Let $(V,\langle,\rangle)$ be a finite dimensional inner product space, $V^{\wedge}$ it exterior algebra, and $\ast$ the Hodge star arising from $\langle,\rangle$. Does there exist any formula to "distribute" the Hodge map over the wedge product. For example, for the de Rham complex we have the Leibniz rule $$ d(w \wedge v) = dw \wedge v + (1)^kw \wedge dv? $$ Is there any analogous formula for $\ast$. I would guess $$ \ast_H(w \wedge v) = \ast(w) \# v \pm w \,\# \ast(v), $$ where $\#$ is contraction.

$\begingroup$ Nope. For one, the degrees of the forms involved don't add up: $\deg *(u\wedge v) = n  \deg u  \deg v$, while $\deg( *u \wedge v) = n  \deg u + \deg v$ and $\deg (u \wedge * v) =n + \deg u  \deg v$. Such a thing would also lead to similar distributive formulas for the $\Lambda$ operator, which would imply that the wedge product of primitive forms were primitive, which it is not in general. $\endgroup$– Gunnar Þór MagnússonApr 3, 2014 at 18:40

1$\begingroup$ Hi Gunnar, yes I see that it obviously does not work, I was just using it as an anology. $\endgroup$– user49105Apr 3, 2014 at 18:51

$\begingroup$ In your suggested formula in the edit, what are the contractions by? Don't get me wrong, I'd love if any such formula existed, but I just don't think it does; it's hard for me to see how its existence wouldn't lead to distributive formulas for $\Lambda$ or $d^*$, which cannot exist. $\endgroup$– Gunnar Þór MagnússonApr 3, 2014 at 19:44

2$\begingroup$ Why can there not exist distributive formulas for $d^*$? $\endgroup$– Tomasz KönerApr 4, 2014 at 7:05
2 Answers
I'm surprised this question has been only partially answered until now. Abel Stern's answer generalizes nicely but it requires thinking about some things that are typically not mentioned in definitions of the Hodge star. This thread is also a duplicate of https://math.stackexchange.com/questions/605448/thehodgeoperatorandthewedgeproduct which is older.
Let $V$ be a finitedimensional oriented inner product space. We think of the $k$forms on $V$ as the space of alternating $k$linear functionals $V^k \to \mathbb R$. The space of $k$linear alternating functionals on $V$ is denoted by various things, I would like to use the notation $\Lambda^k (V^*)$. The star is put there to indicate we are talking about realvalued functions. There is another reason for this notation, in that there is framework where one can talk about abstract wedge and tensor products of abstract vector spaces, and the answer to your question fits naturally in that language.
Precisely, if $V$ and $W$ are vector spaces, and if $f : V \times W \to U$ is a bilinear function, there exists a unique linear functional $\tilde f : V \otimes W \to U$ factoring a certain canonical bilinear map $V \times W \to V \otimes W$. This is called the universal property of the tensor product.
If $f : V \times V \to U$ is alternating and bilinear, the same conclusion holds but replacing $V \otimes W$ with $V \wedge V$. We denote $V \wedge V$ by $\Lambda^2 V$, and iterate this notation $\Lambda^k V = V \wedge V \wedge \cdots \wedge V$.
So in this language, $\Lambda^k(V^*)$ is canonically identified with the dual space of $V \wedge \cdots \wedge V$.
But now you can ask, is there a conflict of notation? $V^*$ is a vector space, so $\Lambda^k(V^*)$ could now be given a 2nd meaning  it is originally the alternating $k$linear functionals on $V^k$ but it is now also $V^* \wedge V^* \wedge \cdots \wedge V^*$. The wedge product of $1$forms gives an alternating $k$linear functional $V^* \times V^* \times \cdots \times V^* \to \Lambda^k(V^*)$ so by the universal property of wedge products of vector spaces, these two objects are canonically identified.
So now for the next step. Contraction is typically viewed as a bilinear operation
$$\Lambda^k(V^*) \times V \to \Lambda^{k1}(V^*)$$
but clearly it makes sense to consider it as a bilinear functional
$$\Lambda^k(V^*) \times \Lambda^j(V) \to \Lambda^{kj}(V^*)$$
We have a canonical isomorphism between $\Lambda^k(V^*)$ and $\Lambda^k(V)^*$ given by the universal property. Let's denote the isomorphism $V \to V^*$ by $c$. This induces an isomorphism $\overline{c} : \Lambda^k(V^*) \to \Lambda^k(V)$.
The Hodge star is therefore the map that takes $\omega \in \Lambda^k(V^*)$ and sends it to the contraction:
$$(*\omega) = Det( \overline{\omega}, \cdot) \hskip 1cm \text{ (my notation for contraction)} $$
Where $Det \in \Lambda^{dim(V)}(V^*)$ is the canonical generator of your topdimensional forms given by the orientation and inner product.
This gives
$$ *(\omega \wedge \eta) = (*\omega)(\overline{\eta}, \cdot) = (1)^{jk}(*\eta)(\overline{\omega}, \cdot)$$
provided $\omega$ is a $j$form and $\eta$ is a $k$form. So this is close to what you were looking for but there's only the one term.
When $w$ is of degree 1, $\ast(v \wedge w) = \iota_{w^\flat} \ast v$ because for the inner product, $\langle w \wedge v, u \rangle = \langle v, \iota_{w^\flat} u \rangle$. This can then be applied repeatedly.
As Ryan Budney explains in far superior detail, the generalization is straightforward. Explicitly, if we define $\iota_{\alpha \wedge \beta} = \iota_\beta \circ \iota_\alpha$ and if we define $w^\flat \in \Lambda^\bullet(V^*) \simeq (\Lambda^\bullet (V))^*$ to be such that $\langle w, u \rangle = w^\flat(u)$, the formula $\langle w \wedge v, u \rangle = \langle v, \iota_{w^\flat} u \rangle $ holds in general, so for all $v,w \in \Lambda^\bullet(V)$, $$\ast(v \wedge w) = \iota_{w^\flat} \ast v$$
Note, by the way, that the Hodge star does not arise from $\langle,\rangle$ alone: one needs a choice of sign for the volume form.