thats ridiculously accurate

@BalarkaSen i guess my point would be we have learnt to capitalize on decay, and thats a pretty disgusting thing

darkest timeline

@BalarkaSen amen

If $f\in M_1(23,\chi_{-23})$ is an eigenform for the Hecke operator $T_2$, is there a way to show $f$ is an eigenform for all $T_m$?

if anyone knows its @EdwardEvans

the physicists torus: $T_2 = S_1 \times S_1$

AGH.

Years ago I knew what a Hecke operator was.

surprised to see the number 23 in there

any good random topology facts about that number @BalarkaSen

from time to time one hears these "random" numbers like "28 exotic spheres in seven dimensions"

And doesn't string theory say we're in 28 dimensions?

google says "In bosonic string theory, spacetime is 26-dimensional, while in superstring theory it is 10-dimensional, and in M-theory it is 11-dimensional. "

close enough

Oh right, 26.

It's all just numerals to me.

@user2103480 No clue haha

@TedShifrin Sorry, I was gone for a bit. Which line bundle are you talking about?

Something is unknown about exotic spheres in dimension 126 or 127

Can't remember what

Oh, go back to the beginnings of my explanations.

Canonical divisor <---> canonical line bundle = cotangent bundle

@BalarkaSen Crazy.

You're writing it in arrows, now Thorgott will have no trouble understanding

Smart, wasn't I?!

this is how we should communicate with him

@Thorgott ^^^ -> -> <- ^

Whoops

@BalarkaSen now you've got 30 lives

loool

It's 26 in bosonic string theory only if you want the Weyl anomaly to vanish (you have to add 26 bosons which contribute each a central charge of 1, to the central charge of -26 to give you c=0 which the anomaly is proportional to)

@Balarka: Michael's answer here is interesting. What's a complex vector bundle with no conjugation? I feel stoooopid.

O(1) + O(1) is trivial as a real bundle but nontrivial as a complex bundle so I suppose it cannot be complexification of a real bundle

Is that wrong? Maybe

You are talking about the $\Bbb C^2$ bundle on $\Bbb P^n$?

Yes

Why is it trivial as an $\Bbb R^4$ bundle?

Actually I meant over $\Bbb P^1$. In that case, $O(1) \oplus O(1) = O(2) \oplus O(0)$, and $O(2)$ is the tangent bundle on $\Bbb P^1$. But as a real bundle that's $TS^2 \oplus \underline{\Bbb R}^2$, which is trivial.

Whoa. Slow down.

Is "canonical line bundle" just a different name for cotangent bundle? In any case, I don't know what that correspondence is supposed to be.

How did you get $O(1)^2 \cong O(2)\oplus O$?

Euler sequence

@Thor: For curves, yes. What is your definition of the canonical divisor?

I explained this more in subsequent comments. You should read everything I wrote.

@EdwardEvans If I have that $f\in M_1(23,\chi_{-23})$ is an eigenfunction for $T_2$, is there a way to show it is an eigenfunction for all $T_m$? I have a feeling there must be a way involving some trickery with the relations between $T_m$'s.

So the sequence splits over $\Bbb R$ is your point, a @Balarka. OK.

Yeah.

I see; that was sneaky clever, Balarka.

I don't think I've ever thought about this issue before. How embarrassing.

it's a divisor associated to a meromorphic 1-form, i.e. $\operatorname{div}(\omega)=\sum\operatorname{ord}_p(\omega)p$

Right. And a meromorphic $1$-form is a meromorphic section of the (holomorphic) cotangent bundle.

So the divisor corresponds <---> to that line bundle.

In general, given any complex line bundle, you can look at the divisor of any of its meromorphic sections. All such divisors are linearly equivalent.

@TedShifrin Thanks! But is this really an example? Why isn't "coordinate wise conjugation" a conjugation for O(1) + O(1)?

I suspect O(1) + O(1) is just the complexification of O(1), treated as a real bundle!

Isn't $L\otimes C \cong L\oplus \bar L$ for a complex line bundle?

Ah, of course.

Yes.

So confuzling.

Agreed

so you're saying there's a genuine correspondence between linear equivalence classes of divisors and line bundles (up to isomorphism, I assume)?

yes

Jawohl.

Something something picard group something noetherian schemes

Oh please. This is very concrete.

I have no clue what this means, but I know it

Explain it to me in full generality, Astyx

ah, interesting

so what does this have to do with Euler characteristic?

I can try

@Astyx I'll be in garbology, you can keep typing stuff there

Euler characteristic is Euler characteristic of the tangent bundle.

Let X be a scheme. The generalization of a line bundle L over X is called an invertible sheaf and is defined as a locally free sheaf of rank 1 (because that's the iff condition for $L \otimes Hom_{O_X}(L, O_X) =O_X$). Invertible sheaves up to iso define a group called the Picard group.

@Thor: Poincaré-Hopf connects to the homological/simplicial definitions.

how do you see $L \otimes Hom_{O_X}(L, O_X) = O_X$?

for a loc. free sheaf of rank $1$

so you're saying the degree of a divisor corresponds to the Euler characteristic of the corresponding line bundle?

That's my famous post, Balarka :) $L\otimes L^* \cong \mathscr O_X$ :P

@TedShifrin yeah but I want Astyx to explain it to me sheaf theoretically or whatever

Yes, @Thor.

I am not thinking, just learning

Isn't the argument the perfect pairing argument plus dimension?

Yeah something like that

wait no, why isn't that nonsense?

it's the map $\phi\otimes m \to \phi(m)$

I don't know how to write down the argument, you should tell me :)

all line bundles have the same Euler characteristic (that of the base space)

wot Thorgott

NOOOO.

spanks Thor and sends him to his room

@Astyx OK, so this is at the level of sections over some open subset. Why isom of sheaves?

??

What is Euler characteristic of a line bundle?

Because you want to consider the sheaf on the whole space and not just locally(?)