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Here is one answer for the question of Descartes in the case $X$ is compact Kahler manifold and the coefficient is $\mathbb{Q}$: Then we can work with coefficient $\mathbb{C}$ as well. This answers uses the de Rham cohomology.

If $X$ is a compact Kahler manifold of complex dimension $n$, then $\Delta X\subset \Delta_X\subset X\times X$ is a cycle of (real) dimension $2n$, hence acts on smooth $2n$ forms on $X\times X$ by integration. Also, Delta has no boundary, hence it is closed (by Stokes theorem). Thus it represents a homology class in $H{2n}(X\times H_{2n}(X\times X)$. Now, $\Delta _X$ \Delta_X$ acts on a form by integration on the restriction of that form on $\Delta _X$. \Delta_X$. Since $\Delta _X$ \Delta_X$ is a complex manifold of dimension$n$, if$\varphi$is a$(p,q)$form on$X\times X$with$p+q=2n$, its restriction to $\Delta _X$\Delta_X$ is nonzero iff $p=q=n$. This shows that $\Delta X$ \Delta_X$ is in $H{n,n}(X)$, H_{n,n}(X)$, and so its Poincare dual represents a cohomology class in $H^{n,n}(X)$.

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(There seems something wrong with Latex. I will repair it later.)

Here is one answer for the question of Descartes in the case $X$ is compact Kahler manifold and the coefficient is $\mathbb{Q}$: Then we can work with coefficient $\mathbb{C}$ as well. This answers uses the de Rham cohomology.

If $X$ is a compact Kahler manifold of complex dimension $n$, then $\Delta _X\subset X\times X$ is a cycle of (real) dimension $2n$, hence acts on smooth $2n$ forms on $X\times X$ by integration. Also, $\Delta X$ Delta has no boundary, hence it is closed (by Stokes theorem). Thus it represents a homology class in $H{2n}(X\times X)$. Now, $\Delta _X$ acts on a form by integration on the restriction of that form on $\Delta _X$. Since $\Delta _X$ is a complex manifold of dimension $n$, if $\varphi$ is a $(p,q)$ form on $X\times X$ with $p+q=2n$, its restriction to $\Delta _X$ is nonzero iff $p=q=n$. This shows that $\Delta X$ is in $H{n,n}(X)$, and so its Poincare dual represents a cohomology class in $H^{n,n}(X)$.

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Here is one answer for the question of Descartes in the case $X$ is compact Kahler manifold and the coefficient is $\mathbb{Q}$: Then we can work with coefficient $\mathbb{C}$ as well. This answers uses the de Rham cohomology. If $X$ is a compact Kahler manifold of complex dimension $n$, then $\Delta _X\subset X\times X$ is a cycle of (real) dimension $2n$, hence acts on smooth $2n$ forms on $X\times X$ by integration. Also, $\Delta X$ has no boundary, hence it is closed (by Stokes theorem). Thus it represents a homology class in $H{2n}(X\times X)$. Now, $\Delta _X$ acts on a form by integration on the restriction of that form on $\Delta _X$. Since $\Delta _X$ is a complex manifold of dimension $n$, if $\varphi$ is a $(p,q)$ form on $X\times X$ with $p+q=2n$, its restriction to $\Delta _X$ is nonzero iff $p=q=n$. This shows that $\Delta X$ is in $H{n,n}(X)$, and so its Poincare dual represents a cohomology class in $H^{n,n}(X)$.