Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Suppose $X$ is a projective variety over $\mathbb C$. I am happy to entertain more or different adjectives — I'm not looking for the most general statement, but rather to understand when and how smooth-manifold intuition leads me astray. I know very little algebraic geometry, and so please forgive and correct me if a statement below is mistaken.

It is rare for a line bundle $\mathcal L \to X$ to have a nowhere-vanishing section, and when it does, there are usually very few (only $\mathbb C^\times$ many). Suppose instead that I ask for a weaker structure than for $\mathcal L$ to have a section, but rather let me ask only that it has a flat connection. My question is:

In algebraic geometry, how often does a line bundle have a flat connection? When it has a flat connection, how many flat connections can it have?

share|improve this question

2 Answers 2

up vote 18 down vote accepted

I presume that your variety $X$ is smooth.

Consider the additive map $\mathrm d\log \colon \mathscr O_X^*\to \Omega^1_X$ that sends $f$ to $\mathrm df/f$. It induces a map $c_1$ in cohomology from $H^1(X,\mathscr O_X^*)$ to $H^1(X,\Omega^1_X)$ — a coherent avatar of the first Chern class. By Hodge Theory, $H^1(X,\Omega^1_X)$ is a subspace of $H^2(X,\mathbf C)$ and the two notions of first Chern class coincide.

A line bundle $\mathscr L$ has a connection if and only if its first Chern class $c_1(\mathscr L)\in H^1(X,\Omega^1_X)$ vanishes. The proof is straightforward: take an open cover $(U_i)$ of $X$, an invertible section $s_i$ of $\mathscr L$ on $U_i$ and the associated cocycle $(f_{ij})$ representing your line bundle in $H^1(X,\mathscr O_X^*)$. A connection $\nabla$ maps $s_i$ to $s_i\otimes\omega_i$, for some 1-form $\omega_i\in H^0(U_i,\Omega^1_X)$. The condition that these $s_i\otimes\omega_i$ come from a global connection on $X$ is exactly the vanishing of $c_1(\mathscr L)$.

It is a non-trivial fact that if $\mathscr L$ has an algebraic connection, then it is automatically flat. Torsten Ekedahl gave an algebraic proof on this thread of MO (Ekedahl also observes that $p$th power of line bundles in characteristic $p$ have an integrable connection), but an analytic proof seems easy. The algebraic connexion $\nabla$ gives rise to a connexion $\nabla+\bar\partial$ on the associated holomorphic line bundle. One checks that the curvature of this connection is a $(2,0)$-form, while it should be a $(1,1)$-form. Consequently, it vanishes.

When non empty, the set of flat connections on a vector bundle $\mathscr E$ is an affine space under $H^0(X,\Omega^1_X\otimes\mathscr E\mathit{nd}(\mathscr E))$, a finite dimensional vector space. In our case, $\mathscr L$ is a line bundle, hence $\mathscr E\mathit{nd}(\mathscr L)$ is the trivial line bundle so that we get $H^0(X,\Omega^1_X)$.

NB. Following the comment of Ben McKay, I edited the last paragraph.

share|improve this answer
2  
Isn't $End(L)$ trivial, since $L$ is a line bundle? –  Ben McKay Mar 8 '13 at 15:09
    
@Ben McKay. You're absolutely right... I'll edit the answer. –  ACL Mar 9 '13 at 0:48
    
A priori, there is no reason for the connection $\nabla +\overline{partial}$ to have curvature of type (1,1) (and hence to be flat). It is the curvature of the Chern connection that satisfies a reality property (if you have fixed an hermitian structure). Also, you mean to write that $\nabla+\overline{\partial}$ is a connection on the complex line bundle corresponding to $\mathscr{L}$, not the holomorphic one. –  Peter Dalakov Mar 10 '13 at 19:37
    
You do have though that all of the Chern classes vanish: the characteristic ring is generated by the Atiyah class, which is zero: see Thm 4 in Atiyah's paper. –  Peter Dalakov Mar 10 '13 at 19:52

A vector bundle with an algebraic connection has to have vanishing all Chern classes, at least in characteristic zero. I remember that this follows from the vanishing of the "Atiyah class", but I don't know the details.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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