# Any reason why K_23(Z) has order 65520?

I'm rereading my notes and they mention that $K_{23}(\mathbb Z) = \mathbb Z/(65520)$

This looks like a good point to stop and ask whether there is any explanation for this $K$-group of integers (23 is just an arbitrary fixed number for this purpose). By "explanation" I mean a reasoning that would allow to find at least some properties of this group in advance of computing it or some intuition behind the result.

Here's one thing I already know:

• non-torsion part of $K(\mathbb Z)$ is $\mathbb Z$ in degrees $0,5,9,13,\dots\$ so $K_{23}(\mathbb Z)$ is pure torsion

Wikipedia says that "The torsion subgroups of $K_{2i+1}(\mathbb Z)$ ... have recently been determined."

Update: I learned from the article by Soule how this number $=2 * 12 * 2730$ where 2730 is the denominator of 12-th Bernoulli number. But the question stands.

-
Do you have a link to a list of the first however many K-groups of Z? I'm sure I've seen one somewhere, but I don't remember where. –  Harrison Brown Dec 31 '09 at 1:08
My guess is that this is because of its appearance in research.att.com/~njas/sequences/A029828 . –  Qiaochu Yuan Dec 31 '09 at 1:12
#Harrison, I didn't find it either but I checked with the known answer for $K_i(\mathbb Z)$. @Qiaochu: this is amazing! Do you think you could post on the possible connection? –  Ilya Nikokoshev Dec 31 '09 at 1:20
The connection with the Bernoulli numbers is well-known for K_n(Z) when n = 2 (mod 4), and I don't think it's a coincidence here either. I found an interesting-looking article by Weibel: math.uiuc.edu/K-theory/0691/KZsurvey.pdf . In general, though, I think you might be asking too much if you want "intuition," considering how notoriously strange homotopy-type stuff can be. –  Harrison Brown Dec 31 '09 at 1:23
Please do post answers even if they are one-liners as above! It will be easier to read and you'll get upvotes. –  Ilya Nikokoshev Dec 31 '09 at 1:28

More generally, if $F$ is a number field with ring of integers $\mathfrak{o}$, and $\zeta_F^\ast(m)$ is the first nonzero coefficient in the Taylor expansion of $\zeta_F$ at $m$, then Lichtenbaum (and Quillen) conjectured that $|\zeta_F^\ast(1-i)|=\frac{\# K_{2i-2}(\mathfrak{o})_{\text{tors}}}{\# K_{2i-1}(\mathfrak{o})_{\text{tors}}}$, times a regulator and some power of 2 (which I believe is not understood in general, although some progress was made on this in Ion Rada's PhD thesis). Hence, odd $K$ groups are related to the denominators of the Bernoulli numbers, and the even ones are related to the numerators. Also, not much cancellation occurs; I think the two $K$-groups can only share factors of 2.

The Voevodsky-Rost theorem might prove the Lichtenbaum conjecture, but I haven't seen anyone come out and say definitely that this is the case.

I don't have much intuition for this, except that the $K$-groups seem to be objects that like to map into étale cohomology groups. In this paper (link to MathSciNet), Soulé constructs Chern class maps from certain $K$-groups to étale cohomology groups. Furthermore, these maps frequently have small (or trivial) kernels and cokernels. I suppose the idea, then, is that $K$-theory is supposed to be a slightly better behaved version of étale cohomology, at least for the purpose of understanding zeta functions.

The rank of $K$-groups of rings of integers was computed by Quillen in the early 70's: it's rank 1 in dimension 0, rank $r_1+r_2-1$ in dimension 1 (Dirichlet's unit theorem), rank 0 in even dimensions $>0$, rank $r_1+r_2$ in dimensions $1\pmod 4$ except 1, and rank $r_2$ in dimensions $3\pmod 4$.

-
Shouldn't there be a regulator factor in the Lichtenbaum conjecture formula (even when F is totally real)? –  Reid Barton Dec 31 '09 at 6:04
Thanks. I fixed it. –  Simon Dec 31 '09 at 15:14
Shouldn't the left hand side of the formula be the absolute value of the leading coefficient of the Taylor series centered at 1-i? –  Bjorn Poonen Dec 31 '09 at 19:04
And on the right hand side, shouldn't the K-groups be replaced by their torsion subgroups? –  Bjorn Poonen Dec 31 '09 at 19:06
Finally, there is more than one abelian group of order 65520, so the formula by itself does not fully answer Ilya's question about why it is Z/65520Z. –  Bjorn Poonen Dec 31 '09 at 19:13