Example 4

— Large triangular group.

Let $n_1, \ldots, n_r$ be integers $\geqslant 1$. Set $p_k = n_1 + \ldots + n_{k-1}$ and $n = p_{r+1} = n_1 + \cdots + n_r$. Let $I_k$ be the set of integers $j$ such that $p_k < j \leqslant p_{k+1}$, and $J$ the union of the $I_k \times I_l$ for $k < l$. Let $G$ be the closed subgroup of $\mathbf{GL}(n, K)$ whose elements are the matrices $(Z_{kl})_{1 \leqslant k \leqslant r, 1 \leqslant l \leqslant r}$ such that: 1) each $Z_{kl}$ is a matrix $(z_{ij})_{i \in I_k, j \in I_l}$ of elements of $K$, with $n_k$ rows and $n_l$ columns; 2) $Z_{kl} = 0$ for $k > l$; 3) $Z_{kk} \in \mathbf{GL}(n_k, K)$ for $1 \leqslant k \leqslant r$.

The formula for block multiplication

shows that $G$ is the topological semi-direct product of the subgroup $D$ of elements $(Z_{kl}) \in G$ such that $Z_{kl} = 0$ for $k \neq l$ and the subgroup $T_J$ of Example 3. Moreover, $D$ is isomorphic to the direct product of the groups $\mathbf{GL}(n_k, K)$ for $1 \leqslant k \leqslant r$.

Let $J'$ be the set of pairs $(j, i)$ for $(i, j) \in J$ and let $H$ be the set of pairs $(i, j) \in [1, n] \times [1, n]$ not belonging to $J'$. Let $Z' = (z_{ij})_{1 \leqslant i \leqslant n, 1 \leqslant j \leqslant n}$ be an element of $G$. By Prop. 14 of §2, No. 9 and the above Examples 1 and 3, one obtains a left Haar measure on $G$ by taking the image of the measure

under the mapping

Now, consider, for $k < l$, the vector space of matrices $Z_{kl} = (z_{ij})_{i \in I_k, j \in I_l}$. It is the direct sum of the $n_l$ subspaces $M_j$ ($j \in I_l$) formed by the matrices such that $z_{ih} = 0$ for $h \neq j$. Each of these subspaces $M_j$ is stable under the mapping $Z_{kl} \mapsto Z_{kk}Z_{kl}$, and the restriction of this mapping to $M_j$ has matrix $Z_{kk}$. Consequently (§ 1, No. 10, Prop. 15) the image of the measure $\bigotimes_{i \in I_k, j \in I_l} dz_{ij}$ under the mapping $Z_{kl} \mapsto Z_{kk}Z_{kl}$ is

A left Haar measure on $G$ is therefore given by

with $q_k = \sum_{k \leq l \leq r} n_l = n - p_k$.

Let us calculate the modulus of $G$, again using Prop. 14 of § 2. The groups $D$ and $T_J$ are unimodular; on the other hand:

where $Z'_{kl} = Z_{kk}Z_{kl}Z_{ll}^{-1}$. Taking into account Example 3, and Prop. 15 of § 1, No. 10, and arguing as above, one sees that if $X = \operatorname{diag}(Z_{11}, \ldots, Z_{rr}) \in D$ then the modulus of the automorphism $Z \mapsto X^{-1}ZX$ of $T_J$ is

therefore

The transposed group $G'$ of $G$ is studied in the same manner. For $G'$, one finds as left Haar measure

and as modulus

If in particular one takes $n_1 = \ldots = n_r = 1$, one finds as group $G$ the group $T(n, K)^*$ of invertible elements of the subalgebra of $M_n(K)$ formed by the matrices $X = (x_{ij})$ such that $x_{ij} = 0$ for $i > j$. This algebra, which we shall denote $T(n, K)$, is called the upper triangular algebra, and the group $T(n, K)^*$ is called the upper large triangular group of order $n$ over $K$. The preceding formulas then take the following form: a left Haar measure on $T(n, K)^*$ is

and the modulus of $T(n, K)^*$ is

For the transpose of $T(n, K)^*$, or lower large triangular group, one finds as left Haar measure

and as modulus