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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 $Z(G) \simeq \mathbb{Z}_2$.

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  • $\begingroup$ 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
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    $\begingroup$ Oh, sorry, somehow I missed the surjectivity assumption. $\endgroup$ – Moishe Kohan Apr 18 at 0:08
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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.

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  • $\begingroup$ 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

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