one dimensional real projective space

Question

$RP^1$ is homeomorphic to $S^1/\sim$, which is in turn homeomorphic to the closed upper semicircle with the endpoints identified.

I cannot imagine how to paste the the antipodal points to make the circle into a semicircle in the 2 dimensional space. Can anyone show me the process of the pasting?

Of course you can bend a closed interval into a circle in the plane, but there's no reason you need to do this all inside the plane. It's fine for this example, but as you progress you will not be able to do that in general. — Ted Shifrin, Aug 06 '20 at 17:39
Does this answer your question? Is the 1-dimension projective line a circle? — brainjam, Aug 06 '20 at 17:41

Vercassivelaunos · Answer 1 · 2020-08-06T17:53:38.343

2

I don't have a picture for you, so you'll have to use your imagination, but here goes: Take the circle, twist it into a figure 8. As a first step, we notice that one of the circles in the figure is the upper semicircle, while the other one is the lower semicircle. We can also notice that the waist consists of two overlapping points, the endpoints of the semicircles. These we identify with each other. Now fold the figure 8 at the waist so the two circles overlap. At every point, there are now two points above one another. Identify them with each other, and you're done.

edited Aug 06 '20 at 17:53

answered Aug 06 '20 at 17:47

Vercassivelaunos

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That's a pretty good way to visualize it too! – rschwieb Aug 06 '20 at 19:21

score 1 · Answer 2 · answered Aug 06 '20 at 17:36

The points in $1$-d projective space correspond to $1$-d subspaces of $\mathbb R^2$.

For a second, let's imagine we are instead modeling it in the unit circle in $\mathbb R^2$. Each subspace (a.k.a line through the origin) cuts the unit circle in two places. We want to associate the line with only a single thing, so we identify the two antipodal points as the same point. So now we have a mapping from the $1$-d subspaces of $\mathbb R^2$ to the equivalence classes of points on the circle (they are all paired up.)

Now, instead of that you could say that you'll just identify a line with the point it intersects the upper semicircle. However, the $x$-axis is still a subspace that hits this semicircle in two points. So you have to do something about that: identify the two endpoints, or else throw one away. Throwing one away seems a little asymmetric, so you might stick with the description with the endpoints identified.

Can anyone show me the process of the pasting?

I'm not really sure if on would consider it "pasting." You could imagine that you're curling the semicircle into a circle, attaching them at the endpoints.

Really you are just looking at an equivalence relation: $(a,b)\sim (c,d)$ if either ($a=c$ and $b=d$) or ($a=-c$ and $b=-d$). The same relation can be applied to the entire unit circle or just the closed semicircle.

Would you mind showing me the the expression of the mapping from the 1-d subspaces of R2 to the equivalence classes of points on the circle? — mathbeginner, Aug 08 '20 at 01:57
You mention that we can identify a line with the point it intersects the upper semicircle. Why we can not identify the x-axis with a point? — mathbeginner, Aug 08 '20 at 02:00
@mathbeginner That’s the rub: in either model it intersects the circle at two points, so the map would not be well defined. The x axis hits the upper semicircle in two points . you identify the points to force it to be well defined. The map is: line maps to pair of antipodal points it cuts on the unit circle. — rschwieb, Aug 08 '20 at 02:49

one dimensional real projective space

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