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I am trying to work through a proof/argument to show that the adjoint representation of a semisimple Lie algebra is completely reducible.

Suppose S denotes an invariant subspace of the Lie algebra, and we pick [itex]Y_i[/itex] in the invariant subspace S. The rest of the generators [itex]X_r[/itex] are such that the natural inner product is [itex](X_r, Y_i) = 0[/itex]. This can be done by some suitable Gram Schmidt orthogonalization if necessary.

To begin with, I argue that the killing metric in this basis is block diagonal. If i denotes an index on Y and r denotes an index on X, then [itex]g_{ir} = 0[/itex] as the Killing form is the natural inner product or its negative depending on whether Y is chosen to be symmetric or antisymmetric. This is OK.

But the following argument is unclear to me

Since S is an invariant subspace, structure constants of the form [itex]{f_{ir}}^{s}[/itex] are zero.

Is it reasonable to expect [X, Y] to be in S as well as its complement? The only way then that this would be possible is if [X, Y] = 0.

The other argument (specious to me) is that Y and X live in different spaces so they must commute. This seems physically reasonable, but I don't see how to argue this mathematically.

Any help would be greatly appreciated. Oh and I should point out, I am learning this from the standpoint of a theoretical physicist, so please feel free to point out mistakes/improvements in the reasoning (or holes in my understanding) from a purely mathematical perspective.

Thanks!

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# The adjoint representation of a semisimple Lie algebra is completely reducible

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