The tangent bundle of $\mathbb{S}^{n}$ is a connected manifold

Question

I have this exercise:

Consider the map $$ F: \mathbb{R}^{n+1} \times \mathbb{R}^{n+1} \to \mathbb{R}^{2},\ \ F(p,q) = (||p||^{2}, \left<p,q\right>). $$ Prove that $F$ is differentiable and $(1,0) \in \mathbb{R}^{2}$ a regular value for $F$. Prove that $M_{n} = F^{-1}(\{(1,0)\})$ is a regular and connected submanifold of $\mathbb{R}^{2n+2}$ of dimension $2n$.

I´ve already solved the first part. That $M_{n}$ is a regular submanifold of $\mathbb{R}^{2n+2}$ follows directly from the first part. Now, I´m trying to show that it must be connected. I can recognize that $M_{n}$ is the tangent bundle of $\mathbb{S}^{n}$. That is, up to a natural identification, it is the set $$ M_{n} = \bigcup_{p\in\mathbb{S}^{n}}T_{p}\mathbb{S}^{n} \equiv \bigcup_{p\in\mathbb{S}^{n}}(\{p\} \times \operatorname{span}(p)^{\perp}). $$ This exercise in my notes is before the tangent bundle has been introduced, so it is thought to be solved using more elementary techniques. I need a hint

Can you show S^n is connected? If you can, what's your current attempt/difficulty in showing the tangent bundle is connected? — dummy, Oct 26 '25 at 16:35
I can´t use anything related to the tangent bundle to prove it. So I can´t use the natural projection for example — MrGran, Oct 26 '25 at 16:37
The answer to your question is a corollary of: Let ${A_α}$ be a collection of connected subspaces of $X$; let A be con. sub. of X. Show that if $A∩A_α≠ ∅$ $∀α$, then $A∪(∪ A_α)$ is connected.. It may thus be closed as a duplicate. — Anne Bauval, Oct 26 '25 at 18:01
@AnneBauval but is there a set intersecting the rest in this case? As far as I know those sets are disjoint — MrGran, Oct 26 '25 at 18:09
@AnneBauval I do not think this is a duplicate of the question you mentioned. This question is a rather special case which admits an answer which makes use of the special situation, see the answer by Didier. As such it may be more helpful to people who understand the specail case. I'd really wish people would be more careful when voting to close questions. — Thomas, Oct 26 '25 at 20:10
It is a special case, as I mentionned. On MSE this is called an "abstract duplicate". The generalization of the linked post, which I took the care to search for, should be more helpful to people who meet analogous situations. — Anne Bauval, Oct 26 '25 at 22:12

Didier · Answer 1 · 2025-10-29T09:27:54.373

To show it is connected, you can show that it is path connected.

Since you recognize the tangent bundle, then you can have the following intuition: each fibre is a vector space, so any point $(p,v)$ can be linked to $(p,0)$. Moreover, the zero section $\{(p,0) \mid p \in S^n\}$ is naturally diffeomorphic to the base $S^n$, so you can join any two points $(p,0)$ to $(q,0)$ by a path, since $S^n$ is connected.

All in one, let $(p,v), (q,w) \in TS^n$. Build a continuous path $\gamma\colon [0,1] \to TS^n$ such that :

$\gamma(t) = (p,(1-3t)v)$ for $t \in [0,1/3]$,
$\gamma(t) = (\sigma(t),0)$ for $t \in [1/3,2/3]$, where $\sigma\colon [1/3,2/3]$ is a continuous path in $S^n$ from $p$ to $q$ (recall that $S^n$ is path connected),
$\gamma(t) = (q, (3t-2)w)$ for $t \in [2/3,1]$.

Then $\gamma$ is continuous and satisfies $\gamma(0) = (p,v)$ and $\gamma(1) = (q,w)$. This easily generalises for any vector bundle over a connected manifold.

In fact, a slight adaptation of this proof generalises to any fibre bundle over a path connected base space with path connected fibres.

The tangent bundle of $\mathbb{S}^{n}$ is a connected manifold

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