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Exercise 2061Topology › Topology of Mn(ℝ) › ConnectednessM1Added 7 Nov 2026

GL(2,ℝ) is not simply connected

#fundamental group#deformation retraction#circle#homotopy

Statement

Show that GL2+(R)={A∈M2(R):det⁡A>0}GL_2^+(\mathbb{R})=\{A\in M_2(\mathbb{R}):\det A>0\} is path connected but not simply connected. Describe a loop that is not null-homotopic.

Hint

Retract GL2+(R)GL_2^+(\mathbb{R}) onto SO(2)SO(2), which is a circle, and use π1(S1)=Z\pi_1(S^1)=\mathbb{Z}.

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Solution

Idea. A space and a deformation retract of it have the same fundamental group.

Step 1: path connected. Proved in Problem 8 (together with Problem 6 for SO(2)SO(2)).

Step 2: SO(2)SO(2) is a circle. The map θ↦Rθ=(cos⁡θ−sin⁡θsin⁡θcos⁡θ)\theta\mapsto R_\theta=\begin{pmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{pmatrix} induces a homeomorphism R/2πZ→SO(2)\mathbb{R}/2\pi\mathbb{Z}\to SO(2) (continuous bijection from a compact space to a Hausdorff space, Problem 6). So SO(2)≅S1SO(2)\cong S^1 and π1(SO(2),I)≅Z\pi_1(SO(2),I)\cong\mathbb{Z}, generated by the loop ℓ(θ)=R2πθ\ell(\theta)=R_{2\pi\theta}, θ∈[0,1]\theta\in[0,1].

Step 3: deformation retraction. By Problem 25, SO(2)SO(2) is a strong deformation retract of GL2+(R)GL_2^+(\mathbb{R}): there is a continuous H:GL2+×[0,1]→GL2+H:GL_2^+\times[0,1]\to GL_2^+ with H(⋅,0)=idH(\cdot,0)=\mathrm{id}, H(⋅,1)=rH(\cdot,1)=r where r:GL2+→SO(2)r:GL_2^+\to SO(2) is a retraction (r∣SO(2)=idr|_{SO(2)}=\mathrm{id}). Explicitly H(A,t)=U(A)((1−t)P(A)+tI)H(A,t)=U(A)((1-t)P(A)+tI) with A=U(A)P(A)A=U(A)P(A) the polar decomposition. Hence the inclusion i:SO(2)→GL2+i:SO(2)\to GL_2^+ induces an isomorphism
i∗:π1(SO(2),I)→ ≅ π1(GL2+,I).i_*:\pi_1(SO(2),I)\xrightarrow{\ \cong\ }\pi_1(GL_2^+,I).

Step 4: conclusion. π1(GL2+(R),I)≅Z≠0\pi_1(GL_2^+(\mathbb{R}),I)\cong\mathbb{Z}\ne0. The loop ℓ(θ)=R2πθ\ell(\theta)=R_{2\pi\theta} is a generator and is not null-homotopic in GL2+GL_2^+.

A direct invariant. The retraction gives an explicit "winding number": to a loop γ\gamma in GL2+GL_2^+ associate the loop r∘γr\circ\gamma in SO(2)≅S1SO(2)\cong S^1; the winding number is the degree. Equivalently, for γ(θ)=(abcd)\gamma(\theta)=\begin{pmatrix}a&b\\c&d\end{pmatrix}, it is the winding number around 00 of the complex number (a+d)+i(c−b)(a+d)+i(c-b), which never vanishes when det⁡γ>0\det\gamma>0 (indeed ∣a+d∣2+∣c−b∣2=(a2+b2+c2+d2)+2det⁡γ>0|a+d|^2+|c-b|^2=(a^2+b^2+c^2+d^2)+2\det\gamma>0).

Remark. For n≥3n\ge3, π1(SO(n))=Z/2\pi_1(SO(n))=\mathbb{Z}/2 (the "plate trick"), so π1(GLn+)=Z/2\pi_1(GL_n^+)=\mathbb{Z}/2: for n=2n=2 the fundamental group is Z\mathbb{Z}, which is exceptional. The universal cover of GL2+(R)GL_2^+(\mathbb{R}) is a non-matrix group.