# Connes cyclic set - construction explaination.

This is from a paper by Scholze and Nikolaus, THH, p145, second equation.

The author defines two categories.

1. $$\Lambda_\infty$$ a fullsubcaregory of Posets with $$\Bbb Z$$ action: Its objects are $$\frac{1}{n}\Bbb Z$$, with its natural ordering.

2. $$\Lambda_p := \Lambda_\infty/ Bp\Bbb Z$$ They said a description of this category is given by:

i) objects are the same as $$\Lambda_\infty$$

ii) the morphism space are quotient out by the relation. $$f \sim \sigma^pf$$ where $$\sigma$$ is a generator of the action of $$\Bbb Z$$ on morphism space. Explicitly $$\sigma^pf=f+p$$.

I get the latter description. However I don't understand what this quotient $$\Lambda_\infty /Bp\Bbb Z$$ means:

a) where is this taking placing?

b) how is this computation done?

Some elaborationi wuold be nice!

• What is $\sigma$? – Berci Nov 13 '20 at 11:36

1. Since you already know what the answer should be, it's maybe not so important where the operation is taking place. But here is one possible context that makes sense.

As explained in the paper, there is an action of $$B\mathbb{Z}$$ on $$\Lambda_\infty$$, i.e., we have a functor $$\alpha: B\mathbb{Z} \times \Lambda_\infty \to \Lambda_\infty$$, which restricts to an action $$\alpha: B(p\mathbb{Z}) \times \Lambda_\infty \to \Lambda_\infty$$. The notation $$\Lambda_\infty / B(p\mathbb{Z})$$ simply means to take the quotient of $$\Lambda_\infty$$ by this action, which can be modeled for example as the coequalizer of $$B(p\mathbb{Z}) \times \Lambda_\infty \overset{\alpha}{\underset{\mathrm{pr}}{\rightrightarrows}} \Lambda_\infty$$ in the category of (small) categories, where $$\mathrm{pr}$$ is the projection onto the second factor.

2. There's not really a "computation" to do here - everything is just a matter of unpacking definitions. Since $$\alpha$$ preserves objects in $$\Lambda_\infty$$, the coequalizer can just be computed as a quotient category of $$\Lambda_\infty$$ by the congruence generated by identifying $$\operatorname{pr}(p\sigma, f) = f$$ with $$\alpha(p\sigma, f) = f + p$$. In other words the objects $$\Lambda_\infty / B(p\mathbb{Z})$$ are the same as $$\Lambda_\infty$$, and the morphisms are equivalence classes of morphisms in $$\Lambda_\infty$$ under the relation $$f \sim f + p$$.

EDIT: What I wrote above is slightly off: $$\alpha$$ doesn't seem to actually be a functor. There may be some way to salvage this, but I don't see an easy way right now, so I'm offering another perspective that hopefully has fewer mistakes.

We think of the action of $$B\mathbb{Z}$$ (really $$\mathbb{Z}$$) on $$\Lambda_\infty$$ as a functor $$F: B\mathbb{Z} \to \mathsf{Cat}$$ that sends the object $$*$$ of $$B\mathbb{Z}$$ to $$\Lambda_\infty$$, and a generator $$\sigma$$ of $$\mathbb{Z} = \operatorname{mor}(B\mathbb{Z})$$ to the functor $$\sigma: \Lambda_\infty \to \Lambda_\infty$$ that is the identity on objects and $$f \mapsto f + 1$$ on morphisms. Then, we set $$\Lambda_\infty / B(p\mathbb{Z})$$ to be the colimit of the composite $$B(p\mathbb{Z}) \to B\mathbb{Z} \xrightarrow{F} \mathsf{Cat}$$.

• By a $B \Bbb Z$ action on $\Lambda_\infty$ this is akin to the axioms of a group actioon right? It doesn't seem to me that simply having a functor $B\Bbb Z \times \Lambda_\infty \rightarrow \Lambda_\infty$ induces a $\Bbb Z$ action on morphism space from (You have an assignment, but not a compatibility one). – Bryan Shih Nov 13 '20 at 23:42
• You're right. While $\alpha$ does satisfy some of the commutative diagrams involved in group actions, it doesn't seem to be a functor after all. I suppose you can still take coequalizers of the set of objects and the set of morphisms separately, but this is rather unsatisfying. As a mea culpa, I've added another approach which may be useful. – JHF Nov 14 '20 at 5:00