The answer to the question in the title is "no". Semisimplicity is an open condition; however, it is not a dense open condition. Indeed, the variety of Lie algebras is reducible. There is one equation which nonsemisimple and only nonsemisimple Lie algebra structures satisfy, namely, that the Killing form Tr(ad(x)ad(y)) is degenerate, just as you writestated in the question. But there is also a system of equations which all semisimple Lie algebra structures satisfy, as do also all reductive and nilpotent Lie algebra structures, but solvable Lie algebra structures in general don't. These are the unimodularity equations Tr(ad(x))=0 for all x in the Lie algebra. These mean that the top exterior power of the adjoint representation is a trivial representation of the Lie algebra, which is obvious for any Lie algebra that coincides with its commutator subalgebra. But the nonabelian 2-dimensional Lie algebra is not unimodular. Hence in any dimension n, the direct sum of the nonabelian 2-dimensional Lie algebra with the abelian (n-2)-dimensional Lie algebra does not belong to the Zariski closure of semisimple Lie algebras.
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The answer to the question in the title is "no". Semisimplicity is an open condition; however, it is not a dense open condition. Indeed, the variety of Lie algebras is reducible. There is one equation which nonsemisimple and only nonsemisimple Lie algebra structures satisfy, namely, that the Killing form Tr(ad(x)ad(y)) is degenerate, just as you write. But there is also a system of equations which all semisimple Lie algebra structures satisfy, as do also all reductive and nilpotent Lie algebra structures, but solvable Lie algebra structures in general don't. These are the unimodularity equations Tr(ad(x))=0 for all x in the Lie algebra. These mean that the top exterior power of the adjoint representation is a trivial representation of the Lie algebra. |
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