Is there a known explicit description of the abelian $2$-group $\mathsf{Ho}(\mathbb{S})\overset{\mathrm{def}}{=}\mathsf{Ho}(QS^0)\cong\Pi_{\leq1}(QS^0)$?
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1$\begingroup$ This question looks a bit cumbersome to me. Are not you asking about the fundamental groupoid of the infinite loop space of the sphere spectrum? $\endgroup$– Fernando MuroCommented Sep 5, 2021 at 8:55
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$\begingroup$ @FernandoMuro Sorry, I was a bit confused. $\endgroup$– EmilyCommented Sep 5, 2021 at 19:08
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1$\begingroup$ The easiest description is that it is the stable quadratic module $\mathbb{Z}\otimes\mathbb{Z}\twoheadrightarrow\mathbb{Z}/(2)\stackrel{0}{\rightarrow}\mathbb{Z}$ $\endgroup$– Fernando MuroCommented Sep 6, 2021 at 8:31
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1$\begingroup$ In case you're not acquainted with stable quadratic modules, if you want to see it as a symmetric monoidal groupoid, then the object set is $\mathbb{Z}$, the automorphism group of an object is $\{\pm1\}$, there are no morphisms other than automorphisms, the tensor product is addition on objects and multiplication on morphisms, the associativity and unitarity constraints are identities, and the commutativity constraint $m+n\rightarrow n+m$ is $(-1)^{mn}$. $\endgroup$– Fernando MuroCommented Sep 6, 2021 at 8:34
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$\begingroup$ @FernandoMuro I had heard of stable quadratic modules before, but at the time I had trouble finding any reference to learn about them. This time however I found a paper you wrote pointing to Baues, so I can finally understand what they are now. Thank you! $\endgroup$– EmilyCommented Sep 6, 2021 at 22:55
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1 Answer
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This is the groupoid given by the 1-truncation $\tau_{\leq 1}(QS^0)$. This groupoid has $\mathbb Z$-many objects (since $\pi_0^s = \mathbb Z$), and each one has automorphism group $C_2$ (since $\pi_1^s = C_2$). The tensor product on objects is given by addition in $\mathbb Z$, and on morphisms by addition in $C_2$. One way to see this is to consider the universal functor $\Sigma \to QS^0$ given by the Barratt-Priddy-Quillen theorem (i.e. the fact that $K(\Sigma) = QS^0$; here $\Sigma$ is the groupoid of finite sets with the disjoint union monoidal structure), and to postcompose with the truncation functor $QS^0 \to \tau_{\leq 1} (QS^0)$; the fact that this functor is symmetric monoidal yields this description of the category. This perspective is discussed a bit more here.
From the description I've given, I suppose it follows that $\tau_{\leq 1} (QS^0)$ splits symmetric monoidally as $\tau_{\leq 1} (QS^0) = \mathbb Z \times BC_2$, (where $\mathbb Z$ is a discrete symmetric monoidal groupoid and $BC_2$ is a 1-object symmetric monoidal groupoid), which is maybe a little surprising. This is not to say that $\tau_{\leq 1} \mathbb S$ splits...
answered Sep 4, 2021 at 22:52
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10$\begingroup$ Hi Tim, the splitting of the category is monoidal but not symmetric monoidal (that would mean you could get a model where the twist automorphism (n+m) -> (m+n) is the identity, when it should be the "(-1)^nm") $\endgroup$ Commented Sep 5, 2021 at 1:03
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3$\begingroup$ @TylerLawson Ah of course! The symmetry is the most interesting part of the story and I didn’t even address it! And it’s a good thing— contrary to what I wrote, if the equivalence were symmetric monoidal, it would give an equivalence of spectra. $\endgroup$ Commented Sep 5, 2021 at 2:01