A couple years ago I wrote a blog post looking at how close the quaternions come to commuting. That is, the post looked at the average norm of *xy* – *yx*.

A related question would be to ask how often quaternions do commute, i.e. the probability that *xy* – *yx* = 0 for randomly chosen *x* and *y*.

There’s a general theorem for this [1]. For a discrete non-abelian group, the probability that two elements commute, chosen uniformly at random, is never more than 5/8 for any group.

To put it another way, in a finite group either all pairs of elements commute with each other or no more than 5/8 of all pairs commute, with no possibilities in between. You can’t have a group, for example, in which exactly 3 out of 4 pairs commute.

What if we have an infinite group like the quaternions?

Before we can answer that, we’ve got to say how we’d compute probabilities. With a finite group, the natural thing to do is make every point have equal probability. For a (locally compact) infinite group the natural choice is Haar measure.

Subject to some technical conditions, Haar measure is the only measure that interacts as expected with the group structure. It’s unique up to a constant multiple, and so it’s unique when we specify that the measure of the whole group has to be 1.

For compact non-abelian groups with Haar measure, we again get the result that no more than 5/8 of pairs commute.

[1] W. H. Gustafson. What is the Probability that Two Group Elements Commute? The American Mathematical Monthly, Nov., 1973, Vol. 80, No. 9, pp. 1031-1034.

Two non-real quaternions commute precisely when they have (modulo +-1) the same “imaginary” part, i.e. what you get when you subtract the real part and normalize. You can also express that using R[q], which is a field isomorphic to C that each non-real quaternion is contained in. q_1 and q_2 commute precisely when R[q_1] = R[q_2].

So for each unital, non-real quaternion q, the set of unital quaternions it commutes with is just the unit circle in R[q].

Assuming we’re looking at S^3 x S^3 to consider pairs of unit quaternions, the total set of commuters looks like the union of {q} x S^1 union S^1 x {q}, as q ranges over all S^3, along with {+-1} x S^3 and S^3 x {+-1}. Which looks to be very sparse to me, so probably of measure zero.

Well, 0 is certainly less than 5/8. :)

The bound is tight, however. There are finite groups where exactly 5/8 of pairs commute, and a finite group is a compact group. I don’t know how tight the bound is, say, for Lie groups.

> Well, 0 is certainly less than 5/8. :)

I genuinely laughed. :)

It is a fascinating result, though. 5/8 is not a number that pops up a lot in fundamental theorems. I wonder if it’s possible to take a finite example that achieves the bound and “stretch a skin over the frame” to get a Lie group?

This is breaking my brain. What intuition can anyone supply?

John Baez has an article on this. I haven’t read it yet, but it’s John Baez so it’s probably good.

https://johncarlosbaez.wordpress.com/2018/09/16/the-5-8-theorem/

This makes me wonder what other values are possible for this probability. Is the set of possible probabilities dense in [0, 5/8], or are there other “forbidden” intervals? For that matter, is it dense in any interval?