Suppose X is a smooth projective variety, say over $\mathbb{Q}$ for simplicity. Let $F$ be a finite extension of $\mathbb{Q}$. Let $\mathrm {Ch}^{r}(X/F)$ denote the Chow group of codimension $r$ algebraic cycles defined over $F$. A conjecture of Tate asserts that the cycle class map from $\mathrm{Ch}^r(X/F)$ to $H^{2r}_{et}(X)(r)$ is injective, with image in the subspace fixed by $\mathrm{Gal}(\overline{\mathbb{Q}}/F)$. In particular, the dimension of $\mathrm{Ch}^r(X/F)$ should be uniformly bounded (by the $2r^{\mathrm{th}}$ Betti number of $X(\mathbb{C})$) as $F$ varies.

On the other hand, let $\mathrm{Ch}^{r}(X/F)_0$ denote the kernel of the cycle class map, which is to say the group of homologically trivial cycles of codimension $r$ modulo rational equivalence. The dimension of this guy is predicted by Beilinson and Bloch to be given as the order of vanishing of $L(s,H^{2r-1}(X/F))$ at its central critical point. Now, the order of vanishing of this L-function can by made to increase very rapidly as $F$ varies; for example, one could (in some circumstances) choose $F$ to be an abelian extension such that each of the twists $L(s,H^{2r-1}(X/\mathbb{Q})\times \chi)$ has root number $-1$ for $\chi$ varying over characters of $\mathrm{Gal}(F/\mathbb{Q})$. When $X$ is an elliptic curve and $r$=1, this phenomenon has been confirmed in a variety of situations: for $\mathbb{Z}/p^{n}\mathbb{Z}$-towers over imaginary quadratic fields (Cornut-Vatsal), for Hilbert class fields of imaginary quadratic fields (Templier), and for towers of Kummer extensions (Darmon-Tian). However, for higher dimensional varieties and higher codimension cycles, the relevant L-functions aren't even well understood.

My question: is there a "conceptual" reason why there should have lots of homologically trivial cycle classes over extensions of the base field? In other words, if you believe certain conjectures about L-functions, then this is not hard to guess, but I am looking for some motivation which avoids L-functions.

(Edited in response to a comment of moonface.)

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    $\begingroup$ You mean "surjective" in the first paragraph, if your coefficients are $\mathbb{Q}_{\ell}(r)$. The last line of your first paragraph is incorrect. $\endgroup$
    – moonface
    Feb 3, 2010 at 16:08
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    $\begingroup$ Just to echo moonface, the injectivity assertion in your statement of the Tate conjecture is false (otherwise the rest of your question would have no content!). Rather Tate conjectures that $Ch^r$ tensored with ${\mathbb Q}_{\ell}$ surjects onto the Galois invariant part of $H^{2r}_{et}(X,{\mathbb Q}_{\ell}(r)).$ $\endgroup$
    – Emerton
    Feb 3, 2010 at 20:06

1 Answer 1


I believe that the question is still slightly mis-stated; see my comment above. (I'm sorry if I've blundered here, but will be set straight if I have, I'm sure!)

To make your question more concrete, one should take $X$ to be an elliptic curve, $r = 1$, and then $Ch(X/F)_0 = E(F)$. So your question becomes, why does $E(F)$ achieve large rank over various extensions. You discussed this in your question, but I don't your discussion fully describes the situation.

There is work of many people, especially Mazur and Rubin (as well as the others you mention) showing that Selmer groups grow in something like the expected way as you enlarge $F$ appropriately. The reason I single out Mazur and Rubin is that they work in a very general context, but they only deal with Selmer groups. One expects that the Mordell--Weil groups grow in the same manner (since one expects Sha to be finite), but this is not known, and no-one knows how to exihibit any points on elliptic curves by theoretical means, other than via Heegner points constructions.

Given this, I think the answer to the higher codimension question will be similar: no-one knows how to make cycles (other than Heegner cycles in certain special contexts). If you replaced the Chow group by some Galois-cohomological stand-in (i.e. some kind of Selmer group), perhaps one could then say something. I don't know of work of that kind, but it may well exist.

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    $\begingroup$ But of course one can certainly force E to have large rank over field extensions; you have tons of points over various quadratic fields, thus (taking epsilon care to avoid linear relations) you can make the rank of E as large as you want over large composita of quadratic fields. I take David's question to be about some version of this crude construction for higher-dimensional cycles. (I guess the construction itself is not very conceptual, but the EXISTENCE of such a construction is conceptual in that you can describe it without doing it...) $\endgroup$
    – JSE
    Feb 4, 2010 at 2:31
  • $\begingroup$ What happens if one tries something similar for higher codimension cycles? Let's say on a surface. Again, we can certainly produce many points on the surface over extension fields of some bounded degree. To make them cohomologically trivial is just a question of combining them into degree zero cycles. What happens at the step where one tries to avoid linear relations? (I'm a bit hazy on what's involved in the epsilon; is just a question of disjoint fields of definition?) $\endgroup$
    – Emerton
    Feb 4, 2010 at 3:13
  • $\begingroup$ For an elliptic curve E over $\mathbb{Q}$, the rank of $E(\mathbb{Q})$ is always infinite. In the higher dimensional cycle case, there are only very few non-trivial examples, mostly due to Chad Schoen, where the rank of the group of homologically trivial cycles over is known to be infinite. (The case of a surface is in some sense understood since homologically trivial cycles correspond to points on Pic^0 or are expected to correspond to points on the Albanese; the interesting case is when the variety has dimension > 2) $\endgroup$
    – naf
    Apr 16, 2010 at 15:33

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