Let us parametrize the set of lattices inside $\mathbb{C}^g$ with the open dense subset $U = \text{GL}_{2g}(\mathbb{R})$ of $\mathbb{R}^{4g^2}$. Does there exist a countable family $(Z_n)_{n \in \mathbb{N}}$ of algebraic real hypersurfaces of $\mathbb{R}^{4g^2}$ such that for every matrix $M \in U \setminus \bigcup_{n \in \mathbb{N}} Z_n$, the only complex subtori of $X = \mathbb{C}^g/\Gamma_M$ are $X$ and $\{0\}$?


1 Answer 1


I think yes. Let $J$ be the complex structure on $\mathbb{R}^{2g}$. Let $L$ be the lattice. Then there is a complex subtorus if and only if $L$ intersects some complex-linear subspace in a full sub-lattice. Thus, there are no complex subtori if and only if for all $x_1,\dots,x_k\in L$ there do not exist $y_1,\dots,y_k\in L$ such that $$\textrm{span}\{x_1,Jx_1,x_2,Jx_2,\dots,x_k,Jx_k\}=\textrm{span}\{x_1,y_1,x_2,y_2,\dots,x_k,y_k\}.$$ This is countably many real-algebraic conditions.

  • $\begingroup$ Some linear independence conditions probably should be thrown in for good measure. Also, clearly $k<g$ as otherwise we would get the whole space. $\endgroup$ Jun 23, 2016 at 3:27

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