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?

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


1 Answer 1


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.

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    $\begingroup$ By the way, there is a blog post presenting this argument at amathew.wordpress.com/2012/10/25/… where the treatment has a higher categorical bent. While many details are missing, it is useful as a conceptual overview of the argument. $\endgroup$ Commented Feb 26, 2013 at 8:17
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    $\begingroup$ Good work Peter! $\endgroup$
    – John Klein
    Commented Feb 27, 2013 at 19:58

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