Induced bilinear form on exterior powers - Towards a global Hodge Star Operator In all constructions of the hodge star operator I've seen so far there was a part where an inner product on the exterior power of the tangent space was defined by the ungodly local formula: 
$$\langle v_1 \wedge \cdots \wedge v_k , w_1 \wedge \cdots \wedge w_k \rangle = det(\langle v_i,w_j \rangle)$$
Although i've been able to verify that it is in fact an inner product I have a terrible aversion to such ad hoc constructions. Therefore i was led to this:
Given a projective (hence locally free) $R$-module $M$ and a nondegenrate bilinear form $g: M \times M \to R$, i.e. one that gives an isomorphism:
$$\varphi: M \to M^* , \varphi: m \to g(-,m)$$
By the non degeneracy, $g$ extends naturally to $M^*$ by: $g(g(-,m_1),g(-,m_2))=g(m_1,m_2)$.
What's the canonical way to extend $g$ to the exterior powers $\bigwedge^k M^*$?
 A: You need that $M$ is finitely presented in addition to projective in order for duality to work nicely. The problem reduces to describing a natural nondegenerate pairing 
$$\wedge^k M \times \wedge^k M^{\ast} \to 1$$
where $1$ denotes the unit module $R$. Equipped with such a pairing, any isomorphism $M \cong M^{\ast}$ gives an isomorphism $\wedge^k M \cong \wedge^k M^{\ast}$, and we can apply the natural isomorphism $\wedge^k M^{\ast} \cong (\wedge^k M)^{\ast}$ coming from the above nondegenerate pairing. 
In fact more generally there is a natural pairing
$$\wedge^n M \times \wedge^k M^{\ast} \to \wedge^{n-k} M$$
as follows: by the universal property of the exterior algebra, derivations $\wedge^{\bullet} M \to \wedge^{\bullet} M$ are freely determined by what they do to the generators $M$. In particular, elements of $M^{\ast}$ are naturally in bijection with derivations of degree $-1$, and again by the universal property of the exterior algebra, this induces an action of $\wedge^{\bullet} M^{\ast}$ on $\wedge^{\bullet} M$ where $\wedge^k M^{\ast}$ acts by "differential operators" of degree $-k$. 
This description has the disadvantage that it doesn't treat $M$ and $M^{\ast}$ symmetrically. There are other things you can say but to my mind they're all a bit unsatisfying for other reasons. One attempt to write down this map ends up writing down $k!$ times this map for reasons I don't really understand. And in positive characteristic there isn't an analogous natural isomorphism $S^k M^{\ast} \cong (S^k M)^{\ast}$ for symmetric powers. There's something funny going on involving invariants vs. coinvariants for the action of $S_n$ on $M^{\otimes n}$ and I haven't sorted it out well in my head. 
