9
$\begingroup$

Given a pointed $(\infty,n)$-category $\mathcal{C}$, one can define the suspension of $\mathcal{C}$, $\Sigma\mathcal{C}$, via the homotopy pushout of $$\ast\leftarrow \mathcal{C}\rightarrow \ast.$$ Dually one can define $\Omega\mathcal{C}$. Can one explicitly identify these $(\infty,n)$-categories in terms of $\mathcal{C}$?

My vague intuition, based on the case $n=0$, says that the $\Omega\mathcal{C}$ should be the endomorphisms of the distinguished object and $\Sigma\mathcal{C}$ should be what you get when you take the free monoidal $(\infty,n)$-category on $\mathcal{C}$, regard it as an $(\infty,n+1)$-category with one object and then invert the $n+1$-morphisms. However, my understanding of (homotopy) limits and colimits in this setting is pretty poor.

Feel free to use any model you wish.

$\endgroup$
  • 2
    $\begingroup$ If by "homotopy" you mean the usual notion then $\Omega C$ is the automorphisms, not the endomorphisms. Directed delooping in $n$-category theory with $n > (\infty,1)$ requires using "lax" generalizations of homotopy pullbacks, such as "comma objects" ncatlab.org/nlab/show/comma+object $\endgroup$ – Urs Schreiber Sep 5 '13 at 10:01
  • 1
    $\begingroup$ By homotopy pullback/pushout, I will take any construction that is equivalent to the derived pullback/pushout constructed using the projective/injective model structures on diagrams in a combinatorial model category modelling $(\infty,n)$-categories. $\endgroup$ – Justin Noel Sep 5 '13 at 10:40
  • $\begingroup$ Do you have a proof that $\Omega\mathcal{C}$ is the automorphisms? In particular, why is it always an $(\infty,0)$-category? $\endgroup$ – Justin Noel Sep 5 '13 at 10:41
  • 6
    $\begingroup$ $\Omega \mathcal{C}$ is the $(\infty,n)$-category of automorphisms not the space of automorphisms (i.e. objects are automorphisms but we allow arbitrary morphisms between them). The reason why these are not all endomorphisms is that $\Delta[1]$ is not an interval object but $E[1]$ (the nerve of the contractible groupoid with two objects) is (e.g. in the Joyal model structure or in the Rezk model structure for $\Theta_n$-spaces). $\endgroup$ – Karol Szumiło Sep 5 '13 at 11:19
  • 1
    $\begingroup$ The pushout of $* \sqcup * \to E[1]$ and $* \sqcup * \to *$ is a homotopy pushout too since $* \sqcup * \to E[1]$ is a cofibration. However, I neglected the fact that the analogous pushout of categories is not preserved by the nerve functor. But that's even better, maps $E[1] \to \mathcal{C}$ classify equivalences in $\mathcal{C}$ by definition so maps $E[1] / (* \sqcup *) \to \mathcal{C}$ classify equivalences with the same source and target i.e. automorphisms. $\endgroup$ – Karol Szumiło Sep 5 '13 at 13:55
4
$\begingroup$

First let me thank Urs, Karol, and Rune Haugseng for helpful comments.

Now note that the inclusion, $i$, of $\infty$-groupoids into $(\infty,n)$-categories has an $\infty$-categorical left adjoint, $L$ (for lack of a better name), and a right adjoint $(-)^\prime$.

Given a pointed $(\infty,n)$-category $\mathcal{C}$, the loop category $\Omega \mathcal{C}$ is defined by the following (homotopy) pullback diagram:

\begin{array}{ccc} \Omega\mathcal{C} & \rightarrow & \ast\\ \downarrow & & \downarrow \\ \ast & \rightarrow & \mathcal{C} \end{array}

Now $\mathcal{C}^{\prime}$ is the maximal sub-$\infty$-groupoid of $\mathcal{C}$ (the core). Since $\ast$ is an $\infty$-groupoid, the inclusion $\ast\rightarrow \mathcal{C}$ factors canonically through $\mathcal{C}^\prime$. By a standard finality argument we see that $\Omega \mathcal{C}$ is equivalent to the homotopy pullback: \begin{array}{ccc} \Omega\mathcal{C} \simeq \Omega^{Top}\mathcal{C}^\prime& \rightarrow & \ast\\ \downarrow & & \downarrow \\ \ast & \rightarrow & \mathcal{C}^\prime \end{array}

Regarding $\mathcal{C}^\prime$ as a space (since it is an $\infty$-groupoid), we see that $\Omega\mathcal{C}$ is equivalent to the space of topological (based) loops on $\mathcal{C}^\prime$ since $i$ preserves (homotopy) limits.

Unraveling this a bit, we see that $\Omega(-)$ is naturally equivalent to $i\Omega^{Top}(-)^\prime$ which is a composite of right adjoints. It follows that the left adjoint, $\Sigma(-)$, is naturally equivalent to $i\Sigma^{Top}L(-)$.

As a consequence, an $(\infty,n)$-category which is a loop category is necessarily a loop space. This shows that the inclusion of spectra objects in $(\infty,0)$-categories (i.e., spectra) into spectra objects in $(\infty,n)$-categories is an equivalence. So the two categories have the same stailizations.

$\endgroup$
  • $\begingroup$ Could you say a bit more about the "standard finality argument"? $\endgroup$ – Marc Hoyois Sep 27 '13 at 15:14
  • $\begingroup$ Oh I see: putting $C'$ instead of $C$ in the lower right corner simply doesn't change the universal property of the pullback in the $(\infty,1)$-category of $(\infty,n)$-categories. $\endgroup$ – Marc Hoyois Sep 28 '13 at 18:31
  • $\begingroup$ Exactly, or one could check that the appropriate comma category is weakly contractible. $\endgroup$ – Justin Noel Oct 1 '13 at 8:31

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.