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

Let $C$ be a compact Riemann surface, and let $U$ be a Zariski open subset in $C.$ Let $L$ be a local system (with coefficients $\mathbb C$ or $\mathbb Q_{\ell}$) on $U.$ For each point $z_i\in C-U,$ let $M_i$ be the monodromy matrix of $L$ around $z_i.$ If we identify $L$ as a representation of $\pi_1(U),$ and let $\gamma_i$ be a small loop in $U$ around $z_i,$ then $M_i$ is the image of $\gamma_i$ under the representation $\rho_L:\pi_1(U,a)\to GL(L_a)$ (where $a$ is a base point and $L_a$ is the fiber of $L$ at $a$).

Here's my question. It seems to me that there should be some relation between (the traces of) these monodromies (as well as the monodromies around loops in $\pi_1(C)$) and (Betti numbers of) the cohomology groups $H^i_c(U,L)$ of $L$ with compact support, but I don't know the precise formulation or reference.

As an example, if $C$ has genus $g$ and $L=\mathbb C$ is constant of rank 1, then $H^1_c(U,\mathbb C)$ has dimension $2g+n-1,$ where $n$ is the number of points at infinity. This number is the rank of $\pi_1(U),$ and each ``canonical generator" of $\pi_1(U)$ has trace 1.

share|improve this question
add comment

1 Answer

up vote 8 down vote accepted

Suppose that $C = \mathbb P^1$ and $U = \mathbb P^1\setminus \{0,\infty\}.$ Then $\pi_1(U)$ is cyclic, freely generated by a loop around $0$.

The local system $L$ is thus given by the vector space $L_a$, equipped with an invertible operator, call it $T$, corresponding to the action of the generator of $\pi_1(U)$. This operator $T$ is the monodromy matrix.

Now $H^1_c(U,L) = $ the space of $T$-invariants of $L_a$, while $H^2_c(U,L) = $ the space of $T$-coinvariants of $L_a$. If you think of the possible Jordan decompositions of $T$, and the fact that the trace is insensitive to the unipotent part, but just depends on the semi-simple part, you'll see that the it's going to be hard to find any interesting relation of the type that you want. (E.g. if $L_a$ is $n$-dimensional, and $T$ acts by the identity, or by a maximally non-trivial unipotent element, in both cases the trace of $T$ is equal to $n$, but in the first case the Betti numbers are also both equal to $n$, while in the second, they are both equal to $1$.)

You might also wonder about the Euler characteristic $H^2_c(U,L) - H^1_c(U,L)$, but this is always equal to (rank $L$) $\cdot \chi(U)$, and so is insensitive to the monodromy matrices.

share|improve this answer
Thanks, Matt. Long ago, de Jong told me that, if U is an affine curve over a finite field, and L ranges over all local systems on U of rank 1, then the dimension h^1_c(U,L) has no upper bound, due to arbitrary ramification at infinity. I'm trying to convice myself why it is so. Is it something special about finite field, like the Artin-Schreier cover? –  shenghao May 3 '10 at 19:01
This is to do with wild ramification at infinity, a phenomenon which you can't see with local systems on Riemann surfaces. The formula for the Euler char. for an $\ell$-adic sheaf on an open curve $U$ is (rank $\mathcal L$)$\cdot \chi(U)$ minus the sum of the Swan conductors at each of the missing points. These Swan conductors measure wild ramification around each of the punctures, and can be arbitrarily large. Thus $H^1_c$ can have arbitrarily large rank in this setting. –  Emerton May 3 '10 at 23:41
Thank you very much, Matt! –  shenghao May 4 '10 at 5:02
add comment

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.