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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.

Proj
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1 Answers1

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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.

KReiser
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  • $h^0(C_{i}, K_{C_{i}}) =g$ is by the definition of geometric genus, right? – Proj Sep 28 '22 at 06:22
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    Sure, that's another way to see that. – KReiser Sep 28 '22 at 12:27
  • @KReiser Do we need the projective assumption ? In my proof which uses $\dim X\times Y=\dim X+\dim Y$, the fact that $X\times Y$ is smooth, the fact that $\Omega_{X\times Y}\cong p_1^\Omega_{X}\oplus p_2^\Omega_{Y}$ and exercise II.5.16.d) from Hartshorne I don't find where the assumption is used and I believe that everything is well defined without it. – raisinsec Dec 11 '23 at 12:30
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    @raisinsec non-projective (irreducible) curves are affine, so there's no need to calculate higher cohomology in that case. – KReiser Dec 11 '23 at 14:58
  • @KReiser I was more talking about the isomorphism $\omega_{X\times Y}\cong p^_1\omega_X\otimes p_2^\omega_Y$. For the inequality I agree with you. Maybe it's off topic, you tell me, but it's one of the first posts I get when asking google why we need projectivity of $X,Y$ for the isomorphism. – raisinsec Dec 11 '23 at 16:09
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    @raisinsec See Hartshorne exercise II.8.3. One set of assumptions that gives you $\omega_{X\times Y} \cong p_1\omega_X\otimes p_2^\omega_Y$ is $X,Y$ smooth over a field; no need for anything beyond that. – KReiser Dec 11 '23 at 16:14