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?

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    $\begingroup$ 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. $\endgroup$ Commented Mar 11, 2021 at 3:08
  • $\begingroup$ @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. $\endgroup$
    – Tim Campion
    Commented Mar 11, 2021 at 3:18

1 Answer 1


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. [1]) 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 $\eta^2 \neq 0$.

[1]: 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).

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    $\begingroup$ 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 $1^{\otimes 4}$. $\endgroup$ Commented Mar 12, 2021 at 0:37
  • $\begingroup$ @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 $S[t^{-1}] = \varinjlim(S \xrightarrow t S \xrightarrow t \cdots)$? $\endgroup$
    – Tim Campion
    Commented Mar 12, 2021 at 16:28
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    $\begingroup$ 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). $\endgroup$ Commented Mar 12, 2021 at 18:21
  • $\begingroup$ 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. $\endgroup$
    – Tim Campion
    Commented Mar 12, 2021 at 18:36
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    $\begingroup$ 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 $1^{\otimes n} \otimes 1^{\otimes n}$. $\endgroup$ Commented Mar 12, 2021 at 19:04

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