Let $K$ be a number field, and $\mathbb{I}_K$ the group of ideles of $K$. The norm $|| \cdot ||: \mathbb{I}_K \rightarrow (0,\infty)$ is given by

$$|| x|| = \prod\limits_v |x_v|_v$$

where $| \cdot |_v$ is the normalized absolute value on $K_v$ so that the product formula holds for elements of $K^{\ast} \subseteq \mathbb{I}_K$. There is an embedding $(0,\infty) \rightarrow \mathbb{I}_K$ given by

$$\rho \mapsto (\rho^{\frac{1}{n}}, ... , \rho^{\frac{1}{n}}, 1, ...)$$

where $\rho$ is only mapped to the infinite places of $K$, and $n = [K : \mathbb{Q}]$. The normalized absolute values on the infinite places ensure that the norm of $\rho$ is $\rho$ itself. If $\mathbb{I}_K^1$ is the kernel of the norm map, then $\mathbb{I}_K$ decomposes into a direct product of topological groups

$$\mathbb{I}_K = \mathbb{I}_K^1 \times (0,\infty)$$

This decomposition is useful in class field theory. For example, we see that open subgroups of $\mathbb{I}_K$ correspond bijectively to open subgroups of $\mathbb{I}_K^1$.

Now, let $G$ be a connected, reductive group over $K$. Then we can define the group $G(\mathbb{A}_K)$ of adelic rational points of $G$. There is no natural norm map $G(\mathbb{A}_K) \rightarrow (0,\infty)$. But is there a meaningful generalization of the above decomposition?

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    $\begingroup$ ...did you ask a question just for the sake of answering it?? $\endgroup$ – Peter Humphries Aug 22 '17 at 17:53
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    $\begingroup$ Yes. When you post a question, there is an option to "share your knowledge Q&A style." I don't keep good track of handwritten notes so I often refer to my answers or questions on stackexchange if I forget something I need. $\endgroup$ – D_S Aug 22 '17 at 17:55
  • $\begingroup$ @D_S Are you trying to share your knowledge ? You are always taking the most theoretical possible approach, which is the best way to need technical terms so that nobody can follow you. Even for Peter it is hard to follow your questions whereas he is the MSE expert on automorphic forms ! $\endgroup$ – reuns Aug 22 '17 at 23:55
  • $\begingroup$ Yes, I hope what I write can serve as a reference to others at the same level as me. I try to make my posts as self contained as possible. $\endgroup$ – D_S Aug 23 '17 at 0:03
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    $\begingroup$ @reuns, I was actually going to say that this answer isn't theoretical enough! :P It only treats the case $K = \mathbb{Q}$. I think I've seen the general case treated somewhere - maybe in Shalika's book on Eisenstein series, or one of Arthur's papers. $\endgroup$ – Peter Humphries Aug 23 '17 at 9:29

Yes. To make life easy, let me assume $K = \mathbb{Q}$, although this procedure should generalize. We replace the norm map with the homomorphism $H_G$ of $G(\mathbb{A})$ into the real Lie algebra $\mathfrak a = \textrm{Hom}(X(G)_{\mathbb{Q}},\mathbb{R})$, defined by

$$H_G(x)(\chi) = \log ||\chi(x)||$$

Here $X(G)_{\mathbb{Q}}$ is the group of rational characters of $G$ which are defined over $\mathbb{Q}$. The kernel $G(\mathbb{A})^1$ of $H_G$ consists of those $x \in G(\mathbb{A})$ for which $\chi(x) \in \mathbb{I}_K^1$ for all rational characters $\chi$ of $G$ which are defined over $\mathbb{Q}$.

Let $A_G$ denote the split component of $G$. The embedding $\mathbb{R} \rightarrow \mathbb{A}, \rho \mapsto (\rho, 1, ...)$ gives a topological embedding $A_G(\mathbb{R}) \subseteq G(\mathbb{A})$, and $A_G(\mathbb{R})$ is isomorphic to a finite product of $\mathbb{R}^{\ast}$s. Then the connected component $A_G(\mathbb{R})^0$ is isomorphic to a finite product of $(0,\infty)$s.

I claim that $G(\mathbb{A})$ is a direct product $G(\mathbb{A})^1 \times A_G(\mathbb{R})^0$.

The least trivial part is the fact that restriction $X(G)_{\mathbb{Q}} \rightarrow X(A_G)$ is an injection whose image is a finite index subgroup of $X(A_G)$. By Smith normal form, we can choose a basis $\gamma_1, ... , \gamma_m$ of $X(A_G)$, and a basis $\chi_1, ... , \chi_m$ of $X(G)_{\mathbb{Q}}$, such that $\chi_i|A_G = d_i \gamma_i$ for some $d_i \geq 1$. The basis $\gamma_1, ... , \gamma_m$ allows us to identify any $a \in A_G(\mathbb{R})$ with the diagonal matrix

$$\begin{pmatrix} \gamma_1(a) \\ & \ddots \\ & & \gamma_m(a) \end{pmatrix}$$

Now given $x \in G(\mathbb{A})$, let $r_i = ||\chi_i(x)||$. Let $a$ be the element of $A_G(\mathbb{R})^0$ corresponding to the matrix

$$\begin{pmatrix} \sqrt[d_1]{r_1} \\ & \ddots \\ & & \sqrt[d_m]{r_m} \end{pmatrix}$$

Then we see that $xa^{-1} \in G(\mathbb{A})^1$. The only thing left to check is that $G(\mathbb{A})^1 \cap G(\mathbb{R})^0$ is trivial, which is obvious.


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