Show that GL2+(R)={A∈M2(R):detA>0} is path connected but not simply connected. Describe a loop that is not null-homotopic.
Hint
Retract GL2+(R) onto SO(2), which is a circle, and use π1(S1)=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)).
Step 2: SO(2) is a circle. The map θ↦Rθ=(cosθsinθ−sinθcosθ) induces a homeomorphism R/2πZ→SO(2) (continuous bijection from a compact space to a Hausdorff space, Problem 6). So SO(2)≅S1 and π1(SO(2),I)≅Z, generated by the loop ℓ(θ)=R2πθ, θ∈[0,1].
Step 3: deformation retraction. By Problem 25, SO(2) is a strong deformation retract of GL2+(R): there is a continuous H:GL2+×[0,1]→GL2+ with H(⋅,0)=id, H(⋅,1)=r where r:GL2+→SO(2) is a retraction (r∣SO(2)=id). Explicitly H(A,t)=U(A)((1−t)P(A)+tI) with A=U(A)P(A) the polar decomposition. Hence the inclusion i:SO(2)→GL2+ induces an isomorphism i∗:π1(SO(2),I)≅π1(GL2+,I).
Step 4: conclusion.π1(GL2+(R),I)≅Z=0. The loop ℓ(θ)=R2πθ is a generator and is not null-homotopic in GL2+.
A direct invariant. The retraction gives an explicit "winding number": to a loop γ in GL2+ associate the loop r∘γ in SO(2)≅S1; the winding number is the degree. Equivalently, for γ(θ)=(acbd), it is the winding number around 0 of the complex number (a+d)+i(c−b), which never vanishes when detγ>0 (indeed ∣a+d∣2+∣c−b∣2=(a2+b2+c2+d2)+2detγ>0).
Remark. For n≥3, π1(SO(n))=Z/2 (the "plate trick"), so π1(GLn+)=Z/2: for n=2 the fundamental group is Z, which is exceptional. The universal cover of GL2+(R) is a non-matrix group.