Differential Geometry – Tangent Bundle of a Sphere T?S? is Diffeomorphic to ?S? × ?S? – ?

Question

Let $\mathbb S^n$ denote the $n$-sphere, which is the smooth manifold consisting of all points in $\mathbb R^{n+1}$ with Euclidean norm one. Recall that the tangent bundle of $\mathbb S^n$, denoted $T \mathbb S^n$, is a $2n$-dimensional smooth manifold. Let $\Delta \subseteq \mathbb S^n \times \mathbb S^n$ denote the diagonal $\Delta := \{(x,x) : x \in \mathbb S^n\}$. By Hausdorffness, $\Delta$ is closed, so $\mathbb S^n \times \mathbb S^n – \Delta$ is an open subset of $\mathbb S^n \times \mathbb S^n$.

In this question about the tangent bundle of the $n$-sphere, (I believe) one of the answers claims that we have a diffeomorphism $$T\mathbb S^n \cong \mathbb S^n \times \mathbb S^n – \Delta.$$

How can we see that this is true? What is an explicit diffeomorphism?

Answer 1

We must be very explicit to prove it.

1. We have $$S^n \times S^n \setminus \Delta = \bigcup_{p \in S^n} \{p\} \times (S^n \setminus \{ p\})$$ which is a smooth submanifold of $\mathbb R^{n+1} \times \mathbb R^{n+1}$.
2. The tangent space $T_pS^n$ can be regarded as the orthogonal complement of $p$, i.e. as $$T_pS^n = \{x \in \mathbb R^{n+1} \mid x \cdot p = 0 \} .$$ Then the "Euclidean" tangent bundle is $$TS^n = \bigcup_{p \in S^n} \{p\} \times T_pS^n = \{(p,x) \in \mathbb R^{n+1} \times \mathbb R^{n+1} \mid p \cdot p = 1, x \cdot p = 0 \} .$$ It is well-known that this is a smooth submanifold of $R^{n+1} \times R^{n+1}$ which is diffeomorphic to the "abstract" tangent bundle of $S^n$.

Our task is to find diffeomorphisms $S^n \setminus \{ p\} \to T_pS^n$ which fit together to a diffeomorphism $S^n \times S^n \setminus \Delta \to TS^n$.

Stereographic projection from $p$ to $T_pS^n$ is the key (see Stereographic projection when the "North/South Pole" is not given by $(0,...,\pm 1)$?). This is given by $$s_p : S^n \setminus \{ p\} \to T_pS^n, s_p(x) = \frac{x- (x \cdot p)p}{1 - x \cdot p} .$$

Let $$\Omega = \{ (p,x) \in \mathbb R^{n+1} \times \mathbb R^{n+1} \mid x \cdot p \ne 1 \} .$$ This is an open subset of $\mathbb R^{n+1} \times \mathbb R^{n+1}$ which contains $S^n \times S^n \setminus \Delta$. Define $$\phi : \Omega \to \mathbb R^{n+1} \times \mathbb R^{n+1}, \phi(p,x) = \left(p,\frac{x- (x \cdot p)p}{1 - x \cdot p}\right).$$ This is a smooth map such that $\phi(S^n \times S^n \setminus \Delta) = TS^n$. It restricts to a smooth map $$\phi' : S^n \times S^n \setminus \Delta \to TS^n .$$

The inverse of $s_p$ is $$s_p^{-1} : T_pS^n \to S^n \setminus \{ p\},s_p^{-1}(y) = \frac{(\lVert y \rVert^2 -1)p +2y }{\lVert y \rVert^2 +1} .$$ Define $$\psi : \mathbb R^{n+1} \times \mathbb R^{n+1} \to \mathbb R^{n+1} \times \mathbb R^{n+1}, \psi(p,y) = \left(p,\frac{(\lVert y \rVert^2 -1)p +2y }{\lVert y \rVert^2 +1} \right) .$$ This is a smooth map such that $\psi(TS^n) = S^n \times S^n \setminus \Delta$. It restricts to a smooth map $$\psi' : TS^n \to S^n \times S^n \setminus \Delta .$$

By construction $\psi' \circ \phi' = id$ and $\phi' \circ \psi' = id$. Hence both maps are diffeomorphisms which are inverse to each other.