Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $a+b=N-k$.

share|improve this question
    
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. –  Donu Arapura Nov 21 '10 at 19:14
    
So yes in the sense you mean. –  Donu Arapura Nov 21 '10 at 19:18
    
The last question is just a rewording of the previous line. Since it is non-essential and apparently confusing I'll delete it. –  Abtan Massini Nov 21 '10 at 19:24
1  
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. –  Spiro Karigiannis Nov 21 '10 at 19:42
1  
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.) –  Donu Arapura Nov 21 '10 at 19:42
show 3 more comments

1 Answer 1

up vote 4 down vote accepted

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 $(p,q)$.

share|improve this answer
    
@Donu: there was a display problem with a math expression. I hope you don't mind that I fixed it. –  Willie Wong Nov 21 '10 at 20:03
    
Willie, I noticed that too. We might have been editing at the same time (?). –  Donu Arapura Nov 21 '10 at 20:05
    
Oh and thanks. Looks better. –  Donu Arapura Nov 21 '10 at 20:06
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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