\section{Introduction}


For nonzero integers $m,n$, let
\begin{align*}
G_{m,n}&=\langle x,y,z\mid x=[z^m,x][z^n,y]\rangle, \\
H_{m,n}&=\langle x,y,z\mid x=[x^m,z^n][y,z]\rangle,
\end{align*}
where the commutator $[x,y]=x^{-1}y^{-1}xy$.
The first family was introduced by Baumslag in 1960s \cite{Ba67,Ba69}, and the second family later \cite{Ba94}. They are examples of parafree groups. A group is called {\it parafree} if it is residually nilpotent and its lower central series of quotients are the same as those of some free group.
It was shown in \cite{FRS97} that $G_{m,1}\ncong G_{1,1}$ for $m>1$ and $G_{m,1}\ncong G_{m',1}$ for distinct prime $m,m'$. Enumerating homomorphisms to finite groups was applied in \cite{BCH04,LL94}, but could only deduce results for finitely many pairs $(m,n)$. Recently, infinitely members in a subfamily of the $G$ family have been distinguished from each other \cite{HK20}. Results on the $H$ family is rarely seen.

There is an isomorphism $G_{m,n}\cong G_{-m,-n}$ given by $x\mapsto x, y\mapsto y, z\mapsto z^{-1}$, so we may always assume $n>0$.
There are isomorphisms $H_{m,n}\cong H_{-m-1,-n-1}$ and $H_{m,n}\cong H_{-m-1,n}$, given respectively by $x\mapsto x, y\mapsto yz^{n+1}x^{m+1}z^{-1}, z\mapsto z$, and $x\mapsto x^{-1}, y\mapsto yz^{1-n}, z\mapsto z^{-1}$.
Note that $H_{-1,n}\cong H_{m,-1}\cong F_2$ (the free group on two generators). So we only need to consider $m,n>0$.

In this paper, we completely solve the isomorphisms by showing
\begin{thm}\label{thm:solution-G}
For any integers $m,m',n,n'$ with $m,m'\ne 0$ and $n,n'>0$, if $G_{m,n}\cong G_{m',n'}$, then $m=m'$ and $n=n'$.
\end{thm}

\begin{thm}\label{thm:solution-H}
For any positive integers $m,m',n,n'$, if $H_{m,n}\cong H_{m',n'}$, then $m=m'$ and $n=n'$.
\end{thm}

This is achieved using character varieties of groups.

Given a finitely presented group $\Gamma$, let $\mathcal{R}^{\rm irr}(\Gamma)$ denote the space of irreducible representations $\Gamma\to{\rm GL}(2,\mathbb{C})$, on which ${\rm GL}(2,\mathbb{C})$ acts by conjugation. For each $\rho\in\mathcal{R}^{\rm irr}(\Gamma)$, its {\it character} $\chi_\rho$ is by definition the function $\Gamma\to\mathbb{C}$ sending $g$ to ${\rm tr}(\rho(g))$. Put $\mathcal{X}^{\rm irr}(\Gamma)=\{\chi_\rho\colon\rho\in\mathcal{R}^{\rm irr}(\Gamma)\}$, and call it the (irreducible) {\it ${\rm GL}(2,\mathbb{C})$-character variety} of $\Gamma$. It is known that $\rho,\rho'\in\mathcal{R}^{\rm irr}(\Gamma)$ are conjugate if and only if $\chi_\rho=\chi_{\rho'}$ (see \cite{Si12} and the references therein), so we may identify the conjugacy class $[\rho]$ with $\chi_\rho$. The morphism $\det_\ast:\mathcal{X}^{\rm irr}(\Gamma)\to\hom(\Gamma,\mathbb{C}^\ast)$ induced by $\det:{\rm GL}(2,\mathbb{C})\to\mathbb{C}^\ast$ is well-defined, and is natural in $\Gamma$.

For each $\rho\in\mathcal{R}^{\rm irr}(\Gamma)$, let $V_\rho=\mathbb{C}^2$, equipped with the $G$-module structure via $\rho$. Then $d^1(\rho):=\dim H^1(\Gamma;V_\rho)$ depends only on $[\rho]$; denote it by $d^1([\rho])$ or $d^1(\chi_\rho)$.
For $k\in\mathbb{N}$, let $\mathcal{J}_k(\Gamma)=\{\chi\in\mathcal{X}^{\rm irr}(\Gamma)\colon d^1(\chi)\ge k\}$. It is a subvariety of $\mathcal{X}^{\rm irr}(\Gamma)$, called the {\it cohomology jump locus} or {\it characteristic variety} in degree 1 and depth $k$ in the literature; see \cite{DP14,DPS09}, etc.

Here is the strategy for proving Theorem \ref{thm:solution-G} and Theorem \ref{thm:solution-H}.
If there exists an isomorphism $\phi:\Gamma\to\Gamma'$, then the following diagram commutes:
\begin{align}
\xymatrix{
&\mathcal{J}_3(\Gamma')\ar[r]^{\phi^\ast}\ar[d]_{\det_\ast}&\mathcal{J}_3(\Gamma)\ar[d]^{\det_\ast} \\
&\hom(\Gamma',\mathbb{C}^\ast)\ar[r]^{\phi^\ast}&\hom(\Gamma,\mathbb{C}^\ast)
}  \label{eq:natural-transformation}
\end{align}
In particular, $\det_\ast(\phi^\ast(\mathcal{C}))\cong\det_\ast(\mathcal{C})$ for each component $\mathcal{C}\subset\mathcal{J}_3(\Gamma')$, and $\det_\ast^{-1}(\phi^\ast(\mathfrak{a}))\cong\det_\ast^{-1}(\mathfrak{a})$ for each $\mathfrak{a}\in\hom(\Gamma',\mathbb{C}^\ast)$.
We are able to well understand the topologies of $\mathcal{J}_3(G_{m,n})$ and $\mathcal{J}_3(H_{m,n})$, and extract sufficiently much numerical information on $m,n$, to deduce $m=m',n=n'$ from the existence of an isomorphism $G_{m,n}\cong G_{m',n'}$ or $H_{m,n}\cong H_{m',n'}$.

In general, the whole character variety $\mathcal{X}^{\rm irr}(\Gamma)$ may be complicated. But $\mathcal{J}_k(\Gamma)$ is of lower dimensional, and is relatively easy to describe. The ${\rm GL}(2,\mathbb{C})$-character variety of a free group has a natural coordinate system via trace functions (see \cite{Dr03} for instance). For a 3-generator group $\Gamma$, we could describe $\mathcal{X}^{\rm irr}(\Gamma)$ as a subvariety of $\mathcal{X}^{\rm irr}(F_3)$. Nevertheless, we choose not to do so; instead, we choose an explicit representative for each conjugacy class of ${\rm GL}(2,\mathbb{C})$-representations, making computations more convenient. In this paper, ``varieties" are almost always treated just as subspaces of $\mathbb{C}^N$ for some $N$.


\begin{nota}
\rm Let $\mathbf{e}$ denote the identity matrix.
For $\lambda,\mu\in\mathbb{C}^\ast, \nu\in\mathbb{C}$, let
\begin{align*}
\mathbf{d}(\lambda,\mu)=\left(\begin{array}{cc} \lambda & 0 \\ 0 & \mu \end{array}\right), \qquad
\mathbf{p}(\lambda)=\left(\begin{array}{cc} \lambda & \lambda \\ 0 & \lambda \end{array}\right).
\end{align*}

For $\mathbf{s}\in{\rm GL}(2,\mathbb{C})$, denote its $(i,j)$-entry by $\mathbf{s}_{ij}$, and its $j$-th column by $\mathbf{s}_{\ast j}$.

For $k\in\mathbb{Z}$ and an element $\mathfrak{r}$ of some ring, let
$[k]_{\mathfrak{r}}=\sum_{j=0}^{k-1}\mathfrak{r}^j$ if $k>0$, $[k]_{\mathfrak{r}}=-\mathfrak{r}^{k}\sum_{j=0}^{|k|-1}\mathfrak{r}^j$ if $k<0$, and let $[0]_{\mathfrak{r}}=0$.

For a finite set $X$, let $\#X$ denote its cardinality.
\end{nota}


\section{The $G$ family}


In this section, let $g={\rm gcd}(m,n)$, the greatest common divisor of $m$ and $n$.
For a positive integer $k$, let $\Lambda_k=\{\exp(2j\pi\sqrt{-1}/k)\colon 1\le j\le k-1\}$.

The relation $x=[z^m,x][z^n,y]$ can be rewritten as
$$z^{n}w^{-1}z^{-m}w^2=y^{-1}z^{n}y, \qquad \text{with}\quad w=z^mx.$$
Given $\mathbf{z},\mathbf{w},\mathbf{y}\in{\rm GL}(2,\mathbb{C})$, there exists a representation $G_{m,n}\to{\rm GL}(2,\mathbb{C})$ sending $z,w,y$ respectively to $\mathbf{z},\mathbf{w},\mathbf{y}$ if and only if
\begin{align}
\mathbf{z}^{n}\mathbf{w}^{-1}\mathbf{z}^{-m}\mathbf{w}^2=\mathbf{y}^{-1}\mathbf{z}^{n}\mathbf{y}; \label{eq:G-relation}
\end{align}
denote this unique representation by $\rho_{\mathbf{z},\mathbf{w},\mathbf{y}}$. It is irreducible if and only if $\mathbf{z},\mathbf{w},\mathbf{y}$ have no common eigenvector.

Suppose $\rho=\rho_{\mathbf{z},\mathbf{w},\mathbf{y}}$ is an irreducible representation.

If $\mathfrak{d}:G_{m,n}\to V_\rho$ is a derivation,
with $\mathfrak{d}(w)=\xi_1, \mathfrak{d}(y)=\xi_2, \mathfrak{d}(z)=\xi_3$,
then
\begin{align*}
\mathfrak{d}(\mathbf{z}^n\mathbf{w}^{-1}\mathbf{z}^{-m}\mathbf{w}^2)
&=\mathbf{z}^n\mathbf{w}^{-1}\mathbf{z}^{-m}(\mathbf{e}+\mathbf{w}-\mathbf{z}^m)\xi_1
+([n]_{\mathbf{z}}+\mathbf{z}^n\mathbf{w}^{-1}[-m]_{\mathbf{z}})\xi_3, \\
\mathfrak{d}(\mathbf{y}^{-1}\mathbf{z}^n\mathbf{y})&=\mathbf{y}^{-1}(\mathbf{z}^n-\mathbf{e})\xi_2+\mathbf{y}^{-1}[n]_{\mathbf{z}}\xi_3.
\end{align*}
Hence the space $\mathfrak{D}_\rho$ of derivations $G_{m,n}\to V_\rho$ can be identified with that of triples $(\xi_1,\xi_2,\xi_3)$ satisfying
\begin{align}
&\mathbf{z}^n\mathbf{w}^{-1}\mathbf{z}^{-m}(\mathbf{e}+\mathbf{w}-\mathbf{z}^m)\xi_1+\mathbf{y}^{-1}(\mathbf{e}-\mathbf{z}^n)\xi_2 \nonumber \\
&\qquad\quad +(\mathbf{z}^n\mathbf{w}^{-1}[-m]_{\mathbf{z}}+[n]_{\mathbf{z}}-\mathbf{y}^{-1}[n]_{\mathbf{z}})\xi_3=0.     \label{eq:relation0}
\end{align}
Under this identification, the subspace of inner derivations
$$\mathfrak{I}_\rho=\{((\mathbf{w}-\mathbf{e})\eta,(\mathbf{y}-\mathbf{e})\eta,(\mathbf{z}-\mathbf{e})\eta)\colon\eta\in\mathbb{C}^2\}\cong \mathbb{C}^2,$$
due to the irreducibility of $\rho$.
According to the basic fact $H^1(G_{m,n};V_{\rho})\cong\mathfrak{D}_\rho/\mathfrak{I}_\rho$ (see \cite{HS97} Chap. VI, Corollary 5.2), $\chi_\rho\in\mathcal{J}_3(G_{m,n})$ if and only if $\dim\mathfrak{D}_\rho\ge 5$.


Multiplied by $\mathbf{z}^m\mathbf{w}\mathbf{z}^{-n}$ on the left, (\ref{eq:relation0}) becomes
$\mathbf{a}\xi_1+\mathbf{c}\xi_3=\mathbf{b}(\mathbf{y}^{-1}\xi_2)$, with
\begin{align*}
\mathbf{a}&=\mathbf{e}+\mathbf{w}-\mathbf{z}^m, \\
\mathbf{b}&=\mathbf{w}^2-\mathbf{z}^m\mathbf{w}\mathbf{z}^{-n}, \\
\mathbf{c}&=\mathbf{z}^m[-m]_{\mathbf{z}}+\mathbf{z}^m\mathbf{w}\mathbf{z}^{-n}[n]_{\mathbf{z}}
-\mathbf{w}^2\mathbf{y}^{-1}\mathbf{z}^{-n}[n]_{\mathbf{z}}.
\end{align*}
Hence $\dim \mathfrak{D}_\rho\ge 5$ if and only if ${\rm rank}(\mathbf{a},\mathbf{b},\mathbf{c})\le 1$.

Regarding (\ref{eq:relation0}), a necessary condition for $\dim \mathfrak{D}_\rho\ge 5$ is $\det(\mathbf{z}^n-\mathbf{e})=0$. Also necessary is $\mathbf{z}^n\ne\mathbf{e}$: otherwise, by (\ref{eq:G-relation}), $\mathbf{w}=\mathbf{z}^m$ so that $\mathbf{a}=\mathbf{e}$.
Therefore, up to conjugacy we may assume $\mathbf{z}=\mathbf{d}(\lambda,\zeta)$ with $\zeta^n=1\ne\lambda^n$, or $\mathbf{z}=\mathbf{p}(\zeta)$ with $\zeta^n=1$.
Suppose
$$\mathbf{w}=\left(\begin{array}{cc} a & b \\ c & d \end{array}\right), \qquad
\mathbf{y}=\left(\begin{array}{cc} a' & b' \\ c' & d' \end{array}\right).$$
with $ad-bc=\lambda^m\zeta^m$, and $t:=a'd'-b'c'\ne 0$.

\begin{rmk} \label{rmk:G-condition}
\rm Now that $\mathfrak{I}_\rho\subset\mathfrak{D}_\rho$, we have
$$\mathbf{c}(\mathbf{z}-\mathbf{e})\eta=\mathbf{a}(\mathbf{e}-\mathbf{w})\eta+\mathbf{b}(\mathbf{e}-\mathbf{y}^{-1})\eta$$
for all $\eta\in\mathbb{C}^2$.
This means, when $\zeta\ne 1$ so that $\mathbf{z}-\mathbf{e}$ is invertible, ${\rm rank}(\mathbf{a},\mathbf{b})\le 1$ has been sufficient,
and when $\zeta=1$, the condition is ${\rm rank}(\mathbf{a},\mathbf{b},\mathbf{c}_{\ast2})\le 1$.
\end{rmk}


\subsection{$\mathbf{z}=\mathbf{d}(\lambda,\zeta)$ with $\zeta^n=1\ne\lambda^{n}$}

Let $\mathbf{v}=\mathbf{z}^n\mathbf{w}^{-1}\mathbf{z}^{-m}\mathbf{w}^2$ for short.
Let $r=a+d$, and introduce $\vartheta_k=\lambda^k-1$ for each $k\in\mathbb{Z}$.

There exists $\mathbf{y}$ with $\mathbf{v}=\mathbf{y}^{-1}\mathbf{z}^n\mathbf{y}$ if and only if ${\rm tr}(\mathbf{v})={\rm tr}(\mathbf{z}^n)$; equivalently,
\begin{align}
\vartheta_n(\lambda^{-m}-\zeta^{-m})rad+(\lambda^{m}\vartheta_n+\zeta^m)a+(\lambda^{n+m}-\vartheta_n\zeta^m)d=(\lambda^n+1)\lambda^m\zeta^m. \label{eq:trace}
\end{align}
We can write $\mathbf{y}\mathbf{v}=\mathbf{z}^n\mathbf{y}$ as
\begin{align}
(a',b')(\mathbf{v}-\lambda^n\mathbf{e})=(c',d')(\mathbf{v}-\mathbf{e})=0.  \label{eq:determine-y}
\end{align}
Write $(\mathbf{a},\mathbf{b})$ explicitly as
$$\left(\begin{array}{cccc}
a+1-\lambda^m & b & (r-\lambda^{m-n})a-\lambda^m\zeta^m & (r-\lambda^{m})b \\
c & d+1-\zeta^m & (r-\zeta^m\lambda^{-n})c & (r-\zeta^{m})d-\lambda^m\zeta^m
\end{array}\right).$$
Clearly, ${\rm rank}(\mathbf{a},\mathbf{b})\le 1$ is equivalent to
\begin{align*}
{\rm rank}\left(\begin{array}{cccc} a+1-\lambda^m & b & \kappa_1 & 0 \\
c & d+1-\zeta^m & 0 & \kappa_2 \end{array}\right)\le 1,
\end{align*}
where
\begin{align*}
\kappa_1&=(\zeta^m\lambda^{-n}-\lambda^{m-n}+\lambda^m-1)a+(\lambda^m-1)d+\zeta^m(\lambda^{-n}-\lambda^{m-n}-\lambda^{m}), \\
\kappa_2&=(\zeta^m-1)a+(\lambda^m-1)d+\lambda^m(1-2\zeta^m).
\end{align*}

If $\kappa_1\ne 0$, then $c=0$, $d=\zeta^m-1$, and $\kappa_2=0$, i.e. $$0=\lambda^m\zeta^m+(\lambda^{m}-1)(\zeta^m-1)+\lambda^m(1-2\zeta^m)=1-\zeta^m=-d,$$
which is absurd.

Thus $\kappa_1=0$. There are two possibilities: $\kappa_2=0$, or $b=a+1-\lambda^m=0$.


\subsubsection{$\kappa_2=0$}


The condition $\det(\mathbf{a})=0$ reads
$$(1-\zeta^{m})a+(1-\lambda^{m})d+(2\lambda^m\zeta^m-\lambda^m-\zeta^{m}+1)=0,$$
which together with $\kappa_2=0$ implies $\zeta^m=1$ so that $\zeta^g=1$, and
$d=\lambda^m\vartheta_m^{-1}$. Then $\kappa_1=0$ becomes $a=\vartheta_n^{-1}$. As a result, $r=\vartheta_{m+n}\vartheta_m^{-1}\vartheta_n^{-1}$,
and
$$\mathbf{v}=\left(\begin{array}{cc}
\lambda^nr-\lambda^{n-m}ra-\lambda^{n-m}d & \lambda^n(1-\lambda^{-m}r)b \\
\lambda^{-m}(1+(1-\lambda^{-m})ra)c & \lambda^{-m}ra+\lambda^{-m}r-a \end{array}\right).$$

It turns out that (\ref{eq:trace}) has been fulfilled.

Observe that $(a,b)(\mathbf{v}-\mathbf{e})=0$, hence $(c',d')\parallel(a,b)$.
Fix the conjugacy indeterminacy of $\rho$ by setting $c'=1$, $d'=\vartheta_nb$.
As can be verified, $b'=b=0$ never occur, so the irreducibility of $\rho_{\mathbf{z},\mathbf{w},\mathbf{y}}$ is always ensured.

When $\zeta\ne 1$, there is no further constraint.
Let $\mathcal{F}_\zeta$ denote the subspace of $\mathcal{R}^{\rm irr}(G_{m,n})$ consisting of $\rho_{\mathbf{z},\mathbf{w},\mathbf{y}}$'s, with
\begin{align}
\mathbf{z}=\left(\begin{array}{cc} \lambda & 0 \\ 0 & \zeta \end{array}\right), \qquad
\mathbf{w}=\left(\begin{array}{cc} \vartheta_n^{-1} & b \\ c & \lambda^{m}\vartheta_m^{-1} \end{array}\right), \qquad
\mathbf{y}=\left(\begin{array}{cc} a' & b' \\ 1 & \vartheta_nb \end{array}\right), \label{eq:Z-W-Y}
\end{align}
such that
\begin{align}
\lambda^n\ne 1, \qquad \lambda^m\ne 1, \label{eq:F-1} \\
bc=\lambda^m(\vartheta_n^{-1}\vartheta_m^{-1}-1), \label{eq:F-2} \\
(a',b')(\mathbf{v}-\lambda^n\mathbf{e})=0, \qquad (a',b')\ne 0. \label{eq:F-3}
\end{align}

\begin{lem}\label{lem:f}
$f(\lambda):=\vartheta_n\vartheta_m-1=\lambda^{n+m}-\lambda^{n}-\lambda^m$ has no multiple root, so the number of distinct roots of $f$ is $\ell_{m,n}:=\max\{m,n,n-m\}$.
\end{lem}
\begin{proof}
Assume $f(\lambda)=f'(\lambda)=0$, i.e. $\lambda^{n+m}=\lambda^{n}+\lambda^m$ and $(n+m)\lambda^{n+m}=n\lambda^n+m\lambda^m$.
Then
$\lambda^m=1-m/n$, $\lambda^n=1-n/m$. Clearly, $m\ne n$.
\begin{itemize}
  \item If $m>n$, then $|\lambda|^n<1$, implying $|\lambda|<1$, so $m/n-1=|\lambda|^m<|\lambda|^n=1-n/m$, which is absurd.
  \item If $0<m<n$, then $|\lambda|^m<1$, implying $|\lambda|<1$, so $1-m/n=|\lambda|^m<|\lambda|^n=n/m-1$, which is also absurd.
  \item If $m<0$, then $|\lambda|^m>1$, implying $|\lambda|<1$, but $|\lambda|^n>0$, a contradiction.
\end{itemize}
\end{proof}

When $\zeta=1$,
\begin{align*}
\mathbf{c}_{\ast2}=\left(\begin{array}{cc}
n\lambda^{m}b+nt^{-1}(r(b'a-a'b)-\lambda^mb') \\ nt^{-1}(r(b'c-a'd)+\lambda^ma')+nw-m \end{array}\right);
\end{align*}
it is parallel to $\mathbf{a}_{\ast2}=(b,d)^t$ if and only if $a'b-b'a=(1+m\lambda^{-m}/n)bt$, so
\begin{align}
b=\frac{(a'd'-b')a}{(1+m\lambda^{-m}/n)t}=\frac{n\lambda^{m}}{(n\lambda^{m}+m)\vartheta_n}.  \label{eq:y}
\end{align}
Hence $\lambda^{m}\ne -m/n$ is required.
Let $\mathcal{F}_1$ denote the subspace of $\mathcal{R}^{\rm irr}(G_{m,n})$ consisting of $\rho_{\mathbf{z},\mathbf{w},\mathbf{y}}$'s given by (\ref{eq:Z-W-Y}), with (\ref{eq:F-1}), (\ref{eq:F-2}), (\ref{eq:F-3}), (\ref{eq:y}) and $\lambda^m\ne -m/n$.





\subsubsection{$b=a+1-\lambda^m=0$}

In this case, $a=\vartheta_m$, $d=\lambda^m\zeta^m\vartheta_m^{-1}$, and $\kappa_1=0$ implies
$f(\lambda)=0$.

As a result, $a=\lambda^{m-n}, d=\lambda^n\zeta^m$. Then (\ref{eq:trace}) turns out to hold.

By direct computation,
$$\mathbf{v}=\left(\begin{array}{cc} 1 & 0 \\ \zeta^{-m}(\lambda^{m-2n}\zeta^{-m}-\vartheta_{-2n})c & \lambda^{n} \end{array}\right).$$
It follows from (\ref{eq:determine-y}) that $d'=0$, and
\begin{align*}
(a',b')\parallel((\lambda^{m-2n}\zeta^{-m}-\vartheta_{-2n})c,\zeta^{m}\vartheta_n).
\end{align*}

Fix the conjugacy indeterminacy by setting $c'=1$.
If $\zeta=1$, then by computation, $\mathbf{c}_{12}=nb'a^2\ne 0$, violating ${\rm rank}(\mathbf{a},\mathbf{b},\mathbf{c}_{\ast2})\le 1$. Hence $\zeta\ne 1$.

For each $\zeta\in\Lambda_n$ and each $\lambda\in f^{-1}(0)$, let $\mathcal{G}_{\zeta}^\lambda$ denote the subspace of $\mathcal{R}^{\rm irr}(G_{m,n})$ consisting of $\rho_{\mathbf{z},\mathbf{w},\mathbf{y}}$'s determined by
$$\mathbf{z}=\left(\begin{array}{cc}  \lambda & 0 \\ 0 & \zeta  \end{array}\right), \qquad
\mathbf{w}=\left(\begin{array}{cc}  \lambda^{m-n} & 0 \\ c & \lambda^{n}\zeta^m \end{array}\right), \qquad
\mathbf{y}=\left(\begin{array}{cc}  a' & b' \\ 1 & 0 \end{array}\right),$$
with $(a',b')$ a nonzero multiple of $((\lambda^{m-2n}\zeta^{-m}-\vartheta_{-2n})c,\zeta^{m}\vartheta_n)$.

When $\zeta^m=1$ (i.e. $\zeta\in\Lambda_g$), $\mathcal{G}_{\zeta}^\lambda$ is contained in $\mathcal{F}_{\zeta}$. Thus what we newly find are $(n-g)\ell_{m,n}$ components, each of which is isomorphic to $\mathbb{C}\times\mathbb{C}^\ast$.


\subsection{$\mathbf{z}=\mathbf{p}(\zeta)$ with $\zeta^n=1$} \label{sec:G-p}


Introduce some more notations for convenience.
Let $\mathbf{s}\lrcorner\mathbf{q}$ denote $\mathbf{s}\mathbf{q}\mathbf{s}^{-1}$.
For $\lambda,\mu\in\mathbb{C}^\ast, \nu\in\mathbb{C}$, let
$$\mathbf{t}(\lambda,\mu;\nu)=\left(\begin{array}{cc} \lambda & \nu \\ 0 & \mu \end{array}\right).$$


If $c=0$, then $(\mathbf{z}^{n}\mathbf{w}^{-1}\mathbf{z}^{-m}\mathbf{w}^2)_{21}=(\mathbf{y}^{-1}\lrcorner\mathbf{z}^n)_{21}$
would imply $c'=0$, too, contradicting the irreducibility of $\rho$. Hence $c\ne 0$.

Since
$(\mathbf{t}(1,1;-ac^{-1})\lrcorner\mathbf{w})_{11}=0$, we may just assume $a=0$ at the beginning.
This also fixes the conjugacy indeterminacy.

Write $(\mathbf{a},\mathbf{b})$ explicitly as
$$\left(\begin{array}{cccc}
1-\zeta^m & b-m\zeta^m & -\zeta^{m}(\zeta^m+mc) & (d-\zeta^m)b+\zeta^mm(nc-d) \\
c & d+1-\zeta^m & (d-\zeta^m)c & d^2-\zeta^{2m}+\zeta^m(nc-d) \end{array}\right).$$
The conditions $\mathbf{a}_{\ast1}\parallel\mathbf{a}_{\ast2}$ and $\mathbf{a}_{\ast1}\parallel\mathbf{b}_{\ast1}$ respectively read
\begin{align*}
(\zeta^m-1)(d+1-\zeta^m)&=\zeta^{2m}+m\zeta^mc, \\
(\zeta^m-1)(d-\zeta^m)&=\zeta^{2m}+m\zeta^mc,
\end{align*}
which imply $\zeta^m=1$, and $c=-1/m$, so that $b=m$. Furthermore, $\mathbf{b}_{\ast2}\parallel\mathbf{a}_{\ast1}$ forces $m=-n$.

Now
$\mathbf{z}^{n}\mathbf{w}^{-1}\mathbf{z}^{-m}\mathbf{w}^2=\mathbf{y}^{-1}\lrcorner\mathbf{z}^{n}$ becomes
$$\left(\begin{array}{cc} 1-d+d^2 & nd^3-nd^2 \\ (1-d)/n & 1+d-d^2 \end{array}\right)
=\left(\begin{array}{cc} 1+nc'd't^{-1} & n(d')^2t^{-1} \\ -n(c')^2t^{-1} & 1-nc'd't^{-1} \end{array}\right),$$
which is equivalent to $t(d-1)=n^2(c')^2$, $d'=nc'd$, forcing $c'\ne 0$ and $d\ne 1$.

When $\zeta\ne 1$, let
$$\mathcal{H}_\zeta=\{\sigma_{a',c',d}\colon a'\in\mathbb{C}, c'\in\mathbb{C}^\ast, d\in\mathbb{C}-\{1\}\}\subset\mathcal{R}^{\rm irr}(G_{m,n}),$$
where
$\sigma_{a',c',d}=\rho_{\mathbf{z},\mathbf{w},\mathbf{y}}$ with
\begin{align*}
\mathbf{z}=\mathbf{p}(\zeta), \qquad
\mathbf{w}=\left(\begin{array}{cc} 0 & -n \\ 1/n & d \end{array}\right), \qquad
\mathbf{y}=\left(\begin{array}{cc} a & na'd-n^2c'(d-1)^{-1} \\ c' & nc'd \end{array}\right).
\end{align*}

When $\zeta=1$,
\begin{align*}
\mathbf{c}=(\mathbf{e}+\mathbf{z}^{-n}\mathbf{w}-\mathbf{w}^2\mathbf{y}^{-1})\mathbf{z}^{-n}[n]_{\mathbf{z}}
=\left(\begin{array}{cc} 0 & (c')^{-1}-n(d+1) \\ \star & \star \end{array}\right)\left(\begin{array}{cc} n & \star \\ 0 & n \end{array}\right),
\end{align*}
where the $\star$'s stand for things that are irrelevant to us, hence $\mathbf{c}_{12}=0$ is equivalent to $nc'(d+1)=1$.
Let
$$\mathcal{H}_1=\{\sigma_{a',c',(nc')^{-1}-1}\colon a'\in\mathbb{C},\ c'\in\mathbb{C}^\ast-\{1/2n\}\}.$$



\subsection{The result}


Recall $f(\lambda)=\lambda^{n+m}-\lambda^n-\lambda^m$ and $\ell_{m,n}=\max\{m,n,n-m\}$ in Lemma \ref{lem:f}.

Identify $\hom(G_{m,n},\mathbb{C}^\ast)$ with $\mathbb{C}^\ast\times\mathbb{C}^\ast$, via $\tau\mapsto(\tau(z),\tau(y))$.

When $m\ne -n$,
$$\mathcal{J}_3(G_{m,n})=\chi(\mathcal{F}_1)\sqcup\bigsqcup_{\zeta\in\Lambda_g}\chi(\mathcal{F}_\zeta)
\sqcup\bigsqcup_{\zeta\in\Lambda_n-\Lambda_g}\bigsqcup_{\lambda\in f^{-1}(0)}\chi(\mathcal{G}_\zeta^\lambda),$$
as decomposed into connected components, where
\begin{align*}
\chi(\mathcal{F}_1)\cong\{\lambda\in\mathbb{C}^\ast\colon\lambda^n\ne 1,\lambda^m\ne 1,\lambda^m\ne-m/n\}\times\mathbb{C}^\ast,
\end{align*}
$\chi(\mathcal{F}_\zeta)$ is a $\mathbb{C}^\ast$-bundle over a complex surface for each $\zeta\in\Lambda_g$, and
$\chi(\mathcal{G}_\zeta^\lambda)\cong\mathbb{C}\times\mathbb{C}^\ast$ for each $\zeta\in\Lambda_n$ and $\lambda\in f^{-1}(0)$.

More intricately,
\begin{align*}
{\det}_{\ast}(\chi(\mathcal{F}_\zeta))
&=\{(\lambda\zeta,t)\colon\lambda\in\mathbb{C}^\ast,\lambda^n\ne 1,\lambda^m\ne 1,t\in\mathbb{C}^\ast\}\subset\hom(G_{m,n},\mathbb{C}^\ast),
\end{align*}
and for each $\mathfrak{a}=(\lambda\zeta,t)\in{\det}_{\ast}(\chi(\mathcal{F}_\zeta))$, the fiber
$${\det}_\ast^{-1}(\mathfrak{a})\cong\begin{cases}
\mathbb{C}^\ast,&f(\lambda)\ne 0, \\
\{(b,c)\in\mathbb{C}\colon bc=0\},&f(\lambda)=0.
\end{cases}$$
Hence the subspace $\{\mathfrak{a}\colon\det_\ast^{-1}(\mathfrak{a})\ncong\mathbb{C}^\ast\}$ consists of $d_{m,n}$ copies of $\mathbb{C}^\ast$.

\medskip

When $m=-n$, the decomposition into connected components is
$$\mathcal{J}_3(G_{-n,n})=(\chi(\mathcal{F}_1)\cup\chi(\mathcal{H}_1))
\sqcup\bigsqcup_{\zeta\in\Lambda_n}(\chi(\mathcal{F}_\zeta)\cup\chi(\mathcal{H}_\zeta)).$$
To us, the most significant point is that
$\{\mathfrak{a}\colon\dim\det_\ast^{-1}(\mathfrak{a})=2\}$ consists of $n-1$ copies of $\mathbb{C}^\ast$.
So we merely sketch the reason for that $\chi(\mathcal{F}_\zeta)\cup\chi(\mathcal{H}_\zeta)$ is connected for each $\zeta$ with $\zeta^n=1$.

Refer to (\ref{eq:Z-W-Y})--(\ref{eq:F-3}) and note that $r=0$. Let $\varrho_{\lambda,a',c'}=\rho_{\mathbf{z},\mathbf{w},\mathbf{y}}$ with
$\mathbf{z}=\mathbf{d}(\lambda,\zeta)$,
\begin{align*}
\mathbf{w}=\left(\begin{array}{cc} \vartheta_n^{-1} & -\vartheta_n^{-2}nc' \\ u^{-1} & -\vartheta_n^{-1} \end{array}\right), \qquad
\mathbf{y}=\left(\begin{array}{cc} a'+\vartheta_n^{-1}u & -a'\vartheta_n^{-1}u-u^2 \\ 1 & -\vartheta_n^{-1}nc' \end{array}\right),
\end{align*}
where $u=nc'(\lambda^n+\lambda^{-n}-1)^{-1}$. Then $\varrho_{\lambda,a',c'}\in\mathcal{F}_\zeta$ for $\lambda$ sufficiently near $\zeta$.
With $\mathbf{h}=\mathbf{t}(-\zeta\vartheta_nu^{-1},\zeta-\lambda;\zeta)$,
we have $\mathbf{h}\lrcorner\mathbf{z}=\mathbf{t}(\lambda,\zeta;\zeta)$,
\begin{align*}
\mathbf{h}\lrcorner\mathbf{w}&=\left(\begin{array}{cc} 0 & -\lambda^{-n}[n]_{\lambda\zeta^{-1}} \\ ([n]_{\lambda\zeta^{-1}})^{-1} & 0 \end{array}\right), \\
\mathbf{h}\lrcorner\mathbf{y}
&=\left(\begin{array}{cc} a'-\vartheta_n^{-1}u & \lambda^{-n}u[n]_{\lambda\zeta^{-1}} \\
u([n]_{\lambda\zeta^{-1}})^{-1} & -\lambda^{-n}\vartheta_nu \end{array}\right).
\end{align*}
Hence $\lim_{\lambda\to\zeta}[\varrho_{\lambda,a',c'}]=[\sigma_{a',c',0}]$ in $\chi(\mathcal{F}_\zeta)\cup\chi(\mathcal{H}_\zeta)$.



\bigskip

\begin{proof}[Proof of Theorem \ref{thm:solution-G}]
Suppose there exists an isomorphism $\phi:G_{m,n}\to G_{m',n'}$. We apply the naturality of $\det_\ast$ as displayed in (\ref{eq:natural-transformation}).

If there exists $\mathfrak{a}\in\hom(G_{m,n},\mathbb{C}^\ast)$ such that $\dim\det_\ast^{-1}(\mathfrak{a})=2$, then $m=-n$, and
$\#\pi_0(\{\mathfrak{a}\colon\dim\det_\ast^{-1}(\mathfrak{a})=2\})=n-1$. This should also occur for $G_{m',n'}$, so $m'=-n'$ and $n'=n$.

Suppose $m\ne -n$ and $m'\ne -n'$. Let $g'={\rm gcd}(m',n')$.
\begin{itemize}
  \item The number of 3-dimensional components of $\mathcal{J}_3(G_{m,n})$ coincides with that of $\mathcal{J}_3(G_{m',n'})$, so $g=g'$.
  \item For each 3-dimensional component $\mathcal{C}\subset\mathcal{J}_3(G_{m',n'})$,
        $$\mathbb{Z}^{n'+|m'|-g'}\cong H_1({\det}_\ast(\mathcal{C}))\cong H_1({\det}_\ast(\phi^\ast\mathcal{C}))\cong\mathbb{Z}^{n+|m|-g},$$
        implying $n+|m|=n'+|m'|$.
        Moreover,
        \begin{align*}
        \ell_{m',n'}&=\#\pi_0(\{\mathfrak{a}\in{\det}_\ast(\mathcal{C})\colon{\det}_\ast^{-1}(\mathfrak{a})\ncong\mathbb{C}^\ast\}) \\
        &=\#\pi_0(\{\mathfrak{b}\in{\det}_\ast(\phi^\ast\mathcal{C})\colon{\det}_\ast^{-1}(\mathfrak{b})\ncong\mathbb{C}^\ast\})=\ell_{m,n}.
        \end{align*}
  \item Comparing the numbers of components which are isomorphic to $\mathbb{C}\times\mathbb{C}^\ast$, we obtain $(n-g)\ell_{m,n}=(n'-g')\ell_{m',n'}$.
\end{itemize}
From all of these, we deduce $n=n'$ and $m=m'$.
\end{proof}


\section{The $H$ family}


Use $u=yz$ to present $H$ as
\begin{align*}
H_{m,n}
\langle x,u,z\mid z^{-n-1}x^{-m}z^nx^{m+1}=u^{-1}z^{-1}u\rangle.
\end{align*}
Given $\mathbf{z},\mathbf{x},\mathbf{u}\in{\rm GL}(2,\mathbb{C})$, there exists a representation $H_{m,n}\to{\rm GL}(2,\mathbb{C})$ sending $z,x,u$ respectively to $\mathbf{z},\mathbf{x},\mathbf{u}$ if and only if
\begin{align}
\mathbf{v}:=\mathbf{z}^{-n-1}\mathbf{x}^{-m}\mathbf{z}^n\mathbf{x}^{m+1}=\mathbf{u}^{-1}\mathbf{z}^{-1}\mathbf{u}; \label{eq:H-relation}
\end{align}
denote this unique representation by $\rho_{\mathbf{z},\mathbf{x},\mathbf{u}}$. It is irreducible if and only if $\mathbf{z},\mathbf{x},\mathbf{u}$ do not share an eigenvector.

Suppose $\rho=\rho_{\mathbf{z},\mathbf{x},\mathbf{u}}$ is irreducible. Clearly, $\mathbf{z}\ne\mathbf{e}$.


If $\mathfrak{d}:H_{m,n}\to V_\rho$ is a derivation, with $\mathfrak{d}(x)=\xi_1, \mathfrak{d}(u)=\xi_2, \mathfrak{d}(z)=\xi_3$,
then
\begin{align*}
\mathfrak{d}(\mathbf{v})&=\mathbf{z}^{-n-1}\mathbf{x}^{-m}\mathbf{g}\xi_1+\mathbf{h}\xi_3,  \\
\mathfrak{d}(\mathbf{u}^{-1}\mathbf{z}^{-1}\mathbf{u})&=\mathbf{u}^{-1}(\mathbf{z}^{-1}-\mathbf{e})\xi_2-\mathbf{u}^{-1}\mathbf{z}^{-1}\xi_3,
\end{align*}
where
\begin{align*}
\mathbf{g}&=\mathbf{z}^n[m+1]_{\mathbf{x}}-[m]_{\mathbf{x}}, \\
\mathbf{h}&=\mathbf{z}^{-n-1}\mathbf{x}^{-m}[n]_{\mathbf{z}}+[-n-1]_{\mathbf{z}}.
\end{align*}
Hence the space $\mathfrak{D}_\rho$ of derivations $H_{m,n}\to V_\rho$ can be identified with that of triples $(\xi_1,\xi_2,\xi_3)$ satisfying
$$\mathbf{u}\mathbf{z}^{-n-1}\mathbf{x}^{-m}\mathbf{g}\xi_1+(\mathbf{e}-\mathbf{z}^{-1})\xi_2+(\mathbf{z}^{-1}+\mathbf{u}\mathbf{h})\xi_3=0.$$

Similarly as in Section 2, $\chi_\rho\in\mathcal{J}_3(H_{m,n})$ if and only if $\dim\mathfrak{D}_\rho\ge 5$, which is equivalent to 
\begin{align}
{\rm rank}(\mathbf{u}\mathbf{z}^{-n-1}\mathbf{x}^{-m}\mathbf{g},\mathbf{e}-\mathbf{z}^{-1},\mathbf{z}^{-1}+\mathbf{u}\mathbf{h})\le 1. \label{ineq:H-rank}
\end{align}
In particular, $\det(\mathbf{z}-\mathbf{e})=0$, so up to conjugacy we may assume $\mathbf{z}=\mathbf{d}(\lambda,1)$ with $\lambda \ne 1$, or $\mathbf{z}=\mathbf{p}(1)$.

As a necessary condition, ${\rm rank}(\mathbf{e}-\mathbf{z}^{-1},\mathbf{z}^{-1}+\mathbf{u}\mathbf{h})\le 1$ is the same as
\begin{align}
(c,d)\mathbf{h}=(0,-1).  \label{eq:H-rank=1}
\end{align}

\begin{rmk} \label{rmk:H-condition}
\rm For a reason similar as in Remark \ref{rmk:G-condition}, (\ref{eq:H-rank=1}) will be also sufficient, as long as $\mathbf{x}-\mathbf{e}$ is invertible.
\end{rmk}

Suppose
$$\mathbf{x}=\left(\begin{array}{cc} a & b \\ c & d \end{array}\right), \qquad
\mathbf{u}=\left(\begin{array}{cc} a' & b' \\ c' & d' \end{array}\right),$$
with $ad-bc=1$, and $a'd'-b'c'\ne 0$.

Let $\mu+\mu^{-1}=r=a+d$. For $k\in\mathbb{Z}$, let
\begin{align*}
\gamma_k=\gamma_k(r)=\begin{cases}
(\mu^k-\mu^{-k})/(\mu-\mu^{-1}), &\mu\notin\{\pm 1\}, \\
k\mu^{k-1}, &\mu\in\{\pm 1\}.
\end{cases}
\end{align*}
We have $\mathbf{x}^k=\gamma_k\mathbf{x}-\gamma_{k-1}\mathbf{e}$.
The following identities are useful:
\begin{align}
\gamma_{k+1}-r\gamma_k+\gamma_{k-1}=0, \nonumber  \\
\gamma_{k+1}^2+\gamma_k^2-r\gamma_{k+1}\gamma_k=1, \label{eq:identity-2} \\
\gamma_{k}^2-\gamma_{k+1}\gamma_{k-1}=1, \nonumber  \\
\gamma_{k+1}\gamma_k-\gamma_{k+2}\gamma_{k-1}=r. \nonumber
\end{align}
An alternative of (\ref{eq:identity-2}) is
\begin{align}
(2-r)\gamma_{k+1}\gamma_k=(1+\gamma_k-\gamma_{k+1})(1+\gamma_{k+1}-\gamma_k). \label{eq:identity-2'}
\end{align}

\begin{rmk} \label{rmk:gamma-r}
\rm If $\gamma_k\gamma_{k+1}\ne 0$, then $r$ is uniquely determined by $\gamma_k,\gamma_{k+1}$.
\end{rmk}

\begin{lem}
$\mathbf{z}\ne\mathbf{p}(1)$.
\end{lem}
\begin{proof}
Assume $\mathbf{z}=\mathbf{p}(1)$.
Similarly as in Section \ref{sec:G-p}, we have $c\ne 0$, and can assume $a=0$, so that $bc=-1$ and $d=r$.

Computing directly,
\begin{align*}
\mathbf{v}&=\left(\begin{array}{cc}
(n+1)n\gamma_{m+1}\gamma_mc^2+(n\gamma_{m+1}^2-n-1)c & \star \\ c-n\gamma_{m+1}\gamma_mc^2 & r-n\gamma_{m+2}\gamma_mc \end{array}\right), \\
\mathbf{g}(\mathbf{x}-\mathbf{e})&=\left(\begin{array}{cc}  \gamma_{m-1}-\gamma_m+n\gamma_{m+1}c & (\gamma_{m+1}-\gamma_m)b+n(\gamma_{m+2}-1) \\
(\gamma_{m+1}-\gamma_m)c & \gamma_{m+2}-\gamma_{m+1}   \end{array}\right),
\end{align*}
where the $\star$ stands for something irrelevant.
The reason for considering $\mathbf{g}(\mathbf{x}-\mathbf{e})$ is that $\mathbf{g}(\mathbf{x}-\mathbf{e})$ has a simpler expression than $\mathbf{g}$.

From ${\rm tr}(\mathbf{v})={\rm tr}(\mathbf{z}^{-1})=2$ we obtain
\begin{align}
(n+1)n\gamma_{m+1}\gamma_mc^2-c=2-r. \label{eq:H-z-np-1}
\end{align}
It follows from $\det(\mathbf{g}(\mathbf{x}-\mathbf{e}))=0$, which is required by (\ref{ineq:H-rank}), that
\begin{align}
1+\gamma_m-\gamma_{m+1}=\frac{2-r}{nc}. \label{eq:H-z-np-2}
\end{align}
\begin{itemize}
  \item If $r=2$, then $\gamma_k=k$, and
        $$\mathbf{g}=\left(\begin{array}{cc} 1-m+nm(m+1)c/2 & mb+n(m+1)(m+2)/2 \\ mc & m+1 \end{array}\right),$$
        so $\det(\mathbf{g})=1-nm(m+1)c/2\ne 0$ by (\ref{eq:H-z-np-1}). This violates (\ref{ineq:H-rank}).
  \item If $r\ne 2$, then
        \begin{align*}
        (2-r+c)(2-r)&\stackrel{(\ref{eq:H-z-np-1})}=(n+1)nc^2\cdot(2-r)\gamma_{m+1}\gamma_m \\
        &\stackrel{(\ref{eq:identity-2'})}=(n+1)nc^2(1+\gamma_{m+1}-\gamma_m)(1+\gamma_{m}-\gamma_{m+1}) \\
        &\stackrel{(\ref{eq:H-z-np-2})}=(n+1)c(1+\gamma_{m+1}-\gamma_m)(2-r).
        \end{align*}
        Hence $$1+\gamma_{m+1}-\gamma_m=\frac{2-r+c}{(n+1)c},$$
        which together with (\ref{eq:H-z-np-2}) implies $2-r=nc$ and $\gamma_{m+1}=\gamma_m$. By (\ref{eq:identity-2}), $nc\gamma_m^2=1$. Then
        $$-n(c')^2(a'd'-b'c')^{-1}=(\mathbf{u}^{-1}\mathbf{z}^{-1}\mathbf{u})_{21}=\mathbf{v}_{21}=0,$$
        so $c'=0\ne d'$. By (\ref{eq:H-rank=1}), $\mathbf{h}_{21}=0$. On the other hand, as we can compute, $\mathbf{h}_{21}=-n\gamma_mc\ne 0$. This is a contradiction.
\end{itemize}
\end{proof}

From now on, assume $\mathbf{z}=\mathbf{d}(\lambda,1)$ with $\lambda\ne 1$. Recall $\vartheta_k=\lambda^k-1$.

Computing directly,
\begin{align*}
\mathbf{v}_{11}&=\lambda^{-1}a+\lambda^{-n-1}\vartheta_n\gamma_{m+1}\gamma_mbc, \\
\mathbf{v}_{12}&=\big(\lambda^{-1}\gamma_{m+1}(\gamma_{m+1}-\gamma_ma)-\lambda^{-n-1}\gamma_m(\gamma_{m+1}d-\gamma_m)\big)b, \\
\mathbf{v}_{21}&=\big(\gamma_{m+1}(\gamma_{m+1}-\gamma_md)-\lambda^{n}\gamma_m(\gamma_{m+1}a-\gamma_m)\big)c, \\
\mathbf{v}_{22}&=d-\vartheta_n\gamma_{m+1}\gamma_mbc.
\end{align*}
The condition ${\rm tr}(\mathbf{v})={\rm tr}(\mathbf{z}^{-1})$ reads
\begin{align}
\lambda^{-1}(a-1)+(d-1)=\lambda^{-n-1}\vartheta_{n+1}\vartheta_n\gamma_{m+1}\gamma_mbc.  \label{eq:H-trace}
\end{align}

We can write $\mathbf{u}\mathbf{v}=\mathbf{z}^{-1}\mathbf{u}$ as
\begin{align}
(a',b')(\mathbf{v}-\lambda^{-1}\mathbf{e})=0, \label{eq:determine-u-1} \\
(c',d')(\mathbf{v}-\mathbf{e})=0. \label{eq:determine-u-2}
\end{align}

Now
\begin{align*}
\mathbf{h}=\left(\begin{array}{cc}  \lambda^{-n-1}([n]_\lambda(\gamma_{m+1}-\gamma_ma)-[n+1]_\lambda) & -n\lambda^{-n-1}\gamma_mb \\
-[n]_\lambda\gamma_mc & n(\gamma_{m+1}-\gamma_md)-n-1 \end{array}\right).
\end{align*}
So (\ref{eq:H-rank=1}) reads
\begin{align}
c'\lambda^{-n-1}\big(\vartheta_n(\gamma_ma-\gamma_{m+1})+\vartheta_{n+1}\big)+d'\vartheta_n\gamma_mc=0, \label{eq:H-1} \\
c'\lambda^{-n-1}\gamma_mb+d'(n(\gamma_md-\gamma_{m+1})+n+1)=1.  \label{eq:H-2}
\end{align}
These two equations uniquely determine $(c',d')$. The second one can be replaced by $\det(\mathbf{h})\ne 0$, which is equivalent to
\begin{align}
(n+1)\vartheta_n(\gamma_ma-\gamma_{m+1})+n\vartheta_{n+1}(\gamma_md-\gamma_{m+1})+(n+1)\vartheta_{n+1}+n\vartheta_n\ne 0. \label{eq:det(h)ne0}
\end{align}


\begin{lem} \label{lem:z=0}
If $bc=0$, then $a=\lambda=d^{-1}$, and $c=0\ne b$.
\end{lem}
\begin{proof}
If $b=0$, then $b'\ne 0$, as required by the irreducibility of $\rho$; by (\ref{eq:determine-u-1}), $\mathbf{v}_{22}=\lambda^{-1}$, and furthermore by (\ref{eq:determine-u-2}), $d'=0$. This contradicts (\ref{eq:H-2}).

Hence $b\ne 0=c$. Due to the irreducibility of $\rho$, we have $c'\ne 0$. Then by (\ref{eq:determine-u-2}), $\mathbf{v}_{11}=1$, so $a=\lambda$.
\end{proof}

The conjugacy indeterminacy of $\rho$ can be fixed by setting $b=1$, and the irreducibility of $\rho$ is ensured as long as $c'\ne 0$.



\subsection{$r=2$} \label{sec:r=2}


Write $a=1+s$ and $d=1-s$. By Lemma \ref{lem:z=0}, $s\ne 0$.

By (\ref{eq:H-trace}),
\begin{align}
\vartheta_{n+1}\vartheta_n(m+1)ms=\vartheta_{n+1}-\vartheta_n.  \label{eq:r=2-1}
\end{align}

In virtue of (\ref{ineq:H-rank}), ${\rm rank}(\mathbf{g},\mathbf{k})\le 1$, where
$$\mathbf{k}=\mathbf{x}^m\mathbf{z}^{n+1}\mathbf{u}^{-1}(\mathbf{e}-\mathbf{z}^{-1})\mathbf{u}
\stackrel{(\ref{eq:H-relation})}=\mathbf{x}^m\mathbf{z}^{n+1}-\mathbf{z}^n\mathbf{x}^{m+1}.$$
By direct computation,
\begin{align*}
&\mathbf{g}=\left(\begin{array}{cc} (m+1)m\vartheta_ns/2+ms+m\vartheta_n+1 & (m+1)m\vartheta_nb/2+mb \\ mc & 1-ms\end{array}\right), \\
&\mathbf{k}_{\ast2}=\left(\begin{array}{cc} -((m+1)\vartheta_n+1)b \\ s \end{array}\right).
\end{align*}
Then $\det(\mathbf{g})=0$ is equivalent to
\begin{align}
(m+1)m\vartheta_ns=2(m+1)\vartheta_n+2. \label{eq:r=2-2}
\end{align}

Actually, (\ref{eq:r=2-1}), (\ref{eq:r=2-2}) are already sufficient to imply (\ref{ineq:H-rank}).

Suppose (\ref{eq:r=2-1}), (\ref{eq:r=2-2}) hold; equivalently,
\begin{align}
(2m+1)\vartheta_{n+1}\vartheta_n+\vartheta_{2n+1}=0, \label{eq:r=2-3} \\
s=\frac{\vartheta_{n+1}-\vartheta_n}{\vartheta_{n+1}\vartheta_n(m+1)m}. \label{eq:r=2-4}
\end{align}

It can be verified that $\mathbf{g}_{\ast1},\mathbf{g}_{\ast2}\parallel\mathbf{k}_{\ast2}$. Clearly, $\mathbf{k}_{\ast2}\ne 0$, and $\det(\mathbf{k})=0$. Hence
${\rm rank}(\mathbf{g},\mathbf{k})\le 1$, which is equivalent to ${\rm rank}(\mathbf{u}\mathbf{z}^{-n-1}\mathbf{x}^{-m}\mathbf{g},\mathbf{e}-\mathbf{z}^{-1})\le 1$.

By (\ref{eq:determine-u-2}), $(c',d')\parallel(\mathbf{v}_{21},1-\mathbf{v}_{11})$.
We have
\begin{align*}
\left(\begin{array}{cc} \mathbf{v}_{21} \\ 1-\mathbf{v}_{11} \end{array}\right)=
\left(\begin{array}{cc} (1-\vartheta_nm-\vartheta_n(m+1)ms)c \\ 1-\lambda^{-1}(1+s)+\lambda^{-n-1}\vartheta_n(m+1)ms^2 \end{array}\right)\ne 0,
\end{align*}
as $\mathbf{v}_{21}=1-\mathbf{v}_{11}=0$ would imply
$$\lambda^n=\frac{3m+1}{3m+2}, \qquad \lambda^{n+1}=\frac{m+1}{m},$$
which is absurd.
It turns out that
$$\mathbf{v}_{21}\lambda^{-n-1}(\vartheta_n(\gamma_ma-\gamma_{m+1})+\vartheta_{n+1})+(1-\mathbf{v}_{11})\vartheta_n\gamma_mc=0,$$
thanks to (\ref{eq:r=2-1}). So (\ref{eq:H-1}) is satisfied,
Furthermore, (\ref{eq:det(h)ne0}) is guaranteed:
\begin{align*}
&(n+1)\vartheta_n(ms-1)-n\vartheta_{n+1}(ms+1)+(n+1)\vartheta_{n+1}+n\vartheta_n \\
=\ &((n+1)\vartheta_n-n\vartheta_{n+1})ms+\vartheta_{n+1}-\vartheta_n \\
\stackrel{(\ref{eq:r=2-1})}=&((n+1)\vartheta_n-n\vartheta_{n+1})ms+\vartheta_{n+1}\vartheta_n(m+1)ms \\
\stackrel{(\ref{eq:r=2-3})}=&\Big(n+\frac{1}{2}\Big)(\vartheta_n-\vartheta_{n+1})ms\ne 0.
\end{align*}
Hence ${\rm rank}(\mathbf{e}-\mathbf{z}^{-1},\mathbf{z}^{-1}+\mathbf{u}\mathbf{h})\le 1$.

Therefore, (\ref{eq:r=2-3}), (\ref{eq:r=2-4}) are sufficient and necessary for (\ref{ineq:H-rank}).



\subsection{$r\ne 2$}

By Remark \ref{rmk:H-condition}, the condition $\dim\mathfrak{D}_\rho\ge 5$ is equivalent to $(c',d')\mathbf{h}=(0,-1)$.

Suppose $\mathbf{v}_{22}\ne 1$. Then $(c',d')$ is a nonzero multiple of $(1-\mathbf{v}_{22},\mathbf{v}_{12})$.
Direct computation reduces (\ref{eq:H-1}) to
\begin{align}
\vartheta_n\gamma_m(a-1)+(\vartheta_n\gamma_{m+1}-\vartheta_{n+1})(d-1)=-\vartheta_{n+1}\vartheta_n\gamma_{m+1}\gamma_mbc.  \label{eq:r-ne2}
\end{align}

\begin{lem}
$\vartheta_n\vartheta_{n+1}\ne 0$, and $(\gamma_{m+1},\gamma_m)\ne(1,0), (0,-1)$.
\end{lem}

\begin{proof}
From (\ref{eq:H-trace}), (\ref{eq:r-ne2}) it is clear that $\vartheta_n\ne0$, and $(\gamma_{m+1},\gamma_m)\ne(1,0), (0,-1)$.

Assume $\vartheta_{n+1}=0$. Then $\vartheta_n=\lambda^{-1}-1$, and by (\ref{eq:H-trace}), (\ref{eq:r-ne2}),
$$a-1+\lambda(d-1)=0, \qquad \gamma_{m+1}=\lambda\gamma_m.$$
The first equation implies
$$a=1+\frac{r-2}{\lambda^{-1}-1}, \qquad d=1+\frac{r-2}{1-\lambda},$$
and the second one together with (\ref{eq:identity-2}) implies
$\gamma_m^2=(\lambda^2-\lambda r+1)^{-1}$.
Then
\begin{align*}
\mathbf{v}_{22}-1&=d-1-\lambda\vartheta_n\gamma_m^2(ad-1) \\
&=\frac{r-2}{1-\lambda}-\frac{1-\lambda}{\lambda^2-\lambda r+1}\Big(r-2-\frac{\lambda}{(1-\lambda)^2}(r-2)^2\Big)=0,
\end{align*}
contradicting the assumption.
\end{proof}

Let $\omega=1+\gamma_m-\gamma_{m+1}$, which is always nonzero by (\ref{eq:identity-2'}).

A comparison of (\ref{eq:r-ne2}) with (\ref{eq:H-trace}) results in
$$\vartheta_n\gamma_m(a-1)+(\vartheta_n\gamma_{m+1}-\vartheta_{n+1})(d-1)=-\lambda^n(a-1)-\lambda^{n+1}(d-1),$$
leading us to
\begin{align}
a=1+(\vartheta_n^{-1}+\gamma_{m+1})(2-r)\omega^{-1}, \qquad
d=1+(\lambda^n\vartheta_n^{-1}+\gamma_{m})(r-2)\omega^{-1}. \label{eq:H-x-w-1}
\end{align}
Then we can convert (\ref{eq:H-trace}) into
\begin{align}
(\lambda^n\gamma_{m+1}-\gamma_m)((\gamma_{m+1}+\gamma_m)\vartheta_{n+1}\vartheta_n+\vartheta_{2n+1})=0. \label{eq:H-CJL0}
\end{align}
Here are the details: the left-hand-side of (\ref{eq:H-trace}) equals
$$(\lambda^{-1}\vartheta_{n+1}\vartheta_n^{-1}+\gamma_m-\lambda^{-1}\gamma_{m+1})\frac{r-2}{\omega},$$
and the right-hand-side equals
\begin{align*}
&\lambda^{-n-1}\vartheta_{n+1}\vartheta_n\gamma_{m+1}\gamma_m(r-2+(a-1)(d-1)) \\
=\ &\lambda^{-n-1}\vartheta_{n+1}\vartheta_n\gamma_{m+1}\gamma_m
\Big(\omega+\frac{(2-r)(\vartheta_n^{-1}+\gamma_{m+1})(\lambda^n\vartheta_n^{-1}+\gamma_m)}{\omega}\Big)\frac{r-2}{\omega} \\
=\ &\lambda^{-n-1}\vartheta_{n+1}\vartheta_n\gamma_{m+1}\gamma_m
\Big(2+\frac{(2-r)\vartheta_n^{-1}(\lambda^n\vartheta_n^{-1}+\gamma_m+\lambda^n\gamma_{m+1})}{\omega}\Big)\frac{r-2}{\omega},
\end{align*}
with (\ref{eq:identity-2'}) applied, so (\ref{eq:H-trace}) becomes
\begin{align*}
&\lambda^n\vartheta_{n+1}\vartheta_n^{-1}+\lambda^{n+1}\gamma_m-\lambda^n\gamma_{m+1} \\
=\ &\vartheta_{n+1}\vartheta_n\big(2\gamma_{m+1}\gamma_m+(1+\gamma_{m+1}-\gamma_m)\vartheta_n^{-1}(\lambda^n\vartheta_n^{-1}
+\gamma_m+\lambda^n\gamma_{m+1})\big) \\
=\ &\vartheta_{n+1}\big(\lambda^n\vartheta_n^{-1}-\vartheta_n^{-1}\gamma_m+\lambda^{2n}\vartheta_n^{-1}\gamma_{m+1}
+(\lambda^n\gamma_{m+1}-\gamma_m)(\gamma_{m+1}+\gamma_m)\big),
\end{align*}
resulting in (\ref{eq:H-CJL0}).
Since $\mathbf{v}_{22}\ne 1$, we have $\lambda^n\gamma_{m+1}\ne\gamma_m$. Thus
\begin{align}
\gamma_{m+1}+\gamma_m+\vartheta_{2n+1}\vartheta_{n+1}^{-1}\vartheta_n^{-1}=0. \label{eq:H-CJL}
\end{align}
As an equation in $r$, it has $m$ solutions for generic $\lambda$.
We should still exclude the solutions with $\lambda^n\gamma_{m+1}=\gamma_m$ or $(\gamma_{m+1},\gamma_m)\in\{(1,0), (0,-1)\}$.

\medskip

Now suppose $\mathbf{v}_{22}=1$, so that $c\ne 0$ (otherwise $d=1=a$).
This combined with (\ref{eq:H-trace}) implies $\vartheta_n\ne 0$ and $d-1=-\lambda^n(a-1)$,
so that
\begin{align}
a=(\lambda^n+1-r)\vartheta_n^{-1}, \qquad d=(\lambda^nr-\lambda^n-1)\vartheta_n^{-1}.  \label{eq:H-x-w-2}
\end{align}
We have $1-\mathbf{v}_{11}=1-\lambda^{-1}\ne 0$, so $(c,d)$ is a nonzero multiple of $(\mathbf{v}_{21},1-\lambda^{-1})$, and (\ref{eq:H-1}) is equivalent to
\begin{align}
\mathbf{v}_{21}\lambda^{-n-1}(\vartheta_n(\gamma_ma-\gamma_{m+1})+\vartheta_{n+1})+(1-\lambda^{-1})\vartheta_n\gamma_mc=0.  \label{eq:H-3}
\end{align}
Then it follows from $\mathbf{v}_{22}=1$ together with $r\ne 2$ that
$$\lambda^n+\lambda^{-n}=r+(\gamma_{m+1}\gamma_m)^{-1}=\gamma_{m+1}\gamma_m^{-1}+\gamma_{m}\gamma_{m+1}^{-1},$$
implying $\gamma_{m}=\lambda^{\pm n}\gamma_{m+1}$.
Note that $\mathbf{v}_{21}=(\gamma_{m+1}-\gamma_m)(\gamma_{m+1}-\lambda^n\gamma_m)c$, which does not vanish by (\ref{eq:H-3}).
Hence $\gamma_{m}=\lambda^{n}\gamma_{m+1}$.

By (\ref{eq:identity-2}), $(\lambda^{2n}-\lambda^{n}r+1)\gamma_{m+1}^2=1$, through which
(\ref{eq:H-3}) can be converted into $\vartheta_{2n}\vartheta_n\vartheta_{n+1}+\vartheta_{2n+1}\vartheta_n\gamma_m^{-1}=0$.
Consequently,
$$\gamma_{m+1}=\frac{-\vartheta_{2n+1}}{\vartheta_{2n}\vartheta_{n+1}}, \qquad
\gamma_m=\frac{-\lambda^n\vartheta_{2n+1}}{\vartheta_{2n}\vartheta_{n+1}}.$$
This is exactly the solution to (\ref{eq:H-CJL}) excluded for $\lambda^n\gamma_{m+1}=\gamma_m$; indeed, (\ref{eq:H-x-w-2}) is consistent with (\ref{eq:H-x-w-1}).

\bigskip
To deal with (\ref{eq:det(h)ne0}), we use (\ref{eq:H-x-w-1}), (\ref{eq:H-CJL}) to compute
\begin{align*}
&(n+1)\vartheta_n(\gamma_ma-\gamma_{m+1})+n\vartheta_{n+1}(\gamma_md-\gamma_{m+1})+(n+1)\vartheta_{n+1}+n\vartheta_n \\
=\ &(2n+1)\Big(\vartheta_{n+1}+\vartheta_n+\frac{(2-r)\gamma_m}{\omega}\Big).
\end{align*}
Hence $\det(\mathbf{h})\ne 0$ if and only if
\begin{align}
(r-2)\gamma_m\ne (\vartheta_{n+1}+\vartheta_n)\omega. \label{eq:exception-1}
\end{align}
If $(r-2)\gamma_m=(\vartheta_{n+1}+\vartheta_n)\omega$, then
\begin{align*}
-(\vartheta_{n+1}+\vartheta_n)\omega\gamma_{m+1}=(2-r)\gamma_{m+1}\gamma_m\stackrel{(\ref{eq:identity-2'})}=\omega(1+\gamma_{m+1}-\gamma_m),
\end{align*}
implying
\begin{align}
\gamma_{m+1}=\frac{\lambda+1-2\lambda^{n+1}}{(\lambda+1)\vartheta_{n+1}\vartheta_n}, \qquad \gamma_m=\frac{2\lambda^{n+1}-(\lambda+1)\lambda^{2n+1}}{(\lambda+1)\vartheta_{n+1}\vartheta_n}. \label{eq:exception-2}
\end{align}



\begin{lem} \label{lem:H-multiple-root}
The polynomial $h_\lambda(r):=\gamma_{m+1}+\gamma_m+\vartheta_{2n+1}\vartheta_{n+1}^{-1}\vartheta_n^{-1}$ has no triple root, and has at most one double root. If the double root $r$ exists, then it satisfies $r\ne\pm 2$, $m\gamma_{m+1}=(m+1)\gamma_m$, and $\gamma_{m+1}\gamma_m\ne 0$.
\end{lem}
\begin{proof}
Write
$$h_\lambda(r)=q(\mu)
=\frac{\mu^{m+1}-\mu^{-m}}{\mu-1}+\vartheta_{2n+1}\vartheta_{n+1}^{-1}\vartheta_n^{-1}.$$
When $\mu\ne\pm 1$,
$$q'(\mu)=\frac{m(\mu^{m+1}-\mu^{-m-1})-(m+1)(\mu^m-\mu^{-m})}{(\mu-1)^2}.$$
If $q(\mu)=q'(\mu)=0$, then
$$q''(\mu)=\frac{m(m+1)(\mu^{m-1}-\mu^{-m-2})}{\mu-1}\ne 0,$$
and $m\gamma_{m+1}=(m+1)\gamma_m$, from which it follows that
$$\gamma_m=-\frac{m\vartheta_{2n+1}}{(2m+1)\vartheta_{n+1}\vartheta_n}, \qquad
\gamma_{m+1}=-\frac{(m+1)\vartheta_{2n+1}}{(2m+1)\vartheta_{n+1}\vartheta_n}.$$
These uniquely determine $r$, as obviously $\gamma_{m+1}\gamma_{m}\ne 0$.

If $h_\lambda(2)=0$, then $\vartheta_{2n+1}\vartheta_{n+1}^{-1}\vartheta_n^{-1}=-2m-1$, and $h_\lambda(r)=(r-2)[m]_\mu[m]_{\mu^{-1}}$,
so $2$ is not a double root of $h_\lambda$.

If $h_\lambda(-2)=0$, then $\vartheta_{2n+1}\vartheta_{n+1}^{-1}\vartheta_n^{-1}=(-1)^{m+1}$, and
$$h_\lambda(r)=(r+2)\frac{(-1)^{m}}{\mu(1-\mu)}[m]_{-\mu}[m]_{-\mu^{-1}},$$
showing that $-2$ is neither a double root of $h_\lambda$.
\end{proof}

\begin{lem}
For there to exist at least one root of $h_\lambda$ with {\rm(\ref{eq:exception-1})} and $(\gamma_{m+1},\gamma_m)\notin\{(1,0), (0,-1)\}$, it is sufficient and necessary that, either $m\ge 3$ is odd, or
$$\begin{cases}
\lambda^{n+1}+\lambda^n\ne 1, &m=1, \\
\lambda^{n+1}+\lambda^n\ne 2, &m\ge 2\text{\ is\ even}.
\end{cases}$$
\end{lem}
\begin{proof}
When $m=1$, the result can be verified straightforwardly.

When $m=2$, $\gamma_m=r$, and $\gamma_{m+1}=r^2-1$.
\begin{itemize}
  \item If $\lambda^{n+1}+\lambda^n=2$, then the two roots of $h_\lambda$ are $-1,0$; the first one satisfies $(\gamma_{3},\gamma_2)=(0,-1)$, and the second one violates (\ref{eq:exception-1}).
  \item If one root $r$ of $h_\lambda$ falls into $(\gamma_{m+1},\gamma_m)\in\{(1,0), (0,-1)\}$, then $r=-1$, so $\lambda^{n+1}+\lambda^n=2$.
  \item If the two roots of $h_\lambda$ coincide, then the double root is $-1/2$ and $\vartheta_{2n+1}=(5/4)\vartheta_{n+1}\vartheta_n$; it can be checked that (\ref{eq:exception-1}) is true.
\end{itemize}

Now suppose $m\ge 3$. 
\begin{itemize}
  \item If $\vartheta_{2n+1}\ne\pm\vartheta_{n+1}\vartheta_n$, then the roots of $h_\lambda$ all satisfy $\gamma_{m+1}\gamma_m\ne 0$; by Lemma \ref{lem:H-multiple-root} and Remark \ref{rmk:gamma-r}, $h_\lambda$ has at least one root satisfying (\ref{eq:exception-1}).
  \item If $\vartheta_{2n+1}=\vartheta_{n+1}\vartheta_n$, i.e. $\lambda^{n+1}+\lambda^n=2$, then $\gamma_{m+1}+\gamma_m=-1$. There are three possibilities:
        \begin{itemize}
        \item $2\nmid m$ and $r=-2$, so that $(\gamma_{m+1},\gamma_m)=(-m-1,m)$;
        \item $\mu^m=-1$ and $\mu\ne \pm1$, which is equivalent to $(\gamma_{m+1},\gamma_m)=(-1,0)$, but then (\ref{eq:exception-1}) does not hold;
        \item $\mu^{m+1}=1$ and $\mu\ne\pm1$, which is equivalent to $(\gamma_{m+1},\gamma_m)=(0,-1)$.
        \end{itemize}
  \item If $\vartheta_{2n+1}=-\vartheta_{n+1}\vartheta_n$, then $\gamma_{m+1}+\gamma_m=1$. There are three possibilities:
        \begin{itemize}
        \item $2\mid m$ and $r=-2$, so that $(\gamma_{m+1},\gamma_m)=(m+1,-m)$;
        \item $\mu^m=1$ and $\mu\ne \pm1$, which is equivalent to $(\gamma_{m+1},\gamma_m)=(1,0)$;
        \item $\mu^{m+1}=-1$ and $\mu\ne\pm1$, which is equivalent to $(\gamma_{m+1},\gamma_m)=(0,1)$, then (\ref{eq:exception-1}) holds except for at most one case when $r=2(\vartheta_{n+1}+\vartheta_n+1)$.
        \end{itemize}
\end{itemize}

In conclusion, if $m\ge 3$ is odd, the existence of $r\in h_\lambda^{-1}(0)$ with {\rm(\ref{eq:exception-1})} and $(\gamma_{m+1},\gamma_m)\notin\{(1,0), (0,-1)\}$ is ensured; if $m$ is even, then exactly $\lambda^{n+1}+\lambda^n\ne 2$ is required.
\end{proof}



\subsection{The result}

Noting
$$\lim\limits_{r\to 2}\frac{2-r}{1+\gamma_m-\gamma_{m+1}}=\lim\limits_{\mu\to 1}\frac{\mu^m(1+\mu^{-1})}{[m+1]_\mu[m]_\mu}=\frac{1}{(m+1)m},$$
we can incorporate the result of Subsection \ref{sec:r=2}.
Let
$$\delta_r=\begin{cases} (2-r)/(1+\gamma_m-\gamma_{m+1}), &r\ne 2, \\ 1/(m+1)m, &r=2. \end{cases} $$


Let $\sigma_{\lambda,r,a_1}=\rho_{\mathbf{z},\mathbf{x},\mathbf{u}}$, with $\mathbf{z}=\mathbf{d}(\lambda,1)$,
\begin{align*}
\mathbf{x}&=\left(\begin{array}{cc} 1+\delta_r(\vartheta_n^{-1}+\gamma_{m+1}) & 1 \\
\delta_r(\delta_r(\gamma_m+\lambda^{2n+1}\vartheta_{n+1}^{-1}\vartheta_n^{-1})-2) & 1-\delta_r(\lambda^n\vartheta_n^{-1}+\gamma_m) \end{array}\right), \\
\mathbf{u}&=\left(\begin{array}{cc} a_1(1-\lambda+\lambda\delta_r(\lambda^n\gamma_{m+1}-\gamma_m)\vartheta_{n+1}^{-1}) & a_1(\lambda^{-n}\gamma_m-\gamma_{m+1}) \\ -\lambda^{n+1}\delta_r\vartheta_{n+1}^{-1}/(2n+1) & 1/(2n+1) \end{array}\right),
\end{align*}


\begin{thm}
The cohomology jump locus $\mathcal{J}_3(H_{m,n})=\chi(\mathcal{F})$, where $\mathcal{F}$ is the subspace of $\mathcal{R}^{\rm irr}(H_{m,n})$ consisting of $\sigma_{\lambda,r,a_1}$'s such that
\begin{align*}
a_1\in\mathbb{C}^\ast, \qquad \vartheta_{n+1}\vartheta_n\ne 0, \\
\gamma_{m+1}+\gamma_m+\vartheta_{2n+1}\vartheta_{n+1}^{-1}\vartheta_n^{-1}=0, \\
(\gamma_{m+1},\gamma_m)\notin\{(1,0),(0,-1)\}, \\
\delta_r\gamma_m+\vartheta_{n+1}+\vartheta_n\ne 0.
\end{align*}
With $\hom(H_{m,n},\mathbb{C}^\ast)\cong\mathbb{C}^\ast\times\mathbb{C}^\ast$ via $\tau\mapsto(\tau(z),\tau(u))$, the morphism
$${\det}_\ast:\mathcal{J}_3(H_{m,n})\to\hom(H_{m,n},\mathbb{C}^\ast), \qquad
[\sigma_{\lambda,r,a_1}]\mapsto\Big(\lambda,\frac{1-\lambda}{2n+1}a_1\Big),$$
has image $R\times\mathbb{C}^\ast$, where $R\subset\mathbb{C}^\ast$ consists of $\lambda$ satisfying
$$\begin{cases}
(\lambda^n-1)(\lambda^{n+1}-1)(\lambda^{n+1}+\lambda^n-1)\ne0, &m=1, \\
(\lambda^n-1)(\lambda^{n+1}-1)(\lambda^{n+1}+\lambda^n-2)\ne0, &m\ge 2\ \text{is\ even}, \\
(\lambda^n-1)(\lambda^{n+1}-1)\ne0, &m\ge 3\ \text{is\ odd}.
\end{cases}$$
\end{thm}

\medskip

\begin{proof}[Proof of Theorem \ref{thm:solution-H}]
Suppose there exists an isomorphism $\phi:H_{m,n}\to H_{m',n'}$.
Then for a generic point $\mathfrak{a}\in\det_\ast(\mathcal{J}_3(H_{m',n'}))$,
$$m'=\#{\det}_\ast^{-1}(\mathfrak{a})=\#{\det}_\ast^{-1}(\phi^\ast(\mathfrak{a}))=m,$$
and from $H_1({\det}_\ast(\mathcal{J}_3(H_{m',n'})))\cong H_1({\det}_\ast(\mathcal{J}_3(H_{m,n})))$ we see $n'=n$.
\end{proof}



