# Reconciling two viewpoints for spectra

As a novice in algebraic topology, I'm trying to grasp the concept of a spectrum. Let me first sketch two motivations.

One motivation goes like this: for singular cohomology of spaces, we have $H^n(X;G)=[X,K(G,n)]$. We observe that these Eilenberg-Mac Lane spaces are such that the maps $K(G,n)\to \Omega K(G,n+1)$ are weak homotopy equivalences. We get the notion of an $\Omega$-spectrum, and realize that any such spectrum gives rise to an (extraordinary reduced) cohomology theory in this way. We can then prove that actually any such cohomology theory on CW complexes arises in this fashion. Thus a spectrum is a cohomology theory (well, an $\Omega$-spectrum actually, but never the matter).

Here's another one. Spectra, whatever that will be, should be the objects of a category where stable phenomena occur. So for instance, Freudenthal's theorem says that for $X$ a finite CW complex, the suspension colimit $colim_k [\Sigma^k X,\Sigma^k Y]$ stabilizes for sufficiently large $k$. We would like this colimit to be a the group of morphisms from a spectrum associated to $X$ to a spectrum associated to $Y$. This will be what the suspension spectra will accomplish: that colimit will be exactly $[\Sigma^\infty X,\Sigma^\infty Y]$.

This is all fine and well, but I don't understand how the same object can come to realize both objectives at the same time. Moreover, both point of views are used at the same time when, for instance, we consider $E^*(E)$ for a spectrum $E$ (which is what we can quickly identify with the group of stable $E$-cohomological operations).

Is there some deep reason behind this? Maybe both motivations are actually the same, if we look at them at a higher level?

• @SanathDevalapurkar: it is sort of ill-expressed. I meant, how is it possible that the same object has two functions which on the face of it are so different: being the same as a cohomology theory, while also being an object concocted to encode stable phenomena. – Bruno Stonek Feb 6 '15 at 16:47
• You might be interested in the notes for the 2011 West Coast Algebraic Topology Summer School by @Aaron here: math.berkeley.edu/~aaron/wcatss/sht1.pdf . In fact, this deep relation that you describe is what leads to the definition of the stable homotopy category (which I think was done by Boardman and Vogt). – user62675 Feb 6 '15 at 19:49
• You may know this already, but the stable homotopy category is something that allows us to study both these phenomena at once (as I mentioned above). There are many different definitions of the stable homotopy category, for example, using $\mathbf{S}$-modules, symmetric spectra, orthogonal spectra, etc. Stable homotopy theory is the study of the stable homotopy category. (See this as well: math.stanford.edu/~carym/stable.pdf) – user62675 Feb 6 '15 at 20:02
• I wrote some "nontechnical" notes that attempt to introduce the stable homotopy category by reasoning along the lines you mention. They have some deficits, but maybe you'll find they address your question: math.berkeley.edu/~ericp/teaching/Fall2013/msri/… . – Eric Peterson Feb 6 '15 at 21:02

Your two objectives are not actually that closely tied together, because there is already a representability theorem in the unstable category. In more detail, we can consider contravariant functors $F$ from connected based spaces to sets, such that
• The natural map $F(\bigvee_{i\in I}X_i)\to\prod_iF(X_i)$ is always bijective;
• $F$ sends homotopy pushout squares to weak pullback squares.
It can be shown that any such functor is represented by some space $BF$. Indeed, this unstable, non-additive version is the original version of Brown representability. Now the main point is just that if $E^*$ is a generalised cohomology theory, then the representing spaces $BE^n$ form an $\Omega$-spectrum. This follows quite transparently from the definitions.