Subalgebra - Maple Help

Query[Subalgebra] - check if a list of vectors defines a Lie subalgebra

Calling Sequences

Query(S, "Subalgebra")

Query(S, parm, "Subalgebra")

Parameters

S       - a list of independent vectors in a Lie algebra $\mathrm{𝔤}$

parm    - (optional) a set of parameters appearing in the list of vectors S.  It is assumed that the set of vectors S is well-defined when the parameters vanish.

Description

 • A list of vectors $S$ defines a basis for a Lie subalgebra if  spanfor all
 • Query(S, "Subalgebra") returns true if the set S defines a subalgebra.
 • Query(S, parm, "Subalgebra") returns a sequence TF, Eq, Soln, SubAlgList.  Here TF is true if Maple finds parameter values for which S is a subalgebra and false otherwise; Eq is the set of equations (with the variables in parm as unknowns) which must be satisfied for S to be a subalgebra; Soln is the list of solutions to the equations Eq; and SubAlgList is the list of subalgebras obtained from the parameter values given by the different solutions in Soln.
 • The program calculates the defining equations Eq for S to be a subalgebra as follows.  First the list of vectors $S$ is evaluated with the parameters set to zero to obtain a set of vectors ${S}_{0}$.  The program ComplementaryBasis is then used to calculate a complement to ${S}_{0}$  The list of vectors  then gives a basis for the entire Lie algebra $\mathrm{𝔤}$.  For each   the bracket is calculated and expressed as a linear combination of the vectors in the basis $B$. The components of $[x,y[$ in $C$ must all vanish for to be a Lie subalgebra.
 • We remark that the equations Eq, which the parameters must satisfy in order for S to be a subalgebra, will in general be a system of coupled quadratic equations.  Maple may not be able to solve these equations or may not solve them in full generality.
 • The command Query is part of the DifferentialGeometry:-LieAlgebras package.  It can be used in the form Query(...) only after executing the commands with(DifferentialGeometry) and with(LieAlgebras), but can always be used by executing DifferentialGeometry:-LieAlgebras:-Query(...).

Examples

 > $\mathrm{with}\left(\mathrm{DifferentialGeometry}\right):$$\mathrm{with}\left(\mathrm{LieAlgebras}\right):$

Example 1.

First initialize a Lie algebra.

 > $L≔\mathrm{_DG}\left(\left[\left["LieAlgebra",\mathrm{Alg},\left[4\right]\right],\left[\left[\left[1,4,1\right],0\right],\left[\left[2,3,1\right],1\right],\left[\left[2,4,2\right],1\right],\left[\left[3,4,3\right],-1\right]\right]\right]\right)$
 ${L}{:=}\left[\left[{\mathrm{e2}}{,}{\mathrm{e3}}\right]{=}{\mathrm{e1}}{,}\left[{\mathrm{e2}}{,}{\mathrm{e4}}\right]{=}{\mathrm{e2}}{,}\left[{\mathrm{e3}}{,}{\mathrm{e4}}\right]{=}{-}{\mathrm{e3}}\right]$ (2.1)
 > $\mathrm{DGsetup}\left(L\right):$

The vectors  do not determine a subalgebra while the vectors do.

 Alg > $\mathrm{S1}≔\left[\mathrm{e2},\mathrm{e3},\mathrm{e4}\right]:$
 Alg > $\mathrm{Query}\left(\mathrm{S1},"Subalgebra"\right)$
 ${\mathrm{false}}$ (2.2)
 Alg > $\mathrm{S2}≔\left[\mathrm{e1},\mathrm{e3},\mathrm{e4}\right]:$
 Alg > $\mathrm{Query}\left(\mathrm{S2},"Subalgebra"\right)$
 ${\mathrm{true}}$ (2.3)

We find the values of the parameters  for which determines a Lie subalgebra.

 Alg > $\mathrm{S3}≔\mathrm{evalDG}\left(\left[\mathrm{e2},\mathrm{e1}+\mathrm{a1}\mathrm{e3}+\mathrm{a2}\mathrm{e4}\right]\right):$
 Alg > $\mathrm{Query}\left(\mathrm{S3},\left\{\mathrm{a1},\mathrm{a2}\right\},"Subalgebra"\right)$
 ${\mathrm{true}}{,}\left\{{0}{,}{-}{{\mathrm{a1}}}^{{2}}{,}{-}{\mathrm{a1}}{}{\mathrm{a2}}\right\}{,}\left[\left\{{\mathrm{a1}}{=}{0}{,}{\mathrm{a2}}{=}{\mathrm{a2}}\right\}\right]{,}\left[\left[{\mathrm{e2}}{,}{\mathrm{a2}}{}{\mathrm{e4}}{+}{\mathrm{e1}}\right]\right]$ (2.4)

There are no values of the parameters   for which determines a Lie subalgebra.

 Alg > $\mathrm{S4}≔\mathrm{evalDG}\left(\left[\mathrm{e2},\mathrm{e3}+\mathrm{a1}\mathrm{e2}+\mathrm{a2}\mathrm{e4}\right]\right):$
 Alg > $\mathrm{Query}\left(\mathrm{S4},\left\{\mathrm{a1},\mathrm{a2}\right\},"Subalgebra"\right)$
 ${\mathrm{false}}{,}\left\{{0}{,}{1}\right\}$ (2.5)