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Example 1.
In this example we illustrate the fact that the result of Inquiry("Indecomposable") does not depend upon the choice of basis for the Lie algebra. First we initialize a Lie algebra.
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| (2.1) |
Now we make a change of basis in the Lie algebra. In this basis it is not possible to see that the Lie algebra is decomposable by examining the multiplication table.
Alg1 >
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| (2.2) |
Both Alg1 and Alg2 are seen to be decomposable.
Alg2 >
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Alg1 >
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Example 2
Here is the simplest example of a solvable Lie algebra which is absolutely decomposable but not decomposable. First we initialize the Lie algebra and display the multiplication table.
Alg2 >
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| (2.5) |
The algebra is indecomposable over the real numbers.
Alg3 >
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The algebra is decomposable over the complex numbers.
Alg3 >
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The explicit decomposition of this Lie algebra is given in the help page for the command Decompose.