Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
1  
For $f_1$, $f_r$ of fixed degree, the possible homogeneous polynomials form a projective space. Presumably the nontrivial ones should form a constructible set, and therefore consists either of almost all or almost none of the set. This should be a satisfactory-enough concept of measure to answer your question. –  Will Sawin Dec 9 '11 at 21:33
7  
To expand on Will's comment, once the degrees of the polynomials are chosen, then Bertini's Theorem says that, outside of a Zariski closed set in the space of coefficients, all such varieties will be complete intersections and hence have degree (= cohomology class in projective space) equal to the product of the degrees. –  Alexander Woo Dec 10 '11 at 7:28
1  
I guess a baby step toward nontriviality is to take a difference of two randomly chosen cycles of fixed degree. I imagine you get a constructible cycle-class-group-valued function on a product of parameter spaces. –  S. Carnahan Dec 11 '11 at 5:37

2 Answers 2

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.

share|improve this answer

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.

share|improve this answer
    
Adding to this: in this particular case, say $X$ has dimension $k \leq r$ and your polynomials $f_1,\ldots,f_r$ have degrees $d_1, \ldots, d_r$. Let $H$ be the class of a hyperplane section on $X$. Then the cycle $Z$ that you construct pairs with $H^{k-r}$ to give the value $d_1 \cdots d_r \cdot \text{deg}(X) \neq 0$. –  Charley Dec 10 '11 at 19:06

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.