# Comparison of two monodromies

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

• I think you need more conditions to ensure connected fibers. – Moishe Kohan Apr 17 at 20:31
• @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? – Francesco Polizzi Apr 17 at 20:38
• Yes, since you require connected fibers over $\Sigma_b$. – Moishe Kohan Apr 17 at 20:44
• 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? – Francesco Polizzi Apr 17 at 21:24
• Oh, sorry, somehow I missed the surjectivity assumption. – 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.

• 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. – Francesco Polizzi Apr 17 at 22:11
• @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. – Will Sawin Apr 17 at 22:26
• 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 $Z(G) \simeq \mathbb{Z}_2$. – Francesco Polizzi Apr 17 at 22:32
• @FrancescoPolizzi I mean, if you let $\gamma$ be trivial, the abelian case is OK. – Will Sawin Apr 17 at 22:47
• 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). – Francesco Polizzi Apr 18 at 7:38