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I'm reading Semi-Riemannian Geometry by Newman - the following theorem

Let $M$ be a smooth manifold, and let $X$ be a vector field on $M$. Then $X\in \mathfrak{X}(M)$ iff for every chart $(U,(x^i))$ on $M$, the components of $X$ are in $C^{\infty}(U)$.

The proof for $\implies$ is given as follows:

Let $X=\sum_i\alpha^i(\partial/\partial x^i)$ be the local coordinate expression of $X$ w.r.t. $(U,(x^i))$, and let $p\in U$. Since $x^i\in C^{\infty}(U)$, (by some previous theorem) there is a function $\tilde x^i\in C^{\infty}(M)$ and a neighborhood $\tilde U\subseteq U$ of $p$ such that $\tilde x^i$ and $x^i$ agree on $\tilde U$. So $$X(\tilde x^i)=\bigg(\sum_j\alpha^j\frac{\partial}{\partial x^j}\bigg)(\tilde x^i)=\sum_j\alpha^j\frac{\partial\tilde x^i}{\partial x^j}=\alpha^i$$ on $\tilde U$. Since $X\in\mathfrak{X}(M)$ and $\tilde x^i\in C^{\infty}(M)$, $X(\tilde x^i)\in C^{\infty}(M)$. So $\alpha^i$ is smooth on $\tilde U$. Since $p$ was arbitrary, $\alpha^i$ is smooth on $U$.

Now a previous result states that the local coordinate expression for a $v\in T_p(M)$ w.r.t. a chart $(U, (x^i))$ covering $p$ is $$v=\sum_i v(x^i)\frac{\partial}{\partial x^i}\bigg|_p$$

I can translate this to $$X_p=\sum_i X_p(x^i)\frac{\partial}{\partial x^i}\bigg|_p=\sum_i X(x^i)(p)\frac{\partial}{\partial x^i}\bigg|_p=\bigg(\sum_i X(x^i)\frac{\partial}{\partial x^i}\bigg)(p) \\\implies X=\sum_i X(x^i)\frac{\partial}{\partial x^i}$$

So basically the local coordinates of a smooth vector field $X$ w.r.t. a chart $(U,(x^i))$ are $X(x^i)$. I'm trying to understand why the proof went through the extra step of considering another chart $(\tilde U, (\tilde x^i))$. Do we run into problems when trying to prove that $X(x^i)$ is smooth? Since both $X$ and $x^i$ are also smooth, it's not immediately obvious to me why we'd have a problem.

Shirish Kulhari
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  • why $\sum_{j}\alpha^{j}\frac{\partial\tilde{x}^{i}}{\partial x^{j}}=\alpha^{i}$ ? – monotone operator Feb 04 '23 at 11:41
  • @monotoneoperator: Because $\tilde x^i$ agrees with $x^i$ on $\tilde U$, and since $\partial x^i/\partial x^j=\delta^i_j$, the result in your first comment follows – Shirish Kulhari Feb 04 '23 at 11:48
  • Thank you very much, it's my default. – monotone operator Feb 04 '23 at 11:50
  • It is just a trick. I think you can do it with one chart, as you point out. We have the chart $(U,\varphi).$ So, using the projections $\pi_j:U\to \mathbb R,$ you get $\frac{\partial}{\partial x^i}\pi_j=\frac{\partial}{\partial r^i}\pi_j\circ \varphi^{-1}=\frac{\partial}{\partial r^i}r^j=\delta_i^j.$ But maybe the best way to do this is to note that $X$ smooth means that it is smooth as a section of the tangent bundle. But $(U,\varphi)$ induces a chart on the tangent bundle defined by the local trivialization. Then, the claim follows: it's just a matter of unwinding the definitions. – Matematleta Feb 04 '23 at 15:31

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