Coordinate free way to construct inner product on exterior powers

Question

Let $V$ be a finite-dimensional vector space over $\mathbb{R}$ equipped with an inner product $\omega(-,-)$. One standard fact is that there is an induced inner product on $\wedge^k V$. For instance, this shows up when you're setting up Hodge theory.

All the constructions I've seen in books construct the inner product on $\wedge^k V$ by first choosing an orthonormal basis $\{e_1,\ldots,e_n\}$ for $(V,\omega)$, and then declaring that the basis $e_{i_1} \wedge \cdots \wedge e_{i_k}$ for $\wedge^k V$ where the $i_j$ range over increasing sequences $1 \leq i_1 < \cdots < i_k \leq n$ is an orthonormal basis for $\wedge^k V$.

This strikes me as pretty unnatural; in particular, you then have to do a calculation to prove that $O(V,\omega)$ acts on $\wedge^k V$ by orthogonal transformations.

Does anyone know a good coordinate-free way to do this? I know a partial solution. Namely, you can view $\omega(-,-)$ as giving an isomorphism $\iota\colon V \rightarrow V^{\ast}$, and we then get an isomorphism $$\wedge^k V \stackrel{\wedge^k \iota}{\longrightarrow} \wedge^k V^{\ast} \cong \left(\wedge^k V\right)^{\ast}.$$ This gives a nondegenerate bilinear form on $\wedge^k V$. However, while it is easy to see that this bilinear form is symmetric, it is not obvious that it is positive-definite (and the calculation you have to do for this is no easier than what I'm trying to avoid!).

Another feature I'd like from a construction is uniqueness: namely, the inner product on $\wedge^k V$ should (up to rescaling) be the unique inner product such that $O(V,\omega)$ acts by orthogonal transformations on $\wedge^k V$.

Answer 1

I'm going to answer my own question, summarizing the comments and adding a little more from my own reflections (marked community wiki, though it doesn't matter since this is not a registered account and I can't earn reputation from it).

The first observation (from Tom Goodwillie) is that it is actually very easy to see that the inner product I wrote down is positive definite. Indeed, it is almost immediate that if $e_1,\ldots,e_n$ is an orthonormal basis for $V$, then the $e_{i_1} \wedge \cdots \wedge e_{i_k}$ form an orthonormal basis for $\wedge^k V$.

The second observation builds on what Igor Khavkine pointed out. Namely, it is obvious from pure thought that the inner product I wrote down is either positive definite or negative definite. This is actually a general phenomena, as follows.

Theorem: Let $G$ be a compact Lie group and let $W$ be a finite-dimensional irreducible real representative of $G$. Then $W$ has a $G$-invariant inner product $\omega(-,-)$, and if $\omega'(-,-)$ is any $G$-invariant symmetric bilinear form on $W$ then there exists some $r \in \mathbb{R}$ such that $\omega'(-,-) = r \cdot \omega(-,-)$. In particular, if $\omega'(-,-)$ is nonzero it is either positive definite or negative definite.

The existence of $\omega(-,-)$ follows from the usual averaging argument (this is where we use the fact that $G$ is compact, which is not used in the rest of the proof). As for $\omega'(-,-)$, since $\omega(-,-)$ is nondegenerate there exists a linear map $f\colon V \rightarrow V$ such that $\omega'(x,y) = \omega(x,f(y))$ for all $x, y \in V$. For $g \in G$, we have $$\omega'(x,y) = \omega'(gx,gy) = \omega(gx,f(gy)) \quad \text{and} \quad \omega'(x,y) = \omega(x,f(y)) = \omega(gx, g f(y))$$ for all $x,y \in V$. From this and the nondegeneracy of $\omega(-,-)$, we see that $f(gy) = g f(y)$ for all $y \in V$. Since $V$ is irreducible, Schur's Lemma implies that there exists some $r \in \mathbb{R}$ with $f(y) = r y$ for all $y \in V$. In particular, $$\omega'(x,y) = \omega(x,f(y)) = \omega(x,r y) = r \cdot \omega(x,y),$$ as desired.