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,"](https://arxiv.org/abs/math-ph/0609082) the author argues that the image of $f$ is the $n$-cusp [hypocycloid](https://en.wikipedia.org/wiki/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](https://johncarlosbaez.wordpress.com/2012/09/11/rolling-circles-and-balls-part-3/#comment-90361). 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](https://books.google.co.za/books?id=eqfgZtjQceYC&lpg=PP1&pg=PA73#v=onepage&q&f=false) 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](https://math.stackexchange.com/questions/1324151/critical-points-of-torus-height-function) 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](https://math.stackexchange.com/questions/732126/regular-values-and-manifolds-with-boundary?rq=1) of $f$, but I have not been able to prove that critical values cannot lie in the interior.



Boundary of the image of a compact manifold in the complex plane
Here is one way to prove the claims about the image of the map $\mathrm{tr}:\mathrm{SU}(n)\to\mathbb{C}$. I'll just outline the steps. For simpicity, I'll always assume $n\ge 3$. (For completeness, observe that $\mathrm{tr}\bigl(\mathrm{SU}(2)\bigr)$ consists of the interval $[-2,2]\subset\mathbb{R}\subset\mathbb{C}$.)


First, some notation: Let $\mathbb{T}\subset\mathrm{SU}(n)$ denote the subgroup consisting of of diagonal $n$-by-$n$ special unitary matrices and let $D(z\_1,\ldots,z\_n)\in\mathbb{T}$ denote the diagonal matrix whose $(j,j)$-entry is $z\_j\in S^1\subset\mathbb{C}$. Of course, $z\_1\cdots z\_n = 1$. Since every matrix in $\mathrm{SU}(n)$ is conjugate to a diagonal matrix,
$$ X\_n = \mathrm{tr}\bigl(\mathrm{SU}(n)\bigr) = \mathrm{tr}\bigl(\mathbb{T}\bigr),$$
and $X\_n\subset\mathbb{C}$ is compact and connected.


Second, it's easy to show that the critical points of $\mathrm{tr}:\mathbb{T}\to\mathbb{C}$ are the diagonal matrices $D(z\_1,\ldots,z\_n)$ for which at most two of the $z\_i$ are distinct. In particular, the interior of $X\_n$ is nonempty and connected (obvious when $n>3$ and true for $n=3$ by inspection) and its 'boundary' (i.e., $X\_n$ minus its interior) is a finite union of analytic arcs consisting entirely of critical values of $\mathrm{tr}$ on $\mathbb{T}$. Since the noncritical points of $\mathrm{tr}$ are open and dense in $\mathbb{T}$, $X\_n$ is the closure of its interior.


I am going to show that a critical point $D(z\_1,\ldots,z\_n)$ is mapped into the interior of $X\_n$ whenever it has two *distinct* eigenvalues, *each* of which has multiplicity at least $2$.


Now, the locus of critical values of $\mathrm{tr}:\mathbb{T}\to\mathbb{C}$, is the union of the $\mathrm{tr}$-images of the curves 
$$
\gamma\_{p,k}(t) = \omega^k\,D(\,\underbrace{e^{i(n-p)t},\ldots,e^{i(n-p)t}}\_{\text{p times}}, \underbrace{e^{-ipt},\ldots,e^{-ipt}}\_{\text{$n-p$ times}}\,)
$$
where $\omega = e^{2\pi i/n}$ is the primitive root of unity, $1\le p<n$, and $0\le k < n$. Note that $\gamma\_{p,k}(t)$ has two distinct eigenvalues as long as $e^{int}\not=1$. 


Suppose now that $p$ and $(n-p)$ are each at least $2$, and consider the two maps
$$
A\_{p,k}(t,s)
=\omega^k\,D(\,\underbrace{e^{i(n-p)t+is},e^{i(n-p)t-is},\ldots,e^{i(n-p)t}}\_{\text{p times}}, \underbrace{e^{-ipt},\ldots,e^{-ipt}}\_{\text{$n-p$ times}}\,)
$$
and
$$
B\_{p,k}(t,s)
=\omega^k\,D(\,\underbrace{e^{i(n-p)t},e^{i(n-p)t},\ldots,e^{i(n-p)t}}\_{\text{p times}}, \underbrace{e^{-ipt+is},e^{-ipt-is},\ldots,e^{-ipt}}\_{\text{$n-p$ times}}\,).
$$
Then
$$
\mathrm{tr}\bigl(A\_{p,k}(t,s)\bigr) = \omega^k\bigl(\,p\,e^{i(n-p)t}+(n{-}p)\,e^{-ipt}-2e^{i(n-p)t}(1-\cos(s))\,\bigr)
$$
while
$$
\mathrm{tr}\bigl(B\_{p,k}(t,s)\bigr) = \omega^k\bigl(\,p\,e^{i(n-p)t}+(n{-}p)\,e^{-ipt}-2e^{-ipt}(1-\cos(s))\,\bigr)
$$
Since $0\le 1-\cos(s) \le 2$ it follows that the points
$$
a\_{p,k}(t,\sigma) = \omega^k\bigl(\,p\,e^{i(n-p)t}+(n{-}p)\,e^{-ipt}
-\sigma e^{i(n-p)t}\,\bigr)
$$
and 
$$
b\_{p,k}(t,\sigma) = \omega^k\bigl(\,p\,e^{i(n-p)t}+(n{-}p)\,e^{-ipt}
-\sigma e^{-ipt}\,\bigr)
$$
all lie in $X\_n$ when $0\le \sigma\le 4$. However, as $t$ varies in the open interval $(2\pi\ell/n,2\pi(\ell{+}1)/n)$ and $\sigma$ varies in the interval $[0,4]$, these two mappings cover different sides of the curve 
$$
g\_{p,k}(t) = \omega^k\bigl(\,p\,e^{i(n-p)t}+(n{-}p)\,e^{-ipt}\,\bigr)
= \mathrm{tr}\bigl(\gamma\_{p,k}(t)\bigr),\quad\text{where} \quad t\in(2\pi\ell/n,2\pi(\ell{+}1)/n).
$$
Hence this curve lies in the interior of $X\_n$, as we wanted to show.


It follows that the only points that could be boundary points of $X\_n$ are the points of the form 
$$
g\_{1,k}(t) = \omega^k\bigl(\,e^{i(n-1)t}+(n{-}1)\,e^{-it}\,\bigr).
$$
But all of these have the same image for all $k$, so the boundary must lie in
$$
g\_{1,0}(t) = e^{i(n-1)t}+(n{-}1)\,e^{-it},
$$
which is the hypocycloid with $n$ vertices, at $n\omega^k$ where $0\le k < n$.


