3
$\begingroup$

It is known that if $G$ is an abelian real Lie group then $\exp :(\mathfrak g,+)\rightarrow G$ is a surjective homomorphism between two abelian groups. Thus, $G\cong \mathfrak g/\ker(\exp)\cong \mathbb R^n/\Gamma\cong \mathbb R^k\times(\mathbb S^1)^{n-k}$ where $\Gamma$ is a discrete subgroup of $\mathbb R^n$.

Now the situation for the complex case is slightly different; Let $G$ be an abelian complex Lie group, then $G\cong \mathbb C^m\times (\mathbb C^*)^n\times C$ where $C$ is a Toroidal group (i.e a complex connected Lie group that contains a connected normal subgroup $H$ which is a complexified torus (i.e., isomorphic to $(\mathbb C^*)^t$ such that $C/H$ is compact).

My question is about mimicking this in the nilpotent case. So let $G$ be a connected real nilpotent Lie group, then $\exp:\mathfrak g\rightarrow G$ is a surjective smooth map between two manifolds. How to prove that $G$ is diffeomorphic to $\mathbb R^m\times (\mathbb S^1)^n$?

Is this also true for the connected complex nilpotent Lie groups. I mean is $G$ holomorphic equivalent to $\mathbb C^m\times (\mathbb C^*)^n\times C$?

$\endgroup$

1 Answer 1

3
$\begingroup$

The question is asked in a possibly confusing way. In the abelian case, you have an isomorphism of complex Lie groups (i.e., both biholomorphic and a group isomorphism) with $\mathbf{C}^m\times(\mathbf{C}^*)\times T$.

In the nilpotent case you can then both ask about the biholomorphic type, or about a result really about complex Lie group classification.

The abelian case is an easy consequence of the fact that the group has to be quotient of its universal cover by a discrete normal subgroup (which has to be central, since any discrete normal subgroup in a connected group is central).

So in the nilpotent case we can mimic this: the connected nilpotent complex Lie groups are quotients of a simply connected complex nilpotent Lie group $G$ by a discrete subgroup $\Gamma$ of its center $Z_G$ (which is isomorphic to $\mathbf{C}^k$ for some $k$).

In the abelian case, one considers the real hull $V$ of $\Gamma$. Then $V\cap iV$ "corresponds" to the $T$ part, and $V+iV$ corresponds to the $\mathbf{C}^n\oplus T$ part. Here we can define these subspaces, but in general we cannot deduce a decomposition of the whole group. So I don't see a better way to state the structural fact than describing as a quotient as above. (See two examples below.)

To get the biholomorphic type (forgetting the group structure) is, however, doable. Indeed, the exponential conjugates the group law of $G$ to the Baker-Campbell-Hausdorff law of its Lie algebra $\mathfrak{g}$. In particular, we have $\exp(z)\exp(x)=\exp(z+x)$ when $z$ is central. So the left action of $Z_G$ in itself it conjugate to the left action of its Lie algebra $\mathfrak{z}$ by addition. Thus $G/\Gamma$ is biholomorphic to $\mathfrak{g}/\Gamma$, which is part of the abelian case since one now considers addition; this answers positively your last question.


First example: $G$ is the 3-dimensional complex Heisenberg group. Then $Z_G$ is 1-dimensional. One can either mod out (a) by $\{0\}$, (b) by a cyclic subgroup of $Z_G$, or (c) by a lattice in $Z_G$. In all the cases, $G/\Gamma$ does not split as a non-trivial direct product (as a complex Lie group). [Actually, the Lie algebra, as a 6-dimensional real Lie algebra, does not split as a direct product, so in none of the cases there is a decomposition even as a 6-dimensional real Lie group] (a) is a single case; (b) is also a single case up to isomorphism, since the automorphism group of the Heisenberg group acts transitively on the center minus 0. (c) Yields may cases, namely the lattice is well-defined modulo similarity, still leaving a continuum of non-isomorphic complex Lie groups (they can be checked to also not be non-isomorphic as real Lie groups, maybe up to pairs).

Second example: start from the direct product $H$ of the latter with $\mathbf{C}$. So the center $Z$ is now isomorphic to $\mathbf{C}^2$, and has a "distinguished" 1-dimensional subspace $D$, namely the intersection with the derived subgroup.

Then $\Gamma$ is isomorphic to $\mathbf{Z}^k$ for some $k\in\{0,1,2,3,4\}$. When $\Gamma$ splits as product of $\Gamma\cap D$ with another subgroup, this yields a product of one of the preceding case with an abelian complex Lie group. I think it happens precisely when the real hull $V$ of $\Gamma$ is such that $\Gamma\cap D$ is a lattice in $V\cap D$. But for $k\ge 2$ there are many other cases, for instance, when $\Gamma$ is a lattice ($k=4$) but has zero (or cyclic) intersection with $V$. Or when $V$ is a real plane ($k=2$) intersecting $D$ in a real line, and $\Gamma$ does not meet $D$.

$\endgroup$
3
  • $\begingroup$ Thank you so much for the wonderful answer. However I got 2 questions from the paragraph "So in the nilpotent case we can mimic this...". What special here for nilpotent groups, isn't every Lie group has a universal covering and the kernel is discrete central? My second question: Is it a fact that the center of simply-connected (nilpotent) group is isomorphic to $\mathbb C^n$? $\endgroup$
    – Ronald
    Aug 7, 2017 at 0:03
  • 1
    $\begingroup$ 1st question: no indeed every connected complex Lie group is quotient of its universal covering by a discrete central subgroup. Nilpotent helps because the structure of simply connected ones is easier to understand. $\endgroup$
    – YCor
    Aug 7, 2017 at 1:21
  • 1
    $\begingroup$ 2nd question: yes. One proof is that since every simply connected nilpotent Lie group $G$ is algebraic, the center $Z$ is Zariski-closed; then $Z/Z_0$ is finite. But $G/G_0$ is torsion-free, hence $Z=Z_0$. Another proof uses the Lie algebra. Actually it also works in the real case. $\endgroup$
    – YCor
    Aug 7, 2017 at 1:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .