CP1 and S^2 are smooth manifolds and diffeomorphic (proof)

Proposition: \mathbb{C}P^1 is a smooth manifold.

Proof:
Define U_1=\{[z^1, z^2]\mid z^1\neq 0\} and U_2=\{[z^1, z^2]\mid z^2\neq 0\}. Also define g_i: U_i\to\mathbb{C} by g_1([z^1, z^2])=\frac{z^2}{z^1} and g_2([z^1, z^2])=\frac{\overline{z^1}}{\overline{z^2}}.

Let f:\mathbb{C}\to\mathbb{R}^2 be the homeomorphism from \mathbb{C} to \mathbb{R}^2 defined by f(x+iy)=(x,y) and define \phi_i: U_i\to\mathbb{R}^2 by \phi_i=f\circ g_i.

Note that \{U_1, U_2\} is an open cover of \mathbb{C}P^1, and \phi_i are well-defined homeomorphisms (from U_i onto an open set in \mathbb{R}^2). Then \{(U_1,\phi_1), (U_2,\phi_2)\} is an atlas of \mathbb{C}P^1.

The transition function \displaystyle \phi_2\circ\phi_1^{-1}: \phi_1(U_1\cap U_2)\to\mathbb{R}^2,
\begin{aligned}  \phi_2\phi_1^{-1}(x,y)&=\phi_2 g_1^{-1}(x+iy)\\  &=\phi_2([1,x+iy])\\  &=f(\frac{1}{x-iy})\\  &=f(\frac{x+iy}{x^2+y^2})\\  &=(\frac{x}{x^2+y^2},\frac{y}{x^2+y^2})  \end{aligned}
is differentiable of class C^\infty. Similarly, \phi_1\circ\phi_2^{-1}:\phi_2(U_1\cap U_2)\to\mathbb{R}^2 is of class C^\infty. Hence \mathbb{C}P^1 is a smooth manifold.

Proposition:
S^2 is a smooth manifold.

Proof:
Define V_1=S^2\setminus\{(0,0,1)\} and V_2=S^2\setminus\{(0,0,-1)\}. Then \{V_1, V_2\} is an open cover of S^2.

Define \psi_1: V_1\to\mathbb{R}^2 by \psi_1(x,y,z)=(\frac{x}{1-z},\frac{y}{1-z}) and \psi_2: V_2\to\mathbb{R}^2 by \psi_2(x,y,z)=(\frac{x}{1+z},\frac{y}{1+z}).

We can check that \psi_1^{-1}(x,y)=(\frac{2x}{1+x^2+y^2},\frac{2y}{1+x^2+y^2},\frac{-1+x^2+y^2}{1+x^2+y^2}). Hence \psi_1 is a homeomorphism from V_1 onto an open set in \mathbb{R}^2. Similarly, \psi_2 is a homeomorphism from V_2 onto an open set in \mathbb{R}^2. Thus \{V_1,V_2\} is an atlas for S^2.

The composite \psi_2\circ\psi_1^{-1}: \psi_1(V_1\cap V_2)\to\mathbb{R}^2 is differentiable of class C^\infty since both \psi_2, \psi_1^{-1} are of class C^\infty. Similarly, \psi_1\circ\psi_2^{-1} is of class C^\infty. Thus S^2 is a smooth manifold.

We can also compute the transition function explicitly:
\displaystyle  \psi_2\psi_1^{-1}(x,y)=(\frac{x}{x^2+y^2},\frac{y}{x^2+y^2}).
Note that \psi_2\psi_1^{-1}=\phi_2\phi_1^{-1}.

Define h:\mathbb{C}P^1\to S^2 by h(\phi_1^{-1}(x,y))=\psi_1^{-1}(x,y) and h(\phi_2^{-1}(x,y))=\psi_2^{-1}(x,y).

We see that h is well-defined since if \phi_1^{-1}(x,y)=\phi_2^{-1}(u,v) then \displaystyle (u,v)=\phi_2\phi_1^{-1}(x,y)=\psi_2\psi_1^{-1}(x,y) so that \psi_2^{-1}(u,v)=\psi_1^{-1}(x,y).

Similarly, we have a well-defined inverse h^{-1}: S^2\to\mathbb{C}P^1 defined by h^{-1}(\psi_1^{-1}(x,y))=\phi_1^{-1}(x,y) and h^{-1}(\psi_2^{-1}(x,y))=\phi_2^{-1}(x,y).

We check that (from our previous workings)
\begin{aligned}  \psi_1 h\phi_1^{-1}(x,y)&=(x,y)\\  \psi_2 h\phi_1^{-1}(x,y)&=\psi_2\psi_1^{-1}(x,y)\\  \psi_1 h\phi_2^{-1}(x,y)&=\psi_1\psi_2^{-1}(x,y)\\  \psi_2 h\phi_2^{-1}(x,y)&=(x,y)  \end{aligned}
are of class C^\infty. So h is a smooth map. Similarly, h^{-1} is smooth. Hence h is a diffeomorphism.

Advertisements

About mathtuition88

http://mathtuition88.com
This entry was posted in math and tagged . Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s