$T\mathbb{S}^n\times \mathbb{R} \cong \mathbb{S}^n \times \mathbb{R}^{n+1}$

Question

Let $\mathbb{S}^n$ be the $n$ dimensional unit sphere, and let $T\mathbb{S}^n$ denote its tangent bundle: $$ T\mathbb{S}^n = \bigsqcup_{p\in\mathbb{S}^n} T_p\mathbb{S}^{n} $$

I would like to show that: $$ T\mathbb{S}^n\times \mathbb{R} \cong \mathbb{S}^n \times \mathbb{R}^{n+1} $$ where here $\cong$ denotes "diffeomorphic". Right now, I think the main property I must use is the fact that $T_p\mathbb{S}^n = \{v \in \mathbb{R}^n : \langle v , p \rangle =0\}$, but outside of this I am having trouble as to how I might attack this. Does anyone have any ideas?

Answer 1

As you say, we can use that fact that $TS^n$ can be identitied with the "Euclidean tangent bundle" of the submanifold $S^n \subset \mathbb R^{n+1}$ given by $$T'S^n = \{ (x,v) \in S^n \times \mathbb R^{n+1} \mid \langle x, v \rangle = 0 \} . $$ It is well-known that this is a smooth submanifold of $\mathbb R^{2n+2}$ and has the structure of a smooth $n$-dimensional vector bundle over $S^n$ with bundle projection $p(x,v) = x$. The fiber over $x$ is $\{x\} \times T'_xS^n$ with the "Euclidean tangent space" $T'_xS^n = \{ v \in \mathbb R^{n+1} \mid \langle x, v \rangle = 0 \}$. Now define $$F : \mathbb R^{n+1} \times \mathbb R^{n+1} \times \mathbb R \to \mathbb R^{n+1} \times \mathbb R^{n+1}, F(x,v,t) = (x,v+tx) ,$$ $$G : R^{n+1} \times \mathbb R^{n+1} \to \mathbb R^{n+1} \times \mathbb R^{n+1} \times \mathbb R , G(x,y) = (x,y - \langle x, y \rangle x ,\langle x, y \rangle) .$$ These are clearly smooth maps. If $(x,v) \in T'S^n$, then trivially $F(x,v,t) \in S^n \times \mathbb R^{n+1}$. If $x \in S^n$, then $\langle x, y - \langle y, x \rangle x \rangle = \langle x,y \rangle - \langle y, x \rangle \langle x, x \rangle = 0$. i.e. $G(x,y) \in T'S^n \times \mathbb R$.

Thus $F$ and $G$ restrict to smooth maps $f : T'S^n \times \mathbb R \to S^n \times \mathbb R^{n+1}$ and $g : S^n \times \mathbb R^{n+1} \to T'S^n \times \mathbb R$. Recalling $\langle x, v \rangle = 0$ and $\langle x, x \rangle = 1$ for $(x,v) \in T'S^n$, we get $$g(f(x,v,t)) = g(x,v + tx) = (x, v + tx - \langle x, v + tx \rangle x, \langle x, v + tx \rangle) = (x,v,t) ,$$ $$f(g(x,y)) = f(x,y - \langle x, y \rangle x,\langle x, y \rangle) = (x,y - \langle x, y \rangle x + \langle x, y \rangle x) = (x,y) .$$ We conclude that $f, g$ are inverse diffeomorphisms which answers your question.

Moreover $f$ is a bundle isomorphism since $f(\{x\} \times T'_xS^n) \subset \{x\} \times \mathbb R^{n+1}$ and the map $f_x : \{x\} \times T'_xS^n \to \{x\} \times \mathbb R^{n+1}$ is a linear bijection. Hence also $g = f^{-1}$ is a bundle morphism.

Answer 2

Consider the correspondence $f:S^n\times\mathbb{R}^n \rightarrow TS^n\times\mathbb{R}$ defined by $f(x,u)=(u-\langle x,u\rangle x,\langle x,u\rangle)$ here $v=u-\langle x,u\rangle x$ is an element of $T_xS^n$ since $\langle v,x\rangle=0$. The map $f$ define an isomorphism of vector bundles.