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 ?


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    $\begingroup$ It's easy enough to define a map from the pullback bundle to the direct sum bundle. It suffices now to prove that on each fiber, the kernel of the map is only the zero element. $\endgroup$
    – Deane
    Apr 5, 2021 at 21:12
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    $\begingroup$ Are you asking a question about what happens at a point or why everything glues together? What is your definition of the normal bundle? If it's a quotient bundle, the question is one step harder. $\endgroup$ Apr 5, 2021 at 22:52
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    $\begingroup$ Let $\iota \colon M \to \mathbb{R}^n$ be the inclusion map. Let $TM$ and $NM$ be the tangent and normal bundles of $M$, respectively. Your question is just: Why is $\iota^*(T\mathbb{R}^n)$ isomorphic to $TM \oplus NM$, is that right? $\endgroup$ Apr 11, 2021 at 12:22
  • $\begingroup$ Thanks all for your comments! Honestly I asked this question when I red this post math.stackexchange.com/questions/114560/… . It seems like it contains an interesting information that I didn't know and I'd like to understand it, however I don't have Further details about it ! $\endgroup$
    – Asma
    Apr 11, 2021 at 16:18

2 Answers 2


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 particuliar formulation: let $p \in M$. Then, as sub-linear spaces of $\mathbb{R}^n$, $$ T_pM \overset{\perp}{\oplus}T_pM^{\perp} = T_p\mathbb{R}^n, $$ and define $\varphi_p: (u,v)\in T_pM \oplus T_pM^{\perp} \to u+v \in \mathbb{R}^n$ this canonical isomorphism. Note that the normal bundle of $M$ in $\mathbb{R}^n$ can be described as: $$ \nu(M) = \bigcup_{p\in M} \{p\}\times T_pM^{\perp} $$ and thus, the vector bundle morphism: \begin{align} \varphi : TM \oplus \nu(M) & \longrightarrow M \times T_p\mathbb{R}^n = 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.

  • $\begingroup$ thanks a lot Didier for your answer! I have just one question: why do we have that $i^*(T \mathbb{R}^n)$ is equivalent to $M × \mathbb{R}^n$ ? $\endgroup$
    – Asma
    Apr 12, 2021 at 14:35
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    $\begingroup$ $T\mathbb{R}^n$ is a trivial bundle $\mathbb{R}^n\times \mathbb{R}^n$. Because $i$ is the inclusion $M\subset \mathbb{R}^n$, $i^* (T\mathbb{R}^n)$ is the restriction of this bundle on the base space $M$, that is, $M \times \mathbb{R}^n$. $\endgroup$
    – Didier
    Apr 12, 2021 at 14:58
  • $\begingroup$ Could you please explicitly define $T_pM^{\bot}$? $\endgroup$
    – user770533
    Dec 2, 2021 at 18:25
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    $\begingroup$ @Cedric It is the orthogonal of $T_pM$ in $\Bbb R^n$ for the standard inner product. $\endgroup$
    – Didier
    Dec 2, 2021 at 18:26
  • 1
    $\begingroup$ @Cedric Exactly. $\endgroup$
    – Didier
    Dec 2, 2021 at 18:34

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.



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