— Special triangular group. Let us take up again the notations at the beginning of Example 4, and consider the subgroup G 1 = G ∩ S L ( n , K ) G_1 = G \cap \mathbf{SL}(n, K) G 1 = G ∩ SL ( n , K ) D 1 = D ∩ S L ( n , K ) D_1 = D \cap \mathbf{SL}(n, K) D 1 = D ∩ SL ( n , K ) T J T_J T J D 1 D_1 D 1 A A A S L ( n r , K ) \mathbf{SL}(n_r, K) SL ( n r , K ) diag ( Z k k ) \operatorname{diag}(Z_{kk}) diag ( Z k k ) Z k k = 1 Z_{kk} = 1 Z k k = 1 k < r k < r k < r
φ : diag ( Z 11 , … , Z r r ) ↦ ( Z 11 , … , Z r − 1 , r − 1 )
\varphi : \operatorname{diag}(Z_{11}, \ldots, Z_{rr}) \mapsto (Z_{11}, \ldots, Z_{r-1, r-1})
φ : diag ( Z 11 , … , Z r r ) ↦ ( Z 11 , … , Z r − 1 , r − 1 ) of
D 1 D_1 D 1 into
G L ( n 1 , K ) × ⋯ × G L ( n r − 1 , K ) \mathbf{GL}(n_1, K) \times \cdots \times \mathbf{GL}(n_{r-1}, K) GL ( n 1 , K ) × ⋯ × GL ( n r − 1 , K ) is surjective and has kernel
A A A . On the other hand,
φ \varphi φ is continuous. Taking into account Lemma 2 of Appendix I,
D 1 / A D_1/A D 1 / A may be identified with
G L ( n 1 , K ) × ⋯ × G L ( n r − 1 , K ) \mathbf{GL}(n_1, K) \times \cdots \times \mathbf{GL}(n_{r-1}, K) GL ( n 1 , K ) × ⋯ × GL ( n r − 1 , K ) . We shall denote by
μ \mu μ the Haar measure of
A A A (cf. Example 6) and by
α = ⨂ k = 1 r − 1 ( ( mod det Z k k ) − n k ⋅ ⨂ i , j ∈ I k d z i j ) ⊗ ′ d μ ( Z r r )
\alpha = \bigotimes_{k=1}^{r-1} \left( (\operatorname{mod} \det Z_{kk})^{-n_k} \cdot \bigotimes_{i,j \in I_k} dz_{ij} \right) \otimes' d\mu(Z_{rr})
α = k = 1 ⨂ r − 1 ( mod det Z k k ) − n k ⋅ i , j ∈ I k ⨂ d z ij ⊗ ′ d μ ( Z r r ) the Haar measure on
D 1 D_1 D 1 such that
α / μ = ⨂ k = 1 r − 1 ( ( mod det Z k k ) − n k ⋅ ⨂ i , j ∈ I k d z i j )
\alpha/\mu = \bigotimes_{k=1}^{r-1} \left( (\operatorname{mod} \det Z_{kk})^{-n_k} \cdot \bigotimes_{i,j \in I_k} dz_{ij} \right)
α / μ = k = 1 ⨂ r − 1 ( mod det Z k k ) − n k ⋅ i , j ∈ I k ⨂ d z ij (§ 2, No. 7, Prop. 10). One then shows as in Example 4 that a left Haar measure on
G 1 G_1 G 1 is given by
mod ( ∏ k = 1 r − 1 ( det Z k k ) n k − q k )
\operatorname{mod} \left( \prod_{k=1}^{r-1} (\det Z_{kk})^{n_k-q_k} \right)
mod ( k = 1 ∏ r − 1 ( det Z k k ) n k − q k ) ⋅ [ ⨂ k = 1 r − 1 ( ( mod det Z k k ) − n k ⋅ ⨂ i , j ∈ I k d z i j ) ⊗ ′ d μ ( Z r r ) ] ⊗ ⨂ ( i , j ) ∈ J d z i j .
\cdot \left[ \bigotimes_{k=1}^{r-1} ((\operatorname{mod} \det Z_{kk})^{-n_k} \cdot \bigotimes_{i,j \in I_k} dz_{ij}) \otimes' d\mu(Z_{rr}) \right] \otimes \bigotimes_{(i,j) \in J} dz_{ij}.
⋅ k = 1 ⨂ r − 1 (( mod det Z k k ) − n k ⋅ i , j ∈ I k ⨂ d z ij ) ⊗ ′ d μ ( Z r r ) ⊗ ( i , j ) ∈ J ⨂ d z ij . Since G 1 G_1 G 1 G G G modulus of G 1 G_1 G 1 G G G
If n r = 1 n_r = 1 n r = 1 A A A G G G
mod ( ∏ k = 1 r − 1 ( det Z k k ) − q k ) ⋅ ⨂ k = 1 r − 1 ( ⨂ i , j ∈ I k d z i j ) ⊗ ⨂ ( i , j ) ∈ J d z i j .
\operatorname{mod}\left( \prod_{k=1}^{r-1} (\det Z_{kk})^{-q_k} \right) \cdot \bigotimes_{k=1}^{r-1} \left( \bigotimes_{i,j \in I_k} dz_{ij} \right) \otimes \bigotimes_{(i,j) \in J} dz_{ij}.
mod ( k = 1 ∏ r − 1 ( det Z k k ) − q k ) ⋅ k = 1 ⨂ r − 1 i , j ∈ I k ⨂ d z ij ⊗ ( i , j ) ∈ J ⨂ d z ij . If one takes n 1 = n 2 = … = n r = 1 n_1 = n_2 = \ldots = n_r = 1 n 1 = n 2 = … = n r = 1 G 1 G_1 G 1 upper special triangular group and its transpose G 1 ′ G'_1 G 1 ′ lower special triangular group . A left Haar measure on G 1 G_1 G 1
mod ( ∏ i = 1 n − 1 z i i i − n − 1 ) ⋅ ( ⨂ i = 1 n − 1 d z i i ) ⊗ ( ⨂ 1 ≤ i < j ≤ n d z i j )
\operatorname{mod} \left( \prod_{i=1}^{n-1} z_{ii}^{i-n-1} \right) \cdot \left( \bigotimes_{i=1}^{n-1} dz_{ii} \right) \otimes \left( \bigotimes_{1 \leq i < j \leq n} dz_{ij} \right)
mod ( i = 1 ∏ n − 1 z ii i − n − 1 ) ⋅ ( i = 1 ⨂ n − 1 d z ii ) ⊗ ( 1 ≤ i < j ≤ n ⨂ d z ij ) and the modulus of
G 1 G_1 G 1 is
mod ( ∏ i = 1 n − 1 z i i 2 i − 2 n ) .
\operatorname{mod} \left( \prod_{i=1}^{n-1} z_{ii}^{2i-2n} \right).
mod ( i = 1 ∏ n − 1 z ii 2 i − 2 n ) . For
G 1 ′ G'_1 G 1 ′ one finds similarly the left Haar measure
mod ( ∏ i = 1 n − 1 z i i n − i − 1 ) ⋅ ( ⨂ i = 1 n − 1 d z i i ) ⊗ ( ⨂ 1 ≤ j < i ≤ n d z i j )
\operatorname{mod} \left( \prod_{i=1}^{n-1} z_{ii}^{n-i-1} \right) \cdot \left( \bigotimes_{i=1}^{n-1} dz_{ii} \right) \otimes \left( \bigotimes_{1 \leq j < i \leq n} dz_{ij} \right)
mod ( i = 1 ∏ n − 1 z ii n − i − 1 ) ⋅ ( i = 1 ⨂ n − 1 d z ii ) ⊗ ( 1 ≤ j < i ≤ n ⨂ d z ij ) and modulus
mod ( ∏ i = 1 n − 1 z i i 2 n − 2 i ) .
\operatorname{mod} \left( \prod_{i=1}^{n-1} z_{ii}^{2n-2i} \right).
mod ( i = 1 ∏ n − 1 z ii 2 n − 2 i ) . Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
Nothing.