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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.2012-03-18

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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.

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    @paulgarrett Great! Thanks! I feel stupid now but better stupid than wrong I guess.2016-04-29