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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.