I am under the impression that a holomorphic line bundle is determined by its first Chern class. I mean this in the sense that $c_1 : H^1(X,\mathcal{O}_X^\ast) \to H^2(X,\mathbb{Z})$ is injective. I also believe that the degree of a line bundle is the same thing as its first Chern class.

Now, the Jacobian variety $J(C)$ of a curve $C$ is the group of degree $0$ lines bundles on it. Obviously, something in my paragraph above must be wrong otherwise the Jacobian of any curve would be trivial. And I think the Jacobian of a genus $1$ curve is famously not trivial.

I can believe that on a genus $0$ curve C one has $J(C) :=\{0\}$. For example, given two distinct points $p,q \in \mathbb{CP}^1$ it holds that $\mathcal{O}_{\mathbb{CP}^1}(p-q) \cong \mathcal{O}_{\mathbb{CP}^1}$. I guess that on an elliptic curve $\mathcal{O}_E(p-q) \ncong \mathcal{O}_E$ but this seems to contradict the fact that a holomorphic line bundle is determined by its $c_1.$

  • $\begingroup$ The degree is a number and the first Chern class is a cohomology class, so the last claim in your first paragraph needs some fixing! $\endgroup$ – Mariano Suárez-Álvarez Feb 22 '18 at 17:51
  • $\begingroup$ So, what is the question? Yes, chern class does not determine holomorphic line bundles (or vector bundles). Just follow the cohomology sequence of the exponential sequence. $\endgroup$ – Mohan Feb 22 '18 at 17:53
  • $\begingroup$ There is a short exact sequence of sheaves on $X$ that looks like $0\to\mathbb Z\to\mathcal O\to\mathcal O^*\to0$, with the last map the exponential. What is its long exact squence? $\endgroup$ – Mariano Suárez-Álvarez Feb 22 '18 at 17:57
  • $\begingroup$ Its the long exact sequence in cohomology gotten from the exponential exact sequence. Here I am using the fact that $Pic(X) = H^1(X,\mathcal{O}_X^\ast)$; since Cech 1-cocycles for $\mathcal{O}_X^\ast$ are the same as transition functions for a line bundle. $\endgroup$ – weissss Feb 22 '18 at 18:56
  • 1
    $\begingroup$ This long exact sequence tells us that a holomorphic line bundle is not determined by its first Chern class, if $h^1(X,\mathcal O_X)\neq 0$. $\endgroup$ – danneks Feb 23 '18 at 7:53

The degree and the first Chern class are connected in the way you remarked. For any smooth projective variety $X$ and line bundle $L$ on $X$, the intersection product wiith any curve class $[C]$ is given by $c_{1}(L).[C]=deg(L|_{C})$. In the curve case, you get of course also a canonical isomorphism $H^{2}(X, \mathbb{Z})\cong \mathbb{Z}$, so the first Chern class is here more or less the same as the degree.

Chern classes are however topological invariants. Therefor it cannot be expected that they classify holomorphic line bundles (in the same way as the genus is not expected to classify all curves). So what the first Chern class does, is it classifies topological line bundles, and in fact completely. $c_1$ gives an isomorphism $Vec^{top}_{1}(X)\cong H^{2}(X, \mathbb{Z})$, where $Vec^{top}_{1}$ is the group of complex line bundles, and $X$ is any topological space with the homotopy type of a CW complex. You can find this in Hatcher's notes on topological K-theory.

So from this perspective you can think about the first Chern class map in the exponential sequence as forgetting the holomorphic structure. And of these holomorphic structures there are many. Indeed, the Jacobian is simply the kernel of the first Chern class map, and it is an algebraic variety of dimension $g$.


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.