# Corepresentability of involutory objects in monoidal $\infty$-categories

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 $$2\leq k\leq\infty$$?

• 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 $E_{k-1}$ Sep 19 '21 at 10:52
• 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 Sep 19 '21 at 10:56
• (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? Sep 19 '21 at 11:26

$$Fun^{\otimes}(\mathbb Z/2, C) \simeq map_{E_1}(\mathbb Z/2, C^\simeq) \simeq map_{E_k}(\mathrm{Ind}_{E_1}^{E_k}\mathbb Z/2, C^\simeq)$$ 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.

• 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?
– Théo
Sep 19 '21 at 21:02
• (Also, I updated the question to correct the E_k vs E_k-1 issue; thanks!)
– Théo
Sep 19 '21 at 21:02
• Théo : 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 Sep 20 '21 at 7:19
• 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 Sep 20 '21 at 7:20
• @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 $\mathbb RP^\infty$ Sep 20 '21 at 7:23