Let $M$ be a compact complex surface, i.e., a complex manifold with complex dimension two. Also, let us denote its Kähler metric by $g$ and its Kähler form by $K$. Let us denote the Kähler class by $[K] \in H^2(M)$. Since $K$ is harmonic with respect to $g$, the Hodge dual of $K$ with respect to $g$, $*_g K$ is also closed. Therefore, $[*_g K] \in H^2 (M)$.

Let us also recall that there is an inner-product defined on the vector space $H^2(M)$. It is given by \begin{equation} \langle \omega, \eta \rangle = \int_M \omega \wedge \eta \end{equation} for $\omega, \eta \in H^2 (M)$.

My questions are the following:

  1. Does $[*_g K]$ only depend on $[K]$? In other words, if the Kähler forms $[K_1]$ and $[K_2]$ of two Kähler metrics $g_1$ and $g_2$ satisfy $[K_1]=[K_2]$, does it follow that $[*_{g_1} K_1] = [*_{g_2} K_2]$?

  2. In case this is true, is there a simple way to get $[*K]$ from $[K]$ using the topological structure of $H^2 (M)$ (such as the inner-product defined above)?

  3. Even if 1. is not true in general, in the case that $M$ is Calabi-Yau (hence a K3 manifold), there is a well-defined map $[K] \mapsto [* K]$, given the complex structure of the manifold is specified --- it is to take the Hodge dual $*$ with respect to the unique Calabi-Yau metric in the Kähler class $[K]$. Is there an answer to the equivalent of question 2. in this case? In other words, can $[*K]$ be obtained from $[K]$ and $[\Omega]$ --- where $\Omega$ is the unique complex two-form on $M$ --- using the topological structure of $H^2 (M)$ in some simple way?

Thank you in advance!


1 Answer 1


(1) Yes. If $M$ is $n$-dimensional and $K$ is the Kahler form of a Kahler metric $g$ (or just the Kahler form of a hermitian metric) we have $$ *_g \frac{K^p}{p!} = \frac{K^{n-p}}{(n-p)!} $$ for any $p = 0,\ldots,n$. For a surface this gives $*_g K = K$. This seems to answer your (2) and (3) also.

This is actually just linear algebra, it doesn't depend at all on the differential structure of $M$. A good reference for this is Huybrechts' book "Complex geometry", Chapter 1, Section 1.2.

  • $\begingroup$ Thank you! I guess I should have been aware of this fact, but wasn't. :) $\endgroup$
    – D. S. Park
    Nov 11, 2013 at 18:55

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