Friday 19 June 2009

lie algebras - In what way do exact sequences of Lie ideals integrate to the category of groups?

Please excuse, very naive question:



Suppose $g$ is a topological Lie algebra over Q and $G$ = $exp(g)$ the associated group



(take free group on formal symbols $exp(X)$, X $in$ $G$, and impose all relations
formally coming from the BCH-formula).



Suppose I have a short exact sequence



$0longrightarrow b_{1}cap b_{2}longrightarrow b_{1}oplus b_{2}longrightarrow glongrightarrow 0$



of g-modules, but special in the sense that



  • $g$ is the full Lie algebra,

  • $b_1$, $b_2$ (and then their intersection) are supposed to be proper Lie ideals (and not just any kind of g-module)

  • and $b_1 + b_2 = g$ ; this can happen if $g$ is weird enough

It seems to me (in some cases)/(always)/(never? ;-) such a sequence should induce
something like an exact sequence



1 -> A -> B -> G -> 1



(exact in the obvious classical sense, clearly groups are not an abelian category...)



where B could be something like the coproduct/free product of the normal
subgroups associated to the Lie ideals b1, b2; and A the subgroup associated
to the intersection of b1 and b2.



Is that true or is it complete nonsense? Is it trivially totally ridiculously false?



[please note, even though it may sound so, I do not want to go in
the direction of 'integrating' g-modules to G-modules, I would like to
transfer Lie algebra decompositions to 'nonlinear' group 'decompositions',
whatever that means....]

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