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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...

  1. 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?

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    $\begingroup$ I am confused. What is in the description $g(\star \omega, \eta) = g(dV,\omega\wedge \eta)$ that is noncanonical? $\endgroup$ Nov 17 '15 at 14:17
  • $\begingroup$ @DenisNardin Is $dV$ the volume form here? Still it's not clear to me how is $g(-,-)$ extended to exterior powers. $\endgroup$ Nov 17 '15 at 14:20
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    $\begingroup$ Yes, as far as I understand the Hodge star depends on a choice of volume form (an "orientation"). Also I think that you may want to use a perfect pairing, rather than just a nondegenerate bilinear form. To extend the pairing I would use the one induced by the standard pairing on $M^{\otimes k}$ (since $\Lambda^kM$ is a quotient of it). If you are interested I could try to write down the details. In which level of generality are you interested in? $\endgroup$ Nov 17 '15 at 14:27
  • $\begingroup$ @DenisNardin Thanks. I'll be satisfied by a construction of a pre-hodge star on a projective finitely presented module of finite rank. By pre-hodge I mean an twisted-hodge star (twisted by $\bigwedge^n M$) such that adding a volume form (in the case of a trivial top exterior power) makes it a hodge star. I'm not interested in the generality just for the sake of it, but rather becasue i'd like to understand it better and stripping away the unnecessary structure is one of the best ways i know. $\endgroup$ Nov 17 '15 at 14:37
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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 $$\star:\bigwedge^kM\to (\det M) \otimes \bigwedge^{n-k}M$$

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  • $\begingroup$ My global constructions via maps between exterior powers was motivated by this precise definition. Since both of these agree locally (their localizations agree) Doesn't it mean that my geometric definition is the same as your "local" one? $\endgroup$ Nov 17 '15 at 15:10
  • $\begingroup$ I think that the definitions are the same, but I always have an hard time parsing everything with more than one index, so don't trust me too much on this. But I do not understand in which sense my definition is local. I only localize to verify that the pairing is perfect, all the rest is done globally. I guess that my point is that you do not need all those messy computations at all: everything is a simple algebraic fact. $\endgroup$ Nov 17 '15 at 15:12
  • $\begingroup$ I see your point. I was searching for a geometrical picture for why the determinant is involved. Top exterior powers of finitely generated sub modules corresponds locally to tangent hyperplanes and the transformation between them induced by the orthogonal projection using the bilinear form is preciesly the determinant which is a very reasonable generalization since it corresponds to the volume of a projected unit box. Just as the inner product correspods to the length of a projected unit vector. Does this sound right to you? $\endgroup$ Nov 17 '15 at 15:23
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    $\begingroup$ Honestly I don't know. I never found those pictures of the exterior powers convincing (expecially because the grassmannian is only a subset of $\mathbb{P}\Lambda^kM$ and not the whole thing) so I'm probably the wrong person to ask. $\endgroup$ Nov 17 '15 at 15:26
  • $\begingroup$ That's enough for me to give up the digging for now. Thanks! $\endgroup$ Nov 17 '15 at 15:28
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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 $D$.

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  • $\begingroup$ I think this is exactly the same construction as in my answer (although I did not emphasize that I did not use the symmetry). Over a general ring (e.g. continous functions on a manifold) however you still need your pairing to be perfect, not only nondegenerate. $\endgroup$ Nov 17 '15 at 15:40
  • $\begingroup$ I should have said that nondegenerate means perfect to me.The finitely generated projective modules of the ring of continuous functions are the modules of continuous sections of a vector bundle, and thus it suffices to work with vector spaces. $\endgroup$ Nov 17 '15 at 15:49

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