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This is probably a very elementary question. I'm trying to get an explicit description of the cochain complex and coboundary maps for Lie algebra cohomology over $\mathbb{Z}$, and more generally, over commutative unital rings that are not fields.

At first, I thought that the explicit descriptions given in these lecture notes would work:

http://www.scholarpedia.org/article/An_introduction_to_Lie_algebra_cohomology/lecture_2

However, it seems that, in their proof that the second cohomology classifies all the Lie algebra extensions, the authors use the fact that the "module extension" aspect of the extension splits. Hence, to specify a Lie algebra extension, they only need to describe the behavior of the Lie bracket. This works well over fields because any module over a field is free, but no longer works for Lie rings over the integers, particularly those with torsion.

Can somebody give either the explicit construction or a reference to an explicit construction of the cochain complex that works for $\mathbb{Z}$? I am aware that the complex can be defined using derived functors, but I am not sure how to translate that description into an explicit cochain complex analogous to the bar complex.

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This is going to be complicated... The analoguous thing for rings is MacLane cohomology, which accounts for extensions of rings that are not split as abelian groups. –  Mariano Suárez-Alvarez Oct 22 '10 at 16:51
    
(Maybe looking at that will be enough to summon the Muses and be inspired for what to do in the Lie context) –  Mariano Suárez-Alvarez Oct 22 '10 at 16:51
    
Mariano, thanks for the pointer. I'll take a look at MacLane cohomology. –  Vipul Naik Oct 22 '10 at 18:51
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2 Answers

Perhaps Weibel's book on homological algebra would help? He seems to work over an arbitrary commutative ring $k$ throughout the chapter on Lie algebras. And for example, when discussing $H^2$ for Lie algebras, he says (p. 235, just after Theorem 7.6.3)

The canonical approach to classifying extensions of groups ... has an analogue only for extensions in which $\mathfrak{g}$ is a free $k$-module... Rather than pursue that method, ... we shall resort to a more functorial method.

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John, thanks. I have been looking through Weibel's book, and knew he was working over arbitrary rings, and that he took a fairly functorial approach. But thanks for highlighting this specific paragraph by Weibel, which I hadn't noticed. –  Vipul Naik Oct 22 '10 at 18:51
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Another reference is Knapp's book on Lie groups, Lie algebras and cohomology.

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