Let L1 and L2 be two Lie algebras.If U(L1)is isomorphic to U(L2)as associative algebra,then L1 is isomorphic to L2 ?
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2$\begingroup$ Since the universal enveloping algebra is a graded ring, all one needs to ensure is whether the isomorphism is graded. The Poincare-Birkhoff-Witt theorem would then be applicable. $\endgroup$– ChaitanyaNov 3, 2017 at 6:09
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4$\begingroup$ A universal enveloping algebra is just filtered, not graded. $\endgroup$– Friedrich KnopNov 3, 2017 at 10:22
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$\begingroup$ Edit: Sorry, I confused 'filtered' with 'graded'. $\endgroup$– ChaitanyaNov 3, 2017 at 10:43
1 Answer
This question is known as the Isomorphism Problem for enveloping algebras and, in general, the answer is negative. The following counterexample is due to Mikhalev, Umbirbaev and Zolotykh. (See A.A. Mikhalev, A.A. Zolotykh: "Combinatorial aspects of Lie superalgebras", CRC Press, Boca Raton, FL, 1995.)
Let $F$ be a field of characteristic $p>2$ and let $L(X)$ be the free Lie algebra generated by $X=\{x,y,z\}$ over $F$. Put $L = L(X)/I$, where $I$ denotes the ideal generated by the element $h = x + [y, z] + (\mathrm{ad} x)^p(z) $ in $L(X)$. Then the Lie algebra $L$ is not free despite the fact that $U(L)$ is freely generated on 2 generators (and so $U(L)$ is isomorphic to the universal enveloping algebra of the 2-generator free Lie algebra).
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$\begingroup$ What about over fields of characteristic zero? $\endgroup$ Nov 6, 2017 at 21:52
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$\begingroup$ I think that in characteristic zero the problem is still open. $\endgroup$ Nov 7, 2017 at 13:59