8
$\begingroup$

Is there a known explicit description of the abelian $2$-group $\mathsf{Ho}(\mathbb{S})\overset{\mathrm{def}}{=}\mathsf{Ho}(QS^0)\cong\Pi_{\leq1}(QS^0)$?

$\endgroup$
5
  • 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? $\endgroup$ Sep 5, 2021 at 8:55
  • $\begingroup$ @FernandoMuro Sorry, I was a bit confused. $\endgroup$
    – Emily
    Sep 5, 2021 at 19:08
  • 1
    $\begingroup$ The easiest description is that it is the stable quadratic module $\mathbb{Z}\otimes\mathbb{Z}\twoheadrightarrow\mathbb{Z}/(2)\stackrel{0}{\rightarrow}\mathbb{Z}$ $\endgroup$ Sep 6, 2021 at 8:31
  • 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 $(-1)^{mn}$. $\endgroup$ Sep 6, 2021 at 8:34
  • $\begingroup$ @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! $\endgroup$
    – Emily
    Sep 6, 2021 at 22:55

1 Answer 1

12
$\begingroup$

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...

$\endgroup$
2
  • 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") $\endgroup$ Sep 5, 2021 at 1:03
  • 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. $\endgroup$
    – Tim Campion
    Sep 5, 2021 at 2:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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