Let $C_1, C_2$ be two smooth projective curves of genus atleast $2$. Let $X:=C_1 \times C_2$. If $P_1, P_2$ are denoted as the first and second projections, then it is known that the canonical bundle relation is given as follows : $K_X := P_1^*(K_{C_1}) \otimes P_2^*(K_{C_2})$.
From this relation is it elementary to see that $h^1(K_X)=h^1(\mathcal O_X) >0$?
Any hint or explaination is appreciated.
Best Answer
If you know the Kunneth formula (ref Stacks or MO), yes! The Kunneth formula says that for quasi-compact schemes $X$ and $Y$ over a field $k$ with affine diagonal (separated implies affine diagonal, for instance) and for $\mathcal{F}$ a quasi-coherent $\mathcal{O}_X$-module, $\mathcal{G}$ a quasi-coherent $\mathcal{O}_Y$-module, we have a canonical isomorphism $$H^n(X\times_{\operatorname{Spec} k} Y, p_1^*\mathcal{F}\otimes_{X\times_{\operatorname{Spec} k} Y} p_2^*\mathcal{G}) \cong \bigoplus_{p+q=n} H^p(X,\mathcal{F})\otimes_k H^q(Y,\mathcal{G}).$$
This gives us that $H^1(X,K_X) \cong H^1(C_1,K_{C_1}) \otimes_k H^0(C_2,K_{C_2}) \oplus H^0(C_1,K_{C_1}) \otimes_k H^1(C_2,K_{C_2})$ in our case. As $H^1(C_i,K_{C_i})\cong k$ by Serre duality and $h^0(C_i,K_{C_i}) = g$ by Riemann-Roch, we have that $h^1(K_X)>0$. Noting that $h^1(K_X)=h^1(\mathcal{O}_X)$ and this is calculated by the same expression by Serre duality, we're done.