Proof The proof is similar to that of [Reference Huang and Zheng8, Lemma 2.4]. Let $a $ be a fixed point on $\partial \Sigma $ and $n=m_{\phi \circ F}(a).$ Because $\phi \circ F$ is analytic a, we can choose a small number $\varepsilon>0$ , so that

$$ \begin{align*}V_\varepsilon = \Sigma \cap \{z: |z-a|<\varepsilon\}\end{align*} $$

is an angular domain with the angle $\alpha $ and

$$ \begin{align*}\phi\circ F(z)-\phi(F(a))=(z-a)^{n} g(z) ,\end{align*} $$

where g is holomorphic on $\{z: |z-a|<\varepsilon \} $ and $g(a)\neq 0$ . We may choose $\varepsilon $ to be small enough, so that

$$ \begin{align*}\phi\circ F(z)-\phi(F(a))= \tilde{g}^{n}(z) ,\end{align*} $$

where $\tilde {g}(z)=(z-a)\sqrt [n]{g(z)}$ is univalent on $\{z: |z-a|<\varepsilon \}$ . Thus, $\phi \circ F$ maps $V_\varepsilon $ to an angular domain with the vertex at $\phi (F (a))$ and with the angle $n\alpha .$ Here, it is allowed that $n\alpha>2\pi $ , as we count multiplicity.

In addition, noting that F maps an inside-neighborhood of $\partial \Sigma $ bijectively to another inside-neighborhood, we have that F maps an inside-neighborhood of a bijectively to an inside-neighborhood W of $F (a)$ . By similar discussions to the above paragraph, $\phi $ maps W to an angular domain with the angle $m_{\phi }(F (a))~\beta .$ Therefore, $\phi \circ F$ maps $V_\varepsilon $ to an angular domain with the angle $m_{\phi }(F (a))~\beta ,$ forcing

$$ \begin{align*}m_{\phi\circ F}(a)~\alpha =m_{\phi}(F (a))~\beta\end{align*} $$

to complete the proof. ▪

Proof Suppose that both g and h are nonconstant functions holomorphic on $ \overline {\Sigma } $ , so that $T_g$ is unitarily equivalent to $T_h$ on $L_a^2(\Sigma )$ . Because proper maps from $\mathbb {D}$ to $\mathbb {D}$ are exactly finite Blaschke products, proper maps from $\Sigma $ to $\Sigma $ are essentially finite Blaschke products, excluding compositions of two conformal maps (one is from $\Sigma $ onto $\mathbb {D}$ , and another is its inverse). Thus, one can get an easy translation of Theorem 2.3 from $\mathbb {D}$ to $\Sigma $ , and fortunately, the maps g and h fulfill the desired conditions (just like the function $\phi $ in $ H^\infty (\mathbb {D})$ ). Thus, we have that there exist a function $\tilde {g}$ in $ H^\infty ( \Sigma )$ and a proper map $\phi _1$ from $\Sigma $ to $\Sigma $ satisfying $ g=\tilde {g}\circ \phi _1 $ and

$$ \begin{align*}\{T_g\}'=\{T_{\phi_1}\}'.\end{align*} $$

Because $T_g$ is unitarily equivalent to $ T_h,$ there is a unitary operator U such that $T_h=UT_g U^*$ . This naturally induces a bijection

$$ \begin{align*} \{T_g\}'' &\longrightarrow \{T_h\}'', \\ S &\longmapsto USU^*. \end{align*} $$

Because $T_{\phi _1}\in \{T_{\phi _1}\}'' =\{T_g\}''$ , and

$$ \begin{align*}\{T_h\}'' \subseteq \{T_z\}'' =\{T_\psi:\psi \in H^\infty(\Sigma)\}'= \{T_\psi:\psi \in H^\infty(\Sigma)\} ,\end{align*} $$

there is a function $\phi _2 \in H^\infty (\Sigma )$ satisfying

(2.1) $$ \begin{align} UT_{\phi_1}U^*= T_{\phi_2} . \end{align} $$

First, we will show that $\phi _2$ must be a proper map. To do so, for each $\lambda \in \Sigma ,$ we have that $\phi _1-\lambda $ has n zeros in $\Sigma $ (n is a constant integer not depending on $\lambda $ ; see [Reference Milnor9, Appendix E] or [Reference Guo and Huang7, Proposition 2.1.2]), $\phi _1-\lambda $ has no zero on $\partial \Sigma ,$ and hence $T_{ \phi _1-\lambda }$ is a Fredholm operator. This implies that $T_{\phi _2-\lambda }$ is also Fredholm. By (2.1),

$$ \begin{align*}\mathrm{Index}\, T_{ \phi_1-\lambda } = \mathrm{Index}\, T_{\phi_2-\lambda} .\end{align*} $$

Thus, for each $\lambda $ in $ \Sigma , \phi _2-\lambda $ has n zeros in $\Sigma $ , counting multiplicity. Because, for $\mu $ not in $\overline {\Sigma }$ , $T_{\phi _1-\mu }$ is invertible, (2.1) gives that $T_{\phi _2-\mu }$ is also invertible. Thus, $\phi _2(z)-\mu $ has no zero in $\Sigma $ for $ \mu \not \in \overline {\Sigma }$ , and thus the image $\phi _2(\Sigma )$ is contained in $\overline {\Sigma }$ . Because $\phi _2$ is an open mapping, we have

$$ \begin{align*}\phi_2(\Sigma )=\Sigma .\end{align*} $$

Let $\sigma :\Sigma \rightarrow \mathbb {D}$ be a Riemann mapping. Then,

$$ \begin{align*}\sigma\circ\phi_2\circ\sigma^{-1} :\mathbb{D}\rightarrow \mathbb{D}\end{align*} $$

is analytic and exactly n-to-one. By the last theorem in [Reference Cowen2], there is a finite Blaschke product B and a univalent function $ \tilde {f} $ on $\mathbb {D}$ satisfying

$$ \begin{align*}\sigma\circ\phi_2\circ\sigma^{-1} =\tilde{f}\circ B.\end{align*} $$

This gives that $\phi _2$ is a proper map with the same image as $\phi _1$ .

Next, we have

$$ \begin{align*} T_h & = UT_gU^* \\ & = U T_{\tilde{g}\circ \phi_1} U^* \\ & = \tilde{g} (UT_{\phi_1}U^*) \\ & = \tilde{g}(T_{\phi_2})\\ & = T_{ \tilde{g}\circ \phi_2 }, \end{align*} $$

to get that $h= \tilde {g}\circ \phi _2 .$

If $\mathcal {V}^*(g,\Sigma )$ is trivial, then $\mathcal {V}^*(\phi _1,\Sigma ) $ is trivial. Using the Riemann mapping $\sigma $ , we have that $\mathcal {V}^*(\sigma \circ \phi _1\circ \sigma ^{-1},\mathbb {D})$ is also trivial. This gives that $\sigma \circ \phi _1\circ \sigma ^{-1}$ is a Blaschke product with order $1$ , and hence the order $\phi _2$ is also $1$ . Thus, we have that both $\phi _1$ and $\phi _2$ are in Aut $(\Sigma )$ to get

$$ \begin{align*} h&=g\circ (\phi_1^{-1}\circ \phi_2)\\ &= g\circ F, \end{align*} $$

where $F=\phi _1^{-1}\circ \phi _2$ . This completes the proof. ▪

Proof of Theorem 1.1 Assume that g is a function holomorphic on $\overline {\Sigma }$ and g has no critical point at vertices of $\overline {\Sigma }$ . Because $\Sigma $ is not equiangular, by [Reference Huang and Zheng8, Theorem 1.2], we have $\mathcal {V}^*(g,\Sigma )=\{T_h, T_h^*\}'=\mathbb {C}I.$ Using Lemma 2.4, we have that there is a function F in Aut $(\Sigma )$ such that

$$ \begin{align*}h =g\circ F.\end{align*} $$

First, we will show that $F $ maps the set of vertices $\{a_1, \ldots , a_n\}$ of $\Sigma $ to itself. If this is not true, without loss of generality, we may assume

$$ \begin{align*}F(b)=a_1\end{align*} $$

for a point b lying on the edge, because Carathéodory’s theorem gives that F is a bijection onto $\partial \Sigma $ . Let $\alpha _1$ be the interior angle at $a_1$ of $\Sigma $ . Because the angle at b formed by edges of $\Sigma $ is $\pi $ , by Lemma 2.1, we have

$$ \begin{align*}m_{g}(a_1)\alpha_1=m_{g\circ F}(b)\pi,\end{align*} $$

where the multiplicity $m_{g\circ F}(b)$ is an integer. Because $g'(a_1)\neq 0,$ we have that the multiplicity $m_{g}(a_1)$ of g at $a_1$ equals $1$ to get

$$ \begin{align*}\alpha_1=m_{g\circ F}(b)\pi.\end{align*} $$

This is a contradiction that interior angles of $\Sigma $ are not equal to neither $\pi $ nor $2\pi $ . Therefore, $F $ maps the set of vertices $\{a_1, \ldots , a_n\}$ of $\Sigma $ to itself.

Next, we will show that $F (z)\equiv z.$ Because $F $ maps the set of vertices $\{a_1, \ldots , a_n\}$ of $\Sigma $ to itself bijectively, by [Reference Huang and Zheng8, Lemma 3.1], either F fixes every vertex of $\partial \Sigma $ or for each j, there is a positive number $k_j$ satisfying

$$ \begin{align*}a_j = F^{k_j}(a_1).\end{align*} $$

In the first case, by Carathéodory’s theorem, we have that three boundary values of a bijective holomorphic mapping from $\Sigma $ to $\Sigma $ determine the mapping uniquely [Reference Rudin11, pp. 290–291] to get

$$ \begin{align*}F(z)=z\end{align*} $$

on $\Sigma $ .

In the latter case, Lemma 2.1 gives

$$ \begin{align*}{\alpha_j} ={m}_j {\alpha}_1\end{align*} $$

for some positive integer ${m}_j$ . Therefore, ${\alpha }_j\geq {\alpha }_1$ . Using [Reference Huang and Zheng8, Lemma 3.1] again, we also have that there is a positive number ${k_j}'$ such that

$$ \begin{align*}{a}_1 = F^{{k_j}'}({a}_{j}),\end{align*} $$

similarly to get ${\alpha }_1 \geq {\alpha }_j$ . This forces that ${\alpha }_j={\alpha }_1.$ Thus, $\Sigma $ is equiangular, which is a contradiction to that $\Sigma $ is not equiangular. So we obtain $F(z)\equiv z$ to complete the proof. ▪

Proof Clearly, we need only to show that for two nonconstant functions g and h holomorphic on $\overline {\Sigma }$ , if $T_g$ is unitarily equivalent to $T_h$ on $L_a^2(\Sigma )$ , then $g=h$ .

By [Reference Huang and Zheng8, Proposition 3.2], we have that the von Neumann algebra $\mathcal {V}^*(h,\Sigma ) $ is trivial. Lemma 2.4 gives that there is a holomorphic function $F\in $ Aut $(\Sigma )$ such that $h=g\circ F$ . Hence, it suffices to prove $F(z)\equiv z$ . To do this, because $\alpha _j$ is not in $\mathbb {Q}\pi $ for each $ j\,(1\leq j\leq q)$ , we observe that the proof of Theorem 1.1 gives that $F $ maps the set of vertices $\{a_1,\ldots ,a_q\}$ of $\Sigma $ to itself. Since the conformality of F (F keeps the orientation on the boundary of $\Sigma $ ), if F fixes some vertex $a_j$ of $\Sigma $ for some j, then F fixes all vertices.

If F does not fix any vertices $a_j$ , by [Reference Huang and Zheng8, Lemma 3.1], we can find $k_j$ in $\{0, 1, \ldots , q-1\}$ such that

$$ \begin{align*}a_{j}=F^{k_j}(a_1).\end{align*} $$

Using Lemma 2.1, by induction, we have that $\alpha _j$ is a rational multiple of $\alpha _1$ . Because the sum $\sum _{j=1}^{q}\alpha _j=(q-2)\pi $ , $\alpha _1$ is a rational multiple of $\pi $ . This is a contradiction. Thus, F must fix some vertices and so $F(z)\equiv z$ . This completes the proof. ▪

Proof Becaue $\Sigma $ is a quadrilateral and $\Sigma $ is not a parallelogram, Theorem 1.4 in [Reference Huang and Zheng8] says that the von Neumann algebra $\mathcal {V}^*(h,\Sigma ) $ is trivial. By an application of Lemma 2.4, one obtains that there exists a holomorphic function $F\in $ Aut $(\Sigma )$ such that $h=g\circ F.$ Then, using the proof of Theorem 1.1 implies that $F $ maps the set of vertices $\{a_1,\ldots ,a_4\}$ (in the anticlockwise direction) of $\Sigma $ to itself. There are three cases to distinguish: $F(a_1)=a_1$ , $F(a_1)=a_2$ , or $a_4$ , $F(a_1)=a_3$ .

Case I: $F(a_1)=a_1$ . In this case, because F keeps the orientation on the boundary of $\Sigma $ , F fixes all vertices. Then, by Carathéodory’s theorem [Reference Rudin11, pp. 290–291],

$$ \begin{align*}F(z)=z\end{align*} $$

on $\Sigma $ .

Case II: $F(a_1)=a_2$ or $a_4$ . It suffices to discuss the case $F(a_1)=a_2$ , and the case $F(a_1)=a_4$ can be finished in a similar way. By conformality of F, $a_1=F^3(a_2)$ and $a_2 = F(a_1) .$ Then, there are two integers k and l such that $\alpha _1=k \alpha _2$ and $\alpha _2=l \alpha _1$ , forcing $\alpha _1=\alpha _2$ . Again by conformality of F, $a_3=F(a_2)$ and $a_4=F(a_3).$ By similar reasoning, $\alpha _4=\alpha _3=\alpha _2=\alpha _1$ , and this leads to that $\Sigma $ is a rectangle, a contradiction.

Case III: $F(a_1)=a_3$ . Then, by conformality of F,

$$ \begin{align*}F(a_1)=a_3, F(a_3)=a_1; F(a_2)=a_4, F(a_4)=a_2.\end{align*} $$

As done in Case II, it follows that $\alpha _1=\alpha _3$ and $\alpha _2=\alpha _4$ , forcing $\Sigma $ to be a parallelogram. This derives a contradiction.

Therefore, $ F(z)=z$ on $\Sigma $ , forcing $g=h $ to complete the proof. ▪

Proof of Theorem 1.3 Let $\Sigma $ be a regular n-polygon centered at $0$ with $3\leq n \leq 6$ . Suppose that both g and $h $ belong to $ H^\infty (\overline {\Sigma })$ , and $g'$ does not vanish at any vertices of $\Sigma $ . We need only to prove that if for some unitary operator U on $L_a^2(\Sigma )$ ,

$$ \begin{align*}U^*T_gU=T_h,\end{align*} $$

then $g(z)=h(\lambda z)$ for some constant $\lambda $ satisfying $\lambda ^n=1$ .

By Lemma 2.4, we get two holomorphic proper maps $\phi _1$ and $\phi _2$ from $\Sigma $ to $\Sigma $ , and a function $\tilde {g}$ in $ H^\infty ( \Sigma )$ such that

$$ \begin{align*}g=\tilde{g}\circ \phi_1 \quad \mathrm{and}\quad h=\tilde{g}\circ \phi_2,\end{align*} $$

and

$$ \begin{align*}U^*T_{\phi_1}U= T_{\phi_2}.\end{align*} $$

Let $\sigma $ be a Riemann mapping from $\Sigma $ to the unit disk $\mathbb {D}$ . Using Theorem 1.5 in [Reference Huang and Zheng8], $\sigma \circ \phi _{i}\circ \sigma ^{-1}$ is a Blaschke product with order less than or equal to $n\leq 6$ for $i=1, 2$ , because $\sigma \circ \phi _{i}\circ \sigma ^{-1}$ is a proper map on the unit disk and $\phi _i$ is a proper map with multiplicity not larger than n. By Theorem 3.1, there is a holomorphic function $F\in $ Aut $(\Sigma )$ such that $\phi _2=\phi _1\circ F $ . Thus, $h=g\circ F.$ To finish the proof, it remains to prove that there is a constant $\lambda $ satisfying $\lambda ^n=1$ and $F(z)= \lambda z$ .

To see this, let $a_1,a_2,\ldots ,a_n $ be the vertices of $\overline {\Sigma }$ in the anticlockwise direction. We claim that F maps vertices onto vertices. Otherwise, there is a point b on the edge such that $F(b)=a_1 $ . Let $\alpha _1$ be the interior angle of $\Sigma $ at the vertex $a_1$ . Because $g'$ does not vanish on any vertices of $\Sigma $ , as in the proof of Theorem 1.1, using Lemma 2.1, we get that $\alpha _1=k\pi $ for some positive integer $k,$ which is a contradiction. Thus, F maps vertices onto vertices. Because F keeps the orientation on $\partial \Sigma ,$ there is an integer s such that F maps $a_1, a_2,\ldots ,a_n $ to $a_{s }, \ldots , a_n,a_1,\ldots , a_{s-1}$ , respectively. Because a conformal map on a simply connected domain is uniquely determined by three boundary values and $\Sigma $ is a regular n-polygon centered at $0,$ there is a constant $\lambda $ such that $\lambda ^n=1$ and $F(z)= \lambda z$ . This completes the proof of Theorem 1.3. ▪

Proof of Theorem 1.2 Theorem 1.2 follows from Proposition 2.6 in the case that the quadrilateral is not a parallelogram. We need only consider the case that the quadrilateral is a parallelogram. Now, we may assume that $\Sigma $ is a parallelogram centered at zero and its vertices are $a_1, \ldots , a_4$ in anticlockwise direction.

It suffices to prove the “only if” part. To do this, we assume that there is a unitary operator U on $L_a^2(\Sigma )$ such that

$$ \begin{align*}UT_{g}U^*= T_{h}.\end{align*} $$

By Lemma 2.4, there exist two proper maps $\phi _1$ and $\phi _2$ from $\Sigma $ to $\Sigma $ , and a function $\tilde {g}$ in $ H^\infty ( \Sigma )$ such that

$$ \begin{align*}g=\tilde{g}\circ \phi_1 ,\, \quad h=\tilde{g}\circ \phi_2,\end{align*} $$

and

$$ \begin{align*}\mathcal{V}^*(g,\Sigma)=\mathcal{V}^*( \phi_1,\Sigma).\end{align*} $$

Let $\sigma $ be a Riemann mapping from $\Sigma $ onto $\mathbb {D}$ , and the finite Blaschke product $ \sigma \circ \phi _1\circ \sigma ^{-1}$ has the same multiplicity (order) as $\phi _1$ , and by Theorem 1.5 in [Reference Huang and Zheng8], order $\sigma \circ \phi _1\circ \sigma ^{-1}\leq 4.$ Similarly, the order of the finite Blaschke product $ \sigma \circ \phi _2\circ \sigma ^{-1}$ is not larger than $4.$ Using these facts and Theorem 3.1, as in the proof of Theorem 1.3, we get that there is a holomorphic function $F\in $ Aut $(\Sigma )$ such that $\phi _2=\phi _1\circ F,$ and hence

$$ \begin{align*}h=g\circ F.\end{align*} $$

Using Lemma 2.1, we obtain that F maps $\{a_1, \ldots , a_4\}$ to $\{a_1, \ldots , a_4\}$ .

If $\Sigma $ is a square, each F in Aut $(\Sigma )$ maps vertices $\{a_1, \ldots , a_4\}$ into $\{a_1, \ldots , a_4\}$ and keeps orientation. Thus, F must equal $F_{k}$ for some k, where

$$ \begin{align*}F_k(z)=i^k z.\end{align*} $$

So we get the desired result.

Let $F_0(z)=z $ and $F_2(z)=-z.$ Because $\Sigma $ is a parallelogram centered at zero, the maps $F_0$ and $F_2$ are members in Aut $(\Sigma )$ , which maps vertices of $\overline {\Sigma }$ to vertices. To complete the proof, we will show that if members in Aut $(\Sigma )$ mapping vertices of $\overline {\Sigma }$ to vertices are different from $F_0, F_2$ , then $\Sigma $ is a square.

Let F be one such member mapping vertices of $\overline {\Sigma }$ to vertices. Because $F(a_1)$ is in $ \{a_1, \ldots , a_4\}$ and F does not equal $F_0$ , $F(a_1)$ does not equal $a_1$ . To derive a contradiction, assume that $F(a_1)= a_3.$ Then, Lemma 3.1 in [Reference Huang and Zheng8] gives that

$$ \begin{align*}F(a_2)= a_4 \quad \mathrm{and}\quad F(a_3)= a_1.\end{align*} $$

On the other hand, the map $F_2$ also has the same boundary values as F, and thus, by the uniqueness of Riemann mapping, we have $F=F_2.$ This is a contradiction, and therefore $F(a_1)$ does not equal $a_3$ . This forces

$$ \begin{align*}F(a_1) = a_2 ~\mathrm{or}~ F(a_1) = a_4 .\end{align*} $$

We will consider only the case that $F(a_1) = a_2$ , because the same argument works for the case that $F(a_1) = a_4.$ In this case, we will show that $\Sigma $ is a square.

First, we show that $\Sigma $ is a rectangle. In this case, we have that F maps $a_1,a_2,a_3$ to $a_2,a_3,a_4$ , respectively. Let $\alpha _i$ be the interior angle of $\Sigma $ at the vertex $a_i$ for $i=1, \ldots , 4.$ Because $g'(a_2)\neq 0,$ the multiplicity $m_{g}(a_2)$ equals $1$ . By Lemma 2.1, we have

$$ \begin{align*}m_{g\circ F}(a_1)\alpha_1=m_{g}(a_2)\alpha_2.\end{align*} $$

This implies that there exists a positive number satisfying

$$ \begin{align*}\alpha_2 = k\alpha_1\geq\alpha_1.\end{align*} $$

Similarly, because $F(a_2)= a_3 $ , we have $ \alpha _3 =k_1\alpha _2$ for some positive integer $k_1$ . This gives that $ \alpha _3 \geq \alpha _2$ . Because $\Sigma $ is a parallelogram, $\alpha _1 =\alpha _3, $ and $\alpha _2=\alpha _4$ . Thus, we have

$$ \begin{align*}\alpha_1 =\alpha_2 =\alpha _3 =\alpha_4,\end{align*} $$

to get that $\Sigma $ is a rectangle.

Next, we will show that $\Sigma $ is a square. Because F maps $a_1,a_2,a_3$ to $a_2,a_3,a_4$ , respectively, F maps each edge of $\Sigma $ to another edge of $\Sigma $ . Then, by iteration applications of the Schwarz reflection principle, we can extend F to be a biholomorphic self-map of a rectangle $\Sigma '$ (four times size as that of $\Sigma $ ), which contains the original rectangle $\Sigma $ , with two edges of $\Sigma $ contained in $\partial \Sigma '.$ The extended map can be extended biholomorphically to a larger rectangle $\Sigma ''$ such that $\partial \Sigma ''$ covers two edges of $\Sigma $ . Because the complex plane can be tiled by rectangles congruent to the rectangle $\Sigma $ , again by the Schwarz reflection principle, F finally extends to be a biholomorphic function on the complex plane $\mathbb {C}$ . Thus, F is extended as a holomorphic automorphism of the complex plane. So

$$ \begin{align*}F(z)=az+b\end{align*} $$

for two complex numbers a and b. Noting that F maps $\Sigma $ to $\Sigma $ , we have that b must be zero. Because F maps $a_1,a_2,a_3$ to $a_2,a_3,a_4$ , respectively, we have $|a|=1$ ; that is, F is a rotation. In order to keep $\frac {\pi }{2}$ -interior angles of the rectangle, we conclude that $a=i^k$ for $k=0, 1, 2, 3.$ This gives that $\Sigma $ is a square to complete the proof. ▪