TITLE: Rotational invariance of cross product
QUESTION [20 upvotes]: I'm looking for a proof that $$ ( Ra \times Rb ) =  R ( a \times b ) $$ where $\times$ is the three-dimensional cross product, and $R$ is a rotational matrix (such that $\det R = 1$ and $R^T  R = I$).
I've already found a proof of it on planetmath, but I'm very new to using Levi-Civita etc. For example, I don't get why $$\epsilon^{imk}  R_{ij} R_{mn} u^j v^n = \epsilon^{iml} \delta_{kl} R_{ij} R_{mn} u^j v^n$$
I'm either looking for a list of identities for the Levi-Civita Symbol which would help me understand this proof or someone who could explain the single steps to me because I'm really quite lost.

REPLY [12 votes]: $\newcommand{\vek}[1]{\boldsymbol{#1}}$
You can use $\vek{a}^T (\vek{b} \times \vek{c}) = \vek{a} \cdot (\vek{b} \times \vek{c}) =  \det (\vek{a}, \vek{b}, \vek{c})$ to prove it. Let $j$ be any component of the vector $R(\vek{a} \times \vek{b})$, then
\begin{align}
\left( R(\vek{a} \times \vek{b}) \right)_j
&= \vek{e}_j \cdot R(\vek{a} \times \vek{b})
\\ &= \vek{e}_j^T R~(\vek{a} \times \vek{b})
\\ &= (R^T \vek{e}_j)^T (\vek{a} \times \vek{b})
\\ &= \det(R^T \vek{e}_j,~ \vek{a}, ~\vek{b})
=  \color{orange}{\det R} ~\det(R^T \vek{e}_j,~ \vek{a}, ~\vek{b})
\\ &= \det(\color{blue}{RR^T} \vek{e}_j,~ R\vek{a}, ~R\vek{b}) 
= \det(\vek{e}_j,~ R\vek{a}, ~R\vek{b})
\\ &= \vek{e}_j \cdot (R\vek{a} \times R\vek{b})
\\ &= (R\vek{a} \times R\vek{b})_j
\end{align}
During the calculation we used $\color{orange}{\det R = 1}$ and $\color{blue}{RR^T = I}$, each once.