In the abelian case compact abelian Lie groups are unique up to diffeomorphism i.e. $(S^1)^n$.
How about the complex case? Are complex tori unique up to holomorphism i.e $(T)^n$ where $T=S^1\times S^1$?
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What does "compact abelian Lie groups are unique up to diffeomorphism" mean? – David C. Ullrich Jun 17 '19 at 20:36
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1I mean if $G$ is a compact abelian Lie group of dimension $n$, then $G\cong(S^1)^n$. – Jun 17 '19 at 20:48
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Ah. Probably you should have said that. – David C. Ullrich Jun 17 '19 at 20:51
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1For a lattice $\Lambda_1 = u\Bbb{Z}+v\Bbb{Z}$ then $\Bbb{C}/\Lambda_1\cong \Bbb{C}/\Lambda_2$ iff $\Lambda_1 = c \Lambda_2$, then $\Lambda_1=(au+bv)\Bbb{Z}+(cu+dv)\Bbb{Z}$ for any $a,b,c,d\in \Bbb{Z},ad-bc = \pm 1$. Letting $t = u/v$ we obtain the moduli space of complex tori is $(\Bbb{C-R}) /PGL_2(\Bbb{Z})= \mathcal{H}/SL_2(\Bbb{Z})$ the first modular curve https://en.wikipedia.org/wiki/Modular_curve – reuns Jun 18 '19 at 04:44
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No. There is a one-dimensional moduli of elliptic curves, for example.
user10354138
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1See this for example. For higher dimension you also have complex tori that are nonprojective. – user10354138 Jun 17 '19 at 20:18