# Is the Hodge star an $SO(n)$-equivariant isomorphism to the dual representation?

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.

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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

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.
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. –  Joe Hannon Mar 18 '12 at 15:23
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. –  Joe Hannon Mar 18 '12 at 16:52
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? –  Joe Hannon Mar 26 '12 at 13:18