Let $X$ be a compact complex n-dimensional manifold and $h$ be a hermitian metric on $X$ and $\omega$ its hermitian form. I define the following condition: $h$ is called a almost $p$-Kahler metric if there exist an integer $p$ (with $1\leq p\leq(n{-}1)$) and a real $(p,p)$-form $\sigma$ such that $d(\omega^{p}+\sigma)=0$.

In the cases in which $p\neq 1,n{-}1$, $d\sigma\neq0$, and $(\omega^{p}+\sigma)\neq \bar{\omega}^{p}$ for some $\bar\omega$, are there any meaningful examples?

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    $\begingroup$ Where does this notion come from? $\endgroup$ – Gunnar Þór Magnússon Dec 24 '11 at 9:10
  • $\begingroup$ What is $\phi$, and what does it have to do with anything? $\endgroup$ – Robert Bryant Dec 24 '11 at 12:59
  • $\begingroup$ what about $\sigma=-\omega^p$? $\endgroup$ – diverietti Dec 24 '11 at 14:06
  • $\begingroup$ @diverietti: I'm guessing that the description of the problem is somewhat mangled, which is why I asked about $\phi$. I suspect that y2011 probably wanted to put some condition on $\sigma$ beyond just being a real $(p,p)$-form. For example, y2011 might want it to be positive (or at least non-negative) on complex $p$-planes. Some kind of inequality such as that might give one access to area bounds for complex submanifolds of $X$, which y2011 could then use in the same way that Gromov used the taming condition to get compactness for the moduli space of curves (i.e., the $p=1$ case). $\endgroup$ – Robert Bryant Dec 24 '11 at 23:44
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    $\begingroup$ @y2011: I cleaned up the question text a bit. Let me know if I mangled your meaning. Assuming that I've understood you correctly, I'll just comment that, without imposing some kind of positivity(?) condition on $\sigma$, you are not likely to get anything interesting because $\omega$ essentially washes out without some restriction. As diverietti remarked, taking $\sigma=-\omega^p$ will make any metric satisfy your condition. If I were you, I'd look at the first nontrivial case, which would be a complex $4$-manifold and with $p=2$, and see whether I could say anything interesting in that case. $\endgroup$ – Robert Bryant Dec 25 '11 at 15:13

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