Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\pi: X \to Y$ be a finite morphism between smooth projective curves over the complex numbers. I would like to known:

(1) what the Gauss-Manin connection with respect to $\pi$ (that is, the connexion corresponding to the local system $\pi_\ast\mathbb{C}$ on $Y$ minus the ramification points) looks like

(2) what kind information does the Grothendieck-Riemann-Roch theorem provide when applied to $\pi$


share|cite|improve this question

I'll take a stab at these:

(1) Well, the Gauss-Manin connection is flat, and all flat connections look alike in local holomorphic trivializations. A better question is what the parallel sections of the local system look like. Here a simple example might be useful.

Let $\pi : \mathbb C \to \mathbb C$ be $z \mapsto z^2$. This is a finite morphism, of the admittedly non-projective affine plane to itself, but it can be extended to a finite morphism on the projective line that is ramified at $0$ and $\infty$ only. The only nontrivial local system associated to this morphism, outside of the ramification points, is a copy if two disjoint $\mathbb C$, one for each point in the preimage of a given point. A parallel section of the associated bundle over $U \subset \mathbb C$ then corresponds to the choice of a square root of $z$ over $U$, and if $U$ is connected this choice of square root does not "jump" between branches, which would correspond to jumping from one point in a preimage of $\pi$ to another.

The case of a general finite morphism should maybe be thought of as similar to this one; parallel sections of the vector bundle associated to the local system correspond to picking a branch of local solutions $x$ of $\pi(x) = y$ when $y$ varies on $Y$.

(2) I haven't worked out the details, but I'm willing to bet good money that we get an extreme overkill proof of the Riemann-Hurwitz formula by applying Grothendieck-Riemann-Roch to the finite morphism $\pi : X \to Y$.

share|cite|improve this answer
NB: There's something subtle going on due to the fact that the fibers of $\pi$ are not connected; in particular the "vector bundle" associated to the local system is actually a disjoint union of line bundles and not an honest vector bundle. Thus the space of sections of this "bundle" is a disjoint union of vector spaces, and not itself a vector space. – Gunnar Þór Magnússon Nov 23 '12 at 11:02

Your Answer


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.