Maths WorkshopMaths problems with full solutions
Exercise 1379Topology › Topology of Mn(ℝ) › Exponential, logarithm and group topologyL3Added 6 Nov 2026

Topology of the groups of upper triangular matrices

#triangular matrices#connected components#contractible#unipotent

Statement

Let UnU_n be the group of upper unitriangular matrices (ones on the diagonal) and BnB_n the group of invertible upper triangular matrices in Mn(R)M_n(\mathbb{R}). Show that:

  1. UnU_n is homeomorphic to Rn(n−1)/2\mathbb{R}^{n(n-1)/2};
  2. BnB_n has exactly 2n2^n connected components, each contractible.
Hint

Use exp⁡\exp on strictly upper triangular matrices for UnU_n, and the factorisation T=D⋅(D−1T)T=D\cdot(D^{-1}T) for BnB_n.

Show solutionHide solution

Solution

1. UnU_n. Let n\mathfrak{n} be the space of strictly upper triangular matrices, a vector space of dimension n(n−1)/2n(n-1)/2 (the free entries above the diagonal). Each N∈nN\in\mathfrak{n} satisfies Nn=0N^n=0 and exp⁡(N)\exp(N) is upper triangular with ones on the diagonal, so exp⁡(n)⊂Un\exp(\mathfrak{n})\subset U_n. Conversely U∈UnU\in U_n is I+NI+N with N∈nN\in\mathfrak{n} and log⁡(I+N)∈n\log(I+N)\in\mathfrak{n} (a polynomial in NN without constant term, so strictly upper triangular). By the previous problem exp⁡:n→Un\exp:\mathfrak{n}\to U_n is a homeomorphism, so Un≅n≅Rn(n−1)/2U_n\cong\mathfrak{n}\cong\mathbb{R}^{n(n-1)/2}.

2. BnB_n: a product decomposition. An invertible upper triangular TT has nonzero diagonal entries di=tiid_i=t_{ii}. Let D=diag(d1,…,dn)D=\mathrm{diag}(d_1,\dots,d_n). Then D−1TD^{-1}T is upper triangular with ones on the diagonal, so T=D⋅UT=D\cdot U with U=D−1T∈UnU=D^{-1}T\in U_n. The map
Bn→D×Un,T↦(D, D−1T)B_n\to\mathcal{D}\times U_n,\qquad T\mapsto\bigl(D,\ D^{-1}T\bigr)
(where D≅(R×)n\mathcal{D}\cong(\mathbb{R}^\times)^n is the group of invertible diagonal matrices) is continuous with continuous inverse (D,U)↦DU(D,U)\mapsto DU. So Bn≅(R×)n×UnB_n\cong(\mathbb{R}^\times)^n\times U_n as topological spaces.

Components. R×=(−∞,0)∪(0,∞)\mathbb{R}^\times=(-\infty,0)\cup(0,\infty) has two components, each homeomorphic to R\mathbb{R}, hence contractible. So (R×)n(\mathbb{R}^\times)^n has 2n2^n components (choose a sign for each diagonal entry), each homeomorphic to (0,∞)n≅Rn(0,\infty)^n\cong\mathbb{R}^n. Multiplying by Un≅Rn(n−1)/2U_n\cong\mathbb{R}^{n(n-1)/2} (connected and contractible) does not change the number of components, and each component of BnB_n is homeomorphic to Rn×Rn(n−1)/2=Rn(n+1)/2\mathbb{R}^n\times\mathbb{R}^{n(n-1)/2}=\mathbb{R}^{n(n+1)/2}, which is contractible.

Check. n=1n=1: B1=R×B_1=\mathbb{R}^\times, 22 components. n=2n=2: B2={(ab0d):ad≠0}B_2=\{\begin{pmatrix}a&b\\0&d\end{pmatrix}:ad\ne0\}, 44 components according to the signs of aa and dd.

Remark. The signs of the diagonal entries are the only topological invariants of BnB_n: there are 2n2^n components, while GLnGL_n has only 22 (connected by rotations, which are not triangular).