15.4.3.2 Lie algebra of the system distribution

Now suppose that a set $ h_1$, $ \ldots $, $ h_m$ of vector fields is given as a driftless control-affine system, as in (15.53). Its associated distribution $ {\triangle }$ is interpreted as a vector space with coefficients in $ {\mathbb{R}}$, and the Lie bracket operation was given by (15.81). It can be verified that the Lie bracket operation in (15.81) satisfies the required axioms for a Lie algebra.

As observed in Examples 15.9 and 15.10, the Lie bracket may produce vector fields outside of $ {\triangle }$. By defining the Lie algebra of $ {\triangle }$ to be all vector fields that can be obtained by applying Lie bracket operations, a potentially larger distribution $ {\cal L}({\triangle})$ is obtained. The Lie algebra can be expressed using the $ \operatorname{span}$ notation by including $ h_1$, $ \ldots $, $ h_m$ and all independent vector fields generated by Lie brackets. Note that no more than $ n$ independent vector fields can possibly be produced.

Example 15..16 (The Lie Algebra of the Differential Drive)   The Lie algebra of the differential drive (15.54) is

$\displaystyle {\cal L}({\triangle}) = \operatorname{span}\{ [\cos\theta \;\; \s...
...;\; 0]^T, \; [0 \;\; 0 \;\; 1]^T, \; [\sin\theta \;\; -\cos\theta \;\; 0]^T\} .$ (15.101)

This uses the Lie bracket that was computed in (15.82) to obtain a three-dimensional Lie algebra. No further Lie brackets are needed because the maximum number of independent vector fields has been already obtained. $ \blacksquare$

Example 15..17 (A Lie Algebra That Involves Nested Brackets)   The previous example was not very interesting because the Lie algebra was generated after computing only one bracket. Suppose that $ X =
{\mathbb{R}}^5$ and $ U =
{\mathbb{R}}^2$. In this case, there is room to obtain up to three additional, linearly independent vector fields. The dimension of the Lie algebra may be any integer from $ 2$ to $ 5$.

Let the system be

$\displaystyle \begin{pmatrix}{\dot x}_1 \\ {\dot x}_2 \\ {\dot x}_3 \\ {\dot x}...
... \end{pmatrix} u_1 + \begin{pmatrix}0 \\ 1 \\ 0 \\ 0 \\ 0 \\ \end{pmatrix} u_2.$ (15.102)

This is a chained-form system, which is a concept that becomes important in Section 15.5.2.

The first Lie bracket produces

$\displaystyle [h_1,h_2] = [ 0 \;\; 0 \;\; -1 \;\; 0 \;\; 0 ]^T .$ (15.103)

Other vector fields that can be obtained by Lie brackets are

$\displaystyle [h_1,[h_1,h_2]] = [ 0 \;\; 0 \;\; 0 \;\; 1 \;\; 0 ]^T$ (15.104)

and

$\displaystyle [h_1,[h_1,[h_1,h_2]]] = [ 0 \;\; 0 \;\; 0 \;\; 0 \;\; 1 ]^T .$ (15.105)

The resulting five vector fields are independent over all $ x \in
{\mathbb{R}}^5$. This includes $ h_1$, $ h_2$, and the three obtained from Lie bracket operations. Independence can be established by placing them into a $ 5 \times 5$ matrix,

$\displaystyle \begin{pmatrix}1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ x_2 & 0 & -1 & 0 & 0 \\ x_3 & 0 & 0 & 1 & 0 \\ x_4 & 0 & 0 & 0 & 1 \\ \end{pmatrix} ,$ (15.106)

which has full rank for all $ x \in
{\mathbb{R}}^5$. No additional vector fields can possibly be independent. Therefore, the five-dimensional Lie algebra is

$\displaystyle {\cal L}({\triangle}) = \operatorname{span}\{h_1,\,h_2,\,[h_1,h_2],\,[h_1,[h_1,h_2]],\,[h_1,[h_1,[h_1,h_2]]]\} .$ (15.107)

$ \blacksquare$

Steven M LaValle 2020-08-14