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?**



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> **1) Can the above construction be tweaked to get an honest bilinear form on $\bigwedge^k M$?**
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Assuming such a tweak is possible we'll now proceed to construct an unoriented analogue 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...



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> 2. **How does the map $\rho$ act on simple elements?**
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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](https://mathoverflow.net/questions/68004/natural-pairings-between-exterior-powers-of-a-vector-space-and-its-dual):


$$(m\_1 \wedge ... \wedge m\_k) \otimes (\theta\_1 \wedge ... \wedge \theta\_k) \to \frac{1}{k!} \sum\_{\sigma \in S\_k} \text{sgn}(\sigma) \prod\_{i=1}^k \theta\_i(m\_{\sigma(i)}).$$


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. 



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> **3. Do these extensions agree? If not can it be fixed?**
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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" $\star: \bigwedge^k M^\* \to \bigwedge^{n-k} M^\*$



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> **4. Does the above map $\star$ coincide with the Hodge star?**
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And finally:



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> **5. Does it have to be so complicated? Or is there a simple construction i'm unaware of?**
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A systematic canonical construction of the Hodge star operator
The first step is to construct a pairing on the modules $\bigwedge^k M$. I will assume that the pairing $g:M\otimes M\to R$ is perfect, that is it induces an isomorphism $M\to \textrm{Hom}(M,R)$. Then we can define a pairing on $\bigwedge^k M$ as follows


$$g(v\_1\wedge\cdots\wedge v\_k, w\_1\wedge \cdots\wedge w\_k) = \det(g(v\_i,w\_j))$$


**CLAIM:** This is a perfect pairing.


Since everything is natural we can work locally and assume that $M$ is free. Then let $m\_1,\dots,m\_k$ be a basis of $M$ and $m\_1^\star,\dots,m\_k^\star$ be the dual basis with respect to $g$. Then it is clear that the matrix representing $g$ in the corresponding bases of $\bigwedge^kM$ is the identity and so $g$ is perfect (note that we are using different basis for the two arguments of $g$).


Using our perfect pairing we identify
$$\textrm{Hom}\_R\left(\bigwedge^{n-k}M,\bigwedge^n M\right)= \left(\bigwedge^{n-k}M\right)^\vee\otimes \bigwedge^n M \cong \bigwedge^{n-k}M\otimes \bigwedge^n M$$


Ok, now we can define our "pseudo-Hodge" operator. We send $\omega\in \bigwedge^k M$ to the map
$$\bigwedge^{n-k}M\to \bigwedge^nM$$
given by $\eta\mapsto \omega\wedge \eta$. By using the previous identification this gives a map
$$\star:\bigwedge^kM\to (\det M) \otimes \bigwedge^{n-k}M$$


