In a lot of places it is mentioned that rational equivalence of cycles on a smooth projective variety is the finest adequate equivalence relation. Assuming that it is an adequate equivalence relation, how to see that this is the finest? Any references for this?
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Among the frequently used equivalence relations (for, say, complex varieties, that is, over $\mathbb C$) there exist the following implications:
$$ \text{Rational}\Rightarrow\text{Algebraic}\Rightarrow\text{Homological}\Rightarrow\text{Numerical}. $$
For a reference see Chapter 19 of Fulton's Intersection Theory.
EDIT. I removed my apparently incorrect assumption that adequate has not been defined.
answered May 17, 2011 at 17:59
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2$\begingroup$ There is a wikipedia article on Adequate Equivalence relations here: en.wikipedia.org/wiki/Adequate_equivalence_relation. Also there is this article by Jannsen "Equivalence Relations on Algebraic Cycles" where he defines adequate equivalence relations. My problem is that to define homological equivalence one needs a cycle class map which should satisfy functoriality conditions. In order to define $f^∗$ for cycles one would need to go modulo some adequate equivalence relation. This condition would be satisfied if Rational implies Homological. $\endgroup$– RexCommented May 17, 2011 at 18:12
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1$\begingroup$ I mean to say that if rational $\Rightarrow$ homological, then the cycle map factors $CH^i(X)\to H^{2i}(X)$. Now we can talk about compatibility with respect to $f^*$. In general, however, it is not clear how to define $f^*$ on the free group on cycles of codimension $i$ i.e. $Z^i(X)$. $\endgroup$– RexCommented May 17, 2011 at 18:19
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4$\begingroup$ It seems to me that using axiom 3 (pushforwards) in the wikipedia article you can see that any adequate equivalence relation for which two points in $P^1$ are equivalent will be such that in general any two rationally equivalent cycles are equivalent. $\endgroup$ Commented May 17, 2011 at 18:42
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3$\begingroup$ Apparently the notion of "adequate equivalence relation" was defined by Samuel. Uwe Jannsen's remarkable Invent. Math. paper from 1992 gives an intuitive version, replacing the pushforward axiom with the requirement that the equivalence relation respect (flat) pullback of cycles. Using that applied to a map to $P^1$, as Tom says, it's easy to see rational equivalence is finest. $\endgroup$ Commented May 17, 2011 at 19:00
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1$\begingroup$ @Rex, to pullback a cycle via a map $f$, first use the moving lemma (or axiom) to make the cycle and the map transverse. Then take the inverse image. You can also use the "pushforward" axiom applied to the graph of $f$, together with the moving axiom, to get a pullback. In more fancy terms, for smooth varieties the operational Chow groups (which are contravariant) are isomorphic to the usual ones, see Fulton, 17.4. $\endgroup$ Commented May 17, 2011 at 19:19