A Peter-Weyl “counter-example”

Let K be a compact Lie group. The Peter-Weyl theorem gives a basis for functions on K. In particular, it tells us that the characters are an orthonormal basis for class functions on K.

Let’s look at K=SU(2). Topologically, K is a three sphere, and the conjugacy classes are latitudinal two spheres. We’ll label the conjugacy classes by the line segment [0, \pi], where \theta labels the conjugacy class of matrices with eigenvalues e^{i \theta} and e^{- i \theta}. The conjugacy class \theta is a sphere of radius proportional to \sin \theta, and hence area proportional to \sin^2 \theta.

The characters of SU(2) are indexed by positive integers, with

\displaystyle{\chi_n(\theta) = e^{- (n-1) i \theta} + e^{- (n-3) i \theta} + \cdots + e^{(n-1) i \theta} = \frac{\sin (n \theta)}{\sin \theta} }.

So, if F is a class function on K, then its inner products with the characters are given by the integrals

\displaystyle{ c_n = \frac{2}{\pi} \int_0^{\pi} F(\theta) \chi_n(\theta) \sin^2(\theta) d \theta }.

Here \sin^2 \theta is the area of the conjugacy class \theta and 2/\pi turns out to be the correct normalization factor.

So, we should expect that

\displaystyle{ F(\theta) = \sum c_n \chi_n(\theta) }.

All of this is pretty standard. So, what would you expect happens if you take F to be 1 on [0, \pi/2] and -1 on [\pi/2, \pi]? Seriously, see if you can guess what peculiar behavior these sums show.

Of course, the integrals are easy to compute. For n odd, c_n vanishes. For n even, we get

\displaystyle{ c_{2m} = (-1)^{m-1} \frac{8 m}{(4 m^2-1) \pi} }.

Here’s the sum of the first 20 terms, plotted together with F:

Looks pretty good. We’re seeing some Gibbs phenomenon from the singularity at \pi/2, that’s to be at expected.

But what’s going on near 0? That’s nowhere near the singularity: F is constant near 0. But our function is way down near 0.361943; nothing like 1. Things get even stranger if we add on one more term:

Here I’ve zoomed into the left hand part, with \theta \in [0, 0.25]. I show the old curve, and the result of adding on one more term. The value at 0 jumps up to 1.63782. Here is the next 10 sums:

Gradually, it looks like the points near 0, but not at 0, are slowly converging to the correct value of 1. But the values are 0 are not converging; they’re oscillating between two values which look like 1-2/\pi and 1+2/\pi. Somehow, that discontinuity way out at \theta = \pi/2 is producing failure of convergence way back at \theta=0.

So, a cute question: Why isn’t this a counter-example to the Peter-Weyl theorem?

And, a more challenging question: How nice does F have to be in order not to get this sort of behavior?

3 thoughts on “A Peter-Weyl “counter-example”

  1. Peter-Weyl says nothing about pointwise convergence for (not even) continuous functions, so it’s not (abstractly) surprising.

    The divergent series I described in this very old post is, I think, equivalent to yours at 0 — up to normalization –. I used a Chebychev-polynomial version of the characters instead, which is why the coefficients look different. The character value for chi_m at 0 is m so that the series you write at x=0 is then very clearly divergent. (It converges in Cesaro mean).

  2. In the first paragraph you meant to say “class functions”! This threw me off at first, because otherwise it’s the (normalised) matrix coefficients that give an o/n basis.

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