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I want to solve the following exercise from M. Spivak's

If $F$ is a differentiable vector field on $M \subseteq \mathbb{R}^n$, show that there is an open $A \supseteq M$ and a differentiable vector field $\tilde{F}$ on $A$ with $\tilde{F}(x)=F(x)$ for $x \in M$. Hint: Do this locally and use partitions of unity.

My attempt:

If $x \in M$ let $U_x$ be an open set and $h_x:U_x \to V_x$ a diffeomorphism as in the definition of a manifold. We can restrict $V_x$ to an open ball around $a=h(x)$. Since a differentiable vector field on a manifold was defined as the pushforward of a differentiable vector field $G$ in Euclidean space, we can extend $F$ locally from $U_x \cap M$ to $U_x$ in the following way:

$$\tilde{F}(x):=F(G(\pi(x)) $$

where $\pi:\mathbb{R}^n \to \mathbb{R}^n$ is the projection to the first $k$ coordinates (here $k$ is the dimension of the manifold $M$).

After extending $F$ locally, I can't see how to combine the extensions using a partition of unity. I'd appreciate any help on that.

P.S. differentiability means smoothness in this case.

Thank you!

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I do assume you know what a paritition of unity is and that there exists one which is subordinate to a given cover. With that in mind look at a locally finite open cover $U_x$ of $M$ by coordinate charts as in your question and associated $F_{U_x}$ (locally extended vector fields) and $\phi_{U_x}$ (paritition of unity). Then let $F = \sum_{U_x}\phi_{U_x} F_{U_x}$

If you feel uneasy because the $F_{U_x}$ are only defined on the corresponding chart note that $\phi_{U_x} F_{U_x}$ makes sense globally since the $\phi$ are supported in the coordinate charts.

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  • $\begingroup$ Thanks Thomas. What I'm having trouble understanding is why this sum gives the extensions locally. Could you please elaborate on that? $\endgroup$ – user1337 May 31 '15 at 10:12
  • $\begingroup$ Sorry, but I don't really get the question. Do you want to know why the construction agrees with the original vector field on $M$? This is because the $\phi$ are a partition of unity, i.e. $\sum \phi_{U_z} = 1$, so whenever $F_{u_k} = F $ (wich is true on $M$) then $\sum \phi_{U_z} F_{U_z}(x) = \sum \phi_{U_z} F(x)= F(x) \sum \phi_{U_z}(x) = F(x) $ $\endgroup$ – Thomas May 31 '15 at 10:59

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