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Look in the bottom for the edited answer: Is possible to prove that we could extend globally a local vector field $X$ defined on a open subset $U$ of a real paracompact manifold $M$? Is my congecture ok?

Let $U$ be the open subset of a smooth manifold $M$ and $X$ the vector field defined on $U$. Wlog there exist $V$ open subset containing $U$, so $\{V,\overline{U}^c\}$ is an open covering of $M$, and for a knowned theorem there exist a countable smooth partition of unity whose dominated the covering; $\{\psi_i\}_{i\in\mathbb{N}}$.

Let $J\subset\mathbb{N}$ such that $\sum_{i\in J}\psi_i(p)=0, \forall p\notin V$, so $\sum_{i\in J}\psi_i(p)=\sum_{i\in\mathbb{N}}\psi_i(p)=1, \forall p\in U$.

Our global smooth vector field is $X(p)\sum_{i\in J}\psi_i(p), p\in U$ and $\underline 0$ for $p\notin V$;

but I have a problem to extend $X$ in $V-U$ because $V$ may not be a domain of chart (in these case it will be trivial) but it could be the non-disjoint union of two domain of charts. Apologize me if I'm writing wrong things The answer to my first question is NO, see the comment. Now I'm asking when is possible to extend a vector field defined on a non-dense subset of a manifold in all the whole manifold?

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    $\begingroup$ I suggest, you first think about functions of one real variable: If $U$ is an open and dense subset of ${\mathbb R}$ and $f: U\to {\mathbb R}$ is a smooth function, does it follow that $f$ has a continuous extension to the entire real line? If you can find an example where an extension is impossible, can you interpret your example as a vector field? $\endgroup$ – Moishe Kohan Oct 15 '19 at 20:38
  • $\begingroup$ There is a tacit hypotesis: $M$ and $U$ is connected... let $frac$ be the mantissa function $frac:\mathbb{R}-\mathbb{N}\to\mathbb{R}$ is a regular function which could be extended in two different ways, right countinuously or left continuously, but I'm thinking that I could make this thanks to the fact that my domain is not connected $\endgroup$ – Elia Oct 15 '19 at 21:28
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    $\begingroup$ Good example, now generalize it to the case when $M=S^1$ or $M=R^2$ and $U$ is the complement to a point. $\endgroup$ – Moishe Kohan Oct 15 '19 at 21:29
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You should look for counter examples (for instance what if $U$ is the whole manifold with only one point removed ?). You might see problems appear (for instance just by taking the open disk in $\mathbb{R}^2$ and by removing its center).

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  • $\begingroup$ If $M=S^2$ and $U=M−\{N,S\}$ the sphere without the "north and south pole", and $X$ is the tangent vector field on $U$, for the hairy ball theorem I can't find a global nonvanishing smooth tangent field on $M$. Are you talking about this? $\endgroup$ – Elia Oct 15 '19 at 20:33
  • $\begingroup$ I was thinking of something even easier. You take the open disk in $\mathbb{R}^2$ centered in 0 and look at it as a manifold (any open set of a real vector space of finite dimension is trivially a manifold). Then taking away its center you can define a vector field on it "rotating around the center" where each vector is of norm 1. Then you cant extend it on the center (try to prove it formally if you want). The idea is that the vector in the tangent space of the center can't be zero (by continuity), and choosing a direction would also contradict smoothness of the vector field. $\endgroup$ – Popyaitte Oct 15 '19 at 21:45
  • $\begingroup$ thanks, yes, your first hint suggest me to think exactly that result... Are there any additional hypotesis to make my congecture generally true? Because in my notes there is an osservation which said that we could extend every vector field defined on a open set of a manifold, to all the manifold. Probably they are wrong or they are incomplete... $\endgroup$ – Elia Oct 16 '19 at 0:52
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Here is the correct statement: Suppose that $M$ is a smooth manifold, $U\subset M$ is an open subset and $X\in {\mathfrak X}(U)$ a (smooth) vector field. Then for every compact $K\subset U$ there exists a smooth vector field $Y$ on $M$ whose restriction to $K$ equals that of $X$. If this is what your notes use, I will write a proof. If not, think of the following example: $M={\mathbb R}^2$, $U=M -\{0\}$, $X=\frac{1}{x^2+y^2}\partial_x$. Then $X$ does not extend to $M$.

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  • $\begingroup$ Thank you, I know how to prove the statement you write with the cut-off/bump function. I'm studying only real manifold, is possible that the problem is only in $\mathbb{C}$? I'm thinking that my notes are incomplete and your statement may be one correct statement... In my notes is written to use the partition of unity, but with them i could only prove that if $U$ is an open (or closed) subset of a chart domain $U_{\alpha}$, so I could extend a vector field on $U$ in a vector field on the whole manifold whose restriction to $U$ equals that of the original vector field on $U$. $\endgroup$ – Elia Oct 16 '19 at 10:01
  • $\begingroup$ @Elia: Complex is irrelevant (see the edit). Having $U$ an open subset in a chart domain also does not help since in my example $U$ is a chart. $\endgroup$ – Moishe Kohan Oct 16 '19 at 14:54
  • $\begingroup$ I mean a open strictly subset of a domain of a chart, because in this case every smooth vector field $X$ on $U$ is $\sum_{i\in\{1,...,n\}} X^i \cdot \frac{\partial}{\partial \varphi_i} $ where $X^i:M\to\mathbb{R}$ are the component of the vector field, and $\{(U_{\alpha},\varphi=(\varphi_1,...,\varphi_n))\}$ is a chart such that $U\subset U{\alpha}$ proper subset, in your example $U=U_{\alpha}$ $\endgroup$ – Elia Oct 16 '19 at 22:59
  • $\begingroup$ @Elia: I suggest you edit your question since I do not know what you mean by a "strictly subset". In my example, ${\mathbb R}^2$ is a chart and $U$ is a strictly smaller subset. $\endgroup$ – Moishe Kohan Oct 16 '19 at 23:01
  • $\begingroup$ I mean a non-dense subset of the chart domain. Your example is the correct answer of my first question, now I'm thinkig what my notes wanna say. Sorry for the misunderstanding $\endgroup$ – Elia Oct 16 '19 at 23:13

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