Orthogonal transformation and vector product 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 :)
 A: 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 $detA=-1$?
Edit:
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:
$$\langle Ax\times Ay, z \rangle = \langle Ax\times Ay, Az' \rangle = [Ax\,\,Ay\,\,Az'] = detA[x\,\,y\,\,z']=detA\langle x\times y, z' \rangle = detA \langle A(x\times y), Az' \rangle = detA\langle A(x\times y), z \rangle .$$
