1
$\begingroup$

I believe there is a reasonable notion of $\text{Ext}^1(G,H)$ in the category of groups (where $G$ and $H$ are groups). Is there a decent reference describing this?

My particular situation involves a nilpotent Lie group $G$ with finitely many components. One may form the short-exact sequence $$1\rightarrow G_0\rightarrow G\rightarrow\pi_0(G)\rightarrow 1,$$ an element of $\text{Ext}^1(\pi_0(G),G_0)$. I am curious about conditions under which this sequence splits. I would appreciate any and all references and suggestions.

Thanks!

$\endgroup$
  • $\begingroup$ ... But to understand the general case (with your assumptions: $G$ virtually connected nilpotent Lie group), you can probably reduce to the case when $G_0$ is a torus and in general, I expect that if $W$ is the maximal (compact) torus in $G_0$ then the nilpotent extensions of $G_0$ by a finite nilpotent group $F$ should be classified by the same object as the central extensions $W$-by-$F$. $\endgroup$ – YCor Apr 12 '13 at 22:07
  • $\begingroup$ By Mostow, $G$ has a maximal compact subgroup $K$ and $KG_0=G$. In particular, if you assume in addition that $G_0$ is simply connected, then $G=G_0\rtimes K$ so your extension is split (just assuming $G_0$ nilpotent). In general, since $G$ is nilpotent and $K$ is compact, the action of $K$ on the Lie algebra of $G$ is unipotent and hence trivial, so $[K,G_0]=1$. Note that there are easy non-split extensions, e.g. with $G_0$ the circle and $G/G_0$ noncyclic group of order 4. $\endgroup$ – YCor Apr 13 '13 at 8:12
1
$\begingroup$

Since $G$ is a Lie group with finitely many components, it has a maximal compact subgroup $K$, unique up to conjugation and $G=KG_0$ (Mostow). Since $G$ is actually nilpotent, $K$ is unique and actually consists of the elements in $G$ contained in a compact subgroup. In particular $K$ is normal in $G$.

From $G$ we can define $K$ as above; conversely if we know $K$ and $G_0$, we know $G$, which is the quotient of $K\times G_0$ by the central subgroup $\{(g,g^{-1}),g\in K_0\}$.

This shows that if $F$ is a finite nilpotent group, the set classifying extensions of $G_0$ by $F$ that are nilpotent is in canonical bijection with the set classifying central extensions of $K_0$ by $F$. The latter is in bijection with $H^2(F,K_0)$, which is a reasonable well-understood object ($K_0$ is a torus).

In particular: 1) If $G_0$ is simply connected we get $G=G_0\times F$ (only the direct extension)

2) Everything boils down to when $G_0$ is a torus. If $S^1$ is the 1-circle, $Q$ is the group of quaternions of order 8 and if we define $G$ as the quotient of $S^1\times Q$ by the diagonal element of order 2, then we get a nontrivial extension (as we see by counting elements of order 2).

NB: you assume explicitly that $G$ is nilpotent. So I don't claim to classify all extensions of $G_0$ by $F$.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Pages 177-190 of here contain a relatively complete description of group extensions, following Schreier, Baer, Eilenber + MacLane, Hochschild, and Serre.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.