Timeline for Complex torus, C^n/Λ versus (C*)^n
Current License: CC BY-SA 2.5
9 events
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| Apr 7, 2010 at 13:53 | comment | added | stankewicz | (unless it's also complex linear) | |
| Apr 7, 2010 at 13:17 | comment | added | stankewicz | @JS : Absolutely. The example to keep in mind is the $n=1$ case where an elliptic curve is described as $\mathbf{C}/\mathbf{Z}\omega_1\oplus\mathbf{Z}\omega_2$ where the $\omega$'s form a basis for $\mathbf{C}$ as a real vector space. This is of course equivalent to the fact that there is a real linear isomorphism from $\mathbf{C}$ to itself sending say $\omega_1$ to $1$ and $\omega_2$ to $i$. Since it's real linear, it's $C^\infty$ but this map clearly doesn't respect the analytic structure. | |
| Apr 7, 2010 at 3:06 | comment | added | Jack Schmidt | Wilber van der Kallen: Oh, I like this. So just like some subgroups of a finite group are normal, and some are not, we are checking whether a specific lattice is "normal" in some special sense. Every lattice is a kernel of a group homomorphism, probably even a C∞ group homomorphism, and you are saying maybe even real Lie group homomorphisms. However, the only lattices I want are the ones that are kernels of complex Lie group homomorphisms, a very special type of "normal". | |
| Apr 7, 2010 at 3:04 | comment | added | Jack Schmidt | stankewicz: So you mean that if I ignore the analytic or complex part of the isomorphism, but keep even all the way up to C∞ diffeomorphism (and group isomorphism), then instead of a very specific lattice with all sorts of fancy ample line bundle type properties, I just get any old lattice, like Z^(2n). In other words, the only way to tell the lattices apart is to keep analytic or complex structure, otherwise they are all the same. | |
| Apr 6, 2010 at 20:49 | comment | added | stankewicz | @Regenbogen : fixed! | |
| Apr 6, 2010 at 20:49 | history | edited | stankewicz | CC BY-SA 2.5 |
added 21 characters in body
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| Apr 6, 2010 at 20:41 | comment | added | Wilberd van der Kallen | When considering $\mathbb C^n/\Lambda$ one should think of it as a complex Lie group, not just as a topological group or a real Lie group. In particular its Lie algebra has the structure of a complex vector space. The exponential map is a map of complex Lie groups. To say that $\mathbb C^n/\Lambda$ is an abelian variety means at least that as a complex Lie group it is isomorphic to an abelian variety. Equivalently, there is a complex Lie group map from $\mathbb C^n$ onto an abelian variety with $\Lambda$ as kernel. | |
| Apr 6, 2010 at 16:32 | comment | added | Regenbogen | "group of rational points of an abelian variety over a field K is only true for global fields." -- Here I take that you mean finite generation, and if so it is true for finitely generated fields by a theorem of Neron. | |
| Apr 6, 2010 at 16:30 | history | answered | stankewicz | CC BY-SA 2.5 |