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I have known for a while that $\mathbb{R}P^1 \simeq S^1$. Indeed, visually it is easy to see why this is the case (any two antipodal points define a line through the origin)

Say we now forget everything we know, and let $F$ be a division ring and let $n \ge 0$. We definie an equivalence relation of $F - \{ 0 \}$ by $x \sim y$ if there exists $\lambda \in F - \{ 0 \}$ with $x = \lambda y$. The quotient set $F^{n+1}-\{ 0 \} / \sim$ is called the F-projective n-space. The class of $x = (x_0,\ldots ,x_n)$ is denoted by $[x] = [x_0, \ldots, x_n] \in FP^n$.

Based on this definition I would like to show $\mathbb{R}P^1 \simeq S^1$

So how do we go from this definition to prove the isomorphism? I am sure it is simple...but I am not really sure where to go. Do I just use the antipodal max $f(x)=-x$? (Note the next part is to extent to show $\mathbb{H}P^1 \simeq S^4$ which is a bit harder to visualise!)

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This is topology and no abstract algebra, right? –  Myself Mar 18 '11 at 3:10
    
Represent $\mathbb{R}P^1$ by a half-sphere with the two end points glued together. Well, that's it, really :) –  Alexei Averchenko Mar 18 '11 at 3:29
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3 Answers

up vote 4 down vote accepted

First, consider the mapping $$\varphi\colon (x,y) \mapsto \left(\frac{x}{\sqrt{x^2+y^2}}, \frac{y}{\sqrt{x^2+y^2}}\right)$$ which maps $\mathbb{R}^2-\{0\}$ to the unit circle $\mathcal{C}$.

If $(x,y)$ and $(z,w)$ map to the same point, then it follows that $x = \lambda z$ and $y=\lambda w$, where $$\lambda = \frac{\sqrt{x^2+y^2}}{\sqrt{z^2+w^2}}.$$

However, it is possible for $(x,y)$ and $(z,w)$ to satisfy $(x,y)=\lambda(z,w)$ and not to map to the same point: indeed, if $(x,y)=\lambda(z,w)$, then it is easy to see that $(x,y)$ maps to $$\frac{\lambda}{|\lambda|}\left(\frac{z}{\sqrt{z^2+w^2}},\frac{w}{\sqrt{z^2+w^2}}\right).$$ So we define an equivalence relation on the unit circle by $\mathbf{x}\sim\mathbf{y}$ if and only if $\mathbf{x}=\pm\mathbf{y}$, and consider $\mathcal{C}^1/\sim$.

Now we consider the composition of $\varphi$ and $\pi\colon \mathcal{C}^1\to\mathcal{C}^1/\sim$. If $(x,y) =\lambda (z,w)$, then $\pi\circ\varphi(x,y) = \pi\circ\varphi(z,w)$ by the discussion above. Conversely, if $\pi\circ\varphi(x,y) = \pi\circ\varphi(z,w)$, then either $\varphi(x,y)=\varphi(z,w)$, in which case $(x,y) = \lambda(z,w)$ with $\lambda$ as above; or else $\varphi(x,y) = -\varphi(z,w)$, hence $$(x,y) = -\frac{\sqrt{x^2+y^2}}{\sqrt{z^2+w^2}}(z,w).$$ So we conclude that $\pi\circ\varphi$ factors through $\mathbb{R}^2-\{(0,0)\}/\equiv$, where $\equiv$ is the equivalence relation on $\mathbb{R}^2=\{(0,0)\}$ given by $\mathbf{x}\equiv \mathbf{z}\Longleftrightarrow \mathbf{x}=\lambda\mathbf{z}$ for some scalar $\lambda$, and induces a bijection $$\frac{\mathbb{R}^2-\{(0,0)\}}{\equiv} \to \mathcal{C}^1/\sim \cong S^1.$$ The maps are all continuous.

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thanks (as always - I feel like a broken record) –  Juan S Mar 18 '11 at 3:36
    
not that it affects the result at all - but should $\lambda = \frac{\sqrt{x^2+y^2}}{\sqrt{z^2+w^2}}$ as written?? –  Juan S Mar 18 '11 at 3:57
    
@Qwirk: Quite right. Thanks. –  Arturo Magidin Mar 18 '11 at 4:00
    
I presume we can use an analogous argument to show that $S^n/\sim \simeq \mathbb{R}P^n$ (where $x\sim y$ if $x = \pm y$)? –  Juan S Mar 21 '11 at 4:36
    
@Qwirk: Yes, it should be doable. –  Arturo Magidin Mar 21 '11 at 8:46
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I suppose you want a homeomorphism between $\mathbb R\mathbb P^1$ and $S^1$. ($S^1$ does not have an algebraic structure, so isomorphism in that sense would not make sense.)

Define $\mathbb R\mathbb P^1\to S^1$ by $[a:b]\mapsto (a/\sqrt{a^2+b^2}, b/\sqrt{a^2+b^2})$ and $S^1\to \mathbb R\mathbb P^1$ by $(a,b)\mapsto [a:b]$. These are inverses to each other and hence bijective. They are both clearly continuous, so we're done.

ps: Why do you skip $\mathbb C\mathbb P^1\sim S^2$?

pps: the proof is the same in all cases including $\mathbb C$ and $\mathbb H$. You just have to figure out the dimension of each division algebra over $\mathbb R$, but it seems you already know that.

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($S^1$ is a real affine variety...) –  Mariano Suárez-Alvarez Mar 18 '11 at 3:25
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Just to have in mind a picture of the procedure of identifying opposite points on the circle, you can think on this: Constructiong $\mathbb{R}\mathbb{P}^1$

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