# Maps between grassmannians with inclusion property

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?

• Shouldn't both '$\in$' be '$\subseteq$'? – LSpice Aug 24 at 1:19
• @LSpice yes thanks I revise it. – Ali Taghavi Aug 24 at 2:05
• 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.) – LSpice Aug 24 at 2:59
• @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. – Ali Taghavi Aug 24 at 3:07
• @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. – Ali Taghavi Aug 24 at 3:15

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).

• @AliTaghavi seems to have shown how to correct my (initially (holomorphic, hence) wrong) answer to a continuous section. – LSpice Aug 24 at 15:37
• Thank you very much for your answer. – Ali Taghavi Aug 25 at 15:21

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.)

• Very interesting!what about the opposit direction? – Ali Taghavi Aug 24 at 3:20
• @AliTaghavi, this answer is wrong (sorry!). Please un-accept it so that I can delete it. – LSpice Aug 24 at 3:20
• 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)? – Ali Taghavi Aug 24 at 4:47
• I unaccept the answer since you asked me so. but i still believe it can be revised . – Ali Taghavi Aug 24 at 5:00
• 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. – LSpice Aug 24 at 13:56