Show that the action of $\mathfrak{so}(3)$ on $\mathbb R^3$, $M_a(x)=a\wedge x$, satisfies $M_{ga}=g M_a g^{-1}$

Question

In the notes on the Lie groups $SU(2)$ and $SO(3)$ found here, the authors state (page 75, before Proposition 2.2) the following.

Consider the $\mathbb R$-linear map $\mathbb R^3\ni a\mapsto M_a\in\mathfrak{so}(3)$ defined by $e_k\mapsto \eta_k$, where $e_k\in\mathbb R^3$ are the elements of the canonical basis of $\mathbb R^3$, $\eta_k\in\mathfrak{so}(3)$ are the corresponding generators: $\exp(i\eta_k)=\mathrm{Rot}(e_k,t)$, and $\mathrm{Rot}(a,t)$ denotes the rotation by angle $t$ around the axis $a\in\mathbb R^3$.

We can then write $M_a(x)=a\wedge x$. This is easy to see, for example writing $\eta_i=\varepsilon_{ijk} (e_j e_k^* - e_k e_j^*)$. Then, they say, given $g\in SO(3)$ and $a,x\in\mathbb R^3$, we have $$M_{ga}(x) = (ga)\wedge x = g(a\wedge (g^{-1}x)) = (g M_ag^{-1})(x). \tag1$$ What's a good way to show (1)?

By direct computation, using $x\wedge y=\sum_{ijk}\varepsilon_{ijk}e_i x_j y_k$, I get $$[(ga)\wedge x]_i = \sum_{jk\ell}\varepsilon_{ijk} g_{j\ell}a_\ell x_k, \\ [g(a\wedge (g^{-1}x))]_i = \sum g_{ij}\varepsilon_{jk\ell} a_k g^{-1}_{\ell m}x_m. $$ Because $g\in SO(3)$ I know that $g^{-1}_{\ell m}=g_{m\ell}$, but even with this I'm not sure how to simplify the expressions to show that they are equal.

I could probably just get the identity from geometrical intuition, observing that it must be that $g(x\wedge y)=(gx)\wedge (gy)$, but I wanted to show it explicitly algebraically here.

Answer 1

This would be one way of proving the statement: Fixing some $r$ and $s$, the expression $$\sum_{i,j,k} \epsilon_{ijk} \cdot g_{ir}g_{js}$$ is the vector product between the $r$-th and the $s$-th column of the matrix $g$. Now, since $g \in SO(3)$, the columns of $g$ form a positively oriented orthonormal basis of $\mathbb{R}^3$, or, in other words, the vector product of two columns gives, up to a sign, the third column, i.e. if we denote the columns of $g$ by $g_1,g_2,g_3$

$$g_1 \times g_2 = - g_2 \times g_1 = g_3, \quad g_2 \times g_3 = - g_3 \times g_2 = g_1, \dots$$ and so on. Using the $\epsilon$-tensor, we can summarize this as

$$\sum_{i,j,k} \epsilon_{ijk} \cdot g_{ir}g_{js} = [g_r \times g_s]_i = \left[\sum_k \epsilon_{rsk} \cdot g_{k}\right]_i = \sum_k \epsilon_{rsk} \cdot g_{ik}. $$

Now, using the identity that you mentioned, and playing around with the indices (recall for example that $\epsilon_{ijk} = \epsilon_{kij} = \epsilon_{jki}$), this will give you the identity you want.

Answer 2

I want to show that, for any $R\in SO(3)$ and $a,b\in\mathbb R^3$, we have $R(a\wedge b) = (Ra)\wedge(Rb)$.

The condition $R\in SO(3)$ is equivalent to the rows of of $R$ being orthonormal and $\det(R)=1$. In terms of the Levi-Civita antisymmetric tensor, this condition reads $$\sum_{ijk}\varepsilon_{ijk}R_{1i}R_{2j}R_{3k}=1.$$ This can be interpreted as the dot product between $(R_{1i})_i$ and $(\sum_{jk}\varepsilon_{ijk}R_{2j}R_{3k})_i$, which can only equal $1$ if the vectors are equal. More generally, we have $R_{i\ell} = \sum_{mn}\varepsilon_{\ell mn}R_{jm}R_{kn}$ where $\varepsilon_{ijk}=1$. We can also write this as $$R_{i\ell} = \frac12 \sum_{jk mn}\varepsilon_{ijk}\varepsilon_{\ell mn} R_{jm}R_{kn}.$$ We therefore have $$[R(a\wedge b)]_i = \sum_{\ell mn}R_{i\ell}\varepsilon_{\ell mn} a_m b_n = \sum_{\ell mnpq} (\varepsilon_{\ell pq}R_{jp}R_{kq}) \varepsilon_{\ell mn} a_m b_n.$$ Using $$\sum_\ell \varepsilon_{\ell mn}\varepsilon_{\ell pq} = \delta_{mp}\delta_{nq} - \delta_{mq}\delta_{np}$$ we then get $$[R(a\wedge b)]_i = \sum_{pq}(R_{jp}R_{kq}a_p b_q-R_{jp}R_{kq}a_q b_b) \\ = (R_j a) (R_k b) - (R_k a) (R_j b) = [(Ra)\wedge (Rb)]_i.$$

We can also perform this calculation in more concise notation writing $a\wedge b=\varepsilon(a\otimes b)$, where we are thinking here of $\varepsilon$ as a linear operator $\mathbb R^3\otimes\mathbb R^3\to \mathbb R^3$. Similarly, $R_i= \varepsilon(R_j\otimes R_k)$, and thus $$[R(a\wedge b)]_i = (\varepsilon(R_j\otimes R_k)) \cdot (\varepsilon(a\otimes b)) = [(Ra)\wedge (Rb)]_i,$$ where we used $\mathrm{Tr}_1[\varepsilon\varepsilon^*] = \mathrm{Id}-\mathrm{Swap}$.