The Hodge star operator in differential geometry is a map from $*: \Omega^{k}(\mathcal{M}) \to \Omega^{m-k}(\mathcal{M})$ where $\Omega^{l}(\mathcal{M})$ is the space of $l$-forms on $m$-dimensional manifold $\mathcal{M}$. I wonder if this operator just like the other Hodge operators like $d$ and $d^{\dagger}$, can have notions of 'determinants' and 'trace'(determinants and trace of operators play an important role in the computation of path-integrals in physics). Also, for an $l$-form $\omega$ we have $$*^2\omega = (-1)^{l(m-l)}\omega$$
If we redefined the $*$-operator as $\star = (-1)^{-l(m-l)/2}*$ so that $\star^2 = \bf{I}$. If we can assign the notion of a functional determinant to the $\star$-operator then can we say that $$(\det\star)^2= \bf{I}$$
Best Answer
$(-1)^{l(m-l)}=(-1)^{l(m-1)}$, so if $m$ is odd then $\star_{m-l} \star_{l} = \bf{I}$, so $\star$ is an involution on the exterior algebra $\Omega(\mathcal{M})$. If $m$ is even then $\star_{m-l} \star_{l} = (-1)^{l}\bf{I}$, so $\star$ viewed as an operator on $\Omega(\mathcal{M})$ satisfies $$\operatorname{det}(\star^2)=\prod_{l=0}^m (-1)^{l\binom{m}{l}}$$ and $$\operatorname{tr}(\star^2)=\sum_{l=0}^m (-1)^{l}\binom{m}{l}$$