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Exercise 2019Topology › Topology of Mn(ℝ) › Exponential, logarithm and group topologyM1Added 6 Nov 2026

The Lie algebra of a closed subgroup is a linear subspace

#Lie algebra#closed subgroup#Lie product formula

Statement

Let H≤GLn(R)H\le GL_n(\mathbb{R}) be a closed subgroup and
h={X∈Mn(R):etX∈H for all t∈R}.\mathfrak h=\{X\in M_n(\mathbb{R}):e^{tX}\in H\ \text{for all }t\in\mathbb{R}\}.
Show that h\mathfrak h is a linear subspace of Mn(R)M_n(\mathbb{R}).

Hint

Closure under scalar multiples is easy. For sums, use the Lie product formula applied to tXtX and tYtY, together with the closedness of HH.

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Solution

Idea. The sum X+YX+Y is approximated through products of elements of HH, and HH is closed.

Contains 00. et⋅0=I∈He^{t\cdot0}=I\in H, so 0∈h0\in\mathfrak h.

Scalar multiples. If X∈hX\in\mathfrak h and c∈Rc\in\mathbb{R}, then for every tt: et(cX)=e(tc)X∈He^{t(cX)}=e^{(tc)X}\in H because tctc is just another real number. So cX∈hcX\in\mathfrak h.

Sums. Let X,Y∈hX,Y\in\mathfrak h and fix t∈Rt\in\mathbb{R}. For every integer k≥1k\ge1, etX/k∈He^{tX/k}\in H and etY/k∈He^{tY/k}\in H (HH contains esXe^{sX} for every real ss), hence the product
gk=(etX/ketY/k)k∈H.g_k=\bigl(e^{tX/k}e^{tY/k}\bigr)^k\in H.
By the Lie product formula applied to A=tXA=tX, B=tYB=tY, gk→et(X+Y)g_k\to e^{t(X+Y)} as k→∞k\to\infty. The limit is invertible (it is an exponential), so it lies in GLnGL_n, and HH is closed in GLnGL_n: et(X+Y)∈He^{t(X+Y)}\in H. This holds for every tt, so X+Y∈hX+Y\in\mathfrak h.

Remark. The hypothesis "closed" is used in the last step. For the non-closed subgroup φ(R)\varphi(\mathbb{R}) of the torus in SO(4)SO(4) the limit et(X+Y)e^{t(X+Y)} need not stay in the subgroup: its closure is the whole torus, whose Lie algebra is two-dimensional, larger than the line generated by the winding direction.