Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $V$ be an oriented inner product space of dimension $n$. The Hodge star operator maps $\Lambda^k V\to \Lambda^{n-k}V$. In particular it maps $V\to \Lambda^{n-1}V.$ $V$ carries a representation of the rotation group $\text{SO}(n)$ (after choosing a basis, perhaps), and $\Lambda^k V$ carries an induced representation, and we might naïvely expect the Hodge star operator to be a morphism of representations so that $(\Lambda^{n-1}g)\cdot *v=*(g\cdot v).$ However for some reason, an isomorphism with the dual space representation seems more appropriate, which I think should look something like $(\Lambda^{n-1}g)\cdot *v=*((g^{-1})^T\cdot v)$. Is this correct? The Hodge dual is an isomorphism between the standard rep and the dual rep? I only suspect it to be so, because that's what I think it would take to make the Laplacian $\Delta=*d*d$ belong to the trivial representation. But I can't figure out how to show it directly. Is there some nice way to see it?

Edit: The Stackexchange software linked me to a related question with an excellent answer by Qiaochu Yuan. He describes the Hodge star as an isomorphism to the exterior power of the dual space, I hadn't seen it that way before, but it lends credence to the idea that you should get a contragredient representation.

share|cite|improve this question
The metric on $V$ extends to a metric on each $\Lambda^k V.$ The Hodge star is an isometry. About representations, you've got me. – Will Jagy Mar 18 '12 at 0:54
up vote 5 down vote accepted

One way to think about this is via the natural (non-degenerate) pairing $\bigwedge^j V \times \bigwedge ^{n-j} V \rightarrow \bigwedge^n V$. The latter is one-dimensional, and an "orientation" is a choice of isomorphism $\bigwedge^n V\rightarrow k$. At this point, the choice of orientation gives a natural isomorphism of $\bigwedge ^{n-j} V$ with the dual of $\bigwedge^j V$, as you speculate.

When $V$ has an inner product $\langle,\rangle$, the associated special orthogonal group $SO(\langle,\rangle)$ acts on each $\bigwedge^j V$ without choosing a basis, etc.

The highest non-vanishing exterior power is necessarily the trivial repn of this orthogonal group, so the duality of $\bigwedge ^{n-j}V$ and $\bigwedge^j V$ is as $SO(\langle,\rangle)$-repns.

share|cite|improve this answer
I think it's starting to come into focus. The Hodge star is the composition of two isomorphisms. One $\Lambda^kV\to (\Lambda^{n-k}V)^*$ which depends on the choice of basis for $\Lambda^nV$, i.e. a volume form (which you get, for example, from orientation + inner product), and another $(\Lambda^{n-k}V)^*\to \Lambda^{n-k}V$ which comes from the inner product. So reason for the transpose is that second isomorphism, which is a raising operator. – ziggurism Mar 18 '12 at 15:23
Yes... except the duality and/or volume form doesn't even need an inner product. – paul garrett Mar 18 '12 at 15:56
Right. The first isomorphism depends only on the volume form, because it gives a basis for $\Lambda^nV$, i.e. an isomorphism to $k$. I made the parenthetical remark because a volume form is not among our given data, but orientation and inner product are, and together they give a volume form. This makes me wonder What the picture looks like if we drop the requirement for an orientation. Presumably we get analogous isomorphisms using a volume pseudoform. – ziggurism Mar 18 '12 at 16:52
Ah. Indeed, there's no necessity of literal volume. No necessity of positive-definiteness in the symmetric bilinear form, either, just non-degeneracy. – paul garrett Mar 18 '12 at 20:25
I thought I understood this, but I got myself confused again when I tried to check it. Because we have an bilinear pairing between $\Lambda^k V$ and $\Lambda^{n-k} V$, we can say that $\Lambda^{n-k} V$ carries the contragredient representation to $\Lambda^k V$. So far so good. But what is this representation? It's given by taking the transpose, with respect to the bilinear form, and here the bilinear form is the wedge product. Isn't this a different transpose operation than that arising from the inner product? Do you see what I mean? – ziggurism Mar 26 '12 at 13:18

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.