TITLE: Naive Z/2-spectrum structure on E smash E?
QUESTION [6 upvotes]: 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 $F$?]

REPLY [3 votes]: Here's a very simple way to obtain $(E \wedge E)^{\Bbb Z_2}$ without having to resort to representations (at least if $E$ is connective). Consider the functor from spectra to spectra given by $$E \mapsto \Sigma^\infty \Omega^\infty E .$$ That is, the suspension spectrum of the zeroth space of $E$.
Let 
$$E \mapsto P_2(E)$$ be its quadratic approximation in the sense of Goodwillie's calculus of homotopy functors (that is, the second stage of the Goodwillie tower).
Then $P_2(E)$ has the homotopy type of $(E \wedge E)^{\Bbb Z_2}$ whenever $E$ is connective.
Incidentally, one can also see that this comes equipped with a fiber sequence
$$
D_2(E) \to P_2(E) \to E
$$
which amounts the tom Dieck's splitting in the case when $E$ is a suspension spectrum.
Here, $D_2(E) = (E\wedge E)_{h\Bbb Z_2}$ is the quadratic construction.
In general, this sequence needn't split.