The meaning of the notation $\frac{\mathrm d \gamma^\chi}{\mathrm d t}(0)$ I'm reading about how a smooth curve is related to a tangent vector from this thread, i.e.,

Given an $m$-dimensional manifold $M$ and $p \in M$, a tangent vector of $M$ at $p$ is any function
$$
v:\{\text{charts of } M \text { around } p\} \rightarrow \mathbb{R}^m, \quad \chi \mapsto v^\chi,
$$
with the property that, for any two charts $\chi$ and $\chi^{\prime}$ around $p$ one has
$$
v^{\chi^{\prime}} = \mathrm d c_{\chi, \chi^{\prime}} (\chi(p)) [v^\chi], \qquad \qquad (1.2)
$$
where $c_{\chi, \chi^{\prime}} :=\chi' \circ \chi^{-1}$ is the change of coordinates from $\chi$ to $\chi^{\prime}$. We denote by $T_p M$ the vector space of all such tangent vectors of $M$ at $p$ (a vector space using the vector space structure on $\mathbb{R}^m$, i.e. $(v+w)^\chi:=v^\chi+w^\chi$, etc).
For $p \in M$, we define
$$
\operatorname{Curve}_p (M) := \{\gamma: (-\epsilon, \epsilon) \to M \text{ smooth such that } \epsilon>0, \gamma (0)=p\}.
$$
Let $\gamma \in \operatorname{Curve}_p (M)$ and $\chi$ a chart around $p$. Let $\gamma^\chi := \chi \circ \gamma$ be the representation of $\gamma$ w.r.t. $\chi$. Consider the operation
$$
\frac{\mathrm d \gamma}{\mathrm d t} (0):\{\text {charts of } M \text { around } p\} \rightarrow \mathbb{R}^m, \quad \chi \mapsto \frac{\mathrm d \gamma^\chi}{\mathrm d t}(0).
$$
Then $\frac{\mathrm d \gamma}{\mathrm d t} (0) \in T_pM$ by chain rule.

My question Because $\frac{\mathrm d \gamma^\chi}{\mathrm d t}(0) = \frac{\mathrm d (\chi \circ \gamma)}{\mathrm d t}(0) \in \mathbb R^m$, I think the notation $\frac{\mathrm d (\chi \circ \gamma)}{\mathrm d t}(0)$ means
$$
\frac{\mathrm d (\chi \circ \gamma)}{\mathrm d t}(0)  := \mathrm d (\chi \circ \gamma)(0)[1]
$$
in the usual sense of Analysis with

*

*$\mathrm d (\chi \circ \gamma)(0)$ is the Fréchet derivative of $\chi \circ \gamma$ at $0$ and thus a continuous linear map from $\mathbb R$ to $\mathbb R^m$,

*$\mathrm d (\chi \circ \gamma)(0)[1]$ is the value of the map $\mathrm d (\chi \circ \gamma)(0)$ at $1$.


Could you confirm if my understanding is fine?

 A: This is just a notational issue. In elementary calculus one customarily writes
$$\frac{df}{dt}(x_0) = f'(x_0) \in \mathbb R$$
for the derivative of a function $f : J \to \mathbb R$ at $x_0 \in J$, where $J$ is an open interval. Similarly, for a function $f : J \to \mathbb R^m$ with coordinate functions $f_i : J \to \mathbb R$ one writes
$$\frac{df}{dt}(x_0) = f'(x_0) \in \mathbb R^m$$
where $f'(x_0)$ has coordinates $f'_i(x_0)$.
The function $\gamma^\chi$ is such a function:
$$\gamma^\chi = \chi \circ \gamma : (-\epsilon', \epsilon') \to \mathbb R^m ,$$
where $\epsilon'$ is small enough so that the $\gamma$ maps $(-\epsilon', \epsilon')$ into the domain of $\chi$. This explains the meaning of
$$\frac{d \gamma^\chi}{dt}(0) .$$
Alternatively one can work with Fréchet derivatives of functions $f : U \to V$ between open $U \subset \mathbb R^n$ and $V \subset \mathbb R^m$. The Fréchet derivative of $f$ at $x_0 \in U$ is a linear map
$$df(x_0) : \mathbb R^n \to \mathbb  R^m .$$
Notation is not standardized; for example you can also find

*

*$df \mid_{x_0}$

*$Df(x_0)$

*$Df \mid_{x_0}$
For $n = 1$ we have in fact
$$\frac{df}{dt}(x_0) = f'(x_0) = df(x_0)[1]. $$
