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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?

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2 Answers 2

up vote 19 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.

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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.

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