A systematic canonical construction of the Hodge star operator

Question

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 tweaked to get an honest bilinear form on $\bigwedge^k M$?

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...

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:

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

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?

Answer 1

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 $$\star:\bigwedge^kM\to (\det M) \otimes \bigwedge^{n-k}M$$

Answer 2

The bilinear form $g$ need not be symmetric. There is a canonical Hodge operator associated to symplectic manifolds. Here is a general construction of a Hodge type operator.

Suppose that $V$ is a finite dimensional real vector space, $\dim V=n$. To construct a Hodge type operator you need two things.

1. A linear isomorphism $D: V\to V^*$.
2. An orientation, i.e., a linear isomorphism $\newcommand{\bR}{\mathbb{R}}$ $\omega: \det V\to \bR$, where $\det V:=\Lambda^{\dim V} V=\Lambda^n V$.

The existence of a linear isomorphism $D: V\to V^*$ is equivalent with the existence of a nondegenerate bilinear map

$$ B: V\times V\to\bR. \tag{1} $$

(The bilinear map $B$ need not be symmetric.)

The duality $D$ induces canonical isomorphisms $ D^{\wedge k}: \Lambda^k V\to\Lambda^k V^*$, $k=1,\dotsc, n$.

We obtain nondegenerate bilinear maps

$$\Xi_k :\Lambda^k V^*\times \Lambda^{n-k} V\to \bR,\;\;\Xi_k(\alpha_k,\beta^{n-k})= \omega\Bigl(\, \bigl(D^{\wedge k}\bigr)^{-1}\alpha_k\wedge \beta^{n-k}\,\Bigr). $$

This nondegenerate bilinear map produces a linear isomorphism

$$H_k: \Lambda^k V^*\to \bigl(\,\Lambda^{n-k} V\,\bigr)^* $$ which is a generalization of the Hodge operator.

More concretely, if we generically denote by $\langle-,-\rangle$ the canonical pairing between a vector space and its dual, then

$$ \langle H_k (\alpha_k), \beta^{n-k}\rangle = \Xi_k(\alpha_k,\beta^{n-k}). $$

If $\underline{e}:=(e_j)$ is a basis of $V$ and $e^j)$ is the $D$-dual basis of $V^*$, $e^j= D(e_j)$. We set

$$ \omega_e:=\omega( e_1\wedge \cdots \wedge e_n)\in\bR\setminus \{0\}. $$

Then

$$ H_k (e^1\wedge \cdots \wedge e^k)= Ce^{k+1}\wedge \cdots \wedge e^n, $$

where the constant $C$ is found from the equality

$$ C \det(\, \langle e_i,e^j\rangle )_{k<i,j\leq n}=\omega_e. $$

Now observe that $$\langle e_i, e^j\rangle = B(e_i,e_j),$$ where $B$ is the bilinear form (1) associated to $D$.