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Given a submanifold of R^n, M, what is the most straightforward way to make a tangent vector field in the geometric sense, say f: M -> R^n, into a vector field g : M -> TM ? Would smoothness be preserved, and why?

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2 Answers 2

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The tangent bundle to $\mathbb R^n$ can be restricted to $M$ and the restriction has a canonical direct sum decomposition $T\mathbb R^n|M=TM\stackrel {\perp}{\oplus} \nu $, where $\nu$ is the norma bundle of the embedding of $M$ into $\mathbb R^n$.
This yields a direct sum decomposition of the global sections of those vector bundles: $\Gamma(M,T\mathbb R^n|M)=\Gamma(M,TM)\oplus\Gamma(M,\nu)$ .
The map you are looking for is just the first projection $pr_1:\Gamma(M,T\mathbb R^n|M) \to \Gamma(M,TM)$. Smoothness is preserved .

Warning
Even if a vector field $X\in \Gamma(M,T\mathbb R^n|M)$ never vanishes, its projection $pr_1(X)\in \Gamma(M,TM)$ may very well vanish somewhere on $M$.
For example, if $M=S^2\subset \mathbb R^3$, there are lots of nowhere vanishing sections $X\in \Gamma(S^2,T\mathbb R^3|S^2)$ (constant ones for example), but the projection of each of them $pr_1(X)\in \Gamma(S^2,TS^2)$ is guaranteed to vanish somewhere on the sphere: you can't comb the hair of a hedgehog!

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A tangent vector field is a vector field. When $M$ is a submanifold of $\mathbb{R}^n$, then each $T_p(M)$ (tangent space of $M$ at $p$, for $p \in M$) can be viewed as a subspace of $\mathbb{R}^n$. In fact, the bundle $TM$ is a subbundle of the rank $n$ trivial bundle $M \times \mathbb{R}^n$.

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