I'm struggling to make sense of the hodge star as a global canonical object. Here are my struggles so far and some questions:

Let $M$ be a finitely generated projective $R$-module (hence locally free and finitely presented). Suppose the $M$ is of rank $n$ and is equipped with a nondegenerate symmetric bilinear form $g(-,-): M \times M \to R$. We first extend the bilinear form to the exterior powers of $M$.

Let $\bigwedge^k m_{j}$,$\bigwedge^k l_{j} \in  \bigwedge^k M$ be given. There's a corresponding "orthogonal projection" map $T_{m,l}: span\{m_j\} \to span\{l_j\}$ defined by: 

$$T_{m,l} : m \mapsto \sum_{j} g(m,l_j)l_j$$

So we obtain a map $T_{m,l} \in Hom(span^k_{j=1}\{m_j\},span^k_{i=1}\{l_i\})$. Denote the induced map on the top exterior powers by $det(T_{m,l})\in$ $Hom(\bigwedge^k span\{m_j\},\bigwedge^k span\{l_i\})$. By local freenes $det(T_{m,l})$ is locally just multiplication by the determinant of the local transformation. 

Define the extended inner product: $<\bigwedge^k m_{j}$,$\bigwedge^k l_{j}> := det(T_{m,l})$. Symmetry and nondegeneracy are inherited from $g(-,-)$. Extend to non simple elements by linearity. 

This was the construction I arrived at when trying to find a cannonincal geometric definition for the extension of an inner product to exterior powers of a vector bundle. My problem now is that by this definition the extended bilinear form is only locally a bilinear form. differential-geometrically there's no real problem because things only need to make sense locally but **what's really happening here?**

> **1) Can the above construction be tweeked to get an honest bilinear form on $\bigwedge^k M$?**

Assuming such a tweek is possible we'll now proceed to construct an unoriented analauge of the hodge star out of the following maps:


$$\phi:\bigwedge^k M \to ( \bigwedge^{n-k}M \to\bigwedge^n M), u \mapsto u \wedge (-)$$

$$\psi:\bigwedge^{n-k} M \otimes\bigwedge^n M\to ( \bigwedge^{n-k}M \to\bigwedge^n M), u\otimes\omega \to <-,u>\otimes \omega$$

Define the "pre-hodge" operator $\rho := \psi^{-1} \circ \phi : \bigwedge^k M \to \bigwedge^{n-k} M \otimes \bigwedge^n M$

Evaluating at an element $\bigwedge_{J} m_j \in \bigwedge^k M$ we get:

$$\rho(\bigwedge_{J} m_j)= \psi^{-1} \circ \phi (\bigwedge_{J} m_j) = \psi^{-1}(\sum_{|I|=n-k} ((\bigwedge_{J} m_j)\wedge (\bigwedge_{I} l_i)) \otimes (\bigwedge_{I} l_i)^*) = \sum_{|I|=n-k} ((\bigwedge_{J} m_j)\wedge (\bigwedge_{I} l_i)) \otimes (\bigwedge_{I} l_i)$$


The summing part is the problem here. It has to be some sum over all $(n-k)-$ collections with signs attached to them but i'm not sure how it should be done...

> 2. **How does the map $\rho$ act on simple elements?**

To apply the above construction to $\bigwedge^k M^*$ there has to be some duality pairing between them. Assume the pariring is as in [the following question by Qiaochu Yuan][1]:

$$(m_1 \wedge ... \wedge m_k) \otimes (\theta_1 \wedge ... \wedge \theta_k) \to \frac{1}{k!} \sum_{\sigma \in S_k} \text{sgn}(\sigma) \prod_{i=1}^k    \theta_i(m_{\sigma(i)}).$$

Now we have two choices for extension to $\bigwedge^k M^*$. We can extend to $M^*$ by the canonincal evaluation pairing and then extend to $\bigwedge^k M^*$ by the above. Or we could first extend to $\bigwedge^k M$ and then extend to $\bigwedge^k M^*$ by the above pairing. 

> **3. Do these extensions agree? If not can it be fixed?**

Finally Supposing $\bigwedge^n M \cong R$. By choosing an orientation we get a trivilization of $\bigwedge^n M$ and composing with $\rho$ gives a map $\bigwedge^k M \to \bigwedge^{n-k} M$ and by the previous paragraph we get a certain "hodge star" $\star: \bigwedge^k M^* \to \bigwedge^{n-k} M^*$

> **4. Does the above map $\star$ coincide with the hodge star?**

And finally:

> **5. Does it have to be so complicated? Or is there a simple construction i'm unaware of?**


  [1]: http://mathoverflow.net/questions/68004/natural-pairings-between-exterior-powers-of-a-vector-space-and-its-dual