# What are the centre and trace of the simplex category?

Definition. The centre of a category $$\mathcal{C}$$ is the set $$\mathrm{Z}(\mathcal{C})$$ defined by \begin{align*} \mathrm{Z}(\mathcal{C}) &\mathbin{\overset{\mathrm{def}}{=}} \int_{A\in\mathcal{C}}\mathrm{Hom}_{\mathcal{C}}(A,A)\\ &\cong \mathrm{Nat}(\mathrm{id}_{\mathcal{C}},\mathrm{id}_{\mathcal{C}}), \end{align*} while the trace of $$\mathcal{C}$$ is the set $$\mathrm{Tr}(\mathcal{C})$$ defined by \begin{align*} \mathrm{Tr}(\mathcal{C}) &\mathbin{\overset{\mathrm{def}}{=}} \int^{A\in\mathcal{C}}\mathrm{Hom}_{\mathcal{C}}(A,A)\\ &\cong \mathrm{End}(\mathcal{C})/\mathord{\sim}, \end{align*} where $$\mathrm{End}(\mathcal{C})$$ is the set of all endomorphisms of $$\mathcal{C}$$ and $$\mathord{\sim}$$ is the equivalence relation on $$\mathrm{End}(\mathcal{C})$$ generated by $$g\circ f\sim f\circ g$$. Here, the quotient map $$\mathrm{tr}\colon\mathrm{End}(\mathcal{C})\to\mathrm{Tr}(\mathcal{C})$$ is called the trace map.

Monoid Structure. The center of $$\mathcal{C}$$ comes with a natural monoid structure given by composition of natural transformations, and every monoidal category structure on $$\mathcal{C}$$ descends to a monoid structure on $$\mathrm{Z}(\mathcal{C})$$ (making it a commutative monoid by Eckmann–Hilton) and $$\mathrm{Tr}(\mathcal{C})$$.

Example. Take a group $$G$$ and view it as a one-object category $$\mathrm{B}G$$. Then, we can show that the centre of $$\mathrm{B}G$$ is the usual group-theoretic centre $$\mathrm{Z}(G)$$ of $$G$$, while the trace of $$\mathrm{B}G$$ is the set of conjugacy classes of $$G$$.

Question. What is the centre and trace of the simplex category $$\Delta$$?

What about the centre and trace of the augmented simplex category $$\Delta_+$$, which comes with a monoidal structure $$\oplus$$, thus making its centre a commutative monoid and its trace a monoid?

• My guess is the center is trivial. Just look at morphisms from the one point total order. I would have to think about trace Apr 20 at 17:52
• Dumb question: but is this definition of "center" a special case of the "center of an adjunction", for the identity adjoint equivalence of the category? ncatlab.org/nlab/show/fixed+point+of+an+adjunction Apr 21 at 20:12
• @hasManyStupidQuestions Nope, the center of $\mathrm{id}_{\mathcal{C}}\dashv\mathrm{id}_{\mathcal{C}}$ is $\mathcal{C}$ itself, since we're looking at the full subcategory of $\mathcal{C}$ spanned by those objects $A$ of $\mathcal{C}$ such that the unit map $A\to\mathrm{id}_{\mathcal{C}}(\mathrm{id}_{\mathcal{C}}(A))$ of this adjunction is an isomorphism, which in this case is all of $\mathcal{C}$. Apr 22 at 1:15

The center is trivial : as Benjamin said in the comments, an endomorphism of the identity is the identity on $$\Delta^0$$, and using the maps $$\Delta^0\to \Delta^n$$ you find that any endomorphism of the identity is the identity.

For the trace, you can use the following fact coming from unwinding the definition : if $$f,g$$ are composable in both orders in $$C$$, then $$tr(fg) = tr(gf) \in Tr(C)$$ where I let $$tr$$ of an endomorphism be the class it defines in $$Tr(C)$$.

Using this, for any endomorphism $$f\in \Delta$$, writing it as a surjection followed by an injection and inducting on size shows that every $$tr(f)$$ is $$tr(id_S)$$ for some $$S\in \Delta$$ (note that any automorphism is the identity). Furthermore, $$f\mapsto$$ the cardinality of the fixed point set of $$f$$ defines a morphism $$Tr(\Delta)\to \mathbb N$$ (this is not obvious, but a fun exercise in combinatorics) which distinguishes all the $$tr(id_S), S\in \Delta$$ so that $$Tr(\Delta)\cong \mathbb N_{>0}$$

• Thank you so much, Maxime! I feel like this is probably easy, but do you see a quick way to show that $\#\mathrm{Im}(f\circ g)=\#\mathrm{Im}(g\circ f)$ for $f\colon[n]\to[m]$ and $g\colon[m]\to[n]$ morphisms of $\Delta$? I ask because $f\mapsto\#\mathrm{Im}(f)$ gives a map $\mathrm{Tr}(\Delta)\to\mathbb{N}_{\geq1}$ serving the same purpose of your fixed points map, but it also seems to be exactly the trace map $\mathrm{tr}\colon\mathrm{End}(\Delta)\to\mathrm{Tr}(\Delta)$, giving a quick description of which "conjugacy class" of $\Delta$ an endomorphism of it lives in. Apr 22 at 1:08
• @Emily: No, the size of the image doesn’t determine an element of the trace. You can falsify the required cyclicity property already with maps $f : [2] \to [3]$, $g : [3 ] \to [2]$. Apr 22 at 8:54
• @PeterLeFanuLumsdaine Thank you, Peter! Apr 22 at 15:47

$$\newcommand{\Fix}{\mathrm{Fix}}\newcommand{\N}{\mathbb{N}}\newcommand{\id}{\mathrm{id}}\newcommand{\End}{\mathrm{End}}\newcommand{\Tr}{\mathrm{Tr}}\newcommand{\tr}{\mathrm{tr}}$$Here are some extra details and explicit computations for determining the trace $$\Tr(\Delta)$$ of $$\Delta$$ as well as the trace map $$\tr\colon\End(\Delta)\to\Tr(\Delta),$$ following Maxime's answer.

## An Explicit Example

Perhaps the general way to proceed is best exemplified by an example: consider the map $$f\colon [4]\to[4]$$ given by

and let's denote it (and maps of $$\Delta$$ in general) by the cleaner notation

To compute $$\tr(f)$$, we can write this map as the composition of a surjection $$s\colon [4]\twoheadrightarrow[2]$$ with an injection $$i\colon[2]\hookrightarrow[4]$$:

Now, since we have $$\tr(f\circ g)=\tr(g\circ f)$$ for any two endomorphisms $$f$$ and $$g$$ of $$\Delta$$ such that $$f\circ g$$ and $$g\circ f$$ are both defined, we see that $$\tr(f)=\tr(i\circ s)=\tr(s\circ i)$$, where $$s\circ i$$ is the map

which is the identity map $$\id_{[2]}\colon[2]\to[2]$$ of $$[2]$$ in $$\Delta$$.

## The General Case

Claim 1. Let $$f\colon[n]\to[n]$$ be an arbitrary endomorphism of $$\Delta$$. We have $$\tr(f)=\tr\big(\id_{[k]}\big)$$ for some $$k\leq n$$.

Proof. If $$f=\id_{[n]}$$, then we are done. Otherwise, we can again decompose it into a surjection $$s\colon[n]\twoheadrightarrow[k]$$ followed by an injection $$i\colon[k]\hookrightarrow[n]$$, where $$k=\#\mathrm{Im}(f)$$ will be strictly less than $$n$$. By the property $$\tr(f\circ g)=\tr(g\circ f)$$ of the trace map, we have \begin{align*} \tr(f) &= \tr(i\circ s)\\ &= \tr(s\circ i), \end{align*} with $$s\circ i\colon[k]\to[k]$$ again an endomorphism of $$\Delta$$. Now, if $$s\circ i=\id_{[k]}$$, we are done. Otherwise, we repeat this process again.

Since $$n$$ is finite, it follows that repeating this process will give us a natural number $$n_0\leq n$$ such that $$\tr(f)=\tr\big(\id_{[n_0]}\big)$$ in finitely many steps.

Next, we aim to show the following:

Claim 2. If $$n\neq m$$, then $$\tr\big(\id_{[n]}\big)\neq\tr\big(\id_{[m]}\big)$$.

To this end, we want we want to construct a map $$\phi\colon\mathrm{End}(\Delta)\to\N_{\geq1}$$ satisfying the following conditions:

1. If $$n\neq m$$, then $$\phi\big(\id_{[n]}\big)\neq\phi\big(\id_{[m]}\big)$$.
2. We have $$\phi(f\circ g)=\phi(g\circ f)$$, so that $$\phi$$ descends to a map $$\widetilde{\phi}\colon\mathrm{Tr}(\Delta)\to\N_{\geq1}$$.

From the example of $$f\colon[4]\to[4]$$ above and similar ones, we might guess that the map $$\sigma\mapsto\#\mathrm{Im}(\sigma)$$ sending an endomorphism of $$\Delta$$ to the cardinality of its image would be our desired map $$\phi$$.

However, as Peter pointed out, the map $$\sigma\mapsto\#\mathrm{Im}(\sigma)$$ fails to satisfy Item (2) above, as the following example shows:

Consider the maps $$f\colon[3]\to[2]$$ and $$g\colon[2]\to[3]$$ given by

Then the maps $$g\circ f\colon[3]\to[3]$$ and $$f\circ g\colon[2]\to[2]$$ are given by

and we have $$\#\mathrm{Im}(g\circ f)=2$$ while $$\#\mathrm{Im}(f\circ g)=1$$.

The map sending $$\sigma\colon[n]\to[n]$$ to the cardinality $$\#\Fix(\sigma)$$ of its set of fixed points does fill both criteria:

1. If $$n\neq m$$, then $$\#\Fix(\id_{[n]})=n+1\neq m+1=\#\Fix(\id_{[m]})$$.

2. We have $$\#\Fix(g\circ f)=\#\Fix(f\circ g)$$, as the maps \begin{align*} f|_{\Fix(g\circ f)} &\colon \Fix(g\circ f) \to \Fix(f\circ g),\\ g|_{\Fix(f\circ g)} &\colon \Fix(f\circ g) \to \Fix(g\circ f) \end{align*} set up a bijection between $$\Fix(g\circ f)$$ and $$\Fix(f\circ g)$$, as:

a) If $$x$$ is a fixed point of $$g\circ f$$, i.e. $$g(f(x))=x$$, then $$f(x)$$ is a fixed point of $$f\circ g$$, since $$f(g(f(x))=f(x)$$.

b) Similarly, if $$y$$ is a fixed point of $$f\circ g$$, then $$g(y)$$ is a fixed point of $$g\circ f$$.

c) For each $$x\in\Fix(g\circ f)$$, we have \begin{align*} [g|_{\Fix(f\circ g)}\circ f|_{\Fix(g\circ f)}](x) &= g(f(x))\\ &= x\\ &= [\id_{\Fix(g\circ f)}](x). \end{align*} so $$g|_{\Fix(f\circ g)}\circ f|_{\Fix(g\circ f)}=\id_{\Fix(g\circ f)}$$.

d) Similarly, $$f|_{\Fix(g\circ f)}\circ g|_{\Fix(f\circ g)}=\id_{\Fix(f\circ g)}$$.

It follows that $$\sigma\mapsto\#\Fix(\sigma)$$ defines a map $$\widetilde{\phi}\colon\Tr(\Delta)\to\mathbb{N}_{\geq1}$$.

Finally, the result.

Since:

1. For all $$\tr(f)\in\Tr(\Delta)$$, there exists some $$n\in\mathbb{N}_{\geq1}$$ such that $$\tr(f)=\tr\big(\id_{[n]}\big)$$ (Claim 1).
2. If $$n\neq m$$, then $$\tr\big(\id_{[n]}\big)\neq\tr\big(\id_{[m]}\big)$$ (Claim 2).

it follows that $$\widetilde{\phi}$$ is a bijection:

1. $$\widetilde{\phi}$$ Is Injective: Given $$\tr(f)$$ and $$\tr(g)$$ with $$\widetilde{\phi}(\tr(f))=\widetilde{\phi}(\tr(g))$$, there exist $$n,m\in\mathbb{N}_{\geq1}$$ such that $$\tr(f)=\tr\big(\id_{[n]}\big)$$ and $$\tr(g)=\tr\big(\id_{[m]}\big)$$ by Claim 1, but since $$\widetilde{\phi}(\tr(f))=\widetilde{\phi}(\tr(g))$$, we must have $$n=m$$ by Claim 2, so $$\tr(f)=\tr(g)$$.
2. $$\widetilde{\phi}$$ Is Surjective: Given $$n\in\N_{\geq1}$$, we have $$\widetilde{\phi}\big(\tr\big(\id_{[n-1]}\big)\big)=n$$.

So $$\Tr(\Delta)\cong\mathbb{N}_{\geq1}$$. Lastly, to determine the trace map

$$\tr\colon\End(\Delta)\to\Tr(\Delta),$$

note that we have $$\phi=\tr\circ\widetilde{\phi}$$ by the universal property of the quotient, and since $$\widetilde{\phi}$$ is a bijection, we have $$\tr=\widetilde{\phi}{}^{-1}\circ\phi$$, which is given by $$\sigma\mapsto\tr\big(\id_{[\#\Fix(\sigma)-1]}\big)$$.