10
$\begingroup$

Given an elliptic curve $E$ over $\mathbb{Q}$, I have read somewhere (But I can't remember exactly where) that the Beilinson conjecture asserts that: The rank of the albelian group $K_{2}(E)$ (the second algebraic K-theory) is equal to the rank of the abelian group of the rational points $E(\mathbb{Q})$.

Conjecture: $\textrm{rk } K_{2}(E)= \textrm{rk } E(\mathbb{Q})$

Question What are some evidences of such conjecture? Is it verified in some known cases?

$\endgroup$
5
$\begingroup$

I would like to add a couple of references to the ones provided in Timo Keller's comment.

First of all, the conjecture is not correctly stated. According to the conjectures, the rank of the curve (which is related to the rank of $K_0(E)$) should equal the order of vanishing of the Hasse-Weil L-function of $E$ at $s=1$, but the rank of $K_2(E)$ should be the order of vanishing of the L-function at $s=0$. The correctly stated conjecture on the rank of $K_2(E)$ is that the rank of $K_2(E/\mathbb{Q})$ should be $1+S$ where $S$ is the number of places with multiplicative/semistable reduction. Alternatively, the rank of $K_2(\mathcal{E})$ should be 1, where $\mathcal{E}$ is a regular proper model over ${\rm Spec}\mathbb{Z}$ of the curve. More generally, for $E/F$ an elliptic curve over a number field $F$, the rank of $K_2(E/F)$ should be $[F:\mathbb{Q}]+S$. This is discussed in the article of Bloch and Grayson linked in Timo Keller's comment:

  • Bloch, S. and Grayson, D., $K_2$ and $L$-functions of elliptic curves: computer calculations, Applications of algebraic $K$-theory to algebraic geometry and number theory, Part I, II (Boulder, Colo., 1983), Contemp. Math., 55 Amer. Math. Soc., Providence, RI, 1986, 79--88.

As far as I know, there is not a single curve for which the conjecture is known. I don't think the full Beilinson conjectures are known for any scheme other than rings of $S$-integers, where they are the classical theorems of algebraic number theory. It seems that this is the only evidence we have. That and the marvelous beauty of the conjectures. Anyway, the conjecture explains the ranks of K-theory resp. motivic cohomology in terms of some analytic invariants which we can actually compute, namely Deligne cohomology. Beilinson's conjectures basically state that Deligne cohomology tells us everything about rational motivic cohomology. (For more precise formulations, look at survey articles on Beilinson's conjectures, such as the ones by Nekovar or Deninger-Scholl.)

Ok, some partial results are known: the work of Goncharov-Levin has produced elements with non-trivial regulator, so we know that the rank must be at least one.

  • Goncharov, A. B. and Levin, A. M., Zagier's conjecture on $L(E,2)$, Invent. Math. 132 (1998), no.2, 393--432.

Analogous results are known in the function field case by the work of Kondo and Yasuda. Again, we know that the rank is at least the predicted rank, by explicit generators.

  • S. Kondo and S. Yasuda. On the second rational K-group of an elliptic curve over global fields of positive characteristic. Proc. Lond. Math. Soc. (3) 102 (2011), no. 6, 1053–1098.

There are further partial results, via K-theory with finite coefficients. A paper of Soulé establishes some results for p-adic K-theory. In particular, he gives examples where the rank of the $p$-adic K-theory is in fact equal to the rank predicted by Beilinson's conjecture. The method is to relate the $p$-adic K-theory to étale cohomology and then compute the rank of the latter.

  • C. Soulé. $p$-adic K-theory of elliptic curves. Duke Math. J. 54 (1987), 249--269.

There is also a paper of Kato. It's mostly concerned with generalizations of the Hasse principle, but there is some application to K-theory of curves over number fields. If you dive into the notation a bit, his Corollary 0.10 states that assuming finiteness of K-groups the conjectures on the ranks of $K_1$ and $K_2$ are equivalent.

  • K. Kato. A Hasse principle for two-dimensional global fields. J. reine angew. Math. 366 (1986), 142-180.

No upper bounds are known, the key problem is the yet unknown finite generation of K-groups...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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