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Let $A=\{z\in\mathbf{C}:1/2<|z|<1\}$ be an open annulus. Let us cover $A$ by 3 open sets: $U_0,U_1$ and $U_2$ which we assume to be all homeomorphic to a 2 dimensional open disc. Moreover, we assume that $U_{ij}:=U_{i}\cap U_j$ are also homeomorphic to two dimensional open discs and that $U_0\cap U_1\cap U_2=\emptyset$. Let $\mathfrak{U}=\{U_0,U_1,U_2\}$.

Let $\mathcal{O}_A$ be the structure sheaf of $A$ where we think of $A$ as a complex analytic manifold. Since $A$, $U_i$'s and $U_{ij}$'s are connected open Riemann surfaces they are Stein manifolds and thus $H^q(A,\mathcal{O}_A)$, $H^q(U_i,\mathcal{O}_A|_{U_i})$ and $H^q(U_{ij},\mathcal{O}_A|_{U_{ij}})$ all vanish for $q\geq 1$. Therefore I would expect $$ \check{H}^{1}(\mathfrak{U},\mathcal{O}_A)\simeq H^1(A,\mathcal{O}_A)=0. $$ Since $$ C^1(\mathfrak{U},\mathcal{O}_A)=\mathcal{O}_A(U_{12})\times \mathcal{O}_A(U_{02})\times \mathcal{O}_A(U_{01}). $$ I would expect every 1-cochain to come from a 0-cochain. So for example, one could take a triple of constant functions (which are clearly holomorphic) for example $$ (1,2,3)\in C^1(\mathfrak{U},\mathcal{O}_A) $$ Thus there should exist $$ (f_0,f_1,f_2)\in \mathcal{O}_A(U_{0})\times \mathcal{O}_A(U_{1})\times \mathcal{O}_A(U_{2}) $$ such that $df=((f_2-f_1)|_{U_{12}},(f_2-f_0)|_{U_{02}},(f_1-f_0)|_{U_{01}})=(1,2,3)$. But this is clearly impossible since $2-1\neq 3$!

Q: So why is Chech cohomology not computing $H^1(A,\mathcal{O}_A)$ here?

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I don't know but could it be that in the last line you need 2-1=3 on the triple intersection, but this is empty so there is no condition? –  Yogesh More Apr 15 at 2:57
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Are you expecting the Cech cohomology of the annulus to be trivial? –  Jon Beardsley Apr 15 at 3:41
    
I mean, I guess what I'm saying is that my initial guess would be that the first cohomology of the annulus would be one dimensional and that you've produced a generator, but I'm not so good at this stuff. –  Jon Beardsley Apr 15 at 3:46
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Jon, don't confuse the first cohomology with constant coefficients with the first cohomology with coefficients in the structure sheaf. Stein spaces are the complex analytic analogue of affine varieties and on them the cohomology of coherent sheaves vanishes. –  Denis Nardin Apr 15 at 4:31
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a similar story holds for "fine sheaves", ie sheaves with a partition of unity. For example the sheaf of smooth (or continuous) functions on any manifold is fine and therefore has no higher cohomology. Sheaf cohomology allows for fairly strange coefficients. –  bananastack Apr 15 at 4:54

2 Answers 2

up vote 16 down vote accepted

Functions $(f_0,f_1,f_2)$ as you requested do in fact exist. Your confusion comes from the fact that you are trying to impose the cocycle condition on the intersection $U_0\cap U_1\cap U_2$, which is empty. That is the $f_i$'s are such that $$ f_2(x) = f_1(x)+1\ \forall x\in U_1\cap U_2$$ $$ f_2(x) = f_0(x)+2\ \forall x\in U_0\cap U_2$$ $$ f_1(x) = f_0(x)+3\ \forall x\in U_0\cap U_1$$ To get a contradiction you'd need an $x$ such that these three things simultaneously hold. But such an $x$ doesn't exists.

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Thanks Denis for clearing away my confusion. –  Hugo Chapdelaine Apr 15 at 13:19

Choosing appropriate branches of $f_j=\frac{1}{2\pi i} \log z$ gives $1$ on the $U_{ij}$ of your choice, and $0$ on the others. Adding multiples of these gives the desired cochain.

On a related note, I found the following very perplexing at the time: $\bar{\partial}$ cohomology is trivial on the annulus as it's a Stein manifold, so there must be a smooth function $f$ with $\bar{\partial}f=\bar{\partial}\theta$, where $\theta$ is the multivalued angle function. But doesn't this imply we can take $f$, add something holomorphic, and get something that's not even continuous (namely $\theta$ away from a branch cut)?

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Thanks for the explicit solution involving the log! –  Hugo Chapdelaine Apr 15 at 13:32

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