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This is an exercise in the first chapter of "Complex Abelian Varieties" by Christina Birkenhake and Herbert Lange. As the title suggests it asks for an example of a complex torus of dimension $\geq 2$ not admitting any nontrivial complex subtorus.

The following is what I have tried for dimension 2. By a change of coordinate over $\mathbb{R}$, we can assume that the lattice is $\mathbb{Z}^4$. The question is equivalent to constructing a $4\times 4$ real matrix $J$ such that $J^2=-\mathrm{Id}$ and $J$ has no nontrivial invariant subspaces that are spanned by vectors with rational coefficients. Let $J$ be such a matrix, then for any nonzero $v_1 \in \mathbb{Q}^4$, we must have $Jv_1 \neq xv_1+yv_2$ for any $v_2\in \mathbb{Q}^4$ and $x,y \in \mathbb{R}$. To achieve this, we note that the field generated by the coordinates of $xv_1+yv_2$ has at most transcendental degree 2 over $\mathbb{Q}$, so if we can choose $J$ with many independent transcental entries such that $Jv$ have high transcendental degree over $\mathbb{Q}$ for all $v\in \mathbb{Q}$, then the problem is solved. But it is at this place I got stuck. Intuitively, I feel this is doable. But I am having trouble to write down such a matrix.

Of course, there are probably other approaches to the problem (the Jacobian of a curve in general position of genus $n\geq 2$, for example).

Thank you.

Edit: I should also assumed that $\det(J)=1$ so it preserves orientation. I think the solution probably look like this: we consider the space of all complex structures on $\mathbb{R}^4$. Each $2$ dimesnional subspace of $\mathbb{Q}^4$ defines a "subvariety" on this space, namely all $J$'s that preserve this subspace. These gives us countably many such "subvarieties". Intuition tells that these should not be able to cover the big space. But I have yet to make everything precise.

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If you see your vector space as a real vector space and you put your lattice generators as the columns of a matrix, you can see your lattice as an element of $GL(2g,\mathbb{R})$. You can then find algebraic relations that your lattice must satisfy in order for there to exist a subtorus... –  Robert Auffarth Mar 7 '12 at 0:13

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I haven't tried this by looking at conditions on the complex structure, but there is an outline in Shafarevich's book on how to do this by finding a condition on the lattice $\Lambda$.

Basically, let $\Lambda = s_1 \mathbb Z \oplus s_2 \mathbb Z \oplus s_3 \mathbb Z \oplus s_4 \mathbb Z$ where $s_i$ are vectors in $\mathbb C^2$ (the details for $n \geq 3$ are similar). Also write $s_i = (\alpha_i,\beta_i)$ for some complex numbers $\alpha_i$ and $\beta_i$.

Let $X = \mathbb C^2 / \Lambda$ be our torus, and suppose that it admits a curve $C$ (not necessarily nonsingular). The curve $C$ defines an entire 2-homology class $[C]$ by integration, and this class is non-zero (integrate a volume form on $X$ over it).

Now, as you surely know a 2-torus is diffeomorphic to $\mathbb R^4 / \mathbb Z^4$, so the 2-homology of $X$ is generated by six classes $S_{ij}$ corresponding to the 2-real-dimensional faces of $\mathbb R^4 / \mathbb Z^4$.

Consider the holomorphic 2-form $\mu = dz \wedge dw$ on $X$, and write $[C] = \sum a_{ij} S_{ij}$ for some integers $a_{ij}$. As $C$ is a curve, it carries no non-zero 2-forms, so $\mu|_C = 0$. Then

$$ 0 = \int\limits_C \mu = \sum a_{ij}(\alpha_i \beta_j - \alpha_j\beta_i), $$

by explicit calculations. You just have to calculate the integral of $\mu$ over $S_{ij}$, this gives the $\alpha_i \beta_j - \alpha_j\beta_i$ terms - the rest is by linearity. (I may be off by a factor of 4, but it doesn't matter).

So, if $X$ admits a curve, then the coefficients of the parameters $s_i$ of the lattice are linearly dependent over $\mathbb Z$. By picking some square roots of prime numbers as parameters, you can find an explicit lattice which does not satisfy this condition. The corresponding torus will thus not admit any complex curve.

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