Normal bundle $\oplus$ tangent bundle is a trivial bundle

differential-geometryvector-bundles

I have the same question as in this post
Sum of normal bundle and tangent bundle.

I'm wondering how to prove that the sum of the normal bundle and the tangent of a submanifold $M \subset \mathbb{R}^n$ is trivial. The answer to this post didn't contain an explanation to the fact that their direct sum is equal to the pullback of tangent bundle over $\mathbb{R}^n$ ? Could someone please explain why this is true or give another proof ?

Thanks

Best Answer

Regarding the comments, it seems question is about this formulation:

Let $i : M \to \mathbb{R}^n$ be the inclusion of a smooth submanifold of $\mathbb{R}^n$, $TM$ be the tangent bundle of $M$ and $\nu(M)$ be its normal bundle defined as: $$ \nu(M) = \{(p,v)\in M \times \mathbb{R}^n \mid \forall w \in T_pM,~ \langle v , w \rangle = 0 \} $$ Then why is the isomorphism $TM \oplus \nu(M) \simeq i^*(T\mathbb{R}^n)$ true?

Here is the answer of that particular formulation: let $p \in M$. Then, as sub-linear spaces of $\mathbb{R}^n$, one has $$ T_pM \overset{\perp}{\oplus}T_pM^{\perp} = T_p\mathbb{R}^n. $$ Define $\varphi_p: (u,v)\in T_pM \oplus T_pM^{\perp} \to u+v \in \mathbb{R}^n$ which is a canonical isomorphism. Note that the normal bundle of $M$ in $\mathbb{R}^n$ can be described as: $$ \nu(M) = \bigcup_{p\in M} \{p\}\times T_pM^{\perp}. $$ Thus, the vector bundle homomorphism: \begin{align} \varphi : TM \oplus \nu(M) & \longrightarrow i^*\left(T\mathbb{R}^n\right) \\ (p, (u,v)) & \longmapsto (p, \varphi_p(u,v)) \end{align} is a vector bundle isomorphism. To conclude, notice that $T\mathbb{R}^n = \mathbb{R}^n \times \mathbb{R}^n$ is canonically trivial, hence, $i^*(T\mathbb{R}^n) \simeq M\times \mathbb{R}^n$ canonically.

Comment: the same proof shows that, if $N \subset (M,g)$ is a submanifold of a Riemannian manifold, then there is a canonical isomorphism $TN \oplus \nu^M(N) \simeq i^*(TM)$ where $\nu^M(N)$ is the normal bundle of $N$ in $M$, that is: $$ \nu^M(N) = \{ (p,v) \in i^*(TM) \mid v \perp^g T_pN\} $$ and $i : N \to M$ is the inclusion map.

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