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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.

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up vote 1 down vote accepted

We can choose a basis in which $\langle\cdot,\cdot\rangle$ is described by a signature matrix of $\pm1$s. The coefficients of a vector in this basis are, up to signs, the scalar products of the vector with the basis vectors. Since $T$ preserves the scalar products, it preserves the coefficients, which means that it's linear.

[Explicit version per request:]

Let $\{b_i\}$ be a basis in which $\langle\cdot,\cdot\rangle$ is represented by a diagonal signature matrix with the first $n_+$ diagonal elements $+1$ and the remaining $n_-$ diagonal elements $-1$, with $n=n_++n_-$ the dimension of the space. Then we can write a vector $v$ as a linear combination $v=\sum_ic_ib_i$ with coefficients $c_i$. Forming the scalar product with $b_j$ yields $\langle v,b_j\rangle=\sum_ic_i\langle b_i,b_j\rangle=\sigma_j c_j$, with $\sigma_j=\pm1$ the $j$-th element of the signature. Thus $c_j=\sigma_j\langle v,b_j\rangle$.

If the images $Tb_i$ of the basis vectors were linearly dependent, the matrix of their scalar products would be singular. However, since $T$ preserves the scalar product, the matrix of their scalar products is the non-singular signature matrix; thus the $Tb_i$ are linearly independent and span the entire space. [This part of the argument was missing in the three-line version.]

Thus we can expand $Tv$ in the basis $\{Tb_i\}$, $Tv=\sum_it_iTb_i$, and because $T$ preserves scalar products, we have $\langle v,b_j\rangle=\langle Tv,Tb_j\rangle=\langle \sum_it_iTb_i,Tb_j\rangle=\sum_it_i\langle Tb_i,Tb_j\rangle=\sum_it_i\langle b_i,b_j\rangle=\sigma_jt_j$ and thus $t_j=c_j$ for all $j$. But that means that the vector $\sum_ic_ib_i$ is mapped to the vector $\sum_ic_iTb_i$, that is, $T$ linearly extends from the $b_i$ to the entire space.

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could you be more explicit? I've tried to write down your argument but I got confused. What do you mean by "T preserves the coefficients"? – Lucas Kaufmann Mar 16 '12 at 15:01
@Lucas: I hope that's clearer? (The original version didn't justify that the images of the basis vectors span the entire space.) – joriki Mar 16 '12 at 15:49
That was precisely the point I wasn't able to justify, that the image of the basis is still a basis. Thanks for clearing it out. – Lucas Kaufmann Mar 16 '12 at 22:25

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