Let $E$ be a spectrum. Then $E \wedge E$ is a $\mathbb{Z}/2$-spectrum in the naivest possible sense, i.e., an object with $\mathbb{Z}/2$-action in the (∞,1)-category of spectra. Can I make it a $\mathbb{Z}/2$-spectrum in the less naive, but still not genuine, sense? (That is, a $\mathbb{Z}/2$-spectrum indexed on the trivial universe.)

I'm thinking of something like the following. I may represent $E$ as an (reduced & continuous) excisive functor from pointed spaces to pointed spaces. Then define

$$G(X) = \mathrm{colim}_{I \times I} \mathrm{Map}(S^{x_1} \wedge S^{x_2}, E(S^{x_1}) \wedge E(S^{x_2}) \wedge X)$$

where $I$ is the category of finite sets and inclusions. Hopefully $G$ is a functor from spaces to $\mathbb{Z}/2$-spaces. If I forget about the $\mathbb{Z}/2$-fixed point set, I can think of it as $E \wedge E$ with its $\mathbb{Z}/2$ action. What spectrum does $G(X)^{\mathbb{Z}/2}$ correspond to? Is there a more familiar name for it? **Edit:** I seem to be getting $E \vee (E \wedge E)^{h\mathbb{Z}/2}$, but without much confidence.

[Leftover part of the question: If so, by my question here I can think of the resulting object as a functor from the opposite of the orbit category of $\mathbb{Z}/2$ to spectra. Unpacking this amounts to giving some spectrum $F$ together with a map $F \to (E \wedge E)^{h\mathbb{Z}/2}$. What is $F$?]

pickingF with such a map, and F (the fixed point object) can be anything. Do you want some relation of F to E? – Tyler Lawson Dec 2 '09 at 6:12