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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) $

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  • $\begingroup$ What is orthogonality in $\mathbb{CP}^n$? $\endgroup$
    – abx
    Commented Aug 16, 2017 at 10:16
  • $\begingroup$ @abx Orthogonality of lines with respect to the standard inner product of $\mathbb{C}^{n+1}$. $\endgroup$
    – Ali Taghavi
    Commented Aug 16, 2017 at 10:17

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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.

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  • $\begingroup$ What is $H_x$? is it a line?How do you choose it, continuously? $\endgroup$
    – Ali Taghavi
    Commented Aug 16, 2017 at 11:35
  • $\begingroup$ It is the orthogonal hyperplane to $x$. I fix an $x$, no need to choose $H_x$ continuously. $\endgroup$
    – Tsemo Aristide
    Commented Aug 16, 2017 at 12:05
  • $\begingroup$ So what does it mean $f([H_x])$? f is defined on the projective space not on grassmanian. $\endgroup$
    – Ali Taghavi
    Commented Aug 16, 2017 at 12:08
  • $\begingroup$ It is the image of $f$ restricted to the projective hyperplane $[H_x]$. $\endgroup$
    – Tsemo Aristide
    Commented Aug 16, 2017 at 12:11
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    $\begingroup$ Or more directly: just choose $n+1$ mutually orthogonal lines in $\mathbb{C}^{n+1}$, and apply $f$ to obtain $n+1$ mutually orthogonal lines in $\mathbb{C}^{m+1}$, which is impossible. $\endgroup$
    – Neil Strickland
    Commented Aug 16, 2017 at 14:42

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