# Extension of vector field implies extension of function

Suppose that I have a well-defined smooth vector field $$Y$$ of the form $$\sum_i f_iX_i$$ on a submanifold $$S$$ of $$M$$ where each $$f_i$$ is a smooth function on $$S$$ and each $$X_i$$ is a smooth vector field on $$S$$.

Suppose also that I know $$f:= \sum_i f_i$$ cannot be continuously extended to all of $$M$$. Does this imply that $$Y$$ can also not be extended to all of $$M$$?

I don't know whether extending $$Y$$ would need to extend $$f$$ in an impossible way. Any help is appreciated.

No. Take $$M=\mathbb R$$, $$S = (0,\infty)$$, $$X_1 = x\partial/\partial x$$, and $$f_1 = 1/x$$. Then $$f = f_1$$ cannot be extended continuously to $$M$$, but $$Y = \partial/\partial x$$ can be.
Note that the key feature of this example is that $$S$$ is not closed in $$M$$. In fact, the hypotheses imply that $$S$$ cannot be a closed embedded submanifold of $$M$$, because if it were, then $$f$$ would necessarily have a continuous (in fact smooth) extension to all of $$M$$. (See Lemma 5.34 in my Introduction to Smooth Manifolds, 2nd ed.)
• Thank you. If the $X_i$ are obtained by extending the first coordinate vector fields to all of $M$ through a partition of unity argument, can we then ensure that $Y$ defined in my post extends only if $f$ does? At least this extra hypothesis would rule out your example. – CuriousKid7 Oct 25 '18 at 19:11
• It depends on what you mean by "extending the first coordinate vector fields through a partition of unity argument." In my example, $X_1$ is a coordinate vector field on $S$ (using $u = \log x$ as a global coordinate on $S$), and it does in fact extend smoothly to all of $M$. ... – Jack Lee Oct 25 '18 at 21:15
• On the other hand, if you can extend $X_1,\dots,X_n$ smoothly to all of $M$ in such a way that they're linearly independent everywhere, then the answer's yes: If $Y$ has a continuous extension to $M$, then it can be expressed uniquely on $M$ as a linear combination of $X_i$'s with continuous coefficients, and thus $f$ has a continuous extension. But extending the $X_i$'s this way is not always possible. – Jack Lee Oct 25 '18 at 21:16
• Thank you again. But to clarify, I just mean $\frac{\partial}{\partial{x}}$ when I say coordinate vector field. Your $X_1$ includes the factor $x$. Could you please clarify this small point? – CuriousKid7 Oct 26 '18 at 2:53
• @CuriousKid7: As I said in my comment, if you use the coordinate function $u = \log x$ on $S$, then my vector field $X_1$ is equal to $\partial/\partial u$. – Jack Lee Oct 26 '18 at 4:55