Does SO(3) preserve the cross product? Let $g\in SO(3)$ and $p,q\in\mathbb{R}^3$. I wondered whether it is true that $$g(p\times q)=gp\times gq$$
I am not sure how to prove this. I guess I will use at some point that the last row $g_3$ of $g$ can be obtained by $g_3=g_1\times g_2$.
But I assume there is an easier proof than writing everything out.
 A: Since both expressions $g(p\times q)$ and $g(p)\times g(q)$ are linear in each of the variables $p,q$, it suffices to check the equality for the nine cases $p,q=e_1,e_2,e_3$, i.e. when $p,q$ are basis vectors. This method is quite often employed when dealing with equality of multilinear functions.
A: Let $v,w\in\mathbb{R}^3$. Suppose that $v=(x_1,x_2,x_3)$, that $w=(y_1,y_2,y_3)$ and that $u=v\times w=(z_1,z_2,z_3)$. Then $v\times w$ is the only vector $u\in\mathbb{R}^3$ such that


*

*$u$ is orthogonal to both $v$ and $w$;

*$\|u\|=\|v\|.\|w\|.\sin\theta$, where $\theta$ is the angle between $u$ and $v$;

*$\begin{vmatrix}x_1&y_1&z_1\\x_2&y_2&z_2\\x_3&y_3&z_3\end{vmatrix}\geqslant0$.


Let $g\in SO(3,\mathbb{R})$. Then


*

*$g(u)$ is orthogonal to both $g(v)$ and $g(w)$ (because $g\in SO(3,\mathbb{R})$);

*$\bigl\|g(u)\bigr\|=\bigl\|g(v)\bigr\|.\bigl\|g(w)\bigr\|.\sin\theta$ (again, because $g\in SO(3,\mathbb{R})$);

*if $g(v)=(x_1',x_2',x_3')$, $g(w)=(y_1',y_2',y_3')$, and $g(u)=(z_1',z_2',z_3')$, then$$\begin{vmatrix}x_1'&y_1'&z_1'\\x_2'&y_2'&z_2'\\x_3'&y_3'&z_3'\end{vmatrix}=\det g\begin{vmatrix}x_1&y_1&z_1\\x_2&y_2&z_2\\x_3&y_3&z_3\end{vmatrix}=\begin{vmatrix}x_1&y_1&z_1\\x_2&y_2&z_2\\x_3&y_3&z_3\end{vmatrix}\geqslant0.$$


Therefore, $g(u)=g(v)\times g(w)$.
A: You may use the scalar triple product formula $r \cdot (p\times q)=\det(r,p,q)$ to prove that
$$
gr \cdot (gp\times gq)=gr \cdot g(p\times q)\tag{1}
$$
($=\det(r,p,q)$) for any vector $r$. Since $g$ is invertible, if $(1)$ holds for every vector $r$, we must have $gp\times gq=g(p\times q)$.
A: If one introduces the cross product in a proper abstract way, this property will be immediate from the definition. First consider any $n$-dimensional real vector space, then each basis has an associated volume form ($n$-linear alternating form), obtained by taking the determinant after expressing the $n$ argument vectors in coordinates with respect to the basis. If the vector space is equipped with an inner product so that it becomes an Euclidean vector space$~E$, then the volume forms of the orthonormal bases for the inner product take two opposite values; choosing one of these volume forms defines the structure of an oriented Euclidean vector space. The automorphism group $O(E)$ of $E$ has a subgroup $SO(E)$ of index$~2$ that fixes the volume forms, and is the automorphism group of the oriented Euclidean vector space.
Now fixing $n-1$ vectors $p_1,\ldots,p_{n-1}$ in an $n$-dimensional oriented Euclidean vector space, there is a linear form $f:v\mapsto\operatorname{Vol}(p_1,\ldots,p_{n-1},v)$, and therefore a unique vector $w$ such that $f(v)=w\cdot v$; by definition, this $w$ is the external product of $p_1,\ldots,p_{n-1}$. For $n=3$, this gives the cross product in an oriented $3$-dimensional Euclidean vector space. Since only the oriented Euclidean structure used in the definition, the external product is automatically compatible with the action of$~SO(E)$; since the (choice of) volume form was actually used in the definition, it is not compatible with the action of$~O(E)$.
A: Two ways of seeing it more directly:


*

*The cross product of two vectors can be expressed in terms of the norms of the vectors and the angle between them, and those properties are preserved by rotations. (Update: As mephistolotl mentioned in the comments, you also need the fact that rotations preserve the orientation. Otherwise, you'll get a  vector of correct length, but different direction.)

*Numerically, you would need to show $\epsilon_{ijk} O_{jl} O_{km} =O_{ip}\epsilon_{plm}$, which follows from orthogonality and the fact that $\det O=1$.

