Could a non-linear transformation be orthogonal?

Question

Could a non-linear transformation be orthogonal?

Or in other words, could the Jacobian matrix of an orthogonal transformation be dependent of the coordinates (does not contain constant values only)?

Answer 1

As Yashas Samaga commented, this should be a math.se quetsion. But since it's on physiscs, I can get away with a not completely rigorous answer.

If I understand you correctly, you are looking for a (nonlinear) map $f:V\to V$ on a vector space $V$ that preserves the inner product, i.e. $$ \left=\left\,,$$ correct?

If you additionally assume $f$ to be differentiable, then the answer is no: Consider $$\left.\frac{\text d}{\text d \epsilon}\left\right|_{\epsilon=0}$$ for a real number $\epsilon$. By the orthogonality, this is $$\left.\frac{\text d}{\text d \epsilon}\left\right|_{\epsilon=0}=\left\,,$$ so it is, in particular, independent of $v$. On the other hand, we can evaluate the derivative directly and obtain $$\left.\frac{\text d}{\text d \epsilon}\left\right|_{\epsilon=0}=\left\,,$$ which does depend on $v$, unless $f'(v)=O$ is constant, in which case $f$ is just $f(v)=O \cdot v$ for an orthogonal matrix $O$.

I guess one can extend this argument to avoid the assumption of differentiablity by considering how $$\left$$ depends on $v$, but I don't have the time now (and in physics, everything is differentiable anyway...).

Answer 2

If the transformation $f : V \to V$ with $V$ real vector space equipped with a symmetric scalar product $\langle\:,\: \rangle$ is surjective the answer is negative: $f$ is necessarily linear.

Let $x,y \in V$ be generic vectors, from $$\langle f(x),f(y)\rangle = \langle x,y\rangle \tag{1}$$ and the linearity of the left argument in the right hand side, you see that $$\langle f(ax+bz),f(y)\rangle =\langle af(x)+bf(z),f(y)\rangle$$ for every reals $a,b$. Namely $$\langle f(ax+bz) -af(x)-bf(z),f(y)\rangle =0 \:.$$ If $f$ is surjective, we can choose $y\in V$ such that $f(y)=f(ax+bz) -af(x)-bf(z)$, so that $$\langle f(ax+bz) -af(x)-bf(z), f(ax+bz) -af(x)-bf(z)\rangle= 0 \:,$$ i.e., $$||f(ax+bz) -af(x)-bf(z)||^2=0$$ As the scalar product is strictly positive, we have that, for every $x,z\in V$ and $a,b \in \mathbb R$, $$f(ax+bz) -af(x)-bf(z) =0$$ which means that $f$ is linear $$f(ax+bz) = af(x)+bf(z)\:.$$

As a final comment I stress that, in view of the polarization identity, the requirement (1) can be weakened into $$||f(x)||=||x|| \quad \forall x \in V\:,$$ which implies (1). This requirement together surjectivity of $f$ implies linearity of $f$.