# As a derivation, a tangent vector is independent of the chart.

Lee's Smooth Manifolds primarily defines tangent vectors as derivations. That is, functions $$v: C^\infty(M) \to \mathbb{R}$$ that satisfy the Leibniz rule.

By appealing to a chart $$(U, \phi)$$, we can write down a derivation $$v$$ concretely as a linear combination of basis vectors given by $$d\phi^{-1}_{\phi(p)} \frac{\partial}{\partial x_i}$$.

When we use a different chart, we of course have a different representation.

In order to compute how the derivation acts on a function $$f \in C^\infty(M)$$, I believe we have the following procedure:

1. Write down $$f \circ \phi^{-1}$$, the coordinate representation of $$f$$.
2. Compute $$v(f) = v(f \circ \phi^{-1} \circ \phi) = d\phi_p (v)(f \circ \phi^{-1})$$. The RHS is simply applying partial derivatives in the ordinary sense, and we are done.

Ah! I seem to have resolved my question in the course of writing it. I wanted to ask why it was obvious that, if you write down $$v$$ with respect to different bases, that computing $$v(f)$$ gets you the same result (i.e. is well defined), but that is just bullet point 2. (right?)

Apologies for frivolous questions! Everything is largely still an indistinct mess of words and symbols for me, so I just want to make sure I understand things in plodding detail. I find a lot of the time that while I can both mechanically compute and regurgitate definitions accurately, I have little understanding of what I am doing all the same.

• Maybe these articles (5 parts) help. It is a mess when it comes to differentiation. physicsforums.com/insights/the-pantheon-of-derivatives-i It is all about perspectives! You always have to figure out what are the variables and what are the parameters of the different perspectives. On manifolds, we additionally have varying coordinates in varying charts. The basic procedure is: go from a chart to $\mathbb{R}^n$ then differentiate and finally return to a chart. Well-definition comes from the requirement that overlapping charts must be compatible. Apr 23 at 23:28

• (1) If you have a tangent vector $$v$$ at $$p$$ then it is correct that - for some smooth $$f$$ defined in a neighbourhood of $$p$$ (for me at least, everything here should be defined in the sheafy way and we should look at germs of functions) - then $$v(f)$$ may be computed with some chart $$(U;\phi)$$ by expressing the derivation $$C^\infty(\Bbb R^n)\overset{\phi^\ast}{\to}C^\infty(U)\overset{v}{\to}\Bbb R$$ in the canonical basis $$\partial_i$$, $$i=1,2,\cdots,n$$ and evaluating this derivation at the function $$f\circ\phi^{-1}$$. By definition, in fact, since $$(v\phi^\ast)(f\circ\phi^{-1})=v(f\circ\phi^{-1}\circ\phi)=v(f)$$. Here $$U$$ should be an open neighbourhood of $$p$$ contained in the domain of definition of $$f$$. To calculate $$v\phi^\ast$$ in the canonical basis, you just need to find the value of $$v(x_j\circ\phi)$$ for $$j=1,2,\cdots,n$$, where "$$x_\bullet$$" denotes the relevant coordinate projection $$\Bbb R^n\to\Bbb R$$.
• (2) What if I want to construct such a $$v$$? Let me take some $$w:=\sum_{i=1}^n h_i\partial_i$$, a derivation on $$C^\infty(\Bbb R^n)$$ in local coordinates, where I have a particular choice of chart $$(U;\phi)$$ in mind, $$U$$ an open neighbourhood of $$p$$; I can certainly define a derivation $$C^\infty(U)\to\Bbb R$$ by $$v=w\circ(\phi^{-1})^\ast$$, and this is well-defined. Of course it is; $$w$$ is an unambiguous function, as is $$\phi$$. If I then applied the procedure of $$(1)$$ to this tangent vector $$v$$, then I could get back the exact same derivation $$w$$ - that's just because $$(\phi^{-1})^\ast\phi^\ast=1$$. The interesting question is what happens when I use a different chart $$(V;\psi)$$. Then you can check $$w$$ would transform according to the Jacobian of the transition map $$\psi\phi^{-1}$$.
• (3) What would not be well defined would be writing $$v=\sum_i h_i\partial_i$$ when you haven't specified what chart you're using. However, once you do, there is no issue. Just be mindful that if you use a different chart, the expression for $$v$$ in terms of the "$$\partial_i$$" would be different.