What is an example of a map not satisfying this rank condition? Definition:
Consider a Lie Group $G$ and a set of right invariant vector fields on $G$, denoted $\Gamma$.
A point $y \in G$ is called normally accessible from a point $x \in G$ by $\Gamma$ if there exist elements $A_1, \ldots, A_k$ in $\Gamma$ and $\hat{t} \in \mathbb{R}^k$ with positive coordinates $\hat{t_1},\ldots,\hat{t_k}$ such that the mapping $F(t_1,\ldots t_k) = e^{t_kA_k}\ldots e^{t_1A_1}x$ as a mapping from $\mathbb{R}^k$ into $G$ satisfies the following conditions:


*

*$F(\hat{t}) = y$

*The rank of the differential $dF$ at $\hat{t}$ is equal to the dimension of $G$.
Source:
This definition is taken from 'Controllability of Invariant Systems on Lie Groups and Homogeneous Spaces' by Yu Sachkov.
My aim: I am trying to gain an intuitive understanding of the second condition.
Questions:


*

*What is an example of a map of this form that does not satisfy condition 2?

*Is there an example of condition 2 being violated for $k = 1$?

 A: Rank
First note that the rank of the differential map $dF$ is the dimension of its image. It is an induced map between the tangent spaces of the domain and range of its associated map, $F$. 
See: https://en.wikipedia.org/wiki/Pushforward_%28differential%29
What is an example of a map of this form that does not satisfy condition 2?
Consider the case where $x$ lives in an $n$-dimensional manifold such that $n \neq 0$, and where we trivially pick $\Gamma$ such that $A_1,\ldots, A_k = \mathbf{0}$. We ask if, in this case, $x$ is normally accessible from $x$. 


*

*$F$ maps all time values in its domain to $x$, so we can find some positive times such that condition $1$ is satisfied. 

*The equivalence classes of curves in $\mathbb{R}^k$ are mapped to the single point $x$. Hence the  rank of the differential map is $0$. We already set $n \neq 0$ therefore condition $2$ is not satisfied.


$x$ is not normally accessible from $x$ by $\Gamma$ in this case.
Is there an example of condition 2 being violated for k=1?
Let $k=1$ and therefore consider the equation:
\begin{equation}
 F(t) = e^{tA}x
\end{equation}
where $t \in \mathbb{R}$; $A \in \Gamma$; $x \in \mathcal{M}$ and $\operatorname{dim}(\mathcal{M}) = n$. Rephrasing condition $2$ we have the question: Is the differential map $dF(t)$ surjective?
First define curves in $\mathbb{R}$:
\begin{equation}
\begin{aligned}
 \sigma_t &: (-\epsilon,\epsilon) \to \mathbb{R} \\
 \sigma_t &: a \mapsto \sigma_t(a)
\end{aligned}
\end{equation}
where $\epsilon > 0$ and $\sigma_t(a)$ are curves such that $\sigma_t(0) = t$.
The tangent map on an equivalence class of curves in $\mathbb{R}$ acts like
\begin{equation}
 dF[\sigma_t(a)] = [e^{tA}x \circ \sigma_t(a)] \quad \text{for any } [\sigma_t(a)] \in T_t\mathbb{R}^k.
\end{equation}
The composition of these two maps is $e^{\sigma_t(a)A}x$, i.e. a curve in $\mathcal{M}$. With one generator the equivalence class of curves will form a one dimensional subvector space in the target tangent space. Hence the differential map will not be surjective for $n > 1$.
