# Homotopy groups of spectra

A spectrum $\mathbf{E}$ consists of a series of pointed spaces $\{E_n | n \in \mathbb{Z}\}$ of pointed spaces (CW-complexes or compactly generated) together with a series of maps $$\sigma_n: \Sigma E_n \rightarrow E_{n+1},$$ where $\Sigma$ denotes the reduced suspension. The homotopy groups of a spectrum for all $k \in \mathbb{Z}$ are defined by $$\pi_k(\mathbf{E})= colim_{n\rightarrow \infty} \pi_{n+k}(E_n),$$ where the inductive system is given by

$$\pi_{n+k}(E_n) \xrightarrow{\Sigma_*} \pi_{n+k+1} (\Sigma E_n) \xrightarrow{(\sigma_n)_*} \pi_{n+k+1} (E_{n+1}).$$ Now a $\Omega$-spectrum is a spectrum where the adjoint maps of the structure maps $E_n \rightarrow \Omega E_{n+1}$ are weak homotopy equivalences.Here $\Omega$ denotes the space of based loops with compact-open topology that is right adjoint to the reduces suspension.

My question concerns the homotopy groups of $\Omega$-spectra. They are supposed to be $\pi_k(\mathbf{E})= \pi_k(E_0)$ for $k \geq 0$ and $\pi_k(\mathbf{E})=\pi_0(E_{-k})$ for $k \leq 0$.

It makes sense that this is true since for any $k \ge 0$ there is isomorphism $\pi_{n+k}(E_n)\cong \pi_{n+k-}(\Omega E_n) \cong E_{n+k-1}(E_{n-1})\cong \dots \cong \pi_k(E_0)$. A smiliar argument can be made for $k \leq 0$. But this is no proof, or is it? So isomorphisms somehow need to be compatible with the colimit. Thank you.

## migrated from mathoverflow.netFeb 13 '17 at 9:51

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• What do you mean "this is no proof"? It seems like a perfectly good proof to me (note that your isomorphisms are exactly the map in the colimit diagram)... – Denis Nardin Feb 11 '17 at 20:06
• That's what I have been trying to understand. Could you maybe elaborate a little on the maps? It makes sense but I don't quite see it. – Hodor Feb 11 '17 at 20:37

You could say that the reason to consider $\Omega$-spectra is precisely that you can calculate the homotopy groups directly, without having to pass to a colimit.
For example, consider the sphere (pre)spectrum $\mathbb S := \Sigma^\infty S^0$. It's the spectrum whose $n^{\text{th}}$ space (for $n\ge 0$) is $S^n$ and whose $n^{\text{th}}$ structure map is the homeomorphism $\Sigma S^n\to S^{n+1}$. (For negative $n$, the spaces and maps are trivial.) The homotopy groups of $\mathbb S$ are the stable homotopy groups of the spheres $\pi_n^S := \operatorname{colim} \pi_{n+k}(S^n)$. The colimit is essential: $\pi_n^S \ne \pi_n(S^0)$, because $\mathbb S$ isn't an $\Omega$-spectrum: $S^n$ is not weakly equivalent to $\Omega S^{n+1}$.
If $E$ is an $\Omega$-spectrum, however, you have a weak equivalence $E_m\cong\Omega^{n-m} E_n$ whenever $m\le n$. By definition, $\pi_{n+k}(E_n) = \pi_k(\Omega^n E_n)$, and since $\Omega^n E_n$ and $E_0$ are weakly equivalent, then $\pi_k(\Omega^n E_n)\cong\pi_k(E_0)$. Thus, $\pi_{n+k}(E_n)\cong\pi_k(E_0)$ for all $n$, so the colimit for determining $\pi_k(E)$ is the colimit of a constant system in $\pi_k(E_0)$ — the maps in the colimit are the identity, because they're $\pi_k$ applied to the weak equivalences $E_m\stackrel\simeq\to \Omega E_{m-1}$. Thus, for $\Omega$-spectra, you can use $\pi_k(E) = \pi_k(E_0)$. (The case for negative $k$ is similar.)