Would somebody be able to prove that the whitney sum of the normal and tangent bundles of a submanifold of $\mathbb{R}^n$ is trivial? Would apreciate a detailed proof...I'm struggling a little. Tina
Sum of normal bundle and tangent bundle.
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You may use the Serre-Swan theorem to prove the result: It says there is an exact equivalence between the category of finite rank projective $C^{\infty}(X)$-modules and maps of modules and the category of finite rank real smooth vector bundles on $X$ and maps of vector bundles.
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 $S1$.
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.
Let $f:M^m\rightarrow \mathbb{R}^n$ be an embedding, let $\tau(M)$ be the tangent bundle of $M$, $\nu (M)$ the normal bundle of the embedding $f$. Then we have the formula
$\tau(M)\oplus\nu(M)\cong f^*\tau(\mathbb{R}^n)$
where $f^*\tau(\mathbb{R}^n)$ denotes the pullback of the tangent bundle of $\mathbb{R}^n$ along $f$ (which you could think of as the tangent bundle of $\mathbb{R}^n$ restricted to $f(M)$).
$\tau(\mathbb{R}^n)$ is canonically isomorphic to $\mathbb{R}^n \times\mathbb{R}^n$, and so its pullback is also trivial.