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Wedderburn's Theorem. Let $A$ be a simple finite $k$-algebra. Then $A$ is a matrix algebra over a finite $k$-algebra $K$ which is a skew field. (Here matrix algebra means $A=M_n(K)$ for some $n$.)

Question : $so(n)=T_{id}SO(n)$ is a matrix algebra over $K$ ?

Community
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HK Lee
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If I am not being mistaken(I remember from reading Antony Knapp's book), I think $so(n)$ is the matrix algebra of $X\in M^{n\times n}$such that $$X+X^{T}=0$$ and the relationship should be immediate from the defining relation of $SO(n)$: $$XX^{T}=1$$by playing with infinitestmals.

Bombyx mori
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Are you referring to the Artin-Wedderburn theorem? The theorem only applies to associative algebras, not Lie algebras.

Ted
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  • There is Ado's theorem in the context of Lie algebras: http://math.stackexchange.com/questions/3031/cayleys-theorem-for-lie-algebras – M Turgeon Feb 18 '14 at 02:38
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    @HeeKwonLee I'm still not sure what you're asking. Are you asking why $so(n)$ isn't a counterexample? – Ted Feb 18 '14 at 02:45
  • Since $so(n) $ is simple, so by Wedderburn's theorem it is a matrix algebra. That is, $ so(n)=M_l (K) $ for some $l $ and $K $. But $so(n)$ is not associative. Hence we can not apply Wedderburn's theorem. – HK Lee Feb 18 '14 at 05:26