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$\newcommand{\sgn}{\mathrm{sgn}}\newcommand{\defeq}{\overset{\mathrm{def}}{=}}$The Barratt–Priddy–Quillen–Segal theorem says that the $\mathbb{E}_\infty$-group completion of the groupoid of finite sets and permutations $$\displaystyle\mathbb{F}\defeq\coprod_{n=0}^\infty\mathrm{B}\Sigma_{n}\simeq\coprod_{n=0}^\infty\mathrm{UConf}_{n}(\mathbb{R}^\infty)$$ is the underlying $\mathbb{E}_\infty$-space $\Omega^\infty\mathbb{S}$ of the sphere spectrum.

Since the additive monoidal structure $\oplus$ of $\mathbb{F}$ satisfies $\sgn(\sigma\oplus\tau)\equiv\sgn(\sigma)+\sgn(\tau)\ \text{mod 2}$ and its symmetry $\sigma_{n,m}\colon n\oplus m\to m\oplus n$ has sign $(-1)^{nm}$, it follows that the symmetric monoidal structure of $\mathbb{F}$ restricts to the subcategory $“\mathbb{F}_\text{even}”$ of $\mathbb{F}$ given by $$\displaystyle\mathbb{F}_\text{even}\defeq\coprod_{\substack{n=0\\n\text{ even}}}^\infty\mathrm{B}\mathrm{A}_{n},$$ where $\mathrm{A}_n$ is the $n$th alternating group. For similar reasons, the additive monoidal structure on the braid groupoid $$\displaystyle\mathbb{B}\defeq\coprod_{n=0}^\infty\mathrm{B}\mathrm{B}_{n}\simeq\coprod_{n=0}^\infty\mathrm{UConf}_{n}(\mathbb{R}^2)$$ ―whose $\mathbb{E}_2$-group completion is given by $\Omega^2S^2$―restricts to the subcategory $“\mathbb{B}_{\mathrm{even}}”$ of $\mathbb{B}$ given by $$\displaystyle\mathbb{B}_{\mathrm{even}}\defeq\coprod_{\substack{n=0\\n\text{ even}}}^\infty\mathrm{B}\mathrm{E}_{n},$$ where $\mathrm{E}_{n}$ is the group of even braids on $n$ strands.

Question. What are the $\mathbb{E}_\infty$-group completions of $\mathbb{F}_{\mathrm{even}}$ and $\mathbb{B}_\mathrm{even}$?

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    $\begingroup$ The group completion theorem tells us that the underlying space of the group completion is $2\mathbb{Z}×BA_\infty^+$. I'm trying to think if we can say something about the corresponding spectrum $\endgroup$
    – Denis Nardin
    Oct 12, 2021 at 6:06
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    $\begingroup$ Do we not just get $BA_\infty=\Omega^\infty S[2,\infty)$? $\endgroup$
    – Neil Strickland
    Oct 12, 2021 at 8:11
  • $\begingroup$ @NeilStrickland After taking plus-constructions, yes. $\endgroup$
    – archipelago
    Oct 12, 2021 at 12:08
  • $\begingroup$ Similarly, $BE_\infty^+$ is the cover of $\Omega^2_0S^2$ that corresponds to the subgroup $2\cdot \mathbb{Z}\le \mathbb{Z}=\pi_1(\Omega_0^2S^2)$. $\endgroup$
    – archipelago
    Oct 12, 2021 at 12:17
  • $\begingroup$ The $\mathbb{E}_\infty$ structure on the group completion corresponds to delooping the space in Nardin's comment to a spectrum. There is a unique homotopy class of spectrum maps $H\mathbb{Z} \to \tau_{\leq 1} S$ inducing multiplication by 2 on $\pi_0$, I would think the relevant spectrum fits in a pullback of the form $H\mathbb{Z} \to \tau_{\leq 1} S \leftarrow S$. $\endgroup$
    – user171227
    Oct 12, 2021 at 13:16

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