Question

It is not too hard to understand the geometric meaning of torsion in homology groups of CW complexes. However, I thought it would be interesting to hear how people describe/think of the geometric meaning of torsion in the homotopy groups of a CW-complex.

Answer 1

Well, the silly answer is that $f:\mathbb{S}^k\to X$ represents a torsion element of order $p$ if $p \cdot f:\mathbb{S}^k\to X$ extends along $\mathbb{S}^k \hookrightarrow D^{k+1}$ to a map $\varphi: D^{k+1}\to X$.

The slightly less silly answer --- the slightly deeper answer, that is --- is that, equivalently, $f$ itself extends to a map $\tilde f: P^{k+1}(p)\to X$. Here $P^{k+1}(p)$ is the CW complex built of a point, a $k$-cell and a $(k+1)$-cell, where the $(k+1)$-cell is attached by a degree-$p$ map ${"p}$; and it's called the $(k+1)$th (cyclic) Moore space of order $p$.

For any $p$ the Moore spaces form a suspension spectrum $P^{k+1}(p) \simeq \mathbb{S}^1\wedge P^k(p)$, and there is in fact a long cofibration sequence $$ \mathbb{S}^1 \overset{"p}{\to} \mathbb{S}^1 \to P^2(p) \to \mathbb{S}^2 \overset{"p}{\to} \dots $$ which gives rise to a long exact sequence of generalized homotopy groups $$ \pi_1(X) \leftarrow \pi_2(X,p) \leftarrow \pi_2(X) \overset{p}{\leftarrow} \pi_2(X) \leftarrow \pi_3(X,p)\leftarrow \cdots $$ which in turn, for $n$ large enough, breaks up into the homotopy Universal Coefficient Theorem $$ 0\leftarrow \mathrm{Tor}(\pi_n(X),\mathbb{Z}/(p)) \leftarrow \pi_{n+1}(X,p) \leftarrow \pi_{n+1}(X)\otimes \mathbb{Z}/(p) \leftarrow 0.$$

The letter $P$ is used here because, at least because they have also been called Peterson spaces, and maybe because $P^2(2)\simeq \mathbb{RP}^2$ is the real projective plane.

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Note that I'm using the letter $p$ because when I worry about these things $p$ is usually prime, but that's not necessary in the above.

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We give up continuing the sequence at the first $\pi_1(X)$ because, without more information, it's not clear that $({"p})_\sharp$ should be a group homomorphism. Someone else can remind me what we can still say about the underlying sets.