Compute the higher homotopy groups \(\pi_i(T^n)\) of the \(n\)-torus \(T^n = (S^1)^n\) for all \(i \ge 2\).
Solution
1. The product of the universal covering maps \(\mathbb R\to S^1\) gives a universal covering map\[ p:\mathbb R^n\longrightarrow T^n=(S^1)^n. \]The total space \(\mathbb R^n\) is contractible.
2. A covering map induces an isomorphism on homotopy groups in every degree \(i\ge2\).
For \(i\ge2\), every based map \(S^i\to T^n\) lifts uniquely after choosing the lift of the basepoint, since \(S^i\) is simply connected. The same lifting statement for homotopies shows that this correspondence induces a bijection on based homotopy classes, hence an isomorphism on \(\pi_i\).
3. Therefore, for every \(i\ge2\),\[ \pi_i(T^n)\cong\pi_i(\mathbb R^n)=0. \]Thus \(T^n\) is aspherical.