Let $X$ be a Kähler manifold. The torsion free part of the singular cohomology $H^n(X,\mathbb{C})$ has a Hodge decomposition $$ H^n(X,\mathbb{C})=\bigoplus_{p+q=n}H^{p,q}(X), $$ where $H^{p,q}(X)$ can either be viewed as the space of de Rham classes of bidegree $(p,q)$ or as the Dolbeault cohomology $H^{q}(X,\Omega^p)$. For any subvariety $Z\subset X$, the fundamental class $[Z] \in H^{2k}(X,\mathbb{C})$ for some $k$.

Since $[Z] \in H^{2k}(X,\mathbb{Z})$, it defines a real class, but how can one prove that it lies in $H^{k,k}(X)$? Why not $[Z] \in H^{k-1,k+1}(X,\mathbb{Z})\oplus H^{k+1,k-1}(X,\mathbb{Z})$?


This is because $Z$ is a complex subvariety of $X$. I think the easiest way to see it is to use local coordinates. First assume $Z$ is smooth to simplify things a little and let $p = \dim_{\mathbb C} Z$.

Let $u$ be some closed $2p$-form on $X$ and let $x$ be a point on $Z$. We can find holomorphic local coordinates $(z_1,\dots,z_n)$ on $X$ such that $Z$ is described by the set $\{ z \mid z_{p+1} = \dots = z_n = 0\}$ in these coordinates. Now write $u = \sum_{a+b = 2p} u_{ab}$ as a decomposition of $(a,b)$-forms. Locally, we have $$ u_{ab} = \sum_{I,J} u_{ab,IJ} d z_I \wedge d\overline z_J, $$ where $I,J$ are multi-indices such that $|I| = a$ and $|J| = b$. But now linear algebra shows that for $dz_I$ to be nonzero once restriced to $Z$, we must have $a = |I| \leq p$. Similarly, we get $b = |J| \leq p$, so $a = b = p$.

For non-smooth $Z$, one can show that the integral over $Z$ of $u$ is equal to the integral over the nonsingular part of $Z$ only. This takes a little work, but is done in one of the first chapters of Demailly's book "Complex analytic and differential geometry". Since the nonsingular part is dense in $Z$ this implies the result.

Now, if $Z$ is just an arbitrary closed smooth subvariety of $X$, there is no reason that it should define a $(p,p)$-class, and indeed it doesn't in general.

  • $\begingroup$ Thanks for the answer. It is not so trivial after all. $\endgroup$ – M. K. Nov 17 '12 at 17:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.