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Let $X$ be a smooth vector field on $S^{n-1}$. Then, does there exist a smooth vector field $\tilde{X}$ on $\mathbb{R}^n$ such that $X$ is the restriction of $\tilde{X}$?

Define $\tilde{X}(x):=||x||X(x/||x||)$ for $x\neq 0$ and $\tilde{X}(0)=0$.

Would it be the one? Without explicitly finding this extension, is there an abstract proof for this?

Ben Grossmann
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Rubertos
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    Yes, your definition works perfectly fine. Why would you want an abstract proof when your explicit construction is completely intuitive? – Ben Grossmann Apr 17 '17 at 16:53
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    It is worth noting that one should show that the limit as $x \to 0$ necessarily exists and is equal to $0$. – Ben Grossmann Apr 17 '17 at 16:54
  • @Omnomnomnom I'm just wondering if there is a (well-known) fundamental reason why this holds! But from your comment, it seems like this problem just asks to construct such $\tilde{X}$. – Rubertos Apr 17 '17 at 16:54
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    Are you sure your vector field is smooth? – Blazej Apr 17 '17 at 16:56
  • @Blazej good catch! I missed that detail. It isn't smooth, as one can see with the map $X(x) = 1$ on $S^0 = {\pm 1}$. A working modification is to take $\tilde X(x) = \beta(|x|) X(x/|x|)$, where $\beta:\Bbb R \to \Bbb R$ is smooth at zero, satisfying $\beta^{(n)}(0) = 0$ and $\beta(1) = 1$. – Ben Grossmann Apr 17 '17 at 17:00
  • Well, a vector field in a closed submanifold of R^n always extends smoothly to the ambient space. If you want, that is the reason (and proving it is not difficult) – Mariano Suárez-Álvarez Apr 17 '17 at 17:01
  • In a very general setting, perhaps this can be related to the existence of a partition of unity. – Ben Grossmann Apr 17 '17 at 17:04
  • Thank you! That is what I was looking for. – Rubertos Apr 17 '17 at 17:07

1 Answers1

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First, your vector field is not smooth. Note that for $X$ given by $X(p)=v$ a constant vector field, we have that $x\mapsto ||x||v$ is not a smooth vector field. So the vector field you mean probably is $$\tilde{X}:=\begin{cases}||x||^2X(x/||x||)&\text{if }x\neq 0\\ 0&\text{if }x=0\end{cases}\text{.}$$ Second, there is no necessity of a more abstract proof, but a more abstract "existencial" proof is possible an valid for all compact submanifolds of $\mathbb{R}^n$. However, one can do such a proof following the following steps (whose details I leave to you):

  1. The extension of a vector field $X$ on a compact submanifold $M$ of $\mathbb{R}^n$ to $\mathbb{R}^n$ follows from the extension of such a vector field to an open neighborhood of it. This is a consequence of the existence of smooth bump functions which are equal to 1 in the closed set and vanish outside the given open set.
  2. For any such compact submanifold $M$ of $\mathbb{R}^n$, we can use the tubular neighborhood $$U_{\varepsilon}(M):=\{x\in\mathbb{R}^n\,|\,d(x,M)\leq \varepsilon\}$$ which for epsilon small enough is diffeomorphic to $M\times (-\varepsilon,\varepsilon)^{n-\dim\,M}$ (as a consequence of the Lebesgue covering lemma and the constant rank theorem). Note that one can even say more and substitute diffeomorphic by (the stronger) isometric.
  3. It is clear that any vector field on $M$ extend to a vector field on $M\times (-\varepsilon,\varepsilon)^{n-\dim\,M}$ and so for $\varepsilon>0$ sufficiently small to a vector field in $U_{\varepsilon}(M)$ (as it would be diffeomorphic to $M\times (-\varepsilon,\varepsilon)^{n-\dim\,M}$) and by point 1 to a vector field on $\mathbb{R}^n$ as desired.

Third, the above proof can be extended easily to closed submanifolds droping the compact assumption (by considering not tubular neighborhoods). However, the above proof is just a generalization for compact submanifolds of the construction of the extension for the sphere.

Josué Tonelli-Cueto
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