The exponential map

The steering problem will be solved by performing calculations on $L_k(y_1,\ldots,y_m)$. The formal power series of $A(y_1,\ldots,y_m)$ is the set of all linear combinations of monomials, including those that have an infinite number of terms. Similarly, the formal Lie series of $L(y_1,\ldots,y_m)$ can be defined.

The formal exponential map is defined for any $p \in A(y_1,\ldots,y_m)$ as

$$e^{p} = 1 + p + \frac{1}{2!} p^2 + \frac{1}{3!} p^3 + \cdots .$$ (15.116)

In the nilpotent case, the formal exponential map is defined for any $p \in A_k(y_1,\ldots,y_m)$ as

$$e^{p} = \sum_{i=0}^k \frac{p^i}{i!} .$$ (15.117)

The formal series is truncated because all terms with exponents larger than $k$ vanish.

A formal Lie group is constructed as

$$G_k(y_1,\ldots,y_m) = \{ e^p \;\vert\; p \in L_k(y_1,\ldots,y_m) \} .$$ (15.118)

If the formal Lie algebra is not nilpotent, then a formal Lie group $G(y_1,\ldots,y_m)$ can be defined as the set of all $e^p$, in which $p$ is represented using a formal Lie series.

The following example is taken from [574]:

Example 15..22 (Formal Lie Groups)   Suppose that the generators $x$ and $y$ are given. Some elements of the formal Lie group $G(x,y)$ are

$$e^x = I + x + \begin{matrix}\frac{1}{2} \end{matrix} x^2 + \begin{matrix}\frac{1}{6} \end{matrix} x^3 + \cdots ,$$ (15.119)

$$e^{[x,y]} = I + [x,y] + \begin{matrix}\frac{1}{2} \end{matrix} [x,y]^2 + \cdots ,$$ (15.120)

and

$$e^{x - y + 3[x,y]} = I + x - y + 3 [x,y] + \cdots ,$$ (15.121)

in which $I$ is the formal Lie group identity. Some elements of the formal Lie group $G_2(x,y)$ are

$$e^x = I + x + \begin{matrix}\frac{1}{2} \end{matrix} x^2 ,$$ (15.122)

$$e^{[x,y]} = I + [x,y] ,$$ (15.123)

and

$$e^{x - y + 3[x,y]} = I + x - y + 3 [x,y] + \begin{matrix}\frac{1}{2} \end{matrix} (x - y)^2 .$$ (15.124)

$\blacksquare$

To be a group, the axioms given in Section 4.2.1 must be satisfied. The identity is $I$, and associativity clearly follows from the series representations. Each $e^p$ has an inverse, $e^{-p}$, because $e^p e^{-p} = I$. The only remaining axiom to satisfy is closure. This is given by the Campbell-Baker-Hausdorff-Dynkin formula (or CBHD formula), for which the first terms for any $p, q \in G(y_1,\ldots,y_m)$ are

$$\operatorname{exp}(p) \operatorname{exp}(q) = \operatorname{exp}(... ...[p,q],p] + \begin{matrix}\frac{1}{24} \end{matrix} [p, [q, [p,q]]] + \cdots ) ,$$ (15.125)

in which $\operatorname{exp}(x)$ alternatively denotes $e^x$ for any $x$. The formula also applies to $G_k(y_1,\ldots,y_m)$, but it becomes truncated into a finite series. This fact will be utilized later. Note that $e^p e^q \not = e^{p + q}$, which differs from the standard definition of exponentiation.

The CBHD formula is often expressed as

$$e^p e^q e^{-p} = \operatorname{exp}\left(\sum_{i=0}^\infty \frac{\operatorname{Ad}^i_p q}{i!}\right) ,$$ (15.126)

in which $\operatorname{Ad}^0_p q = q$, and $\operatorname{Ad}^i_p q = [p,\operatorname{Ad}^{i-1}_p q]$. The operator $\operatorname{Ad}$ provides a compact way to express some nested Lie bracket operations. Additional terms of (15.125) can be obtained using (15.126).