# 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.

*edited below to reflect Igor Rivin's comment*

It is not true in general that for a compact connected manifold, if the pushforward fails to be of maximal rank, this must occur on the boundary of the continuous image of $f$. **Then 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.

isknown that the matrices with two distinct eigenvalues, each of multiplicity greater than 1, have their trace in the interior of the image, not on the boundary. This is not hard to show directly; if you want, I can write it out as an answer. $\endgroup$ – Robert Bryant May 2 '17 at 18:52