Notes

Why spectra, and what stabilisation actually buys you

The Freudenthal suspension theorem says that for an nn-connected space XX, the suspension map

πk(X)⟶πk+1(ΣX)\pi_k(X) \longrightarrow \pi_{k+1}(\Sigma X)

is an isomorphism for k<2n+1k < 2n + 1 and surjective when k=2n+1k = 2n + 1. Iterate it and the homotopy groups stop moving. What is left is the stable homotopy group

πks(X)=colim⁡n→∞πk+n(ΣnX).\pi_k^s(X) = \operatorname{colim}_{n \to \infty} \pi_{k+n}(\Sigma^n X).

This is the whole trick, and it is worth being precise about what it costs.

What you gain

Unstably, [X,Y][X, Y] is a set — pointed, but a set. After stabilisation it becomes an abelian group, and the category of spectra is triangulated: cofibre sequences and fibre sequences agree, so a great deal of homological algebra transfers verbatim. The smash product makes it symmetric monoidal, with the sphere spectrum S\mathbb{S} as unit. Suddenly one can say “module over a ring spectrum” and mean it.

Concretely: the Adams spectral sequence

E2s,t=Ext⁡As,t(H∗(Y;Fp), Fp)⟹πt−s(Y)p∧E_2^{s,t} = \operatorname{Ext}_{\mathcal{A}}^{s,t}(H^*(Y; \mathbb{F}_p),\, \mathbb{F}_p) \Longrightarrow \pi_{t-s}(Y)^{\wedge}_p

computes something genuinely hard — the pp-completed stable homotopy of YY — from something merely difficult, namely Ext⁡\operatorname{Ext} over the Steenrod algebra A\mathcal{A}.

What you lose

Stabilisation is not free. It forgets everything that is not visible after enough suspensions: the Whitehead product dies, and with it much of the unstable information about how XX is glued together. π∗s(S0)\pi_*^s(S^0) is still, famously, not fully known.

Why a physicist should care

Anomalies in quantum field theory are classified by bordism groups, and bordism groups are homotopy groups of Thom spectra — ΩnSO≅πn(MSO)\Omega_n^{SO} \cong \pi_n(MSO). The invariants that a physical theory can actually detect are stable ones, which is a statement about spectra whether or not one chooses to say it that way.

That is roughly the thread I keep pulling on: if the observables are stable invariants, the right home for the question is the stable homotopy category.