Using the CW structure on \(SO(n)\), show that \(\pi_1 SO(n) \approx \mathbb{Z}_2\) for \(n \geq 3\). Find a loop representing a generator, and describe how twice this loop is nullhomotopic.
Solution
Use the CW structure of Proposition 3D.1. For \(n\ge3\), its cells are indexed by strictly decreasing sequences \[ n>i_1>\cdots>i_r>0, \] with cell dimension \(i_1+\cdots+i_r\), together with the single \(0\)-cell.
In dimensions at most \(2\) there are therefore exactly three cells: \[ e^0,\qquad e^1,\qquad e^2. \] The cells \(e^1\) and \(e^2\) are the images of the standard cells of \[ \mathbb{RP}^{n-1}\subset SO(n) \] under Hatcher’s map \[ \rho(v)=r(v)r(e_1). \] The characteristic map of the \(2\)-cell of \(\mathbb{RP}^{n-1}\) restricts on its boundary circle to the double covering \[ S^1\to\mathbb{RP}^1\cong S^1. \] Consequently the \(2\)-cell of \(SO(n)\) is attached to the \(1\)-cell by a map of degree \(2\). Thus the \(2\)-skeleton has fundamental group \[ \pi_1(SO(n)^{(2)}) \cong\langle a\mid a^2=1\rangle \cong\mathbb Z_2. \] Attaching cells of dimension at least \(3\) does not change the fundamental group, so \[ \boxed{\pi_1SO(n)\cong\mathbb Z_2\qquad(n\ge3).} \]
The generator has a concrete geometric description. Let a fixed oriented \(2\)-plane in \(\mathbb R^n\) rotate through angle \[ 0\le\theta\le2\pi \] while the orthogonal complement is fixed pointwise. This gives a loop in \(SO(n)\); it is precisely the loop represented by the \(1\)-cell \(e^1\), hence represents the nonzero element of \(\pi_1SO(n)\).
Traversing this loop twice gives a \(4\pi\) rotation. In the CW structure, the attaching map of \(e^2\) traverses the \(1\)-cell twice, so the characteristic disk of \(e^2\) is an explicit nullhomotopy of this doubled loop. Thus the generator has order exactly \(2\).