Is there any correspondence between Lie rings and Lie groups such that if one proves for a Lie algebra $g$ with ideal $I$ that $g/I$ is a Lie algebra, then the same result holds automatically for the quotient ring?
Thanks in advance AB
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$\begingroup$ You need not prove that the quotient of a Lie algebra by an ideal is a Lie algebra, that's done in books. BTW, what do you mean by 'quotient ring'? $\endgroup$– Fernando MuroApr 30, 2013 at 1:03
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$\begingroup$ I don't understand your question. What quotient ring do you mean? What does a Lie group have to do with anything? $\endgroup$– MTSApr 30, 2013 at 1:51
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$\begingroup$ Take a Lie ring and build a quotient out of it! Here I mean if $g$ is a Lie algebra and $L$ its Lie ring (an associative ring equipped with a bracket operator $[,]$ can be made into a Lie ring), then $L/I$ defines a quotient ring where I is a two-sided ideal in $L$. $\endgroup$– AlirezaApr 30, 2013 at 1:52
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$\begingroup$ Lie ring = universal enveloping algebra? $\endgroup$– Sam GunninghamApr 30, 2013 at 1:55
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$\begingroup$ @Sam Gunningham: Exactly! $\endgroup$– AlirezaApr 30, 2013 at 1:59
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The question is not posed in a clear way but, if I am interpreting correctly, it is enough to recall that the universal enveloping algebra $U(g/I)$ of the Lie algebra quotient $g/I$ is isomorphic to $U(g)/B$, where $B$ is the two-sided ideal of $U(g)$ generated by $I$. (Of course, here $g$ is identified with its isomorphic image in $U(g)$.) This is a rather elementary fact: see e.g. Theorem 1 in Chapter 5 of the book "N. Jacobson: Lie algebras".
answered Apr 30, 2013 at 8:04