I found these thing in an exercise 1.5.6 in the book Differential Geometry of curves and surfaces - Do Carmo.

"Show that the vector product of 2 vectors is invariant under orthogonal transformation with positive determinant. Is the assertion still true if we drop the condition on positive determinant ?"

Denote A is the orthogonal matrix in our orthogonal transformation, I can deduce that $Au\times Av) = A(u\times v)$. Can we call this invariant ? Is there any thing wrong in that exercise ?

I think the orthogonal transformation does not preserve the vector product, because $Au\times Av = \det(A)A(u\times v)$, so when $\det(A) =-1$: $Au\times Av= -A(u\times v)$. So it is never invariant ? Is that true ?

Thanks for your reply :)


I think the problem may not be clear enough when saying what it means by invariant, since when saying something is invariant we mean that the final result doesn't depend on the given transformation we us over the initial objects. For example, the dot product in $\mathbb{R}^n$ is invariant under orthogonal transformations:

$$A(u) \cdot A(v)= (Au)^T(Av) = u^TA^TAv = u^Tv = u\cdot v$$ In the case of cross product you have that $Au \times Av = detA A(u\times v)$ (your formula is incorrect, proof: http://math.ucr.edu/~res/math132/crossproducts.pdf). So the invariance only happens when $detA=1$, which geometrically means $A$ is a rotation over a given axis.

What happens then if $detA=-1$?


Let $x,y,z\in \mathbb{R}^3$ be arbitrary vectors and $A$ an orthogonal matrix. Since $detA\neq 0$ the transformation is onto so there exist a $z'\in \mathbb{R}^3$ such that $z=Az'$. Then, using the notation used in the link:

$$\langle Ax\times Ay, z \rangle = \langle Ax\times Ay, Az' \rangle = [Ax\,\,Ay\,\,Az'] = detA[x\,\,y\,\,z']=detA\langle x\times y, z' \rangle = detA \langle A(x\times y), Az' \rangle = detA\langle A(x\times y), z \rangle .$$

  • $\begingroup$ I have read that proof, but I think there is something wrong in that article. In the 4th line bottom up, how can they deduce $\langle Ax \times Ay, z \rangle = det(A)\langle A(x\times y),z \rangle = det(A)\langle x\times y,z \rangle $. Where is the term $A$ ? $\endgroup$ – genov4 Sep 21 '14 at 3:21
  • $\begingroup$ The proof looks right to me. You're missing the part where they they defined $Az' = z$. $\endgroup$ – hjhjhj57 Sep 21 '14 at 7:22
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    $\begingroup$ Please check it again carefully, there's something wrong in this $\langle Ax\times Ay,z\rangle = det(A)\langle x\times y,z'\rangle = det(A)\langle A(x\times y),Az' \rangle = det(A)\langle A(x\times y),z\rangle$. Then they deduce that $\langle Ax\times Ay,z\rangle = det(A)\langle x\times y,z\rangle$. Where is the term $A$ in $ \langle A(x\times) y,z\rangle$? $\endgroup$ – genov4 Sep 22 '14 at 11:55
  • $\begingroup$ I'll check it as soon as I have some spare time. $\endgroup$ – hjhjhj57 Sep 22 '14 at 13:47
  • $\begingroup$ You're right, the conclusion of the article is wrong, even though the rest of it is correct. I've corrected de post and added the important equalities for the proof. $\endgroup$ – hjhjhj57 Sep 22 '14 at 22:53

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