Mathematics - 2017-07-21 (page 3 of 3)

My main results are calculations for Chern numbers of invariant almost complex structures on some $G_2$ homogeneous spaces (flag manifolds) + a uniqueness result for Kaehler structure on a certain manifold @PVAL

Maks

I'm sorry, my english got a bit rusty D:

Ted Shifrin

@Maks: I mean that you should be able to tell me equations about $a,b,c,d$.

PVAL-inactive

So well have this S^6 business settled soon I guess.

Danu

9:05 PM

@TedShifrin I think they were very happy with the talk. All the PhD students told me it was obvious I could get the position afterwards, which made me feel really good about it. The only time they stopped was to ask whether that rigidity result is "mine" or not :D

Ted Shifrin

Oh, what have you heard, @PVAL?

That's great, @Danu. I think you've learned a lot.

Danu

@PVAL-inactive I found that a complex structure on $S^6$ yields two non-standard complex structures on the quadric in $\Bbb CP^6$

(this was definitely already known to Robert Bryant)

(but I don't think anyone ever wrote it down)

PVAL-inactive

@Ted That was supposed to be a mild joke.

Ted Shifrin

Oh, ok.

Robert knows most everything :P

PVAL-inactive

I know studying G_2 invariant structures was like Chern's idea in trying to kill that problem though.

Ted Shifrin

9:07 PM

Right. But Robert didn't see how to push it through.

Danu

Right... I didn't really focus on $S^6$.

Maks

$ax - cx + dx = 0$ and $by = y$ that's what you mean ?

Ted Shifrin

So how can that hold for all $x,y$, @Maks?

Danu

I just use the fact that one of my spaces fibers over $S^6=G_2/SU(3)$ since it is $G_2/U(2)$ where the $U(2)$ sits inside $SU(3)$.

And the $G_2/U(2)$ is diffeomorphic to the quadric in $\Bbb CP^6$, but also to $\Bbb P(TS^6)\cong \Bbb P(T^*S^6)$.

PVAL-inactive

Can you show these structures aren't Kahler?

Maks

9:09 PM

well , $b = 1$ and $a - c + d = 1$

Danu

They cannot be, since they differ from the standard strcuture and that's the unique Kaehler structure.

The fact that they differ is proven in my thesis via Chern numbers. The uniqueness result is due to Brieskorn (in his thesis).

PVAL-inactive

I didn't know we knew that.

Ted Shifrin

Close, @Maks.

Danu

The statement is $X$ homeo to $Q$ and Kaehler $\implies$ $X$ biholomorphic to $Q$ @PVAL.

Ted Shifrin

@Danu: When you finish, please send me a .pdf of your thesis :)

Danu

9:10 PM

The rigidity result in my thesis is the exact same statement for another manifold (again of the form $G_2/U(2)$ but it's a different $U(2)$ subgroup)

PVAL-inactive

Aren't there lots of 1-parameter families of Kahler surfaces which are homeomorphic but not diffeomorphic?

Danu

Surfaces, yes

Right

I should've said dim>2

Ted Shifrin

Hmm, really? I know about families that are diffeo with diff complex structures ... What's an example, PVAL?

Danu

The Hirzebruch surfaces are the only things that mess it up

Ted Shifrin

Why are they not diffeo?

Danu

9:11 PM

Yeah, wait I'm not sure about that exact claim.

I just know that this Brieskorn result fails in dim 2 because of the Hirzebruch surfaces.

Ted Shifrin

I'm not sure I know this.

Maks

Ups, dont know where that one came from
$a - c + d = 0$

Danu

I also don't know this claim of PVAL's.

Ted Shifrin

Right, OK, @Maks. And the other equation gives you what?

Danu

But it sounded familiar/close to something I did know so I just started talking anyways ;)

Ted Shifrin

9:12 PM

Typical @Danu :D

Danu

Indeed

Mike Miller

@PVAL I assume you mean Stein structures on R^4 or something.

Ted Shifrin

Heya @MikeM. I'm confuzled about names.

Mike Miller

Go on?

PVAL-inactive

I thought there were examples of closed Kahler surfaces which were deformation equivalent which were known to not be isomorphic.

Ted Shifrin

9:13 PM

I met two young people today. One just finished first year at Princeton; the other was your UCLA friend.

Maks

$ b + c + d = 0$
$ ax = x$
Then $b = 1, a = 1, c = 0$ and $d = -1$

Mike Miller

Name @ Princeton?

Ted Shifrin

OK, @Maks. I'll trust you.

PVAL-inactive

and that these were the first examples plugged in to Donaldson's and Freedman's machinery.

Ted Shifrin

@MikeM: He never actually told me, but I assumed he was Justin or Dustin ... :)

Mike Miller

9:15 PM

@PVAL-inactive I think I misread what you said. Deformation equivalent things are diffeomorphic.

The non-diffeomorphic things are not deformation equivwlnt

Ted Shifrin

Yeah, I don't know examples of homeo but not diffeo ... in the complex world.

Maks

Thanks @TedShifrin :D

Danu

@PVAL-inactive Are you talking about these? These are diffeo but not isomorphic as complex manifolds

Ted Shifrin

Anyhow, @MikeM: I didn't know you know the Princeton guy, too.

You're welcome, @Maks :)

Mike Miller

Dustin is the UCLA guy. I thought you were saying I knew the Princeton guy.

@Ted Many examples in 4-manifold land just not deformations of one another.

Ted Shifrin

9:18 PM

Oh, no, I told you I was confuzled. I thought UCLA guy had a different name (Deven)? He told me he'd chatted with me on MSE, but didn't tell me his name here.

Yeah, all the families I know from deformation theory are obviously about deformations, @MikeM.

Mike Miller

There was a Dustin at UCLA who is my friend. Deven was from UCLA and is my boss.

Ted Shifrin

Oh, I didn't know you were doing AoPS now.

Now I'm totally confuzled.

Mike Miller

I've done an online class with them for a year.

PVAL-inactive

I can't find the examples I was looking for.

Ted Shifrin

Very cool. I hadn't known that.

PVAL-inactive

9:22 PM

The family I had in mind might not be deformation equivalent I guess.

I'm pretty certain that many of the first examples of exotic closed 4-manifolds were Kahler things that were distinguished by Kahler means.

So they were obvious things to try.

Ted Shifrin

I wish we had done formal introductions. I must have met Dustin, then. And the Princeton guy knew me from the videos (he had Gunning — 85 years old, he said — for the Honors analysis course).

I don't know that stuff, @PVAL.

PVAL-inactive

Well the Dolgachev surface is certainly Kahler

and I guess that was the first example I can find, but I guess that isn't def. equiv. to E(1) either.

Mike Miller

@PVAL What you're saying now is correct. It's supposed to be elementary that deformations are diffoemorphic I think.

Ted Shifrin

That's the theorem you all learned for your qualifying exam, @MikeM. :)

PVAL-inactive

@Ted enlighten me.

Ted Shifrin

9:33 PM

I thought that's essentially the content of Ehresmann's Theorem for proper submersions.

Mike Miller

Oh.

PVAL-inactive

Yeah I guess that works.

You can double the cobordism and work over base S^1 if you want to be pedantic and just use the compact version.

Mike Miller

Meh.

Daminark

10:40 PM

Hey @Alessandro!

Done with exams?

Alessandro Codenotti

10:52 PM

hi @dami

I'll have one in September, but apart from that I'm free now

Daminark

Woohoo

Daminark

11:06 PM

@Alessandro check our chat

Hey @Semi

Semiclassical

hey-o

Daminark

How's it going?

Typhon

@AkivaWeinberger cool. Sounds good.

@AkivaWeinberger im more busy now. Looking into modular arithmetic as it relates to quadratic rings. In particular I am looking at the modular arithmetic of these rings.

I think some properties might be useful in that.

Semiclassical

reminds me. one of the pieces of number theory / modular arithmetic I never learned about is quadratic reciprocity

at some point i should read up a little on that just for my own edification

Typhon

oh my god

the law of quadratic reciprocity was something that the professor handwaved on the very last day to reveal which primes were and were not equal to norms of certain elements

Maybe I'll stumble onto that.

Semiclassical

11:17 PM

Neat.

I know that's a highlight of classical elementary number theory, but not much more than that.

Typhon

hmm

I've been doing it all on my own.

Semiclassical

yeah.

Typhon

In fact, I have a library of minecraft books I've been sharing with the server.

I decided to play it for a bit this week and well... books. XD

Semiclassical

My main knowledge of modular arithmetic is from personal reading and from a one-semester abstract algebra course.

Typhon

Oh. I know of normal modular arithmetic.

Semiclassical

11:19 PM

But the latter doesn't really get near to quadratic reciprocity.

Typhon

I'm talking about things like $a \equiv b (mod \sqrt{3})$

interestingly enough

Semiclassical

Sure. I'm talking about this: chat.stackexchange.com/rooms/36/mathematics

um what

facepalm

?

11:21 PM

how did I copy-paste that

Typhon

the link goes back to this room

Semiclassical

Quadratic reciprocity

In number theory, the law of quadratic reciprocity is a theorem about modular arithmetic that gives conditions for the solvability of quadratic equations modulo prime numbers. There are a number of equivalent statements of the theorem. One version of the law states that for p and q odd prime numbers, ( p q ) ( q p ) = ( − 1...

Daminark

Do you mean, recursion?

Typhon

hmmm

@Daminark yes, and it forms a mathematical ring.

@Semiclassical interesting. The integers which are norms of elements in Z[sqrt{3}] are the values such that the legendre symbol of (k/3) = 0 or 1.

interesting

very interesting

wait no

only the 1's are guarunteed.

actually nevermind

this only corresponds to ones such that |k| = |N(x)|

Semiclassical

"The early proofs of quadratic reciprocity are relatively unilluminating. The situation changed when Gauss used Gauss sums to show that quadratic fields are subfields of cyclotomic fields, and implicitly deduced quadratic reciprocity from a reciprocity theorem for cyclotomic fields."

I don't really understand that, but it sounds neat.

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