Mathematics - 2016-10-08 (page 3 of 5)

Semiclassical

17:00

@0celo7 You might find this Terry Tao blog post interesting: terrytao.wordpress.com/2007/06/25/…

0celo7

> Thanks for the very nice discussion. I’ve been a fan of ultrafilters since learning about them in high school, although, not being in analysis, I haven’t particularly ever needed to use them.

High school???

Semiclassical

lol

yeah, that's a bit crazy

0celo7

@Adeek I'll go on a typo hunt after emailing my prof

There's probably a few

Balarka Sen

Hmm, I think I have figured out the thing I wanted to.

Semiclassical

oh?

Ted Shifrin

17:03

Howdy @Semiclassic, @Balarka, @0celo.

Semiclassical

howdy

Balarka Sen

Hi @TedShifrin

Semiclassical

oh, @ted, i sent you a ping re: my riemann surface

0celo7

@TedShifrin Hi

Semiclassical

so not as well-behaved as I'd thought

Ted Shifrin

17:05

Yes, @Semiclassic, that $(x,y)\rightsquigarrow (x^2,y^2)$ factorization of the map should say that we have nodes at $(0,0)$ and $(\infty,\infty)$, I bet.

Semiclassical

sounds right.

Danu

Hey @TedShifrin!!

Ted Shifrin

But to adjust the genus formula you need to know the precise nature of the singularity. I'm pretty sure these are just nodes.

Heya @Danu, stranger.

Semiclassical

it's also pretty easy to see that if you've got anything of the form $(x^2+y^2)^2=p(x,y)$ with $p(x,y)$ of degree 3 at most

Balarka Sen

Suppose I have a cobordism $B$ between two manifolds $M$ and $N$. Then $TM$ and $TM$ is stably $TB$ restricted to $M$ and $N$ respectively. That implies $TM$ and $TN$ have the same S-W classes (by using stability of the invariants, and then preservation of pullback), right?

Danu

17:06

I have a bunnnnch of questions, if you're up for it...

Ted Shifrin

Sounds right, @Balarka.

I may or may not be able to help, @Danu, but I'm game for a few minutes.

Danu

Okay, I'll stay with something a bit smaller then

So consider the following:

Semiclassical

then taking the projectivization with coordinates $[x:y:z]$ always gives singularities at $[1:\pm i:0]$

Danu

$H^2(X,\Bbb C)=(H^2(X,\Bbb Z)\otimes_{\Bbb Z}\Bbb R)\otimes_{\Bbb R}\Bbb C$.

Ted Shifrin

@Semiclassic: That can't be right. It depends on $p(x,y)$.

Semiclassical

17:09

that was a bit too quick, yes.

Ted Shifrin

@Danu: Why not just tensor with $\Bbb C$ in one swell foop?

Balarka Sen

@TedShifrin Actually maybe that doesn't quite make sense. What does it even mean to say S-W classes of two vector bundles over different bases are the same?

Semiclassical

i think it's true as long as $p(x,y)$ is a degree 2 polynomial in $x,y$

Danu

@TedShifrin Wait for it ;P

mercio

o.o

Danu

17:09

Anyways, replace the $2$'s by $1$'s, sorry.

Balarka Sen

But I am sure one can fix it a bit so it does make sense.

Semiclassical

since then the projectivization will be $(x^2+y^2)^2=z^4 p(x/z,y/z)=z^2 P(x,y)$ with $P(x,y)$ homogenous of order 2

Danu

Thats okay for me. But now Huybrechts says "this show that $H^1(X,\Bbb Z)\to H^{0,1}(X)$ is injective with discrete image that generates $H^{0,1}(X)$."

Ted Shifrin

@Balarka: What makes sense is S-W numbers (cup to get top classes and evaluate on fundamental class).

Danu

Why $H^{0,1}(X)$ (by the way, I have this decomposition because I'm working with compact Kahler manifolds)

Ted Shifrin

17:10

That's wrong, @Danu.

Danu

Damnit

Sorry

Typo'd

Of course it has to be $\Bbb Z$

(it goes through $H^1(X,\Bbb C)$ though)

Balarka Sen

Yes, now I am starting to see what the title of the next chapter is about. (title is precisely "S-W numbers", and that's all I have read of the next chapter).

Ted Shifrin

Ohhhh ... now it has a better shot. It's because $H^1(X,\Bbb Z)$ injects into $H^1(X,\mathscr O)$.

Mike Miller

@Ted I think he wanted to figure this story out himself.

Danu

Ah

I see

Mike Miller

17:12

I told him to calculate for surfaces. I don't know if he did.

Adeek

oh

hi @TedShifrin

Ted Shifrin

Hi Karim

Balarka Sen

@MikeMiller Tangent bundle of oriented surfaces are stably trivial, nope?

Mike Miller

I asked you to do all surfaces.

Balarka Sen

Oh, hm.

Danu

17:13

So why does $H^1(X,\Bbb Z)$ generate $H^1(X,\mathcal O)$?

Ted Shifrin

Think about rank and dimension, @Danu.

Remember what you know for compact Kähler (that needn't hold in general).

Semiclassical

what I should do now is look at some of the other examples of Riemann surfaces (within the context of the given physics problem) and see if that's consistent

and see if they've got the same sort of structure

Ted Shifrin

Degree 2 seems awfully restrictive, @Semiclassic.

Semiclassical

it does, but it matches the example I had.

hence why I need to look up the other examples I've seen

Ted Shifrin

Okey dokey. I'm still puzzled by my $x^2,y^2$ observation (which I realize needn't hold in general), but maybe I need to sit down and think more.

Danu

17:15

@TedShifrin Eh... $H^1(X,\mathcal O)=H^{0,1}(X)$ which is half dimension of $H^1(X,\Bbb C)$, as is $H^1(X,\Bbb Z)\otimes_{\Bbb Z}\Bbb R$ as a real vector space, since $H^1(X,\Bbb C)=\text{that}\otimes_{\Bbb R} \Bbb C$?

user3925758

Good morning guys.

Ted Shifrin

Right, @Danu.

Danu

I guess that's enough, because you then have an injection into same-dimensional space

Ted Shifrin

Of maximal rank, @Danu.

Danu

What of maximal rank?

Semiclassical

17:16

i mean, I think part of what's going on there is that if you write $(u,v)=(x^2,y^2)$ then $y(x)\,dx=\frac12 \sqrt{\frac{v}{u}}du$

Ted Shifrin

You could include $\Bbb Z$ into $\Bbb C$ and the lattice wouldn't generate.

@Semiclassic: I was thinking merely about singularities, not about the differentials.

Semiclassical

ah.

Ted Shifrin

@Danu The lattice :)

Danu

@TedShifrin Ah, you mean a maximal rank injection?

Ted Shifrin

Think about $\Bbb Z\hookrightarrow \Bbb C$ versus $\Bbb Z\oplus \Bbb Z\hookrightarrow\Bbb C$.

Danu

17:18

Yeah, I see what you mean. The image has not got full rank in the first case

I'm just struggling with how to word it

Ted Shifrin

I worded it :P

Danu

Is "the image is of maximal rank" alright, or not? I'm not sure what noun to put before "of maximal rank" :P

Ted Shifrin

lattice

= discrete subgroup

wowdavers

Question about a proof for the expectation operator: how do I prove |E(X)| ≤ E(|X|)?

Ted Shifrin

@wowdavers: What inequality do you know about integrals involving absolute values? (If you're doing discrete random variables, then the hint is triangle inequality.)

wowdavers

17:22

I know -|X| ≤ X ≤ |X|, and consequently

E(-|X|) ≤ E(X) ≤ E(|X|)

Ted Shifrin

You're almost there.

How do you show $|x|\le 3$?

wowdavers

Having a brain fart on the final step

|X| ≤ 3?

Ted Shifrin

No, I'm just talking simple math. What does a mean to say a number has absolute value at most $3$?

wowdavers

Ah, it's between -3 and 3?

Ted Shifrin

Righto.

Danu

17:27

OK @Ted I guess I'm okay with my understanding of this now. For my next question, you can choose between a "soft" (motivation) question or more technical stuff about the Albanese map.

Ted Shifrin

I don't get to see the questions before I choose?

0celo7

hah

wowdavers

So -E(X) ≤ |E(X)| ≤ E(X) ?

Ted Shifrin

Right, @wowdavers.

Danu

@TedShifrin The latter is rather a chain of questions.

I'll ask the soft question first---probably more fun.

So I have this "Hard Lefschetz Theorem" in my book---but I don't really see (i) what it means (ii) (as a consequence) why I want it

Ted Shifrin

17:28

@wowdavers: So I assume you realized $E(-X) = -E(X)$.

Danu

It gives me an isomorphism $L^{n-k}:H^k(X,\Bbb R)\to H^{2n-k}(X,\Bbb R)$

Ted Shifrin

That's not right, @Danu, is it?

Danu

It is

Secret

Group cohomology

In mathematics, group cohomology is a set of mathematical tools used to study groups using cohomology theory, a technique from algebraic topology. Analogous to group representations, group cohomology looks at the group actions of a group G in an associated G-module M to elucidate the properties of the group. By treating the G-module as a kind of topological space with elements of G n {\displaystyle G^{n}} representing n-simplices, topological properties of the space may be computed, such as the set...

O crap, incomprensible cohomology stuff. Better wait until much later before crunching through this

Danu

And it's weird, because I want to say I already have this from Poincare duality anyways :P

Ted Shifrin

17:29

Oh, right.

wowdavers

How do I come to the conclusion that $|E(X)| \leq E(|X|)$?

Ted Shifrin

@wowdavers: Because of what you said about $|x|\le 3$.

Danu

I mean from PD with universal coefficients (no torsion) I just get $H^k(X,F)\cong H^{2n-k}(X,F)$ for a field $F$ and a [nice] manifold of real dim $2n$

Balarka Sen

@Danu I think the map is important here, not just the fact that it's an isomorphism.

Ted Shifrin

Well, two things, @Danu. (1) is what Balarka just said. That you get the isomorphism by repeatedly intersecting with hyperplanes.

Danu

17:31

@BalarkaSen But I haven't seen it being used at all :\

Mike Miller

@Balarka Compute!

Danu

@TedShifrin I have no idea what that means---no geometric intuition at all for the Lefschetz operator.

wowdavers

Right, I have $ -E(X) \leq |E(X)| \leq E(X) $, but I don't have an absolute value sign inside any of the terms.

Ted Shifrin

(2) You get the Lefschetz decomposition of cohomology into things coming from primitive cohomology.

Danu

@TedShifrin Okay, this has been useful in some things already.

Ted Shifrin

17:31

You will see things later that tell you hypersurfaces (or more generally complete intersections) share certain levels of cohomology with that of their ambient spaces. This is very powerful and important.

@wowdavers: Stop and reread our earlier discussion.

Danu

Okay, so it's supposed to be kinda useless so far?

Ted Shifrin

You have to start studying actual projective manifolds, @Danu. But for a submanifold of $\Bbb P^n$, Poincaré duality tells you that wedging with the Kähler form is (dual to) intersecting with a hyperplane in $\Bbb P^n$. VERY powerful.

Danu

Other question, perhaps a bit stupid-sounding: Why is the Lefschetz decomposition orthogonal? I was able to prove it "by hand" with some trickery in the case I actually encountered but it should be somehow easy to see. Huybrechts says it comes from a formula for $[L^m,\Lambda]$ but I don't see that

Balarka Sen

@MikeMiller Only way I can see how to do it is to look at the oriented double cover. Pullback of $w(S)$ is $1$ by the map on cohomology induced by the covering map... still pondering.

Ted Shifrin

Oh. Something like $\text{im}{T^*} = (\ker T)^\perp$?

Mike Miller

17:34

@Balarka That's worthless. It's a double cover. It's going to kill mod 2 cohomology.

Ted Shifrin

No killing today.

Balarka Sen

Ah, true. Crap.

Mike Miller

What do you know about $w_1$? What do you know about $w_2$?

Ted Shifrin

Interesting. Someone just asked about Poincaré-Miranda (which I'd never heard of) and whether you can deduce Borsuk-Ulam from it. I suspect not. But look at that, @Balarka (and @MikeM if you care).

wowdavers

So $-E(|X|) \leq |E(X)| \leq E(X)$

Ted Shifrin

17:37

You're missing absolute value on the right.

Mike Miller

Trying to get some thinking done this morning.

Ted Shifrin

Answering all these unrelated questions in here is exhausting me. It's time for a nap.

(Plus I'm taking Benadryl to mitigate some of the side-effects of the antibiotics. So it makes me sleepy.)

wowdavers

Sorry, bear with me
I have these two statements which I'm combining:

$-E(|X|) \leq E(X) \leq E(|X|)$
$-E(X) \leq |E(X)| \leq E(X)$

Is that correct so far?

Ted Shifrin

No, @wowdavers. You have just the first line.

Danu

<3 @Ted you're a life-saver

Balarka Sen

17:39

Wiki seems to say it's equivalent to Brouwer fixed point theorem. I don't know the relation between B-T and Brouwer (I think it worked the opposite way, Brouwer is implied by B-U)

Ted Shifrin

But, as you already said, by definition that says $|E(X)|\le E(|X|)$. Done.

Apparently Miranda proved the equivalence, @Balarka.

What's B-T? You mean B-U?

wowdavers

Alright, then where do I go from there?

Balarka Sen

Right, typo.

Ted Shifrin

You go home, @wowdavers. That's what you wanted.

Balarka Sen

@TedShifrin Well, don't answer several unrelated questions at once :)

That's also a possible solution.

Ted Shifrin

17:40

I truly don't think Brouwer can prove B-U, @Balarka.

Balarka Sen

Me neither.

wowdavers

$E(X) \leq E(|X|)$ ?

Sorry if this is frustrating :( Not quite seeing it

Ted Shifrin

No, @wowdavers: $|E(X)|\le E(|X|)$.

wowdavers

Why? Where do the two terms come from? The left term comes from $-E(|X|) \leq E(X)$, yes?

Ted Shifrin

Remember from our discussion of absolute value that we said this is equivalent to $$-E(|X|)\le E(X)\le E(|X|).$$

wowdavers

17:43

Where does the abs. value sign inside the right term come in from?

Ted Shifrin

You said you got what I just typed by integrating $-|X|\le X\le |X|$.

wowdavers

Right

Ted Shifrin

Then if you remember what absolute value means, we're done.

@Danu: Did you understand why wedging with Kähler is P.D. to intersecting with hyperplane?

I'm shocked Huybrechts hasn't told you this.

wowdavers

Great, thanks a lot!

Ted Shifrin

Yippee @wowdavers. :)

wowdavers

17:46

Sorry for that frustration - I've got it now

Danu

@TedShifrin No

@TedShifrin No geometry over hurrr

Ted Shifrin

F*** that, @Danu.

Work in $\Bbb P^n$. What do you know about the Kähler form?

Danu

I agree...

Anyways, I see that $\omega$ will be PD to a hypersurface

Ted Shifrin

But do you know what generates cohomology $H^2(\Bbb P^n,\Bbb Z)$?

Danu

@TedShifrin Locally given by positive definite hermitian matrix blahblah

Osculates to order two

blahblah

@TedShifrin Yes, I know that

Ted Shifrin

17:48

I'm doing Fubini-Study on $\Bbb P^n$.

Danu

Well... Not geometrically :P

Ted Shifrin

You need to integrate the Kähler form over a linear $\Bbb P^1$ sitting in $\Bbb P^n$. What do you get? Everyone needs to do this at least four times.

Danu

I mean, I know $H^*(\Bbb P^n,\Bbb Z)$ is the (truncated) polynomial ring

@TedShifrin So the explicit expression for the Fubini-Study one?

Ted Shifrin

nods

Balarka Sen

@MikeMiller OK, how about I look at the map $f : S \to \Bbb P^2$ which squashes all but one $\Bbb P^2$ component? $w(S) = f^* w(\Bbb P^2) = f^*(1+\alpha+\alpha^2)$. I believe that is $1 + a + b$ where $a$ is the element of $H^1(K)$ which is the generator of the Z/2 factor and $b$ is generator of $H^2(K)$.

Danu

17:50

Okay, for $\Bbb P^1$ I know it gives $1$

Balarka Sen

$S$ is the Klein bottle. Sorry about that, just working out an example.

Danu

I'm going to say it does this always :P

Ted Shifrin

Aha. Yes, by homology.

So that's the same number you get by intersecting with a hyperplane.

Balarka Sen

But it shouldn't be different for other surfaces anyway.

Danu

Right, because the (co)homology of $\Bbb P^n$ agrees with the lower ones in the lower degrees

@TedShifrin Intersecting what?

The linear $\Bbb P^1$?

Ted Shifrin

17:51

We're talking just $\Bbb P^n$.

Yes, the linear $\Bbb P^1$.

Danu

With a transversal hyperplane

Ted Shifrin

Yup.

Danu

(also I have 0 visualization of geometry in $\Bbb P^n$)

Ted Shifrin

So, inductively speaking, you can next deduce what wedging with $\omega^k$ does.

Danu

Should I be thinking about spheres, like for the real case? or not?

Semiclassical

17:52

finally found an example in the literature which isn't (at first glance, anyways) equivalent to the Riemann surface I keep going on about

Ted Shifrin

No, alg. geometers draw the cartesian plane, with a line completing a triangle, that line representing the line at infinity.

Semiclassical

...wait, nope, still has same singularities.

oh well.

Balarka Sen

I think of $\Bbb P^n$ as $\Bbb C^n$, but the rest of $\Bbb P^{n-1}$ someplace I can't see

Danu

I cannot parse that sentence

Balarka Sen

And I can change perspective to see that hyperplane at infinity by changing charts

Ted Shifrin

17:53

I draw it in there ... so the picture is a triangle with all its edges extending ...

Semiclassical

(the case that's been studied in the literature corresponds to period-two resonances; what we were hoping to consider were period-3 instead)

Ted Shifrin

@Danu: I'm pretty sure you'll find pictures in those notes I sent you.

Danu

@TedShifrin A triangle??

Oh, that picture---I've seen something like that before

Ted Shifrin

Hang on. I'll draw it in Illustrator and paste it here.

Happy?

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