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Let $A$ and $B$ are Lie subalgebras of a Lie algebra $L$. $U(A)$, $U(B)$ and $U(L)$ are the universal enveloping algebras of $A$, $B$ and $L$, respectively. Let $[A, B]$ be the Lie subalgebras generated by the set {$[a, b]:a\in A, b\in B$} and $U([A, B])$ be he universal enveloping algebras of $[A, B]$. I want to construct $U([A, B])$ in terms of $U(A)$ and $U(B)$. I guess that $U([A, B])$ is the Hopf subalgebra of $U(L)$ generated by the set {$\sum a_{1}b_{1}S(a_{2})S(b_{2}): a\in U(A), b\in U(B)$}, where $S$ is the antipode of the Hopf algebra $U(L)$. Is that so?

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What happens if $A, B$ are nonzero but centralize each other? It's best to work out a number of typical examples in order to get a precise formulation of the question. (Also, does the characteristic of the underlying field matter, or not?) –  Jim Humphreys May 1 '13 at 13:08
    
@Jim Humphreys: Assume that $k$ is an arbitrary field and $[A, B]=0$. Then $\forall a\in A, b\in B, \sum a_{1}b_{1}S(a_{2})S(b_{2})=\sum a_{1}S(a_{2})b_{1}S(b_{2})=\epsilon(a)\epsilon(b)\in k$. In other words, what I say is true when $k$ is an arbitrary field and $[A, B]=0$. –  sife May 1 '13 at 13:46
    
@sife: Sorry for my hasty reading of your question. But I'm still uncertain about what motivates it, and what examples have been looked at. Does your formulation work in a simple Lie algebra (characteristic 0) for the pair consisting of a Borel subalgebra and its opposite, etc.? –  Jim Humphreys May 1 '13 at 19:30
    
@Jim:If my guess is true, then we can define the "commutator subalgebra" of two Hopf subalgebra $W$ and $T$ to be the Hopf subalgebra generated by the set {$\sum a_{1}b_{1}S(a_{2})S(b_{2}): a\in W, b\in T$}. Because this definition is reasonable. –  sife May 1 '13 at 21:10
    
@Jim:Now Let $L$ be a simple Lie algebra over an algebraic closed field of characteristic 0. Then commutator of the pair consisting of a Borel subalgebra and its opposite is $L$.(I'm not sure whether I misapprehend the last sentence of your second comment). Note that $U([A, B])$ is contained in the Hopf subalgebra of $U(L)$ generated by the set {$\sum a_{1}b_{1}S(a_{2})S(b_{2}): a\in U(A), b\in U(B)$}. Therefore my guess is true in this case. –  sife May 1 '13 at 21:12

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