# Are $C, C^*$ and $T$ the only 1-dimensional complex Lie groups? [closed]

I'm aware that there is a classification of certain kinds of complex Lie groups like semisimple or compact. Is there a classification of the Lie groups in the 1-dimensional case? It seems to me that the only Lie groups are the complex plane ${\mathbb{C}}$, the multiplicative group of nonzero complex numbers ${\mathbb{C}}^*$ and the torus ${\mathbb{T}}$.

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## closed as too localized by Pete L. Clark, Andreas Thom, Todd Trimble♦, Colin Tan, Andres CaicedoNov 28 '10 at 15:35

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Some confusion here. You presumably mean connected Lie groups which are also Riemann surfaces (open or closed). And if you classify them up to holomorphic isomorphism, there is not just one torus. Further you have omitted the complex plane modulo the integers, i.e. a cylinder. –  Charles Matthews Nov 28 '10 at 9:26
I don't think this is a research-level question, so I have voted to close. (@Charles Matthews: $\mathbb{C}^{\times} = \mathbb{C}/\mathbb{2 \pi i \mathbb{Z}}$ -- via the exponential map -- is a topological cylinder and is biholomorphic to your cylinder.) –  Pete L. Clark Nov 28 '10 at 9:46
Pete: You're probably right. What's the standard operating procedure in these circumstances: should I delete my answer? –  Faisal Nov 28 '10 at 9:48
@Faisal: I don't think the question is far enough away from being research level that a nice, simple answer is out of order. I actually voted up your answer, so I certainly don't mind if you keep it. Anyway, let's wait to see what others think. –  Pete L. Clark Nov 28 '10 at 9:50
It sounds like he could have been asking for just the topological types of connected complex Lie groups of dimension 1 over the complex numbers (btw, $\mathbb{C}^\ast$ is a cylinder), but it also sounds as if he might not have been aware that the topology is too coarse to distinguish them complex-analytically. –  Todd Trimble Nov 28 '10 at 12:52

Think about the easier question of classifying one-dimensional complex Lie algebras: there's only one! -- namely, $\mathbb{C}$ with the trivial bracket. The simply connected Lie group with $\mathbb{C}$ as its Lie algebra is $\mathbb{C}$ itself, and thus all other connected complex Lie groups with Lie algebra $\mathbb{C}$ are quotients of $\mathbb{C}$ by a discrete subgroup $\Gamma \subset \mathbb{C}$. This subgroup $\Gamma$ is a lattice in $\mathbb{C}$ and is therefore classified by rank: if $\mathrm{rank} \, \Gamma = 1$, then $\mathbb{C}/\Gamma = \mathbb{C}^\ast$; and if $\mathrm{rank} \, \Gamma = 2$, then $\mathbb{C}/\Gamma$ is a Riemann surface of genus one (and by varying $\Gamma$ all such arise).

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(And of course we could have $\Gamma = 0$, giving rise to $\mathbb{C}$.) This is a nice answer. –  Pete L. Clark Nov 28 '10 at 9:49
Except that when the rank of the discrete subgroup $\Gamma$ is $<2$, it is not called a lattice. –  Chandan Singh Dalawat Nov 28 '10 at 10:06
@Chandan: Technically speaking I agree with you, but I do think that some people would call this a lattice (not of "full rank"). Anyway, I don't think the correctness or the readability of the answer is impaired by using the terminology in this way. –  Pete L. Clark Nov 28 '10 at 10:13
@Pete: Yes I realised about the cylinder, almost* instantly. –  Charles Matthews Nov 28 '10 at 16:05
the answer is "no": if you construct $\mathbb C/(\mathbb Z + \mathbb Z\tau)$ (1 and $\tau$ are $\mathbb R$-linearly independent) you will get an "elliptic curve" that is also a Lie group. For different $\tau$ corresponding elliptic curves may not be equivalent as complex Lie groups