Hodge Star operator and volume form on arbitrary manifold

Question

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 $*\Omega = 1$ ?

tenpercent · Accepted Answer · 2014-02-24 12:01:45Z

1

I've managed to do the following math:

$\Omega \wedge *\Omega = \left(\Omega, \Omega\right)\Omega = \Omega$

$*\Omega $ is scalar, so it seems that my guess was right. Was it?

answered Feb 24, 2014 at 12:01

tenpercent

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$\begingroup$ You're completely right! :) $\endgroup$
– Dragoslav
Commented Feb 24, 2014 at 12:40
1

$\begingroup$ 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. $\endgroup$
– Gil Bor
Commented Feb 24, 2014 at 16:57

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