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
If you take a more general number field , then the Dedekind zeta function of that field is a similar sum or product over the ideals or primes of .
Here is the norm of .
So how can we understand this function? Assume for a moment that is Galois over (that is, if a root of an irreducible polynomial over is in , then all roots of that polynomial are). In this case, we have a Galois group , which acts transitively on the set of prime ideals lying over any integer prime . As all of you who remember group theory know, this means to understand this as a -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 is unramified (and let’s forget that case for now) since all finite extensions of are cyclic. The Frobenius of the prime is a distinguished generator of this cyclic group.
So, if we let denote the order of and denote the order of the Frobenius of , 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
Thus, two number fields which are Galois over 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 , we have a fixed field , and by a famous theorem of Galois, these are all different and every subfield of arises this way.
Is there a similar expression for the Dedekind zeta function of ? Of course, but it’s a bit more complicated, because now the primes lying over might have different norms. Luckily, there’s a way of understanding these in terms of group theory. Two primes lie over the same prime of if and only if they are in the same -orbit on primes of . That is, primes lying over in are in bijection with double cosets where is the subgroup generated by the Frobenius and the norm of is , where is the double coset corresponding to . Thus, our product formula becomes
Now, let’s imagine we had two subgroups such that have the same zeta function (“are arithmetically equivalent”). What would these subgroups have to have in common? For each prime , the distribution of orbit sizes for on and have to be the same. Since each cyclic subgroup shows up as 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 -set , there is a representation , which you can think of as functions from to with the obvious vector space structure and -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 , this is pretty easy
Proposition: The permutation character for on is simply the number of fixed points for each element, i.e.
Proof. The matrix for the action of on is a permutation matrix, with rows and columns corresponding to the elements of , 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 are isomorphic if and only if they have the same character we get the
Theorem. The we have if and only if .
This can be restated a bit more elegantly by noting that if are primitive roots for , that is, , then the stabilizer of in is, of course, . So, we find that (which makes these -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 ” 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).