3
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ $K$ 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 $\mathrm{GL}_n(\mathbb{Z}_p)$. $\endgroup$ – Peter Humphries May 11 at 12:24
  • $\begingroup$ @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 $GL_n(\Bbb{Z}/(k))$) $\endgroup$ – reuns May 14 at 19:54
  • $\begingroup$ @reuns, whoops, yes. $\endgroup$ – Peter Humphries May 14 at 20:18
4
+50
$\begingroup$

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 $w = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$, $$w P(F) = \left\{\begin{pmatrix} 0 & d \\ a & b \end{pmatrix} \in \mathrm{GL}_2(F)\right\}.$$

$\endgroup$
  • $\begingroup$ 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 $P$? $\endgroup$ – Takeshi Gouda May 15 at 14:21
  • 1
    $\begingroup$ 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 $W_n$. $\endgroup$ – Peter Humphries May 15 at 15:41

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.