 DifferentialGeometry/LieAlgebras/Query/CartanSubalgebra - Maple Help

Query[CartanSubalgebra] - check if a list of vectors defines a Cartan subalgebra

Calling Sequences

Query()

Parameters

A        - a list of vectors, defining a subspace of a Lie algebra

options  - one or more of the keyword arguments rank = n (where is a positive integer), algebratype  = "Semisimple" or  algebratype  = "Simple" Description

 • Let be a Lie algebra. A Cartan subalgebra h is a nilpotent subalgebra whose normalizer in g is itself, that is,  .
 • If the Lie algebra  is semi-simple and the rank of the Lie algebra is then any Cartan subalgebra is of dimension and is Abelian. This simplifies checking if a given subspace of vectors is a Cartan subalgebra ( the nilpotent character of h need not be verified). Examples

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

Example 1.

We test if certain subalgebras of are Cartan subalgebras. First define the standard matrix representation for as the space of trace-free matrices.

 > $A≔\mathrm{map}\left(\mathrm{convert},\left[\left[\left[1,0,0\right],\left[0,-1,0\right],\left[0,0,0\right]\right],\left[\left[0,0,0\right],\left[0,1,0\right],\left[0,0,-1\right]\right],\left[\left[0,1,0\right],\left[0,0,0\right],\left[0,0,0\right]\right],\left[\left[0,0,1\right],\left[0,0,0\right],\left[0,0,0\right]\right],\left[\left[0,0,0\right],\left[1,0,0\right],\left[0,0,0\right]\right],\left[\left[0,0,0\right],\left[0,0,1\right],\left[0,0,0\right]\right],\left[\left[0,0,0\right],\left[0,0,0\right],\left[1,0,0\right]\right],\left[\left[0,0,0\right],\left[0,0,0\right],\left[0,1,0\right]\right]\right],\mathrm{Matrix}\right)$
 ${A}{:=}\left[\left[\begin{array}{rrr}{1}& {0}& {0}\\ {0}& {-}{1}& {0}\\ {0}& {0}& {0}\end{array}\right]{,}\left[\begin{array}{rrr}{0}& {0}& {0}\\ {0}& {1}& {0}\\ {0}& {0}& {-}{1}\end{array}\right]{,}\left[\begin{array}{rrr}{0}& {1}& {0}\\ {0}& {0}& {0}\\ {0}& {0}& {0}\end{array}\right]{,}\left[\begin{array}{rrr}{0}& {0}& {1}\\ {0}& {0}& {0}\\ {0}& {0}& {0}\end{array}\right]{,}\left[\begin{array}{rrr}{0}& {0}& {0}\\ {1}& {0}& {0}\\ {0}& {0}& {0}\end{array}\right]{,}\left[\begin{array}{rrr}{0}& {0}& {0}\\ {0}& {0}& {1}\\ {0}& {0}& {0}\end{array}\right]{,}\left[\begin{array}{rrr}{0}& {0}& {0}\\ {0}& {0}& {0}\\ {1}& {0}& {0}\end{array}\right]{,}\left[\begin{array}{rrr}{0}& {0}& {0}\\ {0}& {0}& {0}\\ {0}& {1}& {0}\end{array}\right]\right]$ (2.1)

Calculate the structure equations for these matrices and initialize the resulting Lie algebra.

 > $\mathrm{LD}≔\mathrm{LieAlgebraData}\left(A,\mathrm{sl3}\right)$
 ${\mathrm{LD}}{:=}\left[\left[{\mathrm{e1}}{,}{\mathrm{e3}}\right]{=}{2}{}{\mathrm{e3}}{,}\left[{\mathrm{e1}}{,}{\mathrm{e4}}\right]{=}{\mathrm{e4}}{,}\left[{\mathrm{e1}}{,}{\mathrm{e5}}\right]{=}{-}{2}{}{\mathrm{e5}}{,}\left[{\mathrm{e1}}{,}{\mathrm{e6}}\right]{=}{-}{\mathrm{e6}}{,}\left[{\mathrm{e1}}{,}{\mathrm{e7}}\right]{=}{-}{\mathrm{e7}}{,}\left[{\mathrm{e1}}{,}{\mathrm{e8}}\right]{=}{\mathrm{e8}}{,}\left[{\mathrm{e2}}{,}{\mathrm{e3}}\right]{=}{-}{\mathrm{e3}}{,}\left[{\mathrm{e2}}{,}{\mathrm{e4}}\right]{=}{\mathrm{e4}}{,}\left[{\mathrm{e2}}{,}{\mathrm{e5}}\right]{=}{\mathrm{e5}}{,}\left[{\mathrm{e2}}{,}{\mathrm{e6}}\right]{=}{2}{}{\mathrm{e6}}{,}\left[{\mathrm{e2}}{,}{\mathrm{e7}}\right]{=}{-}{\mathrm{e7}}{,}\left[{\mathrm{e2}}{,}{\mathrm{e8}}\right]{=}{-}{2}{}{\mathrm{e8}}{,}\left[{\mathrm{e3}}{,}{\mathrm{e5}}\right]{=}{\mathrm{e1}}{,}\left[{\mathrm{e3}}{,}{\mathrm{e6}}\right]{=}{\mathrm{e4}}{,}\left[{\mathrm{e3}}{,}{\mathrm{e7}}\right]{=}{-}{\mathrm{e8}}{,}\left[{\mathrm{e4}}{,}{\mathrm{e5}}\right]{=}{-}{\mathrm{e6}}{,}\left[{\mathrm{e4}}{,}{\mathrm{e7}}\right]{=}{\mathrm{e1}}{+}{\mathrm{e2}}{,}\left[{\mathrm{e4}}{,}{\mathrm{e8}}\right]{=}{\mathrm{e3}}{,}\left[{\mathrm{e5}}{,}{\mathrm{e8}}\right]{=}{-}{\mathrm{e7}}{,}\left[{\mathrm{e6}}{,}{\mathrm{e7}}\right]{=}{\mathrm{e5}}{,}\left[{\mathrm{e6}}{,}{\mathrm{e8}}\right]{=}{\mathrm{e2}}\right]$ (2.2)
 > $\mathrm{DGsetup}\left(\mathrm{LD}\right)$
 ${\mathrm{Lie algebra: sl3}}$ (2.3)

Let's check that is semi-simple.

 sl3 > $\mathrm{Query}\left(\mathrm{sl3},"Semisimple"\right)$
 ${\mathrm{true}}$ (2.4)

Test to see if a list of vectors defines a Cartan subalgebra.

 sl3 > $A≔\left[\mathrm{e1},\mathrm{e2}\right]$
 ${A}{:=}\left[{\mathrm{e1}}{,}{\mathrm{e2}}\right]$ (2.5)
 sl3 > $\mathrm{Query}\left(A,"CartanSubalgebra"\right)$
 ${\mathrm{true}}$ (2.6)

Since  has 2 elements, this implies that the rank of   is 2. We can use this information to simplify checking that other subalgebras are Cartan subalgebras

 sl3 > $A≔\mathrm{evalDG}\left(\left[\mathrm{e1}+\mathrm{e6},\mathrm{e2}-2\mathrm{e6}\right]\right)$
 ${A}{:=}\left[{\mathrm{e1}}{+}{\mathrm{e6}}{,}{\mathrm{e2}}{-}{2}{}{\mathrm{e6}}\right]$ (2.7)
 sl3 > $\mathrm{Query}\left(A,\mathrm{rank}=2,\mathrm{algebratype}="Semisimple","CartanSubalgebra"\right)$
 ${\mathrm{true}}$ (2.8)

Here is a 2-dimensional Abelian subalgebra which is not self-normalizing and therefore not a Cartan subalgebra.

 sl3 > $A≔\left[\mathrm{e3},\mathrm{e4}\right]$
 ${A}{:=}\left[{\mathrm{e3}}{,}{\mathrm{e4}}\right]$ (2.9)
 sl3 > $\mathrm{Query}\left(A,"CartanSubalgebra"\right)$
 ${\mathrm{false}}$ (2.10)
 sl3 > $\mathrm{SubalgebraNormalizer}\left(A\right)$
 $\left[{\mathrm{e8}}{,}{\mathrm{e6}}{,}{\mathrm{e4}}{,}{\mathrm{e3}}{,}{\mathrm{e2}}{,}{\mathrm{e1}}\right]$ (2.11)

Example 2.

The notion of a Cartan subalgebra is not restricted to semi-simple Lie algebras. We define a solvable Lie algebra and test to see if some subalgebras are Cartan subalgebras.

 sl3 > $\mathrm{LD}≔\mathrm{_DG}\left(\left[\left["LieAlgebra",\mathrm{alg},\left[5\right]\right],\left[\left[\left[1,4,1\right],a\right],\left[\left[2,4,2\right],1\right],\left[\left[3,4,3\right],1\right],\left[\left[1,5,1\right],1\right],\left[\left[3,5,2\right],1\right]\right]\right]\right)$
 ${\mathrm{LD}}{:=}\left[\left[{\mathrm{e1}}{,}{\mathrm{e4}}\right]{=}{a}{}{\mathrm{e1}}{,}\left[{\mathrm{e1}}{,}{\mathrm{e5}}\right]{=}{\mathrm{e1}}{,}\left[{\mathrm{e2}}{,}{\mathrm{e4}}\right]{=}{\mathrm{e2}}{,}\left[{\mathrm{e3}}{,}{\mathrm{e4}}\right]{=}{\mathrm{e3}}{,}\left[{\mathrm{e3}}{,}{\mathrm{e5}}\right]{=}{\mathrm{e2}}\right]$ (2.12)
 sl3 > $\mathrm{DGsetup}\left(\mathrm{LD}\right)$
 ${\mathrm{Lie algebra: alg}}$ (2.13)
 alg > $\mathrm{Query}\left("Solvable"\right)$
 ${\mathrm{true}}$ (2.14)
 alg > $A≔\left[\mathrm{e4},\mathrm{e5}\right]$
 ${A}{:=}\left[{\mathrm{e4}}{,}{\mathrm{e5}}\right]$ (2.15)
 alg > $\mathrm{Query}\left(A,"CartanSubalgebra"\right)$
 ${\mathrm{true}}$ (2.16)

Any subalgebra which is an ideal cannot be a Cartan subalgebra.

 alg > $A≔\left[\mathrm{e1},\mathrm{e2}\right]$
 ${A}{:=}\left[{\mathrm{e1}}{,}{\mathrm{e2}}\right]$ (2.17)
 alg > $\mathrm{Query}\left(A,"Ideal"\right)$
 ${\mathrm{true}}$ (2.18)
 alg > $\mathrm{Query}\left(A,"CartanSubalgebra"\right)$
 ${\mathrm{false}}$ (2.19)