
> 
> Is there a continuous map $f:\mathbb{C}P^n \to \mathbb{C}P^m$, for some $n>m$
>  which preserve orthogonality?Namely $x\perp y \implies f(x) \perp f(y) $?
> 
> 
> If yes, are there two non homotopic maps with this property?
> 
> 
> 


For a related post see this question


[Continuous maps $f:S^n \to \mathbb{C}P^m$ with $f(x)\perp f(-x) $](https://mathoverflow.net/questions/278828/continuous-maps-fsn-to-mathbbcpm-with-fx-perp-f-x)



Continuous orthogonal preserving maps between projective space
Suppose such a map exists, let $x\in\mathbb{C}^n-\{0\}$, we denote by $[x]$ its class in $\mathbb{C}P^n$. Let $H\_x$ be the orthogonal of $x$, $f([H\_x])$ is orthogonal to $f(x)$, so $f$ induces a map $f\_1:\mathbb{C}P^{n-1}\rightarrow \mathbb{C}P^{m-1}$ which has the same property, recursively, you obtain a map $f\_m:\mathbb{C}P^{n-m}\rightarrow \mathbb{C}P^0$ with the same property, and this is impossible.


