3
$\begingroup$

I am reading pages $132$ and $133$ of Principles of Algebraic Geometry by Griffiths and Harris. They consider a holomorphic line bundle $L \to M$ over a manifold $M$ and an open cover $\left\{ U_{a}\right\}$ associated to the holomorphic trivializations of $L$: $\phi_{a} : L|_{U_{a}}\to U_{a}\times \mathbb{C}$.

Then they explain how the corresponding transition functions $g_{ab} = \phi_{a}\circ\phi^{-1}_{b}$ represent a Čech cocycle in $M$ and that two sets of transition functions are equivalent if they differ by a Čech coboundary. Thus, the set of holomorphic line bundles on $M$ is classified by $H^{1}(M,\mathcal{O}^{\ast})$.

What I don't understand is: at what point the classification of the line bundles stopped depending on the open cover $\left\{ U_{a}\right\}$? As I understand it, what they have shown is that the set of line bundles with trivializations given in terms of the open cover $\left\{ U_{a}\right\}$ is classified by $H^{1}(\left\{ U_{a}\right\},\mathcal{O}^{\ast})$. Only if $\left\{ U_{a}\right\}$ is the finest open cover then I get the classification as elements of $H^{1}(M,\mathcal{O}^{\ast})$. What am I missing?

Edit: How the discussion of Griffiths and Harris can be extended to complex line bundles not necessarily holomorphic?

Thanks.

$\endgroup$

migrated from mathoverflow.net Apr 12 '15 at 4:40

This question came from our site for professional mathematicians.

  • 6
    $\begingroup$ Note that $H^1(M,{\mathcal O}^\times)$ is defined as the direct limit $\lim H^1(\{U_\alpha\},{\mathcal O}^\times)$ over the set of open covers with refinement as arrows between them. $\endgroup$ – Fan Zheng Apr 12 '15 at 2:26
  • $\begingroup$ @Bilateral: Sorry for not updating my answer yet, I will get to it soon. I first had to fix an error in another answer I gave. $\endgroup$ – Michael Albanese Apr 14 '15 at 13:41
3
$\begingroup$

Given a holomorphic line bundle $L$ with trivialising open cover $\mathcal{U} = \{U_{\alpha} \mid \alpha \in A\}$ and trivialisations $\{\phi_{\alpha} \mid \alpha \in A\}$, Griffiths & Harris construct an element $t_L \in H^1(\mathcal{U}, \mathcal{O}^*)$. They then show that $t_L$ does not depend on the choice of trivialisations.

In particular, if $L$ and $L'$ are holomorphic line bundles with trivialising open cover $\mathcal{U}$, then $t_L = t_{L'}$ if and only if $L$ and $L'$ are isomorphic. In order to prove the $\Leftarrow$ direction of the previous statement, you need to use the following fact: if $\mathcal{U}$ is a trivialising open cover for one of two isomorphic holomorphic line bundles, then it is a trivialising open cover for the other.

So far we can say that $H^1(\mathcal{U}, \mathcal{O}^*)$ is the set of isomorphism classes of holomorphic line bundles such that $\mathcal{U}$ is a trivialising open cover. If there is a holomorphic line bundle which isn't trivialised by $\mathcal{U}$, then this is not the same as the set of isomorphism classes of holomorphic line bundles.


The idea now is to pass to finer and finer open covers in an effort to capture all of the holomorphic line bundles. This will be achieved by passing to the direct limit of the vector spaces $H^1(\mathcal{U}, \mathcal{O}^*)$ over all open covers $\mathcal{U}$; i.e. $H^1(X, \mathcal{O}^*)$. Note, by the final sentence in the second paragraph, when passing to finer covers, the isomorphism classes don't get larger, there are just more of them.

I will make the previous paragraph (which is actually what you asked about) more explicit tomorrow as I don't have time to do it now.

$\endgroup$
  • $\begingroup$ thanks, looking forward for the update. You say: "The idea now is to pass to finer and finer open covers in an effort to capture all of the holomorphic line bundles." However, there might be line bundles $L$ for which these finer and finer open covers are not trivialising open covers for them, right? What I am asking I guess is: let $L$ be a line bundle and $\left\{ U_{a}\right\}$ a trivialising open cover. Is every other open cover finer than $\left\{ U_{a}\right\}$ also a trivialising open cover for $L$? $\endgroup$ – Bilateral Apr 12 '15 at 10:19
  • $\begingroup$ @Bilateral: Yes, of course. A trivialization on $U$ restricts to a trivialization on any $V\subset U$. $\endgroup$ – Ted Shifrin Apr 12 '15 at 19:43
  • $\begingroup$ @TedShifrin: Thanks. Then, to consider the space of all vector bundles over a given manifold it is enough to consider the finest trivialising covering? By the way, if you have a minute I am sure you can confirm what I suspect here math.stackexchange.com/questions/1231181/… Thanks. $\endgroup$ – Bilateral Apr 12 '15 at 20:58

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.