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Let $V,W$ be finite-dimensional vector spaces of a field $F$ equipped with non-degenerate symmetric bilinear forms $\langle \cdot,\cdot \rangle_V$ and $\langle \cdot, \cdot \rangle_W$. We call a map $f : V \to W$ orthogonality-preserving if $$\forall v,w\in V: \langle v,w\rangle_V = 0 \implies \langle f(v), f(w)\rangle_W = 0.$$ 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.

Comment. I opened a new question for the situation $V \to V$, where the easy counterexample in the answer below is not possible.

azimut
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  • If a mapping $f:V\to W$ is bijective then for every $w\in W$ there exists a unique $v \in V$ s.t. $w=fv$. Now assume that $\langle w, w' \rangle =0$; we can rewrite this as $\langle fv, fv' \rangle=0$. Since $f$ preserves orthogonality it follows that $\langle v, v' \rangle =0$. Since $v=f^{-1}w$ and $v'=f^{-1}w'$ we can write $$\langle w, w' \rangle =0⇒ \langle f^{-1}w,f^{-1}w'\rangle =0.$$ – Ted Black Dec 14 '24 at 22:49
  • @AnneBauval just saying "its wrong" does not make it so. – Ted Black Dec 14 '24 at 22:55
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    @TedBlack Your conclusion $\langle fv, fv'\rangle = 0 \implies \langle v, v'\rangle = 0$ is not covered by the assumption on $f$. – azimut Dec 15 '24 at 02:00
  • @AnneBauval I sharpened the assumption to $V = W$ now. – azimut Dec 15 '24 at 02:04
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    Please don't significantly change your question after it has been answered in a way that makes the answer(s) no longer relevant. See e.g. this meta discussion and the linked posts there. – anankElpis Dec 15 '24 at 03:43
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    @anankElpis That "relevant answer" should rather have been a comment, to help the asker to improve their question, by an additional assumption preventing from such uninteresting exceptions. – Anne Bauval Dec 15 '24 at 08:31
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    @anankElpis o.k., I will revert the edit and open a new question. My reasoning was that I wanted the question to be interesting, without loopholes in border cases. Therefore in my perception, it was not a "significant change", but an improvement. In any case, you could just have asked for it without downvoting immediately. Please check if you are happy now and remove the downvote. – azimut Dec 15 '24 at 09:04
  • @anankElpis The new question is at https://math.stackexchange.com/q/5011672/61691 Please check. – azimut Dec 15 '24 at 09:17
  • @AnneBauval yes I produced a circular argument. Thank God I did not post this as an answer... – Ted Black Dec 15 '24 at 12:31
  • @AnneBauval no need for this. I want to keep on record that I have posted an erroneous comment, happens to the best of us. – Ted Black Dec 15 '24 at 13:44
  • @TedBlack No need but better imo, to not distract and mislead future readers. – Anne Bauval Dec 15 '24 at 13:47

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The answer is no. Consider a bijective map $f:\mathbb R \rightarrow \mathbb R^2$ with the standard inner products on $\mathbb R, \mathbb R^2$ such that $f(0)=(0, 0)$.

Anne Bauval
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Basics
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  • Do you have in mind what the map might look like? I think there are examples using binary expansion so that $f$ need not be a space filling curve. – The Other Terry Dec 14 '24 at 22:57
  • Here's a solution for constructing a bijection from $\mathbb{R} \to \mathbb{R}^2$: https://math.stackexchange.com/q/183361/65023. Also, if $f(0)\neq (0,0)$ for this example, then simply redefine $f$: find $v$ such that $f(v)=(0,0)$, and set $f(v)=f(0)$ and $f(0)=(0,0)$. – The Other Terry Dec 14 '24 at 23:12
  • This answer should rather have been a comment, to help the asker to improve their question, by an additional assumption preventing from such uninteresting exceptions. – Anne Bauval Dec 15 '24 at 08:32
  • Another example of math stack exchange running smoothly… not. The answer does not make any sense (even grammatically) and yet it was checked! – Ted Black Dec 15 '24 at 10:01
  • @AnneBauval Us mere mortals need a bit more information in the definition of a map. So given $x \in \mathbb{R}$ what is $f(x)$? – Ted Black Dec 15 '24 at 10:45
  • @TedBlack I cannot see any grammatical mistake, and this answer provides a whole family of counterexamples: any (therefore not specified further) bijection $\Bbb R\to\Bbb R^2$ which maps $0$ to $(0,0)$. – Anne Bauval Dec 15 '24 at 10:55
  • @AnneBauval let me ask a silly question if you don’t mind: how can a map from a vector space of lower dimension to a vector space of higher dimension be surjective? – Ted Black Dec 15 '24 at 10:57
  • @TedBlack Read the post carefully ("Note that $f$ is not required to be linear"), and follow (for instance, as there are many other such posts on MSE) the link in my previous comment. See also user65023's second comment above. – Anne Bauval Dec 15 '24 at 11:00
  • Let us continue this discussion in chat. – Ted Black Dec 15 '24 at 11:02
  • @AnneBauval the link you provided is a map from a higher to a lower dimensional Euclidean space. Also, yes obviously the map can be nonlinear, so let’s see an example of a surjective nonlinear map from $\mathbb{R}$ to $\mathbb{R}^2$. That’s what could have been provided in the “answer.” – Ted Black Dec 15 '24 at 11:05
  • Let us continue in chat, as you suggested. – Anne Bauval Dec 15 '24 at 11:07
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    @TedBlack Frankly, I find your tone a bit inappropriate. To be a bit offensive as well: Just cause you don't get it, doesn't render the answer wrong. As soon as the base field $F$ is infinite, the sets $F$ and $F^2$ have the same cardinality, which precisely means that there exists a bijection $F \to F^2$. Then there exists such a bijection with $0 \mapsto 0$, too, and any such bijection will be a counterexample. – azimut Dec 15 '24 at 11:41
  • @azimut answers have to be more than one line hints. In this example the answer has to include such details as (i) one example of how a a bijection would be defined (ii) one example how orthogonality is preserved given such a definition etc. By the way, I never said the answer was wrong. I just said it was too terse to be an answer. – Ted Black Dec 15 '24 at 12:13