Hodge dual on hermitian manifold

Question

Recently I have been reading a number of mathematical physics articles in which different definitions are given for the Hodge dual on a hermitian manifold. For the Hodge operator that transforms a (p,q)-form into a (D-p,D-q)-form I came across the following definition.

$\ast \omega_{p,q} = \frac{(-1)^{\frac{1}{2}D(D+1)}i^D}{p! q! (D-p)!(D-q)!} \bar{\omega}_{i_1 \ldots i_q \bar{i}_1 \ldots \bar{i}_p} \epsilon^{i_1\ldots i_q}{}_{\bar{j}_{q+1} \ldots \bar{j}_D} \epsilon^{\bar{i}_1\ldots \bar{i}_p}{}_{j_{p+1} \ldots j_D} d\xi^{j_{p+1}} \wedge \ldots \wedge d\xi^{j_D} \wedge d\bar{\xi}^{\bar{j}_{q+1}} \wedge \ldots \wedge d \bar{\xi}^{\bar{j}_D}$.

Another definition that I encountered was similar, but had a different prefactor: $\frac{(-1)^{\frac{1}{2} D(D-1) + (D-p)q}i^D}{p! q! (D-p)!(D-q)!}$.

My question is, which prefactor is correct? Moreover, what is the reason for the $i^D$ term? (I also encountered a third definition which has the second prefactor, but without the $i^D$ term)

I hope anyone could help out (despite al the indices :P)

Ygor

PS. D denotes the complex dimension, and $\xi = \frac{1}{\sqrt{2}}(y^{2i-1} + iy^{2i})$

Answer 1

First of all, I think that the Hodge $\ast$-operator takes $(p,q)$-forms to $(D-q,D-p)$-forms. You seem to be talking about the composition of the Hodge $\ast$-operator with complex conjugation.

I will not do the calculation for you, but I will outline it and perhaps you can tell us later which of the expressions is indeed the correct one. (I'm not being purposefully cruel: it's just that I don't work with $\varepsilon$ tensors and hence don't know the answer off the top of my head.)

Let $\theta^a$ be a basis of the $(1,0)$-forms with their complex conjugates $\bar\theta^a$ a basis for the $(0,1)$-form, relative to which the hermitian metric takes the form $$h = \sum_a (\theta^a \otimes \bar\theta^a + \bar\theta^a \otimes \theta^a).$$ If we define the $e^j$ by $$\theta^a = \frac1{\sqrt{2}} ( e^{2a-1} + i e^{2a} )$$ then we see that the $e^j$ are orthonormal with respect to the riemannian metric.

The Hodge $\ast$-operator is easy to define on monomials $e^I = e^{i_1} \wedge \cdots \wedge e^{i_n}$, where $I = (i_1,\cdots,i_n)$ is a multi-index with $1\leq i_1 < \cdots < i_n \leq 2D$. If $I_c=(i_{n+1},\cdots,i_{2D})$ is the complementary multi-index, then $$\ast e^I = (-)^\sigma e^{I_c}$$ where $(-)^\sigma$ is the sign of the permutation $$\sigma = \begin{pmatrix} i_1 & i_2 & \cdots & i_{2D} \cr 1 & 2 & \cdots & 2D \end{pmatrix}.$$

Your mission, Ygor, should you decide to accept it, is to see what $\ast$ does on a monomial $$\theta^{a_1} \wedge \cdots \wedge \theta^{a_p} \wedge \bar\theta^{b_1} \wedge \cdots \wedge \bar\theta^{b_q}.$$

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Perhaps a simpler way is to check for which prefactor the inner product on $(p,q)$-forms defined by $$\alpha \wedge \ast \bar\beta = \left<\alpha,\beta\right> \operatorname{dvol}$$ is the expected one.

Answer 2

I think it can be calculated from the usual definition of a hodge star. The difference is that the Levi-Civita symbol is a pseudotensor, and its transformation properties are a little different from an ordinary tensor. For an orthonormal frame with the same orientation, the Levi Civita symbol is invariant under coordinate transformations (See the Wikipedia page on the Levi Civita symbol). But I don't think the complex coordinates have these properties, and thus we get extra factors upon the coordinate transformation to complex coordinates. Carefully transforming the usual definition of the Hodge star operator to complex coordinates should give you the correct prefactor. Do take everything I've said with a pinch of salt, as I have not proven this myself.