# Clarification in definition tangent vector

Let $$M$$ be a smooth manifold.

A tangent vector of $$M$$ at $$p$$ is an equivalence class $$[\gamma]$$ of smooth curves $$\gamma: (-\epsilon, \epsilon) \to M$$ with $$\gamma(0) =p$$, where $$\gamma_1 \sim \gamma_2$$ means that there is a chart $$\phi$$ around $$p$$ such that $$(\phi\circ \gamma_1)'(0) = (\phi\circ \gamma_2)'(0)$$.

Question: Is $$\epsilon > 0$$ fixed? Or is possible that $$\gamma_1 \sim \gamma_2$$ where the two curves have a different domain? I guess it's the latter, but wanted to be sure.

• Yes, the domains may be different. – Ted Shifrin Oct 8 '19 at 16:40
• How are we even sure that the composition $\phi \circ \gamma_1$ is defined? Nothing says that $\gamma_1(-\epsilon, \epsilon) \subseteq \operatorname{dom} \phi?$ – user661541 Oct 8 '19 at 16:41
• You shrink the $\epsilon$s to make sure the images of both $\gamma_i$ are contained in the chart. After all, all you care about is differentiating at $0$. – Ted Shifrin Oct 8 '19 at 16:42
• Thank you! Much appreciated! – user661541 Oct 8 '19 at 16:47
• So do we need to modify the definition so that it says that there is a chart $\phi$ around $p$ around $p$ such that $\phi \circ \gamma_1, \phi \circ \gamma_2$ are defined and that the derivatives at $0$ are equal, or do we say that there is a chart $\phi$ around $p$ such that $\phi\circ \gamma_1, \phi_\circ \gamma_2$ are defined when restricted to an appropriate smaller domain and such that the derivatives are equal at $0$? – user661541 Oct 8 '19 at 17:01

The intuitive picture of "curves having the same velocity at a given point" is pretty clear, but, as you point out, there are some technical obstacles to making it precise. There are a number of equivalent constructions. Here's one recipe (among many others):

Choose $$p\in M$$.

Choose a chart $$\varphi:U\to \mathbb{R}^m$$ containing $$p$$.

Consider the set of curves $$\gamma:I\to M$$ where $$I\subseteq\mathbb{R}$$ is an interval containing $$0$$ and $$\gamma(0)=p$$.

Define an equivalence as follows: Given $$\gamma_1:I_1\to M$$, and $$\gamma_2:I_2\to M$$ as above, say $$\gamma_1\sim\gamma_2$$ if there are restrictions of domain to open intervals $$J_1\subseteq I_1$$, $$J_2\subseteq I_2$$ containing $$0$$ such that $$\gamma_1|_{J_1}$$ and $$\gamma_1|_{J_2}$$ both map into $$U$$ and $$\left.\frac{d}{dt}\varphi(\gamma_1|_{J_2}(t))\right|_{t=0}=\left.\frac{d}{dt}\varphi(\gamma_2|_{J_2}(t))\right|_{t=0}$$ in the vector calculus sense.

From then, it needs to be checked that these restrictions always exist, and that the equivalence is independent of the choice of restriction, and that the equivalence classes are independent of the choice of chart.

• Thanks for your answer! What do you mean "it must be checked that these restrictions always exist"? – user661541 Oct 9 '19 at 13:50
• That $J_1$ and $J_2$ always exist, i.e. it is always possible to restrict the domain of a curve (in the class of interest) to fall entirely within the chart domain $U$. This ensures that every curve belongs to a nontrivial equivalence class. – Kajelad Oct 9 '19 at 14:14
• Ah, this is because of continuity right? We have that $\gamma: I \to M$ is continuous, so $\gamma^{-1}(U)$ is open (since $U \subseteq M$ is open). Since $0 \in \gamma^{-1}(U)$, this means that there is $\epsilon > 0$ such that $(-\epsilon, \epsilon) \subseteq \gamma^{-1}(U)$ and then $\gamma(-\epsilon, \epsilon) \subseteq U$. – user661541 Oct 9 '19 at 19:59
• Yes. By "curve" I mean a smooth map from $I\subseteq\mathbb{R}$ to $M$.. – Kajelad Oct 9 '19 at 20:05
• Yes, so smooth implies continuity so my argument works? – user661541 Oct 9 '19 at 20:17