Is the inverse of an orthogonality-preserving bijection $V \to V$ orthogonality-preserving?

Question

Let $V$ be a finite-dimensional vector space over a field $F$ equipped with non-degenerate symmetric bilinear form $\langle \cdot,\cdot \rangle$. We call a map $f : V \to V$ orthogonality-preserving if $$\forall v,w\in V: \langle v,w\rangle = 0 \implies \langle f(v), f(w)\rangle = 0.$$ Note that $f$ is not required to be linear.

I wonder if for a bijective orthogonality-preserving map also the direction $\Leftarrow$ is true.

For general orthogonality-preserving maps this property is clearly wrong, a counterexample is given by the all-zero function. But for bijective maps, I could not find a proof neither a counterexample so far.

Remarks

• In this setting, any field $F$ and any non-degenerate bilinear form on $V$ are allowed. In particular, there may be isotropic non-zero vectors $v$, meaning that that $\langle v,v\rangle = 0$. As an example, if $F$ is of characteristic $p \neq 0$, then the vector $(1,1,\ldots,1)\in F^p$ is isotropic with respect to the standard scalar product $\langle x,y\rangle = x_1 y_1 + \ldots + x_n y_n$.

• In this question, an affirmative answer is given for linear maps and the standard scalar product over $F = \mathbb{R}$. The proof is not directly adoptable to the linear case over a general field $F$ because of the presence of isotropic vectors.

• One might also think about the underlying statement of this and this question: Is any (not necessarily linear) orthogonality-preserving map the scalar multiple of a (not necessarily linear) map preserving the scalar product (i.e. $\langle v, w\rangle = \langle f(v), f(w)\rangle$ for all $v,w\in V$, which is stronger than being orthogonality-preserving)? If so, this would answer my question with "yes".

• For the standard scalar product over $F = \mathbb{R}$, any surjective scalar-product-preserving map $f : V \to V$ is linear (even in the infinite dimensional case). This is discussed in the context of this question.

• I had asked the question for maps $V \to W$ before, not being aware that this allows easy counterexamples. First I had modified that question, but it was not well received. Hence I followed the suggestion to open a new question.

Answer 1

Here is a proof that when $f$ is linear it must be "almost orthogonal", thus the converse of orthogality-preservation is true.

Let $\bar f$ be the adjoint, i.e. for all $v,w$ $$ \langle\bar f(v),w\rangle = \langle v,f(w)\rangle. $$ The adjoint is well-defined because of nondegeneracy. Also because of nondegeneracy, we can rephrase orthogonality-preservation as $$ \forall\omega\in V^*.\forall v\in V.\:\omega(v) = 0\implies\omega(\bar f(f(v))) = 0, $$ or in other words $$ \forall\omega\in V^*.\:\bar f(f(\ker\omega)) \subseteq \ker\omega. $$ But all such $\ker\omega$ are the hyperplanes of $V$ containing $0$, so $\bar f\circ f$ must preserve all lines, i.e. $$ \forall v\in V.\exists\alpha\in F.\:\bar f(f(v)) = \alpha v $$ and the only way for this to happen is if $\bar f\circ f$ is a multiple of the identity, i.e. $\alpha$ is constant. Clearly $\alpha\ne0$. It is interesting to note that if $\alpha = \beta^2$ then this also proves that $f/\beta$ is orthogonal.

We now have $$ \langle f(v),f(w)\rangle = 0 \implies \alpha\langle v,w\rangle = 0 \implies \langle v,w\rangle = 0. $$