# Complex torus, C^n/Λ versus (C*)^n

I'm having trouble distinguishing the various sorts of tori.

One definition of torus is the algebraic torus. Groups like SU(2,ℂ/ℝ) and SU(3,ℂ/ℝ) have important subgroups that are topologically a circle and a torus, and I guess those were some of the most important Lie groups so the name torus stuck. Groups like SL(2,ℂ) and SL(n+1,ℂ) have a similar important subgroup isomorphic to ℂ* and (ℂ*)n, so the name torus gets applied to them too. In general, one calls the multiplicative group of an arbitrary field a torus in many situations, sometimes denoting the entire lot of them as Gm.

Another definition of a topological torus is a direct product of circles. A standard way to construct various flat geometries on a torus is to take ℝn and quotient out by a discrete rank n lattice Λ, for instance ℝ/ℤ or ℂ/ℤ[i]. A complex torus is defined analogously as ℂn/Λ where Λ is a rank 2n lattice (since ℂn has real rank 2n).

One reads in various places that every abelian variety is a complex torus, but not every complex torus is an abelian variety. The notation ℂn/Λ is usually nearby.

Is the multiplicative group of the field, Gm or ℂ*, an abelian variety?

In other words, is an algebraic torus over the complexes a complex torus?

Is an abelian variety isomorphic as a group to ℂn/Λ, or just topologically?

My dim memory of elliptic curves was that they were finitely generated abelian groups, but since they are uncountable that doesn't make any sense. Presumably I am thinking of their rational points. However, ℂn/Λ is always an abelian group, so I don't see what the fuss is about deciding when it is an abelian variety. It seems likely to me the group operations are different.

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An abelian variety is projective, so $\mathbb C^*$, which is not even compact, is not one. –  Mariano Suárez-Alvarez Apr 6 '10 at 16:04
If $k$ is a global field, then the group of $k$-points of an abelian variety is finitely generated. Maybe that's what your memory is recalling. –  Mariano Suárez-Alvarez Apr 6 '10 at 16:06
$\mathbb{C}^n/&Lambda;$ is an even-dimensional real torus with a specific (flat) complex structure. $\mathbb{C}*^n$ is a complex(ified) torus. –  some guy on the street Apr 6 '10 at 16:08
A nice book on all this is Complex tori and Abelian varieties, by Olivier Debarre. There you'll find for example the Riemann conditions which are necessary for a torus to be abelian. –  Mariano Suárez-Alvarez Apr 6 '10 at 16:08
Let me try those again... $\mathbb{C}^n/\Lambda$ first; $(\mathbb{C}^*)^n$ second... –  some guy on the street Apr 6 '10 at 16:10

The difference between an Abelian variety and $\mathbb{C}^n/\Lambda$ is that an abelian variety is polarized; that is, it comes with an ample line bundle, which yields an embedding into $\mathbb{P}^m$ for some $m$.

That is, Abelian varieties are projective algebraic, whereas complex tori (in the sense of $\mathbb{C}^n/\Lambda$) are not necessarily.

The fact that we also call $\mathbb{C}^*$ a torus is, to the best of my knowledge, unrelated. It is not an Abelian variety.

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Ah, so the reason a complex torus might not be an abelian variety is not the abelian group part, it is the projective variety part. –  Jack Schmidt Apr 6 '10 at 16:13
Correct. All Abelian varieties are diffeomorphic to S^1 x ... x S^1, and they all have the same group structure, at least over $\mathbb{C}$. I admit I don't know what happens over other fields. –  Simon Rose Apr 6 '10 at 16:16

There are a number of things floating around here.

First among them is the first excellent point that Marino made that the finite generation of group of rational points of an abelian variety over a field K is only true for global fields. So let's say we're working over $\mathbf{C}$, where any positive dimensional variety has uncountably many points.

Second is the other excellent point of Marino that $\mathbf{C}^\times$ is not compact, so it can't fit with the definition of an abelian variety as a complete, connected group variety.

Third, it's much stronger to say that an abelian variety over the complex numbers is $\mathbf{C}^n/\Lambda$ topologically than group-theoretically. But in fact much more is true. Analytically, an abelian variety is isomorphic to $\mathbf{C}^n/\Lambda$. This comes from showing that the exponential map from the tangent space at the identity is in fact surjective, followed by figuring out the kernel. Details on this can be found in Milne's notes on abelian varieties or the first chapter of Mumford's book. In fact, even if we relax down to $C^\infty$ isomorphisms (let alone homeomorphisms) we could say that an abelian variety is isomorphic to $\mathbf{C}^n/\mathbf{Z}^{2n}$.

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"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. –  Regenbogen Apr 6 '10 at 16:32
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. –  Wilberd van der Kallen Apr 6 '10 at 20:41
@Regenbogen : fixed! –  stankewicz Apr 6 '10 at 20:49
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. –  Jack Schmidt Apr 7 '10 at 3:04
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". –  Jack Schmidt Apr 7 '10 at 3:06