# Iwasawa decomposition for $GL_n(\mathbb{A}_{\mathbb{Q}})$. What does $K$ look like?

I am trying to understand the Iwasawa decomposition for $$GL_n(\mathbb{A}_{\mathbb{Q}})$$ and $$GL_n(\mathbb{Q})$$, where $$\mathbb{A}_{\mathbb{Q}}$$ is the adeles. The statement for the case of adeles reads there exists a maximal compact subgroup $$K$$ of $$GL_n(\mathbb{A}_{\mathbb{Q}})$$ such that $$GL_n(\mathbb{A}_{\mathbb{Q}}) = P(\mathbb{A}_{\mathbb{Q}}) K$$ (Here I am taking $$P$$ to be the upper triangular matrices). Could someone please explain to me what does $$K$$ look like in this case?

Also what does $$P(\mathbb{Q}) \backslash GL_n(\mathbb{Q})$$ look like? I would greatly appreciate if someone could explain me both.

Thank you.

• $K$ is the restricted tensor product of the maximal compact of each local field. For the archimedean place, this is just $\mathrm{O}(n)$, while for nonarchimedean places, it is $\mathrm{GL}_n(\mathbb{Z}_p)$. – Peter Humphries May 11 '19 at 12:24
• @PeterHumphries You mean $g_p \in K_p$ for all but finitely many $p$ (personnally I think to $GL_n(\Bbb{A_Q})$ as the group generated by $GL_n(\Bbb{Q})\times 1, 1 \times GL_n(\Bbb{R}), GL_n(\hat{\Bbb{Z}})\times 1$ the latter being the limits of sequences of matrices $A_j \in M_n(\Bbb{Z})$ such that for every $k$, $\lim_{j \to \infty} A_j \bmod k$ converges to an element of $GL_n(\Bbb{Z}/(k))$) – reuns May 14 '19 at 19:54
• @reuns, whoops, yes. – Peter Humphries May 14 '19 at 20:18

## 1 Answer

The standard maximal compact subgroup $$K$$ of $$\mathrm{GL}_n(\mathbb{A}_{\mathbb{Q}})$$ is the restricted tensor product of the maximal compact of each local field. In particular, the maximal compact subgroup $$K_p$$ of $$\mathrm{GL}_n(\mathbb{Q}_p)$$ is $$\mathrm{GL}_n(\mathbb{Z}_p)$$, while the maximal compact subgroup $$K_{\mathbb{R}}$$ of $$\mathrm{GL}_n(\mathbb{R})$$ is the orthogonal group $$\mathrm{O}(n)$$. (Over global fields $$F$$ other than $$\mathbb{Q}$$, there may be complex archimedean places, in which case $$K_{\mathbb{C}} = \mathrm{U}(n)$$ for $$\mathrm{GL}_n(\mathbb{C})$$).

What does this mean? Recall that $$\mathrm{GL}_n(\mathbb{A}_{\mathbb{Q}})$$ consists of elements $$(g_{\mathbb{R}},g_2,g_3,g_5,\ldots)$$ with $$g_{\mathbb{R}} \in \mathrm{GL}_n(\mathbb{R})$$, $$g_p \in \mathrm{GL}_n(\mathbb{Q}_p)$$, and $$g_p \in K_p$$ for all but finitely many $$p$$. Then the maximal compact subgroup $$K$$ of $$\mathrm{GL}_n(\mathbb{A}_{\mathbb{Q}})$$ consists of elements $$(k_{\mathbb{R}},k_2,k_3,k_5,\ldots)$$ with $$k_{\mathbb{R}} \in K_{\mathbb{R}} = \mathrm{O}(n)$$ and $$k_p \in K_p = \mathrm{GL}_n(\mathbb{Z}_p)$$.

For your second question, the standard way to understand $$P(\mathbb{Q}) \backslash \mathrm{GL}_n(\mathbb{Q})$$ is via the Bruhat decomposition. Let $$W_n$$ denote the Weyl group (of order $$n!$$) consisting of all $$n \times n$$ matrices that have precisely one $$1$$ in each row and each column and zeroes elsewhere. (For example, the identity matrix.) The Bruhat decomposition states that for any field $$F$$, $$\mathrm{GL}_n(F) = \bigsqcup_{w \in W_n} P(F) w P(F).$$ Of course, if $$w$$ is the identity matrix $$1_n$$, then $$P(F) w P(F) = P(F)$$. So we observe that $$P(F) \backslash \mathrm{GL}_n(F) = 1_n \sqcup \bigsqcup_{w \in W_n \setminus \{1_n\}} w P(F).$$ For example, for $$n = 2$$ and $$w = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$$, $$w P(F) = \left\{\begin{pmatrix} 0 & d \\ a & b \end{pmatrix} \in \mathrm{GL}_2(F)\right\}.$$

• Thank you very much for this! Is the $W_n$ the matrix of all $n \times n$ with only $n$ 1's and $0$'s as described always? or is this different for different choices of $P$? – Takeshi Gouda May 15 '19 at 14:21
• The Bruhat decomposition depends on the parabolic subgroup $P$; if you take something other than the minimal parabolic (the upper triangular matrices), then the indexing set is a different subset of $W_n$. – Peter Humphries May 15 '19 at 15:41