Proof. This is a simple matter of specifying how to fit the various cases into the framework of Theorem 2.1 by setting the parameters $G$, $L$ and $V$.

1. (I) The group $G$. For (a), (b) and (c) it is $G:=GL(n)$, while in the other two cases it is $G:=U(n)$.

2. (II) The subspace $V\leqslant T^{+}(n)$. All of $T^{+}(n)$ in (a), (b) and (c) for arbitrary (as opposed to semisimple) matrices, and $V:=\{0\}$ in all other cases.

3. (III) The subset $L\subseteq {\mathbb C}^n$. It is $L:={\mathbb C}^n$ in (a), $({\mathbb C}^{\times})^n$ in (b),

   \begin{equation*} \left\{(\lambda_1, \ldots, \lambda_n)\in {\mathbb C}^n\ :\ \prod_{j=1}^n \lambda_j=1\right\} \text{in (c)}, \end{equation*}

   $({\mathbb S}^1)^n$ in (d) and again ${\mathbb C}^n$ in (e).

In cases (a), (b), (c) and (e), where $L$ is a connected complex algebraic variety and $\Delta_L$ a closed algebraic subset thereof, the complement $L\!\setminus\! \Delta_L$ is connected [Reference Cooper and Manning9, Proposition 8.3] so there is nothing further to check with regard to the transitivity hypothesis.

In case (d), $L\!\setminus\!\Delta_L$ is precisely the $n$th configuration space [Reference Ghrist12, Definition 1.1] ${\mathcal C}^n({\mathbb S}^1)$ consisting of $n$-tuples of distinct modulus- $1$ complex numbers. As we will see in Lemma 2.3, ${\mathcal C}^n({\mathbb S}^1)$ is disconnected as soon as $n\ge 3$. The same result, however, shows that the isotropy groups of the connected components are the (transitive) conjugates of the subgroup generated by the cycle $(1\ 2\ \cdots\ n)$.

Proof. Consider $n$ distinct points $z_1, \ldots,z_n\in {\mathbb S}^1$. Traversing the circle counterclockwise will enumerate the points as

\begin{equation*} z_{\tau(1)},\ z_{\tau(2)},\ldots, z_{\tau(n)} \end{equation*}

for some permutation $\tau$, uniquely determined up to right multiplication by powers of $\eta$. Associating the class of $\tau$ in $S_n/\langle{\eta}\rangle$ to the component in question provides a map

(2.3)$${\pi _0}({\mathcal{C}^n}({\mathbb{S}^1}))\overrightarrow {\quad {\text{equivariant for the}}\;{S_n} - {\text{actions on the (co)domain }} } \quad {S_n}/\langle \eta \rangle.$$

That map is surjective because the $S_n$-action on its codomain $S_n/\langle{\eta}\rangle$ is transitive, and the conclusion will follow once we observe that (2.3) is also injective.

It will suffice to prove that the preimage of $\langle{\eta}\rangle\in S_n/\langle{\eta}\rangle$ is a singleton (the argument is not materially different for other cosets). In other words, the goal is to show that

(2.4)\begin{equation} \left\{ (z_1, \ldots, z_n)\in {\mathcal C}^n({\mathbb S}^1)\ :\ z_j\ \text{are ordered counterclockwise increasingly with}\ j \right\} \end{equation}

is (path-)connected. To that end, note that a counterclockwise rotation allows us to assume $z_1=1\in {\mathbb S}^1$, and the resulting subset of (2.4) is

\begin{equation*}\left\{\left(\exp(2\pi ix_1),\;\dots,\;\exp(2\pi ix_n)\right)\;:\;0=x_1\leq\;x_2\leq\cdots\leq\;x_n \lt \;1\right\}.\end{equation*}

This is the image through the continuous map \[{\mathbb{R}^n}\underrightarrow {\quad \exp {{(2\pi i \bullet )}^n}\quad }{({\mathbb{S}^1})^n}\] of the (path-connected) interior of an $(n-1)$-simplex.

Proof. The sets $\{x \in X : g(x) = f_1(x)\},\ldots, \{x \in X : g(x) = f_n(x)\}$ are closed and form a partition of $X$. By the connectedness of $X$, exactly one of them is non-empty and hence equal to all of $X$.

Proof of Theorem 2.1

The ‘if’ part of the theorem is covered by Remark 2.4(1).

Denote by ${\mathbb C}_{\leqslant m}[x]$ the space of complex polynomials of degree $\leqslant m$, equipped with the unique Hausdorff linear topology, compatible with that of ${\mathbb C}$. The function

$${\text{A}}{{\text{d}}_G}{T_{L,V}}\underrightarrow {\quad {k_{\phi ( \cdot )\quad }}}{\mathbb{C}_{m}}[x]$$

is continuous, and on an element with spectrum $(\lambda_1, \ldots, \lambda_n)$ takes one of the finitely many values $\prod_{j=1}^n (x-\lambda_j)^{k_j}$ for non-negative integers $k_1, \ldots, k_n$ with sum $m$. It is enough to prove the claimed equality (2.2) for matrices of the form

(2.5)\begin{equation} T=g \left(diag(\lambda_1, \ldots, \lambda_n)+v\right)g^{-1} ,\quad g\in G,\ (\lambda_1, \ldots, \lambda_n) \in L':=L\!\setminus\!\Delta_L,\ v\in V, \end{equation}

as such matrices (whose eigenvalues are all distinct) form a dense subset of $\text{Ad}_{G}T_{L,V}$. By Lemma 2.7 for $T$ as in (2.5) the characteristic polynomial $k_{\phi(T)}$ depends only on $(\lambda_1, \ldots, \lambda_n)\in L'$: $g$ and $v$ range over the (path-)connected spaces $G$ and $V$, respectively. Furthermore, by the selfsame Lemma 2.7, for every connected component $L'_0$ of $L'$ we have

(2.6)\begin{equation} k_{\phi(diag(\lambda_1, \ldots, \lambda_n))} = \prod_{j=1}^n (x-\lambda_j)^{k_{j,L'_0}}, \quad \forall (\lambda_1, \ldots, \lambda_n)\in L'_0. \end{equation}

Now, because we are furthermore assuming the isotropy group of $L'_0$ in $S_n\cap G$ is transitive and $k_{\phi(\cdot)}$ is invariant under $G$-conjugation, the exponents of (2.6) must in fact all be equal. Indeed, consider $1\leqslant \ell \lt \ell'\leqslant n$ and $\sigma \in S_n\cap G$ fixing $L'_0$ with $\sigma(\ell)=\ell'$. We then have

\begin{equation*} \begin{aligned} (x-\lambda_{\ell'})^{k_{\ell',L'_0}} \cdot \prod_{j\ne \ell'}^n (x-\lambda_j)^{k_{j,L'_0}} &= \prod_{j=1}^n (x-\lambda_j)^{k_{j,L'_0}}\\ &= k_{\phi(diag(\lambda_1, \ldots, \lambda_n))} \quad\text{by (2.6)}\\ &= k_{\phi(\sigma^{-1} diag(\lambda_1, \ldots, \lambda_n)\sigma)} \quad\text{by the G-invariance of}\ k_{\phi(\cdot)}\\ &= k_{\phi(diag(\lambda_{\sigma(1)}, \ldots, \lambda_{\sigma(n)}))} \quad\text{by (2.1)}\\ &= \prod_{j=1}^n \left(x-\lambda_{\sigma(j)}\right)^{k_{j,L'_0}} \quad\text{by}\,\, (2.6)\ \text{again, because}\ \sigma\ \text{fixes}\ L'_0\\ &= (x-\lambda_{\ell'})^{k_{\ell,L'_0}} \cdot \prod_{j\ne \ell}^n (x-\lambda_{\sigma(j)})^{k_{j,L'_0}} \end{aligned} \end{equation*}

and hence $k_{\ell',L'_0} = k_{\ell,L'_0}$. Thus, for each connected component $L'_0$ of $L'$, all exponents $k_{j,L'_0}$ are equal. Denote their common value by $r_{L'_0}$. Since $\sum_{j=1}^n k_{j,L'_0}=m$, we have $m=nr_{L'_0}$. In particular, $n$ divides $m$, and necessarily $r_{L'_0}=m/n$, independently of $L'_0$. Therefore, for every matrix $T$ of the form (2.5),

\begin{equation*} k_{\phi(T)} = \prod_{j=1}^n (x-\lambda_j)^{\frac{m}{n}} = \left(\prod_{j=1}^n (x-\lambda_j)\right)^{\frac{m}{n}} = (k_T)^{\frac{m}{n}}. \end{equation*}

Such matrices form a dense subset of $\text{Ad}_{G}T_{L,V}$, and both sides depend continuously on $T$. Hence (2.2) holds on all of $\text{Ad}_{G}T_{L,V}$.

Proof. The statement for $M_n$ will of course recover that for $GL(n)$, the latter being an open subset of the former. We thus focus on $M_n$.

1. (i) $(\Leftarrow)$ Suppose $\lambda$ is a simple spectrum point for $X\in M_n$. By the continuity of both characteristic polynomials and spectra [Reference Newburgh25, Theorem 3], for small $\varepsilon \gt 0$ and a sufficiently small neighbourhood ${\mathcal U}\ni X$ in $M_n$ every $Y\in {\mathcal U}$ has a unique eigenvalue

   \begin{equation*} \lambda_Y \in \left\{z\in {\mathbb C}\ :\ |z-\lambda| \lt \varepsilon\right\}. \end{equation*}

   This provides us with the desired (plainly continuous) selector: $Y\mapsto \lambda_Y$ for $Y\in {\mathcal U}$.

2. (ii) $(\Rightarrow)$ Assume now that all spectrum points of $X\in M_n$ have algebraic multiplicity $\ge 2$, and that we have a continuous eigenvalue selector

   \begin{equation*} {\mathcal U}\ni Y \mapsto{\quad} \lambda_Y \in {\mathbb C} ,\quad {\mathcal U}\ni X\text{a neighbourhood}. \end{equation*}

   Restricting attention to matrices leaving invariant the (at least two-dimensional, by assumption) generalized $\lambda$-eigenspace [Reference Bronson5, § 9.4]

   \begin{equation*} \bigcup_{m\in {\mathbb Z}_{\ge 0}}\ker\left(\lambda I_n-X\right)^m \end{equation*}
   of $X$, we can assume the spectrum $sp(X)$ is a singleton. Translation by $\lambda I_n\in M_n$ further reduces the problem to nilpotent operators $X$. Because moreover every neighbourhood of a nilpotent operator contains another such with a single Jordan block [Reference Horn and Johnson18, Definition 3.1.1], conjugation turns the problem into its instance for
   \begin{equation*} X := J_n(0) := \begin{pmatrix} 0&1&\phantom{-}&\phantom{-}&\phantom{-}\\ \phantom{-}&0&1&\phantom{-}&\phantom{-}\\ \phantom{-}&\phantom{-}&\ddots&\ddots&\phantom{-}\\ \phantom{-}&\phantom{-}&\phantom{-}&0&1\\ \phantom{-}&\phantom{-}&\phantom{-}&\phantom{-}&0 \end{pmatrix}. \end{equation*}

   This case, in turn, is easily handled as in [Reference Chi-Kwong and Zhang8, Example 2] (the current problem’s version for $n=2$): for any neighbourhood ${\mathcal U}\ni 0$ in ${\mathbb C}$, the $z$-dependent matrix

   \begin{equation*} X_z := \begin{pmatrix} 0&1&\phantom{-}&\phantom{-}&\phantom{-}\\ \phantom{-}&0&1&\phantom{-}&\phantom{-}\\ \phantom{-}&\phantom{-}&\ddots&\ddots&\phantom{-}\\ \phantom{-}&\phantom{-}&\phantom{-}&0&1\\ z&\phantom{-}&\phantom{-}&\phantom{-}&0 \end{pmatrix} \end{equation*}
   with characteristic polynomial $x^n\pm z$ does not admit a continuous eigenvalue selector.

Proof. Consider a fixed diagonal matrix $X=diag(x_1, \ldots, x_n)\in G$ with distinct eigenvalues $x_j$. Its image $\phi(X)$ is a conjugate thereof (by spectrum preservation), so we may as well assume $\phi(X)=X$ for simplicity (and hence $\phi(G)$ consists of diagonal matrices by commutativity preservation). It follows that $\phi(Y)$ simply permutes the entries of $Y$ for any $Y\in G$ with distinct eigenvalues:

\begin{equation*} \left(\forall Y=diag(y_1, \ldots, y_n)\in G\right) \left(\exists \tau\in S_n\right) \quad:\quad \phi(Y) = diag(y_{\tau(1)},\ldots, y_{\tau(n)}). \end{equation*}

As before, we identify the groups of invertible diagonal and unitary diagonal matrices in $GL(n)$ with $({\mathbb C}^{\times})^n$ and $({\mathbb S}^1)^n$, respectively.

1. (i) $G=({\mathbb C}^{\times})^n$ or the determinant-1 tuples therein. That $\tau$ is the identity for all $Y\in G$ with distinct eigenvalues follows from: the continuity of $\phi$, the fact that the space of simple-spectrum elements of $G$ is connected (per [Reference Cooper and Manning9, Proposition 8.3], as remarked in the proof of Corollary 2.2), Lemma 2.7, and the density of such matrices $Y$ in $G$.

2. (ii) $G=({\mathbb S}^1)^n$. By Lemma 2.3, ${\mathcal C}^n({\mathbb S}^1)$ is connected only for $n=1,2$, when the preceding argument applies; we henceforth assume $n \geq 3$. What Lemma 2.7 still delivers is a map sending a component $C\in \pi_0({\mathcal C}^n({\mathbb S}^1))$ to the permutation $\tau=\tau_C\in S_n$ satisfying

   \begin{equation*} \phi(Y) = diag(y_{\tau(1)},\ldots, y_{\tau(n)}) ,\quad \forall Y\in C, \end{equation*}
   and hence a function
   

   Fix a transposition $\sigma=(j\ j')$ and consider convergent tuple sequences

   \begin{equation*} C \supset \textbf{y}_m=(y_{m,i})_{i=1}^n \mathrel{\mathop{\longrightarrow}\limits_{\quad{m \quad}}}{} \textbf{y}=(y_i)_i \mathrel{\mathop{\longleftarrow}\limits_{\quad{m \quad}}}{} \sigma\textbf{y}_m=(y_{m,\sigma^{-1}(i)})_{i=1}^n \subset C' := \sigma C \end{equation*}
   for a connected component $C$ of ${\mathcal C}^n({\mathbb S}^1)$, with the common limit
   \begin{equation*} \textbf{y}\in \overline{C}\cap \overline{C'}\subseteq ({\mathbb S}^1)^n \end{equation*}
   having a single equality between two components (necessarily the $j$th and $(j')$th). Writing $\tau:=\tau_C$ and $\tau':=\tau_{C'}$, we have
   \begin{equation*} (y_{\tau(i)})_i = (y_{\sigma\tau'(i)})_i \ \Longrightarrow{\quad}\ y_{\ell} = y_{\sigma\tau'\tau^{-1}(\ell)} ,\quad \forall 1\leqslant \ell\leqslant n \ \Longrightarrow{\quad}\ \tau'\in \left\{\tau,\ \sigma\tau\right\}. \end{equation*}

   The above assumptions impose some constraints on $C$ and $\sigma=(j\ j')$: elements of $C$ can converge to $\mathbf{y}$ with a single coordinate equality only when that equality is between two entries of $\mathbf{y}$ consecutive in a counterclockwise ordering around the circle: if

   \begin{equation*} y_{m,\tau(1)} ,\quad y_{m,\tau(2)} ,\quad \cdots ,\quad y_{m,\tau(n)} \end{equation*}
   are the respective counterclockwise orderings of the tuples $\mathbf{y}_{m}\in C$, then those of the $\sigma \mathbf{y}_m$ must be obtained from the former by interchanging two indices $\tau(k)$ and $\tau(k+1)$ for some $k\in \left\{1,\ldots, n\right\}$ (identifying $n+1=1$). In other words, if $C$ corresponds via (2.3) to $\tau$, then $\sigma$ must be of the form $\tau (k\ k+1)\tau^{-1}$ for a simple transposition $(k\ k+1)$; in conclusion:
   (3.1)\begin{equation} \left(\forall \theta\in S_n\right) \left(\forall\ \text{simple transposition} \sigma\in S_n\right) \quad:\quad \tau_{\theta\sigma} = \tau_{\theta\sigma\theta^{-1}\cdot\theta} \in \left\{\tau_{\theta},\ \theta\sigma\theta^{-1}\tau_{\theta}\right\}. \end{equation}

   We claim that in fact only the first listed option occurs:

   (3.2)\begin{equation} \left(\forall \theta\in S_n\right) \left(\forall\ \text{simple transposition} \sigma\in S_n\right) \quad:\quad \tau_{\theta\sigma} = \tau_{\theta\sigma\theta^{-1}\cdot\theta} = \tau_{\theta}. \end{equation}

   Note first that this would indeed finish the proof that $\tau_{\bullet}$ is constant, for the simple transpositions generate $S_n$. It thus remains to prove (3.2). Recalling that $\tau_{\bullet}$ is constant on the left cosets of $\langle{\eta:=(1\ 2\ \cdots\ n)}\rangle$, we have $\tau_{\theta}=\tau_{\theta\eta}$. Next, decompose

   (3.3)\begin{equation} \theta\eta = \theta (1\ 2)\cdots (n-1\ n) = \sigma_{n-1}\cdot \sigma_{n-2}\cdots \sigma_1\cdot \theta \end{equation}
   for $\sigma_j$ defined recursively by
   \begin{equation*} \sigma_1:=\theta(1\ 2)\theta^{-1}=\text{Ad}_{\theta}(1\ 2) ,\quad \sigma_j:=\text{Ad}_{\sigma_{j-1}\cdots \sigma_1\cdot \theta}(j\ j+1). \end{equation*}

   By (3.1) we have

   \begin{equation*} \forall j\in \left\{1,\ldots, n-1\right\} \quad:\quad \tau_{\sigma_j\cdots \sigma_1\cdot\theta} \in \left\{ \tau_{\sigma_{j-1}\cdots \sigma_1\cdot\theta} ,\ \sigma_j \cdot \tau_{\sigma_{j-1}\cdots \sigma_1\cdot\theta} \right\}, \end{equation*}
   and hence
   \begin{equation*} \tau_{\theta\eta} = \tau_{\sigma_{n-1}\cdot \sigma_{n-2}\cdots \sigma_1\cdot \theta} = \sigma_{j_k}\cdot \sigma_{j_{k-1}}\cdots \sigma_{j_1}\cdot \tau_{\theta} \end{equation*}
   for some possibly-empty set of indices $j_k \gt \cdots \gt j_1$. As no non-empty product
   \begin{equation*} \sigma_{j_k}\cdot \sigma_{j_{k-1}}\cdots \sigma_{j_1} ,\quad j_k \gt \cdots \gt j_1 \end{equation*}
   can be trivial, the permutations
   \begin{equation*} \tau_{\sigma_j\cdots \sigma_1\cdot\theta} ,\quad j\in \{0,\ldots ,n-1\} \end{equation*}
   must in fact all be equal; in particular,
   \begin{equation*} \tau_{\sigma_1\theta} = \tau_{\theta(1\ 2)}=\tau_{\theta}. \end{equation*}

   That the same conclusion holds for all $(j\ j+1)$ in place of $(1\ 2)$ follows in precisely identical fashion: in (3.3) simply decompose $\eta$ as

   \begin{equation*} \eta = (j\ j+1) (j+1\ j+2)\cdots (n-1\ n)(n\ 1)\cdots \end{equation*}
   in place of the original $(1\ 2)\cdots (n-1\ n)$.

Proof. Fix a subspace $W \leqslant {{\mathbb{C}}}^n$ and consider the unitary operator

\begin{equation*} U_W :=2P_W-I_n= id_W \oplus \,(-id_{W^\perp}) \in U(n). \end{equation*}

Suppose that $U \in U(n)$ and $\lambda\in {\mathbb S}^1$ satisfy

\begin{equation*} \ker(\lambda I_n - U) = W. \end{equation*}

To show that $\Psi$ is well-defined, it suffices to show

(3.5)\begin{equation} \ker(\lambda I_n - \phi(U)) = \ker(I_n - \phi(U_W)). \end{equation}

Indeed, let $S \in U(n)$ be a unitary matrix whose columns consist of the orthonormal eigenbasis for $U$. It easily follows that

\begin{equation*} U, U_W \in S({\mathbb S}^1)^nS^*. \end{equation*}

Proposition 3.2 applied to the maximal torus $G:=S({\mathbb S}^1)^nS^*$ implies that there exists $T_{G} \in GL(n)$ such that

\begin{equation*} \phi(X) = T_{G} X T_{G}^{-1} ,\quad \forall X\in G. \end{equation*}

In particular, we have

\begin{align*} \phi(U) = T_{G} UT_{G}^{-1} {\implies} \ker(\lambda I_n-\phi(U)) & = \ker (T_{G}(\lambda I_n - U)T_{G}^{-1}) = T_{G}(\ker(\lambda I_n - U)) \\ &= T_{G} W. \end{align*}

Exactly the same calculation gives $\ker(I_n-\phi(U_W)) = T_{G} W$ which yields (3.5), so that

(3.6)\begin{equation} \Psi(W)= \ker(I_n-\phi(U_{W}))=T_{G} W. \end{equation}

That $\Psi$ preserves dimensions follows from the map’s expression (3.6), as does inclusion preservation: whenever $W\leqslant W'$ there is a maximal torus $G$ in $U(n)$ containing both $U_W$ and $U_{W'}$, so that $\Psi(W)=T_{G} W\leqslant T_{G} W'=\Psi(W')$.

It remains to verify the continuity of $\Psi$. For a fixed $0 \leqslant d \leqslant n$ let ${\mathcal U}_{d}\subset M_n$ be the collection of diagonalizable matrices $X \in M_n$ such that $\dim\ker(I_n - X) = d$. It is straightforward to check that the map

is continuous (e.g., [Reference Saltman33, Proposition 13.4]). As $U_{W} \in {\mathcal U}_{d}$ varies continuously with $W \in {\mathbb G}(d,V)$, it follows that $\Psi$ is continuous as a composition

of continuous maps.

Proof. This is precisely for the reason just pointed out in defining , i.e., the existence of $G$: for $T=T_G$ attached to $G$ as in (3.6), we have

\begin{equation*} \Psi(W+W')=T(W+W')=TW+TW'=\Psi(W)+\Psi(W'). \end{equation*}

Proof. There are several stages to the proof. First of all, as $\Psi$ preserves dimensions, it restricts to a self-map $\Psi|_{{\mathbb P} V}:{\mathbb P} V \to {\mathbb P} V$.

1. (I) Conclusion, assuming $\Psi|_{{\mathbb P} V}$ is a projective space endomorphism. Suppose, in other words, we have proven (3.7). The image $\Psi({\mathbb P} V)$ of the projective space ${\mathbb P} V\subset {\mathbb G}$ cannot be contained in a line of ${\mathbb P} V$, for by assumption the restriction $\Psi|_{{\mathbb P} V}$ sends lines through the elements of an orthogonal basis to distinct $n=\dim V\ge 3$ (even linearly independent) lines. As we are assuming (in the present portion of the proof) that $\Psi|_{{\mathbb P} V} : {\mathbb P} V \to {\mathbb P} V$ is an endomorphism of ${\mathbb P} V$, it follows from [Reference Faure11, Theorem 3.1] that $\Psi|_{{\mathbb P} V}$ is implemented by a semilinear endomorphism $T: V \to V$:

   \begin{equation*} \Psi(\ell) =span_{{\mathbb C}}\ T\ell, \quad \forall \ell \in {\mathbb P} V, \end{equation*}
   where $T$ is an additive map for which there exists a ring endomorphism $\alpha\in End({\mathbb C})$ such that
   \begin{equation*} T(\mu v)=\alpha(\mu)Tv,\quad \forall \mu \in {\mathbb C},\ v\in V. \end{equation*}

   An arbitrary subspace $W \in {\mathbb G}(d,V)\subset {\mathbb G}$ ( $1 \leqslant d \leqslant n$) can be decomposed as $W=\ell_1 \oplus \cdots \oplus \ell_d$ for some mutually orthogonal lines $\ell_j \in {\mathbb P} V$, so the assumed compatibility with lattice operations under yields

   (3.8)\begin{equation} \Psi(W) = \Psi(\ell_1) + \cdots +\Psi(\ell_d)= span_{{\mathbb C}} TW. \end{equation}

   Obviously $T$ is injective, as $\Psi({\mathbb P} V)\subseteq {\mathbb P} V$. As (by assumption) $\Psi$ is continuous, so is $\alpha$. Indeed (as argued in [Reference Šemrl36, § 3, p. 133]), consider orthogonal non-zero vectors $v,w\in V$. Lattice compatibility under then implies that the lines

   \begin{equation*} {\mathbb C} Tv = \Psi({\mathbb C} v) \quad\text{and}\quad {\mathbb C} Tw = \Psi({\mathbb C} w) \end{equation*}
   are distinct. In particular, $Tv$ and $Tw$ are linearly independent. The map
   
   is continuous, and with it $\alpha$ by the noted linear independence of $Tv$ and $Tw$. That continuity forces $\alpha\in \left\{id,\ \overline{\bullet}\right\}$, so that by (3.8) we have
   \begin{equation*} \forall W\in {\mathbb G} \quad:\quad \Psi(W) = TW, \end{equation*}
   for $T$ either in $GL(V)$ or invertible conjugate-linear.

2. (II) $\Psi|_{{\mathbb P} V}$ is a projective space endomorphism. The dimension preservation of $\Psi$ already constrains the nature of possible counterexamples: the assumed compatibility with implies that $\Psi$ preserves inclusion, so for lines $\ell$, $\ell'$ and $\ell^{\prime\prime}$ with $\ell\leqslant \ell'\bigvee\ell^{\prime\prime}$ we have

   \begin{equation*} \text{lines} \Psi(\ell),\ \Psi(\ell'),\ \Psi(\ell^{\prime\prime})\leqslant \text{2-plane} \Psi\left(\ell'\bigvee\ell^{\prime\prime}\right). \end{equation*}

   All is thus well provided $\Psi$ is injective on ${\mathbb P} V$ (for in that case, if $\ell'\ne \ell^{\prime\prime}$, the distinct lines $\Psi(\ell')$ and $\Psi(\ell^{\prime\prime})$ would have to span the right-hand 2-plane in the above equation). We can always reduce the problem to the case $n=3$, as we will: if $\Psi(\ell)=\Psi(\ell')$ for distinct lines then consider a three-dimensional subspace $V'\leqslant V$ containing $\ell\bigvee\ell'$, and substitute

   • • $V'$ for $V$;

   • • and $R\circ\Psi$ for $\Psi$, where $R\in GL(V)$ sends $\Psi(V')$ back to $V'$ and we identify $R$ with the operator $W\mapsto RW$ it induces on ${\mathbb G}$.

   Suppose, then, that $\Psi(\ell)=\Psi(\ell')$ for $\ell\ne \ell'\in {\mathbb P} V$ and $\dim V=3$. We claim that in this case $\Psi$ is constant on all of ${\mathbb P} \pi$, where $\pi:=\ell\bigvee \ell'$, contradicting the fact that by assumption $\Psi$ sends orthogonal lines to linearly independent lines. The projective line ${\mathbb P}\pi$ being connected, it will be enough to argue that

   (3.9)\begin{equation} \Psi^{-1}\left(\Psi(\ell)\right)\cap {\mathbb P}\pi = \left\{\ell^{\prime\prime}\leqslant \pi\ :\ \Psi(\ell^{\prime\prime})=\Psi(\ell)=\Psi(\ell')\right\} \subseteq {\mathbb P}\pi \end{equation}
   is open (for it is already closed by the continuity of $\Psi$, and non-empty because it contains $\ell$ and $\ell'$ themselves). Without loss of generality, this amounts to showing that along with $\ell'$ the set in question contains an entire neighbourhood thereof in ${\mathbb P}\pi$. Write
   \begin{equation*} \pi':=span\left\{\ell',\ \pi^{\perp}\right\}, \end{equation*}
   so that $\pi$ $\pi'$. We proceed in stages.

   1. (a) The line $\ell'$ belongs to the interior of $\Psi^{-1}(\Psi(\ell))\cap {\mathbb P} \pi'$ regarded as a subspace of ${\mathbb P} \pi'$. Suppose not. There is then a sequence

      \begin{equation*} {\mathbb P} \pi'\!\setminus\! \Psi^{-1}(\Psi(\ell)) \ni \ell_k\mathrel{\mathop{\longrightarrow}\limits^{\text{convergence in}\mathbb{G}}_{k}} B \ell'. \end{equation*}

      Because by assumption $\Psi(\ell_k)\ne \Psi(\ell')=\Psi(\ell)$ for all $k$, we have (as observed earlier in the proof)

      \begin{align*} \forall k\quad:\quad \Psi\left(\ell_k\bigvee \ell'\right) &= \Psi(\ell_k)\bigvee \Psi(\ell') \underline{\underline {\quad \Psi (\ell ') = \Psi (\ell )\quad }} \Psi(\ell_k)\bigvee \Psi(\ell) \\ &= \Psi\left(\ell_k\bigvee \ell\right). \end{align*}

      Now, $\ell_k\bigvee\ell'=\pi'$ for all $k$ on the one hand and $\ell_k\bigvee\ell\mathrel{\mathop{\longrightarrow}\limits_{k}}{}\pi$ on the other, so $\Psi(\pi)=\Psi(\pi')$. On the other hand

      
      and we have a contradiction because $\pi\bigvee\pi'=V$ has dimension $3$, whereas $\Psi(\pi)=\Psi(\pi')$ would make $\Psi(\pi)\bigvee\Psi(\pi')$ a $2$-plane.

   2. (b) The line $\ell'$ belongs to the interior of (3.9). Simply apply the preceding step with altered parameters: $\ell'$ is as before, but substitute any

      \begin{equation*} \ell^{\prime\prime} \in \Psi^{-1}(\Psi(\ell))\cap {\mathbb P} \pi'\setminus \{\ell'\} \end{equation*}
      for $\ell$ and interchange the roles of $\pi$ and $\pi'$.

Proof of Theorem 3.1

First of all, by Corollary 2.2, $\phi$ preserves spectra.

1. (I) ${\mathcal{X}}_n = \mathrm{GL}(n)$. By the continuity of the spectral radius $\rho(\cdot)$ on $M_n$ (see e.g., [Reference Newburgh25, Theorem 3]), the map

   \begin{equation*} F : M_n \to \mathrm{GL}(n), \qquad F(X):= X + (1+\rho(X))I_n \end{equation*}
   is well-defined and continuous. Consider the map
   \begin{equation*}\psi : M_n \to M_n, \qquad \psi(X) := \phi(F(X)) - (1 + \rho(X))I_n.\end{equation*}

   Clearly, $\psi$ is a well-defined, continuous, commutativity- and spectrum-preserving map. Therefore, Theorem A implies that there exists $T \in \mathrm{GL}(n)$ such that $\psi$ is of the form (1.1). By passing to the map $T^{-1}\phi(\cdot)T$ or $(T^{-1}\phi(\cdot)T)^t$, without loss of generality, we can assume that $\psi$ is the identity map on $M_n$. In particular, we have

   (3.10)\begin{equation} \phi(F(X))=F(X), \quad \forall X \in M_n. \end{equation}

   By Proposition 3.2, $\phi$ operates as conjugation by an invertible matrix on every maximal algebraic torus of $GL(n)$ (i.e., conjugate of the subgroup $D(n)\leqslant GL(n)$ of invertible diagonal matrices):

   \begin{equation*} \left(\forall S\in GL(n)\right) \left(\exists R\in GL(n)\right) \left(\forall D\in D(n)\right) \quad:\quad \phi(SDS^{-1}) = RSDS^{-1}R^{-1}. \end{equation*}

   The consequence [Reference Šemrl36, Corollary 2.2] follows:

   (3.11)\begin{equation} \left(\forall X\in GL(n)\right) \left(\forall\text{polynomials}\ p\ :\ p(X)\in GL(n)\right) \quad:\quad \phi(p(X))=p(\phi(X)). \end{equation}

   Now, for a fixed $X\in GL(n)$ we have $F(X)=p_X(X)$ for the polynomial

   \begin{equation*} p_X(\lambda):=\lambda+(1+\rho(X)), \end{equation*}
   and hence (as $\rho(\phi(X))=\rho(X)$)
   \begin{align*} \phi(X)+(1+\rho(X))I_n &= p_X(\phi(X)) \overset{\text{(3.11)}}{=} \phi(p_X(X)) = \phi(F(X)) \overset{\text{(3.10)}}{=} F(X)\\ &= X+(1+\rho(X))I_n. \end{align*}

   We are now done: $\phi(X)=X$, as desired.

2. (II) ${\mathcal{X}}_n = \mathrm{SL}(n)$. Let ${{\mathbb{R}}}_{\leqslant 0}$ denote the set of all non-positive real numbers and let

   \begin{equation*} \sqrt[n]{\cdot}: {{\mathbb{C}}}\setminus {{\mathbb{R}}}_{\leqslant 0} \to {{\mathbb{C}}}\setminus{{\mathbb{R}}}_{\leqslant 0}, \qquad \sqrt[n]{z}=\sqrt[n]{|z|}e^{\frac{i Arg z}{n}} \end{equation*}
   be the principal branch of the complex $n$-th root (which is a continuous map). Denote
   \begin{equation*} \mathrm{GL}(n)_*:=\left\{X \in \mathrm{GL}(n) \ : \ \det X \notin {{\mathbb{R}}}_{\leqslant 0} \right\}. \end{equation*}

   The map

   (3.12)\begin{equation} \zeta : \mathrm{GL}(n)_* \to \mathrm{GL}(n)_*, \qquad \zeta(X) = \sqrt[n]{\det X}\cdot \phi\left(\frac1{\sqrt[n]{\det X}} X\right) \end{equation}
   is a continuous, spectrum- and commutativity-preserving extension of $\phi$ to $GL(n)_* \supseteq SL(n)$. To repeat Step (1), we first replace the map $F$ there by
   \begin{equation*} F_*(X):=X+(n+1)(1+\rho(X))I_n. \end{equation*}

   We claim that $F_*(X)\in GL(n)_*$ for all $X\in M_n$. Indeed, if $\lambda\in sp(X)$, then $|\lambda|\leqslant \rho(X)$, and hence

   \begin{equation*} \operatorname{Re}\big((n+1)(1+\rho(X))+\lambda\big)\ge n\rho(X)+n+1, \qquad \left|\operatorname{Im}\big((n+1)(1+\rho(X))+\lambda\big)\right|\leqslant \rho(X). \end{equation*}

   Therefore

   \begin{equation*} \left|\arg\big((n+1)(1+\rho(X))+\lambda\big)\right| \leqslant \arctan\frac{\rho(X)}{n\rho(X)+n+1} \lt \frac{\pi}{n}. \end{equation*}

   It follows that all eigenvalues of $F_*(X)$ have arguments in $(-\pi/n,\pi/n)$, and consequently $\arg\det F_*(X)\in(-\pi,\pi)$. Thus $\det F_*(X)\notin{{\mathbb{R}}}_{\leqslant0}$, as claimed. Therefore the map

   \begin{equation*} \psi:M_n\to M_n,\qquad \psi(X):=\zeta(F_*(X))-(n+1)(1+\rho(X))I_n \end{equation*}
   is a well-defined, continuous, spectrum- and commutativity-preserving map. Step (1) goes through essentially without change for $\mathrm{GL}(n)_*$: Corollary 2.2 applies in the variant of its case (b) with
   \begin{equation*} L:=\left\{(\lambda_1, \ldots, \lambda_n)\in ({\mathbb C}^{\times})^n\ :\ \prod_{j=1}^n \lambda_j \notin {{\mathbb{R}}}_{\leqslant 0} \right\}; \end{equation*}
   the complement $L\!\setminus\! \Delta_L$ is connected. Indeed, the principal $n$-th root gives a homeomorphism
   \begin{equation*} L \cong ({{\mathbb{C}}}\setminus{{\mathbb{R}}}_{\leqslant0}) \times \left\{ (\mu_1,\ldots,\mu_n)\in({{\mathbb{C}}}^\times)^n\ :\ \prod_{j=1}^n\mu_j=1 \right\}. \end{equation*}

   Under this homeomorphism, the condition $\lambda_i=\lambda_j$ is equivalent to $\mu_i=\mu_j$. Hence $\Delta_L$ corresponds precisely to the product of ${{\mathbb{C}}}\setminus{{\mathbb{R}}}_{\leqslant0}$ with the union of the hypersurfaces $\mu_i=\mu_j$ in the second factor. Since this second factor is isomorphic to $({{\mathbb{C}}}^\times)^{n-1}$, deleting these finitely many proper hypersurfaces leaves a connected complex manifold. We conclude that there exists $T \in \mathrm{GL}(n)$ and $\circ \in \{\text{blank},\ t\}$ such that

   \begin{equation*} \zeta(X) = TX^{\circ}T^{-1}, \quad \forall X \in \mathrm{GL}(n)_* \end{equation*}
   and consequently
   \begin{equation*} \phi(X) = TX^\circ T^{-1}, \quad \forall X \in \mathrm{SL}(n). \end{equation*}

3. (III) ${\mathcal{X}}_n = \mathrm{U}(n)$. Denote by $V\cong {\mathbb C}^n$ the ambient space on which all operators act. The map $\Psi$ constructed in Lemma 3.3 meets the requirements of Proposition 3.5 by Lemma 3.4, so that there exists an invertible $T: V \to V$, either linear or conjugate-linear, such that

   (3.13)\begin{align} \forall W\in{\mathbb G} \quad:\quad & \Psi(W) =\ker(\lambda I_n -\phi(U)) = T(\ker(\lambda I_n -U)) \\ & \text{for any}\ \lambda \in {\mathbb S}^1\ \text{and}\ U \in U(n)\ \text{with}\ W=\ker(\lambda I_n-U). \nonumber\end{align}

   This means the following.

   • • When $\alpha=id$, we have the first option in (1.1):

     \begin{equation*} \phi(U)=TUT^{-1} ,\quad \forall U\in U(n). \end{equation*}

     Indeed, for an arbitrary $U \in U(n)$, Proposition 3.2 and (3.13) say that $\phi(U)$ is the diagonalizable operator with the same spectrum as $U$ and whose $\lambda$-eigenspace (for any $\lambda\in {\mathbb S}^1$) is the image through $T$ of the $\lambda$-eigenspace of $U$. In other words, $\phi(U)$ is precisely $TUT^{-1}$.

   • • When $\alpha$ is conjugation, the analogous argument proves that

     \begin{equation*} \phi(U)=TU^{-1}T^{-1} ,\quad \forall U\in U(n) \end{equation*}
     instead: once more $\phi(U)$ is the diagonalizable operator whose $\lambda$-eigenspaces are the respective images through $T$ of those of $U$, while conjugation by $T$ turns $\lambda$-eigenspaces into $\overline{\lambda}$-eigenspaces respectively (for $\lambda\in {\mathbb S}^1$ so that $\overline{\lambda}=\lambda^{-1}$) by anti-linearity.

     This second variant corresponds to the right-hand branch of (1.1): the conjugate-linear involution attached to the orthonormal basis we have already fixed implicitly in treating operators as matrices has the property that

     \begin{equation*} \forall X\in M_n \quad:\quad JXJ^{-1}=\overline{X} \end{equation*}
     (where for $X=(x_{ij})$, $\overline{X}=(\overline{x_{ij}})$) and hence
     \begin{equation*} \phi(U) = TU^{-1}T^{-1} = (TJ)U^t(TJ)^{-1} \end{equation*}
     because $U$ is unitary (so that $JU^tJ^{-1}=\overline{U^t}=U^{-1}$).

4. (IV) ${\mathcal{X}}_n = N_n$. The conclusion certainly holds for the restriction $\phi|_{U(n)}$ by the preceding step of the proof, so we can assume $\phi|_{U(n)}=id$. We furthermore know from Proposition 3.2 that $\phi$ is of the form (1.1) when restricted to diagonal invertible operators, so also on the algebra of diagonal operators by continuity. This furthermore applies to arbitrary maximal abelian sub-algebras of $N_n$, for these are all conjugate to that of diagonal operators (and one need simply apply Proposition 3.2 with a different choice of orthonormal basis). Commuting normal matrices are simultaneously unitarily diagonalizable [Reference Horn and Johnson18, Theorem 2.5.5], so $\phi$ restricted to any commuting family is additive. An arbitrary normal matrix is a linear combination of commuting unitary matrices (obvious for diagonal matrices, which suffices), hence the conclusion:

   \begin{equation*} \phi|_{U(n)}=id|_{U(n)} \Longrightarrow \phi=id_{N_n}. \end{equation*}

This concludes the proof of the theorem.

Alternative proof of Theorem 3.1: $GL(n)$ and $SL(n)$

Simply observe that the $\phi$ in question, defined on ${\mathcal X}_n$, restricts to such on ${\mathcal X}_{n,ss}$. The conclusion follows from the density of ${\mathcal X}_{n,ss}\subseteq {\mathcal X}_{n}$ for ${\mathcal X}_n\in \left\{GL(n),\ SL(n)\right\}$.

Proof. The issue is quantification ordering: the unitary case of Theorem 3.1 shows that for each $S$ there are such $T=T_S$ and $\circ=\circ_S$, whereas we claim here that the same $\circ$ will do for all $S$. To see this, note that the closed subsets

(3.16)\begin{equation} \left\{S\in GL(n)\ :\ \circ_S=\circ\right\}\subseteq GL(n) ,\quad \circ\in \{\text{blank},\ t\} \end{equation}

partition the connected topological space $GL(n)$. Indeed, the closure and disjointness claims both follow from the characterization of the sets (3.16) as

\begin{equation*}\left\{S\in GL(n)\;:\;\phi\vert_{{\text{Ad}}_SU(n)}\left[\begin{array}{l}\text{preserves multiplication}\ (\text{if}\;\circ=\text{blank})\\\text{reverses multiplication}\ (\text{if}\;\circ=\text{t})\end{array}\right.\right\}.\end{equation*}

Multiplication preservation and reversal are both closed conditions by the continuity of $\phi$, and the two sets cannot overlap because multiplication and its opposite are distinct operations on the non-abelian conjugates of $U(n)$.

Proof. The claim is that (3.17) holds with the same $p$ for all $S$. We will first prove a weaker claim, allowing possibly variable $p$ for different $S$.

1. (I) For each $0\leqslant S\in GL(n)$ (3.17) holds for some $p=p_S\in \{0,-2\}$. For an arbitrary unitary $U \in U(n)$ which commutes with $S$, we have

   \begin{equation*} U=\phi(U)=\phi(SUS^{-1})=TSUS^{-1}T^{-1}=TUT^{-1} \end{equation*}

for some $T=T_S$ by Lemma 3.9: $\circ=\text{blank}$ for $S=I_n$, so must be $\text{blank}$ for all positive $S$ by the (path-connectedness) of $GL(n)_{\ge 0}$. Therefore, $T$ commutes with all unitary matrices commuting with $S$, or equivalently, $T$ belongs to the bicommutant [Reference Blackadar2, § I.2.5.3] of (the positive) $S$. Hence, $T$ must be a polynomial in $S$ by the (finite-dimensional version of the) celebrated von Neumann bicommutant theorem [Reference Blackadar2, Theorem I.9.1.1]:

(3.18)\begin{equation} \exists p_S\in {\mathbb C}[x] \quad:\quad T=p_S(S). \end{equation}

We may as well assume $S$ diagonal: $S=(\lambda_1, \ldots, \lambda_n)$, with the standard basis vectors $(e_1, \ldots, e_n)$ as eigenvectors (as usual, we identify diagonal matrices with the corresponding tuples). It will often also be convenient to assume the tuples $(\lambda_j)=(\lambda_1, \ldots, \lambda_n)$ we work with range over a conveniently chosen dense subset (e.g., they are all distinct): by the continuity of the map

\begin{equation*} GL(n)_{\ge 0} \ni S \mapsto{\quad} \text{Ad}_S U(n)\in \left\{{\text{closed bounded subsets of}}\ GL(n)\right\} \end{equation*}

for the topology on the codomain induced by the Hausdorff distance [Reference Burago, Burago and Ivanov6, Definition 7.3.1] (attached to any metric topologizing $GL(n)$), the sets

(3.19)\begin{equation} \left\{0\leqslant S\in GL(n)\ :\ \phi|_{\text{Ad}_S U(n)}=\text{Ad}_{S^p}\right\}\subseteq GL(n)_{\ge 0} ,\quad p\in \{0,\ -2\} \end{equation}

are closed.

Next, commutativity preservation by $\phi$ entails

(3.20)\begin{equation} \left(\forall U,V\in U(n)\right) \ :\ [U,\ SVS^{-1}]=0 \Longrightarrow [\phi(U),\ \phi(SVS^{-1})] = [U,\ TSVS^{-1}T^{-1}]=0. \end{equation}

We examine (3.20) as applied to the following setup:

• • $V$ is a unitary (hence also self-adjoint) involution with a one-dimensional $(-1)$-eigenspace spanned by $v=\sum_{j=1}^nc_j e_j$.

• • $U$ is also a unitary involution, whose $(-1)$-eigenspace is spanned by $Sv=\sum_{j=1}^n c_j \lambda_j e_j$ and its orthogonal projection $P_{S(v^{\perp})}Sv$ on $S(v^{\perp})$ (where $v^\perp$ is the orthogonal complement of $\{v\}$). Said $(-1)$-eigenspace is thus at most two-dimensional. Generically, the dimension will be precisely 2: it is enough to assume the $S$ non-scalar (as discussed, there is no harm in doing so) and the $c_i$ all non-zero.

As $V$ is a semisimple operator with eigenvalues $\pm 1$, so too is $\text{Ad}_SV$; moreover, the latter’s eigenspaces are the respective images through $S$ of the eigenspaces of $V$:

\begin{equation*} \begin{aligned} \ker\left(I_n+\text{Ad}_SV\right)&=S\ker\left(I_n+V\right)={\mathbb C} Sv\quad\text{and}\\ \ker\left(I_n-\text{Ad}_SV\right)&=S\ker\left(I_n-V\right)=S\left(v^{\perp}\right). \end{aligned} \end{equation*}

By fiat, $U$ leaves these invariant:

• • it scales $Sv$ (spanning the $(-1)$-eigenspace of $\text{Ad}_SV$) by $-1$;

• • also scales by $-1$ the orthogonal projection $v':=P_{S(v^{\perp})}Sv\in S(v^{\perp})$;

• • fixes pointwise $\left({\mathbb C} Sv\oplus {\mathbb C} v'\right)^{\perp}\leqslant S(v^{\perp})$, which coincides with the orthogonal complement of $v'$ in $S(v^{\perp})$;

• • and hence leaves invariant the $1$-eigenspace

  \begin{equation*} S(v^{\perp}) = {\mathbb C} v'\oplus \left({\mathbb C} Sv\oplus {\mathbb C} v'\right)^{\perp} \end{equation*}

of $\text{Ad}_SV$. $U$ thus commutes with $\text{Ad}_SV=SVS^{-1}$ and hence also with $\text{Ad}_{TS}V$ by (3.20). In particular, the $(-1)$-eigenvector

(3.21)\begin{equation}TSv=\sum_{j=1}^nc_j\lambda'_je_j,\quad\lambda'_j:=\lambda_jp(\lambda_j),\quad p=p_S\; \text{as in (3.18)}\end{equation}

of $\text{Ad}_{TS}V$ (unique up to scaling) is also an eigenvector of $U$, with eigenvalue $\pm 1$. We claim that in fact

(3.22)\begin{equation} TSv\in \ker(I_n+U), \end{equation}

i.e., $TSv$ is a $(-1)$-eigenvector of $U$. This is certainly true for $S=(\lambda_j)$ sufficiently close to the identity: $\text{Ad}_{TS}V$ will then be close to $\text{Ad}_SV$ for the latter is in turn close to a unitary and we are assuming $\phi=id$ on unitaries. The $(-1)$-eigenspaces of $\text{Ad}_SV$ and $\text{Ad}_{TS}V$ will thus also be close in the projective space ${\mathbb P} {\mathbb C}^n$, so in particular the $(-1)$-eigenspace of $\text{Ad}_{TS}V$ cannot be contained in the $1$-eigenspace of $U$. To extend this to all $S=(\lambda_j)$ under consideration (i.e., non-scalar diagonal positive matrices), consider the parameter space

(3.23)\begin{equation} {\mathcal P} := \left\{\left((c_j),\ (\lambda_j)\right)\ :\ c_j\in {\mathbb C}^{\times},\ \lambda_j \gt 0\ \text{not all equal}\right\} \end{equation}

for our choices. $U$, $V$ and $v$ depend continuously on the $(c_j)$ and $(\lambda_j)=S$. We have maps

all continuous: $\Omega_{2}$ and $\Omega_{n-2}$ as in the proof of Lemma 3.3 (the portion invoking [Reference Saltman33, Proposition 13.4]), and $\Omega_1$ by the same token, by the continuity of $\phi$ and that of $\text{Ad}_SV$ (as a function of ${\mathcal P}$), since

\begin{equation*} {\mathbb C} TSv=\ker (I_n+\text{Ad}_{TS}V)=\ker(I_n+\phi(\text{Ad}_SV)). \end{equation*}

The loci

\begin{equation*} {\mathcal P}_{\bullet} := \left\{p\in {\mathcal P}\ :\ \Omega_1(p)\leqslant \Omega_{\bullet}(p)\right\} ,\quad \bullet\in \{2,n-2\} \end{equation*}

are closed and partition the connected space ${\mathcal P}$; because we observed above that ${\mathcal P}_2$ is not empty, it must be all of ${\mathcal P}$. Thus, for

\begin{equation*} w:=S^{-1}v=\sum_{j=1}^n c_j \lambda_j^{-1}e_j\in (Sv^{\perp})^{\perp} \end{equation*}

we have $w \in \ker(I_n+U)$, so

\begin{equation*} \sum_{j=1}^n c_j \lambda'_j e_j=TSv\in \ker(I_n+U) = span\left\{Sv,\ w\right\}=span\left\{\sum_{j=1}^n c_j \lambda_j e_j,\ \sum_{j=1}^n c_j \lambda_j^{-1}e_j\right\}. \end{equation*}

In particular, as all $c_1, \ldots , c_n$ are non-zero, we conclude that the tuples $(\lambda_j)$, $(\lambda_j^{-1})$ and $(\lambda'_j)$ are linearly dependent. Now simply substitute the vector

(3.24)\begin{equation} \left(TS\right)^{*-1}v = \sum_{j=1}^n c_j \overline{\lambda'_j}^{-1}e_j \end{equation}

(spanning the orthogonal complement $(TSv^{\perp})^{\perp}$) for $TSv$ in the argument just given:

• • The multiplicative operator $(\bullet)^{*-1}$ preserves multiplicativity and fixes unitaries, so in (3.20) the hypothesis $\left[U,\text{Ad}_SV\right]=0$ and the conclusion $\left[U,\text{Ad}_{TS}V\right]=0$ hold, respectively, if and only if their counterparts $\left[U,\text{Ad}_{S^{-1}}V\right]=0$ and $\left[U,\text{Ad}_{(TS)^{*-1}}V\right]$ do.

• • (3.20) thus goes through with $S^{-1}=S^{*-1}$ and $(TS)^{*-1}$ in place of $S$ and $TS$, respectively, and the discussion leading to (3.22) applies to (3.24) to show that it too must belong to the same $(\leqslant 2)$-dimensional span:

  \begin{equation*} \dim span\left\{(\lambda_j),\ \left(\lambda_j^{-1}\right),\ (\lambda'_j),\ \left(\overline{\lambda'_j}^{-1}\right)\right\}\leqslant 2. \end{equation*}

There are thus $a,b,c,d\in {\mathbb C}$ with

\begin{equation*} \forall 1\leqslant j\leqslant n \quad:\quad \lambda'_j = a\lambda_j+b\lambda_j^{-1} \quad\text{and}\quad {\lambda'_j}^{-1} = c\lambda_j + d\lambda_j^{-1} \end{equation*}

(in the second equation, conjugates can be elided, as $\lambda_j$ are real). This also reads

(3.25)\begin{equation} \exists a,b,c,d\in {\mathbb C} \quad\!:\quad\! \left(a\lambda_j+b\lambda_j^{-1}\right) \left(c\lambda_j + d\lambda_j^{-1}\right) \!=1 \quad\!\text{and hence}\quad\! \left(a\lambda_j^2+b\right) \left(c\lambda_j^2 + d\right) \!=\lambda_j^2 \end{equation}

for all $1 \leqslant j \leqslant n$. Because $n\ge 3$, for a generic choice of $\lambda_j^2$ the only $(a,b,c,d)$ witnessing (3.25) are those that produce a tautological equation (i.e., $\lambda_j^2=\lambda_j^2$). In other words, for a dense set of $(\lambda_j)$ we have

\begin{equation*} \left(b=c=0 \wedge ad=1\right) \vee \left(a=d=0 \wedge bc=1\right) \end{equation*}

and hence by (3.18) and (3.21) we obtain

\begin{equation*} T=T_S=T_{(\lambda_j)}\in {\mathbb C} I_n\cup {\mathbb C} S^{-2}. \end{equation*}

Continuity ensures that this holds for all $S\ge 0$, finishing the proof.

1. (II) The same $p=p_S\in \{0,-2\}$ functions for all $S$. As already observed, the two subsets (3.19) are closed. Their intersection is ${\mathbb R}_{ \gt 0} I_n$, for the sets $\text{Ad}_S U(n)$ and $\text{Ad}_{S^{-1}}U(n)$ are distinct for every non-scalar positive invertible $S$. The intersections of (3.19) with the connected $GL(n)_{\ge 0}\setminus {\mathbb R}_{ \gt 0}I_n$ partition that space, so one of the two must constitute all of $GL(n)_{\ge 0}$.

Proof of Proposition 3.8

Note that $\phi$ is spectrum-preserving by Corollary 2.2.

1. (I) ${\mathcal{X}}_n = \mathrm{GL}(n)_{ss}$. We already know from the unitary branch of Theorem 3.1 that $\phi|_{U(n)}$ is of the form (1.1), so we can recast the claim as

   \begin{equation*} \phi|_{U(n)}=id_{U(n)} \Longrightarrow \phi\in\{id_{GL(n)_{ss}}, \Theta\}. \end{equation*}

   Proposition 3.10 then implies that

   \begin{equation*} \forall S\in GL(n)_{\ge 0} \quad :\quad \phi|_{\text{Ad}_S U(n)}= id \quad \text{or} \quad \phi|_{\text{Ad}_S U(n)}=\text{Ad}_{S^{-2}}\quad\left(\text{exclusively}\right). \end{equation*}

   In either case, the conclusion follows as in the normal-matrix branch of Theorem 3.1: every diagonalizable matrix is a linear combination of mutually commuting elements of $\text{Ad}_SU(n)$ for some $S\in GL(n)$.

2. (II) ${\mathcal{X}}_n = \mathrm{SL}(n)_{ss}$. By Proposition 3.2, $\phi$ restricts to all maximal tori as conjugation, and every semisimple element belongs to some maximal torus. Moreover, as maximal tori are invariant under scaling, $\phi$ preserves multiplication by $n$th roots of unity $\zeta$ (i.e., $\phi(\zeta\bullet)=\zeta \phi$). This means that we can extend $\phi$ canonically to $GL(n)_{ss}$ by scaling:

   (3.26)\begin{equation} \widetilde{\phi}(\eta T):=\eta \phi(T) ,\quad \forall\left(\eta T\in GL(n)_{ss}\ \text{with} T\in SL(n)_{ss}\ \wedge \eta\in {\mathbb C}^{\times}\right). \end{equation}

   The same element of $GL(n)_{ss}$ will be representable as $\eta T$ in several ways, but any two representations $\eta T=\eta' T'$ have $\frac{\eta}{\eta'}$ an $n$th root of unity. The aforementioned preservation of root-of-unity scaling by $\phi$ ensures that (3.26) is consistent, and $\widetilde{\phi}$ is then also continuous and a spectrum/commutativity preserver. This reduces the problem to step (I), hence the conclusion.

1. (1) (a) $\Rightarrow$ (b). If $T$ has at least one eigenvalue $\lambda$ of multiplicity $\ge 2$, we can restrict attention to a two-dimensional subspace $W\leqslant \ker(\lambda I_n-T)$ by working throughout with matrices leaving invariant a decomposition ${\mathbb C}^n=W\oplus W'$. Example 3.12 then applies to produce (even globally) continuous functions $f$ with (3.28) discontinuous at $T|_W=\lambda id_W$.

2. (2) (b) $\Rightarrow$ (a). The simplicity of the spectrum means (by spectrum continuity [Reference Newburgh25, Theorem 3]) that for any convergent sequence $T_k\mathrel{\mathop{\longrightarrow}\limits_{k}}\ T$ and any fixed eigenvalue $\lambda\in sp(T)$ we have

   \begin{equation*}sp(T_k)\ni\lambda_k\left(\text{simple for large}\ k\right)\stackrel{\lbrack}{\longrightarrow}\quad k\quad\rbrack\quad\lambda.\end{equation*}

   The one-dimensional kernels

   \begin{equation*} \ker\left(T_k-\lambda_k I_n\right) = {\text{im}} E_{\lambda_k} \end{equation*}
   converge (in ${\mathbb G}$) to the $\lambda$-eigenspace $\text{im} E_{\lambda}$ of $T$ by yet another application of [Reference Saltman33, Proposition 13.4]. This goes for all (simple!) $\lambda\in sp(T)$, so we also have
   \begin{equation*} \ker E_{\lambda_k} = \bigoplus_{\lambda\ne \lambda'\in sp(T)} \text{im} E_{\lambda'_k} \mathrel{\mathop{\longrightarrow}\limits^{\text{convergence in}\ {\mathbb G}}_{k}}\quad \bigoplus_{\lambda\ne \lambda'\in sp(T)} \text{im} E_{\lambda'} = \ker E_{\lambda}. \end{equation*}

   This yields

   (3.29)\begin{equation} \forall \lambda\in sp(T) \quad:\quad E_{\lambda_k}\mathrel{\mathop{\longrightarrow}\limits_{k}}{\quad} E_\lambda, \end{equation}
   so
   \begin{equation*} f(T_k)=\sum_{\lambda_k\in sp(T_k)}f(\lambda_k) E_{\lambda_k} \mathrel{\mathop{\longrightarrow}\limits_{k}}{\quad} \sum_{\lambda\in sp(T)}f(\lambda) E_{\lambda} = f(T) \end{equation*}
   by (3.29) and the continuity of $f$.

Proof. First observe that if well-defined, $\Theta$ will be defined globally on ${\mathcal{X}}_n$: the semisimple matrices are precisely those conjugate to diagonal ones by some $S\in GL(n)$, and the polar decomposition [Reference Horn and Johnson18, Theorem 7.3.1]

\begin{equation*} S=|S^*|V ,\quad V\in U(n) ,\quad |S^*|=(SS^*)^{\frac{1}{2}}\in GL(n)_{\ge 0} \end{equation*}

renders such a matrix an $|S^*|$-conjugate of a normal one.

As to $\Theta$ being well-defined: for $S,T\in GL(n)_{\ge 0}$ and $M,N\in N_n\cap {\mathcal{X}}_n$ we have

\begin{equation*} \begin{aligned} SNS^{-1} &= TMT^{-1} \Longrightarrow SN^*S^{-1} \\ &= TM^*T^{-1} \quad\left({\textit{Putnam--Fuglede theorem} [16, Problem\ 192]}\right)\\ & \require{mathtools} \stackrel{\text{taking adjoints}}{\Longrightarrow} S^{-1}NS=T^{-1}MT. \end{aligned} \end{equation*}

Clearly, the assignment $\Theta$ is spectrum-preserving. Furthermore, one of the proofs of Putnam–Fuglede (e.g., [Reference Halmos15, § 41, Theorem 2]) can also be adapted to prove that $\Theta$ is also commutativity-preserving: if $SNS^{-1}$ and $TMT^{-1}$ commute then so do $p(SNS^{-1})$ and $p(TMT^{-1})$ for polynomials $p$; apply this to a polynomial

\begin{equation*} p\in {\mathbb C}[x] \quad\text{with}\quad p(\lambda)=\overline{\lambda} ,\quad\forall \lambda\in sp(M)\cup sp(N), \end{equation*}

whence $SN^*S^{-1}=p(SNS^{-1})$ and $TM^*T^{-1}=p(TMT^{-1})$ commute. Taking adjoints then proves that $S^{-1}NS$ and $T^{-1}MT$ do as well.

To assess continuity, observe that for normal $N$ and $S\in GL(n)_{\ge 0}$ we have

(3.30)\begin{equation} \Theta\left(\text{Ad}_S N\right) = \text{Ad}_{S^{-1}}N = \left(\text{Ad}_S N^*\right)^* = \left(\text{Ad}_S f(N)\right)^* = f\left(\text{Ad}_S N\right)^* \end{equation}

with $f(\bullet)$ as in (3.28) for $f:=\overline{\bullet}$ (complex conjugation). The continuity of $\Theta$ thus reduces to that of $f(\bullet)$:

The latter condition fails for $n\ge 3$ by the aforementioned [Reference Niemiec26, Theorem 1.2(A)]:

• • The result applies directly to $GL(n)_{ss}$, where continuity fails because complex conjugation is not holomorphic on the punctured complex plane ${\mathbb C}^{\times}$.

• • While for $SL(n)$ the same conclusion follows from the $GL(n)$ counterpart and the conjugate linearity of (3.28) for $f=\overline{\bullet}$:

  \begin{equation*} f(\alpha T) = \overline{\alpha}f(T) ,\quad \forall T\in GL(n)_{ss} ,\quad \forall \alpha\in {\mathbb C}. \end{equation*}

Proof of Theorem 3.7

That the maps (1.1) meet the requirements of course needs no proof, while Propositions 3.8 and 3.14 give the converse: the latter’s discontinuity claim eliminates the $\Theta$ factor in (3.14).