difference between equivalence relations on algebraic cycles - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T18:10:26Z http://mathoverflow.net/feeds/question/11774 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/11774/difference-between-equivalence-relations-on-algebraic-cycles difference between equivalence relations on algebraic cycles norondion 2010-01-14T19:29:31Z 2010-02-17T19:46:48Z <p>For the definitions of the equivalence relations on algebraic cycles see <a href="http://en.wikipedia.org/wiki/Adequate_equivalence_relation" rel="nofollow">http://en.wikipedia.org/wiki/Adequate_equivalence_relation</a>.</p> <p>I want to know how far away from each other the equivalence relations on algebraic cycles are and what the intuition is for them.</p> <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/11774/difference-between-equivalence-relations-on-algebraic-cycles/11805#11805 Answer by Tony Pantev for difference between equivalence relations on algebraic cycles Tony Pantev 2010-01-15T00:33:00Z 2010-01-15T00:33:00Z <p>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 <em>Griffiths group</em>. 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. </p> http://mathoverflow.net/questions/11774/difference-between-equivalence-relations-on-algebraic-cycles/11813#11813 Answer by jvp for difference between equivalence relations on algebraic cycles jvp 2010-01-15T02:00:37Z 2010-01-15T14:49:57Z <p>I will focus on complex projective varieties. </p> <h2>Codimension one</h2> <p>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 </p> <ul> <li>the group of <strong>rational</strong> equivalence classes is a countable union of abelian varieties;</li> <li>the groups of <strong>algebraic</strong> equivalence classes and <strong>homological</strong> equivalence classes coincide, and are equal to $NS(X)$ a subgroup of $H^2(X,\mathbb Z)$;</li> <li>the group of <strong>numerical</strong> equivalence classes is the quotient of $NS(X)$ by its torsion subgroup.</li> </ul> <h2>Higher codimension</h2> <p>The higher codimension case, as pointed out by Tony Pantev, is considerably more complicate and algebraic and homological equivalence no longer coincide. </p> <p>Concerning rational equivalence, Mumford proved that the Chow group of zero cycles of surfaces admitting non-zero holomorphic $2$-forms are <strong>infinite dimensional</strong>, contradicting a conjecture by Severi. The paper is Mumford, D. Rational equivalence of $0$-cycles on surfaces. J. Math. Kyoto Univ. 9 1968. </p> <h2>Warning</h2> <p>The definitions of rational and algebraic equivalence at <a href="http://en.wikipedia.org/wiki/Adequate%5Fequivalence%5Frelation" rel="nofollow">wikipedia</a> are not correct. I will commment below on the algebraic equivalence.</p> <p>There one can find the following definition.</p> <blockquote> <p>$Z ∼_{alg} Z'$ 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.</p> </blockquote> <p>This is not correct. The correct definition is</p> <blockquote> <p>$Z ∼_{alg} Z'$ if there exists a curve $C$ and a cycle $V$ on $X × C$ flat over C, such that $$V \cap \left( X \times\lbrace c\rbrace \right) -<br /> V \cap \left( X \times\lbrace d\rbrace \right) = Z - Z'$$ for two points $c$ and $d$ on the curve.</p> </blockquote> <p>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 $X_{\infty} + H$.</p> http://mathoverflow.net/questions/11774/difference-between-equivalence-relations-on-algebraic-cycles/14878#14878 Answer by Mikhail Bondarko for difference between equivalence relations on algebraic cycles Mikhail Bondarko 2010-02-10T08:41:15Z 2010-02-10T08:41:15Z <p>You may be interested in the following paper: Nilpotence theorem for cycles algebraically equivalent to zero, by Vladimir Voevodsky <a href="http://www.math.uiuc.edu/K-theory/0041/" rel="nofollow">http://www.math.uiuc.edu/K-theory/0041/</a> (possibly, a newer version exists now).</p> http://mathoverflow.net/questions/11774/difference-between-equivalence-relations-on-algebraic-cycles/15617#15617 Answer by norondion for difference between equivalence relations on algebraic cycles norondion 2010-02-17T19:46:48Z 2010-02-17T19:46:48Z <p>A good reference is also Fulton, Intersection Theory, Chapter 19.</p>