Universal coverings of Steinberg Lie algebras of small characteristic

It is well-known that the second homology group H2(stn(R)) of the Steinberg Lie algebra stn(R) is trivial when n ≥ 5. In this paper, we will work out H2(stn(R)) explicitly for n = 3, 4 which are not necessarily trivial. Consequently, we obtained H2(sln(R)) for n = 3, 4. Introduction Steinberg Lie algebras stn(R) and/or their universal coverings have been studied by Bloch [Bl], Kassel-Loday [KL], Kassel [Ka], Faulkner [F], Allison-Faulkner [AF], Berman-Moody [BM], [G1, 2] and [AG], and among others. They are Lie algebras graded by finite root systems of type Al with l ≥ 2. In most situations, the Steinberg Lie algebra stn(R) is the universal covering of the Lie algebra sln(R) whose kernel is isomorphic to the first cyclic homology group HC1(R) of the associative algebra R and the second Lie algebra homology group H2(stn(R)) = 0. It was shown in [Bl] and [KL] that H2(stn(R)) = 0 for n ≥ 5. [KL] mentioned without proof that H2(stn(R)) = 0 for n = 3, 4 if 1 2 ∈ lies in the base ring K. This was proved (see [G1] 2.63) for n = 3 if 1 6 ∈ K and for n = 4 if 1 2 ∈ K. In this paper, we shall work outH2(stn(R)) explicitly for n = 3, 4 without any assumption on (characteristic of) K. It is equivalent to work on the Steinberg Lie algebras stn(R) of small characteristic for small n. This completes the determination of the universal coverings of the Lie algebras stn(R) and sln(R) as well. More precisely, let K be a unital commutative ring and R be a unital associative Kalgebra. Assume that R has a K-basis containing the identity element (so R is a free K-module). The Lie algebra sln(R) is the subalgebra of gln(R) (the n by n matrix Lie algebra over K with coefficients in R), generated by eij(a) for 1 ≤ i 6= j ≤ n, a ∈ R, where 2000 Mathematics Subject Classification: 17B55, 17B60. Research of the first author was partially supported by NSERC of Canada and Chinese Academy of Science.