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.

  • 1
    $\begingroup$ There is no reason that critical points should only occur in the interior. Just think of the height function on the standard torus - the max and the min are at the boundary of the image interval, the saddles in the interior. $\endgroup$
    – Igor Rivin
    May 2, 2017 at 13:31
  • 1
    $\begingroup$ @Jon Rayner: The critical points are where there are no more than 2 distinct eigenvalues, and, as soon as $n\ge 4$, there will be interior critical values.The boundary is traced out by the image of the elements that have an eigenvalue of multiplicity $n{-}1$, but, for example, when $n=4$, the elements that have two eigenvalues, each of multiplicity 2, go into the interior (except for the $4$ points where those eigenvalues are equal), because these two eigenvalues are conjugate, so the sum is real. Similarly, when $n=5$ with distinct eigenvalues of multiplicites 2 and 3 yield interior images. $\endgroup$ May 2, 2017 at 13:57
  • $\begingroup$ Thanks Igor Rivin, have edited to reflect this @RobertBryant Thanks this is also useful. Any ideas on how one can prove that this is the case, short of calculating all of the critical points and checking which can possibly be boundary candidates? I'm not sure I understood whether you are saying there's a known result about how the multiplicities affect the location of the critical values, or if you are just helping by showing that some critical points lie in the interior in response to what I've written in the attempted solutions. Either way, thanks. $\endgroup$ May 2, 2017 at 15:01
  • $\begingroup$ @JonRayner: Actually, it is known 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. $\endgroup$ May 2, 2017 at 18:52
  • 1
    $\begingroup$ @JonRayner: I looked at Greg's blog, but he doesn't give a proof, just the obvious example when $n=5$ that I alluded to in my answer above. When I have time, maybe this afternoon, I can indicate the proof in an answer. $\endgroup$ May 3, 2017 at 15:24

1 Answer 1


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 $0\le k < n$.

  • $\begingroup$ Thanks, have skim-read, will be able to look this over careful today or tomorrow. $\endgroup$ May 6, 2017 at 3:47

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