# Two-dimensional diffusion orthogonal polynomials ordered by a weighted
degree

S. Yu. Orevkov  Steklov Mathematical Institute, Gubkina 8, Moscow, Russia
IMT, l’université Paul Sabatier, 118 route de Narbonne, Toulouse, France
orevkovmath.ups-tlse.fr

We study the following problem: describe the triplets $(\Omega,g,\mu)$,
$\mu=\rho\,dx$, where $g=(g^{ij}(x))$ is the (co)metric associated with the
symmetric second order differential operator
$\mathbb{L}(f)=\frac{1}{\rho}\sum_{ij}\partial_{i}(g^{ij}\rho\partial_{j}f)$
defined on a domain $\Omega$ of $\mathbb{R}^{d}$ and such that there exists an
orthonormal basis of $\Cal{L}^{2}(\mu)$ made of polynomials which are
eigenvectors of $\mathbb{L}$, where the polynomials are ranked according to
some weighted degree.

In a joint paper with D. Bakry and M. Zani this problem was solved in
dimension 2 for the usual degree. In the present paper we solve it still in
dimension 2, but for a weighted degree with arbitrary positive weights.

## 1\. Introduction

In [3] the following problem is studied (see also [1], [2], [5], [6]):
describe all triples $(\Omega,\mathbb{L},\mu)$ where $\Omega$ is a domain in
$\mathbb{R}^{d}$, $\mathbb{L}$ is an elliptic second order operator of the
form

$\mathbb{L}(f)=\sum_{i,j}g^{ij}(x)\partial_{ij}f+\sum_{i}b^{i}(x)\partial_{i}f$
$None$

with $g^{ij}$ and $b^{i}$ continuous in $\Omega$ and $\mu$ a probability
measure on $\Omega$ which is absolutely continuous with respect to the
Lebesgue measure, and such that there exists a polynomial orthogonal basis of
$\Cal{L}^{2}(\Omega,\mu)$ formed by eigenvectors of $\mathbb{L}$, which is
also a basis (in the algebraic sense) of $\mathbb{R}[x]$,
$x=(x_{1},\dots,x_{d})$. It is clear that in this case $\mathbb{L}$ is
symmetric on the space of polynomials, i.e.,

$\int_{\Omega}f_{1}\mathbb{L}f_{2}\,d\mu=\int_{\Omega}f_{2}\mathbb{L}f_{1}\,d\mu$
$None$

for any two polynomial functions $f_{1}$ and $f_{2}$.

Suppose that $e_{1},e_{2},\dots$ is such a basis, and let $V_{n}$ be the
subspace spanned by $e_{1},\dots,e_{n}$. Then we have an increasing sequence
of $\mathbb{L}$-invariant subspaces $V_{1}\subset V_{2}\subset
V_{3}\subset\dots$ of $\mathbb{R}[x]$ whose union is $\mathbb{R}[x]$, and
$\mathbb{R}[x]$ is dense in $\Cal{L}^{2}(\Omega,\mu)$.

Vice versa, given an increasing sequence of finite-dimensional
$\mathbb{L}$-invariant subspaces of $\mathbb{R}[x]$ whose union is
$\mathbb{R}[x]$, one can always choose an orthogonal polynomial eigenbasis of
$\Cal{L}^{2}(\Omega,\mu)$ provided that $\mathbb{L}$ is symmetric on
polynomials and the space of polynomials is dense in
$\Cal{L}^{2}(\Omega,\mu)$.

It does not seem that this problem can be solved in such generality, without
imposing a condition that the filtration $V_{1}\subset V_{2}\subset\dots$ is
somewhat natural. So, in [3] the above problem was considered with an
additional condition that for any $n$, the space of polynomials of degree
$\leq n$ is invariant under $\mathbb{L}$ and thus occurs among the $V_{i}$’s.
It was also assumed in [3] that, when $\Omega$ is not bounded, (2) holds for
any pair of compactly supported functions (for bounded domains the latter
condition easily follows from the symmetry of $\mathbb{L}$ on the polynomials
combined with the density of $\mathbb{R}[x]$ in $\Cal{L}(\Omega,\mu)$). Under
these assumptions, a complete list of solutions of the above problem is given
in [3] in dimensions $2$. Up to affine transformation of $\mathbb{R}^{2}$,
there is a one-parameter family of bounded domains and also 18 rigid domains
(10 of them are bounded) for which there exists a solution.

Remark 1.1. In dimension 1, the only solutions under the aforementioned
assumptions are the classical systems of orthogonal polynomials: Hermite,
Laguerre, and Jacobi polynomials. They are obtained by Gram-Schmidt
orthogonalization process for the measure densities, respectively,
$Ce^{-x^{2}/2}$ on $\mathbb{R}$, $C_{a}x^{a-1}e^{-x}$ with $a>0$ on
$[0,\infty)$, and $C_{p,q}(1-x)^{p-1}(1+x)^{q-1}$ with $p,q>0$ on $[-1,1]$
(here $C$, $C_{a}$, and $C_{p,q}$ are normalizing constants). The
corresponding operators are $\partial^{2}-x\partial$,
$x\partial^{2}+(a-x)\partial$, and
$J_{p,q}=(1-x^{2})\partial^{2}-\big{(}(p-q)+(p+q)x\big{)}\partial$.

It turns out that the filtration by the usual degree is too restrictive in
dimension $d\geq 2$. Several natural systems of orthogonal polynomial are not
covered. However, they can be obtained by this procedure if one considers a
weighted degree instead (see [1]). As usually, the weighted degree of a
polynomial $P=\sum_{k}a_{k}x^{k}$, $k=(k_{1},\dots,k_{d})$, with real positive
weights $\mathbb{w}=(w_{1},\dots,w_{d})$ is defined as

$\deg_{\mathbb{w}}(P)=\max_{a_{k}\neq 0}(w_{1}k_{1}+\dots+w_{d}k_{d}).$

In this paper (in §6), for any pair of positive weights
$\mathbb{w}=(w_{1},w_{2})$, we give a complete list of two-dimensional
solutions satisfying the condition that for any $n$, the set of polynomials
$P$ with $\deg_{\mathbb{w}}(P)\leq n$ is invariant under $\mathbb{L}$.

Let us give precise definitions. We say that $\Omega\subset\mathbb{R}^{d}$ is
a natural domain if it is a connected open set which coincides with the
interior of its closure.

Definition 1.2. (cf. [1, 3]) Let $\Omega\subset\mathbb{R}^{d}$ be a natural
domain, $\mathbb{L}$ be an elliptic second order differential operator of the
form (1) with coefficients continuous in $\Omega$, and $\mu(dx)=\rho(x)\,dx$
be a probability measure on $\Omega$ such that $\rho$ is continuous in
$\Omega$ and the space of all polynomials is dense in
$\Cal{L}^{2}(\Omega,\mu)$. Let $\mathbb{w}$ be a $d$-tuple of positive real
numbers. We say that the triple $(\Omega,\mathbb{L},\mu)$ is a solution of the
Diffusion Orthogonal Polynomial Problem with weights $\mathbb{w}$
($\mathbb{w}$-DOP problem for short) if, for any $n$, the space
$\\{P\in\mathbb{R}[x_{1},\dots,x_{d}]\mid\deg_{\mathbb{w}}(P)\leq n\\}$
(considered as a subspace of $\Cal{L}^{2}(\Omega,\mu)$) has an orthogonal
basis formed by eigenvectors of $\mathbb{L}$. If in addition the equality (2)
holds for any two smooth functions compactly supported in $\mathbb{R}^{d}$,
then we say that $(\Omega,\mathbb{L},\mu)$ is a solution of the Strong
$\mathbb{w}$-DOP problem ($\mathbb{w}$-SDOP problem).

It is shown in [3, Prop. 2.11] that if $(\Omega,\mathbb{L},\mu)$ is a solution
of the $\mathbb{w}$-SDOP problem, then $\mathbb{L}$ is determined by the
measure density $\rho$ and the cometric $g=(g^{ij})$ (the $g^{ij}$ are the
coefficients in (1)), namely,

$\mathbb{L}(f)=\frac{1}{\rho}\sum_{i,j}\partial_{i}\Big{(}g^{ij}\rho\,\partial_{j}f\Big{)}.$
$None$

We shall then speak also of the triple $(\Omega,g,\rho)$ as a solution of the
$\mathbb{w}$-SDOP problem. Notice that (3) is the Laplace-Beltrami operator
for the riemannian metric $(g_{ij})=g^{-1}$ in the case when $\rho=(\det
g)^{-1/2}$.

As we mentioned above, any solution of the $\mathbb{w}$-DOP problem with a
bounded domain $\Omega$ is a solution of the $\mathbb{w}$-SDOP problem (see
[3, Prop. 2.12]). It is also shown in [3] that any solution of SDOP-problem is
a solution of a certain algebraic problem formulated in terms of the metric
$g^{ij}$ only (AlgDOP-problem; see §2 below). This fact is proven in [3] for
the usual degree but all the proofs extend without changes to any weighted
degree as well. Then, to find all two-dimensional solutions of the weighted
SDOP problem, we follow the same strategy as in [3]. Namely, we first find
solutions of the algebraic problem over $\mathbb{C}$ using some basic
properties of plane complex algebraic curves (see §§4–5), and then we look for
$\Omega$ and $\rho$ (see §6).

All the bounded domains $\Omega$ admitting a solution, appeared already in the
literature except, maybe, one infinite family: the case (B4) with $m\neq n$ in
Theorem 6.2. However orthogonal polynomials are being studied so long time in
so many aspects and the literature is so vast that we cannot be sure that this
family is really new.

## 2\. Some general facts about solutions of weighted DOP/SDOP problem in an
arbitrary dimension

### 2.1. The AlgDOP Problem

Let $\mathbb{w}=(w_{1},\dots,w_{d})$ and let $(\Omega,g,\rho)$ be a solution
of the $\mathbb{w}$-SDOP problem in $\mathbb{R}^{d}$. Let
$\Delta=\det(g^{ij})$. Let $\Cal{P}_{\mathbb{w}}(n;\mathbb{K})$ be the vector
space of polynomials in $x_{1},\dots,x_{d}$ with coefficients in $\mathbb{K}$
whose $\mathbb{w}$-weighted degree is at most $n$. When $\mathbb{K}$ is
$\mathbb{R}$, we write just $\Cal{P}_{\mathbb{w}}(n)$.

Let $I(\partial\Omega)$ be the ideal in $\mathbb{R}[x_{1},\dots,x_{d}]$ of
polynomials vanishing on $\partial\Omega$. The condition that $\Omega$ is a
natural domain implies that $I(\partial\Omega)$ is a principal ideal (because
$U\cap\partial\Omega$ cannot be of codimension $\geq 2$ for any open $U$). Let
$\Gamma$ be a generator of $I(\partial\Omega)$, that is $\Gamma$ is a minimal
polynomial vanishing on the boundary of $\Omega$. In particular, if $\Gamma$
is not identically zero, then it is square-free, i.e., it does not have
multiple factors. By convention we set $I(\varnothing)=\mathbb{R}[x]$, i.e.,
$\Gamma=1$ in the case when $\Omega$ is the whole $\mathbb{R}^{d}$.

######  Proposition \propAlgDOP

(See [3, Thm. 2.21].)

This leads us to the following definition (cf. [3, Definition 3.2]).

Definition 2.2.  Let $\mathbb{K}$ be $\mathbb{R}$ or $\mathbb{C}$ and let
$\mathbb{w}=(w_{1},\dots,w_{d})$ be a $d$-tuple of positive real numbers. A
solution to the Algebraic Counterpart of the $\mathbb{w}$-SDOP Problem over
$\mathbb{K}$ ($\mathbb{w}$-AlgDOP Problem over $\mathbb{K}$ for short) is a
pair $(g,\Gamma)$ where $g=(g^{ij})$ is a symmetric $d\times d$ matrix with
polynomial entries and $\Gamma$ a polynomial such that

Thus Proposition 2.1 implies that if $(\Omega,g,\rho)$ is a solution of the
$\mathbb{w}$-SDOP Problem and $\Gamma$ is a generator of $I(\partial\Omega)$,
then $(g,\Gamma)$ is a solution of the $\mathbb{w}$-AlgDOP Problem over
$\mathbb{R}$ and hence over $\mathbb{C}$.

The following facts immediately follow from the definition.

######  Proposition \propW

Let $\mathbb{w}=(w_{1},\dots,w_{d})$ and
$\mathbb{w}^{\prime}=(w^{\prime}_{1},\dots,w^{\prime}_{d})$ be two $d$-tuples
of positive weights. Suppose that $(g,\Gamma)$ is a solution of the
$\mathbb{w}$-AlgDOP Problem over $\mathbb{K}$, and
$g^{ij}\in\Cal{P}_{\mathbb{w}^{\prime}}(w^{\prime}_{i}+w^{\prime}_{j};\mathbb{K})$
for each $i,j=1,\dots,d$. Then $(g,\Gamma)$ is a solution of the
$\mathbb{w}^{\prime}$-AlgDOP Problem over $\mathbb{K}$.

######  Proposition \propGammaOne

If $(g,\Gamma)$ is a solution of the $\mathbb{w}$-AlgDOP Problem over
$\mathbb{K}$ and $\Gamma_{1}$ is a factor of $\Gamma$, then $(g,\Gamma_{1})$
is also a solution of the $\mathbb{w}$-AlgDOP Problem over $\mathbb{K}$.

If a square-free polynomial $\Gamma$ is given, then it is easy to find all
cometrics $g$ such that $(g,\Gamma)$ is a solution of the $\mathbb{w}$-AlgDOP
problem. Indeed, Condition (A3) of Definition 2.2 (in the form (4)) provides a
system of linear equations for the coefficients of the polynomials $g^{ij}$
and $S^{i}$. In §2.3 we show how to find all $\rho$ for given $\Omega$ and
$g$.

### 2.2. Admissible changes of variables

A $\mathbb{w}$-admissible change of variables is a bijective polynomial
mapping $\Phi:\mathbb{R}^{d}\to\mathbb{R}^{d}$,
$x\mapsto(y_{1}(x),\dots,y_{d}(x))$ with $\deg_{\mathbb{w}}(y_{i})=w_{i}$,
$i=1,\dots,d$. If $d=2$, then $(1,w)$-admissible automorphisms of
$\mathbb{R}^{2}$ for $w=1$ are affine linear transformations, and for $w>1$,
mappings of the form

$(x,y)\mapsto(\alpha x+\beta,\,\gamma y+p(x)),\qquad\alpha\gamma\neq
0,\;\deg(p)\leq w.$ $None$

The following proposition is similar to [3, Prop. 2.5] and we omit its proof.

######  Proposition \propChangeVar

(a). Let $(\Omega,\mathbb{L},\mu)$ be a solution of $\mathbb{w}$-DOP (resp.
$\mathbb{w}$-SDOP) problem and $\Phi$ be a $\mathbb{w}$-admissible change of
variables. Let $\Omega_{1}=\Phi(\Omega)$,
$\mathbb{L}_{1}(f)=\mathbb{L}(f\circ\Phi)\circ\Phi^{-1}$, and
$\mu_{1}(E)=\mu(\Phi^{-1}(E))$. Then $(\Omega_{1},\mathbb{L}_{1},\mu_{1})$ is
also a solution of $\mathbb{w}$-DOP (resp. $\mathbb{w}$-SDOP) problem.

(b). Let $(g,\Gamma)$ be a solution of the $\mathbb{w}$-AlgDOP Problem over
$\mathbb{K}$ and let $\Phi$ be a $\mathbb{w}$-admissible change of variables.
Then $(\Phi_{*}(g),\Gamma\circ\Phi^{-1})$ is also a solution of the
$\mathbb{w}$-AlgDOP Problem over $\mathbb{K}$.

Example 2.6.  Let $d=2$. For any $\mathbb{w}=(w_{1},w_{2})$, the mapping
$\Phi:(x,y)\mapsto(x,-y)$ is $\mathbb{w}$-admissible. Thus, if $(g,\Gamma)$
with $g=\left(\begin{smallmatrix}a&b\\\ b&c\end{smallmatrix}\right)$ is a
solution of the $\mathbb{w}$-AlgDOP Problem over $\mathbb{K}$, then
$(\Phi_{*}(g),\Gamma(x,-y))$ with

$\Phi_{*}(g)=\left(\begin{matrix}a(x,-y)&-b(x,-y)\\\
-b(x,-y)&c(x,-y)\end{matrix}\right)$

is also a solution of the $\mathbb{w}$-AlgDOP problem over $\mathbb{K}$.

Example 2.7.  More generally, let still $d=2$. Any $(1,w)$-admissible change
of variables for $w>1$ is of the form (5) and it is a composition of the
following variable changes $(x,y)\mapsto(X,Y)$:

$T:(x,y)\mapsto(x+\beta,y),\quad H:(x,y)\mapsto(\alpha x,\gamma y),\quad
S:(x,y)\mapsto(x,y+p(x)).$

They transform $g=\left(\begin{smallmatrix}a&b\\\ b&c\end{smallmatrix}\right)$
as follows (see [3, eq. (2.3)]): $T_{*}(g)=g(X-\beta,Y-\beta)$,

$H_{*}(g)=\left(\begin{matrix}a\alpha^{2}&b\alpha\gamma\\\
b\alpha\gamma&c\gamma^{2}\end{matrix}\right)_{\begin{smallmatrix}x=X/\alpha\\\
y=Y/\gamma\end{smallmatrix}}\,\quad
S_{*}(g)=\left(\begin{matrix}a&p^{\prime}a+b\\\
p^{\prime}a+b&(p^{\prime})^{2}a+2p^{\prime}b+c\end{matrix}\right)_{\begin{smallmatrix}\hskip-22.0ptx=X\\\
y=Y-p(X)\end{smallmatrix}}$

### 2.3. From AlgDOP to DOP/SDOP

As we told in §2.1, any solution $(\Omega,g,\mu)$ of the $\mathbb{w}$-SDOP
problem gives a solution $(\Gamma,g)$ of the $\mathbb{w}$-AlgDOP problem where
$\Gamma$ is a generator of the ideal $I(\partial\Omega)$. In this subsection
we discuss how to find all possible $(\Omega,g,\mu)$ from a given
$(\Gamma,g)$.

So, let $(\Gamma,g)$ be a solution of the $\mathbb{w}$-AlgDOP problem over
$\mathbb{R}$. First of all we find all connected components $\Omega$ of
$\mathbb{R}^{d}\setminus\\{\Gamma=0\\}$ such that $g$ is positive definite on
$\Omega$. Then it remains to find all measure densities $\rho$ such that the
operator $\mathbb{L}$ given by (3) is of the form (1) with
$b^{i}\in\Cal{P}_{\mathbb{w}}(w_{i})$ for each $i=1,\dots,d$. By comparing (1)
with (3) we obtain

$b^{i}=\sum_{j}\partial_{j}g^{ij}+\sum_{j}g^{ij}\partial_{j}h,\qquad
h=\log\rho.$

Hence

$\partial_{j}h=\sum_{i}g_{ij}L^{i},\qquad
L^{i}=b^{i}-\sum_{j}\partial_{j}g^{ij},$ $None$

where $(g_{ij})=g^{-1}$. We have
$b^{i}\in\Cal{P}_{\mathbb{w}}(w_{i})\Leftrightarrow
L^{i}\in\Cal{P}_{\mathbb{w}}(w_{i})$. The identities
$\partial_{k}(\partial_{j}h)=\partial_{j}(\partial_{k}h)$ combined with (6)
yield

$\partial_{k}\Big{(}\sum_{i}g_{ij}L^{i}\Big{)}=\partial_{j}\Big{(}\sum_{i}g_{ik}L^{i}\Big{)}$
$None$

which is a system of linear equations for the coefficients of all $L^{i}$.
These observations lead to the following algorithm of finding all solutions
for $\rho$. We start with polynomials $L^{i}$, $\deg_{\mathbb{w}}L^{i}=w_{i}$,
whose coefficients we compute by solving the system of linear equations coming
from (7). The solution may depend on several parameters. Then we find $h$ by
integrating the $L^{i}$’s and set $\rho=\exp(h)$. Finally we choose the values
of the parameters such that $Q(x)\rho(x)$, is integrable over $\Omega$ for any
polynomial $Q$ (when $\Omega$ is bounded, it is enough to demand that
$\int_{\Omega}\rho\,dx<\infty$).

These observations allow us to prove the following fact.

######  Proposition \propOPlus

Let $(\Omega,g,\rho)$ be a solution of the $\mathbb{w}$-SDOP problem in
$\mathbb{R}^{d}$ such that
$g=\operatorname{diag}\big{(}g^{11}(x_{1}),\dots,g^{dd}(x_{d})\big{)}$. Then
there is a partition of the set of indices
$\\{1,\dots,d\\}=J_{1}\sqcup\dots\sqcup J_{m}$,
$J_{k}=\\{j_{k}(1),\dots,j_{k}(d_{k})\\}$, such that:

###### Demonstration Proof

We consider only the case $d=2$. It is not difficult to derive from it the
general case. Let $h=\log\rho$. It is enough to prove that either $g$ is
constant, or $\partial_{12}h=0$ and then $h=h_{1}(x_{1})+h_{2}(x_{2})$ whence
$\rho=\rho_{1}(x_{1})\rho_{2}(x_{2})$. Indeed, the inverse matrix of $g$ is
$\operatorname{diag}(g_{11},g_{22})$ where $g_{ii}=1/g^{ii}$. Then by (7) we
have

$\partial_{12}h=\frac{\partial_{2}L^{1}(x_{1},x_{2})}{g^{11}(x_{1})}=\frac{\partial_{1}L^{2}(x_{1},x_{2})}{g^{22}(x_{2})},\qquad\deg_{\mathbb{w}}L^{i}\leq
w_{i}.$

Therefore $\partial_{12}h$ is a polynomial, hence $g^{ii}$ divides
$\partial_{j}L^{i}$ for $i\neq j$. Thus, if $\partial_{12}h$ is nonzero, we
have
$0\leq\deg_{\mathbb{w}}g^{11}\leq\deg_{\mathbb{w}}\partial_{2}L^{1}=\deg_{\mathbb{w}}L^{1}-w_{2}\leq
w_{1}-w_{2}$ and, symmetrically, $0\leq\deg_{\mathbb{w}}g^{22}\leq
w_{2}-w_{1}$, which implies $w_{1}=w_{2}$ and $\deg_{\mathbb{w}}g=0$, i.e.,
$g$ is constant. ∎

To conclude this subsection, we give some conditions (necessary or sufficient)
on $\rho$ to be compatible with given $g$ and $\Omega$. They are proven in [3]
for the usual degree but the proofs can be easily adapted for any weighted
degree.

######  Proposition \propDOP

(See [3, Thm. 2.21].) Let $(g,\Gamma)$, $\Gamma=\Gamma_{1}\dots\Gamma_{s}$, be
a solution of $\mathbb{w}$-AlgDOP problem over $\mathbb{R}$ (recall that then
$\Gamma$ is squarefree). Let $\Omega$ be a bounded domain such that
$\partial\Omega\subset\\{\Gamma=0\\}$ and $g$ is positive definite in
$\Omega$. Let $a_{1},\dots,a_{s}$ be real numbers such that
$\mu(\Omega)<\infty$ where $\mu$ is the measure with density
$\rho=\prod_{\nu}\Gamma_{\nu}^{a_{\nu}}$ (for example, $a_{\nu}\geq 0$ for
each $\nu$). Then $(\Omega,\mathbb{L},\mu)$ is a solution of $\mathbb{w}$-DOP
problem, where $\mathbb{L}$ is given by (3).

Definition 2.10. A solution $(g,\Gamma)$ of the $\mathbb{w}$-AlgDOP problem is
called maximal (and in this case $\Gamma$ is called a maximal boundary for
$g$) if $\Gamma_{1}$ divides $\Gamma$ for any solution $(g,\Gamma_{1})$. By
Proposition 2.4, a maximal solution is unique for any given $g$.

######  Proposition \propMes

(See [3, Props. 2.15, 2.17].) Let $(\Omega,g,\rho)$ be a solution of
$\mathbb{w}$-SDOP problem in $\mathbb{R}^{d}$ and $\Delta=\det(g)$. Let
$\Gamma$ be the maximal boundary for $g$ and let $\Gamma_{1},\dots,\Gamma_{s}$
be its irreducible (over $\mathbb{C}$) factors. Suppose that each factor
$\Gamma_{k}$ occurs in $\Delta$ with multiplicity $1$, i.e., $\Gamma_{k}^{2}$
does not divide $\Delta$. Then (see Remark 2.12)

$\rho=\Gamma_{1}^{p_{1}}\dots\Gamma_{s}^{p_{s}}\exp(Q)$ $None$

for some $p_{1},\dots,p_{s}\in\mathbb{C}$ and a polynomial $Q$ such that, for
each $j=1,\dots,d$,

$w_{j}\deg_{x_{j}}(Q\Delta)\leq 2w_{1}+\dots+2w_{d}.$ $None$

Remark 2.12. In Proposition 2.11, if $\lambda\Gamma_{k}$,
$\lambda\in\mathbb{C}$, is real for some $k$, we always assume that
$\Gamma_{k}$ is real and positive on $\Omega$; in this case $p_{k}$ is real.
Otherwise $\bar{\Gamma}_{k}$ is also a factor of $\Gamma$ and it must occur in
$\rho$ with the power $\bar{p}_{k}$ because $\rho$ is real. In this case
$\Gamma_{k}^{p_{k}}\bar{\Gamma}_{k}^{\bar{p}_{k}}$ is understood as a single-
valued branch of this function on $\Omega$. Notice that choosing another
single-valued branch we just change the constant term of $Q$.

Remark 2.13. Proposition 2.11 admits the following refinement. We still have
(8) and (9) for a variable $x_{j}$ even when there are multiple factor
$\Gamma_{k}$ of $\Delta$ but they are polynomials in variables $(x_{i})_{i\in
I}$ not including $x_{j}$, that is $j\not\in I$. In this case $Q$ is a
polynomial in $(x_{i})_{i\not\in I}$ whose coefficients are rational functions
in $(x_{i})_{i\in I}$.

######  Corollary \corMes

Under the hypothesis of Proposition 2.11, assume that
$mw_{d}=2(w_{1}+\dots+w_{d})$ and $w_{d}=w_{i}n_{i}$, $i=1,\dots,d$, for some
$m,\,n_{1},\dots,n_{d}$ (e.g. $d=2$ and $(w_{1},w_{2})=(1,2)$). Suppose that
$\deg_{x_{d}}\Delta=m$. Then $Q=\operatorname{const}$ in (8).

###### Demonstration Proof

By an admissible change of variables
$y_{d}=x_{d}+\sum_{i=1}^{d-1}a_{i}x_{i}^{n_{i}}$ and $y_{j}=x_{j}$ for $j<d$,
we may achieve that $w_{j}\deg_{y_{j}}\Delta=2\sum w_{i}$ for each $j$. ∎

Recall that $(g_{ij})=g^{-1}$, hence $g_{ij}=\hat{g}_{ij}/\det g$ with a
polynomial $\hat{g}_{ij}$.

######  Proposition \propMesII

(See [3, Cor. 2.19].) Let
$U=\mathbb{R}\times\mathbb{B}^{d-1}=\\{x_{2}^{2}+\dots+x_{d}^{2}<1\\}$ and
$U_{+}=\mathbb{R}_{+}\times\mathbb{B}^{d-1}=U\cap\\{x_{1}>0\\}$. Let
$M_{1}=\max_{j}\big{(}\lfloor
w_{j}/w_{1}\rfloor+\deg_{x_{1}}\hat{g}_{1j}\big{)}$. Let $(\Omega,g,\rho)$ be
a solution of the $\mathbb{w}$-SDOP problem, $\Delta=\det g$. Suppose that
$U\subset\Omega$ (resp. $U_{+}\subset\Omega$). Then $\deg_{x_{1}}\Delta<M_{1}$
(resp. $\deg_{x_{1}}\Delta<1+M_{1}$).

## 3\. Weighted AlgDOP Problem in $\mathbb{C}^{2}$

### 3.1. Notation

In §§3–5 we study the $\mathbb{w}$-AlgDOP Problem over $\mathbb{C}$ in
dimension 2 for any pair of weights $\mathbb{w}$. It is clear that a
multiplication of $\mathbb{w}$ by a positive number does not change the
problem. Therefore we assume throughout this section that $(g,\Gamma)$, is a
solution of the $\mathbb{w}$-AlgDOP Problem over $\mathbb{C}$ for
$\mathbb{w}=(1,w)$ with a real $w>1$ (the case $w=1$ is already done in [3]).

We denote variables by $(x,y)$ and we set $g=\left(\begin{smallmatrix}a&b\\\
b&c\end{smallmatrix}\right)$. We denote the coefficient of $y^{m}$ in $a(x,y)$
by $a_{m}(x)$ and the coefficient of $x^{k}y^{m}$ by $a_{km}$ and we use
similar notation for $b$ and $c$. Sometimes we write $(a,b,c;\Gamma)$ instead
of $(g,\Gamma)$ when speaking of a solution of the $\mathbb{w}$-AlgDOP
Problem.

As usual, given a polynomial $P(x,y)=\sum p_{km}x^{k}y^{m}$, we define its
Newton polygon $\Cal{N}(P)$ as the convex hull in $\mathbb{R}^{2}$ of the
finite set $\\{(k,m)\mid a_{km}\neq 0\\}$. Condition (A1) of Definition 2.2
means that the Newton polygons of $a$, $b$, $c$, and $\Delta$ are contained in
the polygons shown in Figures 1–2.

$a$ $b$ $c$ $\Delta$

Figure 1 Polygons containing $\Cal{N}(a)$, $\Cal{N}(b)$, $\Cal{N}(c)$,
$\Cal{N}(\Delta)$ for $1<w\leq 2$.

1+w 2w 2+2w

$a$ $b$ $c$ $\Delta$

Figure 2 Polygons containing $\Cal{N}(a)$, $\Cal{N}(b)$, $\Cal{N}(c)$,
$\Cal{N}(\Delta)$ for $w>2$.

### 3.2. Change of the weights and the $(1,\infty)$-AlgDOP Problem

According to Proposition 2.3, any solution of $(1,w)$-AlgDOP Problem with
$1<w\leq 2$ is a solution of the corresponding $(1,2)$-AlgDOP Problem (see
Figure 1). Similarly (see Figure 2) any solution of $(1,w)$-AlgDOP Problem
with $w>2$ is a solution of the corresponding $(1,w^{\prime})$-Problem for any
$w^{\prime}>w$.

So, we say that $(g,\Gamma)$ is a solution of the $(1,\infty)$-AlgDOP Problem,
if it is a solution of the $(1,w)$-AlgDOP Problem for some $w>2$, and a
coordinate change (5) with an arbitrary polynomial $p(x)$ will be called
$(1,\infty)$-admissible.

In the next two sections we find all solutions of the $(1,\infty)$\- and
$(1,2)$-AlgDOP Problems up to $(1,\infty)$\- and $(1,2)$-admissible coordinate
change.

### 3.3. Local branches of the curve $\Gamma=0$

Let $P(x,y)$ be a squarefree polynomial. A local branch of $P$ (or,
equivalently, of the curve $P=0$) is a pair $\gamma=(\xi,\eta)$ of germs at
$0$ of meromorphic functions such that $P(\xi(t),\eta(t))$ is identically
zero. Any meromorphic germ $t\mapsto\gamma(t)=(\xi(t),\eta(t))$ defines a
valuation $v_{\gamma}:\mathbb{C}[x,y]\to\mathbb{Z}\cup\\{\infty\\}$,
$v_{\gamma}(Q)=\operatorname{ord}_{t}Q(\xi(t),\eta(t))$ where
$\operatorname{ord}_{t}(0)=\infty$ and $\operatorname{ord}_{t}f(t)=p_{m}$ if
$f(t)=\sum_{k\geq m}p_{k}t^{k}$ and $p_{m}\neq 0$.

Let $(a,b,c;\,\Gamma)$ be a solution of the $\mathbb{w}$-AlgDOP problem over
$\mathbb{C}$ for some $\mathbb{w}=(1,w)$. Condition (A3) of Definition 2.2
reads

$a\Gamma^{\prime}_{x}+b\Gamma^{\prime}_{y}=L_{1}\Gamma,\qquad
b\Gamma^{\prime}_{x}+c\Gamma^{\prime}_{y}=L_{2}\Gamma.$ $None$

It is easy to check that this condition is equivalent to

$b(\xi,\eta)\dot{\xi}=a(\xi,\eta)\dot{\eta},\qquad
c(\xi,\eta)\dot{\xi}=b(\xi,\eta)\dot{\eta}$ $None$

for any local branch of $\Gamma$. Condition (11) implies that

$v_{\gamma}(a)-v_{\gamma}(b)=v_{\gamma}(b)-v_{\gamma}(c)=\operatorname{ord}_{t}(\dot{\xi})-\operatorname{ord}_{t}(\dot{\eta})$
$None$

if both $\xi(t)$ and $\eta(t)$ are non-constant (see [3, Lemma 3.3]).

The following fact is well-known and immediately follows from the definitions.

######  Lemma \lemNewton

Let $F$ be a polynomial in $(x,y)$ and let
$(p,q)\in\mathbb{Z}^{2}\setminus\\{(0,0)\\}$. A local branch $\gamma$ of $F$
such that $\operatorname{ord}_{t}(\gamma)=(p,q)$ exists if and only if the
vector $(p,q)$ is orthogonal to some edge of $\Cal{N}(F)$ and points from this
edge inward $\Cal{N}(\Gamma)$. ∎

######  Lemma \lemChangeW

Let $(a,b,c;\,\Gamma)$ is a solution of $(1,w)$-AlgDOP Problem with $w>1$.
Suppose that $\Gamma$ is divisible by $x$. Then $a$ and $b$ are divisible by
$x$, hence $\deg_{y}a\leq 1$ and $(a,b,c;\,\Gamma)$ is a solution of the
$(1,\infty)$-AlgDOP Problem.

###### Demonstration Proof

We have $a(0,t)=b(0,t)=0$ by (11), hence $x$ is a factor of $a$ and $b$ whence
$\deg_{y}a\leq 1$ (see Figure 1). ∎

######  Lemma \lemValuA

Let $w>1$. Suppose that $\Gamma$ has a branch $\gamma=(\xi(t),\eta(t))$ such
that $v_{\gamma}(x,y)=\operatorname{ord}_{t}\gamma=(p,q)$ with $q<0<p$. Then
$b_{1}=b_{11}x$ and $a=a_{20}x^{2}$, hence $(g;\,\Gamma)$ is a solution of
$(1,\infty)$-AlgDOP Problem.

###### Demonstration Proof

We have
$\operatorname{ord}_{t}\dot{\xi}-\operatorname{ord}_{t}\dot{\eta}=p-q$, hence
$v_{\gamma}(a)-v_{\gamma}(b)=v_{\gamma}(b)-v_{\gamma}(c)=p-q$ by (12). On the
other hand we have $v_{\gamma}(c)\geq v_{\gamma}(y^{2})=2q$, hence

$v_{\gamma}(b)=v_{\gamma}(c)+p-q\geq p+q.$ $None$

If $b_{01}$ were non-zero, then we would have $v_{\gamma}(b)=v_{\gamma}(y)=q$
which contradicts (13) because $p>0$. Hence $b_{1}=b_{11}x$ (as required), and
this fact implies that

$v_{\gamma}(b)\geq\min(v_{\gamma}(1),v_{\gamma}(xy))=\min(0,p+q),$

thus $v_{\gamma}(a)=v_{\gamma}(b)+p-q\geq\min(0,p+q)+p-q=p+\min(-q,p)>p$.
Since the only monomials which may occur in $a$ are $(y,1,x,x^{2})$ and their
$\gamma$-valuations are $(q,0,p,2p)$, we see that $a=a_{20}x^{2}$. ∎

###### Lemma \lemValu

Let $1<w\leq 2$. Let $\gamma=(\xi(t),\eta(t)$ be a local branch of $\Gamma$,
and $(p,q)=\operatorname{ord}_{t}\gamma=v_{\gamma}(x,y)$.

(a). If $(p,q)=(2,1)$ then $a_{00}=a_{01}=b_{00}=0$, and hence
$\deg_{y}\Delta\leq 2$ and $(a,b,c;\Gamma)$ is a solution of the
$(1,\infty)$-AlgDOP problem.

(b). If $q>0$, then $p\neq 3$.

(c). If $p=-3$ and $-4\leq q\leq-1$, then $\deg b_{0}\leq 2$, $\deg c_{0}\leq
2$, and $\deg c_{1}\leq 1$, hence $(a,b,c;\Gamma)$ is a solution of the
$(1,1)$-AlgDOP problem.

(d). If $q<2p<0$, then $a_{01}=0$, hence $(a,b,c;\Gamma)$ is a solution of the
$(1,\infty)$-AlgDOP problem.

###### Demonstration Proof

(a). If $(p,q)=(2,1)$, then
$\operatorname{ord}_{t}(\dot{\xi},\dot{\eta})=(1,0)$, hence by (12) we have
$v_{\gamma}(b)=v_{\gamma}(c)+1\geq 1$ and
$v_{\gamma}(a)=v_{\gamma}(b)+1=v_{\gamma}(c)+2\geq 2$, and the result follows
from the fact that $1$ (resp. $y$) is the only monomial of $a$ and $b$ of the
$\gamma$-valuation equal to $0$ (resp. to $1$). The condition $a_{01}=0$
implies that $(a,b,c;\Gamma)$ is a solution of the $(1,1)$-AlgDOP problem (see
Figure 2).

(b). Suppose that $p=3$ and $q>0$, Then by the arguments similar to those in
(a), we are going to show that $\deg_{y}a\leq 0$ and hence $\deg_{y}\Delta\leq
2$ which contradicts the condition $\operatorname{ord}_{t}\eta=3$. Notice that
$q\leq\deg_{x}\Delta\leq 6$.

If $q=1$, the proof is the same as in (a).

If $q=2$, then by (12) we have $v_{\gamma}(a)=v_{\gamma}(b)+1$. Since
$v_{\gamma}(a)\neq 1$ and $v_{\gamma}(b)\neq 1$, this is possible only when
$v_{\gamma}(a)\geq 3$. By combining this fact with
$v_{\gamma}(1,y,x,x^{2})=(0,2,3,6)$, we obtain $a_{00}=a_{01}=0$ whence
$\deg_{y}a\leq 0$.

The case $q=3$ is reduced to $q>3$ by a change of variables
$(x,y)\mapsto(x,y+\lambda x)$.

If $q=4$, then $v_{\gamma}(a)\in\\{0,3,4,6\\}$,
$v_{\gamma}(b)\in\\{0,3,4,6,7,9\\}$, $v_{\gamma}(c)\in\\{0,3,4,6,\dots\\}$.
Then one can check that
$v_{\gamma}(c)-v_{\gamma}(b)=v_{\gamma}(b)-v_{\gamma}(a)=1$ (this is (12)) is
possible only when $v_{\gamma}(a)=6$, hence $\deg_{y}a\leq 0$.

If $q=5$, then $v_{\gamma}(a)\in\\{0,3,5,6\\}$,
$v_{\gamma}(b)\in\\{0,3,5,6,8,9\\}$, $v_{\gamma}(c)\in\\{0,3,5,6,8,\dots\\}$.
Then $v_{\gamma}(c)-v_{\gamma}(b)=v_{\gamma}(b)-v_{\gamma}(a)=2$ (this is
(12)) is possible only when $v_{\gamma}(a)=6$, hence $\deg_{y}a\leq 0$.

The case $q=6$ is reduced to $q>6$ by a change of variables
$(x,y)\mapsto(x,y+\lambda x^{2})$.

(c). We need to show that of $x^{3}$, $x^{4}$, and $x^{2}y$ do not occur in
$b$ and in $c$.

If $p=-3$ and $-3\leq q\leq-1$, then (12) implies $v_{\gamma}(c)\geq
v_{\gamma}(b)\geq v_{\gamma}(a)\geq v_{\gamma}(x^{2})=-6$. On the other hand,
$v_{\gamma}(x^{2}y)=-6+q\geq-7$, $v_{\gamma}(x^{3})=-9$,
$v_{\gamma}(x^{4})=-12$, and these are the only monomials which might occur in
$b$ and in $c$ with the respective $\gamma$-valuations.

If $(p,q)=(-3,-4)$, then (12) implies $v_{\gamma}(a)\geq
v_{\gamma}(x^{2})=-6$, $v_{\gamma}(b)=v_{\gamma}(a)-1\geq-7$, and
$v_{\gamma}(c)=v_{\gamma}(b)-1\geq-8$. On the other hand,
$v_{\gamma}(x^{2}y)=-10$, $v_{\gamma}(x^{3})=-9$, $v_{\gamma}(x^{4})=-12$, and
these are the only monomials which might occur in $b$ and in $c$ with the
respective $\gamma$-valuations.

(d). The condition $q<2p<0$ implies $v_{\gamma}(b)\geq v_{\gamma}(xy)=p+q$. It
implies also $p-q>-p$. By (12) we have $v_{\gamma}(a)=v_{\gamma}(b)+p-q$.
Hence

$v_{\gamma}(a)=v_{\gamma}(b)+p-q>v_{\gamma}(b)-p\geq(p+q)-p=q=v_{\gamma}(y).$

Therefore $a_{01}=0$ because $y$ is the only monomial which may occur in $a$
with the $\gamma$-valuation equal to $q$. The condition $a_{01}=0$ implies
that $(a,b,c;\Gamma)$ is a solution of the $(1,1)$-AlgDOP problem (see Figure
2). ∎

## 4\. Solution of the $(1,w)$-AlgDOP Problem in $\mathbb{C}^{2}$ for $w>2$

In this section we give all solutions of the $(1,\infty)$-AlgDOP Problem over
$\mathbb{C}$ up to $(1,\infty)$-admissible change of coordinates. As we
explained in §3.2, this gives all solutions of the $(1,w)$-AlgDOP Problem for
all $w>2$.

######  Lemma \lemA

Let $(a,b,c;\Gamma)$ be a solution of the $(1,w)$-AlgDOP problem over
$\mathbb{C}$ with $w>1$. Suppose that $\deg_{y}\Gamma=2$ and that $\Gamma$ is
not divisible by any non-constant polynomial in $x$. Then $\Gamma$ is monic
with respect to $y$, i.e., $y^{2}$ is the only monomial of $\Gamma$ whose
$y$-degree is $2$.

###### Demonstration Proof

First, remark that $a\neq 0$. Indeed, otherwise $\Delta=b^{2}$, hence $\Gamma$
would be a square-free factor of $b$ which contradicts the condition
$\deg_{y}\Gamma=2$. Suppose that the coefficient of $y^{2}$ in $\Gamma$ is a
non-constant polynomial in $x$. Without loss of generality we may assume that
one of its roots is $0$. Then $\Gamma$ has a branch $\gamma=(\xi(t),\eta(t))$
such that $\operatorname{ord}_{t}\eta<0<\operatorname{ord}_{t}\xi$. Then, by
Lemma 3.3, we have $a=a_{20}x^{2}$ and $b=b_{11}xy+b_{0}(x)$. Since $a\neq 0$,
we may assume that $a=x^{2}$.

If $b_{00}=0$, then $x^{2}$ is a factor of $ac$ and of $b^{2}$, hence $x^{2}$
is a factor of $\Delta$. Since $\Delta$ may not have have monomials
$y^{2}x^{k}$ with $k>2$, it follows that $\Gamma$ is monic in $y$.

Now consider the case $b_{00}\neq 0$. Then the constant term of $\Delta$ is
$-b_{00}^{2}\neq 0$. We have $a=x^{2}$ and $b_{1}=b_{11}x$, hence $x^{2}y^{2}$
is the only monomial of $y$-degree $2$ which may occur in $\Delta$. Since
$\deg_{y}\Delta=2$, we conclude that $\Delta$ cannot be divisible by any non-
constant polynomial in $x$, i.e., $\Gamma=\Delta$.

The coefficients of $y^{2}$ in $\Delta$ and in
$a\Delta^{\prime}_{x}+b\Delta^{\prime}_{y}$ are $d_{22}x^{2}$ and
$2d_{22}(1+b_{11})x^{3}$ respectively where $d_{22}=c_{02}-b_{11}^{2}$. Then
(10) implies $L_{1}=2(1+b_{11})x$. By plugging this back into (10), we obtain
a contradiction. Indeed, we have:

$\begin{split}\Delta&=-b^{2}+O(x^{2})=-b_{00}^{2}-2b_{00}(b_{10}+b_{11}y)x+O(x^{2}),\\\
b\Delta^{\prime}_{y}&=\big{(}b_{00}+O(x)\big{)}\big{(}-2b_{00}b_{11}x+O(x^{2})\big{)}=-2b_{00}^{2}b_{11}x+O(x^{2}),\\\
L_{1}\Delta&=-2b_{00}^{2}(1+b_{11})x+O(x^{2}).\end{split}$

Hence
$a\Delta^{\prime}_{x}+b\Delta^{\prime}_{y}-L_{1}\Delta=2b_{00}^{2}x+O(x^{2})\neq
0$. ∎

######  Lemma \lemSym

Let $\Gamma=y^{2}-p(x)$ and let $(a,b,c;\,\Gamma)$ be a solution of the
$(1,w)$-AlgDOP problem with $w>2$. Then $b_{0}=c_{1}=0$, i.e., $a$ and $c$ are
even with respect to $y$, and $b$ is odd with respect to $y$ (this means that
the corresponding metric is invariant under the symmetry $y\mapsto-y$).

###### Demonstration Proof

Let us set $\hat{a}=a$, $\hat{b}(x,y)=-b(x,-y)$, $\hat{c}(x,y)=c(x,-y)$, and
$\hat{g}=(\hat{a},\hat{b},\hat{c})$. Since $\Gamma$ is symmetric, Proposition
2.5 implies that $(\hat{g};\,\Gamma)$ is also a solution to the $(1,w)$-AlgDOP
problem (see Example 2.6). Hence both $(g;\Gamma)$ and $(\hat{g};\Gamma)$
satisfy the equations (10) and, by linearity,
$\big{(}\frac{1}{2}(g-\hat{g});\,\Gamma\big{)}$ satisfies it as well. Since
$\frac{1}{2}(g-\hat{g})=(0,b_{0},c_{1}y)$, this means
$2yb_{0}=(y^{2}+p(x))L_{1}$ and
$-b_{0}p^{\prime}(x)+2y^{2}c_{1}=(y^{2}+p(x))L_{2}$. The first equation
implies $b_{0}=0$. Plugging this into the second equation, we obtain
$2y^{2}c_{1}=(y^{2}+p(x))L_{2}$. Note that $p\neq 0$ because $\Gamma$ cannot
have multiple factors. Hence the last equations implies $c_{1}=0$, i.e.,
$g-\hat{g}=0$. ∎

######  Proposition \propDegGaTwo

The following is a complete list of solutions $(g;\Gamma)$ of the
$(1,\infty)$-AlgDOP problem over $\mathbb{C}$ up to $(1,\infty)$-admissible
change of variables under condition that $\deg_{y}\Gamma=2$:

###### Demonstration Proof

Let $\Gamma=\Gamma_{0}\Gamma_{2}$ where $\deg_{y}\Gamma_{0}=0$,
$\deg_{y}\Gamma_{2}=2$, and $\Gamma_{2}$ is not divisible by any non-constant
polynomial in $x$. Then $\Gamma_{2}$ is monic in $y$ by Lemma 4.1 combined
with Proposition 2.4, hence it can be reduced to the form
$\Gamma_{2}=y^{2}+p(x)$ by a $(1,\infty)$-admissible change of coordinates.
Since $\Gamma$ cannot have any multiple factor, $p$ is not identically zero.
Moreover, by rescaling the $y$-coordinate, we may set the leading coefficient
of $p$ to any prescribed nonzero complex number. By Lemma 4.2, we have
$a=a(x)$, $b=b_{1}(x)y$, and $c=c_{2}y^{2}+c_{0}(x)$ (with
$c_{2}=c_{02}=\operatorname{const}$). Also $a\neq 0$ (see the beginning of the
proof of Lemma 4.1). By (10) we have

$a(\Gamma_{2})^{\prime}_{x}+b(\Gamma_{2})^{\prime}_{y}=ap^{\prime}+2b_{1}y^{2}=(y^{2}+p)L_{1}.$

Hence $L_{1}=2b_{1}$ and then

$ap^{\prime}=2pb_{1}.$ $None$

The second equation in (10) reads

$b(\Gamma_{2})^{\prime}_{x}+c(\Gamma_{2})^{\prime}_{y}=b_{1}yp^{\prime}+2y(c_{2}y^{2}+c_{0})=(y^{2}+p)L_{2}$

whence $L_{2}=2c_{2}y$ and hence $c_{0}=c_{2}p-\frac{1}{2}b_{1}p^{\prime}$,
i.e., $c=c_{2}\Gamma_{2}-\frac{1}{2}b_{1}p^{\prime}$. By combining this fact
with (14), we see that

$c=c_{2}\Gamma_{2}-b_{1}^{2}p/a.$ $None$

If $b=0$, we have $c=c_{02}\Gamma_{2}$ by (15) and $p=\operatorname{const}$ by
(14). Then we may set $p=-1$ and we arrive to (iii)–(v) with $n=0$ in (iii).

Now let $b\neq 0$. The equation (14) can be rewritten as $(\log
p)^{\prime}=2b_{1}/a$. Since $p$, $a$, and $b_{1}$ are polynomials and
$\deg(a)\leq 2$, we conclude that, up to translation and rescaling the
variable $x$, one of the following three cases occurs.

Case 1. $a=1-x^{2}$.

$\frac{2b_{1}}{a}=\frac{p^{\prime}}{p}=\frac{n}{1+x}-\frac{m}{1-x},\qquad
p=-(1-x)^{m}(1+x)^{n},\qquad m,n>0$

(recall that the leading coefficient of $p$ can be chosen arbitrarily).
Therefore $b_{1}=\frac{1}{2}(n-m)-\frac{1}{2}(n+m)x$. By combining this fact
with (15) we obtain (i) as soon as $\Gamma=-\Gamma_{2}$. So, it remains to
show that $\Gamma$ coincides with $\Gamma_{2}$ up to a constant factor, that
is $\Gamma_{0}=\operatorname{const}$. Indeed, if $\Gamma_{0}$ were non-
constant, by Lemma 3.2 it would be a common factor of $a$ and $b$, but this is
impossible for our explicit form of $a$ and $b$.

Case 2. $a=x$, $p^{\prime}/p=2b_{1}/a=n/x$, $p=x^{n}$, $n\geq 0$, hence
$b_{1}=n/2$. Since $a$ and $b$ are coprime, $\Gamma=\Gamma_{2}$ and we arrive
to (ii) similarly to Case 1.

Case 3. $a=x(x_{0}-x)$, $p^{\prime}/p=2b_{1}/a=-n/(x_{0}-x)$,
$p=-(x_{0}-x)^{n}$, $n\geq 0$, hence $b_{1}=-nx/2$. This yields (iii). ∎

######  Proposition \propDegGaOne

The following is a complete list of solutions $(a,b,c;\Gamma)$ of the
$(1,\infty)$-AlgDOP problem over $\mathbb{C}$ up to $(1,\infty)$-admissible
change of variables under condition that $\deg_{y}\Gamma=1$. Here $a_{10}$,
$a_{20}$, $b_{11}$, $c_{02}$ are constants and $a_{0}$, $b_{1}$, $c_{0}$,
$c_{1}$ are polynomials in $x$; $\deg a_{0}\leq 2$, $\deg b_{1}\leq 1$.

We shall see in §6 that only (iii)–(v) provide a solution of the
$\mathbb{w}$-SDOP problem. Moreover, in cases (iv)–(v), this appears only when
$(a,b,c)$ is $(a_{10}x,0,c_{01}y)$ or $(1-x^{2},0,c_{01}y)$ which corresponds
to the product of Laguerre polynomials or Jacobi and Laguerre polynomials.

###### Demonstration Proof

Let $\Gamma_{1}$ be the irreducible factor of $\Gamma$ of $y$-degree $1$,
i.e., of the form $\Gamma_{1}=\gamma_{1}y-\gamma_{0}$ where $\gamma_{0}$ and
$\gamma_{1}$ are polynomials in $x$ and $\gamma_{1}\neq 0$.

Case 1. $a\neq 0$ and $\gamma_{1}$ is not a constant. By Lemma 3.3, if $x_{0}$
is a root of $\gamma_{1}$, it is also a root of $b_{1}$ and a double root of
$a$. Since $a$ is a non-zero polynomial in $x$ of degree at most $2$, it has
only one double root, hence $\gamma$ also has only one root (maybe multiple)
at $x_{0}$, hence we may assume that $x_{0}=0$ and $a=x^{2}$, $b_{1}=b_{11}x$,
and $\gamma_{1}=x^{n}$, $n\geq 1$. Then $\gamma_{0}/\gamma_{1}$ is a Laurent
polynomial, let us denote it by $p(x)=\sum_{k}p_{k}x^{k}$. By a change of
variable $y=y_{1}+p_{0}+p_{1}x+p_{2}x^{2}+\dots$ we may kill all the
coefficients of non-negative powers of $x$. Thus we have
$p=p_{-n}x^{-n}+\dots+p_{-1}x^{-1}$. The assumption that $\Gamma_{1}$ is
irreducible implies $\gamma_{0}(0)\neq 0$, hence $p_{-n}\neq 0$, i.e.,
$\operatorname{ord}_{x}p=-n$. By rescaling the variable $y$ we may assume
$p_{-n}=1$. The curve $\Gamma_{1}$ is parametrized by $x=x$, $y=p(x)$. Hence
(11) yields

$b(x,p)=x^{2}p^{\prime},\qquad c(x,p)=b(x,p)p^{\prime},\qquad\text{ and thus
}\quad c(x,p)=(xp^{\prime})^{2}.$ $None$

Since $b=b_{11}xy+b_{0}(x)$, the first equation in (16) reads

$b_{11}\big{(}x^{-n+1}+\dots+p_{-2}x^{-1}+p_{-1}\big{)}+b_{0}=-nx^{-n+1}-\dots-2p_{-2}x^{-1}-p_{-1}.$
$None$

Case 1.1. $n\geq 2$. Since $b_{0}$ is a polynomial, we derive from (17) that
$b_{11}=-n$, $b_{0}=-p_{-1}-b_{11}p_{-1}=(n-1)p_{-1}$ (a constant), and
$p_{-n+1}=\dots=p_{-2}=0$, i.e., $p=x^{-n}+p_{-1}x^{-1}$. Then the last
equation in (16) reads

$c_{02}\big{(}x^{-n}+p_{-1}x^{-1}\big{)}^{2}+(x^{-n}+p_{-1})c_{1}+c_{0}=\big{(}nx^{-n}+p_{-1}x^{-1}\big{)}^{2}.$
$None$

By comparing the coefficients of $x^{-2n}$ and then those of $x^{-n-1}$, we
obtain $c_{02}=n^{2}$ and then $p_{-1}=0$. Hence $p=x^{-n}$ and $b_{0}=0$. By
putting $p_{-1}=0$ into (18), we get
$n^{2}x^{-2n}+x^{-n}c_{1}+c_{0}=n^{2}x^{-2n}$, i.e., $c_{1}=-x^{n}c_{0}$. Thus
$(a,b,c)=(x^{2},\,-nxy,\,n^{2}y^{2}-x^{n}c_{0}y+c_{0})$ which corresponds to
(i) with $k=0$ when $\Gamma=\Gamma_{1}$. If $\Gamma$ has another non-constant
factor $\Gamma_{0}$ (a polynomial in $x$), the condition (10) implies that
$\Gamma_{0}$ is a factor of $a$ and $b$ which corresponds to (i) with $k=1$.

Case 1.2. $n=1$ (thus $p=x^{-1}$). Then (17) reads $b_{11}+b_{0}=-1$ and the
last equation in (16) reads $c_{02}x^{-2}+x^{-1}c_{1}+c_{0}=x^{-2}$ whence
$b_{0}=-b_{11}-1$, $c_{02}=1$, and $c_{1}=-xc_{0}$. Hence
$b=b_{11}xy+b_{0}=b_{11}xy-b_{11}-1$ and
$c=c_{02}y^{2}+c_{1}y+c_{0}=y^{2}-xyc_{0}+c_{0}$ which corresponds to (ii)
when $\Gamma=\Gamma_{1}$. Once again, any additional non-constant factor
$\Gamma_{0}$ of $\Gamma$ should be a factor of $a$ and $b$. Then
$\Gamma_{0}=x$ and it divides $b=b_{11}(xy-1)-1$. Therefore $b_{11}=-1$, thus
$b=-xy$. This corresponds to (i) with $n=k=1$.

Case 2. $\gamma_{1}=\operatorname{const}$. Then, up to an admissible change of
variables, we may assume that $\Gamma_{1}=y$. If $\Gamma=\Gamma_{1}$, this
corresponds to (iii). Otherwise, as in the previous cases $\Gamma/\Gamma_{1}$
is a polynomial in $x$ which divides $a$ and $b$ which corresponds to (iv) and
(v).

Case 3. $a=0$. Then $\Gamma$ is a factor of $\Delta=-b^{2}$. By an admissible
change of variables we can reduce $b$ to $y$, $xy$, or $xy-1$. This leads to
(iii), (iv), or (vi). ∎

## 5\. Solution of the $(1,w)$-AlgDOP Problem in $\mathbb{C}^{2}$ for $1<w\leq
2$

In this section we find all solutions of the $(1,2)$-AlgDOP Problem in
$\mathbb{C}^{2}$. They include all solutions of the $(1,w)$-AlgDOP Problem for
any $w$ in the range $1<w\leq 2$ (see §3.2).

### 5.1. Compactification of $\mathbb{C}^{2}$

Let $(a,b,c;\,\Gamma)$ be a solution of the $(1,2)$-AlgDOP Problem in
$\mathbb{C}^{2}$, and let $\Delta=ac-b^{2}$. The Newton polygon of $\Delta$
(and hence of $\Gamma$) is contained in the triangle $[(0,0),(6,0),(0,3)]$
(see Figure 1). Therefore it is natural to consider $\mathbb{C}^{2}$ as the
affine chart $Z\neq 0$ (with coordinates $x=X/Z$, $y=Y/Z^{2}$) of the weighted
projective plane $\mathbb{P}^{2}_{1,2,1}$ which is the quotient of
$\mathbb{C}^{3}\setminus\\{(0,0,0)\\}$ by the equivalence relation
$(X,Y,Z)\sim(\lambda X,\lambda^{2}Y,\lambda Z)$, $\lambda\neq 0$ (we denote
the class of $(X,Y,Z)$ by $[X{:}Y{:}Z]$). This variety is smooth except at the
point $[0\,{:}1{:}\,0]$.

A generic polynomial $P(x,y)$ with $\Cal{N}(P)=[(0,0),(6,0),(0,3)]$ defines an
affine curve $\\{P=0\\}$ in $\mathbb{C}^{2}$ whose closure in
$\mathbb{P}^{2}_{1,2,1}$ is a smooth curve not passing through the singular
point $[0\,{:}1{:}\,0]$, and the linear projection from this point

$\mathbb{P}_{1,2,1}^{2}\setminus\\{[0\,{:}1{:}\,0]\\}\to\mathbb{P}^{1},\quad[X{:}Y{:}Z]\mapsto(X{:}Z),$
$None$

is a 3-fold branched covering of $\\{P=0\\}$ onto $\mathbb{P}^{1}$. This is
why $\mathbb{P}^{2}_{1:2:1}$ is an appropriate compactification of
$\mathbb{C}^{2}$ in our setting. However the coefficients of $\Delta$ are not
necessarily generic and the closure of $\\{\Gamma=0\\}$ may have singularities
and it may pass through $[0\,{:}1{:}\,0]$. To deal with such curves it is
convenient to blow up the point $[0\,{:}1{:}\,0]$. This means that we consider
$\mathbb{C}^{2}$ as an affine chart $(x,y)$ of $\Cal{F}_{2}$ – the Hirzebruch
surface of degree 2 which is the smooth complex surface obtained by gluing
together four copies of $\mathbb{C}^{2}$ with coordinates $(x_{k},y_{k})$,
$k=0,\dots,3$, (where $x_{0}=x$, $y_{0}=y$ are the coordinates on
$\mathbb{C}^{2}$ we started with). The transition functions are:

$\begin{matrix}x_{1}=1/x,\;\,\qquad&x_{2}=x,\;\;\;\,\qquad&x_{3}=\;x_{1}\;\;=1/x,\;\;\lower
3.0pt\hbox{\mathstrut}\\\
y_{1}=y/x^{2},\qquad&y_{2}=1/y,\qquad&y_{3}=1/y_{1}=x^{2}/y.\end{matrix}$
$None$

The set of real points of $\Cal{F}_{2}$ is diffeomorphic to a torus. In Figure
3 we represent it as a rectangle with opposite edges identified. As an
illustration of the chart gluing, we also show in Figure 3 how the closures of
some two curves in $\Cal{F}_{2}$ look like.

Figure 3 Coordinate axes for all the four charts on $\Cal{F}_{2}$ and the
curves $\\{y=1\\}=\\{y_{1}=x_{1}^{2}\\}$ and
$\\{y=-x^{4}-1\\}=\\{y_{3}=-x_{3}^{2}/(1+x_{3}^{4})\\}$.

The projection (19) extends to the projection
$\pi:\Cal{F}_{2}\to\mathbb{P}^{1}$ given in the affine charts by
$(x_{k},y_{k})\mapsto(x_{k}:1)$ for $k=0,2$ and
$(x_{k},y_{k})\mapsto(1:x_{k})$ for $k=1,3$. It is a fibration with fiber
$\mathbb{P}^{1}$. Its restriction to the closure of a curve $\\{P(x,y)=0\\}$
is a branched covering of degree $\deg_{y}P$. The strict transform of
$[0\,{:}1{:}\,0]$ under the blowup (we denote it by $E$) is given by $y_{2}=0$
or $y_{3}=0$ in the respective charts. Its self-intersection is $-2$.

The set of $(1,2)$-admissible changes of variables coincides with the set of
biregular automorphisms of $\Cal{F}_{2}$ preserving $\mathbb{C}^{2}$.

### 5.2. The case when $\Delta$ is irreducible and $\deg_{y}\Delta=3$

Let $(a,b,c;\,\Gamma)$ be a solution of the $(1,2)$-AlgDOP Problem such that
$\Gamma=\Delta=ac-b^{2}$, $\Gamma$ is irreducible, and $\deg_{y}\Gamma=3$. We
assume that $(a,b,c;\,\Gamma)$ cannot be reduced to a solution of the
$(1,1)$-AlgDOP Problem by a $(1,2)$-admissible change of coordinates. Let $C$
be the closure of $\\{\Gamma=0\\}$ in $\Cal{F}_{2}$. We identify
$\mathbb{C}_{2}$ (where the affine curve $\Gamma=0$ sits) with the affine
chart corresponding to the coordinate system $(x,y)$. We shall call it the
main chart. We denote the fiber $\\{x_{1}=0\\}$ by $L_{\infty}$ (see Figure
3). The condition $\deg_{y}\Gamma=3$ implies that $C$ is disjoint from $E$,
and $\pi|_{C}$ is a 3-fold branched covering. Let $\nu:\tilde{C}\to C$ be the
normalization (non-singular model) of $C$. This means $\tilde{C}$ is a smooth
compact Riemann surface of genus $\mathbb{g}$ and $\nu$ a holomorphic mapping
which is injective outside a finite number of points. There is a 1-to-1
correspondence between points of $\tilde{C}$ and local branches of $C$.

The genus formula for $C$ reads

$\mathbb{g}=1+\tfrac{1}{2}C(C+K_{\Cal{F}_{2}})-\sum_{P\in
C}\delta_{P}=4-\sum_{P\in C}\delta_{P}$ $None$

where $\delta_{P}=\delta_{P}(C)$ is the delta-invariant of $(C,P)$, i.e.
$2\delta_{P}=m_{i}(m_{i}-1)$ where $m_{1},m_{2},\dots$ are the multiplicities
of the infinitely near points of $C$ (note that the “$4$” in (21) can be
computed as the number of integral points in the interior of the triangle
$[(0,0),(6,0),(0,3)]$). It is convenient to rewrite (21) in terms of local
branches of $C$ as it is done in [3, §3.2]. Namely, for a point $P\in C$ with
local branches $\gamma_{1},\dots,\gamma_{r}$ we set $n_{P}=\sum_{1\leq i<j\leq
r}\gamma_{1}\cdot\gamma_{j}$. Then we have
$\delta_{P}=n_{P}+\sum\delta(\gamma_{i})$, hence the genus formula (21) takes
the form

$\mathbb{g}=4-n-\sum_{\gamma}\delta(\gamma),\qquad n=\sum_{P\in C}n_{P}$
$None$

where the first sum is over all local branches of $C$.

For a local branch $t\mapsto\gamma(t)=(\xi(t),\eta(t))$ we denote the
ramification index of $\pi\circ\gamma$ by $m_{\pi}(\gamma)$. The number
$m_{\pi}(\gamma)$ can be also defined as the intersection number of $\gamma$
with the fiber of $\pi$ passing through the center of $\gamma$. If
$\operatorname{ord}\xi\geq 0$ then
$m_{\pi}(\gamma)=\operatorname{ord}_{t}(\xi(t)-\xi(0))$. If
$\operatorname{ord}\xi<0$, then $m_{\pi}(\gamma)=-\operatorname{ord}_{t}\xi$.
By Riemann-Hurwitz formula we have

$2-2\mathbb{g}=6-\sum_{\gamma}(m_{\pi}(\gamma)-1).$ $None$

######  Lemma \lemPlu

(a). Let $\gamma$ be a local branch of $C$ at a point $P$. Then
$m_{\pi}(\gamma)\leq 3$ and we have:

(b). The curve $C$ is rational (i.e., $\mathbb{g}=0$) and one of the following
cases occurs (among the numbers $n$, $\alpha_{k}$, $\beta_{k}$, we list only
the non-zero ones):

###### Demonstration Proof

(a). Follows from Lemma 3.4. The only point which maybe needs some comments is
the smoothness of $\gamma$ in the case when $m_{\pi}(\gamma)=3$, and hence
$P\in L_{\infty}$. Up to an admissible change of coordinates we may assume
that $P$ is at $(x_{1},y_{1})=(0,0)$ (see Figure 3). Then
$v_{\gamma}(y_{1})>0$ and the condition $m_{\pi}(\gamma)=3$ means that
$v_{\gamma}(x_{1})=3$, i.e., $v_{\gamma}(x)=-3$. Hence Lemma 3.4(c) implies
that $v_{\gamma}(y)\not\in[-4,-1]$. By (20) we have
$v_{\gamma}(y_{1})=v_{\gamma}(y)-2v_{\gamma}(x)=v_{\gamma}(y)+6$, thus
$v_{\gamma}(y_{1})\not\in[2,5]$. It is easy to see that $v_{\gamma}(y_{1})<6$.
Hence $v_{\gamma}(y_{1})=1$ whence the result.

(b). We have $\beta_{2}+\beta_{3}\leq 1$ and the equations (22) and (23) imply

$4=\mathbb{g}+n+\sum_{k\geq 1}k\alpha_{2k},\qquad
2\mathbb{g}+4=\beta_{2}+2\beta_{3}+\sum_{k\geq 1}\alpha_{2k}.$

The only non-negative solutions are (i)–(vii). ∎

Notice that if a curve in $\mathbb{P}^{2}$ is parametrized by
$t\mapsto\big{(}\xi(t):\eta(t):\zeta(t)\big{)}$, then the projectively dual
curve is parametrized by

$t\mapsto(\dot{\eta}\zeta-\dot{\zeta}\eta:\dot{\zeta}\xi-\dot{\xi}\zeta:\dot{\xi}\eta-\dot{\eta}\xi).$
$None$

######  Lemma \lemParam

The cases (v)–(vii) in Lemma 5.1 are unrealizable. In the other cases the
curve $C$ admits one of the following parametrizations $t\mapsto[X:Y:Z]$ in
the weighted homogeneous coordinates introduced in §5.1:

Figure 4

###### Demonstration Proof

In each of the cases (i)–(vii) we may assume that $C$ is singular at the
origin of the main chart. Then we blow up this point and blow down the strict
transform of the lines $x=0$ and $E$. In coordinates this means that we
consider the curve $C_{1}$ on $\mathbb{P}^{2}$ which is the projective closure
of the affine curve $\Gamma(x,xy)/x^{2}=0$. We consider the homogeneous
coordinates $(X_{1}:Y_{1}:Z_{1})$ on $\mathbb{P}^{2}$ such that
$x=X_{1}/Z_{1}$, $y=Y_{1}/Z_{1}$.

Then $C_{1}$ is a quartic curve tangent to the line $X_{1}=0$ at $(0{:}1{:}0)$
(see Figure 4). If $C$ has the $A_{2k}$ singularity at the origin, then
$C_{1}$ has the $A_{2k-2}$ singularity somewhere on the line $X_{1}=0$ (when
$k>1$) or a simple tangency (when $k=1$). The line $Z_{1}=0$ is the strict
transform of $L_{\infty}$, thus $C_{1}$ has the tangency with it of the same
nature as $C$ has with $L_{\infty}$. Up to a $(1,2)$-admissible change of
coordinates, the curve $C$ is determined by $C_{1}\cup\\{X_{1}Z_{1}=0\\}$. The
weighted homogeneous coordinates (see §5.1) are expressed via
$(X_{1}:Y_{1}:Z_{1})$ as follows:

$[X:Y:Z]=[X_{1}:X_{1}Y_{1}:Z_{1}].$ $None$

Now we separately consider the cases of Lemma 5.1.

Case (i). We assume that the node is at the origin. In this case the curve
$C_{1}$ has singularities $A_{2}$ and $A_{4}$. An irreducible quartic curve
with such singularities is unique up to an automorphism of $\mathbb{CP}^{2}$
and it is autodual (see e.g. [3, Cor. 3.10(iii)]). Such a curve has a single
flex point $P$ (because the dual has a single cusp) and the line $Z_{1}=0$ is
the tangent to $C_{1}$ at $P$. Let $Q$ be the other intersection point of
$C_{1}$ with $\\{Z_{1}=0\\}$ (see Figure 5). Then $Q=(0{:}1{:}0)$ and the line
$\\{X_{1}=0\\}$ is tangent to $C_{1}$ at $Q$. This means that
$C_{1}\cup\\{X_{1}Z_{1}=0\\}$ is uniquely determined up to automorphism of
$\mathbb{P}^{2}$ whence the uniqueness of $C$. Thus it remains to check that
the curve given in the statement of the lemma has the required properties.
Indeed, it has a cusp at $[\frac{32}{27}:\frac{256}{81}:1]$ ($t=0$), an
$A_{4}$ singulariry at $[0{:}0{:}1]$ ($t=\infty$), a node at $[1{:}1{:}1]$
($t=-5\pm 2\sqrt{5}$), and a flex point with tangent $L_{\infty}$ at
$[1{:}0{:}0]$ ($t=-3$).

Figure 5 A quartic curve with $A_{2}$ and $A_{4}$.

Case (ii). In this case the curve $C_{1}$ has three cusps $A_{2}$. The line
$X_{1}=0$ is its bitangent. It is well known that such a curve (tricuspidal
cubic) is unique up to automorphism of $\mathbb{CP}^{2}$, it has only one
bitangent line, and the tangency points $P$ and $Q$ are interchangeable by an
automorphism of $(\mathbb{P}^{2},C_{1})$. Thus $C$ is determined by a choice
of a line $Z_{1}=0$ passing through $P$. So, it depends on one parameter.

The curve $C_{1}$ is projectively dual to a nodal cubic, hence it has two real
forms. We choose the one with the real bitangent (and hence with two complex
conjugate cusps). In some homogeneous coordinates $(X_{2}:Y_{2}:Z_{2})$, it
has a parametrization

$X_{2}=t^{2}(t+1)^{2},\qquad Y_{2}=2t+1,\qquad Z_{2}=(t+1)(3t+1).$

Here the line $X_{2}=0$ is bitangent, the cusps correspond to $t=(-3\pm
i\sqrt{3})/6$ and $t=\infty$. According to the above discussion, we may choose
$X_{1}=X_{2}$, $Y_{1}=Y_{2}$, and $Z_{1}=Z_{2}+\alpha X_{2}$. Plugging this
into (26), we conclude. The roots of $\alpha^{3}-9\alpha^{2}+27\alpha$ are
excluded because they correspond to the cases when the line $L_{\infty}$
passes through a cusp which contradicts our assumptions (see Lemma 5.1(a)).

Case (iii). In this case $C_{1}$ has two cusps and one node. We assume that
one of the cusps is at the origin. Then the line $X_{1}=0$ is bitangent. Let
$C_{2}=\check{C}_{1}$ be the curve projectively dual to $C_{1}$. Using
Plücker’s formulas (in the form given in [3, §3.2] or in [4]), one can check
that the class of curves with two cusps, one node, and at least one bitangent
is stable under the projective duality. Hence $C_{2}$ has two cusps and one
node. Therefore, in some homogeneous coordinates $(X_{3}:Y_{3}:Z_{3})$, it has
a parametrization

$X_{3}=t^{2},\qquad Y_{3}=(t-1)(t-\alpha),\qquad Z_{3}=t^{2}(t-1)(t-\alpha).$
$None$

The cusps correspond to $t=0$ and $t=\infty$. The node corresponds to $t=1$
and $t=\alpha$. Being dual to $C_{2}$, the curve $C_{1}$ is parametrized in
some coordinates by

$t\mapsto\varphi(t)=(X_{2}:Y_{2}:Z_{2})=\big{(}2(t-1)^{2}(t-\alpha)^{2}:\alpha+1-2t:(\alpha+1)t-2\alpha\big{)}.$

(see (24)). The points where $C_{2}$ touches the bitangent line correspond to
the local branches of $C_{3}$ at the node, which are at $\varphi(1)$ and
$\varphi(\alpha)$, thus $X_{1}=X_{2}$. In the coordinates
$(X_{1}{:}Y_{1}{:}Z_{1})$, the tangency points are at $(0{:}0{:}1)$ and at
$(0{:}1{:}0)$. Up to rescaling the coordinate $t$, we may assume that
$\varphi(1)=(0{:}1{:}0)$. Then the line $Z_{1}=0$ is uniquely determined by
the condition that it passes through $(0{:}1{:}0)$ and is tangent to $C_{1}$
which gives

$Z_{1}=\tfrac{\alpha}{2}X_{2}+(\alpha+1)Y_{2}+\alpha^{2}(\alpha+1)Z_{2}.$

The line $Y_{1}=0$ should pass through $(0{:}0{:}1)$. Then we may set
$Y_{1}=Y_{2}+Z_{2}$ and using (26) we obtain the required parametrization of
$C$.

The condition $\alpha\not\in\\{0,1\\}$ is clear from the construction. If
$\alpha=-1$, then $C_{2}$ is a double conic because $X_{3}$, $Y_{3}$, and
$Z_{3}$ are functions of $t^{2}$ (see (27)). If $\alpha$ is a root of
$\alpha^{2}+4\alpha+1$, then one of the cusps is on $L_{\infty}$ which
contradicts Lemma 5.1.

Case (iv). The condition $n=2$ can be attained in three different ways.

Figure 6 Case (iv1) of Lemma 5.2.

Case (iv1): $2A_{2}+2A_{1}$. The curve $C$ has two cusps and two nodes. We
assume that one of the cusps is at the origin. Then $C_{1}$ has one cusp and
two nodes. The line $X_{1}=0$ is bitangent and the line $Z_{1}=0$ is tangent
at a flex point. We choose the line $Y_{1}=0$ to pass through the both
tangency points. Next we choose coordinates $(X_{2}:Y_{2}:Z_{2})$ as shown on
the left hand side of Figure 6 and perform the Cremona transformation
$(X_{2}:Y_{2}:Z_{2})\mapsto(X_{3}:Y_{3}:Z_{3})=(Y_{2}Z_{2}:Z_{2}X_{2}:X_{2}Y_{2})$.
Then the lines $X_{1}=0$ and $Z_{1}=0$ are transformed into lines that we
denote by $L_{0}$ and $L_{\infty}$. The transform of $C_{1}$ is a cuspidal
cubic $C_{3}$ shown on the right hand side of Figure 6 (two complex conjugate
crossings with $Z_{3}=0$ are not shown). The lines $X_{2}=0$ and $Z_{2}=0$
cannot be tangent to the local branches of $C_{1}$ at the nodes because
otherwise $C_{3}$ would have too many tangents in the pencil of lines through
$(0:1:0)$ (see Figure 7).

Figure 7 Unrealizable tangency at $A_{1}$ in Case (iv1).

Let $(X_{4}:Y_{4}:Z_{4})$ be coordinates such that $C_{3}$ is parametrized by
$t\mapsto\varphi(t)=$ $(t^{2}:t^{3}:1)$. Up to rescaling we may assume that
the point of tangency of $C_{3}$ with $L_{0}$ is $\varphi(2)=(4:8:1)$. Then
$C_{3}\cap L_{0}=\varphi(-1)=(1:-1:1)$ (see Figure 6) and the whole
configuration is determined by a choice of the line $Y_{3}=0$ passing through
$\varphi(-1)$, i.e., it is determined by a single parameter $\alpha$ such that
$Y_{3}=(Y_{4}+Z_{4})-\alpha(X_{4}-Z_{4})$. Let $\varphi(t_{1})$ and
$\varphi(t_{2})$ be the other two points of $C_{3}\cap\\{X_{3}=0\\}$. Then
$t_{1}$ and $t_{2}$ are the roots of $(t^{3}+1)=\alpha(t^{2}-1)$ different
from $-1$, i.e., the roots of $t^{2}-(\alpha+1)(t-1)=0$, hence
$t_{2}=t_{1}/(t_{1}-1)$ and $\alpha=(t_{1}^{3}+1)/(t_{1}^{2}-1)$. The
parametrization of $C_{3}$ is

$(X_{3}{:}Y_{3}{:}Z_{3})=\big{(}t^{3}-t_{1}^{3}-3(t^{2}-t_{1}^{2}):(t+1)(t-t_{1})(t-t_{2}):t^{3}-t_{2}^{3}-3(t^{2}-t_{2}^{2})\big{)}$

and after routine computations we obtain the required parametrization of $C$.
If $\alpha\in\\{-3/2,7/2\\}$, this parametrization defines a curve covered by
Case (i).

Case (iv2): $A_{2}+D_{5}$. The curve $C$ has one cusp and a singular point of
type $D_{5}$ (given by $u(u^{2}+v^{3})=0$ is some local curvilinear
coordinates). We assume that the $D_{5}$ singularity is at the origin. We may
also assume that the line $Y=0$ is tangent to its cuspidal local branch and
passes through the point of tangency of $C$ with the line $Z=0$. Then
$\Cal{N}(\Gamma)=[(4,0),(5,0),(0,3),(1,2)]$. Hence after blowing up the origin
we obtain a curve $\mathbb{P}^{2}$ with the Newton polygon
$[(1,0),(2,0),(0,3),(0,2)]$ (cf. Figure 4). This is a cuspidal cubic which has
a simple (quadratic) tangency with the line $X_{1}=0$ and a cubic tangency
with the line $Z_{1}=0$. One easily checks that these conditions uniquely
determine $C\cup\\{X_{1}Z_{1}=0\\}$ up to automorphism of $\mathbb{P}^{2}$.
Hence the curve $C$ is unique up to an admissible change of coordinates. It
remains to observe that the parametrization in the statement of the lemma
gives the required curve when $\alpha=-1$. It has a cusp at $[0:0:1]$ and the
$D_{5}$ singularity at $[4:16:1]$.

Case (iv3): $2A_{2}+A_{3}$. The curve $C$ has two cusps $A_{2}$ and one
tacnode $A_{3}$ (ordinary tangency of two smooth branches). We assume that one
of the cusps is at the origin. Then the curve $C_{1}$ has one cusp $A_{2}$,
one tacnode $A_{3}$ and at least one bitangent (the line $X_{1}=0$). Let us
show that this combination is impossible for quartic curves in
$\mathbb{P}^{2}$. It is convenient to apply Plücker’s equations in the form
given in [4, Thm. 1.3]. In the notation of [4] we have $d=4$, $g=0$,
$n_{v}=2$, $c_{v}=1$, hence $\hat{d}=5$ by [4, Eq. (1.6)]. Then the equations
[4, (1.8)–(1.9)] read $\hat{n}+\hat{c}=6$ and $2\hat{n}+3\hat{c}=16$ whence
$\hat{n}=2$.

It is easy to check that the dual branches of a tacnode form a tacnode of the
dual curve and it contributes $2$ to $\hat{n}$. Hence the dual curve does not
have nodes, i.e., $C_{1}$ does not have bitangents. A contradiction.

Case (v). In this case $C_{1}$ has singularities $A_{2}$ (on the line
$X_{1}=0$) and $A_{4}$. An irreducible quartic curve with such singularities
is unique up to an automorphism of $\mathbb{CP}^{2}$ and it is autodual (cf.
Case (i)). Hence it has a unique flex point because the dual curve has one
cusp $A_{2}$. By the uniqueness, we may assume that $C_{1}$ is a small real
perturbation of real double conic as explained in the remark in the proof of
[3, Cor. 3.10] (see Figure 5). The line $Z_{1}=0$ is the tangent at the flex
point $P$. The line $X_{1}=0$ is the tangent at the transverse crossing $Q$ of
$Z_{1}=0$ with $C_{1}$ and it must pass through the cusp $A_{2}$. We see in
Figure 5 that this is impossible because the arc $A_{2}QA_{4}$ is convex.

Case (vi). We assume that the singularity of $C$ at the origin is $A_{6}$.
Then $C_{1}$ has $A_{2}$ and $A_{4}$ as singularities and the proof is the
same as in Case (v) but with $A_{2}$ and $A_{4}$ exchanged.

Case (vii). We assume that the singularity of $C$ at the origin is $A_{2}$.
Then $X_{1}=0$ is a bitangent and $C_{1}$ has singular points $A_{2}$ and
$A_{4}$. As we mentioned in Case (v), such a curve is autodual, hence it
cannot have a bitangent because the dual curve does not have nodes. A
contradiction. ∎

######  Proposition \propIrred

Let $(g;\,\Gamma)$ be a solution of the $(1,2)$-AlgDOP problem over
$\mathbb{R}$ such that $\Gamma$ is irreducible and $\deg_{y}\Gamma=3$. Then,
up to a $(1,2)$-admissible change of variables, either $(g;\,\Gamma)$ is a
solution of the $(1,1)$-AlgDOP problem, or $\Gamma$ is given by (25) and

$g=\left(\begin{matrix}y+8x-9x^{2}&5(4y-3xy-x^{2})\lower
2.0pt\hbox{\mathstrut}\\\
5(4y-3xy-x^{2})&-25(y^{2}-4xy+3x^{3})\end{matrix}\right).$ $None$

###### Demonstration Proof

Given a parametrization $(X(t),Y(t),Z(t))$ in the weighted homogeneous
coordinates introduced in §5.1, the relation (11) applied to
$\xi(t)=X(t)/Z(t)$, $\eta(t)=Y(t)/Z(t)^{2}$ yields a system of homogeneous
linear equations for the coefficients $a_{ij}$, $b_{ij}$, $c_{ij}$ of the
entries of the cometric $g$ in each of the cases (i)–(iv) of Lemma 5.2. Up to
a constant factor, the only non-zero solution in Case (i) is (28).

In Cases (ii)–(iv) the system of equations polynomially depends on $\alpha$.
It has a non-zero solution if and only if the gcd of the determinants of the
maximal minors vanishes. A straightforward computation shows that this gcd is
a polynomial in $\alpha$ all whose roots are excluded in Lemma 5.2. For
example, in Case (ii), the gcd is a power of $\alpha$ multiplied by a power of
$\alpha^{2}-9\alpha+27$. ∎

Remark 5.4. The solution of the $(1,2)$-DOP Problem given in [1, §8]
transforms (up to a constant factor) into our solution given in Proposition
5.3 by the change of variables

$\theta_{1}=\frac{2+\sqrt{5}}{5}(x-1),\qquad\theta_{2}=-\frac{\sqrt{5}}{125}(4y-5x+1).$

### 5.3. The case when $\Gamma$ has a factor of $y$-degree $2$

######  Proposition \propReducTwo

Let $(g;\,\Gamma)$ be a solution of the $(1,2)$-AlgDOP Problem such that
$\deg_{y}\Gamma=2$. Suppose that it is not a solution of the
$(1,\infty)$-AlgDOP problem. Then, by a $(1,2)$-admissible change of
variables, $(g;\,\Gamma)$ can be reduced either to a solution of the
$(1,1)$-AlgDOP problem or to one of the following three cases:

###### Demonstration Proof

Lemmas 3.2 and 3.3 imply that $\Gamma$ does not have monomials of the form
$x^{k}y^{2}$ with $k>0$. Local branches of $\Gamma$ correspond to edges of
$\Cal{N}(\Gamma)$ (see Lemma 3.1). Hence Lemma 3.4(d) implies that the slope
of any upper edge is steeper than 1:2, hence $\Cal{N}(\Gamma)$ is contained in
the triangle $[(0,0),(4,0),(0,2)]$. This fact combined with Lemma 3.3 (which
means that the affine curve $\Gamma=0$ has no vertical tangent) leaves only
five possibilities for $\Gamma$ up to admissible change of coordinates: the
three cases (i)–(iii) and also $y(y-x)$ and $y^{2}-1$.

In each case, the condition (11) yields a system homogeneous linear equations
for the coefficients of $a$, $b$, and $c$ (the entries of $g$). By solving
these systems we obtain the result. In the last two cases we obtain $a_{1}=0$,
thus these are solutions of the $(1,\infty)$-AlgDOP problem (see Figure 2). In
the other cases we normalize the solutions so that $a_{1}=1$. The parameter
$\mu$ in Case (ii) can be set to $0$ or $1$ by rescaling the coordinates (see
Example 2.7). The conditions $(\alpha,\beta-\beta^{2},\mu)\neq(0,0,0)$ (in
Case (ii)) and $(\alpha,\beta-\beta^{2})\neq(0,0)$ (in Case (iii)) are
equivalent to $\det g\neq 0$. ∎

######  Lemma \lemReduc

Let $(g;\,\Gamma)$ be a solution of $(1,2)$-AlgDOP problem which cannot be
reduced to a solution of $(1,1)$-AlgDOP problem. Let
$\Gamma=y(y-p_{1}(x))(y-p_{2}(x))$. Then the polynomial $p_{1}p_{2}$ has at
most two roots (maybe multiple).

###### Demonstration Proof

Proposition 5.5 applied to $(g,y(y-p_{k}))$ implies that $a_{0}$ vanishes at
the roots of $p_{k}$ (recall that $g^{11}=a_{01}y+a_{0}(x)$). It remains to
prove that $a_{0}$ cannot be identically zero. Indeed, if it is, then $y$
would divide each entry of $g$, hence $y^{2}$ would divide $\det g$ which is
impossible because $\det g=\Gamma$ in our case (see Figure 1) and $\Gamma$ is
squarefree.

######  Proposition \propReducThree

Let $(g;\,\Gamma)$ be a solution of the $(1,2)$-AlgDOP problem over
$\mathbb{C}$ such that $\Gamma$ is reducible and $\deg_{y}\Gamma=3$. Then, up
to a $(1,2)$-admissible change of variables, either $(g;\,\Gamma)$ is a
solution of the $(1,1)$-AlgDOP problem, or $g$ is given by (29) with
$(\alpha,\beta,\mu)=(-18,-3/2,1/2)$ and hence $\Gamma=\Gamma_{1}\Gamma_{2}$
where $\Gamma_{2}=y^{2}-x^{3}$ and $\Gamma_{1}=8y-3x^{2}-6x+1$.

In the latter case the curve $\Gamma=0$ has singularities of the types
$A_{1}$, $A_{2}$, and $A_{5}$ at $(\frac{1}{9},-\frac{1}{27})$, $(0,0)$, and
$(1,1)$ respectively.

###### Demonstration Proof

We have $\Gamma=\Gamma_{1}\Gamma_{2}$ avec $\deg_{y}\Gamma_{k}=k$. By
Proposition 2.4, $(g,\Gamma_{2})$ is also a solution of the $(1,2)$-AlgDOP
problem. Moreover, $(g,\Gamma)$ reduces to a solution of a $(1,w)$-AlgDOP
problem by a $(1,2)$-admissible change if and only if the same is true for
$(g,\Gamma_{2})$. Hence we may assume that $(g,\Gamma_{2})$, is as in
Proposition 5.5.

Case 1. $\Gamma_{2}$ is irreducible. Then $(g,\Gamma_{2})$ is as in
Proposition 5.5(i). A computation shows that $\det g=\Gamma_{1}\Gamma_{2}$
with $\Gamma_{1}=\alpha y-f(x)$, hence $\alpha\neq 0$. Then, for the
parametrization $\xi=t$, $\eta=f(t)/\alpha$ of $\\{\Gamma_{1}=0\\}$, the
equations (11) take the form

$a(\xi,\eta)\dot{\eta}-b(\xi,\eta)\dot{\xi}=6FG/\alpha^{2}=0,\qquad
b(\xi,\eta)\dot{\eta}-c(\xi,\eta)\dot{\xi}=FGH/\alpha^{2}=0,$

where $F=At-3B$, $A=(\alpha-12\beta)(\alpha-9\beta)$,
$B=18+5\alpha\mu-36\beta\mu$, $G=(\beta t+\mu)^{2}-t$, and
$H=(9\beta-\alpha)t+9\mu$. Since $G$ cannot vanish identically, we have
$A=B=0$.

If $\alpha=9\beta$, then $B=-27(2+\beta\mu)$, and we obtain a solution of the
$(1,1)$-AlgDOP problem (the one discussed in [3, §4.10]).

If $\alpha=12\beta$, then $B=-18(3+4\beta\mu)$, and we obtain the announced
solution.

Case 2. $\Gamma_{2}$ is reducible. In this case $\\{\Gamma=0\\}=L_{1}\cup
L_{2}\cup L_{3}$ where $L_{k}=\\{y=p_{k}(x)\\}$, $k=1,2,3$. Let
$P_{1}=L_{1}\cap(L_{2}\cup L_{3})$, $P_{2}=L_{2}\cap(L_{3}\cup L_{1})$, and
$P_{3}=L_{3}\cap(L_{1}\cup L_{2})$. Then Lemma 5.6 implies that each $P_{k}$
has at most two points. By Proposition 5.5, if $k\neq m$, then $L_{k}$ and
$L_{m}$ either are tangent, or cross at two points. Hence, up to a
$(1,2)$-admissible change of coordinates, $\Gamma=y(y-p(x))(y-\lambda p(x))$
where $\lambda\not\in\\{0,1\\}$ and $p(x)$ is $x^{2}$ or $x^{2}-1$. In this
case $g$ is as in Proposition 5.5 (ii) or (iii). Then one easily checks that
(11) is not satisfied for the parametrization $\xi(t)=t$, $\eta(t)=\lambda
p(t)$ of the curve $y=\lambda p(x)$. ∎

## 6\. Solution of the weighted DOP/SDOP problem in $\mathbb{R}^{2}$

### 6.1. Compact solutions

In what follows we give the measure density $\rho$ without the normalizing
constant which is always assumed to be equal to $1/\int_{\Omega}\rho\,dx$.

######  Theorem \thCompTwo

Let $(\Omega,g,\rho)$ be a solution of the $(1,2)$-DOP problem in
$\mathbb{R}^{2}$ which is not a solution of the $(1,\infty)$-DOP problem and
such that $\Omega$ is bounded. Then, up to a $(1,2)$-admissible change of
variables, either it is a solution of the $(1,1)$-DOP problem, or one of the
following cases occurs (see Figure 8):

The solution (B3) reduces to a solution of the $(1,1)$-problem by a
$(1,2)$-admissible change if and only if either $\alpha=4\beta$ (then it is
already so), or $\alpha=4\beta-4$. In the latter case, the variable change is
$(x,y)\mapsto(x,x^{2}-y-1)$ which transforms $g_{(4\beta-4,\beta)}$ into
$-g_{(4-4\beta,1-\beta)}$. We have $g_{(4\beta,\beta)}=-\beta
G^{\prime}_{-1/\beta}$ in the notation of [3, §4.5].

$(0,0)$ $(1,1)$ $\big{(}\frac{32}{27},\frac{256}{81}\big{)}$$(0,0)$
$\big{(}\frac{1}{9},-\frac{1}{27}\big{)}$ $(1,1)$

Dodecahedral quotient Cubic $y^{2}=x^{3}$ and a parabola

Figure 8 The first two domains in Theorem 6.1.

###### Demonstration Proof

According to Propositions 5.3, 5.5, and 5.7, these are the only solutions of
the $(1,\infty)$-AlgDOP problem with bounded components of
$\mathbb{R}^{2}\setminus\\{\Gamma=0\\}$.

The imposed restrictions on $\alpha$ and $\beta$ in Case (B3) are equivalent
to the positive definiteness of $g$. Indeed, we have
$g(0,y)=\operatorname{diag}\big{(}y-\beta+1,\alpha y(1+y)\big{)}$, thus the
positivity of $g$ on $\Omega\cap\\{x=0\\}$ implies $\alpha<0$ and $\beta\leq
0$. Conversely, let $\alpha<0$ and $\beta\leq 0$. Then $g>0$ at
$(0,-\frac{1}{2})$, hence it is enough to show that $\Delta$ does not vanish
in $\Omega$. We have $\Gamma=y\Gamma_{1}$ and
$\Delta:=\det(g)=y\Gamma_{0}\Gamma_{1}$ where

$\Gamma_{0}=\alpha
y+(4\beta-\alpha)(1-\beta)x^{2}+\alpha(1-\beta),\qquad\Gamma_{1}=y-x^{2}+1.$
$None$

Thus it is enough to show that $\\{\Gamma_{0}=0\\}\cap\Omega=\varnothing$. If
$\beta=0$, then $\Gamma_{0}=\alpha\Gamma_{1}$ and we are done. If $\beta>0$,
it is easy to check that the curve $\Gamma_{0}=0$ does not cross
$\partial\Omega$.

The form of the measure density follows from Proposition 2.11 unless $\Delta$
has a multiple factor. This happens only in Case (B3) and, as one can see from
(31), only when $\beta=0$ (then $\Delta=\alpha y\Gamma_{1}^{2}$) or $\beta=1$
(then $\Delta=\alpha y^{2}\Gamma_{1}$). In these two cases one can perform the
computations described in the beginning of §2.3 (in fact, one case is reduced
to the other one by the variable change indicated in the statement of this
theorem). The inequalities for $p$ and $q$ are the integrability conditions
(see [3, Remark 2.28]). ∎

######  Theorem \thCompInf

Let $(\Omega,g,\rho)$ be a solution of the $(1,\infty)$-DOP problem in
$\mathbb{R}^{2}$ such that $\Omega$ is bounded. Then one of the following
cases occurs up to a $(1,\infty)$-admissible change of variables.

###### Demonstration Proof

According to Proposition 4.3, the only solutions of the $(1,\infty)$-AlgDOP
problem with bounded components of $\mathbb{R}^{2}\setminus\\{\Gamma=0\\}$ are
the indicated ones.

Case (B4). We have

$\Delta:=\det(g)=\Gamma_{0}\Gamma,\qquad\Gamma_{0}=\tfrac{1}{4}\Big{(}(n-m)-(n+m)x\Big{)}^{2}-c_{02}(1-x^{2}).$

The condition $c_{02}<0$ is equivalent to the fact that $g$ is positive
definite. Indeed, suppose that $c_{02}\geq 0$. Then

$\Gamma_{0}(x_{0})=-\frac{4c_{02}(c_{02}+mn)}{4c_{02}+(m+n)^{2}}\leq
0\qquad\text{for}\quad x_{0}=\frac{n^{2}-m^{2}}{4c_{02}+(m+n)^{2}}.$

Since $|x_{0}|<1$, we have $(x_{0},0)\in\Omega$ and hence $\Gamma(x_{0},0)>0$.
Therefore $\Delta(x_{0},0)=\Gamma_{0}(x_{0})\Gamma(x_{0},0)\leq 0$, thus $g$
is not positive definite on $\Omega$. Conversely, if $c_{02}<0$, then
$\Gamma_{0}(x)\geq-c_{02}(1-x^{2})>0$ when $|x|<1$, whence
$\Delta|_{\Omega}>0$ which implies $g|_{\Omega}>0$ by Sylvester’s criterion
because $a|_{\Omega}>0$.

The required form of $\rho$ can be derived from Proposition 2.11. Indeed,
$\Gamma$ is maximal for $g$ (by Proposition 4.3) and $\Delta$ is squarefree.
Hence, by Proposition 2.11, $\rho$ is of the required form but with an
additional factor $\exp Q$. Let us show that $Q$ is constant. Set $l=\deg
c_{0}(x)$. Then $(g,\Gamma)$ is a solution of $(2,l)$-SDOP problem. We have
$c_{0}(x)=(1-x)^{m-1}(1+x)^{n-1}\Gamma_{0}$, hence
$l=m+n-2+\deg_{x}\Gamma_{0}$ and we obtain $l=\deg_{x}\Delta-2$. Thus (9) with
$\mathbb{w}=(2,l)$ implies $\deg_{x}Q=0$. On the other hand, $(g,\Gamma)$ is a
solution of $(1,w)$-SDOP problem for some $w>2$, hence (9) for these weights
implies $\deg_{y}Q=0$. Thus $Q=\operatorname{const}$, hence $\rho$ is of the
required form. The inequalities for $p$ and $q$ are the integrability
conditions (see [3, Remark 2.28]).

Case (B5). The proof is almost the same as in Case (B4). We have $\Delta:=\deg
g=\Gamma_{0}\Gamma$ where $\Gamma_{0}=\tfrac{1}{4}n^{2}x-c_{02}(1-x)$. The
condition $c_{02}\leq 0$ is equivalent to $g>0$. Indeed, if $c_{02}>0$, then
$\Gamma(x_{0})=0$ and $0<x_{0}<1$ for
$x_{0}=c_{02}/(c_{02}+\frac{1}{4}n^{2})$, which implies that
$\Delta(x_{0},0)=0$ whence $g|_{\Omega}$ is not positive definite. Conversely,
if $c_{02}\leq 0$, then $\Gamma_{0}(x)>0$ when $0<x<1$, hence
$\Delta|_{\Omega}>0$ whence $g|_{\Omega}>0$ (because $a|_{\Omega}>0$).

The form of $\rho$ is established by the same arguments as in Case (B4). ∎

### 6.2. Non-compact solutions

######  Theorem \thNonCTwo

Any solution $(\Omega,g,\rho)$ of the $(1,2)$-SDOP problem with an unbounded
domain $\Omega$ can be reduced to a solution the $(1,1)$\- or
$(1,\infty)$-SDOP problem by a $(1,2)$-admissible change of variables.

###### Demonstration Proof

Let $\Delta=\deg g$ and $\Gamma$ be the maximal boundary for $g$ (see
Definition 2.10). The integrability condition for the measure $\rho\,dx$
implies that the term $\exp(Q)$ in Proposition 2.11 is non-constant, hence
$\deg_{y}\Gamma\leq\deg_{y}\Delta\leq 2$ (see Corollary 2.14).

Case 1. $\deg_{y}\Gamma=2$. Then $g$ is as in Proposition 5.5. The condition
$\deg_{y}\Delta<3$ implies $\alpha=0$ in all the three cases (i)–(iii). Then,
using the algorithm of §2.3, we find that there is no exponential term in
$\rho$.

Case 2. $\deg_{y}\Gamma=1$. Then $\deg_{y}a=1$ and $\Gamma=y$ up to a
$(1,2)$-admissible coordinate change, because otherwise we have a solution of
the $(1,\infty)$-SDOP problem (see Lemmas 3.2–3.3 and Figure 2). The fact that
$\Gamma=y$ combined with (11) implies that $y$ divides $b$ and $c$. Since
$\deg_{y}a=1$ and $\deg_{y}b\leq 1$, the condition $\deg_{y}(ac-b^{2})<3$
implies $\deg_{y}c=1$, hence $(a,b,c)=(y+a_{0},yb_{1},yc_{1})$ and
$\Delta=(c_{1}-b_{1}^{2})y^{2}+a_{0}c_{1}y$. If $\deg c_{1}\leq 1$, then we
have a solution of the $(1,1)$-SDOP problem, hence $\deg c_{1}=2$.

Case 2.1. $a_{0}\neq 0$. Then $y^{2}$ does not divide $\Delta$, hence, by
Proposition 2.11, $\rho=e^{h}y^{\alpha}$ for a polynomial $h$ such that
$\deg_{y}h\leq 1$, thus $h=yh_{1}(x)+h_{0}(x)$. By (6) we have
$L^{k}\in\Cal{P}_{\mathbb{w}}(w_{k})$, $\mathbb{w}=(1,2)$, where

$L^{1}=(y+a_{0})(yh^{\prime}_{1}+h^{\prime}_{0})+yb_{1}h_{1},\qquad
L^{2}=yb_{1}(yh^{\prime}_{1}+h^{\prime}_{0})+yc_{1}h_{1}.$

The condition $L^{1}\in\Cal{P}_{\mathbb{w}}(w_{1})$ means that
$\deg_{y}L^{1}=0$, $\deg_{x}L^{1}\leq 1$. Hence $h^{\prime}_{1}=0$ (thus
$h_{1}$ is a constant) and $h^{\prime}_{0}=-h_{1}b_{1}$. The integrability
condition implies that $h_{1}<0$ (we assume here that $\Omega=\\{y>0\\}$). Up
to rescaling $y$ we may assume that $h_{1}=-1$, hence $h_{0}^{\prime}=b_{1}$.
Then we have $L^{1}=a_{0}b_{1}$ and $L^{2}=(b_{1}^{2}-c_{1})y$. The condition
$L^{2}\in\Cal{P}_{\mathbb{w}}(w_{2})$ implies that $b_{1}^{2}-c_{1}$ is a
constant and hence $\deg b_{1}=1$ (recall that we have assumed that $\deg
c_{1}=2$). Since $L^{1}=a_{0}b_{1}$ and $\deg_{x}L^{1}\leq 1$, we conclude
that $a_{0}$ is constant. Translating $x$, we may achieve $b_{01}=0$ and we
obtain $(a,b,c)=(y+\alpha,\beta xy,\beta^{2}x^{2}y+\gamma y)$ for some
constants $\alpha,\beta,\gamma$. The change $(x,y)\mapsto(x,y-\frac{1}{2}\beta
x^{2})$ transforms it into $(y+\frac{1}{2}\beta x^{2}+\alpha,-\alpha\beta
x,\alpha\beta^{2}x^{2}+\gamma y)$ which is a solution of the $(1,1)$-SDOP
problem (cf. [3, §6(2ii)]).

Case 2.2. $a_{0}=0$. Then $\Delta=y^{2}\Delta_{0}$,
$\Delta_{0}=c_{1}-b_{1}^{2}$. We proceed as in the beginning of §2.3. Let

$L^{1}=p(x)=p_{0}+p_{1}x,\qquad L^{2}=q_{1}y+q(x)$ $None$

($p_{0},\,p_{1},\,q_{1}$ are constants). Then the equation (7) takes the form
$e_{2}y^{2}+e_{1}y+e_{0}=0$ where $e_{2}=q_{1}\Delta_{0}^{\prime}$. The
equation $e_{2}=0$ yields $c_{1}-b_{1}^{2}=\operatorname{const}$. Then we
reduce to the $(1,1)$-problem by the same variable change as in Case 2.1 (cf.
[3, §6(2iii)]).

Case 3. $\deg_{y}\Gamma=0$. Then $\Omega$ contains a vertical strip
$\Omega_{0}$. We have $\deg_{y}a\leq 1$. If $\deg_{y}a=0$, then this is a
solution of the $(1,\infty)$-SDOP problem (see Figure 2). If $\deg_{y}a=1$,
then $a$ vanishes at some point $P\in\Omega_{0}$, thus $g$ cannot be positive
definite in $P$. ∎

######  Theorem \thNonCInf

Let $(\Omega,g,\rho)$ be a solution of the $(1,\infty)$-SDOP problem in
$\mathbb{R}^{2}$ with an unbounded $\Omega$. Then one of the following cases
occurs up to a $(1,\infty)$-admissible change of variables.

###### Demonstration Proof

Let $w_{\min}$ be the minimal number $w$ such that $w\geq 1$ and
$(\Omega,g,\rho)$ is a solution of the $(1,w)$-SDOP problem. If $w_{\min}=1$,
the result can be derived from the classification in [3, §§5–6]. So, we assume
that $w_{\min}>1$. Let $\Delta=\deg g$ and $\Gamma$ be the maximal boundary
for $g$ (see Definition 2.10). Then $\partial\Omega\subset\\{\Gamma=0\\}$. We
have $\deg_{y}\Gamma\leq\deg_{y}\Delta\leq 2$ (see Figure 2).

Case 1. $\deg_{y}\Gamma=2$. Then $(\Gamma,g)$ realizes one of Cases (i)–(v) of
Proposition 4.3. In Case (iii) with $n=0$ and in Cases (iv)–(v) we obtain (U2)
by Proposition 2.8. In Case (i), the same arguments as in the proof of Theorem
6.2 show the absence of the exponential factor in $\rho$, as well as in Case
(iii) with $x_{0}=1$, $n>0$.

In Case (iii) of Proposition 4.3 with $x_{0}=0$, $n>0$, the above arguments
(based on Proposition 2.11) do not apply literally, because $\Delta$ is no
longer squarefree, but they give the result being combined with [3, Corollary
2.19]. Let us give however a self-contained proof of the unrealizability of
this case. We follow the algorithm from §2.3. Let $h=\log\rho$. Rewrite $g$
replacing $x$ by $-x$ and changing the sign:

$g=\left(\begin{matrix}x^{2}&\tfrac{1}{2}nxy\lower 5.0pt\hbox{\mathstrut}\\\
\tfrac{1}{2}nxy&\tfrac{1}{4}n^{2}x^{n}-c_{02}\Gamma_{2}\end{matrix}\right),\qquad\Gamma_{2}=x^{n}-y^{2},\qquad\Delta=\big{(}\tfrac{1}{4}n^{2}-c_{02}\big{)}x^{2}\Gamma_{2}.$

Let $L^{i}$ be as in (32). Then (7) reads
$np_{0}y-nxq(x)+2x^{2}q^{\prime}(x)=0$. Hence $p_{0}=0$ and $q(x)=Cx^{n/2}$
($C=0$ when $n$ is odd). Plugging the obtained solution into (6) and
integrating $h^{\prime}_{x}$ and $h^{\prime}_{y}$, we obtain
$\rho=C_{1}x^{\alpha}(x^{n}-y^{2})^{\beta}$ when $n$ is odd, and
$\rho=C_{1}x^{\alpha}(x^{n/2}+y)^{\beta_{1}}(x^{n/2}-y)^{\beta_{2}}$ when $n$
is even, with some constants $C_{1}$, $\alpha$, $\beta$, $\beta_{1}$,
$\beta_{2}$. Thus $\rho$ is not integrable on any component of
$\mathbb{R}^{2}\setminus\\{\Gamma=0\\}$.

In Case (ii) of Proposition 4.3, we obtain (U1). Indeed, we have
$(a,b,c)=(x,\frac{1}{2}ny,\frac{1}{4}n^{2}x^{n-1}-c_{02}\Gamma)$ and
$\Delta=\Gamma_{0}\Gamma$ where $\Gamma_{0}=\frac{1}{4}n^{2}-c_{02}x$ and
$\Gamma=x^{n}-y^{2}$. It is clear that $n>0$ because if $n=0$, then $\Delta$
vanishes in each component of $\\{\Gamma=0\\}$ which implies that $g$ cannot
be positive definite in $\Omega$. The same argument leads to our choice of the
real form of $\Gamma$ and to the choice of $\Omega$ (up to the variable change
$x\mapsto-x$). The form of $\rho$ follows from Proposition 2.11. Indeed, let
$Q$ be as in (8). Since $(\Omega,g,\rho)$ is a solution of $(1,w)$-SDOP
problem for some $w>2$, we deduce from (9) that $\deg_{y}Q=0$. But
$(\Omega,g,\rho)$ is also a solution for $w=\frac{1}{2}\deg c_{0}$ (thus $w=n$
if $c_{02}\neq 0$ and $w=n-1$ if $c_{02}=0$), and we have
$\deg_{x}\Delta=w+1$. Thus (9) implies $\deg_{x}Q\leq 1$. The inequalities for
$p$, $q$, and $\lambda$ are the integrability conditions (see [3, Remark
2.28]), and $c_{02}\leq 0$ is equivalent to $\Gamma_{0}|_{\Omega}>0$ and hence
to $g|_{\Omega}>0$. Thus we obtain (U1).

Case 2. $\deg_{y}\Gamma=1$. Then $(\Gamma,g)$ realizes one of Cases (i)–(vi)
of Proposition 4.4.

Case 2(i). Then $k=1$ because otherwise $\Gamma$ is not $g$-maximal. We have
$\Gamma=x\Gamma_{1}$, $\Gamma_{1}=x^{n}y-1$,
$\Delta=-x^{2}c_{0}(x)\Gamma_{1}$, thus $c_{0}\neq 0$. As above, we follow
§2.3 to find $\rho$. Let $L^{i}$ be as in (32). Then (7) takes the form

$\Big{(}(np+q_{1}x)xc^{\prime}_{0}+(p_{0}+2np+2q_{1}x)nc_{0}\Big{)}y+nxqc_{0}+x^{2}(qc_{0}^{\prime}-q^{\prime}c_{0})=0,$

Equating the coefficient of $y^{0}$ to zero, we obtain $q=C_{1}x^{n}c_{0}$
(here and below $\alpha,\,\beta,\,C_{1},\,C_{2},\,C_{3}$ are some constants).
Equating the coefficient of $y^{1}$ to zero, we obtain two solutions: the
first one is $p_{0}=0$, $q_{1}=-np_{1}$ (then $c_{0}$ is an arbitrary
function); the second one is $c_{0}=C_{2}(np+q_{1}x)/x^{2n+1}$. The first
solution yields $\rho=C_{3}x^{\alpha}\Gamma_{1}^{\beta}$ which contradicts the
integrability condition. The second solution is irrelevant because it cannot
be a nonzero polynomial.

Case 2(ii). We have $\Gamma=xy-1$ and
$\Delta=(xy+1-x^{2}c_{0}-b_{11}^{2}\Gamma+2b_{11})\Gamma$. Then $c_{0}\neq 0$
(otherwise $w_{\min}=1$) and $b_{11}\neq-1$ (otherwise $g$ is as in 2(i),
hence $\Gamma$ is not $g$-maximal). Hence $\Gamma^{2}$ does not divide
$\Delta$, thus we can apply Proposition 2.11 and write $\rho$ in the form (8).
Then (9) with $\mathbb{w}=(1,1+\deg c_{0})$ implies $\deg_{x}Q=0$ and (9) with
$\mathbb{w}=(1,w)$, $w\gg 0$, implies $\deg_{y}Q=0$. Hence $\rho$ is not
integrable.

Case 2(iii). We have $(a,b,c;\Gamma)=(a_{0},b_{1}y,c_{2}y^{2}+c_{1}y;y)$,
$\Delta=(a_{0}c_{2}-b_{1}^{2})y^{2}+a_{0}c_{1}y$. If $\deg c_{1}<2$, then
$w_{\min}=1$, thus $\deg c_{1}\geq 2$. Suppose that $\deg a\geq 1$. Then
$\deg_{x}ac\geq 3>\deg_{x}b^{2}$ whence $\deg_{x}\Delta=\deg_{x}ac$ and we
obtain a contradiction with Proposition 2.15 for $w=\deg c_{1}$. Thus $0\neq
a=\operatorname{const}$ and we may set $a=1$.

We have $ac_{1}\neq 0$, hence $y$ does not divide $\Delta$. Then Proposition
2.11 with $\mathbb{w}=(1,w)$, $w\gg 0$, implies that $\rho=y^{p}e^{h}$ with
$\deg_{y}h=2-\deg_{y}\Delta$. If $\deg_{y}\Delta=2$, this contradicts the
integrability condition, hence $\deg_{y}\Delta=1$. Then $c_{2}=b_{1}^{2}$, in
particular $b_{1}=\operatorname{const}$, thus $(a,b,c)=(1,\beta
y,\beta^{2}y^{2}+c_{1}y)$, $\beta\in\mathbb{R}$. Let $h=h_{1}(x)y+h_{0}(x)$.
Then $\deg_{y}L^{1}=0$ for
$L^{1}=ah^{\prime}_{x}+bh^{\prime}_{y}=h^{\prime}_{1}y+h^{\prime}_{0}+\beta
yh_{1}$, hence $h^{\prime}_{1}=\beta h_{1}$. If $\beta\neq 0$, we obtain
$h_{1}=0$ (since $h_{1}$ is a polynomial). If $\beta=0$, the condition
$\deg_{(1,w)}L^{2}\leq w$ for
$L^{2}=bh^{\prime}_{x}+ch^{\prime}_{y}=yc_{1}h_{1}$ again implies $h_{1}=0$
because $\deg c_{1}>0$. Thus $\deg_{y}h=0$ which contradicts the integrability
condition.

Case 2(iv). Then
$(a,b,c;\Gamma)=(x\tilde{a}_{0},\,b_{11}xy,\,c_{2}y^{2}+c_{1}y;\,xy)$,
$\Delta=(\tilde{a}_{0}c_{2}-b_{11}x)xy^{2}+ac_{1}y$. As in Case 2(iii), we
obtain $\deg c_{1}\geq 2$, $\tilde{a}_{0}=1$, $y$ does not divide $\Delta$,
hence (see Remark 2.13) $\deg_{y}\Delta=1$, that is $c_{2}=b_{11}^{2}x$. Thus
we arrive to $(a,b,c)=(x,\beta xy,\beta y^{2}+c_{1}y)$. Then
$\rho=x^{p}y^{q}e^{h}$ with $h=h_{1}y+h_{0}$ where $h_{k}$ are rational
functions of $x$. The rest of the proof is as in Case 2(iii).

Case 2(v). We obtain (U2) by Proposition 2.8.

Case 2(vi). $a=0$ is impossible for a positive definite $g$.

Case 3. $\deg_{y}\Gamma=0$, i.e., $\Gamma=\Gamma(x)$ is a polynomial in $x$
only. Then $\Omega=I\times\mathbb{R}$ for a finite or infinite (from either
side) interval $I$. Further, $\rho$ is of the form (8) (see Remark 2.12) where
$\deg_{y}Q\geq 2$ by the integrability condition, hence $\deg_{y}Q=2$ and
$\deg_{y}\Delta=0$ by (9); we write $Q=h_{2}y^{2}+h_{1}y+h_{0}$ where $h_{k}$
are rational functions of $x$ and $h_{2}\neq 0$. Then (see (6))

$\begin{matrix}\deg_{\mathbb{w}}L^{1}=1\quad\text{for}\quad
L^{1}=a(h_{2}^{\prime}y^{2}+h_{1}^{\prime}y+h_{0}^{\prime})+b(2h_{2}y+h_{1}),\\\
\deg_{\mathbb{w}}L^{2}=w\quad\text{for}\quad
L^{2}=b(h_{2}^{\prime}y^{2}+h_{1}^{\prime}y+h_{0}^{\prime})+c(2h_{2}y+h_{1}).\end{matrix}$
$None$

Since $\deg_{y}\Gamma=0$, (11) implies that $\Gamma$ divides $a$ and $b$, thus
$\deg_{x}\Gamma\leq 2$. Since $\Omega\subset\\{\Gamma=0\\}$, Proposition 2.15
implies

$\deg_{x}\Delta\leq\deg_{x}\Gamma-1+\max\big{(}\,\lfloor
w_{\min}\rfloor+\deg_{x}b,\,1+\deg_{x}c\,\big{)}.$ $None$

Case 3.1. $\deg_{x}\Gamma=2$. Then, up to rescaling, $a=\Gamma$ and
$b=b_{0}=\tilde{b}(x)\Gamma$ for some polynomial $\tilde{b}$. Then the
variable change $y\mapsto y-p(x)$ with $p^{\prime}=\tilde{b}$ (see Example
2.7) makes $b=0$. Then $\deg_{y}\Delta=0$ implies $c=c_{0}$. Then (33) yields
$h_{2}^{\prime}=0$ and $2h_{2}c_{0}=\operatorname{const}$. Since $h_{2}\neq
0$, we conclude that $c_{0}=\operatorname{const}$, hence we obtain (U2).

Case 3.2. $\Gamma=x$. Then $a=x\tilde{a}$ and $b=x\tilde{b}=x(\beta
y+\tilde{b}_{0}(x))$ ($\beta=b_{11}$) by (11), hence the coefficient of
$y^{2}$ in $\Delta$ is $ac_{2}-x^{2}\beta$. Since $\deg_{y}\Delta=0$, $a\neq
0$, and $c_{2}=\operatorname{const}$, we have either $a=a_{20}x^{2}$ (then we
may assume $a_{20}=1$) or $\beta=0$.

Case 3.2.1. $a=x^{2}$. We have $\Delta=x^{2}d(x)$ for some polynomial
$d=d(x)$, so we may write $c=\tilde{b}^{2}+d$. If $\deg\tilde{b}_{0}\leq 1$,
then either (34) fails or $w_{\min}=1$. Thus we assume that
$\deg\tilde{b}_{0}\geq 2$. The change of variables $y\mapsto y+\lambda x^{n}$
transforms $b$ into $b+\lambda(n-\beta)x^{n+1}$ (see Example 2.7). Hence we
may kill all coefficients of $b_{0}$ unless $\beta=n\in\mathbb{N}$ in which
case we kill all of them except $b_{n+1,0}$. Since $\deg\tilde{b}_{0}\geq 2$,
we then assume that $\tilde{b}=ny+\alpha x^{n}$, $n\geq 2$.

Then (see (33)) the coefficient of $y^{2}$ in $L^{1}$ is
$x^{2}h_{2}^{\prime}+2nxh_{2}$, hence $h_{2}^{\prime}=-2nxh_{2}$ whence
$h_{2}=Cx^{-2n}$ which contradict the facts that $h_{2}\neq 0$ and $h_{2}$ is
a rational function with denominator at most $x$ (see [3, Prop. 2.15]).

Case 3.2.2. $\beta=0$. Then the condition $\deg_{y}\Delta=0$ implies
$c_{2}=c_{1}=0$, i.e., $g$ does not depend on $y$. By the change of variables
$y\mapsto y+p(x)$ we may achieve that $\deg_{x}\tilde{b}<\deg_{x}\tilde{a}$.
If $\deg\tilde{a}=0$, then $b=0$ and the proof is the same as in Case 3.1. If
$\deg\tilde{a}=1$ and $b\neq 0$, we obtain a contradiction with (34).

Case 3.3. $\Gamma$ is constant. Then $\Omega=\mathbb{R}^{2}$. Since
$a|_{\Omega}>0$, we have up to translation $a=x^{2}+1$ or $a=1$.

Case 3.3.1. $a=x^{2}+1$. Recall that $\deg_{y}\Delta=0$, hence
$(x^{2}+1)c_{2}=b_{1}^{2}$ and $ac_{1}=2b_{1}b_{0}$ whence
$b_{1}=c_{2}=c_{1}=0$. By a variable change $y\mapsto y-p(x)$ we may achieve
that $\deg b\leq 1$ (see Example 2.7) and we obtain a contradiction with (34).

Case 3.3.2. $a=1$. If $b_{1}=0$, the proof is as in Case 3.1. If $b_{1}\neq
0$, then (33) for $L^{1}$ implies $h_{2}^{\prime}=-b_{1}h_{2}$ which is
impossible for nonzero polynomials. ∎

Remark 6.5. There is a small mistake in [3, §4.2]. It is erroneously claimed
that each cometric solution on the square $[-1,1]^{2}$ is proportional to
$\operatorname{diag}(1-x^{2},1-y^{2})$. The correct answer is
$g=\operatorname{diag}\big{(}\alpha(1-x^{2}),\beta(1-y^{2})\big{)}$ with
arbitrary positive $\alpha$ and $\beta$. All the corresponding riemannian
metrics $(g_{ij})=g^{-1}$ are pairwise non-isometric.

## 7\. Realization of solutions as images of the Laplace operator

Let $M_{1}$ and $M_{2}$ be smooth manifolds and $\Phi:M_{1}\to M_{2}$ be a
smooth mapping which is a submersion at a generic point of $M_{1}$. Let
$\Cal{L}_{1}$ and $\Cal{L}_{2}$ be differential operators on $M_{1}$ and on
$\Omega=\Phi(M_{1})$ respectively. We say that $\Cal{L}_{2}$ is the image of
$\Cal{L}_{1}$ through $\Phi$ if $\Cal{L}_{2}(f)=\Cal{L}_{1}(f\circ\Phi)$.
Notice that the image of $\Cal{L}_{1}$ through $\Phi$ may or may not exist,
moreover, it does not exist for generic $\Cal{L}$ and $\Phi$ unless $\Phi$ is
injective. However it does exist when $\Cal{L}_{1}$ and $\Phi$ are invariant
under an action of a group $G$ on $M_{1}$, and then $\Phi$ identifies $\Omega$
with the orbit space $M_{1}/G$.

For example, for half-integer $p$ and $q$, the Jacobi operator $J_{p,q}$ on
the interval $(-1,1)$ (see Remark 1.1) is the image of the Laplace operator
${\lx@ams@boldsymbol@{\Delta}}_{\mathbb{S}^{n}}$ on the sphere
$\mathbb{S}^{n}$, $n=2p+2q-1$, through the mapping

$(x_{1},\dots,x_{2p},y_{1},\dots,y_{2q})\mapsto x_{1}^{2}+\dots+x_{2p}^{2}$

(see [3, §2.1], the end of [3, §4.1], and references therein).

In [3, §4] similar interpretations are given for many values of parameters of
each compact solution of $(1,1)$-DOP problem. They are interpreted as images
of the Laplace (Casimir) operators on $\mathbb{S}^{n}$, $\mathbb{R}^{n}$,
$SO(n)$, or $SU(n)$.

Each quotient of $\mathbb{S}^{2}$ or $\mathbb{R}^{2}$ by a reflection group
can be identified with a compact domain $\Omega$,
$\partial\Omega\subset\\{\Gamma(x,y)=0\\}\subset\mathbb{R}^{2}$, so that
$(\Omega,\mathbb{L},\Gamma^{-1/2}dx)$ is a solution of $(1,w)$-DOP problem
where $\mathbb{L}$ is the image of
${\lx@ams@boldsymbol@{\Delta}}_{\mathbb{S}^{2}}$ or
${\lx@ams@boldsymbol@{\Delta}}_{\mathbb{R}^{2}}$ through the quotient map.
Under this identification, the vertices of the fundamental polygon of the
group action are in bijection with singular points of $\partial\Omega$ so that
the angle $\pi/n$ corresponds to the singularity $A_{n-1}$ (given by
$u^{2}=v^{n}$ in some local coordinates). Explicit formulas for the mappings
$\mathbb{S}^{2}\to\mathbb{R}^{2}$ and $\mathbb{R}^{2}\to\mathbb{R}^{2}$ which
realize the operators $\mathbb{L}$ as images of the Laplace operator are given
in [3] and [1]; see detailed references in Table 1.

Angles  Boundary of $\Omega$ $w$ Reference  2,2,2,2  Rectangle (see also
Remark 6.5)  1  [3,§4.2]

$\mathbb{R}^{2}$ 2,4,4  Parabola with two tangents  1  [3,§4.7]

3,3,3  Deltoid: $x=2\cos\theta+\cos 2\theta$, $y=2\sin\theta-\sin 2\theta$ 1
[3,§4.12]

2,3,6  Cubic $y^{2}=x^{3}$ with a cubically tangent parabola  2  [3, end of
§4.12]  —  Circle  1  [3,§4.3]

2,2  Coaxial parabolas: $y^{2}=(1-x^{2})^{2}$ 1  [3,§4.5]

$n,n$ $y^{2}=(1-x^{2})^{n}$, $n\geq 2$ $n$ [1,§6]: $\Omega_{1}^{(n)}$

2,2,2  Triangle  1  [3,§4.4]

$\mathbb{S}^{2}$ 2,2,3  $(y^{2}-x^{3})(x-1)=0$ 1  [3,§4.9]

2,2,4  $y(y-x^{2})(x-1)=0$ 1  [3,§4.6]

2,2,$n$ $(y^{2}-x^{n})(x-1)=0$, $n\geq 2$ $n$ [1,§6]:
$\Omega_{3}^{(n)}\\!\\!,\Omega_{6}^{(n)}$

2,3,3  Swallow tail: $\text{discrim}_{t}(t^{4}-t^{2}+xt+y)=0$ 1  [3,§4.11]

2,3,4  Cubic $y^{2}=x^{3}$ with a tangent line  1  [3,§4.10]

2,3,5  Dodecahedral quotient, see Theorem 6.1(B1)  2  [1,§8]: $\Omega_{21}$

Table 1 Quotients of $\mathbb{R}^{2}$ and $\mathbb{S}^{2}$ by reflection
groups; $a,b,\dots$ in the “Angles” cell means: the angles of the fundamental
domain are $\frac{\pi}{a},\frac{\pi}{b},\dots$

If $\mathbb{L}$ lifts to ${\lx@ams@boldsymbol@{\Delta}}_{\mathbb{S}^{2}}$ or
${\lx@ams@boldsymbol@{\Delta}}_{\mathbb{R}^{2}}$, then the corresponding
metric is of constant non-negative curvature. Moreover, the classification
obtained in [3] implies that the curvature is constant and non-negative for
all solutions with $\det(g)$ of maximal degree. For maximal degree solutions,
Lev Soukhanov [5], [6] proved the constancy of curvature (but not its non-
negativity) not using the classification. He also proved a generalization of
this result to an arbitrary dimension.

The curvature in solution (B4) in Theorem 6.2 is constant if and only if $m=n$
and $c_{02}=-n^{2}$. In (B5) it is constant if and only if
$c_{02}=-\frac{1}{4}n^{2}$. If $m\neq n$ in (B4), the curvature cannot be
constant because there are no biangles of constant curvature with different
angles. For $m=n$ in (B4) and for (B5), the curvature is

$\frac{-\lambda
n^{2}\big{(}2(n^{2}+\alpha)x^{k}+\alpha\big{)}}{\big{(}(n^{2}+\alpha)x^{k}-\alpha\big{)}^{2}},\qquad(k,\lambda,\alpha)=\begin{cases}(2,2,c_{02}),&\text{(B4),
$m=n$},\\\ (1,\frac{1}{2},4c_{02}),&\text{(B5)},\end{cases}$

which is evidently non-constant when $\alpha\neq-n^{2}$ (cf. [3, §4.5]). The
similarity of the formulas for the curvature is not occasional: up to
adjusting the constants, (B5) is the image of (B4) through
$(x,y)\mapsto(x^{2},y)$.

According to our classification, the only bounded domains which admit
solutions but are not covered by Table 1 are those in Theorem 6.2(B4) for
$m\neq n$. The simplest one is the nodal cubic $y^{2}=x^{2}-x^{3}$ which
corresponds to $(m,n)=(1,2)$. This solution is realized in [3, §4.8] as the
image of ${\lx@ams@boldsymbol@{\Delta}}_{\mathbb{S}^{3c}}$, $c=1,2,4,8$. The
realization for $c=1$ immediately extends to any $(m,n)$ as follows (it could
be interesting to do the same for $c=2,4,8$). We consider $\mathbb{S}^{3}$ as
the unit sphere in $\mathbb{C}^{2}$ with coordinates $(z_{1},z_{2})$. Then the
image of $\frac{1}{4}{\lx@ams@boldsymbol@{\Delta}}_{\mathbb{S}^{3}}$ through

$(z_{1},z_{2})\mapsto(X,Y)=\big{(}\,|z_{1}|^{2},\,\operatorname{Re}(z_{1}^{n}\bar{z}_{2}^{m})\,\big{)}$

is an operator on the domain bounded by the curve $(1-X)^{m}X^{n}-Y^{2}=0$.
Its image through the affine change
$(X,Y)\mapsto(x,y)=(\,2X-1,\,2^{(m+n)/2}Y\,)$ is the operator corresponding to
(B4) with $p=q=\frac{1}{2}$ and $c_{02}=-\frac{1}{4}(m+n)^{2}$.

## References

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