Why does the defining action of $\mathfrak{so}(3)$ conincide with the adjoint action?

Question

The real Lie algebra $\mathfrak{so}(3)$ consists of infinitesimal rotations, i. e., skew-symmetric operators $\mathbb{R}^3\to\mathbb{R}^3$. Given an orientation on $\mathbb{R}^3$, the vector space of such operators is in canonical correspondence with $\mathbb{R}^3$ itself, by sending each infinitesimal rotation to the corresponding angular velocity vector. In a more fancy language, you can view a skew-symmetric operator as a bivector and then use Hodge star to turn it into a vector. Let's denote this correspondence by $\sigma$.

Now, we have two actions of $\mathfrak{so}(3)$ on $\mathbb{R}^3$. One is the defining action $v\mapsto Av$. And another one is the adjoint action of $\mathfrak{so}(3)$ on itself, which, thanks to $\sigma$, we can view as action on $\mathbb{R}^3$: $v\mapsto \sigma([A,\sigma^{-1}v])$. I understand that the two actions coincide; the action of $\sigma^{-1}w$ on $v$ is just $w\times v$ in both cases. And this is, of course, not very difficult to prove in coordinates.

My question is, what is the underlying reason for this coincidence? Is it possible to prove it in a coordinate-free way, ideally, without computations at all? Also, there are two inner products on $\mathbb{R}^3$, the original one and the one coming from the Killing form on $\mathfrak{so}(3)$. The same question applies, is there a coordinate-free way to see they coincide?

Answer 1

This is a write-up of the details of the answer by Qiaochu Yuan in the comments.

The group $SO(V)$ naturally acts on $V$, but also on each of the $\Lambda^k(V)$. Namely, viewing $\omega\in \Lambda^k$ as a $k$-linear form on $V^\star$, the action is given by $(O.\omega)(u_1,\dots,u_k)=\omega(O^\star u_1,\dots,O^\star u_k).$ Differentiating, this leads to the action of $\mathfrak{so}(3),$ by $$ (A.\omega)(u_1,\dots,u_k)=\omega(A^\star u_1,\dots,u_k)+\dots+\omega(u_1,\dots,A^\star u_k). $$

• For $k=1$, this is the defining action of $\mathfrak{so}(V)$.
• For $k=2$, this gives the adjoint action once we intertwine by the canonical isomorphism $\mu:\Lambda^2(V)\to \mathfrak{so}(V)$ defined by $(\mu \omega)(u,v)=\omega(u,v^\star)$, where $v\mapsto v^\star$ is the isomorphism of $V$ and $V^\star$, and we view an operator as an element of $V\otimes V^\star$. Indeed, we have, for $v\in V$ and $u\in V^\star$, \begin{multline} \mu(A.\omega)(u,v)=\omega(A^\star u;v^\star)+\omega(u,A^\star v^\star)=\omega(A^\star u;v^\star)-\omega(u,(A v)^\star). \end{multline} The last expression is just $(A\circ\mu\omega)(v)-(\mu\omega\circ A)(v)$ paired to the covector $u$.
• when $k=\dim V$, the action of $SO(V)$ preserves the volume form $\mathcal{Vol}\in \Lambda^{\dim V}(V)$. That is to say, $A.\mathcal{Vol}=0$ for any $A\in\mathfrak{so}(V)$.
• For $\dim V=3$, the last observation gives, for any $v\in V$ and $u_{1,2}\in V^\star$, $$ \mathcal{Vol}((A v)^\star,u_1,u_2)=-\mathcal{Vol}(A^\star v^\star,u_1,u_2)=\mathcal{Vol}(v^\star,A^\star u_1,u_2)+\mathcal{Vol}(v^\star, u_1,A^\star u_2). $$ The left-hand side is the two-form $\star (Av)$, evaluated on $(u_1,u_2)$, while the right-hand side is the form $A.(\star v)$, evaluated on the same vectors. This indeed shows that the Hodge star $\star$ intertwines the action of $\mathfrak{so}(3)$ on $V$ and $\Lambda^2(V)$, and hence $\sigma^{-1}=\mu\circ\star$ intertwines the defining and the adjoint action.

(Thus far, no representation theory was needed. I would need to think a bit more about the Killing form, though.)