# de Rham cohomology class of diagonal

I post again a question I asked in the post by Descartes:

Since this is the topic on diagonal, I like to ask a question: Let $X$ be a compact Kahler manifold of complex dimension $n$, and let $\Delta _X\subset X\times X$ be the diagonal of $X$. We denote by $\{\Delta _X\}$ its cohomology class in $H^{2n}(X\times X)$ (here I consider the cohomology with complex coefficients). In fact, we know that $\{\Delta _X\}$ lives in $H^{n,n}(X\times X)$. Let $\pi _1,\pi _2:X\times X\rightarrow X$ be the projections. By Kunneth's theorem and Hodge decomposition, we know that $H^{n,n}(X\times X)=\sum _{p+r=q+s=n}\pi _1^*(H^{p,q}(X))\otimes \pi _2^*(H^{r,s}(X))$, thus $\{\Delta _X\}$ lives in this direct sum.

My question is this: Sometime ago, I discussed with one mathematician about the possibility that $\Delta _X$ lives in $\sum _{p+r=n}\pi _1^*(H^{p,p}(X))\otimes \pi _2^*(H^{r,r}(X))$, but he said that this is not true for a general compact Kahler manifold, but we had little time to discuss that he could produce a specific example for the claim. Do any of you know of one such example?

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Take $X$ a smooth compact curve of genus $>0$. The diagonal can be written as $\sum \pi_1^* e_i \smile \pi_2^* e'_i$ where $e_i$ form a basis of the total cohomology and $e_i'$ form a Poincar\'e dual basis. So the $\pi_1^*H^1\smile\pi_2^* H^1$-component of the diagonal is nonzero, and so the diagonal does not lie in $\pi_1^* H^{0,0}\smile \pi_2^* H^{1,1} +\pi_1^* H^{1,1}\smile \pi_2^* H^{0,0}$.
I see. Thanks a lot. So the same argument works if $H^{p,q}(X)\not= 0$ for some $p\not= q$. – anonymous Sep 1 '11 at 20:45