# The push forward $\gamma_{*}(\partial t)$ of $\partial t \in T_{t_0}\mathbb{R}$ is the tangent vector $\gamma'(t_0)$

I am trying to understand the following: Let $$\gamma: \mathbb{R} \supset I \to M$$ be a smooth curve (M is a smooth manifold). Then the push forward $$\gamma_{*}(\partial t)$$ of $$\partial t \in T_{t_0}\mathbb{R}$$ is the tangent vector $$\gamma'(t_0) \in T_{\gamma(t_0)}M$$, where $$\gamma'(t_0) \sim t \mapsto \gamma(t-t_0)$$.

We defined the pushforward of a function $$F:N \to M$$ at the point $$p$$ as $$F_{*}([\gamma]) = [F \circ \gamma]$$, where $$l \in [\gamma]$$ if $$\gamma(0)=l(0)=p$$ and $$\gamma_{\alpha}'(0) = l_{\alpha}'(0)$$ for any (thus every) given chart $$\phi_{\alpha}$$ and $$l_{\alpha}= \phi_{\alpha} \circ l$$.

Now if I follow this definition, then $$\gamma_{*}(\partial t) = \gamma_{*}([\partial t]) = [\gamma \circ \partial t]$$. This should be easy, but I have no idea how to go on. Can anyone give me a hint?

• Hint : What is a curve whose tangent vector is $\partial_t$ ? Commented May 8, 2022 at 21:24
• Oh, is it meant to be $\gamma_{*}(\partial t) = \gamma_{*}[t] = [\gamma(t)]=\gamma'(t)$, and the notation is to write the curves in $[]$ and the derivatives of the curves without $[]$? So $[t]=\partial t$? Commented May 8, 2022 at 21:55
• Does this answer your question? Clarifying notation of derivative in differential geometry Commented May 9, 2022 at 0:36
• @ Kurt G. Your link explains the theory in common, but it is not explained why the push forward maps $\partial t$ to $\gamma'$ there, either. What I was missing was that $\partial t$ is the tangent vector of the equivalence class $t \mapsto t + t_0$, which was explained in the answer of SolubleFish. Commented May 9, 2022 at 10:39

From your notations, I am assuming that you are defining a tangent vector as an equivalence classes of curves. With this definition, $$\partial_t \in T_{t_0}\mathbb R$$ is the equivalence class of $$t\mapsto t+ t_0$$.
Given a smooth map $$f:N\to M$$ and a tangent vector $$X = [\gamma] \in T_p N$$, the pushforward is defined as the equivalence class of the composite $$F\circ \gamma$$ ie : $$f_*[\gamma] = [f\circ\gamma]$$
In our case, we have $$\gamma:\mathbb R\to M$$ and : $$\gamma_*\partial t = \gamma_*[t\mapsto t+t_0] =[t\mapsto \gamma(t+t_0)] = \gamma'(t_0)$$
• Thanks for your clear answer! The $-$ must but be a typo then. Commented May 9, 2022 at 10:41