Mathematics - 2016-10-17 (page 4 of 4)

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Ted Shifrin

19:00

If $\langle\cdot,\cdot\rangle$ is a hermitian inner product, what is $\langle c\cdot,c\cdot\rangle$?

Danu

norm squared

times

Ted Shifrin

Right.

Danu

I suck at linear algebra :D :D

Also I honestly think my shitty notation is making things worse

0celo7

@Danu lol

Ted Shifrin

So the hermitian inner product transforms by the norm squared of the transition functions (or inverse thereof).

Now you know how transition functions of the dual work.

Danu

19:01

So divide by the norm squared

Or multiply by inverse times its complex conjugate (?) if it's not a line bundle I guess

Ted Shifrin

OK ... Sort it out.

What we're saying generalizes to higher rank bundles, but then you need hermitian matrices, etc.

Danu

Yeah, that should be pretty simple in principle. Let's not dwell on it.

Ted Shifrin

I'm done.

Danu

not too worried about that

@TedShifrin I'm sorry for not being useful today :P

Ted Shifrin

Just takes time to write things out carefully. What's next before I leave?

Mary Star

19:05

Let $K=\mathbb{Q}(a^2)$, then $[\mathbb{Q}(a^2):\mathbb{Q}]=\def (x^2-2x-1)=2$. Is this correct? @TedShifrin

Danu

Well, I should still figure out why this is the same as the restriction from the trivial bundle---but perhaps I should do that myself.

Ted Shifrin

Right, @MaryStar.

Mary Star

Great!! Thank you!! :-) @TedShifrin

Danu

I will leave you be for now---thanks for telling me to look at the transition functions

Ted Shifrin

Okey dokey. I'll be back later.

Danu

19:06

Have a nice day!

By the way, I'm probably-definitely going to try to do my thesis project on complex geometry stuff

Mike Miller

I'm not sure how it happened but when I openeed the room right now I had everybody but Danu on ignore.

Danu

LOL

I feel honored

Mike Miller

Of course, I want the opposite.

Danu

(or caught in mod power abuse...)

Cry evrytiem

You killed the mood @MikeMiller

Balarka Sen

don't worry, I'm on his hypothetical ignore list too

Mike Miller

19:16

0celo7

I'm surprised when I'm not on it

Danu

::crickets::

Most of the Time

►

Balarka Sen

I like his "Stuck inside the mobile with the Memphis Blues again". Just sayin'

Danu

@MikeMiller what'd you think about Dylan winning the Nobel prize?

@BalarkaSen There is so much of his music that I like... :)

Mike Miller

good

Danu

19:20

:) Me too

I think it's a good gesture

I'm really getting interested in characteristic classes now

Balarka Sen

what're you reading

Danu

I read about the "Atiyah class" in Huybrechts today, which is supposed to contain all characteristic classes of a holomorphic line bundle (?!)

Mike Miller

Never read that formalism before but it makes some flavor of sense.

Balarka Sen

sanity check, $H^*(\text{Grass}(n, \infty); \Bbb Z/2)$ is just the polynomial ring over $\Bbb Z/2$ in $n$ variables of each grade upto $n$, yes?

Mike Miller

Why do you think that?

Danu

19:25

WHY WOULD YOU THINK THAT

(no idea)

Mike Miller

I was hoping for an answer as opposed to silence.

Balarka Sen

@MikeMiller: For no particular reason. I want to compute that ring, and wanted to check if I remembered it right.

Mike Miller

That's correct.

I don't know how you intend to compute it.

Balarka Sen

I think it should be doable by the Leray-Hirsch.

Or maybe there's something simpler. I vaguely remember the splitting principle.

Danu

Principle?

Balarka Sen

19:35

If I can write every vector bundle as direct sum of line bundles, there should be a classifying map to $\Bbb{RP}^\infty \times \cdots \times \Bbb{RP}^\infty$

Mike Miller

You can't, but ok.

Danu

That'd be a nice result, if true

Balarka Sen

Huh, maybe I misremembered Ted.

Mike Miller

It's obviously false, man. Take $S^2$.

Balarka Sen

duh

Mike Miller

19:36

The result is that you can always pull it back to some other manifold over which it does split.

Balarka Sen

ah, k

Danu

@MikeMiller A small elaboration, perhaps? :P

Balarka Sen

every line bundle on S^2 is trivial

Danu

...but $TS^2$ is not, I guess (Hairy ballz).

Balarka Sen

real line bundle

@Danu yes, sure

Mike Miller

19:49

There are a good number of nontrivial bundles over $S^2$. Some detected by $w_2$, the rest detected by $e$.

@BalarkaSen I don't remember how to do it off the top of my head. All your favorite authors prove it.

These are the sort of results I'm usually glad to just take for granted.

Balarka Sen

I am curious if there is a geometrically obvious homology ring isomorphism $(\Bbb{RP}^\infty)^n \to G(n, \infty)$.

Mike Miller

Oh, do it inductively. $O(n+1)/O(n) = S^n$. $BO(1) = \Bbb{RP}^\infty$. Use the Serre spectral sequence on $S^n \to BO(n+1) \to BO(n)$ with $\Bbb F$ coefficients.

@BalarkaSen That's false!!!

You should go to bed when you're making statements like that... The degrees of the generators of the polynomial rings are completely different

Balarka Sen

oops

Eh, I guess I'm done with math for today.

Mike Miller

Yes, Serre gives it quickly as long as you also know its multiplicative structure.

What I said is nonsense above. Oh well.

1 hour later…

SAWblade

21:30

holy shit guys i am two layers deep into this problem with generating functions

i am breathing capital sigmas lol

DHMO

@SAWblade ?

SAWblade

i'm doing a quite involved problem xD

DHMO

@SAWblade what is it?

SAWblade

Combinatorics. P:<

DHMO

@SAWblade what is the question?

1 hour later…

DemCodeLines

23:02

Anyone mind explaining to me how to find all the solutions of an equation on Z17 (integers mod 17)?

Ted Shifrin

@DemCodeLines: What kind of equation?

0celo7

@TedShifrin Have you heard of strong differentiability?

Ted Shifrin

I dunno. What does it mean?

0celo7

Semiclassical

hi chat

Ted Shifrin

23:13

Hi @Semiclassic

Semiclassical

kind've funny how my day works sometimes

Ted Shifrin

No, 0celo. It seems to be working with chords near $a$ rather than secants from $a$ to nearby points.

But I've never seen that before.

0celo7

@TedShifrin yep

Semiclassical

in the morning, i talked with my advisor about research stuff including riemann surfaces (though not in depth)

and then I spent the afternoon doing intro lab stuff with capacitors and voltage sources

Ted Shifrin

What's an example of a differentiable function that's not strongly differentiable?

0celo7

23:15

@TedShifrin According to my prof, they are "very rare and very pathological"

Semiclassical

so yeah, that's a bit of a difference

Ted Shifrin

Most of us have had such lives, @Semiclassic.

Semiclassical

yeah

0celo7

@TedShifrin It's in between differentiable and $C^1$

Ted Shifrin

Teaching, in general, is at a different level from research, @Semiclassic.

Semiclassical

23:16

well, certainly

and it's not a surprise by any means, just thought it was kind've funny

Ted Shifrin

@0celo: I was going to ask if my favorite is an example.

Not kind've, @Semiclassic. kind'f :P

Mike Miller

Who cares?

Samuel Yusim

hello

caring is usually a bad idea

0celo7

@TedShifrin the way you check is if the first order Taylor remainder is Lipschitz

Locally Lipschitz

Ted Shifrin

@0celo: Well, in this case, the Taylor remainder (at $0$) is the function itself.

I think it is an example.

Because of the oscillations we can choose $1/x$ to be an odd multiple of $\pi/2$ and $1/y$ an even multiple of $1/\pi$.

@0celo: Yup, if you choose $x$ and $y$ carefully, you don't go to 0 fast enough.

DemCodeLines

23:30

@TedShifrin x^2 + 1 = 0

Ted Shifrin

Aha, @DemCodeLines. Do you need some solutions or all?

DemCodeLines

all solutions, but you don't need to do it out. I think the way to find some solutions would be the same as the way to find all

Ted Shifrin

Well, you should be able to find 2 solutions very quickly. Then you need to decide if there can be any more.

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