# The equivalence class of a tangent vector does not depend on the choice of a chart $(U,\varphi)$.

Let $$M$$ be a smooth manifold of dimension $$n$$, and $$x\in M$$. I want to double check that I understand the definition of the tangent vector space $$T_xM$$ using velocity vectors correctly.

• We define $$T_xM$$ as a set of tangent vectors at $$x$$ where each tangent vector is the equivalence class of curves $$\gamma$$ that goes through point $$x$$ and share the same tangent vector. In other words, $$T_xM=\{\gamma^{\prime}(0)|\gamma:I\to M\text{ is a smooth path}\}$$ where $$\gamma^{\prime}(0)$$ is the equivalence class of curves $$\gamma:I\to M$$, and we say that $$\gamma_1\sim\gamma_2$$ iff $$(\varphi\circ \gamma_1)^{\prime}(0)=(\varphi\circ \gamma_2)^{\prime}(0)$$ for some choice of open chart $$(U,\varphi)$$ of $$M$$.

So, I have the following question: at this step, we need to show that our equivalence class does not depend on the choice of the chart $$(U,\varphi)$$, correct? In other words, let's take two charts $$(U,\varphi)$$ and $$(V,\psi)$$, then we want to show that $$\text{if }(\varphi\circ \gamma_1)^{\prime}(0)=(\varphi\circ \gamma_2)^{\prime}(0)\text{, then }(\psi\circ \gamma_1)^{\prime}(0)=(\psi\circ \gamma_2)^{\prime}(0)$$ i.e. if the curves are equivalent via one chart, then they are equivalent in another chart.

emphasized textTo show this, we need just to observe that $$(\psi\circ \gamma_1)^{\prime}(0)=(\psi\circ\varphi^{-1}\circ\varphi\circ\gamma_1)^{\prime}(0)=d(\psi\circ\varphi^{-1})_0((\varphi\circ\gamma_1)^{\prime}(0))=d(\psi\circ\varphi^{-1})_0((\varphi\circ\gamma_2)^{\prime}(0))=(\psi\circ \gamma_2)^{\prime}(0)$$ where $$\psi\circ\varphi^{-1}$$ is a diffeomorphism since $$\varphi$$ and $$\psi$$ are diffeomorphisms. Finally, as a step $$(1)$$, we defined a tangent space $$T_xM$$ as a set of equivalence class and we showed that those equivalence classes does not depend on the choice of the chart. So, for the step $$(2)$$, we need to show that $$T_xM$$ is actually a vector space.

• So what is your question? You correctly proved that the equivalence class does not depend on the choice of the chart - as in the title. Sep 19, 2020 at 11:14
• @PaulFrost, I just want to see if I understand correctly this way of defining the tangent space of a smooth manifold. There are a lot of different ways to define the tangent space. My goal is to understand the definition using the derivations and compare it to the other definitions. Thank you for your comment! Sep 20, 2020 at 20:08
• For point $(2)$ see my answer to math.stackexchange.com/q/3798409. Sep 20, 2020 at 23:36
• Sep 20, 2020 at 23:43