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We can characterise $\mathbb{Z}$ and $\mathbb{Z}/2$ as the corepresenting abelian groups of the functors \begin{align*} \mathsf{Forget} &\colon \mathsf{Ab} \to \mathsf{Sets},\\ \mathrm{Inv} &\colon \mathsf{Ab} \to \mathsf{Sets} \end{align*} given by $(A,\cdot_A,1)\mapsto A$ and $A\mapsto\mathrm{Inv}(A)\overset{\mathrm{def}}{=}\left\{a\in A\ \middle|\ a^2=1_A\right\}$.

A similar approach in the $\infty$-world gives the $\mathbb{E}_\infty$-groups $QS^0$ and $\Omega Q\mathbb{RP}^\infty$. Passing to spectra via the equivalence between $\mathbb{E}_\infty$-groups and connective spectra, we obtain the sphere spectrum $\mathbb{S}$ corresponding to $QS^0$ and a spectrum $E$ corresponding to $\Omega Q\mathbb{RP}^\infty$.

(One possible name for $E$ might be "$\mathbb{S}/2$" since it satisfies an analogous universal property to that of $\mathbb{Z}/2$, corepresenting "involutory objects". However, that notation already usually denotes the mod 2 Moore spectrum, so let's write $E$ for it instead.)

For comparison, their first $8$ homotopy groups are as follows: $$ \begin{aligned} \pi_0(\mathbb{S}) &\cong \mathbb{Z},\\ \pi_1(\mathbb{S}) &\cong \mathbb{Z}/2,\\ \pi_2(\mathbb{S}) &\cong \mathbb{Z}/2,\\ \pi_3(\mathbb{S}) &\cong \mathbb{Z}/24,\\ \pi_4(\mathbb{S}) &\cong 0,\\ \pi_5(\mathbb{S}) &\cong 0,\\ \pi_6(\mathbb{S}) &\cong \mathbb{Z}/2,\\ \pi_7(\mathbb{S}) &\cong \mathbb{Z}/16\times\mathbb{Z}/3\times\mathbb{Z}/5, \end{aligned} \quad\quad \begin{aligned} \pi_0(E) &\cong \mathbb{Z}/2,\\ \pi_1(E) &\cong \mathbb{Z}/2,\\ \pi_2(E) &\cong \mathbb{Z}/8,\\ \pi_3(E) &\cong \mathbb{Z}/2,\\ \pi_4(E) &\cong 0,\\ \pi_5(E) &\cong \mathbb{Z}/2,\\ \pi_6(E) &\cong \mathbb{Z}/16\times\mathbb{Z}/2,\\ \pi_7(E) &\cong \mathbb{Z}/2\times\mathbb{Z}/2\times\mathbb{Z}/2. \end{aligned} $$ (The ones for $E$ are taken from Liulevicius; see also MO 230790.)

What (homotopy associative, homotopy commutative, $\mathbb{A}_k$-, $\mathbb{E}_k$-, or $\mathbb{E}_\infty$-) ring spectra structures, if any, are there on $E$?

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    $\begingroup$ S/2 doesn't have a unital multiplication. In general, S/p does have an A_{p-1}-structure, but not an A_p-structure. $\endgroup$
    – skd
    Commented Sep 21, 2021 at 1:27
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    $\begingroup$ "$\mathbb S / 2$" would standardly denote the mod 2 Moore spectrum, which is the same as $\Sigma^{\infty-1} \mathbb R \mathbb P^2$... but you've defined "$\mathbb S/2$" to be $\Omega Q (\mathbb R \mathbb P^\infty)$... first of all $Q(X)$ usually means $\Omega^\infty \Sigma^\infty X$, so what you've written is a space and not a spectrum, thought the (unstable) homotopy groups of $\Omega Q \mathbb R \mathbb P^\infty$ are the same as the (stable) homotopy groups of $\Sigma^{\infty-1} \mathbb R \mathbb P^\infty$. Still, that's a different spectrum from the mod 2 Moore spectrum. Could you clarify? $\endgroup$
    – Tim Campion
    Commented Sep 21, 2021 at 2:04
  • $\begingroup$ @TimCampion I was writing "$\mathbb{S}/2$" to mean the connective spectrum associated to the $\mathbb{E}_\infty$-group $\Omega Q\mathbb{RP}^\infty$ because it satisfies an analogous universal property to the one for $\mathbb{Z}/2$: while morphisms of monoids from $\mathbb{Z}$ and $\mathbb{Z}/2$ to a monoid $A$ are the same as invertible/involutory elements of $A$, symmetric monoidal functors from $QS^0$ and $\Omega Q\mathbb{RP}^\infty$ to $\mathcal{C}$ are the same as invertible/involutory objects of $\mathcal{C}$. I've edited the question to use a less confusing notation. Thanks, Tim! $\endgroup$
    – Emily
    Commented Sep 21, 2021 at 2:31
  • $\begingroup$ @skd Sorry for the confusing notation! I didn't mean the mod 2 Moore spectrum $\endgroup$
    – Emily
    Commented Sep 21, 2021 at 2:31
  • $\begingroup$ $(-)^\times$ is a pretty weird name for what seems to be the forgetful functor from abelian groups to sets... $\endgroup$
    – Achim Krause
    Commented Sep 21, 2021 at 5:42

1 Answer 1

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There are no left-unital multiplications on E. If there were, then for any element $x$ in $\pi_n(E)$, we would have $x+x = 1 \cdot x + 1 \cdot x = (1+1) \cdot x = 0$ because all elements in $ \pi_0 E$ are 2-torsion. This is not satisfied by the homotopy groups in your table.

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  • $\begingroup$ Amazing! Thanks! $\endgroup$
    – Emily
    Commented Sep 21, 2021 at 3:38

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