Does the following map from the circle to the projective space of $\mathbb{R}^2$ admit a continuous section?

Asked Jun 02 '20 at 18:28

Active Jun 02 '20 at 18:36

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Does the map $S^1\to P(\mathbb{R}^2)$ defined by $\vec{v} \mapsto \mathbb{R}\vec{v}$ admit a continuous section. Where:

$S^1:=\{ (x,y) \in \mathbb{R}^2 : x^2+y^2=1 \}$ denotes the circle, $P(\mathbb{R}^2)$ denotes the projective space, and a section of a map $\pi:X\to Y$ is a map $\sigma : Y \to X$ such that $\pi \circ \sigma = id_Y$.

I believe that that it doesn't admit a continuous section, but I'm not sure since I haven't been able to prove or disprove.

edited Jun 02 '20 at 18:36

Ted Shifrin

115,160

asked Jun 02 '20 at 18:28

solrak

2

Does it help to think of $\Bbb P(\Bbb R^2)$ as $S^1/{\sim}$ where $x\sim -x$? – Ted Shifrin Jun 02 '20 at 18:39
Not much, really. Already tried thinking it like that but with no success in proving or disproving. – solrak Jun 02 '20 at 23:00
1

Starting at $[(1,0)]$. the section must assign either $(1,0)$ or $(-1,0)$. When we "go around" $\Bbb RP^1$ and end up at $[(-1,0)] = [(1,0)]$ the section must have the opposite value, if it's continuous. – Ted Shifrin Jun 02 '20 at 23:06
That helps, Thanks!! – solrak Jun 03 '20 at 03:21

Does the following map from the circle to the projective space of $\mathbb{R}^2$ admit a continuous section?

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