wne, wner

 

  
Purpose 		 

Compute the right-hand side of the Wei-Norman formula:
$\sum_{k=1}^{r}X_k u_k = \dot{S}(t)S^{-1}(t)$, 
where $S(t) = \prod_{k=1}^{r} e^{g_k(t)X_k}$ and
$\dot{S}(t)=\frac{dS(t)}{dt}$.





Syntax 		 
   e:=wne(r,max_bracket_order,X);


e:=wner(r,max_bracket_order,X,B,lbt);




Description 		

Computes the right-hand side of the Wei-Norman formula:




$$\sum_{k=1}^{r}X_k u_k = \dot{S}(t)S^{-1}(t)~$$ 
(40)


where $u_k$ are scalar time-dependent functions,
i.e. $u_k:t\rightarrow \mathbb{R}$, $X_k$ are indeterminate elements
independent of time that define a basis for an arbitrary Lie algebra,
and with 




$$S(t) = \prod_{k=1}^{r} e^{g_k(t)X_k}$$ 
(41)


and 




$$\dot{S}(t)=\frac{dS(t)}{dt}=\sum_{i=1}^r
\dot{g}_i(t)\prod_{j=1}^{i-1}e^{g_j X_j}X_i\prod_{j=i}^r e^{g_j X_j}$$ 
(42)


The Lie algebra generated by the $X_k$ is required to be finite
dimensional.


The above expressions for $S(t)$ and $\dot{S}(t)$, together with the
exponential formula given in equation (9) allow to
express the right-hand side of the Wei-Norman equation as:




$$\dot{S}(t)S^{-1}(t)=
\sum_{i=1}^{r}\dot{g}_i(t)
\prod_{j=1}^{i-1}e^{g_j ad_{X_j}}X_i~$$ 
(43)


Hence, this procedure can be implemented in a simple way in terms of the
functions ead, eadr and pead, peadr.



wner, reduces the Lie brackets in the expression
to elements in the PHB, and further simplifies the expression 
according to the supplied Lie bracket table.  If the Lie bracket table
is an empty list or set, no additional simplifications are carried
out.





Arguments 		
 $r$  $\parbox{0.64\textwidth}{Dimension of the Lie
 algebra basis. In part...
...\ corresponds to the number of Lie brackets in the PHB
 which are independent.}$


 		 max_bracket_order


 		 		 $\parbox{0.64\textwidth}{The
 maximum bracket order in the exponentia...
...ion of
 $e^{ad_x}$; see \texttt{ead}, \texttt{eadr} on
 page~\pageref{ss:ead}.}$


 		 $X$ $\parbox{0.64\textwidth}{Label or name for the
 (indeterminate) eleme...
...xttt{wner(r,max\_bracket\_order,B,B,lbt)}, where \texttt{B}
 contains the PHB.}$




 		 wner additionally requires:


 		 $B$		 $\parbox{0.64\textwidth}{A Philip Hall basis.}$


 		 $lbt$		 $\parbox{0.64\textwidth}{A Lie bracket t...
...n empty list or set if no
 dependencies between the brackets in the PHB exist.}$




Examples 		 

The examples below consider the P. Hall basis B for a
nilpotent Lie algebra of degree four generated by three indeterminates
f0, f1 and f2, obtained in the example for
the function phb on page .  The maximum
bracket order in the series expansion of the exponential that will be
considered in the next examples is equal to three (i.e. the degree of
nilpotency minus one, so that the resulting brackets are always
contained in the PHB).

In the following two examples assume that the actual basis for the Lie
algebra is of dimension 4.  Notice that in the first example has an
empty Lie bracket table, i.e. all the elements in B are
independent.

> w2r:=wner(r,max_bracket_order,B,B,{});

  w2r := dg1~ f0~ + dg2~ f1~ + (dg2~ g1~ + dg4~) (f0~ &* f1~)

                        2
         + (1/2 dg2~ g1~  + dg4~ g1~) (f0~ &* (f0~ &* f1~)) +

                     3               2
        (1/6 dg2~ g1~  + 1/2 dg4~ g1~ )

        (f0~ &* (f0~ &* (f0~ &* f1~))) + dg3~ f2~

         + dg3~ g2~ (f1~ &* f2~)

                       2
         + 1/2 dg3~ g2~  (f1~ &* (f1~ &* f2~))

                       3
         + 1/6 dg3~ g2~  (f1~ &* (f1~ &* (f1~ &* f2~)))

         + dg3~ g1~ (f0~ &* f2~)

                       2
         + 1/2 dg3~ g1~  (f0~ &* (f0~ &* f2~))

                       3
         + 1/6 dg3~ g1~  (f0~ &* (f0~ &* (f0~ &* f2~)))

         + dg4~ g3~ (f2~ &* (f0~ &* f1~))

                       2
         + 1/2 dg4~ g3~  (f2~ &* (f2~ &* (f0~ &* f1~)))

         + dg4~ g2~ (f1~ &* (f0~ &* f1~))

                       2
         + 1/2 dg4~ g2~  (f1~ &* (f1~ &* (f0~ &* f1~)))

         + dg4~ g1~ g2~ (f1~ &* (f0~ &* (f0~ &* f1~)))




		

A similar calculation to the previous one, but with the assumption
that the bracket $B_6=[f_1,f_2]$ is zero, yields: 
> w5r:=wner(r,max_bracket_order,B,B,{B[6]=0});

  w5r := dg1~ f0~ + dg2~ f1~ + (dg2~ g1~ + dg4~) (f0~ &* f1~)

                        2
         + (1/2 dg2~ g1~  + dg4~ g1~) (f0~ &* (f0~ &* f1~)) +

                     3               2
        (1/6 dg2~ g1~  + 1/2 dg4~ g1~ )

        (f0~ &* (f0~ &* (f0~ &* f1~))) + dg3~ f2~

         + dg3~ g1~ (f0~ &* f2~)

                       2
         + 1/2 dg3~ g1~  (f0~ &* (f0~ &* f2~))

                       3
         + 1/6 dg3~ g1~  (f0~ &* (f0~ &* (f0~ &* f2~)))

         + dg4~ g3~ (f2~ &* (f0~ &* f1~))

                       2
         + 1/2 dg4~ g3~  (f2~ &* (f2~ &* (f0~ &* f1~)))

         + dg4~ g2~ (f1~ &* (f0~ &* f1~))

                       2
         + 1/2 dg4~ g2~  (f1~ &* (f1~ &* (f0~ &* f1~)))

         + dg4~ g1~ g2~ (f1~ &* (f0~ &* (f0~ &* f1~)))

As expected, the difference w2r-w5r shown below contains only
terms of the bracket $B_6=[f_1,f_2]$ which was assumed to be zero in the
simplification of w5r.
                       3
         + 1/6 dg3~ g2~  (f1~ &* (f1~ &* (f1~ &* f2~)))

                       2
         + 1/2 dg3~ g2~  (f1~ &* (f1~ &* f2~))

         + dg3~ g2~ (f1~ &* f2~)



		

Note that an actual computation in Maple of 'w2r-w5r;'
returns an expression that contains terms, such as:
 (-dg1~ + dg1~) f0 +
 + (-dg2~ g1~ - dg4~ + dg2~ g1~ + dg4~) (f0~ &* f1~) + ...

that are not simplified to zero because each time wner is 
invoked, different variables dg1, dg2, etc., are
created at each time (i.e. the variables have the same name, but they 
do not correspond to the same instance of a unique variable in the 
Maple space, in fact they are instances of different variables).
Although this does not seem so far 
to cause significant problems, its is worth to mention that if one
wishes to make comparisons between expressions it would probably be
convenient to modify the routines wne and wner so
that they declare variables g and dg in the global
space as unique instances.





See Also 		 

ead, eadr, pead, peadr.





References 		 

See [32] and references therein for an explanation on the
derivation of the Wei-Norman equations.