# Hodge Star operator and volume form on arbitrary manifold

I guess this is the question on definitions, however, I haven't managed to find a clear answer to this question:

Suppose we have a manifold, there is metric tensor, so we can use it to calculate Hedge Star operator on differential forms.

Let $\Omega$ be the volume form.

Is it true, that $*\Omega = 1$ ?

$\Omega \wedge *\Omega = \left(\Omega, \Omega\right)\Omega = \Omega$
$*\Omega$ is scalar, so it seems that my guess was right. Was it?
• It is important to add the assumption that the manifold is oriented in order to define "the" volume form $\Omega$ associated with the metric. Switching the orientation amounts to changing $\Omega$ to $-\Omega$. Then you can just follow the definition of the Hodge operator, as you did correctly in your answer. – Gil Bor Feb 24 '14 at 16:57