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? 



de Rham cohomology class of diagonal
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 $\pi\_1^\* H^{0,0}\smile \pi\_2^\* H^{1,1} +\pi\_1^\* H^{1,1}\smile \pi\_2^\* H^{0,0}$.


