# extending a vector field defined on a closed submanifold

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?

• The support of $w_i$ is going to have more than just points of $S$ so how is $w_i X$ defined on $M$? Mar 12, 2012 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 $\mathbb{R}$ Mar 12, 2012 at 20:49
• @ Eric: you're right, this way $w_iX$ is not defined outside S.. Mar 12, 2012 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. Mar 12, 2012 at 21:24
• You mean the second sentence to to say `Let $X$ be a vector field on $S$'? Mar 12, 2012 at 22:08

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!
• Could you please explain what $x_*$ is? Jun 8, 2014 at 16:58
• $x_*$ is the map on vector fields induced by the diffeomorphism $x$. Jun 16, 2015 at 20:43
• Why do you need $x(U)$ to be a ball? Oct 23, 2018 at 16:38
• @CuriousKid7 So we can have $(x_1,\dots,x_k,0,\dots,0)\in x(U)$ for any $(x_1,\dots,x_n)\in x(U)$. Jul 25, 2020 at 6:13