Boundary of the image of a compact manifold in the complex plane

Question

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.

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 $0\le k < n$.