There are some more interesting things to do, however. For example, there’s a unique probability measure on SU(n) that’s translation-invariant. This pushes forward via the trace map to a probability measure on the interior of the n-cusped hypocycloid. This gives the ‘probability distribution on traces of matrices in SU(n)’.

The real part or imaginary part of the trace then becomes a real-valued random variable, and we can compute its moments: its mean, its variance, and so on. There will be a nice combinatorial way to do this.

The case of SU(3) was discussed on Mathoverflow by Qiaochu Yuan. In this case the moments of twice the real part of the trace form this sequence:

1, 0, 2, 2, 12, 30, 130, 462, 1946, 7980, 34776, 153120, 694056, 3194334, 14971242, …

which is sequence A151366 in the Online Encyclopedia of Integer Sequences. The nth term here is the number of walks with n steps start and end at the origin and stay within where at each step you either go one step north, south, east, west, northwest or southeast. In other words, you can move any of these ways:

(-1, 0), (-1, 1), (0, -1), (0, 1), (1, -1), (1, 0)

Qiaochu also works out the moments of the imaginary part of the trace.

So, this just goes to show that there’s a lot of math here…

So he’s exploiting a lot of nice symmetries of the problem, but I wish he’d given a more careful account of the things he’s relying on to obtain the result.

and then says that we can find the boundary by eliminating all but one variable with the “zero-derivate conditions”: that the partial derivatives of the trace with respect to are all zero.

It follows that the first of the are equal (say to ), and for the remaining two terms we have:

which lets us write the trace as:

For a phase this is just the familiar formula for the hypocycloid with n cusps.

It’s worth stressing that in all the discussions we’ve had about the boundary, we’ve set all the independent phases to be equal there, while the final phase — the product of their reciprocals — is different, whereas Kaiser ends up with the same overall collection of phases but the last two of them are swapped compared to our scheme.

But why is Kaiser allowed to choose the partial derivates of the trace with respect to to be zero on the boundary? He uses this condition without explaining it, but I guess the idea is that we have a compact manifold without boundary of real dimension being projected onto the complex plane, and where the manifold projects to the boundary of its shadow the linearised map has to change from having an -dimensional kernel to an -dimensional kernel, so you can choose coordinates there such that of the coordinate vectors lie in the kernel.

“What about E₇ and E₈?”

and

“Why did Kaiser draw pictures for Spin(4n+2) but not other spin groups?”

The answer can be found here:

• John Baez, Division algebras and quantum theory.

The author here recalls that every irreducible representation of a group is either ‘real’, ‘complex’ or ‘quaternionic’ in a certain sense: this is Dyson’s ‘three-fold way’. And he writes:

SU(2) is not the only compact Lie group with the property that all its irreducible continuous unitary representations on complex Hilbert spaces are real or quaternionic. For a group to have this property, it is necessary and sufficient that every element be conjugate to its inverse. All compact simple Lie groups have this property except those of type A_n for n > 1, D_n with n odd, and E₆ (see Bourbaki [20]).

If a representation of a Lie group is ‘real’, the trace of any group element in that representation is a real number, so we don’t get a pretty picture of the set of allowed traces: it’s just some interval in the real line.

If it’s ‘quaternionic’ I bet the trace of any group element is again real. I don’t see why right off the bat, but I know an example. The spin-1/2 representation of SU(2) is ‘quaternionic’, and the trace of an SU(2) element in this representation is a real number lying on the interval [-2,2]. Indeed, this interval is the line segment traced out (pardon the pun) by the Tusi couple!

So, it’s only the truly ‘complex’ representations where the traces give a pretty picture, and—translating notations—the only compact simple Lie groups with representations of this sort are SU(n) for n > 2, Spin(4n+2) for n > 1, E₆.

That’s why we see SU(6), Spin(14) and E₆ in the pictures above, but not Spin(12) or G₂ or F₄ or E₇ or E₈.

Fortunately (from the viewpoint of someone just wanting the truth to be known), he seems to have gone further and done traces of all the compact simply-connected Lie groups in certain famous irreps (though not all irreps). It’s neat how the traces of Spin(4n+2) elements in the spinor rep lie in a curve a bit like an astroid, but which gets skinnier and skinnier as n increases:

It’s also neat how E₆ in one of its two complex 27-dimensional reps gives traces lying in a curve a bit like a deltoid, but skinnier:

• N. Kaiser, Mean eigenvalues for simple, simply connected, compact Lie groups, 28 September 2006.

Abstract: We determine for each of the simple, simply connected, compact and complex Lie groups SU(n), Spin(4n+2) and E₆ that particular region inside the unit disk in the complex plane which is filled by their mean eigenvalues. We give analytical parameterizations for the boundary curves of these so-called trace figures. The area enclosed by a trace figure turns out to be a rational multiple of $\pi$ in each case. We calculate also the length of the boundary curve and determine the radius of the largest circle that is contained in a trace figure. The discrete center of the corresponding compact complex Lie group shows up prominently in the form of cusp points of the trace figure placed symmetrically on the unit circle. For the exceptional Lie groups G₂, F₄ and E₈ with trivial center we determine the (negative) lower bound on their mean eigenvalues lying within the real interval [-1,1]. We find the rational boundary values -2/7, -3/13 and -1/31 for G₂, F₄ and E₈ respectively.

This famous result is usually stated in this form: “there’s no retraction of the disk onto its boundary”. And the usual proof uses a bit of algebraic topology, though Milnor found one using calculus.

This famous result is often used as a lemma to prove Brower’s fixed point theorem.