#The Question#

Consider the trace of an $n \times n$ unitary matrix with determinant 1

\begin{align} f: SU(n) &\rightarrow \mathbb{C}\\ U \mapsto \text{tr}\, U &= \sum\limits_{i=1}^{n-1} z_i + \frac{1}{z_1 \cdots z_{n-1}} \end{align}

where the $z_i$ are the eigenvalues of $U$ and we have used $\det U =1$ to write $z_n$ in terms of the other eigenvalues, without loss of generality.

In section 3 of the paper "Mean eigenvalues for simple, simply connected, compact Lie groups," the author argues that the image of $f$ is the $n$-cusp hypocycloid.

A critical step in the argument relies on the statement that on the boundary of the image, we can set $n-2$ of the partial derivatives of $f$ equal to zero, that is

\begin{align} \frac{\partial f}{\partial z_1} = \cdots = \frac{\partial f}{\partial z_{n-2}} = 0 \end{align}

Why is it true that imposing this condition gives the boundary of the image of $f$? I'm currently trying to use this argument for a generalization of $f$ (determining the image of sums and products of traces of $SU(n)$ matrices by first finding the boundary of the image).

Attempt at a solution 1#

Confusion over this argument in the paper was mentioned in the comment section of this blog post. Greg Egan writes:

"I guess the idea is that we have a compact manifold without boundary of real dimension $n-1$ 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 $(n-3)$-dimensional kernel to an $(n-2)$-dimensional kernel, so you can choose coordinates there such that $n-2$ of the coordinate vectors lie in the kernel."

"Generically there will be some choice of coordinates where the derivatives on the boundary vanish for all but one coordinate, but for a more general function than the trace that coordinate system need not line up with the phases.

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."

Is what Greg writes true? I wasn't able to make it rigorous myself, thinking that the tangent space on the boundary of $f(SU(n)) \subset \mathbb{C}$ is still $2$ dimensional. Maybe someone can recommend some resources on the topic of the tangent space at the boundary of the continuous image of a compact connected manifold.

Attempt at solution 2#

Let $n = 3$ for simplicity. If we instead think of $f$ in this case as

\begin{align} \widetilde{f}: U(1) \times U(1) &\rightarrow \mathbb{C}\\ (\theta_1, \theta_2) &\rightarrow e^{i \theta_1} + e^{i \theta_2} + e^{-i( \theta_1 + \theta_2)} \end{align}

then with respect to charts $(V_1, \theta_1, \theta_2)$ at some $p \in U(1) \times U(1)$ and the obvious charts (projecting real and imaginary parts) on $\mathbb{C}$, the pushforward/differential/Jacobian is given by

\begin{align} J(p) = \left( \begin{array}{cc} -\sin (\text{$\theta $1})-\sin (\text{$\theta $1}+\text{$\theta $2}) & -\sin (\text{$\theta $2})-\sin (\text{$\theta $1}+\text{$\theta $2}) \\ \cos (\text{$\theta $1})-\cos (\text{$\theta $1}+\text{$\theta $2}) & \cos (\text{$\theta $2})-\cos (\text{$\theta $1}+\text{$\theta $2}) \\ \end{array} \right) \end{align}

Then we can see that the pushforward/differential/Jacobian is not of maximal rank at $p$ if $\theta_1 = \theta_2$, which maps out the hypocycloid.

Is there any reason why points where the pushforward fails to be of maximal rank must occur on the boundary of the continuous image of a compact connected manifold? If not, what additional assumption is needed? I know that in this situation the regular values of $f$ must lie in the interior of the image of $f$, but I have not been able to prove that critical values cannot lie in the interior.