Let S and N be two pseudo-Riemannian manifolds. Is it true that
$$\star_{S\times N}(\theta\wedge\omega)=\star_{S}\theta\wedge\star_{N}\omega,$$
where $\theta$ is a form on S and $\omega$ is a form on N?

In the Riemannian case, $*$ can be defined by $\alpha\wedge *\alpha=volume$. So then your formula seems right, at least up to sign.
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Donu ArapuraDec 7 '11 at 13:46

I guess first of all that your manifolds are oriented, right? Then, how do you define the Hodge * operator in the pseudo-riemannian case? Is it still an isometry?
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diveriettiDec 7 '11 at 15:57

As Donu says, the formula is correct up to signs. More precisely if $\theta \in \Omega^p(S)$ and $\omega \in \Omega^q(N)$ with $\dim S = m$ and $\dim N = n$, then $$ \star_{S \times N} (\theta \wedge \omega) = (-1)^{q(m-p)} \star_S \theta \wedge \star_N \omega $$
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José Figueroa-O'FarrillDec 8 '11 at 1:58

Thank you very much! But I obtain the following $$\star_{S\times N}(\theta\wedge\omega)=\frac{p!q!}{(p+q)!}(-1)^{q(m-p)}\star_{S}\theta\wedge\star_{N}\omega.$$
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NataliaDec 8 '11 at 13:17

Here I suppose that $\theta\wedge\omega=Alt(\theta\otimes\omega),$ $Alt T=\frac{1}{p!}\Sigma_{\sigma\in S_{p}}(sgn \sigma) T^{\sigma},$ $$\langle u_{1}\otimes... \otimes u_{p},\tilde{u}_{1}\otimes... \otimes \tilde{u}_{p}\rangle=\langle u_{1},\tilde{u}_{1}\rangle...\langle u_{p},\tilde{u}_{p}\rangle,$$ where $u_{1},...u_{p},\tilde{u}_{1},...,\tilde{u}_{p}$ are 1-forms. Then I get $$\langle\theta_{1}\wedge\omega_{1},\theta_{2}\wedge\omega_{2}\rangle=\frac{p!q!}{(p+q)!}\langle\theta_{1},\theta_{2}\rangle\langle\omega_{1},\omega_{2}\rangle$$ therefore the coefficient appear. Is it true?
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NataliaDec 8 '11 at 13:20