The Hodge star operator $\ast$ acts on the differential forms of a differential manifold sending $\Omega^{k}$ to $\Omega^{N-k}$. If the manifold is complex, then for $p+q=k$, does $\ast$ map $\Omega^{p,q}$ into some $\Omega^{a,b}$, where $a+b=N-k$.
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asked Nov 21, 2010 at 18:50
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$\begingroup$ The last question is ambiguous, but I think the 2nd sentence is also a question, so you should put a question mark after it. You can find the answer in many places, e.g. Griffiths-Harris page 66. $\endgroup$– Donu ArapuraCommented Nov 21, 2010 at 19:14
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$\begingroup$ So yes in the sense you mean. $\endgroup$– Donu ArapuraCommented Nov 21, 2010 at 19:18
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$\begingroup$ The last question is just a rewording of the previous line. Since it is non-essential and apparently confusing I'll delete it. $\endgroup$– Abtan MassiniCommented Nov 21, 2010 at 19:24
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2$\begingroup$ Since $∗$ is a real operator, to be more precise, you should say that after complexifying the space of forms, and extending $∗$ to be complex linear, then indeed $∗$ maps $\Omega^{p,q}$ to $\Omega^{N-q,N-p}$. Frequently, it is preferable to use $\bar *$, which is the composition of $∗$ with complex conjugation, to map $\Omega{p,q}$ to $\Omega{N−p,N−q}$. This way, we have $\alpha \wedge \bar * \beta = g(\alpha, \beta) vol$, where $g$ is the Hermitian metric. $\endgroup$– Spiro KarigiannisCommented Nov 21, 2010 at 19:42
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1$\begingroup$ Our comments seemed to have crossed. Yes, p 82 not 66, and yes. (Be warned that some authors use a different convention, where $N-p,N-q$ get switched. It's a question of whether $*$ is linear or antilinear.) $\endgroup$– Donu ArapuraCommented Nov 21, 2010 at 19:42
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1 Answer
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As Abtan requested, I'm converting my comments to an answer:
Suppose that $X$ is an $N$ (complex) dimensional complex manifold endowed with a Hermitean metric, or equivalently a Riemannian metric g satisfying $g(JX,JY)=g(X,Y)$, where $J$ is the complex structure. Let $*$ denote the $\mathbb{C}$-antilinear extension of the Hodge star operator to complex valued forms (some people -- including me -- prefer to write this as $\overline{*}$ as Spiro points out in the comments). Then as one finds on page 82 of Griffiths and Harris, $$*\Omega^{pq}\subset \Omega^{N-q,N-p}$$ where I'm following the notation in the question and writing $\Omega^{pq}$ for the space of $C^\infty$ forms of type $(p,q)$.
answered Nov 21, 2010 at 20:01
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$\begingroup$ Isn't the point that the antilinear extension sends p,q to N-p,N-q in contrast to what is written in your answer? $\endgroup$– TomoCommented Sep 20, 2016 at 16:46