calculate a right or a central extension of a Lie algebra - Maple Programming Help

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LieAlgebras[Extension] - calculate a right or a central extension of a Lie algebra

Calling Sequences

Extension(AlgName1, A, AlgName2)

Extension(AlgName1, ${\mathbf{β}}$, AlgName2)

Parameters

AlgName1 - a name or string, the name of the Lie algebra to be extended

A        - a transformation, defining derivation of $\mathrm{𝔤}$

$\mathrm{β}$        - a closed 2-form

AlgName2 - a name or string, the name to be given to the Lie algebra extension

Description

 • Let $\mathrm{𝔤}$ be a Lie algebra and let be a derivation on $\mathrm{𝔤}$. Then the right extension of by is the Lie algebra  with Lie bracket

] = (for all and

The right extension $k$ is said to be trivial if splits as a Lie algebra direct sum where is isomorphic to $\mathrm{𝔤}$. The extension is trivial precisely when is an inner derivation.

 • Let be a Lie algebra and let $\mathrm{β}$ be a closed 2-form on $\mathrm{𝔤}$. Then the central extension of by $\mathrm{β}$ is the Lie algebra  with Lie bracket

] = for all and

The extension is said to be trivial if splits as a Lie algebra direct sum where is isomorphic to $\mathrm{𝔤}$. The extension is trivial precisely when $\mathrm{β}$ is exact, that is, $\mathrm{\beta }$ = $d\left(\mathrm{η}\right)$.

 • Extension computes a right extension when its second argument is a matrix and a central extension when the second argument is a 2-form. The procedure returns the Lie algebra data structure for the extended algebra. The structure equations for the extension are displayed.
 • The command Extension is part of the DifferentialGeometry:-LieAlgebras package. It can be used in the form Extension(...) only after executing the commands with(DifferentialGeometry) and with(LieAlgebras), but can always be used by executing DifferentialGeometry:-LieAlgebras:-Extension(...).

Examples

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

Example 1.

Calculate two right extensions and show that the first is trivial and the second is not. First initialize the Lie algebra Alg1 and display the multiplication table.

 > $\mathrm{L1}≔\mathrm{_DG}\left(\left[\left["LieAlgebra",\mathrm{Alg1},\left[4\right]\right],\left[\left[\left[1,4,1\right],2\right],\left[\left[2,3,1\right],1\right],\left[\left[2,4,2\right],1\right],\left[\left[3,4,2\right],1\right],\left[\left[3,4,3\right],1\right]\right]\right]\right)$
 ${\mathrm{L1}}{:=}\left[\left[{\mathrm{e1}}{,}{\mathrm{e4}}\right]{=}{2}{}{\mathrm{e1}}{,}\left[{\mathrm{e2}}{,}{\mathrm{e3}}\right]{=}{\mathrm{e1}}{,}\left[{\mathrm{e2}}{,}{\mathrm{e4}}\right]{=}{\mathrm{e2}}{,}\left[{\mathrm{e3}}{,}{\mathrm{e4}}\right]{=}{\mathrm{e2}}{+}{\mathrm{e3}}\right]$ (2.1)
 > $\mathrm{DGsetup}\left(\mathrm{L1}\right):$

Here are two derivations we shall use to make right extensions.

 Alg1 > $\mathrm{A1}≔\mathrm{Adjoint}\left(\mathrm{e1}-\mathrm{e2}+2\mathrm{e3}-\mathrm{e4}\right);$$\mathrm{A2}≔{\mathrm{Derivations}\left("Outer"\right)}_{1}$
 ${\mathrm{A1}}{:=}\left[\begin{array}{rrrr}{2}& {-}{2}& {-}{1}& {2}\\ {0}& {1}& {1}& {1}\\ {0}& {0}& {1}& {2}\\ {0}& {0}& {0}& {0}\end{array}\right]$
 ${\mathrm{A2}}{:=}\left[\begin{array}{cccc}{1}& {0}& {0}& {0}\\ {0}& \frac{{1}}{{2}}& {0}& {0}\\ {0}& {0}& \frac{{1}}{{2}}& {0}\\ {0}& {0}& {0}& {0}\end{array}\right]$ (2.2)

Use the matrix A1 to make a right extension.

 Alg1 > $\mathrm{L2}≔\mathrm{Extension}\left(\mathrm{Alg1},\mathrm{A1},\mathrm{Alg2}\right)$
 ${\mathrm{L2}}{:=}\left[\left[{\mathrm{e1}}{,}{\mathrm{e4}}\right]{=}{2}{}{\mathrm{e1}}{,}\left[{\mathrm{e1}}{,}{\mathrm{e5}}\right]{=}{2}{}{\mathrm{e1}}{,}\left[{\mathrm{e2}}{,}{\mathrm{e3}}\right]{=}{\mathrm{e1}}{,}\left[{\mathrm{e2}}{,}{\mathrm{e4}}\right]{=}{\mathrm{e2}}{,}\left[{\mathrm{e2}}{,}{\mathrm{e5}}\right]{=}{-}{2}{}{\mathrm{e1}}{+}{\mathrm{e2}}{,}\left[{\mathrm{e3}}{,}{\mathrm{e4}}\right]{=}{\mathrm{e2}}{+}{\mathrm{e3}}{,}\left[{\mathrm{e3}}{,}{\mathrm{e5}}\right]{=}{-}{\mathrm{e1}}{+}{\mathrm{e2}}{+}{\mathrm{e3}}{,}\left[{\mathrm{e4}}{,}{\mathrm{e5}}\right]{=}{2}{}{\mathrm{e1}}{+}{\mathrm{e2}}{+}{2}{}{\mathrm{e3}}\right]$ (2.3)

Initialize this Lie algebra. Since it was constructed using an inner derivation, it should be a trivial extension. This we check using the Decompose command.

 Alg1 > $\mathrm{DGsetup}\left(\mathrm{L2}\right)$
 ${\mathrm{Lie algebra: Alg2}}$ (2.4)
 Alg2 > $\mathrm{Query}\left(\mathrm{Alg2},"Indecomposable"\right)$
 ${\mathrm{false}}$ (2.5)

Repeat these computations using the outer derivation A2.

 Alg2 > $\mathrm{L3}≔\mathrm{Extension}\left(\mathrm{Alg1},\mathrm{A2},\mathrm{Alg3}\right)$
 ${\mathrm{L3}}{:=}\left[\left[{\mathrm{e1}}{,}{\mathrm{e4}}\right]{=}{2}{}{\mathrm{e1}}{,}\left[{\mathrm{e1}}{,}{\mathrm{e5}}\right]{=}{\mathrm{e1}}{,}\left[{\mathrm{e2}}{,}{\mathrm{e3}}\right]{=}{\mathrm{e1}}{,}\left[{\mathrm{e2}}{,}{\mathrm{e4}}\right]{=}{\mathrm{e2}}{,}\left[{\mathrm{e2}}{,}{\mathrm{e5}}\right]{=}\frac{{1}}{{2}}{}{\mathrm{e2}}{,}\left[{\mathrm{e3}}{,}{\mathrm{e4}}\right]{=}{\mathrm{e2}}{+}{\mathrm{e3}}{,}\left[{\mathrm{e3}}{,}{\mathrm{e5}}\right]{=}\frac{{1}}{{2}}{}{\mathrm{e3}}\right]$ (2.6)

Initialize this right extension. Since it was constructed using an outer derivation, it should be not be a trivial extension. This we check using the Decompose command.

 Alg1 > $\mathrm{DGsetup}\left(\mathrm{L3}\right)$
 ${\mathrm{Lie algebra: Alg3}}$ (2.7)
 Alg3 > $\mathrm{Query}\left(\mathrm{Alg3},"Indecomposable"\right)$
 ${\mathrm{true}}$ (2.8)

Example 2.

Calculate two central extensions and show that the first is trivial and the second is not. First initialize the Lie algebra Alg4 and display the multiplication table. Now display the exterior derivatives of the 1-forms for Alg1.

 Alg3 > $\mathrm{L4}≔\mathrm{_DG}\left(\left[\left["LieAlgebra",\mathrm{Alg4},\left[4\right]\right],\left[\left[\left[2,4,1\right],1\right],\left[\left[3,4,3\right],1\right]\right]\right]\right)$
 ${\mathrm{L4}}{:=}\left[\left[{\mathrm{e2}}{,}{\mathrm{e4}}\right]{=}{\mathrm{e1}}{,}\left[{\mathrm{e3}}{,}{\mathrm{e4}}\right]{=}{\mathrm{e3}}\right]$ (2.9)
 Alg3 > $\mathrm{DGsetup}\left(\mathrm{L4}\right):$
 Alg4 > $\mathrm{MultiplicationTable}\left(\mathrm{Alg4},"ExteriorDerivative"\right)$
 ${d}{}\left({\mathrm{θ1}}\right){=}{-}{\mathrm{θ2}}{}{\bigwedge }{}{\mathrm{θ4}}$
 ${d}{}\left({\mathrm{θ2}}\right){=}{0}{}{\mathrm{θ1}}{}{\bigwedge }{}{\mathrm{θ2}}$
 ${d}{}\left({\mathrm{θ3}}\right){=}{-}{\mathrm{θ3}}{}{\bigwedge }{}{\mathrm{θ4}}$
 ${d}{}\left({\mathrm{θ4}}\right){=}{0}{}{\mathrm{θ1}}{}{\bigwedge }{}{\mathrm{θ2}}$ (2.10)

Define a pair of 2-forms and check that they are closed.

 Alg4 > $\mathrm{β1}≔\mathrm{θ2}&wedge\mathrm{θ4};$$\mathrm{β2}≔\mathrm{θ1}&wedge\mathrm{θ4}$
 ${\mathrm{β1}}{:=}{\mathrm{θ2}}{}{\bigwedge }{}{\mathrm{θ4}}$
 ${\mathrm{β2}}{:=}{\mathrm{θ1}}{}{\bigwedge }{}{\mathrm{θ4}}$ (2.11)
 Alg4 > $\mathrm{ExteriorDerivative}\left(\mathrm{β1}\right),\mathrm{ExteriorDerivative}\left(\mathrm{β2}\right)$
 ${0}{}{\mathrm{θ1}}{}{\bigwedge }{}{\mathrm{θ2}}{}{\bigwedge }{}{\mathrm{θ3}}{,}{0}{}{\mathrm{θ1}}{}{\bigwedge }{}{\mathrm{θ2}}{}{\bigwedge }{}{\mathrm{θ3}}$ (2.12)

Use ${\mathrm{β}}_{1}$ to make a central extension.

 Alg4 > $\mathrm{L5}≔\mathrm{Extension}\left(\mathrm{Alg4},\mathrm{β1},\mathrm{Alg5}\right)$
 ${\mathrm{L5}}{:=}\left[\left[{\mathrm{e2}}{,}{\mathrm{e4}}\right]{=}{\mathrm{e1}}{+}{\mathrm{e5}}{,}\left[{\mathrm{e3}}{,}{\mathrm{e4}}\right]{=}{\mathrm{e3}}\right]$ (2.13)

Initialize this Lie algebra. Since the form is exact, this central extension is trivial.

 Alg4 > $\mathrm{DGsetup}\left(\mathrm{L5}\right):$
 Alg5 > $\mathrm{Query}\left(\mathrm{Alg4},"Indecomposable"\right)$
 ${\mathrm{true}}$ (2.14)

Now make the central extension using ${\mathrm{β}}_{2}.$ This extension is indecomposable.

 Alg4 > $\mathrm{L6}≔\mathrm{Extension}\left(\mathrm{Alg4},\mathrm{β2},\mathrm{Alg6}\right)$
 ${\mathrm{L6}}{:=}\left[\left[{\mathrm{e1}}{,}{\mathrm{e4}}\right]{=}{\mathrm{e5}}{,}\left[{\mathrm{e2}}{,}{\mathrm{e4}}\right]{=}{\mathrm{e1}}{,}\left[{\mathrm{e3}}{,}{\mathrm{e4}}\right]{=}{\mathrm{e3}}\right]$ (2.15)
 Alg4 > $\mathrm{DGsetup}\left(\mathrm{L6}\right):$
 Alg6 > $\mathrm{Query}\left(\mathrm{Alg6},"Indecomposable"\right)$
 ${\mathrm{true}}$ (2.16)