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Let $M$ be a smooth manifold, $X$ be a vector field on $M$ and $f\in C^{\infty}(M)$ be a smooth function on $M$. As obvious as it may sound, by $f\in C^{\infty}(M)$, I'm interpreting this as a map $M\to\mathbb{R}$ that sends any point $p\in M$ to only one unique real number $f(p)$ (because it's a map).

In this lecture, it's mentioned that $X$ is a map $C^{\infty}(M)\to C^{\infty}(M)$, which is the source of my confusion. This means that $Xf\in C^{\infty}(M)$, which in turn means it's a map that assigns each point $p\in M$ to a unique real number $(Xf)(p)$.

As an example, if I take $X=\partial_i$, then given a point $p$, $(\partial_if)(p)$ is the directional derivative of $f$ at $p$ in the direction of the $i$-th coordinate curve, which in turn depends on the chart we're choosing at $p$. The value of $(\partial_if)(p)$ is chart-dependent; the $(\partial_if)$ map fails to assign a unique real value to the point $p$.

So what's going on here? How do I reconcile this contradiction?

Edit: Also, the components of a vector field $V^i$ have a similar behavior. $V^i(p)$ is chart-dependent so it's not exactly a $C^{\infty}(M)$ map either. What kind of objects are $V^i$'s and $\partial_if$'s exactly?

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    That's because in general $\partial_i$ is NOT a vector fields as defined in the lecture. $\partial_i$ is defined only in a local coordinates. – Arctic Char Jul 31 '20 at 10:52
  • @ArcticChar: Ah okay! And the components of a vector field $V^i$ - by the same logic they are not $C^{\infty}(M)$ functions. What kind of objects are the $V^i$'s and $\partial_if$'s? I have also edited my question to include this query. Will be glad to accept your answer if you're interested in writing it down as an answer :) – Shirish Kulhari Jul 31 '20 at 11:26
  • Fredric Schuller is, absolutely, my favorite lecturer. So glad to see other people finding his content! – Taylor Rendon Jul 31 '20 at 12:41
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    @TaylorRendon: Yeah! His lectures are really good especially for those who are comfortable with learning abstract notions first and then see how those notions apply for particular cases. – Shirish Kulhari Jul 31 '20 at 12:46

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In the lecture it appears that "vector fields" implicitly refers to smooth vector fields. (A vector field is smooth if its component functions are smooth in all coordinate charts.)

Given a chart $\varphi:U\to\mathbb{R}^n$, the component functions of a smooth vector field $V^i$ can be thought of either as an element of $C^\infty U$ (a local section) or as an element of $C^\infty(\varphi(U))$ (a coordinate representative). It's common to identify points with their coordinate representations, so in practice there isn't much difference: writing $V^i(p)$ would refer to the former while $V^i(x^1,\dots,x^n)$ would refer to the latter.

However, it's a common abuse of notation (especially in GR) to use local coordinate representations as stand-ins for global objects. For instance, given a smooth function $f$ and a smooth vector field $V$, we could write $$Vf=V^i\partial_i f$$ which (pedantically) means that at any point $p\in M$, $Vf(p)$ is equal to $(V^i\partial_i f)(p)$ with respect to some (and thus any) coordinate chart conatianing $p$. Equivalently, using any coordinate chart to define the right side, both sides agree on their common domain.

This also gives a quick way of showing that $Vf$ is smooth: the right side is smooth because partial derivatives and products of smooth functions are smooth. This argument works because everything we're talking about is local; we can establish smoothness on each neighborhood separately. When dealing with global statements like integration or solving PDEs, this abuse of notation becomes much more dangerous.

Kajelad
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  • You said that we can establish smoothness on each neighborhood separately. Since it's a smooth manifold, I'm guessing chart smoothness of $V^i$ and $\partial_i$ extends to smoothness of $Vf$ on the entire manifold? – Shirish Kulhari Jul 31 '20 at 12:35
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    Yes: For a fixed chart $\varphi$ with domain $U$, the equality $Vf=V^i\partial_i f$ only holds on $U$, and so we can only conclude $Vf$ is smooth on $U$. However, the same argument applies to every other chart, and since we can cover the manifold with charts, we can conclude $Vf$ is smooth on all of $M$. – Kajelad Jul 31 '20 at 12:41