Example 1: Simplification of Lie algebraic expressions

To explain some of the capabilities of the package we consider a few examples. To simplify the following expression $x:=[\alpha X_3,[\alpha X_2, (\alpha+\beta^2) X_1 ] ] )$, in which $\alpha$ and $\beta$ are considered to be symbolic scalars, the function y:=simpLB(x) is invoked and returns the result: $(\alpha^3+\alpha^2\beta^2) [X_3,[X_2,X_1]]$, as well as, but separately, the scalar part of it, $(\alpha^3+\alpha^2\beta^2)$, and the Lie monomial $[X_3,[X_2,X_1]]$. Such an answer form facilitates further calculations; for example when the expression needs to be rewritten in terms of elements of the basis $B$. The latter can be accomplished by subsequently invoking the function phbize(y[3]), which acts on the third argument of the result. Another example, where skillful simplification is essential, is provided by the composition of exponential mappings $e^{Z_1}e^{Z_2}=e^{Z_3}$, with $Z_1$ and $Z_2$ declared as two simple Lie polynomials: $Z_1=a_1X_1+a_2X_2+a_3X_3$, $Z_2=b_1X_1+b_2X_2+b_3X_3$, and with $a_i$, $b_i$, $i=1,2,3$, declared as symbolic scalars. Employing the CBH formula in Dynkin's form, (3), a truncation of the series for $Z_3$ involving brackets up to order $n=4$ is obtained by invoking first the function cbhexp($Z_1$,$Z_2$,n). This produces a complicated expression involving 231 Lie products of indeterminates, which are further simplified by executing the function reduceLB($Z_3$,$B$). This reduces $Z_3$ into its expression in the Hall basis $B$ which, in this particular case, counts only 29 components. The first 12 terms of the final result are shown below:

$$\begin{eqnarray*} Z_3 &\texttt{:=}& (a_1 +b_1)X_1 + (a_2 +b_2)X_2 + (a_3 +b_3... ... -b_3 a_1 b_2 -a_3 a_2 b_1 +b_1 a_2 b_3)[X_3,[X_1,X_2]] +\ldots \end{eqnarray*}$$