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$\newcommand{\Tr}{\mathrm{Tr}}$Lately I've been trying to gather more examples of centres and traces, hoping to write a comprehensive treatment on those on Clowder.

One of the examples I've been trying to understand better is the category of divisibility posets $\mathcal{D}$, which is defined as follows.

  • For each $n\in\mathbb{N}_{\geq1}$, let $D_n$ be the poset consisting of the set $\{1,\ldots,n\}$ together with the preorder $\preceq$ given by declaring $k\preceq\ell$ if $k$ divides $\ell$.
  • Then, let $\mathcal{D}$ be the full subcategory of the category $\mathsf{Pos}$ of posets and monotone maps spanned by the posets $D_n$ for all $n\in\mathbb{N}_{\geq1}$.

Recalling the definitions, the center of $\mathcal{D}$ is defined as \begin{align*} \mathrm{Z}(\mathcal{D})&\overset{\scriptstyle\mathrm{def}}{=}\int_{D_n\in\mathcal{D}}\mathrm{Hom}_{\mathcal{D}}(D_n,D_n)\\ &\cong\mathrm{Nat}(\mathrm{id}_{\mathcal{D}},\mathrm{id}_{\mathcal{D}}), \end{align*} while the trace of $\mathcal{D}$ is defined as \begin{align*} \mathrm{Tr}(\mathcal{D})&\overset{\scriptstyle\mathrm{def}}{=}\int^{D_n\in\mathcal{D}}\mathrm{Hom}_{\mathcal{D}}(D_n,D_n)\\ &\cong\left.\left(\coprod_{D_n\in\mathrm{Obj}(\mathcal{D})}\mathrm{Hom}_{\mathcal{D}}(D_n,D_n)\right)\middle/\mathord{\sim}\right., \end{align*} where $\sim$ is the equivalence relation generated by $f\circ g\sim g\circ f$. Finally, the quotient map $\mathrm{tr}$ from $\coprod_{D_n\in\mathrm{Obj}(\mathcal{D})}\mathrm{Hom}_{\mathcal{D}}(D_n,D_n)$ to $\mathrm{Tr}(\mathcal{D})$ is called the trace map of $\mathcal{D}$.

Looking at naturality with respect to the maps $[k]\colon D_1\to D_n$ given by $1\mapsto k$ shows the centre of $\mathcal{D}$ to be trivial. However, as is usual with centres and traces, computing the trace seems to be much, much harder.

Question. Is there a neat description of the trace of $\mathcal{D}$ and the trace map?

(By "neat" here I'm thinking of something like how conjugacy classes of symmetric groups may be described as Young diagrams.)


Here are some miscellaneous notes:

Numerical Experiments. The trace of the category $\mathcal{D}_m$ spanned by the divisibility posets $D_1,\ldots,D_m$ is described for $1\leq m\leq8$ here.

In particular, the cardinality of $\mathrm{Tr}(\mathcal{D}_m)$ for $1\leq m\leq8$ is $(1,2,4,6,11,21,39,59)$, for which OEIS gives nothing.

For example, the equivalence classes in $\mathrm{Tr}(\mathcal{D}_3)$ are the following, where e.g. $(1,3,2)$ denotes the map $D_3\to D_3$ sending $(1,2,3)$ to $(1,3,2)$:

Class 1:
  D_1 -> D_1: (1,)
  D_2 -> D_2: (1, 1)
  D_2 -> D_2: (2, 2)
  D_3 -> D_3: (1, 1, 1)
  D_3 -> D_3: (1, 1, 2)
  D_3 -> D_3: (1, 3, 1)
  D_3 -> D_3: (2, 2, 2)
  D_3 -> D_3: (3, 3, 3)

Class 2:
  D_2 -> D_2: (1, 2)
  D_3 -> D_3: (1, 1, 3)
  D_3 -> D_3: (1, 2, 1)
  D_3 -> D_3: (1, 2, 2)
  D_3 -> D_3: (1, 3, 3)

Class 3:
  D_3 -> D_3: (1, 2, 3)

Class 4:
  D_3 -> D_3: (1, 3, 2)

Numerical evaluation suggests that increasing $m$ does not add new identifications between endomorphisms of $D_k$ and $D_\ell$ for $k,\ell\leq m$, i.e. if $f\colon D_k\to D_k$ and $g\colon D_\ell\to D_\ell$ are not equal in $\mathrm{Tr}(\mathcal{D}_n)$, then they remain not equal in $\mathrm{Tr}(\mathcal{D}_{n+1})$. In other words, the natural map $$\mathrm{Tr}(\mathcal{D}_n)\to\mathrm{Tr}(\mathcal{D}_{n+1})$$ seems to be injective. (Take this with a grain of salt however; I was only able to run these experiments for very small $n$.)

Fixed Points and Invariants. One strategy for computing traces is to find invariants under the relation $f\circ g\sim g\circ f$.

In our case, the number of fixed points of an endomorphism of $\mathcal{D}$ is one such invariant (by the same argument given in MO 469617).

However, this does not give a complete characterization of the trace map, since e.g. the maps $(1,1,1)$ and $(1,3,2)$ in $\mathcal{D}_3$ have the same number of fixed points but belong to different elements of $\mathrm{Tr}(\mathcal{D}_3)$.

Relation to Conjugacy Classes. I believe there's an isomorphism $$\mathrm{Tr}(\mathcal{D}_n)\cong\mathrm{Cl}(\mathrm{End}(D_n)),$$ where $\mathrm{Cl}(\mathrm{End}(D_n))$ denotes the set of ($\sim^*_p$) conjugacy classes of $\mathrm{End}(D_n)$.

Proof. Let $n,k\in\mathbb{N}$ with $k<n$. We have a commutative diagram in $\mathcal{D}_{n}$ of the form

where $p$ is given by $f$ composed with the inclusion $D_k\to D_n$ and $q$ is defined as follows:

  • If $i\leq k$, then $q(i)=i$.
  • If $i\geq k$, then we have $i=ab$ for some $a,b\in\mathbb{N}$ with $a\in D_k$ and $b$ as small as possible. We then define $q(i)=a$.

With this definition, if $i\mid j$, then $q(i)\mid q(j)$, so $q$ is indeed a morphism of posets. The existence of this commutative diagram gives $[f]=[p\circ q]$ in $\mathrm{Tr}(\mathcal{D}_n)$, so the map $\mathrm{Tr}(\mathcal{D}_n)\to\mathrm{Cl}(D_n)$ sending $f$ to $p\circ q$ is well defined. Injectivity and surjectivity of this map then follow from the fact that the relation $\sim$ on $\mathcal{D}_n$ restricts to the ($\sim^*_p$) conjugacy relation in $\mathrm{End}(D_n)$.

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  • $\begingroup$ Ignoring all the category theory stuff, and focusing on the order theory: your "divisibility posets" are just products of finite chains. $\endgroup$ Commented Nov 9 at 2:27
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    $\begingroup$ @SamHopkins actually that was claimed in the post too, but seems to be incorrect: for instance, $D_6$ contains the prime number $5$, which is not of the form $2^{a_2} \cdot 3^{a_3}$. The colimit as $n \to \infty$ is an "infinite-dimensional cube", but $D_n$ is not a cube. However, this mistake does not seem to affect the computations for $n \leq 8$, if I am reading them correctly. $\endgroup$ Commented Nov 9 at 2:32
  • $\begingroup$ @R.vanDobbendeBruyn oh whoops I'm used to $D_n$ only consisting of divisors of $n$. $\endgroup$ Commented Nov 9 at 2:36
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    $\begingroup$ Is it true for every morphism of the category generated by D1,…,Dn in general is equivalent to an endomorphism of Dn? For you D1,D2,D3 is basically character equivalence or generalized conjugacy for the endomorphism monoid of Dn. For n=3 you have the conjguacy classes of the group of units and then one class for the rank 2 idempotent and the rank 1 idempotent. $\endgroup$ Commented Nov 9 at 20:20
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    $\begingroup$ @BenjaminSteinberg Yep! I think we have an isomorphism between $\mathrm{Tr}(\mathcal{D}_n)$ and what Araújo–Kinyon–Konieczny–Malheiro call the $\sim^*_p$ conjugacy classes of $\mathrm{End}(D_n)$ here. So to understand the trace and the trace map "all" we need to do is understand $\mathrm{Cl}(\mathrm{End}(D_n))$ for all $n$. $\endgroup$
    – Emily
    Commented Nov 11 at 18:39

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