Let $V$ be a finite dimensional real vector space, and $g$ a nondegenerate symmetric positive definite bilinear form (i.e. scalar product) on it. Let's take a $f: V\to V$ map. According to the Mazur-Ulam theorem, if $f(0)=0$ and $g(fu,fv)=g(u,v)$ for all $u,v\in V$, then $f$ is linear.

Now let's take instead of $g$ a $\omega$ skew-symmetric nondegenerate bilinear form (i.e. symplectic form). Is it true, that if $f(0)=0$ and $\omega(fu,fv)=\omega(u,v)$ for all $u,v\in V$, then $f$ is linear?


1 Answer 1


Let $B : V \times V \to V$ be any non-degenerate bilinear form on a real vector space $V$. We claim that if $f : V \to V$ preserves $B$ (in the sense that $B(fu, fv) = B(u,v)$ for all $u,v \in V$) and satisfies $f(0)=0$, then $f$ is linear (in fact, a linear isomorphism).

Let us first establish a weak form of linearity. Let $u,v,w \in V$ and $\lambda, \mu \in \mathbb{R}$, then we compute

$$ \begin{align}B(fu, \lambda fv + \mu fw) &= \lambda B(fu,fv) + \mu B(fu, fw) \\ &= \lambda B(u,v) + \mu B(u, w) = B(u, \lambda v + \mu w) \\ &= B(fu, f(\lambda v + \mu w)) .\end{align}$$

If we knew $f$ to be surjective, the linearity of $f$ would then follow from the non-degeneracy of $B$, since $fu$ could be any element in $V$. However, we do not here really need $fu$ to be chosen arbitrary in $V$, but only in a basis of $V$. Indeed, if $span\{v_1, \dots, v_n\} = V$, then the non-degeneracy of $B$ implies that $B(u,v_i) = 0$ for all $1 \le i \le n$ if and only if $u = 0$.

So we need at last to prove that $\mathrm{Im} f$ do contain a basis for $V$. This follows from the existence of "standard" bases on $V$ with respect to $B$ (If $B$ were a scalar product, "standard" could mean "orthonormal" ; If $B$ were a symplectic form, "standard" could mean "symplectic".) Indeed, since $f$ preserves $B$, it transforms a "standard" basis into another one.

Note that the Mazur-Ulam theorem has a stronger form than the one you stated : If $(V, \| \cdot \|)$ is a normed real vector space and if $f : V \to V$ is a norm-preserving map such that $f(0)=0$, then it is a linear map. In this more general context, one does not have the bilinear structure of a scalar product from which the linearity of $f$ might be inferred.


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