Derivation of hodge dual of different bivectors

Question

So, I read that the Hodge dual operator acts in the following way (you can find the following in https://en.wikipedia.org/wiki/Hodge_star_operator#Geometric_explanation): $$\alpha \ \wedge \ \star \beta \ = \ (\alpha |\beta)w$$ where $w=\sqrt{det(g)}dx^1 \wedge dx^2 \wedge...\wedge dx^n $, where n is the dimension of the vector space. My question is how it would be possible to carry out the inner product with 2-forms, for example, in order to find their Hodge dual. How is this inner product defined? Does inner product always return a scalar? It is supposed a euclidean space if it makes the question more concrete.

Answer 1

Yes, inner products always return scalars. If you have an inner product in a vector space $V$, then that gives an isomorphism $V\to V^*$, as $v\mapsto \langle v,\cdot \rangle$. Using this isomorphism, we get an inner product $\langle\cdot,\cdot\rangle_*$ on the dual space $V^*$ (space of $1$-forms on $V$).

Now, suppose $1\leq k\leq n$, and that we have a wedge-product of 1-forms $\alpha=\alpha^1\wedge \cdots \wedge \alpha^k$ and $\beta=\beta^1\wedge\cdots \wedge \beta^k$. Then, we define \begin{align} \langle\alpha,\beta\rangle:=\det \bigg(\langle\alpha^i,\beta^j\rangle\bigg). \end{align} In words, use the inner product on $V$ to induce one on $V^*$, and then form a $k\times k$ matrix of inner products of 1-forms, and then take the determinant. This defines an inner product on "basic" $k$-forms. You now extend bilinearly to define it on all pairs of $k$-forms.

Of course, there are several details to be verified. Firstly, one should really use the universal property of the exterior power to show that such a bilinear extension is possible. Next, it still has to be verified that this map really is an inner product on the space of $k$-forms. For the details, one source is Loomis and Sternberg's Advanced Calculus, section 7.8 (there they also consider the more general case of pseudo-inner products, though they refer to that simply as "scalar product").