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I am dealing with smoothness issues for which I do not even know a successful approach, so any help or reference would be welcome.

Let $M$ be a manifold, $E\to M$ a smooth vector bundle, and let $\omega_1,\omega_2$ be smooth sections of $E$ (I am primarily interested in symmetric $(2,0)$-tensors).

  1. Assume that $\omega_1$ and $\omega_2$ are linearly indepent over $C^\infty(M)$. Then is it true that if $A_1, A_2$ are continuous functions such that $A_1\omega_1+A_2\omega_2$ is smooth, then $A_1, A_2$ must be smooth as well?
  2. Under the same assumptions on $\omega_1$, $\omega_2$, suppose now that $A_1$, $A_2$ are continuous and defined only on an open dense set $U\subset M$. Then, if $A_1\omega_1+A_2\omega_2$ (in principle only defined on $U$) can be extended to a smooth section on the whole of $M$, is it true that $A_1$, $A_2$ can be extended to smooth functions on the whole of $M$?

If there is even a reference to the type of results that could help in this situation, they would be highly appreciated!

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The answer to both questions is negative.

2 is a little simpler to see. Consider a trivial vector bundle of rank 2 on $\mathbb R$. Let $\omega_1=e_1$ and let $\omega_2 = x e_2$. Then $\omega_1 + x^{-1} \omega_2=e_1+e_2$ is defined on the open set $x\neq 0$ but extends to all of $\mathbb R$. However $x^{-1}$ does not extend to all of $\mathbb R$.

For 1, we just need to enhance the construction. Let $\omega_1=e_2, \omega_2=e^{-1/|x|} e_2$, $A_1 = 0, A_2 = |x|$. Certainly $A_2$ is continuous but not smooth. However, $A_1 \omega_1 + A_2 \omega_2 = |x| e^{- 1/|x|} e_2$ is smooth since $e^{-1/|x|}$ vanishes to arbitrarily high order.

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  • $\begingroup$ Thanks for the response! It seems that the counterexample to 1 hinges on the fact that $\omega_2$ vanishes to all orders. Would there be a counterexample if the $\omega_i$ were analytic? $\endgroup$ Commented Oct 23 at 9:35
  • $\begingroup$ @CuriousUser No. You'd want to fix a global analytic structure on your vector bundle for that to make sense. Then work locally and fix a basis for the vector bundle. If all the $2\times 2$ minors of the expression of $\omega_1,\omega_2$ in that basis vanish then $\omega_1$ and $\omega_2$ are linearly dependent everywhere by something that is locally a ratio of analytic functions, and thus, I think, globally a ratio of smooth functions, although I'm not completely sure of the step. If the 2 times 2 minor is nonvanishing then invert that 2 times 2 submatrix to conclude. $\endgroup$
    – Will Sawin
    Commented Oct 23 at 11:33
  • $\begingroup$ @CuriousUser The key point (other than that local-to-global issue) is that a smooth function divided by an analytic function, if it's defined at all, is still smooth. $\endgroup$
    – Will Sawin
    Commented Oct 23 at 11:35

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