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Everything is in the title, I didn't try so much but I hope someone can share some hints for understanding the objects mentionned in terrytao's blog/Tate’s proof of the functional equation, how to get the intuition (in the simplest case) for :

  • the adele ring $\mathbb{A}$ of $K$

  • the non-archimedean absolute value $|.|_p$ and the corresponding measures

  • $\int_{\mathbb{A}^\times} f(x)|x|^sd^\times x$, for example how to translate this into an explicit series/integral ?

I know the proofs of the functional equation for $\zeta(s)$, the Fourier transform in many settings, and number fields (but not really the completion of $K$ with respect to $|.|_p$)

I guess the main point is that $\mathbb{A}$ is self-Pontryagin dual, but how is it obvious ?

reuns
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  • This Fourier analysis on number fields might help – reuns Feb 06 '17 at 14:25
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    You should read the first chapter of Goldfeld and Hundley. It's extremely readable! – Peter Humphries Feb 06 '17 at 15:57
  • @PeterHumphries It is there. The p-adic integration and Fourier transform looks really weird. Is it complicated to explain in short the main ideas, for example why $\zeta_K(s)$ becomes similar to $\zeta_\mathbb{Q}(s)$ under this framework ? – reuns Feb 06 '17 at 16:31
  • I think that chapter does a good job of showing how the p-adic Fourier transform is analogous to the Fourier transform on $\mathbb{R}$. Though it looks weird initially, $p$-adic integration is very easy to do once you use the fact that $\mathbb{Q}p^{\times} = \bigsqcup{n = -\infty}^{\infty} p^n \mathbb{Z}_p^{\times}$, so instead of continuous integrals on $\mathbb{R}$, you're instead dealing with countably infinite sums plus integration on $\mathbb{Z}_p^{\times}$. – Peter Humphries Feb 06 '17 at 19:43
  • The advantage of the adèles is that all number fields and function fields $K$ become very similar. In particular, for things like zeta functions and $L$-functions (i.e. Tate's thesis), you decompose the global $L$-function into a product of local $L$-functions, one at each prime: this is just the local factor at $p$ in the Euler product! In this sense, global fields are all alike in that they have countably many prime ideals, which can be ordered to be nondecreasing in norm. – Peter Humphries Feb 06 '17 at 19:46
  • Tate's thesis basically says that using the adèles, you can prove the functional equation for an $L$-function by defining the $L$-function as a Mellin transform of a function on the adèles, then use the fact that the adèle ring is the restricted tensor product of local fields to write that Mellin transform as the infinite product of local integrals. Each of these local integrals is the local $L$-function. So this Mellin transform is just the Euler product. Then you use the Poisson summation formula and do the same thing again to get the functional equation (well, more or less). – Peter Humphries Feb 06 '17 at 19:51
  • @PeterHumphries Yes sorry, this is what I read everywhere. But why does it work, where is the magic (why the integration, Fourier transform, Pontryagin duality/ Poisson summation formula is manageable in this setting ?) – reuns Feb 06 '17 at 19:52
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    Well, the usual proof of the functional equation uses the Poisson summation formula in the form of the modularity of the theta function. This is just repackaging that proof in a more complex notation. The advantage is that this proof then generalises nicely and works for other $L$-function and other global fields. – Peter Humphries Feb 06 '17 at 19:54
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    The key point is that there is nothing new in the proof. It is just a repackaging. Paragraph 3 (and the whole) of this is a must read! – Mathmo123 Feb 06 '17 at 20:01
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    In answer to your first and third bulllets, the point of the Adeles is to allow us view all places simultaneously, and on an equal footing. With that in mind, the most natural way to view the integral is as the restricted product of the local integrals. That is, in a sense, the whole point. – Mathmo123 Feb 06 '17 at 20:03
  • @Mathmo123 Of course there is something new when $K$ is a number field with non-trivial ideal class group, because $\sum_{I \subset \mathcal{O}K} N(I)^{-s}$ isn't so easy to manipulate, at first. And for $\zeta\mathbb{Q}$, we prove the functional equation by using its additive properties (the dirac comb distribution is its own Fourier transform), while here it seems we use the prime factorization property (plus the additive property in some local fields?) which is completely different – reuns Feb 06 '17 at 20:15
  • @user1952009 The ideal class group lurks in the background of the adeles (it is $K^\times\backslash\mathbb A^\times/\hat{\mathcal O}_K$), but never needs to be brought out. The adelic formalism brings simplicity and clarity, particular in viewing an $L$-function via its Euler product. But if you unpack all the details, there isn't much that is formally new. – Mathmo123 Feb 06 '17 at 20:20
  • @Mathmo123 $K^\times\backslash\mathbb A/\widehat{\mathcal O}_K$ is a complete mystery to me. Is it the main step, to understand this object ? – reuns Feb 06 '17 at 20:21
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    @user1952009 I don't think so. In the classical proof, the class group is an annoying hurdle to deal with, whereas in the adelic proof, it just lurks unseen and isn't explicitly used. In terms of understanding the ideal class group (which should be $K^\times\backslash\mathbb A^\times/K_\infty\hat{\mathcal O}K^\times$), given any idele $(u\pi_v^{k_v})_v$, we get an ideal $\prod_v\mathfrak p_v^{k_v}$. The kernel of this map $\mathbb A^\times\to I_K$ is $K\infty\times\hat{\mathcal O}^\times_K$. Quotienting by $K^\times$ is the same as quotienting by the principal ideals. – Mathmo123 Feb 06 '17 at 20:33

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