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Every subalgebra of su (N) is reductive. The same holds true for other compact Lie groups as well. The easiest way to prove this is to use the following result: if a Lie algebra has a representation V in which its trace form (defined by (a,b) = trV (ab) is nondegenerate, then the Lie algebra is reductive. This is proved, e.g., in Sec. 5.7 of. But since the trace form in the fundamental representation of su (N) is negative definite, its restriction to any subalgebra is also negative definite.
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Every subalgebra of su (N) is reductive. The same holds true for other compact Lie groups as well. The easiest way to prove this is to use the following result: if a Lie algebra has a representation V in which its trace form (defined by (a,b) = trV (ab) is nondegenerate, then the Lie algebra is reductive. This is proved, e.g., in Sec. 5.7 of. But since the trace form in the fundamental representation of su (N) is negative definite, its restriction to any subalgebra is also negative definite.
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This is known as a Levi decomposition (see, e.g.,).
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Because the Hilbert space carries a representation of g= /k gk it carries a representation of G= /k Hk through exponentiation and therefore H= /k Hk, where Hk is a representation for Gk and gk (see, e.g.,).
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