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Let $T=V/\Gamma $ be a complex torus; so $V$ is a finite-dimensional complex vector space and $\Gamma$ a lattice in $V$. Moreover I have a positive definite hermitian form $H$ on $V$ such that the real part of $H$ takes integral values on $\Gamma $.

If $\dim V=1$, the existence of such a form is equivalent to saying that the elliptic curve $T$ has complex multiplication. What can one say if $\dim V>1$?

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  • $\begingroup$ To construct such an abelian variety, you can take any integral lattice (so an integral vector space), tensor it with $\mathbb R$, and choose a complex structure agreeing with the Euclidean structure on that vector space. So for each integer lattice there is one connected moduli space of dimension $g(g-1)$. $\endgroup$
    – Will Sawin
    Nov 16, 2013 at 4:27
  • $\begingroup$ You probably mean $\frac{1}{2}g(g-1) $? If your lattice is $L$, you are looking for decompositions $L\otimes \Bbb{C}=V\oplus \bar{V}$, with $V$ totally isotropic for the extended quadratic form. The dimension of such isotropic spaces is $\frac{1}{2}g(g-1) $. $\endgroup$
    – abx
    Nov 16, 2013 at 7:23

1 Answer 1

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Some minor disagreement has arisen in the comments as to `what' an elliptic curve is. Concretely, to perhaps an algebraic geometer the quotient of $\mathbb{C}$ by the lattice $\mathbb{Z}+\mathbb{Z} \sqrt{-D}$ is a very reasonable and simple elliptic curve. However I'm prone to dismiss it (or specifically, rescale it to have covolume one) since $diag(1, \sqrt{D})$ does not lie in $SL_2\mathbb{R}$ but rather $GSL_2$. Likewise, I am prone to only admit principally polarized abelian varieties, i.e. arising from $Sp_{2g}$ and not $GSp_{2g}$. This unfortunately invalidates my perspective on the OP's question, since he/she appears open to arbitrary covolume lattices and my comments below are (i see now) specific to being in the unimodular (normalized covolume one) setting. This confuses the discussion because the CM is not scale-invariant, e.g. the torus $\mathbb{Z} D^{1/4} + \mathbb{Z} iD^{3/4}$ has $\mathbb{Q}(i)$-CM, versus $\mathbb{Q}(\sqrt{-D})$-CM in the case of $\mathbb{Z}[\sqrt{-D}]$.

I'd like to make an extended comment. As I see it, the positive definite hermitian form arising from an elliptic curve (or, principally polarized abelian variety) has the form $H=g_j+i\omega$, where $\omega$ is the standard symplectic form $\omega(x,y)={}^txj_{std}y$ on $\mathbb{R}^{2g}$ and $g_j(x,y):=\omega(jx,y)$ for $j$ an $Sp_{2g}\mathbb{R}$ conjugate of $j_{std}$. In otherwords the data of a complex elliptic curve (or ppav) amounts to an $Sp_{2g}\mathbb{R}$-conjugate of the standard almost complex structure $j_{std}=\begin{pmatrix} 0 & -I_g \\ I_g & 0 \end{pmatrix}$.

In the special case of ppavs, the hypothesis that the real part of $H=H_j$ be integral on the lattice $\mathbb{Z}^{2g}$ (i.e. $j\mathbb{Z}^{2g} \subset \mathbb{Z}^{2g}$) is equivalent to having $j\mathbb{Q}^{2g} \subset \mathbb{Q}^{2g}$. This is because if $j\mathbb{Q}^{2g} \subset \mathbb{Q}^{2g}$, then $j\mathbb{Z}^{2g}$ is a sublattice of some $\frac{1}{N}\mathbb{Z}^{2g} \subset \mathbb{Q}^{2g}$ on which the symplectic form $\omega$ (i.e. imaginary part of $H=H_j$) is integral and unimodular. But $\mathbb{Z}^{2g}$ is the maximal submodule of $\mathbb{Q}^{2g}$ on which this occurs. In otherwords, we know $j$ is defined over $\mathbb{Q}$.

Now the claim that I want to make is this: if $j$ is defined over $\mathbb{Q}$ then the cyclic subgroup $\{ \exp j\theta\}_{\theta}$ of $Sp_{2g}\mathbb{R}$ is defined over $\mathbb{Q}$, where $\exp$ refers to matrix exponential. This would establish that a ppav has CM (in sense of hodge theory) if $j$ is defined over $\mathbb{Q}$. Now the converse of this statement is not clear (and i don't believe true). In an earlier version of this `answer' i wrongly said they were equivalent. In fact $Sp_{2g}\mathbb{R}$ has $\mathbb{Q}, \mathbb{R}, \mathbb{C}$-rank all equal to $g$, and $e^{j\theta}$ is merely a one-dimensional torus. This explains somewhat how the equivalence of $j$ rational and CM degenerates in higher dimension.

Hodge structures (in the sense, say, of Griffiths/Green/Kerr) on ppav's consist of nonconstant linear representations $\rho:\mathbb{S}^1 \to Sp_{2g}\mathbb{R}$. The image of $i \in \mathbb{S}^1$ actually determines the representation, with $\rho(i)$ giving an almost complex structure $j$ and extending to all of the unit circle via the exponential formula $\exp j\theta=(\cos\theta) I_{2g}+(\sin\theta) j$. I first learned of this formula right here on MO thanks to R. Bryant! The so-called Mumford-Tate group $MT(\rho)$ of the representation is the minimal algebraic subgroup of $Sp_{2g}\mathbb{R}$ defined over $\mathbb{Q}$ and containing the image of the representation $\rho(\mathbb{S}^1)$. From this point-of-view, CM is characterized as those representations (i.e. almost complex structures) whose MT-group is a (necessarily compact!) algebraic torus defined over $\mathbb{Q}$ in $Sp_{2g}\mathbb{R}$.

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  • $\begingroup$ You have to be a bit careful. If you want both the real and imaginary part to literally be integers, not only is it a CM variety, it is isogenous to a direct sum of elliptic curves with CM by $\mathbb Q(i)$! This is because not only do you know that MT is a torus, you know "which" torus it is. If you just demand they be integers after scalling, you get a direct sum of elliptic curves with CM by $\mathbb Q(\sqrt{-D})$ for some $D$ depending on the scaling factor. This is because MT is a one-dimensional torus. $\endgroup$
    – Will Sawin
    Nov 16, 2013 at 5:05
  • $\begingroup$ Another way to say this is that you have a rational matrix $j$ which is the complex structure up to a scalar, so $j^2=-D$ for some $D$, and so you make the lattice tensored with $\mathbb Q$ a vector space over $\mathbb Q(j)=\mathbb Q(\sqrt{-D})$. Then all endomorphisms of this vector space will give endomorphisms of the abelian variety, up to multiplication by an integer to clear the denominator. However this relies on the form having both real and imaginary parts integral, rather than just real. $\endgroup$
    – Will Sawin
    Nov 16, 2013 at 5:07
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    $\begingroup$ Thanks to both of you, but I insist that I want only the real part of $H$ to be integral on the lattice. I expect that most of these complex tori are non algebraic. $\endgroup$
    – abx
    Nov 16, 2013 at 6:25
  • $\begingroup$ @WillSawin, it seems i have misread complex torus' as meaning principally polarized' complex torus, i.e. having $j^2=-1$. However, I don't understand your first point that the ppav's i'm describing are isogenous to direct sums of elliptic curves with $Q(i)$ CM. Elliptic curves with CM (besides the `square' elliptic curve at $\tau=i$) are not isogenous to $Q(i)$ CM curves, i.e. do not have the standard square lattice as sublattice. $\endgroup$
    – JHM
    Nov 16, 2013 at 13:39
  • $\begingroup$ If you take an elliptic curve with lattice $ \mathbb Z[\sqrt{-D}]$, then a Hermitian form whose imaginary part is integral must have the norm $N z \overline{w} / \sqrt{D}$ for some integer $N$. A Hermitian form whose real part is intetgral must have the form $M z \overline{w}$ for some integer $M$. These are inconsistent. If the principal polarization is not equal to the form with integral real part, I don't see how your argument works. $\endgroup$
    – Will Sawin
    Nov 16, 2013 at 16:35

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