Take the 2-minute tour ×  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?
share|improve this question In the Riemannian case, $*$ can be
defined by $\alpha\wedge *\alpha=volume$. So then your formula
seems right, at least up to sign. –  Donu Arapura Dec 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? –  diverietti
Dec 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 $$
\star_{S \times N} (\theta \wedge \omega) = (-1)^{q(m-p)}
\star_S \theta \wedge \star_N \omega $$ –  José
Figueroa-O'Farrill Dec 8 '11 at 1:58 Thank you very much! But I
obtain the following $$\star_{S\times
N}(\theta\wedge\omega)=\frac{