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Given a homology class $\alpha \in H_k(X,\mathbb{Z})$ on a variety $X$, is there a way to determine if there exists an irreducible subvariety $Y \subset X$ that has that class, i.e. $[Y] = \alpha$?

For practical reasons, assume that $X$ is a toric variety, so that one has a good handle on the cohomology ring of $X$. For example, if $X = \mathbb{P}^2$, with $H \in H_2(\mathbb{P}^2,\mathbb{Z})$ the hyperplane class, and $\alpha = 2H$, then a possible $Y$ would be the curve defined by $x y -z^2 =0$, where $x,y,z$ are the usual projective coordinates. Could this information - the existence of such a curve - solely be deduced from the properties (degree, intersection numbers, ect.) of $\alpha$?

To narrow down the problem further: Say I work in a smooth, complete two-dimensional toric variety $X$, where I know that $Pic(X)$ is generated by the toric divisors. These are the classes the zero loci $x_i=0$, where $x_i$ are the coordinates associated to the vertices of the fan of X. Now if I take an linear combination $\alpha = \sum_i \lambda_i [x_i]$, is there a irreducible subvariety in this class?

Many thanks in advance!

EDIT: Are there also necessary conditions for the existence of irreducible representatives? I.e. can I say what the circumstances are under which there is never a irreducible variety in a given divisor class.

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    $\begingroup$ An example: take an abelian variety $A$ of dimension $g$ with a principal polarization $\theta \in H^2(A,\mathbb{Z})$. The class $\dfrac{\theta^{g-1}}{(g-1)!} $ in $H^{2g-2}(A,\mathbb{Z})\cong H_2(A,\mathbb{Z})$ is representable by an irreducible curve if and only if $A$ is a Jacobian. If you could find a criterion as you suggest, you would have solved the notoriously difficult Schottky problem. $\endgroup$
    – abx
    Apr 28, 2015 at 16:09
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    $\begingroup$ Also, you would probably want to ask this differently. Say, not for a general homology class, but say a class in $\mathrm{Pic}\, X$. Otherwise, besides solving the Schottky problem you are also in danger of proving the Hodge Conjecture. $\endgroup$
    – Sándor Kovács
    Apr 28, 2015 at 16:44
  • $\begingroup$ For a complex torus, the canonical homomorphism $\wedge^* H^1(A,\mathbb{Z})\rightarrow H^*(A,\mathbb{Z})$ is an isomorphism. A principal polarization gives a unimodular alternating form on $H^1(A,\mathbb{Z})$, and the corresponding class $\theta \in H^2(A,\mathbb{Z})\cong \wedge^2 H^1(A,\mathbb{Z})$ is $e_1\wedge f_1+\ldots +e_g\wedge f_g$, for any choice of a symplectic basis $(e_1,\ldots ,e_g;f_1,\ldots ,f_g)$ of $H^1(A,\mathbb{Z})$. Since $(e_i\wedge f_i)^2=0$ one gets that $e^{\theta }=\prod (1+e_i\wedge f_i)$ is integral, hence each $\frac{\theta ^p}{p!} $ is integral. $\endgroup$
    – abx
    Apr 28, 2015 at 19:36
  • $\begingroup$ Oh my, I hope the men from Clay Institute won't be knocking at my door tomorrow, this would be an awkward conversation... Bad jokes aside, I further specified my problem (see original post). Is it now solvable? $\endgroup$
    – moep
    Apr 28, 2015 at 23:16
  • $\begingroup$ For divisors on surfaces you can solve the problem in many cases. First, if the divisor is ample, then by Bertini's theorem you can make it irreducible. Second, if the base locus of the divisor contains a curve, unless your divisor is equal to that curve you can't make it irreducible. Third, if the linear system factors through a curve, you can only make it irreducible if the degree is $1$. There are certainly explict ways to compute when these things happen on toric varieties $\endgroup$
    – Will Sawin
    Apr 29, 2015 at 3:01

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