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?
 A: 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 $p<k$, 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 $J'\subset I$.
