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Edit: According to the comment of L. Spice I changed the inclusion sign to the subset sign.

Is there a continuous map $f:\mathbb{C}P^3 \to \textrm{Gr}_{\mathbb{C}}(2,4)$ with $x\subset f(x)$? What about a map $g$ in the opposite direction with $g(x)\subset x$? What about a holomorphic version ($f$ or $g$ holomorphic)? What about a generalization about such maps between arbitrary grassmannian spaces?

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    $\begingroup$ Shouldn't both '$\in$' be '$\subseteq$'? $\endgroup$ – LSpice Aug 24 at 1:19
  • $\begingroup$ @LSpice yes thanks I revise it. $\endgroup$ – Ali Taghavi Aug 24 at 2:05
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    $\begingroup$ Note that giving a map $f$ as in the question is equivalent to giving a self-map of $P^3\mathbb C$ without fixed points. (I don't know off the top of my head if such a thing exists, but I'll bet someone does.) $\endgroup$ – LSpice Aug 24 at 2:59
  • $\begingroup$ @LSpice the only complex projective space with fixed point property are $\mathbb{C}P^{2k}$. Please read the revise history of this question. I had changed $\mathbb{C}P^2$ to CP^3, based on the same reason you mentioned. $\endgroup$ – Ali Taghavi Aug 24 at 3:07
  • $\begingroup$ @LSpice I guess fixed point free maps in odd dimension is constructed linearly with a combination of 90 degree rotation and complex conjugation. But for quaternioun all projective space have FPP (both odd and even). This is proved in Hatcher book. $\endgroup$ – Ali Taghavi Aug 24 at 3:15
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I think there is no holomorphic such map. Consider the incidence variety $Z=\{(p,\ell)\in \mathbb{P}^3\times \mathbb{G}(2,4)\,|\, x\in\ell\} $. The projection $p:Z\rightarrow \mathbb{P}^{3}$ is a $\mathbb{P}^2$-bundle, in fact it is the projective tangent bundle to $\mathbb{P}^3$. You are asking for a section of this bundle; that would give a line bundle $M$ on $\mathbb{P}^3$ which is a subbundle of the tangent bundle $T_{\mathbb{P}^3}$. Computing $c_3$ one sees that this line bundle must be $\mathcal{O}_{\mathbb{P}^3}(2)$; but $H^0(T_{\mathbb{P}^3}(-2))$ is zero, so $\mathcal{O}_{\mathbb{P}^3}(2)$ does not inject into $T_{\mathbb{P}^3}$.

I do not know if there exists a continuous section (contrary to what I wrote before editing).

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  • $\begingroup$ @AliTaghavi seems to have shown how to correct my (initially (holomorphic, hence) wrong) answer to a continuous section. $\endgroup$ – LSpice Aug 24 at 15:37
  • $\begingroup$ Thank you very much for your answer. $\endgroup$ – Ali Taghavi Aug 25 at 15:21
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The map $f \mathrel: \ell \mapsto \ell \oplus \ell'$, where $\ell' = \mathbb C\cdot\overline{(b, -a, d, -c)}$ when $\ell = \mathbb C\cdot(a, b, c, d)$, satisfies your first condition.

(I originally had a version without the complex conjugation, which doesn't work because $\ell' = \ell$ when $\ell = \mathbb C\cdot(1, i, 1, i)$. Fortunately @AliTaghavi pointed out how to fix it. The candidate without the conjugation would have been holomorphic, hence contradicted @abx's answer.)

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  • $\begingroup$ Very interesting!what about the opposit direction? $\endgroup$ – Ali Taghavi Aug 24 at 3:20
  • $\begingroup$ @AliTaghavi, this answer is wrong (sorry!). Please un-accept it so that I can delete it. $\endgroup$ – LSpice Aug 24 at 3:20
  • $\begingroup$ I do not see why it is wrong. Every fixed point free map $f$ works. you can revise it. i think we need conjugation too $\bar{b}, -\bar{a},\bar{d}, -\bar{c}$. Yes?any way your effort really helped me. Am I mistaken to think that your $f$ can be revised?any way should I un-accept(after all)? $\endgroup$ – Ali Taghavi Aug 24 at 4:47
  • $\begingroup$ I unaccept the answer since you asked me so. but i still believe it can be revised . $\endgroup$ – Ali Taghavi Aug 24 at 5:00
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    $\begingroup$ I agree with your correction; thanks. Actually it's a good thing, since the version without conjugation would have been holomorphic, hence contradicted @abx's nice answer. $\endgroup$ – LSpice Aug 24 at 13:56

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