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

One definition of torus is the algebraic torus. Groups like $SU(2,\mathbb{C})$ and $SU(3,\mathbb{C})$ 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 $SL(2,\mathbb{C})$ and $SL(n+1,\mathbb{C})$ have a similar important subgroup isomorphic to $\mathbb{C}^\ast$ and $(\mathbb{C}^\ast)^n$, 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 $\mathbb{G}_m$.

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 $\mathbb{R}^n$ and quotient out by a discrete rank $n$ lattice $\Lambda$, for instance $\mathbb{R}/\mathbb{Z}$ or $\mathbb{C}/\mathbb{Z}[i]$. A complex torus is defined analogously as $\mathbb{C}^n/\Lambda$ where $\Lambda$ is a rank $2n$ lattice (since $\mathbb{C}^n$ 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 $\mathbb{C}^n/\Lambda$ is usually nearby.

Is the multiplicative group of the field, $\mathbb{G}_m$ or $\mathbb{C}^\ast$, 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 $\mathbb{C}^n/\Lambda$, 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, $\mathbb{C}^n/\Lambda$ 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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    $\begingroup$ An abelian variety is projective, so $\mathbb C^*$, which is not even compact, is not one. $\endgroup$ Apr 6, 2010 at 16:04
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    $\begingroup$ 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. $\endgroup$ Apr 6, 2010 at 16:06
  • $\begingroup$ $\mathbb{C}^n/Λ$ is an even-dimensional real torus with a specific (flat) complex structure. $\mathbb{C}*^n$ is a complex(ified) torus. $\endgroup$ Apr 6, 2010 at 16:08
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    $\begingroup$ We're really talking about spaces here, not rings, so if you want to "tensor" something, that's where you'd need to be working. The "$\times R$" you see is akin to the way $\mathbb{C}$ is like $\mathbb{R}\times\mathbb{R}$; and we can talk of analytically continuing functions on the unit circle over the complex torus as well --- this is analogous to analytic continuation of functions on $\mathbb{R}$, but we might think sooner of Laurent series in this new setting, for example $\endgroup$ Apr 6, 2010 at 16:31
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    $\begingroup$ As you have seen, the terminology "algebraic torus", though common, can be confusing. It is usually used for an algebraic group which is, over the algebraic closure, isomorphic to the direct sum of n copies of the multiplicative group. "Complex torus" is usually used for a compact, complex Lie group, necessarily then C^n modulo a lattice. But an "algebraic torus over C" is very reasonably called a "complex torus" and many complex tori -- those with Riemann forms, e.g. all elliptic curves -- are algebraic! I prefer the term "linear torus" for the (C^*)^n guy. $\endgroup$ Apr 6, 2010 at 18:19

4 Answers 4

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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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    $\begingroup$ 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. $\endgroup$ Apr 6, 2010 at 16:13
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    $\begingroup$ 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. $\endgroup$
    – Simon Rose
    Apr 6, 2010 at 16:16
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Let $ M=\mathbb C^n/Δ$ be a complex torus, then $M$ is abelian if and only if there is an integral closed positive $(1,1)$-current $\omega$ on $M$ and a point $p\in M $ such that $\omega−\epsilon\Sigma ^n_{i=1}dz_i∧d\bar z_i\geq0$ on $U$ in the sense of currents for some neighborhood $U$ of $p$ and for some positive constant number $\epsilon$, where $(z_1,⋯,z_n)$ is a coordinate system on $U$.

Note that the definition of positivity of current(introduced by Lelong) is different with positivity of a form

See Smoothing of currents and Moišezon manifolds. Several complex variables and complex geometry, by Ji, Shanyu

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    $\begingroup$ Riemann criterion characterizing abelian varieties:A complex torus $X = \mathbb C^n/Γ$ ( $Γ$ a lattice of $\mathbb C^n$) is an abelian variety if and only if there exists a positive definite hermitian form $h$ on $\mathbb C^n$ such that $$Im (h(γ_1, γ_2))\in\mathbb Z$$ for all $γ_1, γ_2 ∈ Γ.$ . See (14.5) Corollary of www-fourier.ujf-grenoble.fr/~demailly/manuscripts/agbook.pdf $\endgroup$
    – user21574
    Nov 25, 2017 at 19:19
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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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  • $\begingroup$ "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. $\endgroup$
    – Regenbogen
    Apr 6, 2010 at 16:32
  • $\begingroup$ 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. $\endgroup$ Apr 6, 2010 at 20:41
  • $\begingroup$ @Regenbogen : fixed! $\endgroup$
    – stankewicz
    Apr 6, 2010 at 20:49
  • $\begingroup$ 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. $\endgroup$ Apr 7, 2010 at 3:04
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    $\begingroup$ @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. $\endgroup$
    – stankewicz
    Apr 7, 2010 at 13:17
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The reason why ${\mathbb C}^*$ is called torus is clear by looking at it real points. It could be a split real torus ${\mathbb R}_+^*$ or a compact torus $U_1({\mathbb C}) \cong {\mathbb R}/{\mathbb Z}$.

In general, a torus in a real Lie group looks like $({\mathbb R}_+^*)^n \times {\mathbb R}^m/{\mathbb Z}^m$. The split part $({\mathbb R}_+^*)^n$ is a bit like $({\mathbb C}^*)^n$. The compact part ${\mathbb R}^m/{\mathbb Z}^m$ is a topological torus.

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  • $\begingroup$ The torus $\mathbb C^\times$ can occur in a real Lie group, but it is not of the form you quote. (In fact your definition of torus seems strange, since, according to it, all non-compact tori are disconnected.) Perhaps you are thinking algebraically, so that the torus is really $\operatorname{GL}_1^n \times \ker(\operatorname{Res}_{\mathbb C/\mathbb R}\operatorname{GL}_1 \to \operatorname{GL}_1)^m$; but then again $\mathbb C^\times$, or rather $\operatorname{Res}_{\mathbb C/\mathbb R} \operatorname{GL}_1$, is not of this form (it is a quotient). $\endgroup$
    – LSpice
    Jul 17, 2019 at 2:01
  • $\begingroup$ Agree! Is it better now? $\endgroup$
    – Bugs Bunny
    Jul 17, 2019 at 9:04
  • $\begingroup$ Yep! Although I just note here in the comments—no need to edit—that the non-compact tori you're defining aren't the same as the real points of the tori in a linear Lie group (which would be the disconnected groups that you wrote down initially). $\endgroup$
    – LSpice
    Jul 17, 2019 at 10:44

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