Construction
U a = S 1 − ( 0 , 1 ) U_a = S^1 - (0,1) U a = S 1 − ( 0 , 1 ) φ a ( u , v ) = u 1 − v \varphi_a(u,v) = \frac{u}{1-v} φ a ( u , v ) = 1 − v u U b = S 1 − ( 0 , − 1 ) U_b = S^1 - (0,-1) U b = S 1 − ( 0 , − 1 ) φ b ( u , v ) = u 1 + v \varphi_b(u,v) = \frac{u}{1+v} φ b ( u , v ) = 1 + v u and R P 1 \R P^1 R P 1
U 1 = { [ x : y ] ∣ x ≠ 0 } U_1 = \{[x:y] \mid x\ne0\} U 1 = {[ x : y ] ∣ x = 0 } φ 1 ( [ x : y ] ) = y x \varphi_1([x:y]) = \frac{y}{x} φ 1 ([ x : y ]) = x y U 2 = { [ x : y ] ∣ y ≠ 0 } U_2 = \{[x:y] \mid y\ne0\} U 2 = {[ x : y ] ∣ y = 0 } φ 1 ( [ x : y ] ) = x y \varphi_1([x:y]) = \frac{x}{y} φ 1 ([ x : y ]) = y x The diffeomorphism ψ : R P 1 → S 1 \psi : \R P^1 \to S^1 ψ : R P 1 → S 1
ψ ( p = [ x : y ] ) = { φ a − 1 ∘ φ 1 if p ∈ U 1 φ b − 1 ∘ φ 2 if p ∈ U 2 \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} ψ ( p = [ x : y ]) = { φ a − 1 ∘ φ 1 if p ∈ U 1 φ b − 1 ∘ φ 2 if p ∈ U 2 Given [ x : y ] [x:y] [ x : y ] k = y / x k = y/x k = y / x ( u , v ) (u,v) ( u , v ) S 1 S^1 S 1
k = u 1 − v ⟹ k ( 1 − v ) = u ⟹ k 2 ( 1 − v ) 2 = u 2 ⟹ k 2 ( 1 − v ) 2 + v 2 = 1 ⟹ k 2 ( 1 − 2 v + v 2 ) + v 2 = 1 ⟹ ( k 2 + 1 ) v 2 − 2 k 2 v + k 2 = 1 ⟹ ( k 2 + 1 ) v 2 − 2 k 2 v + ( k 2 − 1 ) = 0 k = \frac{u}{1-v}
\implies k(1-v)=u \\
\implies k^2(1-v)^2=u^2 \\
\implies k^2(1-v)^2 + v^2=1 \\
\implies k^2(1-2v+v^2) + v^2=1 \\
\implies (k^2+1) v^2 - 2k^2v + k^2=1 \\
\implies (k^2+1) v^2 - 2k^2v + (k^2-1)=0 k = 1 − v u ⟹ k ( 1 − v ) = u ⟹ k 2 ( 1 − v ) 2 = u 2 ⟹ k 2 ( 1 − v ) 2 + v 2 = 1 ⟹ k 2 ( 1 − 2 v + v 2 ) + v 2 = 1 ⟹ ( k 2 + 1 ) v 2 − 2 k 2 v + k 2 = 1 ⟹ ( k 2 + 1 ) v 2 − 2 k 2 v + ( k 2 − 1 ) = 0 Now we can use quadratic formula to solve v v v v ≠ 1 v \ne 1 v = 1
v = 2 k 2 ± 4 k 4 − 4 ( k 2 + 1 ) ( k 2 − 1 ) 2 ( k 2 + 1 ) = 2 k 2 ± 2 k 4 − ( k 2 + 1 ) ( k 2 − 1 ) 2 ( k 2 + 1 ) = k 2 ± k 4 − ( k 2 + 1 ) ( k 2 − 1 ) k 2 + 1 = k 2 ± k 4 − ( k 4 − 1 ) k 2 + 1 = k 2 ± 1 k 2 + 1 v = \frac{
2k^2 \pm \sqrt{4k^4 - 4(k^2+1)(k^2-1)}
}{2(k^2+1)}
=
\frac{2k^2\pm2\sqrt{k^4-(k^2+1)(k^2-1)}}{2(k^2+1)} \\
=
\frac{k^2\pm\sqrt{k^4-(k^2+1)(k^2-1)}}{k^2+1} \\
=
\frac{k^2\pm\sqrt{k^4-(k^4-1)}}{k^2+1}
=
\frac{k^2\pm1}{k^2+1} v = 2 ( k 2 + 1 ) 2 k 2 ± 4 k 4 − 4 ( k 2 + 1 ) ( k 2 − 1 ) = 2 ( k 2 + 1 ) 2 k 2 ± 2 k 4 − ( k 2 + 1 ) ( k 2 − 1 ) = k 2 + 1 k 2 ± k 4 − ( k 2 + 1 ) ( k 2 − 1 ) = k 2 + 1 k 2 ± k 4 − ( k 4 − 1 ) = k 2 + 1 k 2 ± 1 However, v ≠ 1 v \ne 1 v = 1
v = k 2 − 1 k 2 + 1 v = \frac{k^2-1}{k^2+1} v = k 2 + 1 k 2 − 1 By this we can see
u = 2 k k 2 + 1 u = \frac{2k}{k^2+1} u = k 2 + 1 2 k Therefore, φ a − 1 \varphi_a^{-1} φ a − 1
φ a − 1 ( k ) = ( 2 k k 2 + 1 k 2 − 1 k 2 + 1 ) \varphi_a^{-1}(k)
=
\begin{pmatrix}
\frac{2k}{k^2+1}\\
\frac{k^2-1}{k^2+1}
\end{pmatrix} φ a − 1 ( k ) = ( k 2 + 1 2 k k 2 + 1 k 2 − 1 ) The similiar reasoning can show
φ b − 1 ( k ) = ( 2 k k 2 + 1 1 − k 2 k 2 + 1 ) \varphi_b^{-1}(k)
=
\begin{pmatrix}
\frac{2k}{k^2+1}\\
\frac{1-k^2}{k^2+1}
\end{pmatrix} φ b − 1 ( k ) = ( k 2 + 1 2 k k 2 + 1 1 − k 2 ) Hence, for U 1 ∩ U 2 U_1 \cap U_2 U 1 ∩ U 2 x ≠ 0 ∧ y ≠ 0 x \ne 0 \land y \ne 0 x = 0 ∧ y = 0
φ a − 1 ∘ φ 1 = φ b − 1 ∘ φ 2 \varphi_a^{-1} \circ \varphi_1
=
\varphi_b^{-1} \circ \varphi_2 φ a − 1 ∘ φ 1 = φ b − 1 ∘ φ 2 By component, first check u u u
2 y / x y 2 / x 2 + 1 = 2 y / x ( x 2 + y 2 ) / x 2 = 2 y x 2 ( x 2 + y 2 ) x = 2 x y x 2 + y 2 and 2 x / y x 2 / y 2 + 1 = 2 x / y ( x 2 + y 2 ) / y 2 = 2 x y 2 ( x 2 + y 2 ) y = 2 x y x 2 + y 2 \frac{2 y/x}{y^2/x^2 + 1}
= \frac{2 y/x}{(x^2+y^2)/x^2}
= \frac{2 y x^2}{(x^2+y^2)x}
= \frac{2 xy}{x^2+y^2}
\\
\text{and}
\\
\frac{2 x/y}{x^2/y^2 + 1}
= \frac{2 x/y}{(x^2+y^2)/y^2}
= \frac{2 xy^2}{(x^2+y^2)y}
= \frac{2 xy}{x^2+y^2} y 2 / x 2 + 1 2 y / x = ( x 2 + y 2 ) / x 2 2 y / x = ( x 2 + y 2 ) x 2 y x 2 = x 2 + y 2 2 x y and x 2 / y 2 + 1 2 x / y = ( x 2 + y 2 ) / y 2 2 x / y = ( x 2 + y 2 ) y 2 x y 2 = x 2 + y 2 2 x y Then check v v v
( y / x ) 2 − 1 ( y / x ) 2 + 1 = y 2 − x 2 y 2 + x 2 and 1 − ( x / y ) 2 ( x / y ) 2 + 1 = y 2 − x 2 x 2 + y 2 \frac{(y/x)^2-1}{(y/x)^2+1}
= \frac{y^2-x^2}{y^2+x^2}
\\
\text{and}
\\
\frac{1-(x/y)^2}{(x/y)^2+1}
= \frac{y^2-x^2}{x^2+y^2} ( y / x ) 2 + 1 ( y / x ) 2 − 1 = y 2 + x 2 y 2 − x 2 and ( x / y ) 2 + 1 1 − ( x / y ) 2 = x 2 + y 2 y 2 − x 2 So ψ \psi ψ
ψ − 1 ( p = ( u , v ) ) = { φ 1 − 1 ∘ φ a if p ∈ U a φ 2 − 1 ∘ φ b if p ∈ U b \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} ψ − 1 ( p = ( u , v )) = { φ 1 − 1 ∘ φ a if p ∈ U a φ 2 − 1 ∘ φ b if p ∈ U b where φ 1 − 1 ( a ) = [ 1 : a ] \varphi_1^{-1}(a) = [1 : a] φ 1 − 1 ( a ) = [ 1 : a ] φ 2 − 1 ( b ) = [ b : 1 ] \varphi_2^{-1}(b) = [b : 1] φ 2 − 1 ( b ) = [ b : 1 ] u , v u,v u , v ψ \psi ψ