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Let $(X,\omega)$ be a compact Kahler manifold, and let $\alpha$ and $\beta$ be smooth $(1,1)$-forms on $X$ that are harmonic (with respect to $\omega$). I can consider each of my $(1,1)$-forms as an antilinear vector bundle morphism $T_X \to \overline T_X^*$. Then I can fabricate a new $(1,1)$-form on $X$ by setting $F = \alpha \circ \omega^{-1} \circ \overline \beta$.

Question: Is the form $F$ harmonic?

I was hoping there was some general theory available to answer this quickly, but I haven't found anything. Calculations in local coordinates also quickly degenerated into filth.

For motivation, if you take $Tr_\omega(F)$ then you get the value of the scalar product on $(1,1)$-forms induced by $\omega$ of $\alpha$ and $\beta$. Thus knowing that $F$ is harmonic lets one conclude that the map $x \mapsto \langle \alpha(x),\beta(x) \rangle_\omega$ is constant.

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up vote 8 down vote accepted

By your last sentence, this is clearly not true in general.

Let $X$ be a $K3$ surface and let $\omega$ be a Kähler form on $X$ whose associated metric is Ricci flat. Let $H$ denote the space of the real-valued harmonic $(1,1)$-forms $\alpha$ that satisfy $\omega\wedge\alpha = 0$, i.e., the primitive $(1,1)$-forms. Then $H$ is a real vector space of dimension $19$, and, just by algebra, one has an identity of the form $\alpha\wedge\beta = -\langle\alpha,\beta\rangle\ \omega^2$.

Now, if $\langle\alpha,\beta\rangle$ were constant for all $\alpha,\beta\in H$, then one could easily find $3$ elements $\alpha_1,\alpha_2,\alpha_3\in H$ such that $\langle\alpha_i,\alpha_j\rangle = \delta_{ij}$, and it would follow that these $3$ forms were a basis at each point of the anti-self-dual $2$-forms on $X$. But then, any $\beta\in H$ that satisfied $\langle\alpha_i,\beta\rangle=0$ for $i = 1,2,3$ would have to vanish identically. In particular, the dimension of $H$ would be 3, not 19.

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