Suppose $f,g:X \rightarrow Y$ are finite morphisms between connected smooth curves over $\mathbb{C}$, with $Y$ of genus at least $2$.

If $f$ and $g$ induce the same morphism $H^*(Y,\mathbb{C}) \rightarrow H^*(X,\mathbb{C})$, does $f=g$?


Yes. Since $Y$ embeds into its Jacobian $B$, it is enough to prove the statement for pairs of maps to an abelian variety $f, g\colon X\to B$ sending a base point $x\in X$ to $0\in B$. Every such map factors uniquely through the Albanese variety $A$ of $X$, so we reduce further to the case of pairs of maps $f, g\colon A\to B$ between abelian varieties (sending $0$ to $0$). Every such map is necessarily a group homomorphism, and is uniquely determined by what it does on $\pi_1 = H_1$, or on $H^1(-, \mathbf{C})$.

| cite | improve this answer | |
  • 3
    $\begingroup$ It seems to me that you have not used that the genus of $Y$ is at least $2$ (which is certainly a necessary condition). $\endgroup$ – naf Apr 10 '19 at 4:15
  • 1
    $\begingroup$ @ulrich I think this condition is used implicitly in the first step. The argument after that point provides the equality $f=\alpha \circ g$ where $\alpha:B \rightarrow B$ is a translation, but genus $>1$ is needed to then conclude that $f=g$. I did this using that $\alpha$ induces both the identity on cohomology and an automorphism of $Y$, and using the Lefschetz fixed point formula (maybe Piotr had something else in mind). $\endgroup$ – rj7k8 Apr 10 '19 at 16:22
  • $\begingroup$ @ulrich You are right, the argument works literally if $f$ and $g$ satisfy $f(x)=g(x)$. Without this assumption, it shows that $f$ and $g$ differ by a translation in $B$. So it remains to show that if $Y$ is a curve of genus $>1$ embedded in its Jacobian $B$, then for every $b\in B$, $Y\cap (b+Y)$ is finite. Otherwise, $Y = b+Y$ and you can argue as rj7k8 above, but I guess there should be a direct argument using Riemann-Roch. $\endgroup$ – Piotr Achinger Apr 10 '19 at 16:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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