find a Cartan subalgebra of a Lie algebra - Maple Programming Help

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LieAlgebras[CartanSubalgebra] - find a Cartan subalgebra of a Lie algebra

Calling Sequences

CartanSubalgebra()

CartanSubalgebra(alg)

CartanSubalgebra(N)

Parameters

alg   - name or string, the name of an initialized Lie algebra

N     - a list of vectors, defining a nilpotent subalgebra

Description

 • Let be a Lie algebra. A Cartan subalgebra h is a nilpotent subalgebra whose normalizer in g is itself, that is,  . If g is a semi-simple Lie algebra, then every Cartan subalgebra h is Abelian and (see Adjoint) is a semi-simple linear transformation for every  (that is, $\mathrm{ad}\left(x\right)$ is diagonalizable over C). Cartan subalgebras are not unique. However, if g is a semi-simple Lie algebra, then any two Cartan subalgebras of g are related by an automorphism of g Let n be a nilpotent subalgebra of g and let  be the generalized null space of n. Then there always exists a Cartan subalgebra If is a regular element, then the generalized null space of $x$ is a Cartan subalgebra.
 • The procedure CartanSubalgebra returns a list of vectors whose span is a Cartan subalgebra.
 • The procedure CartanSubalgebra implements the algorithm for calculating Cartan subalgebras presented in W. A. De Graaf: Lie Algebras: Theory and Algorithms.

Examples

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

Example 1

We calculate the Cartan subalgebra for the 8-dimensional Lie algebra of 3x3 trace-free matrices. The structure equations are obtained using the SimpleLieAlgebraData command.

 > $L≔\mathrm{SimpleLieAlgebraData}\left("sl\left(3\right)",\mathrm{sl3},\mathrm{labelformat}="gl",\mathrm{labels}=\left[E,\mathrm{θ}\right]\right)$
 ${L}{:=}\left[\left[{\mathrm{e1}}{,}{\mathrm{e3}}\right]{=}{\mathrm{e3}}{,}\left[{\mathrm{e1}}{,}{\mathrm{e4}}\right]{=}{2}{}{\mathrm{e4}}{,}\left[{\mathrm{e1}}{,}{\mathrm{e5}}\right]{=}{-}{\mathrm{e5}}{,}\left[{\mathrm{e1}}{,}{\mathrm{e6}}\right]{=}{\mathrm{e6}}{,}\left[{\mathrm{e1}}{,}{\mathrm{e7}}\right]{=}{-}{2}{}{\mathrm{e7}}{,}\left[{\mathrm{e1}}{,}{\mathrm{e8}}\right]{=}{-}{\mathrm{e8}}{,}\left[{\mathrm{e2}}{,}{\mathrm{e3}}\right]{=}{-}{\mathrm{e3}}{,}\left[{\mathrm{e2}}{,}{\mathrm{e4}}\right]{=}{\mathrm{e4}}{,}\left[{\mathrm{e2}}{,}{\mathrm{e5}}\right]{=}{\mathrm{e5}}{,}\left[{\mathrm{e2}}{,}{\mathrm{e6}}\right]{=}{2}{}{\mathrm{e6}}{,}\left[{\mathrm{e2}}{,}{\mathrm{e7}}\right]{=}{-}{\mathrm{e7}}{,}\left[{\mathrm{e2}}{,}{\mathrm{e8}}\right]{=}{-}{2}{}{\mathrm{e8}}{,}\left[{\mathrm{e3}}{,}{\mathrm{e5}}\right]{=}{\mathrm{e1}}{-}{\mathrm{e2}}{,}\left[{\mathrm{e3}}{,}{\mathrm{e6}}\right]{=}{\mathrm{e4}}{,}\left[{\mathrm{e3}}{,}{\mathrm{e7}}\right]{=}{-}{\mathrm{e8}}{,}\left[{\mathrm{e4}}{,}{\mathrm{e5}}\right]{=}{-}{\mathrm{e6}}{,}\left[{\mathrm{e4}}{,}{\mathrm{e7}}\right]{=}{\mathrm{e1}}{,}\left[{\mathrm{e4}}{,}{\mathrm{e8}}\right]{=}{\mathrm{e3}}{,}\left[{\mathrm{e5}}{,}{\mathrm{e8}}\right]{=}{-}{\mathrm{e7}}{,}\left[{\mathrm{e6}}{,}{\mathrm{e7}}\right]{=}{\mathrm{e5}}{,}\left[{\mathrm{e6}}{,}{\mathrm{e8}}\right]{=}{\mathrm{e2}}\right]{,}\left[{\mathrm{E11}}{,}{\mathrm{E22}}{,}{\mathrm{E12}}{,}{\mathrm{E13}}{,}{\mathrm{E21}}{,}{\mathrm{E23}}{,}{\mathrm{E31}}{,}{\mathrm{E32}}\right]{,}\left[{\mathrm{θ11}}{,}{\mathrm{θ22}}{,}{\mathrm{θ12}}{,}{\mathrm{θ13}}{,}{\mathrm{θ21}}{,}{\mathrm{θ23}}{,}{\mathrm{θ31}}{,}{\mathrm{θ32}}\right]$ (2.1)

Initialized the Lie algebra.

 > $\mathrm{DGsetup}\left(L\right)$
 ${\mathrm{Lie algebra: sl3}}$ (2.2)

Find a Cartan subalgebra.

 sl(3) > $\mathrm{CSA}≔\mathrm{CartanSubalgebra}\left(\right)$
 ${\mathrm{CSA}}{:=}\left[{\mathrm{E11}}{,}{\mathrm{E22}}\right]$ (2.3)

We can check that this subalgebra is Abelian (and hence nilpotent) and self-normalizing.

 sl3 > $\mathrm{Query}\left(\mathrm{CSA},"Abelian"\right)$
 ${\mathrm{true}}$ (2.4)
 sl3 > $\mathrm{SubalgebraNormalizer}\left(\mathrm{CSA}\right)$
 $\left[{\mathrm{E22}}{,}{\mathrm{E11}}\right]$ (2.5)

These properties can also be checked with the Query command

 sl3 > $\mathrm{Query}\left(\mathrm{CSA},"CartanSubalgebra"\right)$
 ${\mathrm{true}}$ (2.6)

For the split real forms of the simple Lie algebras, a Cartan subalgebra can always be found consisting of diagonal matrices in the standard representation.

 sl3 > $\mathrm{StandardRepresentation}\left(\mathrm{sl3},\mathrm{CSA}\right)$
 $\left[\left[\begin{array}{rrr}{1}& {0}& {0}\\ {0}& {0}& {0}\\ {0}& {0}& {-}{1}\end{array}\right]{,}\left[\begin{array}{rrr}{0}& {0}& {0}\\ {0}& {1}& {0}\\ {0}& {0}& {-}{1}\end{array}\right]\right]$ (2.7)

Example 2

Other Cartan subalgebras for can be found with the second calling sequence.

 sl3 > $\mathrm{CartanSubalgebra}\left(\left[\mathrm{E11}+3\mathrm{E31}\right]\right)$
 $\left[{\mathrm{E11}}{+}{3}{}{\mathrm{E31}}{,}{\mathrm{E22}}{+}\frac{{3}}{{2}}{}{\mathrm{E31}}\right]$ (2.8)

Example 3

The Cartan subalgebra of a nilpotent Lie algebra g is g itself. Retrieve the structure equations for a nilpotent Lie algebra from the DifferentialGeometry library.

 sl3 > $\mathrm{LD3}≔\mathrm{Retrieve}\left("Winternitz",1,\left[5,5\right],\mathrm{alg3}\right)$
 ${\mathrm{LD3}}{:=}\left[\left[{\mathrm{e2}}{,}{\mathrm{e5}}\right]{=}{\mathrm{e1}}{,}\left[{\mathrm{e3}}{,}{\mathrm{e4}}\right]{=}{\mathrm{e1}}{,}\left[{\mathrm{e3}}{,}{\mathrm{e5}}\right]{=}{\mathrm{e2}}\right]$ (2.9)
 sl3 > $\mathrm{DGsetup}\left(\mathrm{LD3}\right):$

Check that the algebra is nilpotent.

 alg3 > $\mathrm{Query}\left("Nilpotent"\right)$
 ${\mathrm{true}}$ (2.10)
 alg3 > $\mathrm{CartanSubalgebra}\left(\right)$
 $\left[{\mathrm{e1}}{,}{\mathrm{e2}}{,}{\mathrm{e3}}{,}{\mathrm{e4}}{,}{\mathrm{e5}}\right]$ (2.11)

Example 4

We find the Cartan subalgebra for a solvable Lie algebra. Retrieve the structure equations for a solvable Lie algebra from the DifferentialGeometry library.

 alg3 > $\mathrm{LD4}≔\mathrm{Retrieve}\left("Winternitz",1,\left[5,34\right],\mathrm{alg4}\right)$
 ${\mathrm{LD4}}{:=}\left[\left[{\mathrm{e1}}{,}{\mathrm{e4}}\right]{=}{a}{}{\mathrm{e1}}{,}\left[{\mathrm{e1}}{,}{\mathrm{e5}}\right]{=}{\mathrm{e1}}{,}\left[{\mathrm{e2}}{,}{\mathrm{e4}}\right]{=}{\mathrm{e2}}{,}\left[{\mathrm{e3}}{,}{\mathrm{e4}}\right]{=}{\mathrm{e3}}{,}\left[{\mathrm{e3}}{,}{\mathrm{e5}}\right]{=}{\mathrm{e2}}\right]$ (2.12)
 alg4 > $\mathrm{DGsetup}\left(\mathrm{LD4}\right)$
 ${\mathrm{Lie algebra: alg4}}$ (2.13)

Check that the algebra is solvable.

 alg4 > $\mathrm{Query}\left("Solvable"\right)$
 ${\mathrm{true}}$ (2.14)
 alg4 > $\mathrm{CSA4}≔\mathrm{CartanSubalgebra}\left(\right)$
 ${\mathrm{CSA4}}{:=}\left[{\mathrm{e4}}{,}{\mathrm{e5}}\right]$ (2.15)
 alg4 > $\mathrm{Query}\left(\mathrm{CSA4},"CartanSubalgebra"\right)$
 ${\mathrm{true}}$ (2.16)

Example 5.

We find the Cartan subalgebra for a Lie algebra with a non-trivial Levi decomposition. Retrieve the structure equations for such a Lie algebra from the DifferentialGeometry library.

 alg4 > $\mathrm{LD5}≔\mathrm{Retrieve}\left("Turkowski",1,\left[7,4\right],\mathrm{alg5}\right)$
 ${\mathrm{LD5}}{:=}\left[\left[{\mathrm{e1}}{,}{\mathrm{e2}}\right]{=}{2}{}{\mathrm{e2}}{,}\left[{\mathrm{e1}}{,}{\mathrm{e3}}\right]{=}{-}{2}{}{\mathrm{e3}}{,}\left[{\mathrm{e1}}{,}{\mathrm{e4}}\right]{=}{\mathrm{e4}}{,}\left[{\mathrm{e1}}{,}{\mathrm{e5}}\right]{=}{-}{\mathrm{e5}}{,}\left[{\mathrm{e2}}{,}{\mathrm{e3}}\right]{=}{\mathrm{e1}}{,}\left[{\mathrm{e2}}{,}{\mathrm{e5}}\right]{=}{\mathrm{e4}}{,}\left[{\mathrm{e3}}{,}{\mathrm{e4}}\right]{=}{\mathrm{e5}}{,}\left[{\mathrm{e4}}{,}{\mathrm{e5}}\right]{=}{\mathrm{e6}}{,}\left[{\mathrm{e4}}{,}{\mathrm{e7}}\right]{=}{\mathrm{e4}}{,}\left[{\mathrm{e5}}{,}{\mathrm{e7}}\right]{=}{\mathrm{e5}}{,}\left[{\mathrm{e6}}{,}{\mathrm{e7}}\right]{=}{2}{}{\mathrm{e6}}\right]$ (2.17)
 alg4 > $\mathrm{DGsetup}\left(\mathrm{LD5}\right)$
 ${\mathrm{Lie algebra: alg5}}$ (2.18)

Check that the Levi decomposition is non-trivial.

 alg5 > $\mathrm{LeviDecomposition}\left(\right)$
 $\left[\left[{\mathrm{e4}}{,}{\mathrm{e5}}{,}{\mathrm{e6}}{,}{\mathrm{e7}}\right]{,}\left[{\mathrm{e1}}{,}{\mathrm{e2}}{,}{\mathrm{e3}}\right]\right]$ (2.19)

Calculate the Cartan subalgebra.

 alg5 > $\mathrm{CartanSubalgebra}\left(\mathrm{alg5}\right)$
 $\left[{\mathrm{e1}}{,}{\mathrm{e7}}\right]$ (2.20)