Let $C$ be a (quasi-projective) variety over an algebraically closed field $k$ and let $k(C)$ be its field of rational functions. Then for any birational map $\sigma: C \dashrightarrow C$ there is an automorphism of $k(C)$ that induces an automorphism of the Galois cohomology group $H^1(k(C),A)$ for any $Gal(k(C)^{alg}/k(C))$-module $A$.

I am looking for some clues on what techniques might help to understand the action of the group of birational automorphisms of $C$ on the cohomology group.

In particular, I would like to know how to decide, given a class $h \in H^1(k(C),A)$ and an automorphism $\sigma: C \dashrightarrow C$, if $\sigma(h)=h$ or if $\sigma(h)=g$ where $g$ is some other cohomology class.

To make things concrete, suppose $C$ is a curve and let $A=\mu_n$, $n\neq \mathrm{char}\ k$. In this case we know by Kummer theory that $H^1(k(C),\mu_n)=k(C)^\times/(k(C)^\times)^n$. If I want to find out if $\sigma(h)=h$ for a cohomology class $h$, are there any options other than writing down the equations that define $C$ and $\sigma$ and trying to see explicitly if $\sigma(h)$ and $h$ belong to the same coset of $(k(C)^\times)^n$? Is there a more ``geometric'' way of looking at things?


1 Answer 1


since $\left(k(C)^\times\right)^n)$ certainly includes $k$, $\sigma(h)/h$ lies in it if and only if its divisor class is an $n$-fold multiple of a principal divisor.

One can easily compute the divisor class of $\sigma(h)$ by pulling back (or pushing forward?) the divisor class of $h$ along the automorphism $\sigma$.

There are two obstructions to being an $n$-fold multiple of a principal divisor. The first is local: the order at any point must certainly be a multiple of $n$. The second is global: if that is satisfied, then the divisor class of $D/n$ must be trivial, that is, it must generate a principal line bundle.

So the order of a point, mod $n$, must be constant on orbits of $\sigma$, and if this holds one constructs a line bundle, which must be trivial.

Is that geometric enough?

  • $\begingroup$ Dear Will, this is certainly geometric enough to force me out of the block! I somehow did not realise that I am looking at the map that $\sigma$ induces on the divisors. $\endgroup$ Sep 3, 2012 at 21:28

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