt{n+1}+\sqrt n}))\sim \log (1+\frac1{\sqrt{n+1}+\sqrt n}) \sim \frac1{\sqrt{n+1}+\sqrt n}\sim \frac1{2\sqrt n}$. Hence your original series is asymptotically equivalent to $\sum \frac {(-1)^n}{2\sqrt n}$

Mr. Feynman

20:35

But to be honest I got used to work with indices too, although I don't plan to keep doing it forever. The problem is when it's difficult to translate something that only seems a notation game into real differential geometry

Alessandro Terminiello

@SineoftheTime ok but

but so the step of seeing if the log is > 0 is useless?

Mr. Feynman

The thing I'm having most trouble with in GR are lightlike hypersurfaces (i.e. hypersurfaces on which the restriction of the metric tensor is zero). Do you happen to know any differential geometry book discussing those?

Being "generated" by geodesic, introducing a notion of integration via a "transverse" metric are all things that I don't find to be compelling reading GR books

leslie townes

alessandro, the key is that the log thing [eventually] has constant sign. it isn't so important whether it's [eventually] positive or [eventually] negative, because a sign difference like that would not affect convergence or the application of the test.

Alessandro Terminiello

@SineoftheTime

Mr. Feynman

Sorry **on which the normal is null (the metric tensor is *not zero)

Sine of the Time

20:40

@AlessandroTerminiello it depends on how do you want to solve it. Comparison test and then Leibniz seems the fastest route

Alessandro Terminiello

@leslietownes uh ok

@SineoftheTime ok but so log > 0 is useless?

what method would that be?

leslie townes

it sounds like you're working through someone else's solution? it might help to ask that person. they would maybe be more familiar with what you have and haven't seen, or are "allowed to use"

Sine of the Time

by comparison test, you need to apply Leibniz to $\frac1{\sqrt n}$

Alessandro Terminiello

ok, what if i resolve it as you are saying? How should i do

Sine of the Time

is $\sum \frac{(-1)^n}{\sqrt n}$ convergent?

Alessandro Terminiello

20:45

no?

Sine of the Time

why?

Alessandro Terminiello

i have to use leibniz

i read bad before

its -1

i read 1

sorry

yep

it converges for leibniz

Sine of the Time

correct

Alessandro Terminiello

because 1) lim a_n tends to 0 , 2) a_n is decresing

Sine of the Time

what can you say about the original problem now?

Alessandro Terminiello

20:50

the key is that the log thing [eventually] has constant sign. it isn't so important whether it's [eventually] positive or [eventually] negative, because a sign difference like that would not affect convergence or the application of the test.

what this mean

@leslietownes

@SineoftheTime the series converges by asymptotic comparison

Sine of the Time

yes

leslie townes

it sounds like you were working through a solution where someone is checking the hypotheses of a theorem about series of the form (-1)^n b_n or (-1)^{n+1} b_n. checking that the sequence b_n has eventually constant sign would be part of applying this kind of theorem.

Alessandro Terminiello

yes it can be

leslie townes

if you don't know at all why someone would be worried about whether b_n would be positive in analyzing a series of the form sum (-1)^n b_n, i suggest asking whoever presented what you describe as " the step of seeing if the log is > 0" why that is worth doing or necessary

it sounds like this "step" was not your idea

i'm hesitant to suggest more detail because i don't know what techniques you are or are not familiar with

Alessandro Terminiello

in reality, the teacher told me something like this

leslie townes

20:55

there are really two questions, is the series convergent (it seems like the answer is yes), and how can we see that the series is convergent using tests that are known to you. the second part is trickier than the first

the reason books have hypothesis like "b_n > 0" in theorems about sum (-1)^n b_n is that if the sequence b_n does not have constant sign, you may have no control over what the sign of (-1)^n b_n is, and in particular, sum (-1)^n b_n might not be an "alternating series"

for a theorem like that, it isn't so important that b_n be always positive, or even eventually positive. it is important that b_n eventually not change sign

so that (-1)^n b_n is something that alternates in sign

Alessandro Terminiello

so leibniz I can always use it if I have a (-1)^n * something?

leslie townes

that theorem likely has other hypotheses that might be difficult to verify in this case. a standard way of showing that a sequence b_n of positive numbers is eventually decreasing is to find a function of a real variable f(x) with b_n = f(n) and to use calculus to show that the sign of f'(x) is eventually negative

i would start on an easier example of a problem of this type before doing this one

Alessandro Terminiello

ok thank you

Ted Shifrin

@Mr.Feynman That's more mathematicians than physicists. My impression is that the truly mathematical theoretical physicists form a very small percentage of physicists.

monoidaltransform

If $G$ is a compact Lie group with biinvariant metric $g$ and $H$ is a closed subgroup, then $H$ is an embedded submanifold of $G$ and so $i^{*}g$ restricted to $H$ is bi-invariant too. Now my question is, and this might be stupid, if $d_{g}$ is the induced Riemannian distance on $G$ does $d_{g}$ and $d_{i^*g}$ coincide on $H$?

Ted Shifrin

21:05

@leslie Interesting point. If $a_n = b_n + O(1/n^2)$ and $b_n>0$ decreases, then Leibniz should work for $\sum (-1)^n a_n$. But I do not recall discussing this in a class.

Mr. Feynman

@TedShifrin That is indeed true. It's what I aim to be nonetheless

Ted Shifrin

@monoidal As with any Riemannian manifold/submanifold, iff $H$ is totally geodesic.

I.e., the second fundamental form vanishes.

You can work that out all in terms of the Lie algebra information.

monoidaltransform

If $H$ is connected, then its totally geodesic?

leslie townes

ted: yes, one thing we are running into here is that none of those "alternating series test" theorems are "if and only if" and even when such a test does apply, it might not be the simplest approach. so, maybe a gap between what the chat might come up with and what a student is supposed to do using a textbook theorem as a black box

Balarka Sen

@Ted: Silly question, suppose $f : \Bbb H^2 \to \Bbb H^3$ is an isometric embedding in the Riemannian sense (pullback of induced Riemannian metric in the codomain under embedding agrees with the Riemannian metric in the domain). Does that mean image of $f$ is necessarily totally geodesic?

This is not true for $S^2$ in $\Bbb R^3$, certainly.

leslie townes

21:10

i have not seen a textbook exercise introduce the idea of applying an alternating series test to only 'part' of a convergent series, even if that's easier than analyzing the behavior of the 'full' series

Ted Shifrin

@monoidaltransform NO, not remotely.

@leslie I've seen books that totally overlook the fact that $a_n$ fails to be decreasing, but it is eventually and that's all that matters.

Alessandro Codenotti

long time no see @Balarka how's life?

Balarka Sen

Hi @AlessandroCodenotti. I'm alright, how about you?

Ted Shifrin

@BalarkaSen I don't see why that should be the case. As I just said to monoidal, it is about the second fundamental form.

Alessandro Terminiello

hi im good thanks

Ted Shifrin

21:13

Hi, demonic !!

Alessandro Terminiello

ah

i think you tag the wrong alessandro : |

Ted Shifrin

Any two people with the same first 4 letters get tagged.

Last names are irrelevant.

Balarka Sen

Corrected.

Alessandro Terminiello

ah ok

Balarka Sen

@TedShifrin True, maybe I should compute. It's hard to imagine.

Alessandro Terminiello

21:14

can i ask the last thing about series

Ted Shifrin

If you think about the Lorentz hyperbolic models, then you can see it in terms of the slicing hyperplane @Balarka

It's just like for spheres in spheres.

Alessandro Terminiello

$$\sum_{{n=2}}^{\infty} a^n \tan\left(\frac{\sqrt{n^2+n} - \sqrt{n^2+1}}{3^n+n^3}\right) \sim \frac{1}{2}\left(\frac{a}{3}\right)^n$$ , i need to use D'Alembert criterion right?

Balarka Sen

Ah, that's smart!!

I love that

leslie townes

ted: related to that, the business of not minding the lower index of summation when spamming out a list of textbook exercises about sum n = "1" to infinity of 1/(n(n-5)), or 1/ln(ln(ln(n))), or whatever

Ted Shifrin

@leslie That's why I usually don't write the lower index :D

@BalarkaSen It happens, by chance, once in a while :)

monoidaltransform

21:16

Ah right ofcourse. Going back to my original question, out of curiosity, what if $H$ is also simple and compact, then since bii-invariants metrics on simple compact Lie groups are unique up to scalar then does that mean that $d_{i^*g}=ad_g$ for some scalar a?

Alessandro Codenotti

All good @Balarka, I started a new job about a month ago. I'm back in Italy now and I'm much happier than when I lived in Germany

Xander Henderson

Too many Alessandros. @AlessandroTerminiello, pick new name. @AlexanderGruber, you too. It is getting confusing. :P

Ted Shifrin

I don't know without some serious thought, monoidal.

Alex Gruber gets priority. He's been around foreverest.

Alessandro Terminiello

give me a new name

i will change

Balarka Sen

@AlessandroCodenotti That's nice!

leslie townes

21:20

Leibniz :)

Ted Shifrin

Don Giovanni :)

Xander Henderson

Dom Pérignon :)

Sine of the Time

Xander :)

Xander Henderson

@SineoftheTime banned.

Sine of the Time

:(

Balarka Sen

21:23

Xander is like a scuffed version of Alexander

2

Alessandro Terminiello

.

Ted Shifrin

@monoidal I really think you have to look at the structure constants relative to the splitting $\mathfrak h\oplus\mathfrak k$.

Xander Henderson

@BalarkaSen It is the alcohol free version.

Alessandro Terminiello

$$\sum_{{n=2}}^{\infty} a^n \tan\left(\frac{\sqrt{n^2+n} - \sqrt{n^2+1}}{3^n+n^3}\right) \sim \frac{1}{2}\left(\frac{a}{3}\right)^n$$ , i need to use D'Alembert criterion right?

i changed my name

Xander Henderson

@XanderHenderson 'cause there's no ale.

3

Alessandro Codenotti

21:32

@BalarkaSen math wise the project here is pretty cool even though I have to (re)learn a bunch of algebraic topology and category theory

Some might even consider this a plus

Balarka Sen

Oh that sounds interesting

Thorgott

@AlessandroCodenotti the #1 Münster hater

Alessandro Codenotti

The basic idea goes back to the very beginning of AT but it was not explored much. People wanted to think about cohomology groups (and other algebraic invariants) as topological groups, but this is kind of terrible because coboundaries are almost never closed in the cocyles, so the topology on the quotient is all messed up

Xander Henderson

Yay! My new glasses are ready! Gonna go pick them up after office hours end in 20 minutes.

Anyone here need help with precalc? None of my students have shown up... :/

Alessandro Codenotti

So the solution turns out to be to think about those invariants as "groups with a Polish covers" that is groups explicitely presented as the quotient of a Polish group (the cover, cocyles for example) by a Polishable subgroup (coboundaries) and to say that a morphism between Polishable subgroups is a definable morphism, in the sense that it lifts to a Borel map between their covers

The category of abelian groups with a Polish cover turns out to be the canonical completion to an Abelian category of the category of Polish abelian groups (whatever that means) so they are apparently great objects to work with

And they give somewhat finer invariants than just the cohomology groups as discrete groups

Thorgott

21:39

"polishable" lol

Xander Henderson

Is that "polishable" or "Polishable"?

Thorgott

do these have a reasonable homotopy theory

Alessandro Codenotti

I know nothing about the homotopy side of this whole business

Thorgott

cause what you really wanna do, alg-topologically speaking, if you wanna work finer than cohomology is consider the cochain complex

Alessandro Codenotti

I barely know anything about the classical version of homotopy theory

Thorgott

21:40

as object in some appropriate $\infty$-category, most likely

Balarka Sen

uh oh the infinity pill is dropped

Thorgott

i say "homotopy", you say "coherent"

Alessandro Codenotti

@AlessandroCodenotti a morphism between groups with a Polish cover*

Balarka Sen

@Thorgott I dont think this carries more information than homology in general, tbf.

The appropriate infinity-category is just going to be the dg derived category of chain complexes

Thorgott

it should, even without all the fancy terminology

like you get higher cup products on the cochain level

the cup-1 stuff or whatever

Balarka Sen

21:42

Thats not just the chain complex

Thorgott

we can always add structure

Balarka Sen

Yeah then you're thinking of C*(X) as an E_infty-operad or whatever

E_infty-algebra* rather

Thorgott

ye, something like that

Balarka Sen

@AlessandroCodenotti Sounds cool. BTW, I still want to know the answer to the "2-dimensional dendrite" question I asked you long back

Alessandro Codenotti

What was the question? I forgot

Balarka Sen

21:56

I think it was the following: define a "tree 2-complex" to be a cell complex which is locally homeomorphic to either R^2, (R^2 with a half-plane glued by its edge along the x-axis) or (R^2 with a half-plane glued by its edge along the x-axis, and a half plane glued by its edge along the y-axis (Rmk. these glued half-planes do not intersect in the interior)).

Can one define a "2-dimensional dendrite" as an appropriate inverse limit of "tree 2-complexes"?

This was motivated by a comment you made which I think shows this is indeed true in dimension 1.

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