Let $V$ be a finite dimensional real vector space and $\langle\cdot,\cdot \rangle$ be a positive definite scalar product in $V$.
It is well know that if a map $T:V \to V$ preserves $\langle\cdot,\cdot \rangle$ (i.e. $\langle Tv,Tw \rangle = \langle v,w \rangle$ for all $v,w \in V$) then $T$ must be linear. One way of seing this is by computing
$||x+y-z||^2 = ||x||^2 + ||y||^2 + ||z||^2 + 2\langle x,y\rangle - 2\langle x,z\rangle - 2\langle y,z\rangle$ $||Tx+Ty-Tz||^2 = ||Tx||^2 + ||Ty||^2 + ||Tz||^2 + 2\langle Tx,Ty\rangle - 2\langle Tx,Tz\rangle - 2\langle Ty,Tz\rangle$.
Comparing both sides we get $||Tx+Ty-Tz|| = ||x+y-z||$ for all $x,y,z \in V$ and setting $z=x+y$ we obtain $T(x+y)=Tx + Ty$. By continuity its not hard to show that $T(\lambda x) = \lambda Tx$ for $\lambda \in \mathbb R, x \in V$.
Now let $g:V \times V \to \mathbb R$ be an indefinite scalar product (that is, a non-degenrate symmetric bilinear form) and $T:V \to V$ a map preserving $g$, i.e., $g (Tv,Tw) = g(v,w)$ for all $v,w \in V$.
Is it true that $T$ must be linear in this case?
The above computation no longer aplies, since vectors with zero norm need not to be zero anymore. The only thing one can say is that $T(x+y)-Tx - Ty$ is light-like for every $x,y \in V$.
I think I managed to write a proof using the fact that $T$ is $C^1$ and surjective. Does anyone know if these conditions are necessary? I wasn't able to find any counter example.