# Why is $f \in C^{\infty}(U)$ when working with this basis of $T_pM$

In Lee's Introduction to smooth manifolds page $$60$$ he writes:

First, suppose that $$M$$ is a smooth manifold and let $$(U,\varphi)$$ be a smooth coordinate chart on $$M$$. Then $$\varphi$$ is, in particular, a diffeomorphism from $$U$$ to an open subset $$\widehat{U} \subseteq \mathbb{R}^n$$. Combining propositions $$3.9$$ and $$3.6(d)$$ we see that $$d\varphi_p:T_pM \to T_{\varphi(p)} \mathbb{R}^n$$ is an isomorphism.

He then mentions the derivations $$\left.\frac{\partial}{\partial x^1}\right\vert_{\varphi(p)},...,\left.\frac{\partial}{\partial x^n}\right\vert_{\varphi(p)}$$ that are a basis for $$T_{\varphi(p)}\mathbb{R}^n$$. So the preimages of these under $$d\varphi_p$$ are a basis of $$T_pM$$ when we identify $$T_pM \cong T_pU$$ using the isomorphism $$d \iota_p$$, where $$\iota: U \to M$$ is the inclusion and similarly $$T_{\varphi(p)}U \cong T_{\varphi(p)}\mathbb{R}^n$$ and abuse notation by writing $$d \varphi_p$$ instead of the composition of these functions. The symbols $$\frac{\partial}{\partial x^i}\vert_p$$ are then defined by $$\left.\frac{\partial}{\partial x^i}\right\vert_p=(d\varphi_p)^{-1}\left(\left.\frac{\partial}{\partial x^i}\right\vert_{\varphi(p)}\right)=d(\varphi^{-1})_{\varphi(p)}\left(\left.\frac{\partial}{\partial x^i}\right\vert_{\varphi(p)}\right)$$ where again $$\varphi$$ is used to denote the composition of $$\varphi$$ with the inclusions. However, then he says

Unwinding the definitions, we see that $$\partial/\partial x^i \vert_p$$ acts on a function $$f \in C^{\infty}(U)$$ by $$\left.\frac{\partial}{\partial x^i}\right\vert_{p}f=\left.\frac{\partial}{\partial x^i}\right\vert_{\varphi(p)}(f \circ \varphi^{-1})$$

$$(1)$$ I wondered why the first centered equation holds. $$(d\varphi_p)^{-1}$$ should mean $$(d\widehat{\iota}_{\varphi(p)} \circ d\varphi_p \circ (d \iota_p)^{-1})^{-1}$$ so does $$d(\varphi_p)^{-1}$$ mean $$(d\widehat{\iota}_{\varphi(p)})^{-1} \circ d(\varphi_p)^{-1} \circ d\iota_p$$ where $$\widehat{\iota}: \varphi(U) \to \mathbb{R}^n$$ and $$\iota:U \to M$$?

$$(2)$$ Why is $$f \in C^{\infty}(U)$$ and not in $$C^{\infty}(M)$$? We identified in such a way that the preimages are supposed to be a basis for $$T_pM$$, thus being derivations $$C^{\infty}(M) \to \mathbb{R}$$. Is there anotoher identification used here implicitly? If so, why?

• when $f\in C^\infty(M)$, then by restriction, $f\in C^\infty(U)$ as well. But the $\frac{\partial}{\partial x^i}$ are only definend on $U$. Commented Jan 26, 2023 at 16:43
• @Thomas Why are they only defined on $U$? $d\varphi_p$ is defined on $T_pM$. Commented Jan 26, 2023 at 16:46
• ok, more precisely, they are only defined on the restriction of the tangent bundle to $U$, which is the union of the $T_pM$ with $p$ ranging over $U$. As for the why: the chart $\varphi$ is only defined on $U$. Commented Jan 26, 2023 at 18:46
• @Thomas I think my mistake lies around the first centered equation, where I write that $\varphi$ is supposed to mean $\varphi$ composed with the inclusions, so a function $M \to \mathbb{R}^n$. But this doesnt make any sense since the inclusions are not bijective in general. However, I then don't see why the inversion commutes with the differential. $(d\varphi_p)^{-1}$ should mean $(d \iota \circ d \varphi \circ (d\iota)_p)^{-1}$, if I am not mistaken. And this is a function $T_pM \to T_{\varphi(p)}\mathbb{R}^n$ as written in the first quote in the post. Commented Jan 26, 2023 at 19:30
• Yes. On $T_pM$ with $\mathbf{ p\in U}$. Again, $\varphi$ is only defined on $U$. You wrote that down yourself: "$\varphi$ is, in particular, a diffeomorphism from $U$ to ..." Commented Jan 26, 2023 at 20:17

What's going on here is that I'm making use of the identification between $$T_pU$$ and $$T_pM$$ given by Proposition 3.9, as explained at the bottom of page 56 and the top of page 57.
• Thanks. So when writing out the compositions explicitly as I did in $(1)$, you pretty much leave out the composition with $(d\iota_p)^{-1}$, right? However, strictly speaking, the $\frac{\partial}{\partial x^i}\vert_p$ should then be a basis of $T_pU$ and are only a basis of $T_pM$ when using the identification with $(d\iota_p)^{-1}$ as well, shouldnt they? In particular, the derivations of proposition 3.15 are composed with $(d\iota_p)^{-1}$. I hope i got this right. Again, thanks. Commented Jan 27, 2023 at 13:26
• @user3118: Yes. This is analogous to the way we use a coordinate chart $(U,\varphi)$ to identify points in $U$ with their images in Euclidean space (see the bottom of page 15). This kind of identification happens all the time in mathematics -- for example, complex numbers are commonly defined as ordered pairs of real numbers with a certain multiplication operation. Once that's done, there's a canonical embedding of $\mathbb R$ into $\mathbb C$ given by $x \mapsto (x,0)$. But we usually just identify elements of $\mathbb R$ with their images in $\mathbb C$, and don't mention the embedding. Commented Jan 27, 2023 at 15:52
• @user3118: Yes, in principle you could use $f\in C^\infty(M)$ and apply the partial derivative operator to the composition $\widehat f = f\circ \phi^{-1}$. Commented Jan 28, 2023 at 21:24