Topology of the groups of upper triangular matrices
Statement
Let be the group of upper unitriangular matrices (ones on the diagonal) and the group of invertible upper triangular matrices in . Show that:
- is homeomorphic to ;
- has exactly connected components, each contractible.
Hint
Use on strictly upper triangular matrices for , and the factorisation for .
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Solution
1. . Let be the space of strictly upper triangular matrices, a vector space of dimension (the free entries above the diagonal). Each satisfies and is upper triangular with ones on the diagonal, so . Conversely is with and (a polynomial in without constant term, so strictly upper triangular). By the previous problem is a homeomorphism, so .
2. : a product decomposition. An invertible upper triangular has nonzero diagonal entries . Let . Then is upper triangular with ones on the diagonal, so with . The map
(where is the group of invertible diagonal matrices) is continuous with continuous inverse . So as topological spaces.
Components. has two components, each homeomorphic to , hence contractible. So has components (choose a sign for each diagonal entry), each homeomorphic to . Multiplying by (connected and contractible) does not change the number of components, and each component of is homeomorphic to , which is contractible.
Check. : , components. : , components according to the signs of and .
Remark. The signs of the diagonal entries are the only topological invariants of : there are components, while has only (connected by rotations, which are not triangular).