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

Question

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

Answer 1

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.

Answer 2

You may use the Serre-Swan theorem: It says there is an equivalence between the category of finite rank projective $C^{\infty}(X)$-modules and the category of finite rank real smooth vector bundles on $X$.

If $di: T(X) \rightarrow i^*T(Y)$ is the tangent mapping and $N_X(Y):=coker(di)$, the Serre-Swan theorem proves the following: Let

$$di^*: T(X)^* \rightarrow (i^*T(Y))^*$$

be the corresponding map of projective $C^{\infty}(X)$-modules. Since $T(X)^*$ is projective and $di^*$ is injective it follows $di^*$ splits and you get an isomorphism

$$S1.\text{ }T(X)^* \oplus N_X(Y)^* \cong (i^*T(Y))^*$$

and $(i^*T(Y))^*$ is a trivial $C^{\infty}(X)$-module of rank $n$ since $T(Y)$ is the trivial vector bundle of rank $n$. It follows there is an isomorphism of vector bundles

$$T(X) \oplus N_X(Y) \cong (i^*T(Y)).$$

Note: If $R$ is a commutative unital ring and $P$ is a finite rank projective $R$-module it follows for any surjection $\phi: M \rightarrow N$ of left $R$-modules and any map $g:P \rightarrow N$ there is a lift $g^*: P \rightarrow M$ with $g^* \circ \phi=g$. A similar result holds dually for injections. Apply this result to the map $di^*$ to get the splitting in $S1$.

Example: If $\phi: R^n \rightarrow P \rightarrow 0$ is a surjection with $P$ a projective $R$-module, there is a section $s:P \rightarrow R^n$ with $p \circ s= Id_P$ the identity map. Let $\psi:= s \circ p$. It follows

$$ \psi \circ \psi = s \circ p \circ s \circ p = s \circ p = \psi,$$

hence $\psi \in End_R(R^n)$ is an idempotent. It follows $R^n \cong ker(\psi)\oplus im(\psi) \cong Q \oplus P$ where $Q:=ker(\phi)$. Since $T(X)^*$ and $N_X(Y)^*$ are projective modules this argument implies the splitting in $S1$.

This type of reasoning proves the result in general: If

$$0 \rightarrow E \rightarrow F \rightarrow G \rightarrow 0$$

is an exact sequence of finite rank vector bundles on $X$ it follows

$$S2.\text{ }0 \rightarrow E^* \rightarrow F^* \rightarrow G^* \rightarrow 0$$

is an exact sequence of projective $C^{\infty}(X)$-modules. It follows $S2$ splits. Hence there is an isomorphism $F\cong E\oplus G$.

Note: The Serre-Swan theorem is a classical result in differential geometry relating finite rank vector bundles on manifolds to finite rank projective modules on commutative rings. Similar result hold for complex holomorphic vector bundles on complex manifolds and algebraic vector bundles on algebraic varieties.

https://en.wikipedia.org/wiki/Serre%E2%80%93Swan_theorem