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

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

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 $$T_p M$$.

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.

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.

## 2 Answers

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.

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 $$\bar{f} = \sum_{i=1}^n g_i(p) \bar{x}_i.$$

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 $$(G_p / S_p)^*$$.

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.

• It is more or less the proof that I have in mind. Good job filling the details! Let me know if something remains unclear. – Dante Grevino Dec 11 '18 at 16:02
• Thank you @DanteGrevino :) I think I've finally got the hang of the definitions now. – Brahadeesh Dec 11 '18 at 16:03
• I think there is a typo. $m_p \cap S_p = {m_p}^2$ You have written $m_p + S_p = {m_p}^2$ – Error 404 Dec 12 '18 at 16:21
• @Error404 Thanks, corrected :) – Brahadeesh Dec 12 '18 at 16:25
• @Error404 In most contexts, yes, but in the problem I described in the "Further motivation" section I want to be able to say that $\frac{\partial}{\partial x_1}\big|_{q},\dots,\frac{\partial}{\partial x_n}\big|_{q}$ is a basis of $T_q M$ for all points $q \in U$, where $(U,\varphi)$ is a chart around $p$, and $\varphi$ describes coordinates $x_1,\dots,x_n$ in $U$ with $p \equiv 0$. I cannot do this without the general definition of $T_p M$ as $(G_p / S_p)^*$. – Brahadeesh Dec 13 '18 at 10:21