Invariant Lagrangian in the first homology of a cover of a surface

Question

If $\tilde{S}\to S$ is a Galois cover of a surface of genus $g$ and group $G$, a well-know theorem of Chevalley-Weil tells that, as a $\mathbb{Q}[G]$-module, one has

$$H_1(\tilde{S},\mathbb{Q})=\mathbb{Q}^2\oplus\mathbb{Q}[G]^{2g-2}$$

I would like to understand the intersection form with respect to the $G$-action in particular I wonder whether it should always exist a $G$-invariant Lagrangian $L\subset H_1(\tilde{S},\mathbb{Q})$ (which means triviality in a convenient Witt group).

For instance it exists if the fundamental class of $S$ vanishes in $H_2(G,\mathbb{Z})$ because then, there exists a $3$-manifold $M$ bounding $S$ and a compatible covering $\tilde M\to M$ such that $L=\ker(H_1(\tilde S,\mathbb{Q})\to H_1(\tilde{M},\mathbb{Q}))$.

With some pain, I can prove it exists when $\mathbb{Q}$ is replaced with $\mathbb{R}$. Over $\mathbb{Q}$ (or $\mathbb{Q}_p)$, I am stuck.

Answer 1

Here are some thoughts, based on ideas from surgery theory. I am not an expert on these things though, and may well be wrong.

If we fix the finite group $G$, there is a Mischenko--Ranicki "symmetric signature" map $$\Omega_n(BG) \overset{\sigma}\to L^n(\mathbb{Z}[G])$$ from the $n$th oriented bordism of $BG$ to the $n$th symmetric $L$-theory of the ring with involution $\mathbb{Z}[G]$. To a map $f : M^n \to BG$ this records, roughly, the intersection form on the chain complex $C_*(\widetilde{M})$ of $\mathbb{Z}[G]$-modules: symmetric $L$-theory consists of "cobordism classes" of such symmetric forms.

This is not the same as your construction, in that it does not focus on the middle homology. However, I think itmay be related to your question as there is a natural map $$L^n(\mathbb{Z}[G]) \to L^n(\mathbb{Q}[G])$$ and for $n$ even the latter, as $\mathbb{Q}$ contains $1/2$, are isomorphic to the Witt groups of $(-1)^{n/2}$-symmetric forms over $\mathbb{Q}[G]$, by doing "algebraic surgery below the middle dimension".

So I think your question, asked for fixed $G$ and for all Galois $G$-covers of surfaces, is equivalent to asking whether the composition $$H_2(G; \mathbb{Z}) = \Omega_2(BG) \overset{\sigma}\to L^2(\mathbb{Z}[G]) \to L^2(\mathbb{Q}[G])$$ is zero.

This reformulation seems to give some cheap partial results. For example, using that every oriented surface admits a spin structure, I think one can show that in this particular degree the map $\sigma$ lands inside the image of the symmetrisation map $L_2(\mathbb{Z}[G]) \to L^2(\mathbb{Z}[G])$ from quadratic $L$-theory. The quadratic $L$-groups have been studied more, as they are the home of the surgery obstruction, and in particular it seems to be known that for finite $G$ their torsion subgroups are 2-primary. In particular, this would seem to imply that if $H_2(G;\mathbb{Z})$ has no 2-torsion then the answer to your question is yes.

This may well be overkill, or may be wrong. I suggest you ask an expert in $L$-theory, such as Ian Hambleton or Wolfgang Lück.

Answer 2

Finally, I found with Jean Barge a proof that such a Lagrangian always exists, see https://arxiv.org/abs/2202.00311. The proof starts exactly as in Oscar's answer and ends using classification results for irreducible rational representations of 2-groups.