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I've been studying particle physics, and studying Lie algebra using physics text book doesn't give me enough information, so I'm asking my question here.

Given a Lie algeba $\mathcal{A}$ where product is $[\,\cdot\, , \,\cdot\,]$, is there always a product $*$ so that $[X,Y] = X*Y - Y*X$?

Travis Willse
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Henry
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    No, but you can embed $\mathcal{A}$ in an algebra called universal enveloping algebra $U(\mathcal{A})$ such that this is true. – mez Sep 21 '14 at 03:30
  • @mez Thank you very much! – Henry Sep 21 '14 at 03:32
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    Moreover, one can embed any finite dimensional $\mathcal{A}$ into some $gl(n,k)$ for some $n$, this is Ado's theorem, see http://en.wikipedia.org/wiki/Ado%27s_theorem. Note that $U(\mathcal{A})$ is infinite dimensional if $\mathcal{A}\ne 0$. – orangeskid Sep 21 '14 at 04:21
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    The answer is yes if you just require a bilinear product (just pick $XY=[X,Y]/2$). On the other hand if we require $$ to be associative (or even "left-symmetric"), there are counterexamples. I'm not sure what are the smallest examples I guess there is no associative product for $\mathfrak{sl}_2$. – YCor Sep 30 '14 at 22:03

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