jump to navigation

Zeta function relations and linearly equivalent group actions August 29, 2007

Posted by Ben Webster in Number theory, papers, representation theory.
trackback

Since we’ve already had one post on the relationship between group theory and algebraic number theory, I don’t see any reason to stop, so I thought I would write about some old research I did as an undergraduate (6 years ago! yeesh).

Now, every number field has a zeta function (often called the Dedekind zeta function), generalizing the Riemann zeta function. Now, I hope you will all remember that the Riemann zeta function (which is the zeta function of the rationals) is defined by

\displaystyle{\zeta(z)=\sum_{n\in \mathbb{Z}_{>0}}n^{-z}=\prod_{p \text{ prime}}(1-p^{-z})^{-1}}.

If you take a more general number field K, then the Dedekind zeta function of that field is a similar sum or product over the ideals \mathcal{I}_K or primes \mathcal{P}_K of K.

\displaystyle{\zeta_K(z)=\sum_{\mathfrak n\in \mathcal{I}_K}N(\mathfrak n)^{-z}=\prod_{\mathfrak p \in \mathcal{P}_K}(1-N(\mathfrak p)^{-z})^{-1}}.

Here N(\mathfrak n) is the norm of \mathfrak n.

So how can we understand this function? Assume for a moment that K is Galois over \mathbb Q (that is, if a root of an irreducible polynomial over \mathbb Q is in K, then all roots of that polynomial are). In this case, we have a Galois group G, which acts transitively on the set of prime ideals lying over any integer prime p. As all of you who remember group theory know, this means to understand this as a G-set we just need to find the stabilizer of a single prime in the Galois group. Using some basic arguments from basic algebraic number theory, you can see that this stabilizer has to be cyclic as long as p is unramified (and let’s forget that case for now) since all finite extensions of \mathbb F_p are cyclic. The Frobenius of the prime p is a distinguished generator of this cyclic group.

So, if we let a denote the order of G and f_p denote the order of the Frobenius of p, and we group primes of $K$ according to which integer prime they lie over, we end up with an expression for the zeta function of the form

 

\displaystyle{\zeta_K(z)=\prod_{p \text{ prime}}(1-p^{f_p})^{-a/f_p}}.

Thus, two number fields which are Galois over \mathbb Q will have the same zeta function if and only if the Frobeniuses (it’s tempting to say “Frobenii,” but Scott would beat me to a bloody pulp with a Latin text book. The correct German plural, meanwhile, seems to be Frobenius-Automorphismen, so no help there) of each prime with respect to both have the same order. Remarkably, it turns out that this requires the fields to be the same.

But, of course, most number fields are not Galois. For every subgroup H\subset G, we have a fixed field L=K^H, and by a famous theorem of Galois, these are all different and every subfield of K arises this way.

Is there a similar expression for the Dedekind zeta function of L? Of course, but it’s a bit more complicated, because now the primes lying over p might have different norms. Luckily, there’s a way of understanding these in terms of group theory. Two primes \mathfrak p_1, \mathfrak p_2 lie over the same prime \mathfrak q of L if and only if they are in the same H-orbit on primes of K. That is, primes lying over p in L are in bijection with double cosets H\backslash G /F(p) where F(p) is the subgroup generated by the Frobenius and the norm of \mathfrak q is p^{\#HgF(p)}, where HgF(p) is the double coset corresponding to \mathfrak q. Thus, our product formula becomes

\displaystyle{\zeta_L(z)=\prod_{p \text{ prime}}\left(\prod_{HgF(p)\in H\backslash G /F(p)} (1-p^{\#HgF(p)})^{-1}\right)}.

Now, let’s imagine we had two subgroups H_1, H_2 such that L_1=K^{H_1},L_2=K^{H_2} have the same zeta function (“are arithmetically equivalent”). What would these subgroups have to have in common? For each prime p, the distribution of orbit sizes for F(p) on G/H_1 and G/H_2 have to be the same. Since each cyclic subgroup shows up as F(p) for some prime $p$ (by the Chebotarev Density Theorem), this must hold true for all cyclic subgroups.

Exercise: This condition is equivalent to the number of fixed points for each cyclic subgroup being the same (hint: relate the number of orbits of a given size an element has to the number of fixed points of its powers).

This condition can be restated a second time, in a slightly surprising way: for any G-set X, there is a representation \mathbb Q[X], which you can think of as functions from X to \mathbb Q with the obvious vector space structure and G-action.

Now, whenever we have a representation that we find “in the wild,” the first thing we should do is try to calculate its character. Luckily, with \mathbb Q[X], this is pretty easy

Proposition: The permutation character for \mathbb Q[X] on G is simply the number of fixed points for each element, i.e.

\mathrm{tr}(g,\mathbb Q[X])=\#X^g.

Proof. The matrix for the action of g on \mathbb Q[X] is a permutation matrix, with rows and columns corresponding to the elements of X, with a 0 on the diagonal entry if the corresponding point is not fixed, and a 1 if it is. Q.E.D.

So, plugging in the results of our exercise above, and the fact that representations over \mathbb Q are isomorphic if and only if they have the same character we get the

Theorem. The we have \zeta_{L_1}(z)=\zeta_{L_2}(z) if and only if \mathbb Q[G/H_1]\cong \mathbb Q[G/H_2].

This can be restated a bit more elegantly by noting that if \alpha_1,\alpha_2 are primitive roots for L_1,L_2, that is, L_i=\mathbb Q[\alpha_i], then the stabilizer of \alpha_i in G is, of course, H_i. So, we find that G/H_i\cong G\cdot \alpha_i (which makes these G-set look more intrinsic).

Now, this is all pretty old stuff; perhaps the definitive treatment is by one of my favorite people, Robert Perlis, in a paper called “On the equation \zeta_{K}(z)=\zeta_{K'}(z)” from 1977. But next time, we’ll talk about some more recent work on how to find such pairs of subgroups which aren’t conjugate (it ain’t easy).

About these ads

Comments

1. Isabel - August 29, 2007

“The correct German plural, meanwhile, seems to be Frobenius-Automorphismen, so no help there”

As it should be! When did “Frobenius” become the noun here? I can’t think of any other mathematical objects where the discoverer’s name is used in this way. You don’t talk about the “Dedekind of a number field”, for example.

2. Ben Webster - August 29, 2007

I can’t think of any other mathematical objects where the discoverer’s name is used in this way.

Personally, I’m pretty happy to say “a p-Sylow of G” instead of a “Sylow p-subgroup,” and suspect I could come up with other examples, especially from spoken English (I feel like I’ve heard people say “Riemann zeta” in place of “Riemann zeta function”). Over all, this strikes me as a fairly standard maneuver for an English speaker, possibly falling under the general heading of metanymy.

3. Theo - August 29, 2007

Of course, there are Legendrians and Lagrangians in symplectic geometry, and every physicists has heard of the Higgs.

4. Blake Stacey - August 29, 2007

Do “Laplacian” and “D’Alembertian” count?

5. Allen Knutson - August 29, 2007

I hear Schubert calculus people use “Schuberts” to refer both to Schubert cycles and to Schubert polynomials, depending on context. (Sometimes even “Schubs”!) Likewise they’ll use “Demazures” to refer to either Demazure modules or their characters. Certainly I hear people refer to “the Steinberg” and not only to the Steinberg variety. And on and on.

6. Blake Stacey - August 29, 2007

I feel marginally ashamed at not thinking of “Higgs” (kudos to Theo). Along the same lines, the bare names “Majorana” and “Dirac” are — fairly commonly, I believe — used as adjectives. “If the neutrino were not Majorana. . .” It’s also pretty common to call boundary conditions “Neumann” or “Dirichlet”. This isn’t quite the same as the “p-Sylow”, “Frobenius” or “Riemann zeta” cases mentioned above, but I wonder if it’s a related speech pattern.

(I’ve heard “Riemann zeta” for “Riemann zeta function”, too.)

7. Noah Snyder - August 29, 2007

Cartan. Borel.

8. Kenny - August 29, 2007

Do those all get used as nouns? “Any two disjoint analytic sets are separated by a borel”? Lots of these examples either seem to add “-ian” to a name, or use the bare name as an adjective. Ben wants to use “Frobenius” as a noun.

9. Ben Webster - August 29, 2007

“Borel” is short for “Borel subgroup,” (as opposed to all the objects in measure theory with Borel’s name attached to them) and yes, Lie groups types use it as a noun constantly. For example, you can find it used twice in this sense in my most recent paper, and nobody (*nobody*) says “Borel subgroup.”

10. Noah Snyder - August 29, 2007

People might not *say* “Borel subgroup,” but I think a lot of older mathematicians think that when you write you should write out “Borel subgroup.” Ivan made us change every use of “Cartan” to “Cartan subalgebra” in this paper. Similarly, Lenstra told me that “irrep” is too informal to be used in a paper.

11. Ben Webster - August 29, 2007

I feel like “Borel” and “irrep” are different levels of informality. I probably wouldn’t use “irrep” in a paper, but as I’ve shown, have no compunctions about “Borel” or “Cartan.”

12. Scott - August 29, 2007

By the way, the analysis Borel is Émile Borel. The representation theory Borel is Armand Borel, which is more amenable to nounification, e.g., Any element of \mathfrak{g} lies in A. Borel.

13. Alon Levy - August 30, 2007

Hausdorff?

The adjectives “Open,” “irreducible,” “closed,” etc. are all used as if they were nouns.

Also, Scott, it’s interesting that you use the word “Nounification” instead of “nominalization”…

14. Isabel - August 31, 2007

In response to all the people who have come up with examples — thanks.

As various people have pointed out, a word like “Laplacian” (which is short for Laplacian operator, I guess?) is different from “Laplace”; for some reason the bare noun bothers me a lot more. Perhaps it’s just a silly mental picture I have, of the actual person Laplace (feel free to substitute Frobenius, Borel, or any of the other names that have been used here) sitting there on the page. Using the adjective form indicates that it’s something that has to do with that person, not the actual person.

This seems like a subtlety of the English language that might not exist to the same degree in other languages; perhaps the internationalization of mathematics in general has something to do with this? But I am not a linguist.

This use of people’s names as nouns for mathematical objects associated with them might just be a particular pet peeve of mine, though, as it seems that it doesn’t bother the rest of you. That’s okay. (It also strikes me as odd that people are saying it’s correlated with age; it doesn’t seem that way to me.)

By the way, “Riemann zeta” doesn’t bother me; “Riemann” is still acting as an adjective, with “zeta” the noun here. (Which raises another question — would I be okay with “Riemann Z” if Riemann had chosen a Latin letter instead of a Greek one? I’m not sure.)

15. J. Ellenberg - September 2, 2007

The fact that different permutation reperesentations can be isomorphic as abstract representations — or more generally that there are nontrivial linear relations between permutation characters — is what drives the beautiful recent paper by Tim and Vlad Dockhitser, “On the Birch-Swinnerton-Dyer conjectures modulo squares.”

As for mathematicians being used as nouns: topologists routinely talk about “a pseudo-Anosov,” which Anosov might have found strange. Worse still is the habit of number theorists of referring to “Frobenii,” which is presumably not how Frobenius referred to his immediate family.

16. Laplacian spectra and linearly equivalent G-sets « Secret Blogging Seminar - September 19, 2007

[...] in differential geometry, representation theory. trackback So, to build on my previous post about zeta functions and group actions, let me talk about a second application of linearly equivalent actions, this time in differential [...]

17. Bruce Bartlett - July 19, 2008

Hi Ben and secret bloggers,

I found this discussion on Google after searching for “linearly equivalent G-sets”; I’m sorry I missed it the first time :-)

I have a problem: Say G is a finite group. Is there some kind of “counting fixed-points” invariant of finite G-sets which will distinguish them?

I ask because recently I had convinced myself that the “double character” distinguishes G-sets: for each pair of commuting elements g,h in G, it counts the number of simultaneous fixed points of g and h.

Then I realized I had fooled myself (at one point I had assumed what I was trying to prove :-().

I mean, for representations of groups there is this cool thing which distinguishes them: for each g in G, compute the trace Tr(g). If they’re different, then the reps are different.

I know of no invariant (i.e. consisting of some kind of function on G or G^2 or G^3 or …) of this form which will distinguish G-sets.

Can you help? Also, do you know anything about the “double character” invariant of G-sets? Whether it is indeed more powerful than the permutation character; or examples of G-sets which it doesn’t distinguish.

18. Bruce Bartlett - July 19, 2008

Woops… I should of course add that I mean “distinguish” above in the strong sense; i.e. some invariant of G-sets which, if it is different for X and Y, then X is not isomorphic to Y, and if it is the same, then X is isomorphic to Y.

19. Ben Webster - July 19, 2008

Bruce,

I don’t have time to go into detail right now, but two G-sets are isomorphic if and and only if each *subgroup* has the same number of fixed points on X and Y. This is all in my paper on the subject, Small linearly equivalent $G$-sets and a construction of Beaulieu

20. Scott Carnahan - July 19, 2008

Bruce,

It seems unlikely to me, but only because the set of abelian subgroups of G generated by two elements doesn’t sound like a very natural collection of subgroups (unless you’re interested in classifying G-torsors on an elliptic curve). Your question reminded me of Theorems A and C in a paper by Hopkins, Kuhn, and Ravenel (just Google Scholar the names), but I couldn’t tell if they would give you the reconstruction you want. On the other hand, if they did, I’d be surprised if you really needed complex oriented descent to prove your case of Euler characteristics of 0-manifolds.

Ben,

I’m having trouble finding the fact you mentioned in your paper. Could you point to it?

21. Ben Webster - July 20, 2008

Scott-

Yeah, I lied. I wrote that comment in a big hurry to catch a train, and my brain failed me.

On the other hand, the proof is really easy: If each subgroup fixes the same number of points on X as it does on Y, then there must be a subgroup which is the stabilizer of a point in X and in Y, thus isomorphic subrepresentations. Remove these and induct on the size of X.

What this result shows is that commuting elements is not even close to good enough. For every subgroup H of G, there are G-sets where every subgroup not conjugate to H has the same number of fixed points, and H does not (this is just simple linear algebra).

In fact, commuting elements isn’t even good enough to get isomorphic representations over fields of positive characteristic (this actually is in that paper) because you also need p-groups and cyclic extensions of them.

22. Bruce Bartlett - July 20, 2008

Thanks Ben and Scott. This gives me a lot of food for thought. That’s really interesting – its the subgroups that count. Ben, I’m also struggling a bit to find that point where you mention this in your paper. Looked up Hopkins, Kuhn and Ravenel theorems A and C – now I will try to unravel what those statements mean…

23. Carnival of Mathematics XVI « Learning Computation - April 30, 2010

[...] bit more high-brow, I’d recommend Ben Webster’s post at the Secret Blogging Seminar on Zeta function relations and linearly equivalent group actions. Ben mentions that the subject of this post comes from some research he did as an undergrad, and it [...]


Sorry comments are closed for this entry

Follow

Get every new post delivered to your Inbox.

Join 715 other followers

%d bloggers like this: