Are $C, C^*$ and $T$ the only 1-dimensional complex Lie groups? - MathOverflow [closed] most recent 30 from http://mathoverflow.net 2013-06-19T14:27:13Z http://mathoverflow.net/feeds/question/47573 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/47573/are-c-c-and-t-the-only-1-dimensional-complex-lie-groups Are $C, C^*$ and $T$ the only 1-dimensional complex Lie groups? Colin Tan 2010-11-28T09:19:19Z 2010-11-28T10:08:31Z <p>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 ${\mathbb{T}}$.</p> http://mathoverflow.net/questions/47573/are-c-c-and-t-the-only-1-dimensional-complex-lie-groups/47575#47575 Answer by Faisal for Are $C, C^*$ and $T$ the only 1-dimensional complex Lie groups? Faisal 2010-11-28T09:34:35Z 2010-11-28T09:34:35Z <p>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 <em>connected</em> 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 <code>$\mathrm{rank} \, \Gamma = 1$</code>, then $\mathbb{C}/\Gamma = \mathbb{C}^\ast$; and if <code>$\mathrm{rank} \, \Gamma = 2$</code>, then $\mathbb{C}/\Gamma$ is a Riemann surface of genus one (and by varying $\Gamma$ all such arise).</p> http://mathoverflow.net/questions/47573/are-c-c-and-t-the-only-1-dimensional-complex-lie-groups/47577#47577 Answer by zroslav for Are $C, C^*$ and $T$ the only 1-dimensional complex Lie groups? zroslav 2010-11-28T10:08:31Z 2010-11-28T10:08:31Z <p>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</p>