rotational invariance of cross product

Theorem
Let R be a rotational $3\times 3$ matrix, i.e., a realmatrix with $det\text{𝐑}=1$ and ${\text{𝐑}}^{-1}={\text{𝐑}}^T$.Then for all vectors $\text{𝐮},\text{𝐯}$ in ${ℝ}^3$,

$$\text{𝐑}\cdot (\text{𝐮}\times \text{𝐯})=(\text{𝐑}\cdot \text{𝐮})\times (\text{𝐑}\cdot \text{𝐯}).$$

Proof.Let us first fix some right hand oriented orthonormal basis in ${ℝ}^3$.Further, let $\{u^1,u^2,u^3\}$ and $\{v^1,v^2,v^3\}$ be the componentsof u andv in that basis. Also, in the chosen basis, we denote the entriesof R by $R_{ij}$. Since R is rotational, we have$R_{ij}{R}_{kj}={\delta }_{ik}$ where ${\delta }_{ik}$ is theKronecker delta symbol. Here we use the Einstein summation convention.Thus, in the previous expression, on the left hand side, $j$ should be summedover $1,2,3$. We shall use theLevi-Civita permutation symbol $\epsilon$ to write the cross product.Then the $i$:th coordinate of $\text{𝐮}\times \text{𝐯}$ equals${(\text{𝐮}\times \text{𝐯})}^i={\epsilon }^{ijk}{u}^j{v}^k$.For the $k$th component of $(\text{𝐑}\cdot \text{𝐮})\times (\text{𝐑}\cdot \text{𝐯})$ wethen have

${((\text{𝐑}\cdot \text{𝐮})\times (\text{𝐑}\cdot \text{𝐯}))}^k$$=$${\epsilon }^{imk}{R}_{ij}{R}_{mn}{u}^j{v}^n$

$=$${\epsilon }^{iml}{\delta }_{kl}{R}_{ij}{R}_{mn}{u}^j{v}^n$

$=$${\epsilon }^{iml}{R}_{kr}{R}_{lr}{R}_{ij}{R}_{mn}{u}^j{v}^n$

$=$${\epsilon }^{jnr}det\text{𝐑}{R}_{kr}{u}^j{v}^n.$

The last line follows since${\epsilon }^{ijk}{R}_{im}{R}_{jn}{R}_{kr}={\epsilon }^{mnr}{\epsilon }^{ijk}{R}_{i1}{R}_{j2}{R}_{k3}={\epsilon }^{mnr}det\text{𝐑}$.Since $det\text{𝐑}=1$, it follows that

${((\text{𝐑}\cdot \text{𝐮})\times (\text{𝐑}\cdot \text{𝐯}))}^k$$=$$R_{kr}{\epsilon }^{jnr}{u}^j{v}^n$

$=$$R_{kr}{(\text{𝐮}\times \text{𝐯})}^r$

$=$${(\text{𝐑}\cdot \text{𝐮}\times \text{𝐯})}^k$

as claimed. $\mathrm{\square }$