What is the probability for a random algebraic cycle to be homologically trivial? Someone recently asked me about the Hodge conjecture. As I understand it the conjecture asserts the existence of many non trivial algebraic cycles. The difficulty comes from the fact that we don't have a process on how to define "interesting" algebraic cycles. 
So I asked myself a very naive question. What is the probability for a random algebraic cycle to be homologically trivial?
If $X\subset \mathbb{P}^n$ be a smooth variety, we can chose random homogenous polynomials $f_1,\ldots, f_r$ and consider an algebraic sub-variety $Z = \{x \in X ~|~ f_1 = \ldots = f_r = 0\}$. Is there anything interesting we can say about the random cohomology class $cl(Z)$ (after imposing some conditions obviously)? Is there some good reference about probabilistic treatment of algebraic cycles? If not, is such an approach considered as doomed for fail and for which reasons? Difficulty to define a good parameter space and non trivial probability measure on it comes to mind.     
 A: That's interesting. No, I've not seen anything like that. Here are some thoughts, however.
For each algebraic cohomology class $\gamma\in H^*(X,\mathbb{Z})$, consider the set
of cycles $S_\gamma=\lbrace Z\mid cl(Z)=\gamma\rbrace$. My first instinct would be
to imagine that the events $S_\gamma$ are equally likely. But then the probably
measure of each would have to be zero, since it's countably additive, which is not so interesting for you. So you would want to modify this requirement somehow, but I'm not sure how.
Instead looking at cycle classes in singular cohomology, which is what I assume you meant,
you can look in  real Deligne cohomology. This is an extension of a discrete group by a torus,
and you do have interesting measures on the torus part...
Added:
What I meant was look at the homologically trivial cycles $Z^p(X)_{hom}$, on a smooth projective variety $X$, there is the Griffiths-Abel-Jacobi map to a torus
$\alpha:Z^p(X)_{hom}\to J^p(X)$. If you give the torus the Haar measure, then the
situation seems a bit more interesting (to me). Although I have no idea what can be done
with this.
A: If $X \subset \mathbb{P}^N$ is a projective variety and $Z$ is a nonempty effective algebraic cycle, then the cohomology class of $Z$ is always nonzero. Proof: Let $\dim X = n$ and let $\dim Z = k$. Then, by Bertini, a generic $\mathbb{P}^{N-k}$ in $\mathbb{P}^N$ meets $Z$ in a finite number of points. Let $H$ be such a $\mathbb{P}^{N-k}$ and let $W = X \cap H$. Then the classes of $W$ and $Z$ have nontrivial cup product in $H^{\ast}(X)$. The key point here is that, when two algebraic varieties of complementary dimensions meet in a set of dimension zero, that intersection always has positive multiplicity.
So nothing like the construction you suggest in your final paragraph can give trivial classes -- you need to either work on a non-projective variety or allow negative terms in your cycles.
