My question is similar to this, but not identical. I believe the following to be true, but I'd like a reference.

Given (quasicoherent?) sheaves of $\mathcal O_X$ modules $E$ and $F$ on a projective variety $X$, $$ H^n(X, E\otimes F) \cong \bigoplus_{p+q=n} H^p(X, E) \otimes H^q(X, F). $$

Is this true, and what is a good reference or counterexample? If true, is quasicoherent necessary?


No, it is false, consider e.g. $E=F=\mathcal{O}_X$ on a curve $X$ of genus one.

(Here is something that is true though: let $p$ and $q$ be the two projection maps $X \times X \to X$. Then $H^\bullet(X\times X,p^\ast E \otimes q^\ast F) \cong H^\bullet(X,E) \otimes H^\bullet(X,F)$. There are lots of basic variants of this theorem and all of them are called the Kunneth formula.)

  • $\begingroup$ Why is it false for $\mathcal O_X$? The only non-vanishing cohomology is $H^0(X,\mathcal O_X)$, so doesn't the formula hold? $\endgroup$ Dec 2 '10 at 18:50
  • $\begingroup$ @James: No, $H^1(X,\mathcal O_X)$ is one-dimensional. $\endgroup$
    – Matt E
    Dec 2 '10 at 18:55
  • $\begingroup$ Oh wow, I thought $H^m(X, \mathcal O_X)=0$ for all $m>0$!! $\endgroup$ Dec 2 '10 at 19:07
  • 3
    $\begingroup$ That's really far from true. :) You can relate the groups $H^m(X,\mathcal{O}_X)$ to more familiar cohomology groups using a tiny bit of Hodge theory. Basically the cohomology group $H^n(X,\mathbb C)$ has a filtration where all the graded pieces look like $H^p(X,\Omega^q_X)$ for $p+q = n$. In particular $H^m(X,\mathcal{O}_X)$ gives you a piece of the cohomology group $H^m(X,\mathbb C)$. Simplest example: on a curve of genus $g$, the first cohomology is made up out of two pieces, $H^1(X,\mathcal{O}_X)$ and $H^0(X,\Omega^1_X)$. Both have dimension $g$. $\endgroup$ Dec 2 '10 at 19:42

It is false. Take $E = \mathcal O_{\mathbb P^n}(2)$ and $F= \mathcal O_{\mathbb P^n}(-2)$. Then, $$ H^0(\mathbb P^n, \mathcal O_{\mathbb P^n}) = H^0(\mathbb P^n, E\otimes F) \ne H^0(\mathbb P^n, \mathcal O_{\mathbb P^n}(2)) \otimes H^0(\mathbb P^n, \mathcal O_{\mathbb P^n}(-2)). $$


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