# The Hodge star operator and the wedge product: $\alpha \wedge (\star \beta)$

According to Wikipedia,

The Hodge star operator on a vector space $$V$$ with an inner product is a linear operator on the exterior algebra of $$V$$, mapping $$k$$-vectors to $$(n-k)$$-vectors where $$n=\dim V$$, for $$0 ≤ k ≤ n$$. It has the following property, which defines it completely: given two $$k$$-vectors $$α$$, $$β$$, $$\alpha \wedge (\star \beta) = \langle\alpha,\beta \rangle\,\omega$$ where $$\langle \cdot,\cdot \rangle$$ denotes the inner product on $$k$$-vectors and $$\omega$$ is the preferred unit $$n$$-vector.

But suppose instead that $$\alpha$$ is a $$k$$-vector and $$\beta$$ is a $$p$$-vector, with $$k\ne p$$. Then is there a well-known formula for $$\alpha \wedge (\star \beta)$$ like there is when $$k=p$$? I think $$\langle\alpha,\beta \rangle$$ is not defined in this case, correct?

Of course, you need an orientation on your vector space $$V$$ to get "the preferred unit $$n$$-vector." You're correct, that you only have an inner product when your two "vectors" live in the same vector space, i.e., are both $$k$$-vectors for the same $$k$$. I tend to think of this formula as a convenient way to define the inner product, rather than as a way to define the Hodge star operator.
Of course, when $$p, we have $$k+(n-p)>n$$ and so $$\alpha\wedge\star\beta = 0$$. When $$p>k$$, I would just expand $$\alpha$$ and $$\star\beta$$ in terms of the orthonormal basis $$e_1,\dots,e_n$$. Remember that the set $$e_I=e_{i_1}\wedge\dots\wedge e_{i_k}$$ with $$I$$ increasing give an orthonormal basis for $$\Lambda^k V$$. So you will get nonzero contributions to $$\alpha\wedge\star\beta$$ when you have terms of the form $$e_I\wedge e_J$$, with $$J\cap I = \emptyset$$, and so this means that the complementary set $$J'\subset I$$. Thus, for a given monomial $$a_Ie_I$$ expression in $$\alpha$$, you want to look at terms $$b_{J'}e_{J'}$$ in $$\beta$$ with $$J'\subset I$$.
• Thanks Ted. So, to your knowledge, there is no simple formula for when $p>k$; one is best of working in coordinates, correct? – Doubt Dec 6 '18 at 3:33