7
$\begingroup$

Two (number) fields are arithmetically equivalent if their Dedekind zeta functions are the same. It is known that any two arithmetically equivalent fields are not necessarily isomorphic; Prasad (http://www.math.tifr.res.in/~dprasad/refined-equiv.pdf) gives an example of two non-isomorphic, arithmetically equivalent fields whose idele class groups are isomorphic.

Under Langlands Philosophy, the structure of absolute Galois groups of number fields can be understood through $n$-dimensional automorphic representations of adele rings. My question is:

Do there exist two non-isomorphic, arithmetically equivalent number fields whose $n$-dimensional automorphic representations of their adele rings are isomorphic? Here, $n>1$ is assumed.

Note: I asked this question in https://math.stackexchange.com/questions/1548060/arithmetically-equivalent-number-fields-and-langlands-program

$\endgroup$
9
  • 4
    $\begingroup$ I don't understand the question. Which n-dimensional representations? One of them? All of them? What does it mean for 2 different representations of two different groups to be isomorphic? $\endgroup$
    – eric
    Nov 28, 2015 at 19:50
  • 1
    $\begingroup$ Isomorphic as what? Clearly they're isomorphic as Banach spaces. [usually one fixes a central character but this is the least of our worries at this point] $\endgroup$
    – eric
    Nov 28, 2015 at 20:04
  • 2
    $\begingroup$ I agree with eric that the quantifiers in you question are not clear. Anyway, a theorem of Neukirch-Uchida states that any number field is determined by its absolute Galois group: en.wikipedia.org/wiki/Neukirch%E2%80%93Uchida_theorem. This suggests to me that all the automorphic data associated to a number field should determine it (though of course I can't make this more precise). $\endgroup$ Nov 28, 2015 at 20:12
  • 1
    $\begingroup$ The problem with this approach is that you might have to assume Langlands' philosophy :-) $\endgroup$
    – eric
    Nov 28, 2015 at 20:14
  • 1
    $\begingroup$ The isomorphism from Prasad's paper comes from the fact that there is isomorphism of algebraic groups $\mathrm{R}_{F/\mathbb{Q}} \mathbb{G}_m \cong \mathrm{R}_{F'/\mathbb{Q}} \mathbb{G}_m$ over $\mathbb{Q}$ (here $\mathrm{R}_{F/\mathbb{Q}}$ denotes the Weil restriction). Perhaps you could obtain an isomorphism of the associated general linear groups is a similar way...? $\endgroup$ Nov 28, 2015 at 20:30

1 Answer 1

4
$\begingroup$

This isn't an answer, it's just too long to be a comment.

I think the question might be this. Take two number fields $K$ and $L$ such that $K$ and $L$ are not isomorphic, but $K$ and $L$ have isomorphic adele rings. This can happen! It's a fact from group theory that there's a finite group $G$ and two non-conjugate subgroups $H_1$ and $H_2$ such that $\mathbb{Z}[G/H_1]$ and $\mathbb{Z}[G/H_2]$ are isomorphic as representations of $G$, and realising everything as a Galois group over the rationals gives $K$ and $L$; for details see Dipendra Prasad's paper linked to in the question.

For such number fields, we can fix an isomorphism between the adele rings, which is completely horrible and non-canonical (e.g. we could just take two primes which split completely in all the fields in question and then just permute some of the factors in one of the adele rings but not the other, so fixing this isomorphism is an extremely un-natural and non-canonical thing to do, and there are also uncountably many ways to do it, none of which seem to me to be better than any other) and we get an induced isomorphism between $GL_n(\mathbb{A}_K)$ and $GL_n(\mathbb{A}_L)$ which is probably also completely non-canonical and random.

Having fixed this isomorphism, we can ask the following completely un-natural question: fix a ccharacter $\psi$ of $\mathbb{A}_K^\times$ and we get an induced character $\psi'$ of $\mathbb{A}_L^\times$. Does there exist some $n>1$ and an isomorphism $L^2(GL_n(K)\backslash GL_n(\mathbb{A}_K),\psi)\cong L^2(GL_n(L)\backslash GL_n(\mathbb{A}_L),\psi')$ which intertwines the actions of the isomorphic groups $GL_n(\mathbb{A}_K)$ and $GL_n(\mathbb{A}_L)$? The reason the isomorphism doesn't follow from what we have already is that $GL_n(K)$ sits in the adele group in a different way to $GL_n(L)$ so the spaces are most definitely not identified by all the choices we've made so far.

So I can't answer this question, and of course I've already made it pretty clear that I don't think it's a natural question, because although the adele rings are isomorphic, there is no canonical isomorphism between them. However, as Daniel Loughran said, if you somehow had this for all $n$ at once then one might attempt to deduce that the conjectural Tannakian category of representations of the global Langlands groups of $K$ and $L$ are isomorphic, and this would imply that the absolute Galois groups of $K$ and $L$ were isomorphic (these being the connected components of the Langlands groups) and then by Neukirch-Uchida $K$ and $L$ would be isomorphic. This is no good really though because firstly one needs the isomorphisms for all $n$, secondly one needs more than isomorphisms, one needs isomorphisms compatible with a non-existent tensor product, and thirdly one needs the Langlands program. On the other hand it's not true in general that two different discrete groups give two different spectra -- because as is well known we can't hear the shape of a drum. So this question seems theoretically hard but also perhaps very unnatural.

$\endgroup$

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.