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. 



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


