# How does it End?

A recent project has forced my colleague and me to take a rather abstract approach to dynamical systems, and the following definition arose naturally in that context.

Let $\mathcal{C}$ be a category. Its endomorphism category $\text{End}(\mathcal{C})$ is defined as follows. The objects are the endomorphisms $f:x \to x$ of $\mathcal{C}$ and a morphism from $f:x \to x$ to $g: y \to y$ is a morphism $h:x \to y$ in $\mathcal{C}$ so that the commuting relation $h \circ f \equiv g \circ h$ holds in $\mathcal{C}$.

That's all we really need for the project, but now a perverse question springs to mind. Note that $\text{End}$ itself defines a dynamical system. After all, we have $\text{End}(\text{End}(\mathcal{C}))$ and so on by iterating the process. I'm interested in the asymptotic behavior. So, here are some questions:

Which categories are fixed points of $\text{End}$, i.e., for which $\mathcal{C}$ is $\text{End}(\mathcal{C})$ equivalent to $\mathcal{C}$?

Slightly more interesting is the question of when $\text{End}(\mathcal{C})$ is homotopy equivalent to $\mathcal{C}$. And while I suspect that the answer to the following question is no, I was unable to prove it:

Is it possible to have non-trivial periodic orbits, for instance, non-equivalent categories $\mathcal{C}$ and $\mathcal{D}$ so that $\text{End}(\mathcal{C})$ is equivalent to $\mathcal{D}$ and vice-versa?

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Clearly any category whose only endomorphisms are identities is fixed. Do you have other examples? –  Benjamin Steinberg Jul 9 '13 at 16:17
I'm curious how this question arose. It doesn't seem like a natural question since you haven't chosen a natural functor either way. –  Qiaochu Yuan Jul 9 '13 at 17:24
@ViditNanda, I don't understand your example. Let $G$ be a group, viewed as a 1-object category. Then $End(G)$ has objects $G$ and morphisms $h\colon a\to b$ if $h$ conjugates $a$ to $b$. So the isomorphism classes of objects of $End(G)$ are in bijection with conjugacy classes. Hence $End(G)$ is never equivalent to $G$ for a non-trivial group (even an abelian one). If $G$ is abelian, then $End(G)$ seems to me to be the coproduct (in Cat) of $|G|$ copies of G. –  Benjamin Steinberg Jul 9 '13 at 18:56
Note that the full subcategory of End(C) on the identity arrows is isomorphic to C in general so there is a natural arrow from C to End(C). –  Benjamin Steinberg Jul 9 '13 at 19:33
Are there any finite categories which are fixed points and have non-identity endomorphisms? I guess not. –  Benjamin Steinberg Jul 10 '13 at 13:13


Iterating, let $\BN^{\oplus\infty}$ denote the category with one object, whose morphisms are the abelian monoid (under addition) of finite (but arbitrarily long) sequences of nonnegative integers. There is an isomorphism $\BN \times \BN^{\oplus\infty}$, giving an isomorphism of categories $\End(\hom(\BN^{\oplus\infty},C)) = \hom(\BN^{\oplus\infty},C)$. So this gives a class of examples.

There are other examples. For instance, I could have used $\BN^{\times\infty}$, the countable direct product of $\BN$ with itself, instead of $\BN^{\oplus\infty}$. Then I'd still have $\BN \times \BN^{\times \infty} = \BN^{\times \infty}$, and so $\End(\hom(\BN^{\times\infty},C)) = \hom(\BN^{\times\infty},C)$ for any category $C$.

I had hoped to prove that the first class of fixed points was universal. Here is the argument that I had hoped to use: Suppose we are given an equivalence of categories $F: \hom(\BN,D) \to D$. Then by currying we do get a sequence of equivalences $F_k: \hom(\BN^{k+1},D) \to \hom(\BN^k,D)$ for any finite $k$. We can try to compute the limit of this system in two different ways. On the one hand, it is a system of equivalences, and so its limit is equivalent to every other term in the system. On the other hand, $\BN^{\oplus\infty}$ is a colimit, and colimits in the first variable of $\hom(,)$ become limits, and so it is very tempting to think that $\underset{\leftarrow k}{\lim\limits} \hom(\BN^k,D) = \hom(\BN^{\oplus \infty},D)$. In this way, you could hope to conclude that $D \simeq \hom(\BN^{\oplus\infty},D)$.

Unfortunately, I think that this argument has no chance of working. (The following is based on a conversation with William Johnson.) For, suppose that it did. Then there would be, for any category $C$, a natural equivalence of categories $\hom(\BN^{\oplus \infty}\times \BN^{\times \infty},C) \simeq \hom(\BN^{\times \infty},C)$. By the Yoneda lemma, this would imply that $\BN^{\oplus\infty} \times \BN^{\times \infty} \simeq \BN^{\times \infty}$, or, equivalently, $\newcommand\NN{\mathbb N}\NN^{\oplus\infty} \times \NN^{\times \infty} \cong \NN^{\times \infty}$ as commutative monoids.

I claim that this is impossible. It is easier to work with groups, so let's allow subtractions. Then I claim that $\newcommand\ZZ{\mathbb Z} \ZZ^{\oplus \infty} \times \ZZ^{\times \infty} \not\cong \ZZ^{\times \infty}$. The proof is as follows:

Suppose that you have any linear map $\ZZ^{\oplus \infty} \to \ZZ^{\otimes \infty}$. You can write this as an $\infty\times \infty$ matrix, where the first column is where $e_1 = (1,0,\dots)$ goes, the second column is where $e_2 = (0,1,0,\dots)$ goes, and so on. (The matrix may have non-zero entries everywhere, if you want.) I will try to implement the usual Gaussian elimination algorithm to diagonalize the matrix. Suppose temporarily that the upper left corner of the matrix is $1$. Then by modifying $e_i \mapsto e_i' = e_i - \# e_1$, where $\#$ is some matrix entry, I can clear the rest of the first row. This manipulation is simply the precomposition of the matrix with some automorphism of $\ZZ^{\oplus\infty}$.

Similarly, arbitrary permutations of the columns are also implemented by automorphisms of $\ZZ^{\oplus\infty}$. What happens if the first row never has a $1$ in it? Well, $\ZZ$ is a PID; let $g$ denote the principal generator of the ideal ideal generated by all the entries in the first row, i.e. the GCD of the first row. But $g$ is by definition some finite linear combination of the entries in the first row, so again basic column operations allow us to get $g$ into the upper left corner. Finally, every entry in the first row is divisible by $g$, and so we can use elementary column operations to clear the rest of the first row.

Now look at the second row. Don't worry about the first column, but clear the rest. Etc. At the end of the day, you get some matrix $X : \ZZ^{\oplus\infty} \to \ZZ^{\times \infty}$ in column-echelon form.

Let $f_n = (1,2,6,\dots,n!,0,0,\dots) \in \ZZ^{\oplus \infty}$. This gives a sequence of terms $Xf_1,Xf_2,\dots \in \ZZ^{\times \infty}$. Because $X$ is in column-echelon form, it is clear that this sequence has a limit $Xf_\infty \in \ZZ^{\times \infty}$ (indeed, the first $n$ terms of $Xf_\infty$ agree with the first $n$ terms of $Xf_n$).

Now look at the image $[Xf_\infty]$ of $Xf_\infty$ in the quotient $\ZZ^{\times \infty} / (X \ZZ^{\oplus\infty})$. I claim first that $[Xf_\infty] \neq 0$. Indeed, suppose that $Xf_\infty = Xh \in \ZZ^{\oplus \infty}$ for some finite sequence $h\in \ZZ^{\oplus \infty}$. Say that $h$ is all zeros after the first $m$ terms. Recall that a pivot in $X$ is the first non-zero entry in any column. Let $n$ by the row for the $m$th pivot. Since $X$ is an injection, we can reconstruct $h$ from the first $n$ entries of $Xh$. But these first $n$ entries are precisely the first $n$ entries of $f_n$, and so $h$ and $f_n$ agree to their first $m$ spots, i.e. $h = f_m$. But again since $X$ is an injection, $X f_m \neq X f_{m+1}$, differing in the $(n+1)$th spot. So this proves that $Xf_\infty \not\in \ZZ^{\oplus\infty}$.

On the other hand, I claim that $[Xf_\infty]$ is infinitely divisible in the abelian group $\ZZ^{\times \infty} / (X \ZZ^{\oplus\infty})$. Indeed, it is divisible by $n$, because $f_{N} - f_n$ is for $N > n$, and $[Xf_\infty] = [X(f_\infty - f_n)]$. We have therefore constructed a non-zero infinitely-divisible element of $\ZZ^{\times \infty} / (X \ZZ^{\oplus\infty})$.

On the other hand, $(\ZZ^{\oplus \infty} \times \ZZ^{\times \infty})/ \ZZ^{\oplus\infty} = \ZZ^{\times \infty}$ does not contain any non-zero infinitely-divisible elements. Therefore $\ZZ^{\oplus \infty} \times \ZZ^{\times \infty} \not\cong \ZZ^{\times \infty}$, completing the proof.

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I think it is easy to show that these two groups have different cardinality Hom sets into Z. –  Benjamin Steinberg Jul 9 '13 at 21:18
@BenjaminSteinberg: Great, since I'm worried my argument has a flaw anyway. –  Theo Johnson-Freyd Jul 10 '13 at 2:43
The $F_k$ are induced from $F$, not from any functor $B \mathbb{N}^k \to B \mathbb{N}^{k+1}$, so there is no diagram of the $B\mathbb{N}^k$ to use for the colimit in the argument $\lim \hom(B\mathbb{N}^k,D) = \hom(\mathop{\mathrm{colim}} B\mathbb{N}^k, F)$. –  Omar Antolín-Camarena Jul 10 '13 at 22:34
@OmarAntolín-Camarena: I agree, and in any case the argument is bound to fail somewhere. The functor I have in mind, though, is $F_k = hom(\BN,F_{k-1})$, and so all of the data of $F$ is concentrated in one piece of it. So it's not a priori impossible that the limit would be something explicit. Of course, it isn't what I wanted it to be, but I do think that failure is a posteriori, and not obvious a priori. –  Theo Johnson-Freyd Jul 11 '13 at 3:18

Here are some easy examples within groupoids.$\DeclareMathOperator\End{End}$

For a group $G$, $\End(BG)$ is equivalent to $\coprod_{x\in I} BC_x$ (which is skeletal) where $I$ is a set containing one representative of each conjugacy class of $G$ and $C_x$ denotes the centralizer of $x$ in $G$ (in fact, $\End(BG)$ is isomorphic to $\coprod_{x \in I} BC_x \times E_{[G:C_x]}$ where $E_n$ is the groupoid on $n$ objects and a unique isomorphisms between every pair of them). From this we see, for example, that:

1. The only finite groupoids that are fixed points for $\End$, or even belong to an $\End$-cycle, are the finite discrete categories: when you apply $\End$ to a groupoid $C$ you always get a copy of $C$ in the result and, unless, $C$ is discrete, you get a groupoid with a strictly larger skeleton.
2. Among groupoids there are no cycles that are not simply fixed points (again, because $BC$ always contains a copy of $C$, so the multiplicity of any given group in $\End^n(C)$ is a monotone function of $n$).
3. Given any $\mathcal{G}$ is any class of groups such that whenever $x\in G \in \mathcal{G}$ then also $C_x \in \mathcal{G}$, we can form the groupoid $C := \coprod_{G \in \mathcal{G}} \coprod_{\mathbb{N}} BG$ and this will be an $\End$-fixed point. In particular, for any Abelian group $G$, $\coprod_{\mathbb{N}}BG$ is a fixed point. (And notice too, that the equivalence $C \simeq \End(C)$ is not canonical in these examples.)
4. As a simple, explicit example, $\End(BS_3) = BS_3 \sqcup B\mathbb{Z}/2 \sqcup B\mathbb{Z}/3$, and $\End(BG) = |G|\cdot BG$ for any Abelian $G$ (as was already mentioned by Benjamin Steinberg --and where $n \cdot C$ is the coproduct of $n$ copies of the category $C$), so we get that $\End^n(BS_3) = BS_3 \sqcup (2^n-1)\cdot B\mathbb{Z}/2 \sqcup \frac{1}{2}(3^n-1) \cdot B\mathbb{Z}/3$. (And similarly, for any fixed group, working out the formula for the $n$-th iterate $\End^n(BG)$ is just a a matter of patience.)

ADDED: It might be worth mentioning that when you stick to just groupoids, the question is very homotopy theoretical: equivalence is the same as homotopy equivalence and $\End$ is just the free loop space functor. In other words, the question for groupoids becomes "study the dynamical system given by the free loop space functor on the homotopy category of 1-types". A reasonable analogue of this question for $\infty$-groupoids would be to study the free loop space functor on the homotopy category (of all spaces) as a dynamical system. An analogue of the above description of $\End(BG)$ works for topological groups too: the free loop space $\mathop{Map}(S^1,BG)$ is the homotopy quotient $G/G$ where $G$ acts on itself by conjugation — this is analogous because the description given above of (the isomorphism type of) $\End(BG)$ for discrete $G$ is just the action groupoid of $G$ acting on itself by conjugation.

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Thank you, these are extremely interesting observations! –  Vidit Nanda Jul 10 '13 at 18:34