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Let M be a differentiable (smooth) manifold, and S a closed submanifold. Let X be a vector field on S. Prove that X is the restriction of a vector field Y defined on M. I tried this way but i'm not completely sure: i choose a covering (maybe finite if with closed we intend compact) of S with $U_j$ open in M. Then i consider a partition of unity $w_i$ subordinate to this covering (can I?). Then i have $Z=\sum_i w_iX$ is a vector field extending X in the open covering. Then i use a result that states that i can extend on the whole M a vector field defined in an open set of M. Is this correct?

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The support of $w_i$ is going to have more than just points of $S$ so how is $w_i X$ defined on $M$? – Eric O. Korman Mar 12 '12 at 20:33
Simple remark : you can't always extend a vector field defined on an open subset of a manifold. Take $\frac{1}{x}\frac{\partial}{\partial x}$ on $\mathbb{R}^*$ : it can't be extented to $\mathbb{R}$ – Selim Ghazouani Mar 12 '12 at 20:49
@ Eric: you're right, this way $w_iX$ is not defined outside S.. – balestrav Mar 12 '12 at 21:22
@Selim Ghazouani: you're right, the result i cited was about constructing a vector field defined on M which is equal to the given vector field in an open subset on the given set. – balestrav Mar 12 '12 at 21:24
You mean the second sentence to to say `Let $X$ be a vector field on $S$'? – treble Mar 12 '12 at 22:08
up vote 7 down vote accepted

Take a submanifold chart $x:U\rightarrow x(U) \subseteq \mathbb R^n$ such that $x(S\cap U) = \{x^{k+1}=\dots x^n=0\}$ (edit: and $x(U)$ is a ball)).

Then you can define a vector field $Z$ on $x(U)$, by $$ Z(x_1,\dots,x_n):=x_* (X(x^{-1}(x_1,\dots,x_k,0,\dots,0))) $$

Transport back to $M$ and get $x^{-1}_* Z$ on $U$ and then use a partition of unity argument!

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Could you please explain what $x_*$ is? – Wauzl Jun 8 '14 at 16:58
$x_*$ is the map on vector fields induced by the diffeomorphism $x$. – Alex G. Jun 16 '15 at 20:43

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