# Is every $\otimes$-invertible object "coherently sym-central"?

Let $$\mathcal C$$ be a symmetric monoidal $$\infty$$-category, and let $$L \in \mathcal C$$ be a $$\otimes$$-invertible object. Then the braiding $$L \otimes L \to L \otimes L$$ is simply multiplication by $$\dim L$$, where $$\dim L$$ is some involution on the unit object $$I$$.

Thus the unique symmetric monoidal functor $$\Sigma \to \mathcal C$$ sending $$1 \mapsto L$$ ($$\Sigma$$ is the 1-groupoid of finite sets, i.e. the free symmetric monoidal $$\infty$$-category on an object) descends -- at the level of the homotopy category $$ho(\mathcal C)$$! -- through the canonical functor $$\Sigma \to S$$, where $$S$$ is a certain symmetric monoidal 1-groupoid with the same objects as $$\Sigma$$, but $$Aut_S(n) = C_2$$ for all $$n \in \mathbb N$$ (the functor $$\Sigma \to S$$ is defined by taking the sign of a permutation).

Let us say that an object $$L \in \mathcal C$$ is homotopy sym-central if the functor $$\Sigma \to \mathcal C \to ho(\mathcal C)$$ extends along $$\Sigma \to S$$ in a symmetric monoidal way, and coherently sym-central if the functor $$\Sigma \to \mathcal C$$ extends along $$\Sigma \to S$$ in a symmetric monoidal way. Thus any invertible $$L$$ is homotopy sym-central.

Question: Let $$\mathcal C$$ be a symmetric monoidal $$\infty$$-category and $$L \in Pic(\mathcal C)$$ a $$\otimes$$-invertible object. Is $$L$$ necessarily coherently sym-central?

• Without loss of generality $\mathcal{C} = Pic(\mathcal{C})$ is the $0$th space of a connective spectrum $X$. You're asking if every map from the sphere spectrum into $X$ factors through the $1$-truncation of the sphere spectrum. No; for example, the identity map from the sphere spectrum to itself does not admit such a factorization. Mar 11, 2021 at 3:08
• @JacobLurie Ah— that makes sense. So you’re saying that the group completion of the symmetric monoidal groupoid S is the 1 truncation of the sphere spectrum. That’s very believable just from looking at homotopy groups but I don’t quite see why it’s so. Mar 11, 2021 at 3:18

Just to confirm Jacob Lurie's comment above (EDIT: And the following has been corrected -- a previous version fell for a classic blunder as pointed out by Jacob Lurie below): the group completion of $$S$$ is $$\Omega^\infty \tau_{\leq 1} \mathbb S$$ as an infinite loop space. We can see this using a group completion lemma:

Lemma: (cf. ) Let $$C$$ be an $$E_\infty$$ space, and let $$t \in \pi_0 C$$. Then

1. The localized $$E_\infty$$ space $$C[t^{-1}]$$ agrees with the localization $$t^{-1} C$$ of $$C$$ with respect to $$t$$ as a $$C$$-module.

2. Moreover, let $$C_\infty = \varinjlim(C \xrightarrow t C \xrightarrow t \cdots)$$. Then $$t^{-1} C_\infty = t^{-1} C$$.

3. Therefore $$C_\infty = t^{-1} C$$ if and only if $$C_\infty$$ is a $$C[t^{-1}]$$-module.

Proof: (1) follows by the Yoneda lemma: on the category of $$C[t^{-1}]$$-modules, $$t^{-1} C$$ and $$C[t^{-1}]$$ both corepresent the forgetful functor to spaces. (2) holds because $$t: t^{-1} C \to t^{-1}C$$ is invertible. For (3), "only if" is obvious; "if" follows because $$C[t^{-1}]$$ modules are (by definition!) local with respect to the map $$t: C \to C$$ and hence with respect to transfinite composites thereof.

Corollary: The group completion $$K(S)$$ of $$S$$ is $$\Omega^\infty \tau_{\leq 1} \mathbb S$$.

Proof: Let $$t: S \to S$$ be the functor given by tensoring with $$1$$. Then in the notation of the lemma, $$S_\infty$$ is easily seen to have a similar description to $$S$$ but with objects $$\mathbb Z$$ instead of $$\mathbb N$$, and by inspection $$t$$ acts invertibly on $$S_\infty$$. So by the lemma, we have $$K(S) = S[t^{-1}] = t^{-1}S = S_\infty$$. This category looks a lot like $$\Omega^\infty \tau_{\leq 1} \mathbb S$$, and in fact we can see that they are the same because $$\Omega^\infty \tau_{\leq 1} \mathbb S$$ is a Picard 1-category, so the canonical functor from $$K(\Sigma) = \Omega^\infty \mathbb S$$ extends along $$K(\Sigma) \to K(S)$$. The extension is obviously a bijection on objects, and hits the involution on $$1$$ which generates the category symmetric monoidally so it is full as well. Since the hom-sets are finite, it is also faithful and thus an equivalence of categories.

Corollary: The universal functor $$\Sigma \to K(\Sigma) = \Omega^\infty \mathbb S$$ does not factor through $$K(S)$$, and hence does not factor through $$S$$.

Proof: If it did, that would be to say that $$\tau_{\leq 1} \mathbb S$$ splits off of $$\mathbb S$$, but it can't; for example $$\eta^2 \neq 0$$.

: This version of the group completion theorem was based on Prop 6 in an expository note by Thomas Nikolaus, "The group completion theorem via localizations of ring spectra", Prop 6. The note is available from Nikolaus' website; here's a direct link which will directly download the pdf, <1 MB).

• The corollary is correct but the proof given doesn't actually show it (the formula $K(S) = S[t^{-1}]$ is false in general, for example it is false when $S = \coprod_{n} B\Sigma_n$ is the free $E_{\infty}$-space on one generator). You need some additional input to draw the conclusion: for example, you can use the group completion theorem and the fact that the components of $S$ are nilpotent spaces, or you could argue abstractly that $K(S) = S[t^{-1}]$ using that the braiding of $1^{\otimes 2}$ with itself is the identity as an automorphism of $1^{\otimes 4}$. Mar 12, 2021 at 0:37
• @JacobLurie Thanks! In the notation of the present version of my answer, I believe you are saying that the formula $S[t^{-1}]= S_\infty$ is not automatic; since 1-truncated spaces are closed under filtered colimits, this formula holding in general would contradict the Barratt-Priddy-Quillen theorem. The first argument you suggest is similar in spirit to what I have written now. I don't quite follow the second argument you suggest -- are you saying that if $t \in \pi_0 S$ is coherently sym-trivial, then $S[t^{-1}] = \varinjlim(S \xrightarrow t S \xrightarrow t \cdots)$? Mar 12, 2021 at 16:28
• This is why (for example) the plus construction appears in the definition of algebraic K-theory but not in the definition of topological K-theory. Permutation matrices are not equal to the identity, but they belong to the identity component of $\mathrm{GL}_n(\mathbf{C})$ (and to the identity component of $\mathrm{GL}_n(\mathbf{R})$ in the case of even permutations). Mar 12, 2021 at 18:21
• Maybe I should say explicitly that I'm relying on abstract nonsense to guarantee the existence of objects $t^{-1} C$ and $C[t^{-1}]$ with these defining universal properties -- only $C_\infty$ is given an explicit construction. Mar 12, 2021 at 18:36
• Ah, sorry; deleted my previous comment because I was misunderstanding the notation. To prove that $C_{\infty}$ has the desired universal property, it suffices to show that $t$ acts invertibly on it. The "obvious" attempt to prove this will work if the braiding automorphism of $1 \otimes 1$ is (homotopic to) the identity. But since you are free to trade $t$ to $t^n$, it's also true if the braiding automorphism is (homotopic to) the identity on $1^{\otimes n} \otimes 1^{\otimes n}$. Mar 12, 2021 at 19:04