3
$\begingroup$

I have the following situation: let $X$ be a projective complex manifold and let $f \in H^1(X,\mathcal{M}^{\times})$. So $f$ defines something like a line bundle with meromorphic transition functions.

We can identify the poles of $f$ which is a divisor $D \subset X$. Then the restriction of $f$ to $X \setminus D$ gives an element $$f\rvert_{X \setminus D} \in H^1(X\setminus D,\mathcal{O}^{\times}) = \mathrm{Pic}(X\setminus D)$$ which defines a line bundle $L$ on $X \setminus D$.

My question is the following: if $s$ is a section of the line bundle $L$ on $X \setminus D$, does there exist a geometric object $W$ (a line bundle, gerbe, ...) on the whole of $X$ which extends $L$ in a nice way, i.e. the section $s$ extends to a section on $W$, and if yes, what are the requirements for $f$ and $s$?

I know that this is not a very precise question, but maybe there is some known theory that handles such situations.

Remark: In my situation, the cocycle $f$ comes from a group 1-cocycle $e \in H^1(\pi_1(X),H^0(\tilde{X},\mathcal{M}^{\times}))$ where $\tilde{X}$ is the universal covering of $X$.

Edit: I found a similar construction that I would like to generalize: in Brylinski's "Loop Spaces, Characteristic Classes and Geometric Quantization" they discuss the magnetic monopole which is characterized by a line bundle $[L] \in H^2(\mathbb{R}^3\setminus\{0\},\mathbb{Z})$. They then use the canonical isomorphism $$H^2(\mathbb{R}^3\setminus\{0\},\mathbb{Z}) \to H^3(\mathbb{S}^3,\mathbb{Z})$$ to construct a gerbe. I guess my question is the following: is there something like a canonical isomorphism $$H^1(X\setminus D,\mathcal{O}^{\times}) \to H^2(X,\mathcal{O}^{\times})$$ so that the holomorphic line bundle induces a holomorphic gerbe?

Improve this question
$\endgroup$
6
  • 4
    $\begingroup$ I believe that cohomology group is the quotient of the Picard group of $X$ by the group of "Cartier divisor classes." So if $X$ is projective, for instance, then this cohomology group is trivial. However, it is nontrivial for the Hopf manifold. $\endgroup$
    – Jason Starr
    Commented May 20, 2023 at 23:37
  • $\begingroup$ In my case, $X$ is projective, and hence, the element $f \in H^1(X,\mathcal{M}^{\times})$ is trivial. Even more: the section $s$ trivializes the object. But $s$ is a section of a meromorphic bundle, so it has zeroes and poles. So the question remains, what is the obstruction for $s$ to be a section of a line bundle, etc.? $\endgroup$
    – KuSi
    Commented May 22, 2023 at 8:49
  • 1
    $\begingroup$ In my situation, I'm given a group cocycle $e \in H^1(\pi_1(X),H^0(\tilde{X},\mathcal{M}^{\times}))$, and I know that there exists a divisor $D \subset X$ such that for any $\gamma \in \pi_1(X)$, the function $e_\gamma$ has poles only poles along $\pi^{\ast} D$ where $\pi \colon \tilde{X} \to X$ is the universal covering. Therefore, I believe it makes sense to have a well-defined divisor for $f$. $\endgroup$
    – KuSi
    Commented May 22, 2023 at 13:20
  • 2
    $\begingroup$ @Jason: it seems to be a common misbelief that the sheaf $\mathcal{M}^{\times}$ is acyclic whenever $X$ is projective. In "The sheaf of nonvanishing meromorphic functions in the projective algebraic case is not acyclic" by Chen, Kerr, and Lewis, it is proven that this statement is false. Even more: it is almost the converse as it holds if and only if $\mathrm{dim} X = 1$. $\endgroup$
    – KuSi
    Commented May 22, 2023 at 21:32
  • 1
    $\begingroup$ I completely agree with you! The group $H^2(X,\mathcal{M}^\times)$ is often nonzero and captures part of the Brauer group of $X$. However, the short exact sequence from $\mathcal{O}^\times$ to $\mathcal{M}^\times$ to the "sheaf of Cartier divisors" shows that $H^1(X,\mathcal{M}^\times)$ is the quotient that I mentioned. $\endgroup$
    – Jason Starr
    Commented May 23, 2023 at 11:20

0

Reset to default

You must log in to answer this question.