TITLE: do spectra have diagonal maps?
QUESTION [9 upvotes]: Topological spaces have diagonal maps $X \rightarrow X \times X$ and $X \rightarrow X \wedge X$, and suspension spectra also have diagonal maps $\Sigma^\infty X \rightarrow \Sigma^\infty(X \wedge X) \cong (\Sigma^\infty X) \wedge (\Sigma^\infty X)$. What about general spectra? (i.e. symmetric spectra, S-modules, or any other convenient definition.) I always assumed you could, but I haven't thought through it carefully. And if not, can we still get a cup product on $E^*(X)$ when $E$ and $X$ are spectra?

REPLY [13 votes]: The existence of an $E_\infty$-diagonal is an obstruction for equipping a spectrum $E$  with the structure of a suspension spectrum. Conversely, in
Klein, J.R.: Moduli of suspension spectra. Trans. Amer. Math. Soc. 357  (2005), 489–507
I showed that the existence of a suitably defined notion of $A_\infty$-diagonal on $E$ 
is equivalent to equipping $E$ with the structure of a suspension spectrum provided we are in the metastable range. Here "metastable" means $E$ is $r$-connected (for $r \ge 1$) and is weak
equivalent to a cell spectrum of dimension $\le 3r+2$. 
There are various elementary ways of defining the notion of $A_\infty$-diagonal, 
but they in the end
amount to the existence of a map $\delta: E \to (E\wedge E)^{\Bbb Z_2}$ (for a suitably
defined version of the smash product), which is a homotopy section to the map 
$(E\wedge E)^{\Bbb Z_2} \to E$ which is given by passing from categorical to geometric fixed points. The way I do this in the paper is the use the second stage of the Taylor tower
of the functor $E \mapsto \Sigma^\infty \Omega^\infty E$; this second stage turns out to be a model for $(E\wedge E)^{\Bbb Z_2}$.