it's good to ask a lot

When a book says "all higher sheaf cohomology groups", do they mean $\geq 1$ or $\geq 2$?

geq 1

are you looking at defn of acyclic

Yes :P

I'm trying to understand the following sentence:

Using the $\bar\partial$-Poincare lemma and the fact that the sheaves $\mathcal A^{p,q}_X$ are acyclic, we see that Dolbeault cohomology computes the cohomology of the sheaf $\Omega^p_X$.

I don't see how it follows :P

it's just the general fact that the cohomology of a sheaf is the same as the cohomology of an acyclic resolution of that sheaf, and you have the resolution $\Omega^p_X \to \mathcal A^{p,0}_X \to \dots$

that 'general fact' also eg immediately implies the de Rham theorem

So when I learned sheaf cohomology I learned the Cech way

that's fine

I don't know this acyclic resolution stuff, and Huybrechts doesn't introduce it

the above is still true

Can I see this stuff the Cech way?

everything is algebra once you have a definition and basic properties of sheaf cohomology, not really anything about 'the cech way'

let $\mathcal F$ be a sheaf, and $0 \to \mathcal F \to \mathcal S_0 \to \mathcal S_1 \to \dots$ an exact sequence of sheaves where $\mathcal S_i$ are acyclic

okay

btw I should have "$\mathcal A^{p,0}_X$" up above

Corrected :P

So yeah that's just inclusion the first map and then it's exact as a sequence of sheaves by the $\bar\partial$-Poincare lemma

then if $\mathcal Z_i$ is the kernel of each arrow in that sequence, we know the sequences $0 \to \mathcal Z_i \to \mathcal S_i \to \mathcal Z_{i+1}\to 0$ are exact

that's precisely what it means that the sheaves are acyclic

and $\mathcal Z_0 = \mathcal F$

now use these exact sequences in cohomology. $H^p(\mathcal F) = H^{p-1}(\mathcal Z_1) = H^{p-2}(\mathcal Z_2) = \dots = H^1(\mathcal Z_{p-1})$

(b/c $\mathcal S_i$ is acyclic)

okay...

do you follow so far

That's very helpful

Yes

now use the LES in cohomology again - $H^1(\mathcal Z_{p-1}) = H^0(\mathcal Z_p)/dH^0(\mathcal S_p)$, which is precisely the same as $$\text{ker}(d: \mathcal S_p \to \mathcal S_{p+1})/\text{im}(d: \mathcal S_{p-1} \to \mathcal S_p)$$

So why is $\mathcal A^{p,q}$ acyclic?

it's a fine sheaf

don't know what that is

Ah

I can use partitions of unity

I saw this used to show that the sheaf of meromorphic functions has no 1-st degree cohomology in Forster's class

I guess that generalizes

Ok, so actually I don't get why that sequence with the $\mathcal A$'s is a resolution @Mike

that literally is the delbar lemma

Oh, it is. I thought so and then convinced myself it wasn't :P

Yeah, durr

Okay, so I'll try to run through your above argument in detail on my own

I'll report back to you if anything more comes up.

i think i have some typos flying around but i imagine you can fix them

@MikeMiller I'm sure I'll run into them ^^

@MikeMiller So here $\mathcal Z_{i+1}$ is, by exactness (of the resolution) equal to the image of the $i$-th map, right?

mhm

That confuses me a bit

That sequence in general

So you're saying that acyclicity is precisely what makes e.g. $\mathcal A^{p,0}_X/\Omega^p_X$ be the image of the first $\bar\partial$-map

sure, it's what gets me those successive isomorphisms with simpler cohomology groups

& the short exact sequences

@MikeMiller So I can't see how to get this sequence.

So the first step (that it's injective) follows from the exactness of the resolution

I don't understand the second map

you don't understand what it is or why it's exact there

the fact that the first map is injective is bc $\mathcal Z_i$ is defined to be a subsheaf of $\mathcal S_i$

why it's exact

nothing to do with exactness, it's just an inclusion

@MikeMiller Sure---I just thought about it as the image of the previous map

no reason to

But I guess that's not right because images are not always nice

the exactness of the second arrow also has nothing to do with exactness of the resolution. it's exact because the kernel of the map $\mathcal S_i \to \mathcal S_{i+1}$ is defined to be $\mathcal Z_i$

which is obviously the image of the inclusion of $\mathcal Z_i$

the exactness (surjectivity) of the third arrow is where you use that the resolution is exact

@MikeMiller Yes, but this is not the map $\mathcal S_i\to \mathcal S_{i+1}$. It's to a subsheaf of $\mathcal S_{i+1}$.

Why is the kernel not bigger? Acyclicity?

Ah, exactness

The image of $\mathcal S_i$ is already contained in $\mathcal Z_{i+1}$

@Danu You are not using any assumptions here whatsoever, other than that it's a complex of sheaves. The map has image contained in $\mathcal Z_{i+1}$.

So all I'm doing is restricting the codomain.

@MikeMiller Yeah, exactly

@MikeMiller I just didn't see immediately that the restriction doesn't change anything for the map

just coboundaries are cocycles

I mean, that's just a general thing

if a map between mathematical objects has image lying inside another mathematical object, then we can restrict without really changing anything

in other words, the map "factors" through the subobject

Yeah, yeah

I forgot about

i.e. that the image is contained in the next kernel

Thanks a ton for explaining that to me!!!

Okay, so everything else worked out fine @Mike

@Danu !!!

G'night, @MikeM ... happy almost birthday :)

Danu, did you ever sort out the transition functions for $\mathscr O(D)$? I haven't thought about it since ... I've spent the last 24+ hours trying to get a new computer up and running. Finally have done it.

Hi @TedShifrin!! Nice to see you again

@TedShifrin Yeah, I think I was right about it.

I don't :)

Balarka also agreed with me (I made him take a look :P)

@TedShifrin Did you see the last message I sent you about it?

I should just look at my notes.

The point is that he's obviously wrong on one hand of the equality

So the other has to be wrong too, for it to match again

(though I first discovered the not-so-obvious side being wrong)

I do agree that canonical bundle and det of tangent bundle are inverses :P

OK, let me think for a moment.

@TedShifrin I only found out about that when my aunt texted me today.

So is the fact that Dolbeault cohomology computes sheaf cohomology of $\Omega^p$ what is called the Dolbeault theorem?

Yes, @Danu.

You only found out = you prefer to forget? @MikeM

I don't think much about it.

That's what people my age say!

Meh, it kicks in earlier for our generation :P

@Danu: Part of the problem here is that my transition functions go from $j$ trivialization to $i$ trivialization. Is that your convention?

@TedShifrin Yup.

Okey dokey.

@Ted: I'll be spending it learning infty categories, apparently.

$\varphi_{ij}=\varphi_i\circ\varphi_j^{-1}$

OK, Danu, we agree on that.

wow i just found out that there's an exact formula for partition(n)

that really blows my mind xD

So, @Danu, you had $\phi_{ij} = f_i/f_j$?

I see @SAW is one week further into his Budapest experience :)

@TedShifrin yes.

OK, @Danu. I apologize. You be right, as usual :)

As does Huybrechts when he actually defines $\mathcal O(D)$. He contradicts himself later on

@TedShifrin Okay, that's good to hear :D

Didn't learn this from Budapest, learned it from trawling through some of the high-voted Stackexchange questions. xD

My professor even said Wednesday that partition(n) is unknown. xD

You mean trolling, @SAW?

@SAWblade You mean number of ways of partitioning $n$?

what a gross word, ted, and yes, danu

Fun fact: It is very closely related to the free boson partition function in string theory

(partition function! :D)

given that they both contain partition i would imagine so xD

And then there are the partition functions of statistical thermo.

Yeah

They're the same notion

OK, @Danu, so what's the question to stump me tonight?

It's funny that the focus on partition functions kinda disappears in quantum mechanics/QFT but then in strings it completely reappears

Like... everything is about partition functions!

@TedShifrin Oh, I didn't do much work today... So nothing much yet. I had something small

Méchant garçon :P

In one proof, Huybrechts suddenly starts using notation which I think is about inverse image sheaves n stuff which I don't know about

So maybe you'd like to explain the proof to me? :)

Relevant part:

I tried to just be like... Okay I have a map between the bundles and hurr durr somethingsomething holomorphic blahblah it induces a map between sections

Is that going to work, or do I really have to do ^^ that stuff

$f^{-1}\Omega_Y$ is the notation Huybrechts uses for the inverse image sheaf (?) in the appendix

@TedShifrin I'm not sure what you mean, btw. Did you really mean to say "mean boy"? Or did you mean "bad boy"? :P

It's just like the pullback bundle, yes. Nothing esoteric. $f^{-1}$ is more complicated.

Méchant means naughty :P

@TedShifrin Means "mean" to me

Hmm ... not to me. I wonder if that's changed in French in the last decade. Where's @Hippa or @Ramanewbie when you need 'em?

Hmm :P

The examples I was taught involved bullying of schoolchildren :P

It's odd that he's using the resolution to argue it. I would have argued that it's just pullback of holomorphic 1-forms on open sets. Maybe I'd better check his definitions.

I still remember the line "Méchant garçon" from one of my favorite movies of all time, called Mon Oncle (directed by and starring Jacques Tati). An older woman clucks that line at kids who are being mischievous.