ConceptComplete

Higher Homotopy Groups

The higher homotopy groups πn(X)\pi_n(X) for n2n \geq 2 generalize the fundamental group by using maps from higher-dimensional spheres instead of loops. Unlike π1\pi_1, these groups are always abelian.


Definition

Definition

Let XX be a pointed topological space with basepoint x0x_0. The nn-th homotopy group of XX is πn(X,x0)=[(Sn,s0),(X,x0)]\pi_n(X, x_0) = [(S^n, s_0), (X, x_0)] the set of basepoint-preserving homotopy classes of maps f:(Sn,s0)(X,x0)f : (S^n, s_0) \to (X, x_0). Equivalently, viewing Sn=In/InS^n = I^n / \partial I^n, πn(X,x0)={f:InXf(In)=x0}/\pi_n(X, x_0) = \{f : I^n \to X \mid f(\partial I^n) = x_0\} / \simeq where \simeq denotes homotopy relative to In\partial I^n.

The group operation is defined by "concatenation in the first coordinate": ([f]+[g])(t1,t2,,tn)={f(2t1,t2,,tn)0t11/2g(2t11,t2,,tn)1/2t11([f] + [g])(t_1, t_2, \ldots, t_n) = \begin{cases} f(2t_1, t_2, \ldots, t_n) & 0 \leq t_1 \leq 1/2 \\ g(2t_1 - 1, t_2, \ldots, t_n) & 1/2 \leq t_1 \leq 1 \end{cases}

Definition

A space XX is nn-connected if πk(X)=0\pi_k(X) = 0 for all knk \leq n. A space is simply connected if π0(X)=0\pi_0(X) = 0 (path-connected) and π1(X)=0\pi_1(X) = 0. A weakly contractible space has πn(X)=0\pi_n(X) = 0 for all n0n \geq 0.


Commutativity

Theorem9.1Abelianness of Higher Homotopy Groups

For n2n \geq 2, the group πn(X,x0)\pi_n(X, x_0) is abelian. That is, for any two maps f,g:SnXf, g : S^n \to X, [f]+[g]=[g]+[f][f] + [g] = [g] + [f]

The proof uses the Eckmann-Hilton argument: for n2n \geq 2, there are two independent directions in which to concatenate, and interchanging them provides a homotopy between f+gf + g and g+fg + f.

ExampleHomotopy groups of spheres

The computation of πn(Sm)\pi_n(S^m) is one of the central problems in algebraic topology. Key known values include:

  • πn(Sn)Z\pi_n(S^n) \cong \mathbb{Z} for all n1n \geq 1 (generated by the identity map)
  • π3(S2)Z\pi_3(S^2) \cong \mathbb{Z} (generated by the Hopf fibration)
  • πn+1(Sn)Z/2Z\pi_{n+1}(S^n) \cong \mathbb{Z}/2\mathbb{Z} for n3n \geq 3
  • πn+2(Sn)Z/2Z\pi_{n+2}(S^n) \cong \mathbb{Z}/2\mathbb{Z} for n2n \geq 2

The full computation remains an open problem; the structure is far more complex than homology.


RemarkDifficulty of computation

Unlike homology, homotopy groups are extremely difficult to compute. There is no general algorithm for computing πn(X)\pi_n(X), and the homotopy groups of even the simplest spaces (like spheres) exhibit intricate and unpredictable patterns. This difficulty reflects the fundamentally non-linear nature of homotopy theory.