00:00 - 18:0018:00 - 00:00

12:05 AM
Inclusion maps $A\subset \mathbb C$ are analytic. right?
My argument:- Let $i_A$ be the inclusion map. It is nothing but the restriction $I:\mathbb C \to \mathbb C$(Identity map.
Since, $I$ is analytic. satisfy C-R equation.
$i_(z)=z$. Here $u(x,y)=x$ and $v(x,y)=y$. So u and $v$ are continuously differentiable

yes, yes, as long as $A$ is an open subset

Hence, Inclusion maps are analytic.
Where is the mistake in the proof without taking $A$ to be open?

You need an open set to talk about differentiability.
What's the definition of analytic?

A function is analytic at a point. then, there is an open set containing that point where function is analytic everywhere in that open set.
if function is analytic in some domain. then, it is analytic everywhere in that domain.

So you need an open set.

12:15 AM
But Here I didn't use the domain as an open connected set.

I don't care about connected. You are not paying attention to what I said.

Can't there is an analytic function from closed disk to itself?

No. That means it's actually analytic on a larger open set.

identity map?

@TedShifrin then what are we classifying when we classify Aut(D) using Schwarz lemma?

12:16 AM
If you pay attention, they'll talk about analytic on the open disk and continuous up to the boundary.
The open disk?

oh yeah you're right

It happens by mistake sometimes.

oh and D is highly related to Fourier series yay

Well, actually, @Leaky, that would probably be $\bar D$ :)
hi Demonark

If I keep stumbling upon the nerd chat I fear I may risk becoming a nerd myself in the future
How's it going?

12:30 AM
That risk took root 4 years ago.

:0

it's hip to be square
took root 33 years ago :P

I'm not following

huh huh
hips don't lie

12:38 AM
Oh Ted, don't tell people this but I think at some point in the near future I might actually want to learn algebraic geometry

What am I supposed to do with this secret information?

time to float: Suppose $T:\Bbb{R}^n \rightarrow \Bbb{R}^n$ be a linear map which satisfies $T^2 = T -I_{n}$, then is $T$ invertible?

Can $0$ be an eigenvalue of $T$?

cool gotcha

Or, more directly, can $Tx=0$ for a nonzero vector $x$?

12:41 AM
it cant be
s oits invertible!

so it would appear

it would appear?

I went to a talk a couple days ago that I told you about and while I didn't follow well, it has finally convinced me that the stuff is cool. I'm sorta wondering, aside from commutative algebra, what do you think one should know going in?

T(I-T)=I

My personal bias is to know some classical stuff on Riemann surfaces/curves. As you know I don't want to do everything with just reams of algebra.
That'll do, too, Leaky.

12:43 AM
On the internet (read: reddit) there's back and forth about whether folk should know complex geometry going in, so I was wondering your take on that in particular, but also in general.

Complex geometry is one of the approaches to algebraic geometry. That's the world I lived in (although I took 2 years from Hartshorne).

Yeah cool leaky this also gives what the inverse of $T$ should be

Of course, Demonark, I don't necessarily know what the reddit debaters meant by complex geometry.
Oh, it's a @Balarka.

Hi @Ted, @Dami

Hey Balarka, how's it going?

12:46 AM
a @Balarka, I presume you're in touch with Mike. He hasn't even messaged me on FB ....

just woke up
Yeah I am, @Ted.

Well, say hi.

What time is it there? Has your sleep schedule been repaired for this week?

Done, @Ted. :)

Getting up at 6 in the morning is better than staying up all night, Demonark.

12:48 AM
@Daminark 6 AM. Went to sleep at 11 AM. Fixed as fuck.

Nice!

I'm glad to see that Balarka is working on obscene alliteration.

lol

I guess officially that's consonance, but everyone calls it alliteration even with consonants.

I think Mike still has an account at MO.

12:52 AM
I would hope so, now that he's an adult. :)

:-)

@TedShifrin It's easy to tell you were a language nerd back in the days

Did I ever stop, a @Balarka? :)
I think languages/linguistics are fascinating ...

I should learn one or two other languages to be honest. I have to learn French at some point, I am sure.
Too much math in that language that hasn't been translated yet
Most papers of Thom I think

I might have read some Thom in French.

12:56 AM
Cool!

In fact, I think that to prove reading competence in French in grad school, Mo Hirsch might have given us Thom to read and translate. More words than symbols, but that didn't bother me.

3
"French Language"

@TedShifrin Damn haha that makes perfect sense

LOL

This is from Springer

12:59 AM
Still funny, Demonark.

Oh no for me that makes it even better
In contrast to some random list on the internet or something. Gotta petition to ICM to use this categorization

I think most of math uses a bit more specific classification scheme (not including French language).

@TedShifrin Mike says hi!

He just did to me.
I didn't scold him too much.

Haha.
I'll probably forever be in this chat. I will vanish during semester but in the vacations I'll be all over this place

1:09 AM
The personality of chat has changed a lot, missing anon, Pedro, Mike, and others.

What is even Pedro up to
Or anon for that matter. I see him on main sometimes

He's doing well in grad school in Ireland.

With some other different account

I have no idea what happened to anon. I wish I knew.

1:10 AM
He's just busy with his job. He shows up once in a while.
But he was never a presence like the others.
At least, not since I showed up 5 years ago or whatever it was.

@TedShifrin Oh, I forgot he's in Ireland

I hear from him occasionally. He seems to be doing great. I had hoped to visit, but I am not going to Europe this summer.

Daniel Fishcer has disappeared also.

well, he disappeared ages ago ... as soon as he was elected moderator.

All he had to do was get into office, collect the health benefits, and run

1:14 AM
Him, robjohn, and anon were huge back in the day.

Yup.
I guess I won't correct skull's crummy syntax.

not to mention Asaf

Judging by Pedro's publications, he's a full blown homological algebraist now

Yes, Balarka, so 'twould appear.
@Balarka, does your starred comment suggest that in India there's as much drunken partying among college students as there is in the US? That surprises me.

I was reading a book, called "Academically Adrift", and I was surprised to see that what I am seeing in the student culture has been happening for decades in US
So I suppose yes

1:24 AM
Yup.

Was it like that when you were in college, Ted?

Yes, sure, to some extent.
More in the fraternities and sororities than in the dorms. Now it's more pervasive, I think.

I think this is a relatively new phenomenon in India.
As we try to emulate the "Western norm" without understanding what that is, I mean

I figured that religion (be it Hinduism, Buddhism, or Muslim) in your part of the world would make students a bit different, Balarka.

Oh but Hinduism is like, old. And historically it was way more liberal than the rest as well. Buddhism basically doesn't exist in any form in the current times.
So I think we on effect of Hinduism is not far off from people on effect of Christianity in the West. There's no religious ethics that's ingrained.
There's religious hate. That's a problem!

1:29 AM
in the US, people just claim to be religious in order to discriminate and hate. Yeah.

@BalarkaSen whatever u do just don’t learn mine it’s stupid and it sucks

heya @Eric. Someone on FB just told me I shouldn't believe the news media when I said Brazil's president was a raging homophobe.

what a thermonuclear take

I didn't engage in an argument cuz I don't know much.

well they were wrong and u were right so don’t worry bout it lol

1:34 AM
it looks like she's actually from Brazil (friend of a friend).

yeah and brazilians elected a raging homophobe

@TedShifrin I like the simile between US and India right now, even though the former is the rich capitalist first world and the latter is the pathetic third world. Both our societies are a wreck at this point, both of us have elected a neofascist government, both of us think of the country as a religious sanctuary of one particular religion(Islam is somehow a problem for both of us), and we're doing the best we can to convert democracy into imperialism of the industry
@ÉricoMeloSilva i need to read latin america literature though, thankfully most of them are translated

all the good ones are spanish anyway lmao (our modernists are cool im lying)

I'm still very much an atheist, but having a Muslim boyfriend is making me think about stuff a bit more.

1:41 AM
@TedShifrin this is actually a moon logic take from this person though bc the media (both in the anglophone and lusophone world) is largely uncritical of that absolute POS bc he’s great for transnational business interests, at best they’re tepidly criticizing his neofash style, but by and large the media has no interest in being honest about how bad he is (i guess cause he’s not bad for you unless you’re working class or marginalized in some way)

have you ever been married? (don't answer if you want)

Makes perfect sense to me, Eric.
Hell no, skull.

2:06 AM
Wild things.

Hey @anakhro, @Alessandro

Hi @BalarkaSen

@BalarkaSen I finally found a proof that H^1(R,M) is a Hilbert manifold

Oh? @Ryan

I am trying to understand a theorem which seems to fail to make sense to me. I feel like I am missing something or not noticing something.

2:08 AM
@BalarkaSen I've been on a wild jorney to understand these spaces better

Intuitively I always thought that's because you can exponentiate to H^1(R, R^n)
That's how they prove C^infty(M, N) is Banach, eg

Yeah it works in that case because the maps are AC

@anakhro What's the theorem?
Ah I see @Ryan

Understanding the structure of $W^{1,p}(M^m,N^n)$ is really, really hard
it depends on $p$, $m$, $n$, and the homotopy type of $N$

Yeah you told me

2:10 AM
The rabbit hole goes deeper
If $M$ has a boundary then you want to understand how to restrict maps to it
and then the converse of this problem

"Any two contact embeddings $j_i$, $i = 0,1$, of $B_{st}$ into a connected contact manifold $(M,\xi)$ (with $\dim B_{st} = \dim M = 2n + 1$) are contact isotopic."

Sobolev obstruction theory

Just the analog of the "Disc theorem"

It was a hot topic in the 90s

for contact topology.

2:11 AM
Oh ok
What's the subtlety?

@RyanUnger Crazy

well it's embarassing because the paper says it's easy

This definitely should not mean that any Seifert surface for a Legendrian unknot in a 3-manifold is isotopic to another.
Oh, of course, dim B is 3.

but it raised another interesting question so the email isn't wasted

2:14 AM
Seifert surface of an unknot is the disk not the ball.
So of course not.

Right, @anakhro
Disc theorem is about isotopy class of embedded disks of the same dimension as the manifold, of course
"connected sum is well defined"

crazy

Can you take connected sums in the contact category?
I guess there's extra care you need to take to get the distributions to match along the boundaries of the balls

up to contactomorphism, yes.

2:20 AM
@RyanUnger is that you, 0celo?

uh are you trying to ban me

No.

then yes
otherwise no
if youre lying

I just noticed the raccoon display picture.

@BalarkaSen basically I'm trying to understand the following
if you take a 3-manifold with $\pi_3\ne 0$ and look at a nontrivial homotopy class
you can think of maps $\sigma:S^3\to M$ as maps of $S^2\times I$ with the ends mapped to points
so you look at things like $\max_{t\in I}Area(\sigma(\cdot,t))$
and then you minimize this over all $\sigma$ in the homotopy class
so it's like Morse theory for the Area functional

2:25 AM
Makes sense; some quantitative homotopy theory thing going on

Pretty neat idea.

Gromov does this for the simplicial volume

then there's a claim that if $\pi_3\ne 0$, then this min-max area is positive for any nontrivial class
I've been trying to understand that for some number of weeks now
It's like a grown-up version of "small maps are nullhomotopic"
Problem is, small in area might mean nothing

Cool

You need a lot of theory to make this work
and this is a toy model for what Marques-Neves did/are doing
Toy model but also different. Instead of using harmonic maps they use GMT
@BalarkaSen those guys look at the space $\mathcal Z_n(M^{n+1},\Bbb Z/2)$
closed hypersurface currents with Z/2 coefficients

2:29 AM
whatever that means

it's a theorem of Almgren that this is homotopy equivalent to $\Bbb RP^\infty$
so you expect the Area function, if it were a Morse function, to have lots of critical points
it's not a Morse function but you can still make this work
@BalarkaSen just think of Lipschitz maps of closed n-manifolds

Generically not a Morse function?

@anakhro ah well there's a genericity condition in some of their papers. I don't know yet if this is where it comes in
Recently that was removed

Ah.

@BalarkaSen so you can define "eigenvalues" for the Area functional in this manner
and there's an associated universal growth law
@BalarkaSen I believe there's a GMT proof of the fact that H_2 in M^4 can be repped.
It's probably just GMT-ifying what you told me
Using Almgren's Big Theorem

2:34 AM
Interesting, what's the theorem

Let $\Sigma^k$ be a homologically area minimizing integral current in $M^n$. Then $\Sigma^k$ is regular up to a set of dimension $k-2$.
So for 2 in 4 you get point singularities, which can be resolved.
So much, much more complicated than what you said haha

Sounds analogous to un/stable manifold theorem from Morse theory

Almgren's paper was 1700 pages

o my

It was improved considerably
I think now it's a couple hundred

2:37 AM
It's neat seeing that happen with stuff.
People collaborating and making things look better and easier to understand.

Someone needs to do it with the Poincare conjecture
It's nowhere near 1700 pages but there's some parts that are inelegant still
Classification of finite simple groups haha

Is the Ricci flow part any good?
Hamilton's part.

Perelman did Ricci flow too.
Everything that Hamilton did is great
The whole proof rests on some fucking black magic though. The Hamilton--Ivey pinching estimate.

Maybe you can help?

It's really the only place where $n=3$ matters
Because in $n=3$, $\Lambda^2V^3\cong V$
All the curvature tensors carry the same info in 3 dims
@anakhro Peter Topping has an alternative version that's less magical
But it's weaker
Does what you want for the PC tho

2:46 AM
Wonder if anyone taught it as the subject of a course somewhere and has lecture notes for it
In the dover Kosinski edition there is an appended part about the Poincare conjecture.

Oh it's possble to understand the proof
But there's some serious strokes of genius in the middle
And then the finale is a technical crapshoot that has nothing to do with Ricci flow

Technical stuff is sometimes the most intriguing stuff.
For my thesis I had to read through this rather intriguing paper (in the same sense as above) by Sotomayor.

The intriguing stuff is that minimal surfaces behave nicely under Ricci flow
The crapshoot is that finding minimal surfaces is really hard

But why it is really hard must be interesting

Yeah, but the point is that it has nothing to do with Ricci flow
from the perspective of proving the Poincare conjecture, it's a big technical lemma

2:50 AM
Yeah.

basically you want to show that the flow of your homotopy sphere ends after finitely many surgeries
the only topology you have is $\pi_3$ so you have to leverage that somehow
hence what I was talking about earlier
what would be really cool would be to show that a homotopy 3-sphere has positive scalar curvature

Pretty neat. I will have to look into this more later this summer. Thesis is bogging me down. I don't want to work on it anymore and I am sick of it.
Anything sounds better than working on my thesis.

I know techniques for showing something doesn't have positive scalar curvature.
The other way around seems like madness.
Lawson-Yau tried to prove this

3:38 AM
0

Consider the minimization problem described this paper. Let $f_{\lambda}$ be the minimizer. As a part of extending my work, I am able to show the following facts $$\lim_\limits{\lambda \to 0}\|f_{\lambda}\| = 0$$ and $$\lim_\limits{\lambda \to \infty}\|f_{\lambda}\| = 0$$ My problem now is (as ...

1 hour later…
4:48 AM
Is there a special name for Dedekind cuts on a free monoid with prefix order?

5:13 AM
$\sum _{_{n=1}}^∞\frac{1}{n^{1+\frac{1}{n}}}$ How do I check the convergence of this series?
I applied Cauchy's nth root test. No use:(
Can I able to compare with some convergent or divergent series?
I think I can do by using limit comparison test. right?
comapring with $\sum _{_{n=1}}^∞\frac{1}{n}}$

5:53 AM
I don't know why but many books remove the necessary definition of a manifold being hausdorff. Hausdorffness is a very important condition. It removes in a certain sense all the wonky topologies you can give to it. But books don't even mention it. Do they inherently assume that the space they are working with is a subspace of R^n?

Sure, many books, including Guillemin-Pollack, define manifolds as subspaces of R^n
non-Hausdorff locally Euclidean spaces are not that important to pay attention to. They definitely do come up in literature (leaf spaces of foliations) but they're not the central object of study
Neither are non second countable guys
A manifold is a Hausdorff second countable locally Euclidean space. Period.

That is true. But like if I don't mention Hausdorffness then when it comes to the smoothness condition of the transition function, I can define my open sets in a certain way which will make a lot of not smooth things seem smooth. That's why I thought that the Hausdorff condition should be put. But again if you just say it as a subspace of R^n its all fine.

6:12 AM
Yo
My talk starts in about 90 minutes

Good luck! What's it on?

Same thing as the math talk I did
Basically stuff about quantum mechanics/ Bell’s inequality via polytopes and the elliptope
(3-elliptope: Set of 3-by-3 symmetric positive-semidefinite matrixes with ones on the diagonal)
Some easy little linear algebra there

Ooo seems interesting. All the best @Semiclassical

Thx
I can probably send the slides along after. Not going to try to do that right now tho lol

Sure. I will try to understand stuff. I am also doing some quantum mechanics stuff so yeah.

6:19 AM
@Semiclassical Aha

6:30 AM
(There’s a slide which could probably be entitled “obvious to anyone who knows about PSD matrices”)

7:08 AM
Phir Ek baar

8:04 AM
Morning all

hi pal

Morning @Skullpatrol

12

I woke up this morning to find that I suffered massive drop in points across network because of lots of users were removed . What happened to those users? Was there a database cleanup?

8:28 AM
RIP :-(

9:20 AM
Hello. If one wants to compute the integral of a two form, say $\omega = y dx\wedge dy + z dy\wedge dz$ on the submanifold given by $z=16-x^2-y^2$ and $z>0$ of $\Bbb R^3$, how one approach this?

Do I reduce this to a classic calculus type double integral somehow? Perhaps something like $$\int \int (16-x^2-y^2)y dxdy + \int\int (16-x^2-y^2)z dydz,$$
with the appropriate regions (and appropriate orientations by placing an orientation on the submanifold)? Or something else entirely

@F.White Do you know Stokes' theorem?

Yep, but can I use that in the case this form isn't exact?

What can you say about your submanifold? And $\omega$?
Stokes' theorem is a general statement about integrals of a form and it's exterior derivative over a manifold with boundary, not just something which is to be used only when you see an exact form.

9:37 AM
If $z$ is strictly greater than zero, my manifold is without boundary right? and if my form $\omega$ was exact (and it were $z\geq 0$), with $d\gamma = \omega$, I would be able to integrate the 1-form $\gamma$over the boundary circle
My submanifold is orientable, and my form is closed
I guess my region is closed and bounded and hence compact too, is the other thing you wanted me to observe

You're working with the wrong dimensions. Your submanifold doesn't have boundary, but is bounded by a domain. Namely, consider the interior of the paraboloid.
@F.White Not quite, you have $z > 0$ there. It's not closed, but forget about that for now.
Just suppose that you were given the submanifold $S = \{(x, y, z) \in \Bbb R^3 : z = 16 - x^2 - y^2\}$.
Good observation that $\omega$ is closed.

$S$ is closed (but unbounded) by the regular level set theorem, and my thing is an open submanifold of that I suppose

Yes.
That's all fine. Do you see why $S$ is bounded by something?

Sorry, I'm not sure what you mean. It ($S$) seems to be unbounded from below (in the z-dimension), or you mean something else?

I mean $S = \partial V$ where $V$ is the solid paraboloid, given by $\{(x, y, z) \in \Bbb R^3 : z < 16 - x^2 - y^2\}$.
It's boundary of a certain manifold with boundary

9:52 AM
Ohhhh

Now use Stokes :)

Thank you!

(1) I should have written $V = \{(x, y, z) \in \Bbb R^3 : z \leq 16 - x^2 - y^2\}$ up there (2) Remember to not forget that your submanifold is $S \cap \{z > 0\}$ when writing a solution for your problem (but that isn't a problem).

10:23 AM
Hi @MatheinBoulomenos

2 hours later…
12:22 PM
It appears that @TobiasKildetoft vanished into the "void" :P

He is working in industry now
so probably very busy lol

Thanks for the info, pal.
I'll always remember him for this classic remark:
May 11 at 16:52, by Tobias Kildetoft
If you lecture long into the void, the void starts lecturing back.

1:16 PM
I don't even know what that remark might mean.

2 hours later…
2:50 PM
I am not completely gone. But it is true that I rarely have time to hang out in chat these days.

3:17 PM
Thanks for dropping by pal @TobiasKildetoft :-)

3:50 PM
I have a question about sparse matrices on the Wikipedia page (en.wikipedia.org/wiki/Sparse_matrix#cite_note-5). Under the definition of IA, it says "Thus, the first m elements of IA store the index into A of the first nonzero element in each row of M..". I just can't make sense of that. It seems like this differs from so-called row pointers that I'm familiar with in a sparse matrix. The elements in IA add up the non-zero elements of M, but then I don't understand the thing with indices?
Scroll up or down in the article to "Compressed sparse row (CSR, CRS or Yale format)"

4:39 PM
Welcome back professor @TedShifrin

5:04 PM
hola
bois
the set of all purely imaginary numbers is closed, right?
in the metric space $(\mathbb{C},d)$, where $d$ is the standard Euclidean norm
@BalarkaSen the manifold must be orientable(!?)

5:18 PM
@SubhasisBiswas What does purely imaginary mean again?

Having $0$ real part, I guess. In which case it follows from the real part map being continuous.

@Thorgott yes. $0$ real part. My answer is right then?

Yes, it is closed.

now, what if I exclude the origin from this set?

Consider $(i/n)_n$

5:25 PM
will it become neither open, nor closed?
not open, since it is a line in two dimension, and there does not exist any circular nbd around any point that is entirely contained in that set
@Thorgott yes

Indeed

:D
what are you studying right now?

@SubhasisBiswas Sure.

Nothing rn, was working on my assignments earlier

@N.Maneesh the series diverges. Compare with $\sum {1/n}$
$\displaystyle\frac{\frac{1}{n}}{\frac{1}{n^{1+\frac{1}{n}}}}={n^{\frac{1}{n}}}$. Now, $\lim_{n \to \infty} n^{1/n}=1$
@Thorgott are you in your master's degree?

5:48 PM
I wish

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