## December 18, 2019

### Random Permutations (Part 10)

#### Posted by John Baez Groupoids generalize sets in an obvious way: while elements of a set can only be the same or different, objects of a groupoid can be ‘the same’ — that is, isomorphic — in more than one way.

Less obviously, groupoids also let us generalize the concept of cardinality. While the cardinality of a finite set is a natural number, the cardinality of a finite groupoid can be any nonnegative rational number!

This suggests that probabilities in combinatorics could secretly be cardinalities of groupoids. Indeed, Poisson distributions naturally show up this way. For a good self-contained introduction, see:

Now I want to explain how groupoid cardinality can be used to prove some important (and well-known) facts about random permutations.

A while back we saw that many questions about random permutations could be answered using just this one fact: the expected number of cycles of length $k$ in a random permutation of an $n$-element set is $1/k$, at least if $1 \le k \le n$.

Last time we proved a stronger fact. To state this, we treat the number of cycles of length $k$ in a permutation of an $n$-element set as a random variable, say $C_{n,k}$. Then we have this: the first $n$ moments of $C_{n,k}$ match those of a Poisson distribution with mean $1/k$.

We actually did more: we described all the moments! We did this by working with ‘factorial moments’. The $p$th factorial moment of a random variable $x$ is the mean of

$x^{\underline{p}} = x(x-1)(x-2) \, \cdots \, (x-p+1)$

Then we have this: the first $n$ factorial moments of $C_{n,k}$ match those of the Poisson distribution of mean $1/k$, while the rest vanish. Or if you prefer to leave Poisson distributions out of it:

$E(C_{n,k}^{\underline{p}}) = \left\{ \begin{array}{ccl} \displaystyle{\frac{1}{k^p}} & \mathrm{if} & p k \le n \\ \\ 0 & \mathrm{if} & p k \gt n \end{array} \right.$

The proof of this was a fairly easy calculation, which however involved some miraculous cancellations and provided no compelling story about why the answer turns out to be so simple.

I now think I think I’ve found that story. It doesn’t connect to Poisson distributions — not yet, anyway. But it does explain the meaning of the answer $1/k^p$. It’s the cardinality of a groupoid.

### A quick sketch

The idea is quite simple when you know the underlying concepts. We’ll show that $E(C_{n,k}^{\underline{p}})$ is the cardinality of a certain groupoid. When $n \lt k p$ this groupoid has no objects, so its cardinality is zero. Otherwise, this groupoid is equivalent to a product of a groupoid with cardinality $1$ and $p$ copies of the group $\mathbb{Z}/k$, thought of as a one-object groupoid. Thus

$E(C_{n,k}^{\underline{p}}) = \frac{1}{k^p}$

The nice thing is that each copy of $\mathbb{Z}/k$ is naturally associated to a cycle of length $k$ in our random permutation. It’s no coincidence that $\mathbb{Z}/k$ is a ‘cyclic group’.

What is this groupoid, whose cardinality is $E(C_{n,k}^{\underline{p}})$? Its objects are $n$-element sets equipped with a permutation that in turn is equipped with an ordered $p$-tuple of disjoint cycles of length $k$. There’s an obvious concept of isomorphism between two such gadgets, so we get a groupoid of them. And then we just think about this groupoid a bit, and everything follows.

But let me slow down and start from the beginning.

### Review of groupoid cardinality

To compute the cardinality of a groupoid $X$, we go through its isomorphism classes of objects, pick one representative $x$ of each class, count the morphisms from $x$ to itself, take the reciprocal… and add up all these reciprocals. In a formula:

$|X| = \sum_{[x] \in \underline{X}} \frac{1}{|\mathrm{Aut}(x)|}$

Here $\underline{X}$ is the set of isomorphism classes of $X$, and $\mathrm{Aut}(x)$ is the automorphism group of $x$, an arbitrary representative of some isomorphism class.

To make sure the sum converges, we can play it safe and assume our groupoid is finite. We don’t need to do this: for example, for the groupoid of finite sets the sum converges to $e$. But today I’ll only work with groupoids that are either finite or equivalent to one that’s finite. (Equivalent groupoids clearly have the same cardinality.)

One way to get a groupoid is by taking a set $X$ with a group $G$ acting on it. Then we get a groupoid $X/\!/G$ where an object is an element $x \in X$ and a morphism from $x$ to $x'$ is a group element $g \in G$ with $g x = x'$. We compose these morphisms in the obvious way: by multiplication in $G$.

We call $X/\!/G$ the weak quotient of $X$ by $G$. In the ordinary quotient $X/G$, we declare $x$ and $x'$ equal when there’s a group element $g$ with $g x = x'$. In the weak quotient, we merely put in an isomorphism between $x$ and $x'$. Thus, the weak quotient retains more information.

Here’s one of my favorite equations:

$|X/\!/G| = |X|/|G|$

It says that to understand division, we can use groupoids. If you take a 5-element set and mod out by a 2-element group, you don’t get a set with $5/2$ elements. But if you take the weak quotient, you get a groupoid with cardinality $5/2$.

### Groupoid cardinality and probability theory

If you have a group $G$, how can you find a set that it acts on? One way is to use $G$ itself. $G$ acts on itself in three interesting ways: by left translations, by right translations, and by conjugation. No matter which method you choose, you’ll get

$|G/\!/G| = 1$

If $G$ acts on itself by left or right translation, every object of $G/\!/G$ is isomorphic to every other object in a unique way. Thus, $G/\!/G$ is equivalent to the groupoid with one object and one morphism. Groupoid cardinality is invariant under equivalence. So it’s no surprise that $|G/\!/G| = 1$ in this case.

It’s more interesting when $G$ acts on itself by conjugation. Then the weak quotient $G/\!/G$ has many different isomorphism classes, one for each conjugacy class of $G$. The fact that $|G/\!/G|$ equals $1$ then says this: if we run through the conjugacy classes of $G$, pick a representative of each one, take the reciprocal of the cardinality of its centralizer, and sum up these reciprocals, we get $1$.

But the fun really starts when we consider a subset $X \subseteq G$ that’s invariant under conjugation. The probability that a random element of $G$ lies in $X$ is $|X|/|G|$. But this equals $|X/\!/G|$. So, we can compute probabilities using groupoid cardinality!

Here, of course, we are treating $G$ as a probability measure space where each element has measure $1/|G|$.

More generally, suppose $\pi \colon X \to G$ is a map of $G$-sets, where $G$ acts on itself by conjugation. Then we get a function $f \colon G \to \mathbb{N}$ that counts the points of $X$ that map to each element of $G$:

$f(g) = |\pi^{-1}(g)|$

We can think of $f$ as a random variable, since it’s a function on a probability measure space, namely $G$. The expected value of $f$ is then the groupoid cardinality of $X/\!/G$. Here’s why:

$\begin{array}{ccl} E(f) &=& \displaystyle{ \frac{1}{|G|} \sum_{g \in G} f(g) } \\ \\ &=& \displaystyle{ \frac{1}{|G|} \sum_{g \in G} |\pi^{-1}(g)| } \\ \\ &=& |X|/|G| \\ \\ &=& |X/\!/G| \end{array}$

This may seem like an impractical way to compute expected values. But it really works wonders in some cases, namely when $X/\!/G$ is equivalent to a groupoid whose cardinality we can compute by other means.

And that’s what we’ll see now.

### The calculation

The key fact is this: since the function

$C_{n,k} \colon S_n \to \mathbb{N}$

tells us the number of cycles of length $k$ in a permutation of an $n$-element set,

$C_{n,k}^{\underline{p}} = C_k (C_k-1)(C_k-2) \, \cdots \, (C_k-p+1)$

tells us the number of ordered $p$-tuples of distinct cycles of this sort. And since distinct cycles are the same as disjoint cycles, we can replace the word ‘distinct’ with ‘disjoint’ here — and we’ll do that from now on.

The expected value $E(C_{n,k}^{\underline{p}})$ is thus the number of ordered $p$-tuples of disjoint cycles of length $k$ of a permutation, summed over all permutations, divided by $n!$.

But this is the cardinality of a groupoid!

Let $X$ be the set whose elements are of this form: a permutation $\sigma \in S_n$ together with an ordered $p$-tuple $(c_1, \dots, c_p)$ of disjoint cycles of length $k$. The group $S_n$ acts on $X$ in obvious way: it acts by conjugation on itself, and the cycles go along for the ride. There’s a map

$\pi \colon X \to S_n$

that forgets the cycles and only keeps the permutation $\sigma$. This is clearly a map of $S_n$-sets. So, we can use the results described in the last section!

We get a function $f \colon S_n \to \mathbb{N}$ that counts the points of $X$ that map to each element of $S_n$:

$f(\sigma) = |\pi^{-1}(\sigma)|$

And this function is just what I’m calling $C_{n,k}^{\underline{p}}$. For each permutation $\sigma$, it gives the number of ordered $p$-tuples of distinct cycles of length $k$.

It follows that the expected value we’re trying to compute is a groupoid cardinality:

$E(C_{n,k}^{\underline{p}}) = |X/\!/S_n|$

But this groupoid $X/\!/S_n$ is equivalent to one I find more charismatic: the groupoid $\mathrm{Cyc}(n,k,p)$ of $n$-element sets equipped with a permutation that is itself equipped with a chosen $p$-tuple of disjoint cycles of length $k$. The morphisms in this groupoid are the obvious ones. (At this point I’m speeding up, assuming you can think like a category theorist.)

In other words, instead of fixing our ‘favorite’ $n$-element set, which is sort of ridiculous, we use all of them.

Now, it’s easy to see that an $n$-element set equipped with this structure is the same as a $p$-tuple of cyclically ordered $k$-element sets together with the ‘remainder’: a set with $n - p k$ elements equipped with an arbitrary permutation. Thus, we have

$\mathrm{Cyc}(n,k,p) \simeq \mathrm{Cyc}(k)^p \times \mathrm{Perm}(n-k p)$

Here $\mathrm{Cyc}(k)$ is the groupoid of cyclically ordered $k$-element sets, and $\mathrm{Perm}(n - k p)$ is the groupoid of sets with $n - k p$ elements equipped with a permutation.

But $\mathrm{Cyc}(k)$ is equivalent to the groupoid $B(\mathbb{Z}/k)$, which has one object and the cyclic group $\mathbb{Z}/k$ as automorphisms. That’s because all objects of $\mathrm{Cyc}(k)$ are isomorphic, and the symmetry group of a cyclically ordered $k$-element set is $\mathbb{Z}/k$. An easy calculation shows that

$|B(\mathbb{Z}/k)| = \frac{1}{k}$

On the other hand,

$\mathrm{Perm}(n-k p) \simeq S_{n-k p} /\!/S_{n - k p}$

where the symmetric group is acting on itself by conjugation. So, we get

$|\mathrm{Perm}(n-k p)| = 1$

We thus get

$\begin{array}{ccl} E(C_{n,k}^{\underline{p}}) &=& |\mathrm{Cyc}(n,k,p)| \\ \\ &=& |\mathrm{Cyc}(k)^p \times \mathrm{Perm}(n-k p)| \\ \\ &=& |\mathrm{Cyc}(k)|^p \cdot | \mathrm{Perm}(n-k p)| \\ \\ &=& (1/k)^p \cdot 1 \\ \\ &=& \displaystyle{ \frac{1}{k^p} } \end{array}$

Voilà!

### Afterthoughts

Experts in category theory will note that I’m vacillating between using various large groupoids, like $\mathrm{Perm}(n-k p)$, and various finite groupoids that are equivalent to these, like $S_{n-k p} /\!/S_{n - k p}$. In my own mind, I barely notice the distinction — I flit back and forth depending on what’s more convenient at the moment. But when I try to explain what I’m doing, I feel some need to talk about this, and then everything becomes more lengthy. It’s possible I could have given a more efficient treatment, but I wanted to show you a way of thinking.

I also introduced a bit more notation than strictly necessary. When you think of a group $G$ as a one-object groupoid, people often call that groupoid $B G$. The reason is that if we turn this groupoid into a space, it becomes a space known as the classifying space of $G$, also denoted $B G$. But the groupoid $B G$ is also a weak quotient:

$B G \cong 1/\!/G$

where $G$ is acting on the $1$-element set in the only possible way. So, it should be no surprise that

$|B G| = \frac{1}{|G|}$

Also, we have

$\mathrm{Cyc}(k) \cong \mathrm{Cyc}(k,k,1)$

In other words, a $k$-element cyclically ordered set is the same as a $k$-element set equipped with a permutation having one chosen cycle of length $k$.

Also, we have

$\mathrm{Perm}(m) \cong \mathrm{Cyc}(m,k,0)$

In other words, an $m$-element set equipped with a permutation is the same as an $m$-element set equipped with a permutation having no chosen cycles of length $k$. (It can have lots of cycles of length $k$; we’re just not choosing any of them.)

So, I didn’t need to talk about $B(\mathbb{Z}/k)$ or $\mathrm{Cyc}(k)$ or $\mathrm{Perm}(n - k p)$. But I thought that reducing notation to the absolute minimum would make things harder to understand, not easier.

Posted at December 18, 2019 5:43 PM UTC

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