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.

José Carlos Santos
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5 Answers5


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 $gp\times gq=g(p\times q)$.

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    I know that this proof should fail if $g \in O(3)$, but it's not clear to me where it does. – Michael Seifert Nov 25 '17 at 03:15
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    @MichaelSeifert If $g\in O(3)$, you have $\det(gr,gp,gq)=\det(g)\det(r,p,q)$. Therefore the scalar triple product is only preserved if $\det(g)=1$. Otherwise you flip a sign in the determinant and the cross product. – Joonas Ilmavirta Nov 25 '17 at 05:00

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 $\det O=1$.
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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.

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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

  1. $u$ is orthogonal to both $v$ and $w$;
  2. $\|u\|=\|v\|.\|w\|.\sin\theta$, where $\theta$ is the angle between $u$ and $v$;
  3. $\begin{vmatrix}x_1&y_1&z_1\\x_2&y_2&z_2\\x_3&y_3&z_3\end{vmatrix}\geqslant0$.

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

  1. $g(u)$ is orthogonal to both $g(v)$ and $g(w)$ (because $g\in SO(3,\mathbb{R})$);
  2. $\bigl\|g(u)\bigr\|=\bigl\|g(v)\bigr\|.\bigl\|g(w)\bigr\|.\sin\theta$ (again, because $g\in SO(3,\mathbb{R})$);
  3. 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, $g(u)=g(v)\times g(w)$.

José Carlos Santos
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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$~O(E)$.

Marc van Leeuwen
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