« Home

RP1\R P^1 is diffeomorphic to S1S^1 [math-0015]

Construction

  1. Ua=S1−(0,1)U_a = S^1 - (0,1) and φa(u,v)=u1−v\varphi_a(u,v) = \frac{u}{1-v}
  2. Ub=S1−(0,−1)U_b = S^1 - (0,-1) and φb(u,v)=u1+v\varphi_b(u,v) = \frac{u}{1+v}

and RP1\R P^1 use

  1. U1={[x:y]∣x≠0}U_1 = \{[x:y] \mid x\ne0\} and φ1([x:y])=yx\varphi_1([x:y]) = \frac{y}{x}
  2. U2={[x:y]∣y≠0}U_2 = \{[x:y] \mid y\ne0\} and φ1([x:y])=xy\varphi_1([x:y]) = \frac{x}{y}

The diffeomorphism ψ:RP1→S1\psi : \R P^1 \to S^1 defined as

ψ(p=[x:y])={φa−1∘φ1 if p∈U1φb−1∘φ2 if p∈U2\psi(p=[x:y]) = \begin{cases} \varphi_a^{-1} \circ \varphi_1 \text{ if } p\in U_1 \\ \varphi_b^{-1} \circ \varphi_2 \text{ if } p\in U_2 \end{cases}

Given [x:y][x:y] we have a ratio k=y/xk = y/x, we want to recover a point (u,v)(u,v) on S1S^1, how to get this inverse map? The idea is

Broken pipe (os error 32)

Now we can use quadratic formula to solve vv (and remind that v≠1v \ne 1):

Broken pipe (os error 32)

However, v≠1v \ne 1 hence

v=k2−1k2+1v = \frac{k^2-1}{k^2+1}

By this we can see

u=2kk2+1u = \frac{2k}{k^2+1}

Therefore, φa−1\varphi_a^{-1} is

φa−1(k)=(2kk2+1k2−1k2+1)\varphi_a^{-1}(k) = \begin{pmatrix} \frac{2k}{k^2+1}\\ \frac{k^2-1}{k^2+1} \end{pmatrix}

The similiar reasoning can show

φb−1(k)=(2kk2+11−k2k2+1)\varphi_b^{-1}(k) = \begin{pmatrix} \frac{2k}{k^2+1}\\ \frac{1-k^2}{k^2+1} \end{pmatrix}

Hence, for U1∩U2U_1 \cap U_2 (i.e. x≠0∧y≠0x \ne 0 \land y \ne 0) we have

φa−1∘φ1=φb−1∘φ2\varphi_a^{-1} \circ \varphi_1 = \varphi_b^{-1} \circ \varphi_2

By component, first check uu:

Broken pipe (os error 32)

Then check vv:

Broken pipe (os error 32)

So ψ\psi is indeed well defined, and smooth on it domain, the inverse defined as

ψ−1(p=(u,v))={φ1−1∘φa if p∈Uaφ2−1∘φb if p∈Ub\psi^{-1}(p=(u,v)) = \begin{cases} \varphi_1^{-1} \circ \varphi_a \text{ if } p\in U_a \\ \varphi_2^{-1} \circ \varphi_b \text{ if } p\in U_b \end{cases}

where φ1−1(a)=[1:a]\varphi_1^{-1}(a) = [1 : a] and φ2−1(b)=[b:1]\varphi_2^{-1}(b) = [b : 1] because the equivalence. These maps are smooth on u,vu,v and agree each other (inverse the ratio because the position) when meet hence well defined. Hence ψ\psi is a diffeomorphism.