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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

$0\longrightarrow b_{1}\cap b_{2}\longrightarrow b_{1}\oplus b_{2}\longrightarrow g\longrightarrow 0$

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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You aren't requiring that the map $b_1 \oplus b_2 \to g$ be a Lie algebra morphism, rather just that it's a map of $g$-modules, which I think presents a problem. The derivative of the exact sequence of groups

$1 \to A \to B \to G \to 1$

should be the one you started with for it to be reasonably related. But the fact that $B \to G$ is a group homomorphism means that $b_1 \oplus b_2 \to g$ (given by $(x,x') \mapsto x+x'$) is a Lie algebra morphism. This is equivalent to asking that $b_1$ and $b_2$ commute with one another since we need $[(x,x'), (y,y')] = [x,y] + [x',y']$ for all $x,y \in b_1$ and $x',y' \in b_2$. In this case we could take $B = B_1 \times B_2$ where $B_i = \exp(b_i)$ for $i=1,2$, and $A$ would be the kernel of the map $B_1 \times B_2 \to G$ given by $(b,b') \mapsto bb'$, I guess.

I think there may be an issue with putting a differentiable structure on the free product of two Lie groups, but I've never seen this mentioned before.

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very fair point. I was blind, it doesn't work :-( Well, but thanks alot for pointing this out. – olli_jvn May 3 '10 at 8:50

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