I'm trying to show that the tangent bundle, $TS^n$ of the n-sphere $S^n$ is diffeomorphic to the set $\sum z_i^2 = 1$ in $\mathbb{C}^{n+1}$.
It's relatively straightforward to see that the tangent bundle of the sphere can be identified with:
$$TS^n = \{ (x_0,…,x_n,y_0,…,y_n) : x_i,y_i \in \mathbb{R}, \sum x_i^2 = 1, \sum x_i y_i = 0 \}$$
Now to show this diffeomorphism I tried the natural thing of writing $z_j = x_j + iy_j$ but now we have $\sum z_j^2 = 1 – \sum y_i^2$ so it only lies in the required subspace if we restrict the tangent spaces of the sphere. I'm wondering how to write down a different map that does this?
I'm also a little concerned about how to show such a map is a diffeomorphism, how could I show that the identification I've made above as the tangent bundle embedded in $\mathbb{R}^{2(n+1)}$ is smooth? It's probably obvious but I'm struggling to see it!
Thanks for any help
Best Answer
Let $Q\subseteq\mathbf C^{n+1}$ be the affine quadric defined by the equation $\sum z_i^2=1$. The map $$ f\colon TS^n\rightarrow Q $$ defined by $$ z=f(x,y)=x\sqrt{1+||y||^2}+y\sqrt{-1} $$ does the job, where $||y||^2=\sum y_i^2$. Indeed, one has $f(x,y)\in Q$ since $$ \sum_{i=0}^n z_i^2=\sum_{i=0}^n x_i^2(1+||y||^2)-y^2_i+2x_iy_i\sqrt{-1}\sqrt{1+||y||^2}=\\ 1+||y||^2-||y||^2+2\sqrt{-1}\sqrt{1+||y||^2}\sum x_iy_i=1, $$ for $(x,y)\in TS^n$.
The map $f$ is a diffeomorphism since its inverse is $$ g\colon Q\rightarrow TS^n $$ defined by $$ g(z)=\left(\frac{x}{\sqrt{1+||y||^2}}, y\right), $$ where $z=x+y\sqrt{-1}$. One has $g(z)\in TS^n$ since $$ ||x||^2-||y||^2=1 $$ and $$ 2\sqrt{-1}\sum x_iy_i=0 $$ for $z\in Q$.