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The group $\mathbb{Z}/2$ corepresents the functor $\mathrm{Inv}\colon\mathsf{Mon}\to\mathsf{Sets}$ sending a monoid $A$ to its set of involutory elements (those satisfying $a^2=1_A$).

A similar story is true for $\mathbb{Z}$ and invertible elements, but let's instead tell it in the $\infty$-setting: namely, the $\infty$-category of $\mathbb{E}_1$-monoidal functors $\mathbb{Z}_\mathsf{disc}\to\mathcal{C}$ is just $\mathsf{Pic}(\mathcal{C})$, and thus $\mathbb{Z}_\mathsf{disc}$ corepresents the functor $$\mathsf{Pic}\colon\mathsf{Mon}_{\mathbb{E}_1}(\mathsf{Cats}_{\infty})\to\mathcal{S}$$

However, replacing

  • $\mathsf{Mon}_{\mathbb{E}_1}(\mathsf{Cats}_\infty)$ by $\mathsf{Mon}_{\mathbb{E}_\infty}(\mathsf{Cats}_\infty)$, the $\infty$-category of symmetric monoidal $\infty$-categories;
  • $\mathcal{S}$ by $\mathsf{Grp}_{\mathbb{E}_\infty}(\mathcal{S})$;

changes the corepresenting object from $\mathbb{Z}_{\mathsf{disc}}$ to the sphere spectrum $\mathbb{S}$. Similarly, if we pass to $\mathbb{E}_k$ rather than $\mathbb{E}_{\infty}$, we get $\Omega^kS^k$ instead of $\mathbb{S}$.

Now, define an involutory object of a monoidal $\infty$-category $\mathcal{C}$ to be a strong monoidal functor $(\mathbb{Z}/2)_{\mathsf{disc}}\to\mathcal{C}$. By definition, $(\mathbb{Z}/2)_{\mathsf{disc}}$ corepresents the functor $$\mathsf{Inv}\colon\mathsf{Mon}_{\mathbb{E}_1}(\mathsf{Cats}_{\infty})\to\mathcal{S}$$ sending $\mathcal{C}$ to $\mathsf{Inv}(\mathcal{C})\overset{\mathrm{def}}{=}\mathsf{Fun}^\otimes((\mathbb{Z}/2)_{\mathsf{disc}},\mathcal{C})$.

Question. Is the functor $$\mathsf{Inv}\colon\mathsf{Mon}_{\mathbb{E}_{k}}(\mathsf{Cats}_\infty)\to\mathsf{Grp}_{\mathbb{E}_{k-1}}(\mathcal{S})$$ corepresentable by an $\mathbb{E}_{k}$-monoidal category for $2\leq k\leq\infty$?

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    $\begingroup$ I think if $C$ is $E_k$-monoidal and $X$ is $E_1$-monoidal, then $Fun^\otimes(X,C)$ is $E_{k-1}$-monoidal, not $E_1$-monoidal (think of $k=1$: what is a monoidal structure on $Fun^\otimes(C,D)$ for $C,D$ barely monoidal ? on $Alg(D)$ ? ); the case of $X=\mathbb Z$ is special, because $Fun^\otimes(\mathbb Z,C)\to C$ is the inclusion of a full sub-groupoid which is stable under tensor products. Your question still makes sense though, if you replace the second $E_k$ with an $E_{k-1}$ $\endgroup$
    – Maxime Ramzi
    Commented Sep 19, 2021 at 10:52
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    $\begingroup$ Also, the lifting of $Inv$ to $Ab$ makes $\mathbb Z/2$ into a co-abelian group in $CMon$ - which is not a surprise, everyone is a co-(commutative monoid) in $CMon$, and the corresponding shear map is just the shear map $\endgroup$
    – Maxime Ramzi
    Commented Sep 19, 2021 at 10:56
  • $\begingroup$ (Re the original question): Couldn't I take the pushout in E_k-spaces of pt<---Free(x)--->Free(y) where x goes to y^2? $\endgroup$
    – Dylan Wilson
    Commented Sep 19, 2021 at 11:26

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$$Fun^{\otimes}(\mathbb Z/2, C) \simeq map_{E_1}(\mathbb Z/2, C^\simeq) \simeq map_{E_k}(\mathrm{Ind}_{E_1}^{E_k}\mathbb Z/2, C^\simeq)$$ where $\mathrm{Ind}_{E_1}^{E_k}$ denotes the left adjoint to the forgetful functor.

So $Inv$ is representable, and the natural $E_{k-1}$-structure (see my comments for why I wrote $E_{k-1}$ and not $E_k$ - it is possible that in this special case too we could get $E_k$, but I don't see a reason why, and what I wrote works for any $E_1$-space $X$) on this space gives $\mathrm{Ind}_{E_1}^{E_k}\mathbb Z/2$ a natural co-$E_{k-1}$-structure (in $E_k$-spaces - with the coproduct monoidal structure).

Now does the space $\mathrm{Ind}_{E_1}^{E_k}\mathbb Z/2$ have a reasonably concrete description ? I think it's something like a bar construction $Bar(E_1,E_k, \mathbb Z/2)$ so you can get an explicit description involving the space of little $k$-disks, but I'm not entirely sure you can get much better. I'd love to hear about a better description.

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  • $\begingroup$ Thanks, Maxime! Would it be okay to ask a few questions? 1) Why do we have $Fun^\otimes(\mathbb{Z}/2,C)\cong map_{E_1}(\mathbb{Z}/2,C^\simeq)$? 2) Why did you choose the notation $\mathrm{Ind}^{E_k}_{E_1}$? 3) Is there a reason to expect $\mathrm{Ind}^{E_k}_{E_1}(\mathbb{Z}/2)$ to be an $E_k$-group? $\endgroup$
    – Emily
    Commented Sep 19, 2021 at 21:02
  • $\begingroup$ (Also, I updated the question to correct the E_k vs E_k-1 issue; thanks!) $\endgroup$
    – Emily
    Commented Sep 19, 2021 at 21:02
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    $\begingroup$ Emily : 1) it's because any monoidal morphism between strong monoidal functors out of a grouplike monoidal category is invertible. But in fact, even if that weren't so, because you want it to land in $S$ you would have to take $map$ rather than $Fun$ (and the $C^\simeq$ is because $\mathbb Z/2$ is a groupoid). 2) I'm viewing it as some form of extension of scalars, so it's a similar notation as for extension of scalars along a ring map $A\to B$: $\mathrm{Ind}_A^B$. 3) Yes: for any $E_1$-group $G$ this will be the case $\endgroup$
    – Maxime Ramzi
    Commented Sep 20, 2021 at 7:19
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    $\begingroup$ Indeed, $map_{E_k}(\mathrm{Ind}(G), X) = map_{E_1}(G,X) = map_{E_1}(G,X^{inv}) = map_{E_k}(\mathrm{Ind}(G), X^{inv})$ where $X^{inv}$ is the full sub-$E_k$-groupoid on invertible elements $\endgroup$
    – Maxime Ramzi
    Commented Sep 20, 2021 at 7:20
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    $\begingroup$ @skd : it's almost that, although I think there's a shit: it should be $\Omega^k\Sigma^{k-1}B\mathbb Z/2$, because $map_{E_k}(\Omega^k\Sigma^{k-1} BX, Y) = map_{E_{k-1}}(\Omega^{k-1}\Sigma^{k-1} BX, BY) = map_{S_*}(BX, BY) = map_{E_1}(X,Y)$. That being said, I agree with you that I don't see why there would be a more explicit description, except maybe for replacing $B\mathbb Z/2$ with $\mathbb RP^\infty$ $\endgroup$
    – Maxime Ramzi
    Commented Sep 20, 2021 at 7:23

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