The meaning of the notation $\frac{\mathrm d \gamma^\chi}{\mathrm d t}(0)$

Question

I'm reading about how a smooth curve is related to a tangent vector from this thread, i.e.,

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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.

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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?

Answer 1

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 $$\frac{df}{dt}(x_0) = f'(x_0) = df(x_0)[1]. $$