Let us consider a smooth projective curve $\Sigma_b$ of genus $b$, and assume that there is a surjective group homomorphism $$\varphi \colon \pi_1 (\Sigma_b \times \Sigma_b - \Delta) \to G,$$ where $\Delta \subset \Sigma_b \times \Sigma_b $ is the diagonal and $G$ is a finite group. Assume moreover that the homomorphisms $$\psi_i \colon \pi_1(\Sigma_b-\{p \}) \to G, \qquad i=1,\, 2,$$ induced by the inclusion of the two factors, are also surjective.

Then, by Grauert-Remmert Extension Theorem, there exist a holomorphic $G$-cover $X \to \Sigma_b \times \Sigma_b$, branched over the diagonal, that, after composing with the first (or the second) projection, yields a Kodaira fibration, namely, a non-isotrivial holomorphic fibration $f \colon X \to \Sigma_b$ with smooth and connected fibres. In fact, everything turns out to be algebraic.

Calling $g$ the fibre genus of $f$, the fundamental group $\pi_1(X)$ sits into a split exact sequence $$1 \to \pi_1(\Sigma_g) \to \pi_1(X) \to \pi_1(\Sigma_b) \to 1,$$ that gives, by conjugation, monodromy representations $$\rho \colon \pi_1(\Sigma_b) \to \operatorname{Out} \, \pi_1(\Sigma_g), \quad \bar{\rho} \colon \pi_1(\Sigma_b) \to \operatorname{Aut} \, H^1(\Sigma_g, \, \mathbb{Q}).$$

On the other hand, the homomorphism $\psi_1$ (or $\psi_2$) realizes the fibre $\Sigma_g$ of $f$ as a $G$-cover $\Sigma_g \to \Sigma_b$, branched over exactly one point, and so there is a short exact sequence $$1 \to \pi_1(\Sigma_g) \to \pi_1(\Sigma_b-\{p\})^{\operatorname{orb}} \to G \to 1.$$ Here the group in the middle is the quotient of $\pi_1(\Sigma_b-\{p\})$ by the normal closure of the subgroup $\langle \gamma^s \rangle$, where $\gamma$ is a generator that loops around $p$ and $s$ is the order of $\psi_1(\gamma) \in G$. This sequence gives, by conjugation, two monodromy representations $$\rho_G \colon G \to \operatorname{Out} \, \pi_1(\Sigma_g), \quad \bar{\rho}_G \colon G \to \operatorname{Aut} \, H^1(\Sigma_g, \, \mathbb{Q}).$$

Question. How are $\bar{\rho}$ and $\bar{\rho}_G$ related? In particular, is it true that their invariant subspaces in $H^1(\Sigma_g, \, \mathbb{Q})$ are equal (or, at least, that they have the same dimension)?

Edit. Will Sawin's answer shows that the invariant subspace of $\bar{\rho}$ always contains the invariant subspace of $\bar{\rho}_G$.

Can one provide some conditions ensuring that also the reverse inclusion holds?

Note that, since I am assuming non-trivial ramification of the $G$-cover $X \to \Sigma_b \times \Sigma_b$, then $\psi_1(\gamma) \in G$ is non-trivial. If this can help, in my specific situation $G$ is an extra-special group of order $32$ and $\psi_1(\gamma)$ is the generator of the center $Z(G) \simeq \mathbb{Z}_2$.

  • $\begingroup$ I think you need more conditions to ensure connected fibers. $\endgroup$ – Moishe Kohan Apr 17 at 20:31
  • $\begingroup$ @MoisheKohan: why? A connected $G$-cover $X \to Y$ is equivalent to the datum of a group epimorphism $\pi_1(Y-B)\to G,$ where $B$ is the branch locus. Or am I missing something? $\endgroup$ – Francesco Polizzi Apr 17 at 20:38
  • $\begingroup$ Yes, since you require connected fibers over $ \Sigma_b$. $\endgroup$ – Moishe Kohan Apr 17 at 20:44
  • $\begingroup$ Sorry, probably I do not understand. The fibre of $f \colon X \to \Sigma_b$ over a point $p \in \Sigma_b$ is the preimage in $X$, via $X \to \Sigma_b \times \Sigma_b$, of the corresponding fibre of $\Sigma_b \times \Sigma_b \to \Sigma_b$. Now, this preimage is the curve corresponding to the group homomorphism $$\psi_1 \colon \pi_1(\Sigma_b-\{p \}) \to G,$$ and this is connected because I am assuming that $\psi_1$ is onto. Since one fibre is connected, all of them are so. Where am I wrong? $\endgroup$ – Francesco Polizzi Apr 17 at 21:24
  • 1
    $\begingroup$ Oh, sorry, somehow I missed the surjectivity assumption. $\endgroup$ – Moishe Kohan Apr 18 at 0:08

The $\tilde{\rho}$-invariants contain the $\tilde{\rho}_G$-invariants, at least.

The map $\Sigma_g \to \Sigma_b$ defines a pullback map $ H^1( \Sigma_b, \mathbb Q) \to H^1(\Sigma_g, \mathbb Q) $ (i.e. cohomology is a contravariant functor).

The image of this pullback map is the $G$-invariants in $H^1(\Sigma_g, \mathbb Q)$. The image of this pullback map consists of $G$-invariants as a formal consequence of the fact that $G$ acts by automorphisms of $\Sigma_g$ over $\Sigma_b$. It consists of all the $G$-invariants because, given any $1$-form whose cohomology class is $G$-invariant, we can average to get a $1$-form that is itself $G$-invariant, which necessarily descends from $\Sigma_b$. (Or take the trace / integration map $H^1(\Sigma_g, \mathbb Q) \to H^1(\Sigma_b, \mathbb Q)$, divide by $|G|$, and pull back.)

We can check that this pullback map is invariant under the monodromy action of $\Sigma_g$. To do this, one way is to factor $\Sigma_g \to \Sigma_b$ as $\Sigma_g \to X \to \Sigma_b \times \Sigma_b \to \Sigma_b$, with the last map projection onto the left factor, and to note that the image of $H^1(X, \mathbb Q) \to H^1(\Sigma_g, \mathbb Q)$ is the $\pi_1(\Sigma_b)$-invariants of $H^1(X, \mathbb Q)$.

So the image of this map is contained in the monodromy invariants.

  • $\begingroup$ Thank you for the nice answer. Can you see any condition on the finite group $G$ implying equality? For instance, in my case $G$ is extra-special. $\endgroup$ – Francesco Polizzi Apr 17 at 22:11
  • $\begingroup$ @FrancescoPolizzi I currently can't see how to do any case except for abelian groups, which perhaps suggests that extra-special groups are not far off, but I don't see how to extend it yet. $\endgroup$ – Will Sawin Apr 17 at 22:26
  • $\begingroup$ Abelian groups cannot occur, since I want ramification on the diagonal, hence the image of the element $\gamma$, that is a non-trivial commutator in $\pi_1(\Sigma_g \times \Sigma_g - \Delta)$, must give a non-trivial commutator in $G$. For extra-special $p$-groups, this means that the image of $\gamma$ must lie in the center $Z(G) \simeq \mathbb{Z}_p$. Maybe, for extra-special $2$-groups one can say something more precise about the monodromies, since in that case the image of $\gamma$ is the unique generator of the center $Z(G) \simeq \mathbb{Z}_2$. $\endgroup$ – Francesco Polizzi Apr 17 at 22:32
  • $\begingroup$ @FrancescoPolizzi I mean, if you let $\gamma$ be trivial, the abelian case is OK. $\endgroup$ – Will Sawin Apr 17 at 22:47
  • $\begingroup$ Ah, ok. In my situation, since I am assuming actual ramification, then $\gamma$ is non-trivial (and moreover I have that $G/\langle \gamma \rangle$ is elementary abelian). $\endgroup$ – Francesco Polizzi Apr 18 at 7:38

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.