difference between equivalence relations on algebraic cycles - MathOverflow

difference between equivalence relations on algebraic cycles

2010-01-14T19:29:31Z

For the definitions of the equivalence relations on algebraic cycles see http://en.wikipedia.org/wiki/Adequate_equivalence_relation.

I want to know how far away from each other the equivalence relations on algebraic cycles are and what the intuition is for them.

My impression is that rational equivalence gives much bigger Chow groups than algebraic equivalence, and that algebraic equivalence, homological equivalence and numerical equivalence are quite tight together.

Take for example an elliptic curve. We have $CH^1(E) = \mathbb{Z} \times E(K)$, algebraic equivalence (take $C = E$) $\mathbb{Z}$ = numerical equivalence.

Answer by Tony Pantev for difference between equivalence relations on algebraic cycles

2010-01-15T00:33:00Z

It is indeed true that rational equivalence gives bigger groups of cycles than say algebraic equivalence. However algebraic equivalence is also far away from homological equivalence. In complex geometry people often study a basic invariant of a variety $X$ called the Griffiths group. By definition the Griffiths group $Gr(X)$ is the group of cycles homologous to zero (in the classical topology) modulo cycles algebraically equivalent to zero. Griffiths originally showed that this group can contain non-torsion elements, and Clemens showed that it can happen that $Gr(X)\otimes \mathbb{Q}$ is infinite dimensional as a rational vector space. People have studied Griffiths groups quite a bit and have proven some great theorems about them. For instance Voisin showed that the Griffiths group of a Calabi-Yau threefold which is general in its moduli is infinitely generated.

Answer by jvp for difference between equivalence relations on algebraic cycles

2010-01-15T02:00:37Z

I will focus on complex projective varieties.

Codimension one

The situation in codimension one is considerably simpler than in higher codimensions. Codimension one rational equivalence classes are parametrized by $Pic(X)= H^1(X,\mathcal O_X^{\ast})$ while algebraic equivalence classes are parametrized by the Neron-Severi group of $X$, which can be defined as the image of the Chern class map from $Pic(X)$ to $H^2(X,\mathbb Z)$. It follows that in codimension one

the group of rational equivalence classes is a countable union of abelian varieties;
the groups of algebraic equivalence classes and homological equivalence classes coincide, and are equal to $NS(X)$ a subgroup of $H^2(X,\mathbb Z)$;
the group of numerical equivalence classes is the quotient of $NS(X)$ by its torsion subgroup.

Higher codimension

The higher codimension case, as pointed out by Tony Pantev, is considerably more complicate and algebraic and homological equivalence no longer coincide.

Concerning rational equivalence, Mumford proved that the Chow group of zero cycles of surfaces admitting non-zero holomorphic $2$-forms are infinite dimensional, contradicting a conjecture by Severi. The paper is Mumford, D. Rational equivalence of $0$-cycles on surfaces. J. Math. Kyoto Univ. 9 1968.

Warning

The definitions of rational and algebraic equivalence at wikipedia are not correct. I will commment below on the algebraic equivalence.

There one can find the following definition.

$Z ∼_{alg} Z'$ if there exists a curve $C$ and a cycle $V$ on $X × C$ flat over C, such that $$V \cap \left( X \times\lbrace c\rbrace \right) = Z \quad \text{ and } \quad V \cap \left( X \times\lbrace c\rbrace \right) = Z' $$ for two points $c$ and $d$ on the curve.

This is not correct. The correct definition is

$Z ∼_{alg} Z'$ if there exists a curve $C$ and a cycle $V$ on $X × C$ flat over C, such that $$V \cap \left( X \times\lbrace c\rbrace \right) -
V \cap \left( X \times\lbrace d\rbrace \right) = Z - Z' $$ for two points $c$ and $d$ on the curve.

To construct an example of two algebraically equivalent divisors which do not satisfy the wikipedia definition let $X$ be a projective variety with $H^1(X,\mathcal O_X) \neq 0$ and take a non-trivial line-bundle $\mathcal L$ over $X$ with zero Chern class. If $Y = \mathbb P ( \mathcal O_X \oplus \mathcal L)$ then $Y$ contains two copies $X_0$ and $X_{\infty}$ of $X$ ( one for each factor of $\mathcal O_X \oplus \mathcal L$ ) which are algebraically equivalent but can't be deformed because their normal bundles are $\mathcal L$ and $\mathcal L^{\ast}$. This does not contradict the second definition because for sufficiently ample divisors $H$ it is clear $X_0 + H$ can be deformed into $X_{\infty} + H$.

Answer by Mikhail Bondarko for difference between equivalence relations on algebraic cycles

2010-02-10T08:41:15Z

You may be interested in the following paper: Nilpotence theorem for cycles algebraically equivalent to zero, by Vladimir Voevodsky http://www.math.uiuc.edu/K-theory/0041/ (possibly, a newer version exists now).

Answer by norondion for difference between equivalence relations on algebraic cycles

2010-02-17T19:46:48Z

A good reference is also Fulton, Intersection Theory, Chapter 19.