A distribution15.8 expresses a set of vector fields on a smooth manifold. Suppose that a driftless control-affine system (15.53) is given. Recall the vector space definition from Section 8.3.1 or from linear algebra. Also recall that a state transition equation can be interpreted as a vector field if the actions are fixed and as a vector space if the state is instead fixed. For $ U = {\mathbb{R}}^m$ and a fixed $ x \in X$, the state transition equation defines a vector space in which each $ h_i$ evaluated at $ x$ is a basis vector and each $ u_i$ is a coefficient. For example, in (15.54), the vector fields $ h_1$ and $ h_2$ evaluated at $ q
= (0,0,0)$ become $ [1 \;\; 0 \;\; 0]^T$ and $ [0 \;\; 0 \;\; 1]^T$, respectively. These serve as the basis vectors. By selecting values of $ u \in {\mathbb{R}}^2$, a 2D vector space results. Any vector of the form $ [a \; \; 0 \;\; b]^T$ can be represented by setting $ u_1 = a$ and $ u_2 = b$. More generally, let $ {\triangle}(x)$ denote the vector space obtained in this way for any $ x \in X$.

The dimension of a vector space is the number of independent basis vectors. Therefore, the dimension of $ {\triangle}(x)$ is the rank of $ H(x)$ from (15.56) when evaluated at the particular $ x \in X$. Now consider defining $ {\triangle}(x)$ for every $ x \in X$. This yields a parameterized family of vector spaces, one for each $ x \in X$. The result could just as well be interpreted as a parameterized family of vector fields. For example, consider actions for $ i$ from $ 1$ to $ m$ of the form $ u_i = 1$ and $ u_j = 0$ for all $ i \not = j$. If the action is held constant over all $ x \in X$, then it selects a single vector field $ h_i$ from the collection of $ m$ vector fields:

$\displaystyle {\dot x}= h_i(x) .$ (15.63)

Using constant actions, an $ m$-dimensional vector space can be defined in which each vector field $ h_i$ is a basis vector (assuming the $ h_i$ are linearly independent), and the $ u_i \in {\mathbb{R}}$ are the coefficients:

$\displaystyle u_1 h_1(x) + u_2 h_2(x) + \cdots + u_m h_m(x) .$ (15.64)

This idea can be generalized to allow the $ u_i$ to vary over $ X$. Thus, rather than having $ u$ constant, it can be interpreted as a feedback plan $ \pi : X
\rightarrow U$, in which the action at $ x$ is given by $ u = \pi (x)$. The set of all vector fields that can be obtained as

$\displaystyle \pi _1(x) h_1(x) + \pi _2(x) h_2(x) + \cdots + \pi _m(x) h_m(x)$ (15.65)

is called the distribution of the set $ \{h_1,\ldots,h_m\}$ of vector fields and is denoted as $ {\triangle }$. If $ {\triangle }$ is obtained from an control-affine system, then $ {\triangle }$ is called the system distribution. The resulting set of vector fields is not quite a vector space because the nonzero coefficients $ \pi_i$ do not necessarily have a multiplicative inverse. This is required for the coefficients of a vector field and was satisfied by using $ {\mathbb{R}}$ in the case of constant actions. A distribution is instead considered algebraically as a module [469]. In most circumstances, it is helpful to imagine it as a vector space (just do not try to invert the coefficients!). Since a distribution is almost a vector space, the $ \operatorname{span}$ notation from linear algebra is often used to define it:

$\displaystyle {\triangle}= \operatorname{span}\{h_1,h_2,\ldots,h_m\} .$ (15.66)

Furthermore, it is actually a vector space with respect to constant actions $ u \in {\mathbb{R}}^m$. Note that for each fixed $ x \in X$, the vector space $ {\triangle}(x)$ is obtained, as defined earlier. A vector field $ f$ is said to belong to a distribution $ {\triangle }$ if it can be expressed using (15.65). If for all $ x \in X$, the dimension of $ {\triangle}(x)$ is $ m$, then $ {\triangle }$ is called a nonsingular distribution (or regular distribution). Otherwise, $ {\triangle }$ is called a singular distribution, and the points $ x \in X$ for which the dimension of $ {\triangle}(x)$ is less than $ m$ are called singular points. If the dimension of $ {\triangle}(x)$ is a constant $ c$ over all $ x \in X$, then $ c$ is called the dimension of the distribution and is denoted by $ \dim({\triangle})$. If the vector fields are smooth, and if $ \pi $ is restricted to be smooth, then a smooth distribution is obtained, which is a subset of the original distribution.

Figure 15.15: The distribution $ {\triangle }$ can be imagined as a slice of the tangent bundle $ T(X)$. It restricts the tangent space at every $ x \in X$.

As depicted in Figure 15.15, a nice interpretation of the distribution can be given in terms of the tangent bundle of a smooth manifold. The tangent bundle was defined for $ X = {\mathbb{R}}^n$ in (8.9) and generalizes to any smooth manifold $ X$ to obtain

$\displaystyle T(X) = \bigcup_{x \in X} T_x(X) .$ (15.67)

The tangent bundle is a $ 2n$-dimensional manifold in which $ n$ is the dimension of $ X$. A phase space for which $ x = (q,{\dot q})$ is actually $ T({\cal C})$. In the current setting, each element of $ T(X)$ yields a state and a velocity, $ (x,{\dot x})$. Which pairs are possible for a driftless control-affine system? Each $ {\triangle}(x)$ indicates the set of possible $ {\dot x}$ values for a fixed $ x$. The point $ x$ is sometimes called the base and $ {\triangle}(x)$ is called the fiber over $ x$ in $ T(X)$. The distribution $ {\triangle }$ simply specifies a subset of $ T_x(X)$ for every $ x \in X$. For a vector field $ f$ to belong to $ {\triangle }$, it must satisfy $ f(x) \in {\triangle}(x)$ for all $ x \in X$. This is just a restriction to a subset of $ T(X)$. If $ m = n$ and the system vector fields are independent, then any vector field is allowed. In this case, $ {\triangle }$ includes any vector field that can be constructed from the vectors in $ T(X)$.

Example 15..7 (The Distribution for the Differential Drive)   The system in (15.54) yields a two-dimensional distribution:

$\displaystyle {\triangle}= \operatorname{span}\{ [\cos\theta \;\; \sin\theta \;\; 0]^T, \; [0 \;\; 0 \;\; 1]^T\} .$ (15.68)

The distribution is nonsingular because for any $ (x,y,\theta)$ in the coordinate neighborhood, the resulting vector space $ {\triangle}(x,y,\theta)$ is two-dimensional. $ \blacksquare$

Example 15..8 (A Singular Distribution)   Consider the following system, which is given in [478]:

\begin{displaymath}\begin{split}\begin{pmatrix}{\dot x}_1  {\dot x}_2  {\dot...
...egin{pmatrix}x_1  x_1  0  \end{pmatrix} u_3 . \end{split}\end{displaymath} (15.69)

The distribution is

$\displaystyle {\triangle}= \operatorname{span}\{h_1,h_2,h_3\}.$ (15.70)

The first issue is that for any $ x \in {\mathbb{R}}^3$, $ h_2(x) = h_1(x) x_2$, which implies that the vector fields are linearly dependent over all of $ {\mathbb{R}}^3$. Hence, this distribution is singular because $ m = 3$ and the dimension of $ \Delta(x)$ is $ 2$ if $ x_1 \not = 0$. If $ x_1 =
0$, then the dimension of $ \Delta(x)$ drops to $ 1$. The dimension of $ {\triangle }$ is not defined because the dimension of $ \Delta(x)$ depends on $ x$. $ \blacksquare$

A distribution can alternatively be defined directly from Pfaffian constraints. Each $ g_i(x) = 0$ is called an annihilator because enforcing the constraint eliminates many vector fields from consideration. At each $ x \in X$, $ {\triangle}(x)$ is defined as the set of all velocity vectors that satisfy all $ k$ Pfaffian constraints. The constraints themselves can be used to form a codistribution, which is a kind of dual to the distribution. The codistribution can be interpreted as a vector space in which each constraint is a basis vector. Constraints can be added together or multiplied by any $ c \in {\mathbb{R}}$, and there is no effect on the resulting distribution of allowable vector fields.

Steven M LaValle 2012-04-20