# Dimension of the space of invariant quadratic differentials in Galois covers

Let $f: X \rightarrow Y$ be a Galois cover of with $X$ and $Y$ algebraic curves over $\mathbb{C}$. I want to compute the dimension of the subspace of $G$-invariants in $H^{0}(X,\omega^{\otimes2})$ (or by Serre duality $H^{1}(X,T_{X})$). The point is that I always get this dimension to be zero, which I know is not true, because at least I knew a proof that for $Y=\mathbb{P}^{1}$ and $G$ cyclic it must be equal to $s-3$, where $s$ is the number of branch points. My argument is as follows:

We have the exact sequence of tangent bundles $0\rightarrow T_{X/Y}\rightarrow T_{X}\rightarrow f^{*}T_{Y} \rightarrow0$. The associated long exact sequence gives: $...\rightarrow H^{1}(X,T_{X/Y})^{G}\rightarrow H^{1}(X,T_{X})^{G}\rightarrow H^{1}(Y,f_{*}\mathcal{O}_{X}^{G}\otimes T_{Y})=H^{1}(Y, T_{Y})$. The last assertion because of the projection formula and $f_{*}\mathcal{O}_{X}^{G}= \mathcal{O}_{Y}$. But for $Y=\mathbb{P}^{1}$, $H^{1}(Y,T_{Y})=0$ and since $T_{X/Y}$ is with finite support (the ramification points) we have that $H^{1}(X,T_{X/Y})=0$ , this forces $H^{1}(X,T_{X})^{G}=0$. Can someone tell me where I am making a mistake. You can always assume that $Y=\mathbb{P}^{1}$ and that $G$ is cyclic.

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Your short exact sequence is wrong. Unless the morphism $f$ is etale (which it is not when $Y$ equals $\mathbb{P}^1$), then the induced morphism $T_X\to f^*T_Y$ is not surjective.
Dear Jason, thanks alot. Do you know a reference for the computation of $H^{1}(X,T_{X})^{G}$? I desperately need to know the answer for $G$ abelian. For $G$ cyclic I know that this dimension is $s-3$ (s: number of branch points) does this also hold for $G$ abelian? –  Darius Math Jul 27 '13 at 13:23
You may get what you need from various papers on the equivariant Riemann-Roch theorem applied to your G-linearized sheaf $T_X$. For example, look at the papers of Borne and/or Ellingsrud-Lonstead.