6
$\begingroup$

Let $X$ be a complex compact manifold, and write $\mathcal{O}_X$ for the sheaf of holomorphic functions on $X$. Let $\mathcal{O}_X^{\times}$ be the subsheaf consisting of holomorphic functions. These are both sheaves of abelian groups. If we identify $H^1(X, \mathcal{O}_X^{\times})$ with the Picard group $\text{Pic}(X)$, then we can regard classes of $H^1(X, \mathcal{O}_X^{\times})$ geometrically as iso classes of line bundles on $X$.

Surely, one way to obtain from an abstract class $[\alpha] \in H^1(X, \mathcal{O}_X^{\times})$ the line bundle back is via the Cech cocycle data containing the whole glueing information about the associated line bundle up to isomorphism.

Recently I saw a different construction associating to an abstract class $[\alpha] \in H^1(X, \mathcal{O}_X^{\times})$ a representing line bundle and I would like to learn more about the background of this construction and why this gives finally a line bundle which coincides with that one obtained from Cech cycle. Does it has a name under which it can be looked up in literature? What is it's geometrical origin? Is there some construction from (differential) topology motivating it?

Here the construction: We take some injective resolution $1 \to \mathcal{O}_X^{\times} \to \mathcal{I}^0 \to \mathcal{I}^1 \to ... $ Then a $[\alpha] \in H^1(X, \mathcal{O}_X^{\times})$ is represented by construction by a section $\alpha \in H^0(X, \mathcal{I}^1)$ becoming zero in $H^0(X, \mathcal{I}^2)$.

We consider the bundle $\mathcal{M}$ obtained as fibre product

$$ \require{AMScd} \begin{CD} \mathcal{M} @>{} >> \mathbb{Z}_X \\ @VVV @VV{\alpha}V \\ \mathcal{I}^0 @>{}>> \mathcal{I}^1 \end{CD} $$

where $\mathbb{Z}_X \to \mathcal{I}^1$ is induced by $\alpha$ via $1 \vert _U \to \alpha \vert _U $. Since $ \mathcal{I}^0 $ contains $\mathcal{M}$ and $\mathcal{O}_X^{\times}$ as abelian subsheaves, there is a action $\mathcal{O}_X^{\times} \times \mathcal{M} \to \mathcal{M}$ by $\mathcal{O}_X^{\times}$, which restricts on $\mathcal{O}_X^{\times}$ to the usual multiplication $\mathcal{O}_X^{\times} \times \mathcal{O}_X^{\times} \to \mathcal{O}_X^{\times}$.
It's well defined because for $U \subset X$ the sections $\mathcal{M}(U)$ are given as sections in $\mathcal{I}^0(U)$ mapping to $\alpha \vert _U$ and $\mathcal{O}_X^{\times} \to \mathcal{I}^0 \to \mathcal{I}^1$ is exact.

Then we obtain an invertible sheaf $\mathcal{L}$ as $(\mathcal{M} \times \mathcal{O}_X)/\mathcal{O}_X^{\times}$ by moding out the diagonal action.

The claim is that this coincides with the line bundle obtained from class $[\alpha]$ via Cech cycle. Why? The construction reminds me on algebraic analoga of topological associated bundle. The aim behind the construction is obviously to exchange the fiber $\mathcal{O}_X^{\times}$ by the "right" fiber $\mathcal{O}_X$ giving the resulting object the desired structure of a line bundle. (This fiber replacing business is essentially for what the construction of associated bundle good for. Morally one wants to keep the same base space & in certain sense the same twisting behaviour, but replace the fiber)

But I not see from this construction why this one gives up to isomorphism the same line bundle one obtains from the Cech cocycle data.

$\endgroup$
3
  • $\begingroup$ This does not seem to be literally correct. Consider the simplest case $\alpha=0$. We get $\mathcal M=\mathbb Z_X\times\mathcal O_X^\times$, and the quotient you get is $\mathbb Z_X\times\mathcal O_X$, if I am not mistaken. In general, you will get something like $\bigoplus_{n\in\mathbb Z}\mathcal L^{\otimes n}$, not $\mathcal L$ itself. $\endgroup$
    – Z. M
    2 days ago
  • $\begingroup$ @Z.M: could you sketch the idea how to see in case of $[\mathcal{L}]=\alpha \neq 0$, ie not representing the trivial line bundle, that $(\mathcal{M} \times \mathcal{O}_X)/\mathcal{O}_X^{\times}$ has the form $\bigoplus_{n\in\mathbb Z}\mathcal L^{\otimes n}$? Note that after taking quotient on level of presheaves, we have to sheafify it, and this can lead to structure of local sections which strongly differs from the 'naive' quotient, even if the stalks not change when proceeding to sheafification $\endgroup$
    – JustusC
    2 days ago
  • 1
    $\begingroup$ It should have been a coproduct in sheaves of sets, not a direct sum as I wrote. $\endgroup$
    – Z. M
    yesterday

1 Answer 1

2
$\begingroup$

Let $\mathcal M$ be the sub-sheaf of $\mathcal I^0$ of sections mapping to (the restrictions of) $\alpha$. (I suppose that's what the OP has intended anyways.) That's easily seen to be an $\mathcal{O}_X^\times$-torsor. By Stacks Project Tag 0A6G, if $X=\bigcup_i U_i$ is an open cover trivialising this torsor, say via sections $s_i\in\mathcal M(U_i)$, then the Čech cocycle defined by the $s_{ij}\in\mathcal O_{U_{ij}}^\times$ such that $s_{ij}(s_i|_{U_{ij}})=s_j|_{U_{ij}}$ does indeed represent $[\alpha]\in H^1(\mathcal O_X^\times)$.

Moreover, let $\mathcal L$ be the quotient sheaf of the action of $\mathcal O_X^\times$ on $\mathcal M\times \mathcal O_X$ which acts on sections as $s.(m,z)=(sm,s^{-1}z)$ – that's not quite the diagonal action, but exactly what you should be familiar with from the associated bundle construction. $\mathcal L$ is trivialised over the same open cover by the classes $[s_i,1]$ of the sections $(s_i,1)\in \mathcal M(U_i)\times\mathcal O_X(U_i)$ and by construction, $s_{ij}[s_i,1]=[s_i,s_{ij}]=[s_{ij}s_i,1]=[s_j,1]$, i.e., the cocycle yields the transition functions of the associated line bundle.

By the way, the construction works quite generally, cf. Stacks Project Tag 040E.

$\endgroup$
3
  • $\begingroup$ I'm a little bit confused about the point that one can regard $ \mathcal{M} $ as subsheaf of $\mathcal{I}_0$. Take for example the class $\alpha=0 $ as suggested by @Z.M. in the comment above. Then $\mathcal M$ should be equal $\mathbb Z_X\times\mathcal O_X^\times$ and this not looks like something beeing subsheaf of $\mathcal{I}_0$. Do you see where the mistake appears in the considerations? $\endgroup$
    – JustusC
    2 days ago
  • 1
    $\begingroup$ What I'm saying is that we shouldn't be taking $\mathcal I^0\times_{\mathcal I^1}\mathbb Z_X$. To avoid the issue mentioned by Z.M, I'm proposing a different definition of $\mathcal M$ just to make the statement correct that its sections (or germs) are those of $\mathcal I^0$ mapping to $\alpha$. $\endgroup$
    – Ben
    yesterday
  • 1
    $\begingroup$ Another way to put this is that the morphism $\mathbb Z_X\to\mathcal I^1$ in your fiber square should be replaced by $1_X\xrightarrow{\alpha} \mathcal I^1$. Otherwise, you will get the union of all the torsors corresponding to the integral multiples of $\alpha$, which is why in the end you would be getting the sum of all the integral powers of the line bundle you're after. $\endgroup$
    – Ben
    yesterday

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.