Let $X$ be a connected, based CW complex. Then the James splitting
of $\Sigma\Omega\Sigma X$ gives, in particular, a weak equivalence of spectra
$$
\Sigma^{\infty} \Omega\Sigma X\_+ \quad \simeq \quad \Sigma^{\infty} (S^0 \vee X \vee X^{[2]} \vee X^{[3]} \vee \cdots ) ,
$$
where $X^{[n]}$ denotes the $n$-fold smash product of $X$ with itself. Each side of this equivalence has the structure of $A\_\infty$-ring spectrum (the structure on the right side is induced by
concatenation $X^{[n]} \wedge X^{[m]} \cong X^{[m+n]}$, and the right side can be seen as the tensor algebra over the sphere spectrum on the points of $X$). 


Now, my understanding$^\dagger$ is that the Cartan formula for Hopf invariants
implies that the above splitting is multiplicative up to homotopy.


**Question:** Can the above splitting be 
enriched to an equivalence of $A\_\infty$-rings?


If so, can anyone provide me with a reference?


$^\dagger\tiny \text{From being once upon a time in Bill Richter's orbit.}$



On the stable splitting of loops on a suspension
Hi John, this is not a complete proof, but it should give the idea of one. There is a clever proof of the splitting of suspension spectra of $\Omega^n\Sigma^n X$ due to Ralph Cohen in his
paper "Stable proofs of stable splittings". His proof is also given in section VII.5 of
LMS "Equivariant stable homotopy theory". It applies more simply to give the splitting that you are interested in. Modernizing using EKMM Section II.4 (or symmetric or orthogonal spectra), the right side of your equivalence is equivalent to the free $A\_\infty$ ring spectrum generated by $\Sigma^{\infty} X$. With minor finagling (about base points in particular), the freeness will give a map of $A\_{\infty}$ ring spectra from right to left. It should be an equivalence by inspection of the cited proof.


