# The Hodge $*$-operator and the wedge product

On every Riemannian manifold $M$, we can consider the Hodge $*$-operator, which is characterised by the following formula: $$a \wedge *b = (a,b)\nu.$$ Here $a$ and $b$ are smooth forms on $M$, $(\ ,\ )$ is a metric on $\wedge T^*\!M$ and $\nu$ is the volume form with respect to the Riemannian metric.

My question: Is a formula of $*(a \wedge b)$ known?

I suspect that we can have a formula like "$*(a \wedge b)=(*a)\wedge(*b)$" or "$*(a \wedge b) = *a \wedge b \pm a \wedge *b$". Of course these formulae never hold. (Look at the degree.)

• I put an answer to your question in the MO thread: mathoverflow.net/questions/162366/… – Ryan Budney Mar 25 '15 at 23:00
• @RyanBudney Thank you. Your formula is simpler than I expected, even though the contraction is non-trivial. – H. Shindoh Mar 26 '15 at 10:56

## 1 Answer

A toy example. Given an orthonormal basis $e_1,\dots,e_n$ for the vector space $V$, then $\nu=e_1\wedge\dots\wedge e_n$ and

$$*(e_1\wedge e_2)=e_3\wedge\dots\wedge e_n\in \bigwedge^{n-2}V,$$

with $*e_1=e_2\wedge\dots\wedge e_n$ and $*e_2=-e_1\wedge e_3\wedge\dots\wedge e_n$ both in $\bigwedge^{n-1}V$. Introducing the insertion operator $i_\bullet: \bigwedge^{k}V\rightarrow \bigwedge^{k-1}V$ we arrive at

$$*(e_1\wedge e_2)=\frac{1}{2}\left( \underbrace{i_{e_2}(\underbrace{*e_1}_{\in \bigwedge^{n-1}V}}_{\in \bigwedge^{n-2}V})-i_{e_1}(*e_2)\right).$$

I do not if this is what you were searching for, but I hope it helps (a bit).

• For related formulae, I refer to the "Three dimensional example" with $\star (\mathbf u \wedge \mathbf v )=\mathbf {u \times v}\,,\quad\star (\mathbf u \times \mathbf v ) = \mathbf u \wedge \mathbf v \,,$ in en.wikipedia.org/wiki/Hodge_dual – Avitus Dec 13 '13 at 20:33
• Thanks for the formula. It could suggest that we can not have a simple formula (or may need a special notation). – H. Shindoh Dec 15 '13 at 20:26
• You are welcome. No "simple" formula (like Jacobi identity, as you pointed out) is probably true. I found interesting the "dual" relationship between contraction and Hodge, though. The 3-dim case in the above comment is something non trivial. – Avitus Dec 15 '13 at 20:56