I'm trying to show that the tangent space $T_xX$ of a manifold $X$ at $x$ is well-defined. Could you have a check on my proof?

First, I recall related definitions to remove ambiguity.

  • A subset $X \subseteq \mathbb{R}^{N}$ is called a smooth $n$-dimensional manifold if $\forall x \in X$, $\exists$ a diffeomorphism $\varphi: U \to V$ such that $V$ is open in $X$, $U$ is open in $\mathbb{R}^{n}$, and $x \in V$. Then $\varphi$ is called a local parameterization of $V$. The inverse $\varphi^{-1}$ is called a local coordinate system, or chart, on $V$.

  • Let $X \subseteq \mathbb R^N$ be a $n$-dimensional smooth manifold and $x \in X$. Let $\varphi: U \to V$ be a local parameterization around $x \in V$. The continuous linear map $\mathrm d \varphi_{\varphi^{-1}(x)} : \mathbb R^n \to \mathbb R^N$ is the Fréchet derivative of $\varphi$ at $\varphi^{-1}(x) \in U$. The tangent space of $X$ at $x$, denoted by $T_xX$, is defined as the image of $\mathrm d \varphi_{\varphi^{-1}(x)}$, i.e., $$T_xX := \operatorname{im} \left (\mathrm d \varphi_{\varphi^{-1}(x)} \right ).$$

Theorem: Let $X \subseteq \mathbb R^N$ be a $n$-dimensional smooth manifold and $x \in X$. Then the tangent space $T_xX$ is well-defined, i.e. it's independent from the choice of local parameterization.

Proof: Let $\varphi_1: U_1 \to V_1$ and $\varphi_2: U_2 \to V_2$ be two local parameterizations around $x$. Let $y_1 := \varphi^{-1}_1(x)$ and $y_2 := \varphi^{-1}_2(x)$. Let $V := V_1 \cap V_2$, $U_1 := \varphi_1^{-1}(V)$, and $U_2 := \varphi_2^{-1}(V)$. Then $V$ is open in $X$. Also, $U_1$ and $U_2$ are open both in $U$ and in $\mathbb R^n$. Also, $y_1 \in U_1$ and $y_2 \in U_2$.

Let $\phi_1: U_1 \to V$ and $\phi_2: U_2 \to V$ be the restrictions of $\varphi_1$ on $U_1$ and $\varphi_2$ on $U_2$ respectively. Then they are also local parameterizations around $x$. Let $\psi := \phi_1^{-1} \circ \phi_2$. Then $\psi^{-1} \circ \psi = \operatorname{id}_{U_2}$ and $\psi \circ \psi^{-1} = \operatorname{id}_{U_1}$. Notice that $\psi$ is a diffeomorphism from $U_2$ to $U_1$, so $\mathrm d \psi (y_2)$ and $\mathrm d \psi^{-1} (y_1)$ are bijective and thus automorphisms of $\mathbb R^n$. This in turn implies $$\operatorname{im} (\mathrm d \phi_2 (y_2) \circ \mathrm d \psi^{-1} (y_1)) = \operatorname{im} (\mathrm d \phi_2 (y_2)).$$

On the other hand, $\phi_1 = \phi_2 \circ \psi^{-1}$. By chain rule, $$\mathrm d \phi_1 (y_1) = \mathrm d \phi_2 (y_2) \circ \mathrm d \psi^{-1} (y_1).$$

The result then easily follows.


1 Answer 1


Your proof is correct. Just a minor suggestion: You use the fact that if $\varphi : U \to V$ is a local parameterization around $x \in V$ and $\phi : U' \to V'$ is a local local parameterization around $x \in V'$ which is a restriction of $\varphi$, then $d\varphi_{\varphi^{-1}(x)} = d\phi_{\phi^{-1}(x)}$. This is quite obvious but should perhaps be mentioned.

  • $\begingroup$ Thank you so much. That point is important. It justifies the transfer from original parameterizations to new parameterizations whose images are the same. This in turn allows us to obtain a diffeomorphism from $U_1$ and $U_2$. All the good things arise from this diffeomorphism. $\endgroup$
    – Akira
    Nov 22, 2021 at 18:31

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