Wait, not surface groups, a bigger class

$\langle a_1, b_1, \cdots, a_g, b_g | ([a_1, b_1] \cdots [a_g, b_g])^n = 1 \rangle$

Fundamental groups of surface with a cone point of order n

oh

that is very cool where do you find more information on those things

I know one (well, both) of the authors personally, that's all

I don't know anything about this stuff

@BalarkaSen really cool when I get more into complex geometry we will discuss more

I don't know complex geometry

I am just trying to minimize my time here on chat because it is distracting

I see

Oh

It's still open if there are Kahler groups which are not $\pi_1$ of smooth complex projective varieties

I think it is true I think it is always possible to find khaler groups which are $\pi_1$ of a smooth complex projective variety

I have a hunch that it is true

@MikeMiller

If we don't do that, we don't know that $f(\lambda)=\lambda$. It could be $\lambda\mu^k$.

You better read more than just the theorem in your book!

I honestly did

You'll find that they define $T_0$ somewhere.

MAINS REAL BAD

what does that mean, PVAL?

Main is real bad.

what in particular?

Main what

There have been a few good questions of late.

Some terrible upvoted answer.

Including the one I got wrong and you taught me about.

fun question inspired by one on main: which permutations in $S_{mn}$ correspond to permutation matrices which are kronecker products $A\otimes B$ of an $m\times m$ permutation matrix and an $n\times n$ permutation matrix.

Oh ... yeah.

heya @anon :)

heya

I ain't gonna post it. I hope the user deletes it promptly.

it's not obvious to me that the tensor product gives you a permutation matrix, anon.

it does

I mean it's obviously got only 0s and 1s ...

Oh, I see.

Is it just any permutation that doesn’t mix the two sets?

nope

definitely not

Hmmm

two sets?

He's thinking of $\{1,\dots,m\}$ and $\{1,\dots,n\}$ as two distinct sets.

Oh, yeah, I see why that doesn’t work

but $S_{mn}$, not $S_{m+n}$

Yeah

it is an internal copy of $S_m\times S_n$. (form an $m\times n$ array, have $S_m$ permute rows and $S_n$ permute columns.) it can be formed as the knit product of (the diagonal copy of $S_m$ inside $S_m^n$ inside $S_m\wr S_n$ inside $S_{mn}$) and (the second factor of $S_n$ in the wreath product $S_m\wr S_n$ inside $S_{mn}$), and similarly as a copy inside $S_n\wr S_m$ inside $S_{mn}$.

...oook

I had a colleague at UGA years ago who did lattice ordered groups and taught me wreath product, but I never beremember it.

@TedShifrin so to get $S_m\wr S_n$, have an $m\times n$ array, have $n$ copies of $S_m$ act on the $n$ columns independently, then have $S_n$ permute the columns as they are.

oh, $n$ copies of $S_m$. Yikes.

I keep trying to think of it as $S_{m+n}$ which is of course wrong

Oh, I see why I’m going wrong—I keep thinking direct sum

Similar: $C_m\wr S_n$ is like having $n$ copies of $C_m$ acting on a cryptex with $n$ characters and an alphabet of $m$ letters, and then also being able to rearrange the bands in the cryptex.

yeah, I understand, anon ...

Hello handsome folks

@KasmirKhaan how did you know?

So if G acts transitivly on a some Set S

I don’t, but uh

G/stab(x) is in bijective correspondance with The set S right?

I think I’m okay with that

@anon Know what? ><

that feel when you click your message to edit, but it gets bumped out of the way at the last second

@KasmirKhaan that we are handsome

window for the joke has passed

it was a failure

@KasmirKhaan in the case the action is transitive, yes

there is a G-equivariant bijection G/Stab(x)->Orb(x)

@Kasmir: We've been over this literally 50 times.

@anon Haha I say stuff that dont offend anyone =p like if i said hi "ugly" folks , like no one would say " hi am ugly and i know it "

or even in the case of Lie groups, Stab(x)->G->Orb(x) is a fiber bundle

Hmm

I did not mean here orbit of S

I mean litterly the Set S

the way i see it is , G/ stab(x)

yes, if G acts on S transitively, then S=Orb(x) is the orbit of every x in S

Oh right <<

Stupid of me :D

But tell me if i view this right

The way I see it , G/stab(x) , give us a cosets, that each coset act uniqly of elements of S

morning

@anon You'd need the Lie group to act properly, however, I think

@TedShifrin Ted I know we did talk about this , But am still amazed how much one can get from actions

hi Faust.

Otherwise you get foliated by orbits which get asymptotically close to each other

I just found out that Groups of prime Power have no trivial center :D

Balarka, go ununsleep!

Working on it

@KasmirKhaan prime power

sleep is overrated

@BalarkaSen dunno what you mean by asymptotically close to each other

@anon Thanks :D

One last Question guys

in proving sylow

we have a Group actiong on its own subgroups?

or what kind of action is it

conjugation

yes, G acts on the set of p-Sylow subgroups

hmm

So G has order mp^i

and it act on its p-sylow subgroups

I ll keep that in mind :D, I did all sylow theorems Before without understanding what an action is

Now its all make sense

@anon Sorry, I'm thinking about G --> M --> M/G

ah, yes

did not even knew that a Group can act on its subsets and subgroups Before -__-

morning

@MatheiBoulomenos @anon thanks guys ! :D ill keep working now and come later with more handsome Q's :)

@Faust morning faust

how u been long time no see?

@Kasmir: Nonsense. You certainly know that a group can act on itself by left multiplication.

@TedShifrin yes ofc, but did not consider acting on its subsets

@TedShifrin or subgroups

That was kinda new to me

well, sometimes, yes.

that was the heart of sylow and other questions i did not solve on the book

i was thinking at level 1

elements level

I wish one of you guys were my teacher :D

That is Ted anon and mathei :D

Anyway kasmir is back to work now !

You have my book, Kasmir. It knows more than I do.

I try to use it as much as possible , but the geometry parts scares me off =p

at least i understood homomorphism theorems from there :D

and Ill use it for ring theory part :)

ill start with that next week, exam in 3 weeks =p

@AkivaWeinberger In that presentation of the trefoil group the longitude is kinda gross

fun question popped into my head: let $G$ be a finite group and $H\le G$ a subgroup. find the size of $B=\{aHb:a,b\in G\}$ in terms of the sizes $|G|,|H|,|N_G(H)|$.

I don't feel like working today

I am just gonna take the day off

@anon trying to solve it now :d

@anon: So you're thinking in terms of $(aHa^{-1})b$.

It's a cool question.

@KasmirKhaan might be too difficult for someone new to group actions

hi anon

although you have all the tech if you know what a normalizer is

Karim: Today is already tonight :P

hey

@TedShifrin that's useful in the proof yes

@TedShifrin Yeah haha

@MikeMiller Really? In $\langle x, y | x^2 = y^3 \rangle$ the longitude is just $y$ isn't it.

@TedShifrin tonight off I guess I am getting 12 days off starting from tomorrow. so I am gonna focus on research. But feel kinda mentally exhausted today.

Karim: You don't need to apologize or make excuses. We all need a break from time to time. For me it was always cooking ... :)

haha yeah

@MikeMiller is it true that every knot in $S^3$ bounds an immersed disk?

@TedShifrin For me it is running and talking to people here haha

@TedShifrin Hey btw today I came to nice intuition about sheaves

@AnubhavMukherjee Those are ribbon knots, no?

so sheaves essentially are some kinda of generalizations of VB

@anon nice problem

vector bundle

Oh I guess not

not really the right way to think of 'em, Karim

only if they're locally free sheaves

Ribbon knots bounds immersed disks with specific singularities

Oh so I guess only for coherent sheaves?

the "ribbon singularities"

@BalarkaSen are you thinking of seifert surfaces?

I see

Karim: That's closer, yes. There are all sorts of dimension jumping phenomena with the dimensions of stalks of sheaves.

@anon can I choose b to be a' ?

@BalarkaSen exactly...

@anon a ^-1

@TedShifrin I see

@Kasmir: You could. But you might also not.

But see my remark above.

@KasmirKhaan the set $B$ collects all of the sets $aHb$, including those where $b=a^{-1}$ and those where $b\ne a^{-1}$.

Karim: There are things called skyscraper sheaves, supported on proper subvarieties.

@BalarkaSen see people down voted my answer and said this math.stackexchange.com/questions/2513068/…

@anon okay :) ill give it a try :D

@anon Well, no. Ribbon knots bound disks with self-intersections of a nice sort.

@TedShifrin I see so stalks are kinda like fibers in the case of VB

For example, if you have a divisor $D$ on $X$, you have $$ 0\to \mathscr O_X\to\mathscr O_X(D)\to\mathscr O_X(D)\big|_D\to 0.$$

yes

No, Karim. These are living over a subvariety and the sheaf has 0 stalk most places.

Totally not like vector bundles.

@TedShifrin I see ok I am just trying to get a geometrical intuition for those things

@BalarkaSen no, pass 2 abelianizaion

I do no have kes beween R and U for some reason

Karim: You can think of the exact sequence I wrote as taking residues at the points of $D$ (when $X$ is a curve).

oh oh

that is cool

I guess you can't have a knot (unless it's an unknot) bounding an immersed disk with self intersections set in the interior

That breaks Dehn's lemma

I don't understand PAVAL's comment

And it is more annoying that someone downvoted

Weird, Balarka: I always expected the self-intersections to be in the interior.

To what post, @Anubhav?

@TedShifrin math.stackexchange.com/questions/2513068/…

here

I'm guessing that PVAL and others think you're not quite correct. I don't know this area at all, but I'm surmising you didn't completely understand what you thought you heard in the lecture.

PVAL's idea should work.

Reversing crossings to get the unknot and then capping that off by the disk should give such an immersion

@TedShifrin I was just wondering if Stiefel whitney classes or Chern classes appear in algebraic geometry btw ? today we saw few things about first chern class.

Chern classes everywhere

in pretty much every paper I ever wrote

@TedShifrin I agreed that it could be wrong, but after editing my answer, someone down voted...

oh

Can I pass a question to all of you?

Stiefel-Whitney is more topologists' game, are you have to be working with mod-2 cohomology.

Oh- don't ask to ask, I'm gonna ask it then.

I see. @TedShifrin It is nice to see different geometry fields interacting.

@Anubhav It's easy to prove the candidate PVAL suggests is an immersion. Locally it's like inclusion of an arc on $S^1$ cross $(-\epsilon, \epsilon)$ in $\Bbb R^3$

Right now I like thinking analytically and geometrically about things, but maybe that will change as I dig deeper into algebraic geometry. My supervisor keeps mentioning that many slick proofs comes from spectral sequences.

The singularities are also transverse singularities; think about what happens when you reverse a crossing. Locally near a crossing it should look like two intersecting planes in R^3

How can I prove a 3x3 matrix with only entries down its main diagonal, a_11 = a, a22 = b+c, a33 = b-c is a basis for all 3x3 real diagonal matrices? I reckon what it's boiling down to is whether b+c and b-c can be expressed as arbitrary, unrelated constants. If I can prove that, then I can say it is. But how can I?

@BalarkaSen I can see singular disks, but I am not sure how to convert it into a immersion

I think of algebra though more as a tool than anything else.

@sangstar: Write down the equations and solve 'em.

@MikeMiller Oh yikes

Right down which equations and solve for what?

yes.. May by a littly homotopy, i can make the first derivative non-zero @BalarkaSen

F = b+c, g = b-c?

You're asking if you can solve $b+c=x$, $b-c=y$ for $b$ and $c$.

I scrolled back to read about sets of random, commuting Hermitian amtricies. And instead I found out that math chat is kink-shame-free

Right, you end up with something like x + y = 2b

I'll take it. But not exatly what I was going for.

But that doesn't tell me anything abrupt

So what's $b$?

@KevinDriscoll It's all good here

@KevinDriscoll wait what?

We are into weird stuff

(x+y)/2

Done. Now do it for $c$.

knot theory, projecting onto subspace, converging to limits...

@TedShifrin So today while teaching calculus lab I wanted to teach a differential as a section of the cotangent bundle

haha

@anon Its not too far up when semiclassical was talking about these matricies

Ah!! x - y = 2c

Or substitute what you already know for $b$, yes, @sangstar. Either way. Of course, you can use basic linear algebra and use matrices to do this, too.

c = (x-y)/2

Karim: I always talked about tangent-line or tangent-plane approximation and never used the word differential (unless I was teaching differential forms).

I could, but now I've shown I can express b and c using arbitrary constants, but using the same TWO arbitrary constants

Huh?

You've shown you can get any diagonal matrix as a linear combination.

oh yeah @TedShifrin I watched most of your lectures

Karim: But in a usual Calc I or Calc III class I don't use "differential." I talk about tangent-line (-plane) approximations.

That class you watched was super-fancy.

b and c are both expressed in terms of x and y, yet they're still arbitrary constants?

(x and y are arbitrary constants, that is)

You want to write the matrix with diagonal entries $a,x,y$ as a combination of your matrices. How do you do it?

I see @TedShifrin It provided me with so much intuition

But it's not the way to teach a standard engineering/science calculus class, Karim.

I take any number for a, and for x and y I have to pick a number for b and a number for c, and then x and y are automatically evaluated

yeah @TedShifrin I am happy one of my students told me I teach better than prof.

@anon Think I found the solution :D

Sometimes it's easier to be more effective in smaller discussion classes.

@anon or at least very Close to :D

I only understand what "section of the cotangent bundle" means by analogy to the cases from $\mathbb{R}^n$ that I'm used to

@KevinDriscoll so a cotangent bundle is defined as follows

First of all given a tangent space

@sangstar: You have to show that you can combine matrices of the form you said (for appropriate $a,b,c$) to get a diagonal matrix with entries $a,x,y$. And you have to show you can do so uniquely.

@Adeek Ya Im familiar with how it works

@KevinDriscoll if you understand how it works in $R^n$ then generalizing to abstract manifolds is easy

@TedShifrin Yeah Dehn's lemma says if you have an immersed disk in a 3-manifold with singularity set in the interior, there is an embedded disk with the same boundary as the immersed chap

@Ted Shifrin Could you explain what you mean exactly by uniquely?

Of course, but that's the point. I know how it works in $\mathbb{R}^n$ without any of the fancy words.

To say vectors form a basis means that every vector can be written uniquely (in exactly one way) as a linear combination of them.

Ted !

Did you see anon question ?

Oh, interesting, @Balarka.

Or should I post it ?><

Yes, @Kasmir, I saw it. I commented on it, remember?

oh okay :D

what I got so far

Don't tell me ... tell anon.

I mean the thing with manifolds is just you do multivariable analysis locally and patch things up using bump functions to path from local picture to global picture

G acts on G/H

grrrrrrrrr

:(

but that is with $C^{\infty}$ manifolds though

heh

Anon :D

complex manifolds don't have this

okay let me tell you what I got so far :D

there isn't analytic bump functions

G acts on G/H , by conjugation

Karim: Did you do an exercise that you can't think of tangent spaces as derivations without smooth? With $C^k$ you get infinite dimensions.

@KasmirKhaan does it?

baH(ba)^-1

hmm i Think it does :D

and the stabilizer of H

gH is an element of G/H, which if we apply x->axa^-1 to, we get agHa^-1, which may not be a left coset of H

oh no I didn't do the exercise @TedShifrin where is it localted in the book ?

I have the book

No, not my book. Your manifolds course.

This is way too fancy for my course.

I think my confusion is that, yes, we've defined arbitrary constants x = b + c and y = b - c. Then, we expressed b and c in terms of those arbitrary constants. But I would think, in order for b and c to be truly unbound, b would have to be defined by different arbitrary constants, and not the same x,y ones that c has

also, I'm not talking about G/H={aH: a in G}, I'm talking about B={aHb: a,b in G}

oh okay I Think I need to use an other name for those cosets

yes yes

@sangstar: You're confused. I give you $a,x,y$. You tell me which values of $a,b,c$ will work and what matrices to add.

If we have this : baH (ba)^-1

@TedShifrin A proof that tangent vectors are better off when thought as actual tangent vectors instead of fancy algebraic shit

@TedShifrin I mean intuitively that makes sense

@Adeek It does?

yeah

not algebraic, Balarka, analytic.

the stablizer of that , would be the normalizer of H in G

@Adeek How so?

It's perfectly reasonable to think of tangent directions as directional derivatives in those directions.

ba*H = ba (H) (ba)^-1

@KasmirKhaan the stabilizer of ba(H)(ba)^-1? what are a and b?

I don't understand what you're doing

@TedShifrin But derivations do not always come from the directional derivative operator, is the issue here

@anon Hmm I know its not very clear, but what do we call something like aHb ?

doesn't matter, call it a biset

They do with smoothness. It's unsmoothness that's unnatural.

I mean the tangent vectors are just arrows emanating at a point p. So if you think infinitesimally then one way of thinking of tangent vectors is just as ted said as directional derivatives and as you varry the directional derivative you get all the arrows

Too many functions.

Yeah, but you need smooth functions, Karim. That's the point.

The values for a that work are any real number, b is any x, y, such that b = (x+y)/2 and c is any x,y such that c=(x-y)/2. However, when we choose x and y be and c are automatically evaluated

It's a wonderful exercise.

okay we take that biset, muliply by b of the left and a on the right @anon

@Adeek That doesn't explain why it's false in $C^k$

*b and c

For $k < \infty$

Correct, @sangstar.

@KasmirKhaan you mean the biset (ab)H(ab)^-1? what are a and b?

Karim: I've posted that exercise here before. Should I do so again?

@TedShifrin yeah very nice exercise

yeah @TedShifrin

But does the fact that picking any x and y automatically evaluate b and c take away from the freedom to choose any entries for our diagonal matrix?

that would be nice