TITLE: Prove $R(a \times b) = Ra \times Rb$ given $R \in \mathcal{SO}(3)$ and $a,b \in \mathbb{R}^3$.
QUESTION [6 upvotes]: Here, $R$ is a proper rotation matrix, and $(\times)$ is the cross-product.
I already found 3 ways of proving this, all have problems, and I am requesting an elegant approach.
(1) Direct calculation, $R = R_{\mathbf{u},\theta}$ has an explicit expression, where $\mathbf{u}$ is the vector about which the rotation of $\theta$ is carried out. It is essentially possible to calculate both sides and compare. Inelegant.
(2) Using antisymmetric matrices: $Ra \times Rb=S(Ra)Rb=RS(a)R^\top Rb=RS(a)b=R(a\times b)$.
 My issue with this is that the equality I am trying to prove, is used to prove $RS(a)R^\top=S(Ra)$. And so using this feels circular.
(3) $Ra \times Rb=\|Ra\|\|Rb\|\sin(Ra,Rb)\mathbb{\hat{u}_1}$ and $a \times b=\|a\|\|b\|\sin(a,b)\mathbb{\hat{u}_2}$.

Here, essentially $\|Ra\|$ should equal $\|a\|$ since $R$ only affects orientation.

Because the relative orientation does not change, the $\sin(\cdot)$ term should be equal.

Likewise, $\mathbb{\hat{u}_1}$ and $\mathbb{\hat{u}_2}$ intuitively I know are equal but I am having a hard time expressing it.

Lastly, I have no idea how to bridge $(a \times b)$ to $R(a \times b)$.

I intuitively see it, and perhaps $\det R = 1$ might be useful, but I feel it is hard to write.
Please, a fourth approach is welcome, and insight is always appreciated.

REPLY [2 votes]: Here's a component-wise proof. Note that I'm using the Einstein convention: If an index appears twice in a product, it is summed over. For example
$$(Rv)_i = R_{ij}v_j$$
actually means
$$(Rv)_i = \sum_{j=1}^3 R_{ij} v_j$$
because $j$ appears twice in $R_{ij}v_j$.
Also in case you don't know those symbols,
$$\delta_{ij} = \begin{cases} 1 & i=j\\ 0 & \text{otherwise} \end{cases}$$
and
$$\epsilon_{ijk} = \begin{cases}
  1 & ijk \in \{123,231,312\}\\
 -1 & ijk \in \{132,213,321\}\\
  0 & \text{otherwise}
\end{cases}.$$
Then we can easily check that
$$(a\times b)_i = \epsilon_{ijk}a_j b_k.$$
The equation we want to prove is
$$R(a\times b)=Ra\times Rb.$$
Now since $R$ is invertible, we can multiply both sides with $R^{-1}$ and get an equivalent equation:
$$a\times b=R^{-1}(Ra\times R_b)$$
Thus we get:
\begin{aligned}
&&(a\times b)_i &= (R^{-1}(Ra\times Rb))_i\\
\iff&&(a\times b)_i &= (R^{-1})_{ij}(Ra\times Rb)_j\\
\iff&&(a\times b)_i &= R_{ji}(Ra\times Rb)_j && \text{orthogonality of $R$}\\
\iff&&\epsilon_{ikl}a_kb_l &= R_{ji} \epsilon_{jmn}(Ra)_m(Rb)_n\\
\iff&&\epsilon_{ikl}a_kb_l &= R_{ji} \epsilon_{jmn}R_{mk}a_kR_{nl}b_l\\
\iff&&\epsilon_{ikl}a_kb_l &= (\epsilon_{jnm} R_{ji} R_{mk} R_{nl}) a_k b_l
\end{aligned}
Thus we have to prove that
$$\epsilon_{ikl} = \epsilon_{jnm} R_{ji} R_{mk} R_{nl} \tag{*}$$
Now $\epsilon$ is completely characterized by the following two properties:

$\epsilon_{123} = 1$
Exchange of two indices changes the sign.

So let's check both properties for the right hand side of the above equation (note that the free indices are $i$, $k$, and $l$):

$ \epsilon_{jmn} R_{j1} R_{m2} R_{n3} \stackrel?= 1$
The left hand side is the determinant of $R$, which by assumption is $1$, so this
equation is indeed true.
Exchange of two indices: Let's without loss of generality exchange the last two indices, $§k$ and $l$.
Clearly due to the commutativity of multiplication, we have
$$\epsilon_{jnm} R_{ji} R_{ml} R_{nk} = \epsilon_{jnm} R_{ji} R_{nk} R_{ml}$$
Now the indices of $\epsilon$ in this term are summed over, therefore we can just
rename them without changing anything. In particular, we exchange the names $m$
and $n$:
$$\epsilon_{jnm} R_{ji} R_{nk} R_{ml} = \epsilon_{jnm} R_{ji} R_{mk} R_{nl}$$
Now we exchange the last two indices of the $\epsilon$ factor:
$$\epsilon_{jnm} = -\epsilon_{jmn}$$
Putting all together, we then get
$$\epsilon_{jnm} R_{ji} R_{ml} R_{nk} = -\epsilon_{jmn} R_{ji} R_{mk} R_{nl}$$
Thus exchanging $k$ and $l$ indeed changes the sign.

Thus we see that equation $(*)$ indeed holds, and thus $R(a\times b)=Ra\times Rb$.