Bourbaki

Example 3

Tag 09H7, permanent. Cite it as [Bourbaki, Tag 09H7].

Integration, VII, p. 62, § 3. Applications and examples

Strict triangular group.

Let [1,n][1, n] be the set of integers mm such that 1≤m≤n1 \leq m \leq n. Let JJ be a subset of [1,n]×[1,n][1, n] \times [1, n] satisfying the following conditions: 1) if (i,j)∈J(i, j) \in J then i<ji < j; 2) if (i,j)∉J(i, j) \notin J then, for every integer kk such that i<k<ji < k < j, at least one of the two pairs (i,k)(i, k) and (k,j)(k, j) does not belong to JJ.

Let TJT_J be the set of matrices Z=(zij)1≤i≤n,1≤j≤nZ = (z_{ij})_{1 \leq i \leq n, 1 \leq j \leq n} with elements in KK, such that zii=1z_{ii} = 1, and zij=0z_{ij} = 0 if i≠ji \neq j and (i,j)∉J(i, j) \notin J. This is a closed subset of GL(n,K)\mathbf{GL}(n, K). The mapping Z↦(zij)(i,j)∈JZ \mapsto (z_{ij})_{(i, j) \in J} is a homeomorphism of TJT_J onto KsK^s (where ss denotes the number of elements of JJ). If Z′=(zij′)∈TJZ' = (z'_{ij}) \in T_J, then Z′Z=(zij′′)Z'Z = (z''_{ij}) with

zij′′=zij+zij′+∑i<h<jzih′zhjfor i<j, z''_{ij} = z_{ij} + z'_{ij} + \sum_{i < h < j} z'_{ih} z_{hj} \quad \text{for } i < j,
zij′′=0 for i>j,zii′′=1, z''_{ij} = 0 \text{ for } i > j, \quad z''_{ii} = 1,

whence Z′Z∈TJZ'Z \in T_J. If TJT_J is identified with KsK^s, then the mapping Z↦Z′ZZ \mapsto Z'Z (for fixed Z′Z') is identified with an affine mapping, and its determinant is 1, as one sees by ordering the pairs (i,j)∈J(i, j) \in J lexicographically and applying the following lemma:

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Cites

Nothing.

Transcribed from the printed volume. Not translated.