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.)

  • $\begingroup$ 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. $\endgroup$ – CuriousKid7 Oct 25 '18 at 19:11
  • $\begingroup$ 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 $M$. ... $\endgroup$ – Jack Lee Oct 25 '18 at 21:15
  • $\begingroup$ 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. $\endgroup$ – Jack Lee Oct 25 '18 at 21:16
  • $\begingroup$ 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? $\endgroup$ – CuriousKid7 Oct 26 '18 at 2:53
  • 1
    $\begingroup$ @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 $\partial/\partial u$. $\endgroup$ – Jack Lee Oct 26 '18 at 4:55

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