de Rham cohomology class of diagonal - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T05:29:35Z http://mathoverflow.net/feeds/question/74283 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/74283/de-rham-cohomology-class-of-diagonal de Rham cohomology class of diagonal anonymous 2011-09-01T19:31:59Z 2011-09-01T19:51:19Z <p>I post again a question I asked in the post by Descartes:</p> <p>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. </p> <p>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? </p> http://mathoverflow.net/questions/74283/de-rham-cohomology-class-of-diagonal/74285#74285 Answer by algori for de Rham cohomology class of diagonal algori 2011-09-01T19:50:23Z 2011-09-01T19:50:23Z <p>Take $X$ a smooth compact curve of genus $>0$. The diagonal can be written as <code>$\sum \pi_1^* e_i \smile \pi_2^* e'_i$</code> where $e_i$ form a basis of the total cohomology and $e_i'$ form a Poincar\'e dual basis. So the <code>$\pi_1^*H^1\smile\pi_2^* H^1$</code>-component of the diagonal is nonzero, and so the diagonal does not lie in <code>$\pi_1^* H^{0,0}\smile \pi_2^* H^{1,1} +\pi_1^* H^{1,1}\smile \pi_2^* H^{0,0}$</code>.</p>