3
$\begingroup$

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?

$\endgroup$
2
  • 9
    $\begingroup$ 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. $\endgroup$ 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
    Mar 11, 2021 at 3:18

1 Answer 1

3
$\begingroup$

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).

$\endgroup$
7
  • 1
    $\begingroup$ 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 $1^{\otimes 4}$. $\endgroup$ 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
    Mar 12, 2021 at 16:28
  • 1
    $\begingroup$ 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). $\endgroup$ 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
    Mar 12, 2021 at 18:36
  • 1
    $\begingroup$ 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 $1^{\otimes n} \otimes 1^{\otimes n}$. $\endgroup$ Mar 12, 2021 at 19:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.