What is an example of a map not satisfying this rank condition?

Question

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:

1. $F(\hat{t}) = y$

2. 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 $k = 1$?

Answer 1

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 $n > 1$.