1
$\begingroup$

Let $ M $ be a smooth manifold, and let $ X\colon M\to \mathrm TM $ be a vector field on $ M $.

If we pick a chart $ (U,\psi\colon U\to \underline U\subset \mathbb R^n) $ we can define a smooth local frame $ \partial_1,\dots,\partial_n\colon U\to \mathrm TM $ by $$ \partial_j(p) = \Bigl(p,\left.\frac{\partial}{\partial\phi^j}\right|_p\Bigr) $$ for every $ p\in U $, where $ \partial/\partial\phi^j|_p $ is the basis vector on $ \mathrm T_pM $ induced by the chart $ (U,\phi = (\phi^1,\dots,\phi^n)) $.

We can thus write $$ X(p) = \sum_{j = 1}^n a^j(p)\partial_j(p) $$ for $ p\in M $, where $$ a^j(p) = X_p(\phi_p^j) $$ is the $ j $-th coordinate of the vector $ X_p\in \mathrm T_pM $, where $ X(p) = (p,X_p) $.

I'm trying to show that the functions $$ a^j\colon U\to \mathbb R $$ are smooth, but I got lost in the formalism. Why are they smooth?

$\endgroup$
1
  • $\begingroup$ If you apply a smooth vector field (as a derivation) to a smooth function, you get a smooth function. $\endgroup$ Commented Oct 16, 2023 at 14:28

1 Answer 1

Reset to default
1
$\begingroup$

Here's an outline:

A chart $(\psi, U)$ for $M$ induces a chart $(\tilde{\psi}, \pi^{-1}(U))$ for $TM$, where $\pi: TM \to M$ is the projection map. Express $X: M \to TM$ in local coordinates using these charts. The resulting map between Euclidean spaces is smooth since $X$ is smooth. In particular, the last $n$ coordinates will be smooth - use this to conclude that your $\alpha^j$ are smooth.

If this doesn't help, see propositions 3.18 and 8.1 in Lee's book on smooth manifolds.

$\endgroup$
2
  • $\begingroup$ Thank you so much, it was easier than I thought! $\endgroup$ Commented Oct 17, 2023 at 12:42
  • $\begingroup$ @GeometriaDifferenziale you're welcome! $\endgroup$ Commented Oct 17, 2023 at 13:18

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.