A generalization of $\mathbb{I}_K = \mathbb{I}_K^1 \times (0,\infty)$ 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?
 A: 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.
