I am reading Do Carmo's "Differential Geometry of Curves and Surfaces", and there is an exercise in it that asks us to prove that the cross product (vector product) is invariant under orthogonal transformation.

Now I was rusty on my Linear Algebra so I had to lookup solutions after a lot of trying. I came across the following which I am struggling to understand:

Let $\rho$ be an orthogonal transformation and let $u,v \in \mathbb{R}^3$ be vectors, then we need to show $\rho(u \wedge v) = \rho u \wedge \rho v$.

\begin{equation} \begin{split} \rho u \wedge \rho v & = \rho(u_1 e_1 + u_2 e_2 + u_3 e_3) \wedge \rho(v_1 e_1 + v_2 e_2 + v_3 e_3) \\ & = (u_1 \rho e_1 + u_2 \rho e_2 + u_3 \rho e_3) \wedge (v_1 \rho e_1 + v_2 \rho e_2 + v_3 \rho e_3) \\ & = (u_1 v_2 - u_2 v_1) \rho e_1 \wedge \rho e_2 + (u_1 v_3 - u_3 v_1) \rho e_1 \wedge \rho e_3 + (u_2 v_3 - u_3 v_2) \rho e_2 \wedge \rho e_3 \\ & = \rho(u_1 v_2 - u_2 v_1) e_3 + \rho (u_1 v_3 - u_3 v_1)e_2 + \rho (u_2 v_3 - u_3 v_2) e_1 \\ & = \rho (u_2 v_3 - u_3 v_2, u_1 v_3 - u_3 v_1, u_1 v_2 - u_2 v_1) \\ & = \rho (u \wedge v) \end{split} \end{equation}

I can understand some steps and here is my understanding of those steps (corresponding to line numbers of RHS of the equations above):

1: Rewrite $u,v$ using their components.

2: Each component is transformed by $\rho$ for e.g., $\rho (u_1 e_1) = u_1 \rho e_1$ etc.

3: I can see some pattern here, but I don't know how to justify it rigorously. I tried reading orthogonal matrices and their properties, but I couldn't justify this step based on those.

4: $\rho e_1 \wedge \rho e_2 = \rho e_3$ and so on..

5: Can factor out the transformation from individual components to the whole vector.

6: Previous statement has the formula for vector product $u \wedge v$ in the parentheses.

I need help in understanding step 3. How did we get from 2 to 3?


  • $\begingroup$ The wedge product statement is correct, but it is not correct for the vector (cross) product. See this for the correct formula and derivation. $\endgroup$ Commented Jun 2, 2021 at 1:36

2 Answers 2


You expand the brackets in a similar way to multiplying normal algebra, but you need to know the following:

$e_1 \wedge e_1=0$

$e_2 \wedge e_2=0$

$e_3 \wedge e_3=0$

$e_2 \wedge e_1= -e_1 \wedge e_2$

$e_3 \wedge e_1= -e_1 \wedge e_3$

$e_3 \wedge e_2=-e_2\wedge e_3$

  • $\begingroup$ I actually tried that and I ended up with $u_1 \rho e_1 \wedge v_2 \rho e_2 - v_1 \rho e_1 \wedge u_2 \rho e_2 + u_1 \rho e_1 \wedge v_3 \rho e_3 - v_1 \rho e_1 \wedge u_3 \rho e_3 + u_2 \rho e_2 \wedge v_3 \rho e_3 - v_2 \rho e_2 \wedge u_3 \rho e_3$. So I came pretty close to step 3, but I don't know how to proceed from here. $\endgroup$
    – Sun
    Commented Jun 1, 2021 at 22:57
  • $\begingroup$ That is, $u_1 \rho e_1 \wedge v_2 \rho e_2 - v_1 \rho e_1 \wedge u_2 \rho e_2 = (u_1 v_2 - v_1 u_2) \rho e_1 \wedge \rho e_2$ and so on, but why? $\endgroup$
    – Sun
    Commented Jun 1, 2021 at 22:59

If I may.

Take $\mathbf v_1,\dots,\mathbf v_{n-1}$ in $\mathbb R^n$. Put $\mathbf y = \mathbf v_1\times\cdots\times\mathbf v_{n-1}$. For $R\in SO(n)$, \begin{align*} \det(R\mathbf v_1,\dots,R\mathbf v_{n-1},R\mathbf x) &= \det(R[\mathbf v_1,\cdots,\mathbf v_{n-1},\mathbf x])\\ &= \det R\det(\mathbf v_1,\dots,\mathbf v_{n-1},\mathbf x)\\ &= \langle\mathbf y,\mathbf x\rangle\\ &= \langle R\mathbf y, R\mathbf x\rangle. \end{align*} Now the result follows from the definition of cross product and the fact that $R$ is an epimorphism (actually, an isomorphism).

Note that I've supposed $\det R=1$. For the case $\det R=-1$ the result doesn't hold. Take for example $R\colon(x,y,z)\mapsto(y,x,z)$; then $$ R(\mathbf e_1\times\mathbf e_2) = - (R\mathbf e_1\times R\mathbf e_2). $$

  • $\begingroup$ See the link in my comment last year. In fact, you're wrong about the wedge product statement; it's only when you use the correspondence to the cross product (pseudo-)vector that the sign of the determinant is relevant. $\endgroup$ Commented Mar 16, 2022 at 0:28
  • $\begingroup$ @TedShifrin Thanks. I think you've noticed that I've used $\wedge$ instead of $\times$ for the cross product (defined as the realizer of $\langle\mathbf y,\mathbf x\rangle=\det(\mathbf v_1,\dots,\mathbf v_{n-1},\mathbf x)$). Could you tell me what's my mistake? $\endgroup$ Commented Mar 16, 2022 at 1:03
  • 2
    $\begingroup$ Well, the point of my answer that I linked is that the actual wedge product gives a fine formula with linear transformations: $T(v_1\wedge\dots\wedge v_k) = Tv_1\wedge\dots\wedge Tv_k$. The usage of $\wedge$ for $\times$ leads to this precise confusion in this precise situation. The real issue, as a physicist would explain it, is that the cross-product is actually not a vector. It's not unusual for DoCarmo to stick to this "European" notation, but it truly leads to horrid confusion. Please do me the courtesy of reading my actual post. $\endgroup$ Commented Mar 16, 2022 at 1:11
  • $\begingroup$ Ok. I need to learn more to understand how $\wedge$ is defined. I'm changing to the $\times$ notation if that was the issue. $\endgroup$ Commented Mar 16, 2022 at 1:19

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