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Example 1.
We initialize three different Lie algebras and print their multiplication tables.
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Alg3 >
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Alg1 is solvable and therefore the radical is the entire algebra.
Alg3 >
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Alg3 >
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Alg2 is semisimple and therefore the radical is the zero subalgebra.
Alg1 >
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Alg1 >
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Alg3 has a non-trivial Levi decomposition.
Alg1 >
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Alg1 >
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It is easy to see that in this last example the Levi decomposition is not unique. First we find the general complement to the radical using the ComplementaryBasis program.
Alg1 >
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Next we determine for which values of the parameters for which he subspace SS0 is a Lie subalgebra. We find that .
Alg3 >
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| (2.6) |
Alg3 >
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