If $(U,\varphi)$ is a coordinate chart around $p \in M$, where $M$ smooth manifold, then how does $\varphi$ induce coordinates on $T_p M$?

Question

I am studying differential topology and I have some trouble understanding how coordinates are induced on the tangent space at any point.

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Let $M$ be an $n$-dimensional smooth manifold, and let $p \in M$. Let $(U,\varphi)$ be a coordinate chart around $p$, with coordinates $x_1,\dots,x_n$. The tangent space $T_p M$ is defined to be $(\mathfrak{m}_p/\mathfrak{m}_p^2)^*$, where $\mathfrak{m}_p$ is the $\Bbb{R}$-algebra of germs at $p$ that vanish at $p$.

Now, we have the following lemma (whose proof I omit).

Lemma. If $x_1,\dots,x_n$ are coordinates around $p \in M$ and $f \in \mathfrak{m}_p$, then there exist germs $g_1,\dots,g_n$ at $p$ such that $f = \sum_i x_i g_i$.

Using this we can show that the images of the coordinates in $\mathfrak{m}_p/\mathfrak{m}_p^2$ form a basis for this vector space. Indeed, we have the following:

1. $f \in \mathfrak{m}_p^2 \iff g_i \in \mathfrak{m}_p$ for all $i = 1,\dots,n$, since clearly $x_i \in \mathfrak{m}_p$ for all $i = 1,\dots,n$.
2. If $f = \sum x_i g_i \in \mathfrak{m}_p$, then $f = \sum x_i (g_i - g_i(0)) + \sum g_i(0) x_i \implies \bar{f} = \sum g_i(0) \bar{x}_i$. Here, the bar indicates the image in $\mathfrak{m}_p/\mathfrak{m}_p^2$.

Let $\frac{\partial}{\partial x_1},\dots,\frac{\partial}{\partial x_n}$ be the dual basis of $\bar{x}_1,\dots,\bar{x}_n$. Then, we say that the coordinates $x_1,\dots,x_n$ around $p$ induce the coordinates $\frac{\partial}{\partial x_1},\dots,\frac{\partial}{\partial x_n}$ on $T_p M$.

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If my understanding so far is correct, what this means is that I can talk about the coordinates $\frac{\partial}{\partial x_1},\dots,\frac{\partial}{\partial x_n}$ on $T_p M$ induced by $\varphi$ only when the coordinates around $p$ given by $\varphi$ are such that $\varphi(p) = 0$. Otherwise, in the above construction it is no longer true that the coordinates $x_1,\dots,x_n$ define germs at $p$ that vanish at $p$.

So, in particular this means that although $(U,\varphi)$ defines coordinates around every point $p' \in U$, $\varphi$ does not induce coordinates on $T_{p'} M$ if $\varphi(p') \neq 0$.

Is this true? I would appreciate any helpful comments and answers in this regard.

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Further motivation.

This line of thinking emerged while I was trying to solve the following problem.

Exercise. Let $f : M \to N$ be a smooth map between smooth manifolds. Show that the set of critical points of $f$ form a closed subset of $M$.

The natural method in my opinion is to show that the complement, that is the set of regular points, is open in $M$. So if $p \in M$ is a regular point, then $df_p : T_p M \to T_{f(p)} N$ is of maximal rank. If suitable coordinate charts $(U,\varphi)$ and $(V,\psi)$ are chosen around $p$ and $f(p)$, respectively, defining coordinates $x_1,\dots,x_m$ and $y_1,\dots,y_n$, respectively, then $df_p$ is represented by the Jacobian of $\tilde{f} \equiv \psi \circ f \circ \varphi^{-1}$ at the point $\varphi(p)$, with respect to the bases $\frac{\partial}{\partial x_1},\dots,\frac{\partial}{\partial x_m}$ on $T_p M$ and $\frac{\partial}{\partial y_1},\dots,\frac{\partial}{\partial y_n}$ on $T_{f(p)} N$. Now, I know that for every $\varphi(p')$ in a small enough neighbourhood around $\varphi(p)$, $\mathrm{Jac}(\tilde{f})_{\varphi(p')}$ is also of maximal rank.

At this point, I would like to be able to say that $\mathrm{Jac}(\tilde{f})_{\varphi(p')}$ represents the map $df_{p'} : T_{p'} M \to T_{f(p')} N$ with respect to the bases $\frac{\partial}{\partial x_1},\dots,\frac{\partial}{\partial x_m}$ on $T_{p'} M$ and $\frac{\partial}{\partial y_1},\dots,\frac{\partial}{\partial y_n}$ on $T_{f(p')} N$.

If I could, then I would be done. But, I went back to the definition of $\frac{\partial}{\partial x_1},\dots,\frac{\partial}{\partial x_m}$ as I've understood it above, and I'm not so sure that I am correct in saying so.

Answer 1

There are several equivalent definitions of the tangent space. Consider $C_p(M)$ the $\mathbb{R}$-algebra of germs at $p$ and define the tangent space of $M$ at $p$ as the dual space of the quotient $C_p(M)/(\mathfrak{m}_p^2+\mathbb{R})$, where the copy of $\mathbb{R}$ in the denominator is the vector subspace of constant functions. With this definition, the value of $\varphi(p)$ does not matter.

Answer 2

Let $G_p$ be the set of germs at $p$. Let $(U,\varphi)$ and $(V,\psi)$ be coordinate charts around $p$, with coordinates $x_1,\dots,x_n$ on $U$ and $y_1,\dots,y_n$ on $V$. If $f : M \to \Bbb{R}$ is a smooth function such that $$ \frac{\partial f}{\partial x_i}\bigg|_{p} = 0 \quad \text{ for all } \quad i = 1,\dots,n,\tag{$*$} $$ then $$ \frac{\partial f}{\partial y_i}\bigg|_{p} = 0 \quad \text{ for all } \quad i = 1,\dots,n. $$ Also, if $f_1,f_2 : M \to \Bbb{R}$ are smooth functions that agree on a small neighbourhood around $p$, then we have $$ \frac{\partial f_1}{\partial x_i}\bigg|_p = \frac{\partial f_2}{\partial x_i}\bigg|_p \quad \text{ for all } \quad i = 1,\dots,n. $$ Let $S_p$ be the subspace of stationary germs at $p$, that is, those germs at $p$ for which ($*$) hold. By the above remarks, $S_p$ is a well-defined subspace of $G_p$.

Now, suppose $f \in G_p$ and $x_1,\dots,x_n$ are coordinates around $p$ with $(\alpha_1,\dots,\alpha_n)$ the coordinates of $p$. Then, by the above Lemma, we have that $$ f(x) = f(p) + \sum_{i=1}^n (x_i - \alpha_i) g_i(x) = f(p) - \sum_{i=1}^n \alpha_i g_i(p) + \sum_{i=1}^n (x_i - \alpha_i) (g_i(x) - g_i(p)) + \sum_{i=1}^n g_i(p) x_i. $$ The first two terms of the right-most expression are constant germs and hence stationary germs. The third term is also a stationary germ, as can be seen by differentiating and applying the chain rule. Hence, if $\bar{f}$ denotes the image of $f$ in $G_p / S_p$, we have that $$ \bar{f} = \sum_{i=1}^n g_i(p) \bar{x}_i. $$

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So, we can still make sense of $\bar{x}_1,\dots,\bar{x}_n$ and its dual basis $\frac{\partial}{\partial x_1},\dots,\frac{\partial}{\partial x_n}$ when $p$ does not have coordinates $(0,\dots,0)$. Note that $G_p/S_p \cong \mathfrak{m}_p/(\mathfrak{m}_p \cap S_p)$ by the second isomorphism theorem, because $G_p = \mathfrak{m}_p + S_p$. Moreover, $\mathfrak{m}_p \cap S_p = \mathfrak{m}_p^2$. So, there is no trouble in defining the tangent space $T_p M$ to be $(G_p / S_p)^*$.

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I obtained some supplementary course notes that briefly mention $T_p^* M := G_p/S_p = \mathfrak{m}_p / \mathfrak{m}_p \cap S_p$, and $T_p M := (T_p^* M)^*$. I have only elaborated this in greater detail in my answer.

Also, the expression for $f(x)$ that I have obtained here shows that $S_p = \mathfrak{m}_p^2 + \Bbb{R}$, where $\Bbb{R}$ denotes the subspace of constant germs. So, this answer can also be considered an elaboration of the answer by @DanteGrevino.