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\title{TRANSFORMING THE HEUN EQUATION TO THE\\HYPERGEOMETRIC EQUATION:\\
I.~POLYNOMIAL TRANSFORMATIONS}
\author{Robert S. Maier\thanks{Depts.\ of Mathematics and Physics, University of Arizona,
Tucson AZ 85721. This research was partially supported by NSF grants
PHY-9800979 and PHY-0099484.}}
\begin{document}
\maketitle
\begin{abstract}
The reductions of the Heun equation to the hypergeometric equation by
rational changes of its independent variable are classified.
Heun-to-hypergeometric transformations are analogous to the classical
hypergeometric identities (i.e., hypergeometric-to-hypergeometric
transformations) of Goursat. However, a transformation is possible only~if
the singular point location parameter and normalized accessory parameter of
the Heun equation are each restricted to take values in a discrete set.
The possible changes of variable are all polynomial. They include
quadratic and cubic transformations, which may be performed only~if the
singular points of the Heun equation form a harmonic or an equianharmonic
quadruple, respectively; and several higher-degree transformations.
\end{abstract}
\begin{keywords}
Special functions, Heun equation, hypergeometric equation, hypergeometric
identities, Lam\'e equation, Clarkson--Olver transformation.
\end{keywords}
\begin{AMS}
33E30, 34M35, 33C05.
\end{AMS}
\pagestyle{myheadings}
\thispagestyle{plain}
\markboth{ROBERT~S. MAIER}{TRANSFORMING THE HEUN EQUATION}
\section{Introduction}
\label{sec:intro}
Consider the class of linear second-order differential equations on the
Riemann sphere~$\mathbb{CP}^1$ which are Fuchsian, i.e., have only regular
singular points~\cite{Hille76}. Any such equation with exactly three
singular points can be transformed to the hypergeometric equation by
appropriate changes of the independent and dependent variables. Similarly,
any such equation with exactly four singular points can be transformed to
the Heun equation~(\cite{Erdelyi53}, Chap.~15; \cite{Ronveaux95,Snow52}).
Solutions of the Heun equation are much less well understood than
hypergeometric functions~\cite{Arscott81}. No~general integral
representation for them is known, for instance. Most work on solutions of
the Heun equation has focused on solutions of special cases, such as the
Lam\'e equation~\cite{Erdelyi53,Hille76}. The parameters of the Heun
equation include four characteristic exponent parameters, a~singular point
location parameter, and a global accessory parameter, so there is a large
parameter space to investigate. This is in~contrast to the hypergeometric
equation, which has only three parameters.
Solutions of Heun equations, such as the Lam\'e equation, are occasionally
used in mathematical modeling. However, it~is difficult to carry~out
practical computations involving them. An~explicit solution to the
two-point connection problem for the general Heun equation is not
known~\cite{Schafke80a}, though the corresponding problem for the
hypergeometric equation has a classical solution. Determining when the
solutions of a Heun equation are expressible in closed form in~terms of
more familiar functions would obviously be useful: it~would facilitate the
solution of the two-point connection problem, and the computation of Heun
equation monodromies. A~significant result in this direction was obtained
by Kuiken~\cite{Kuiken79}. It~is sometimes possible, by performing a
quadratic change of the independent variable, to transform the Heun
equation to the hypergeometric equation, and thereby express its solutions
in~terms of hypergeometric functions. Kuiken's quadratic transformations
are not so well known as they should~be. The useful monograph edited by
Ronveaux~\cite{Ronveaux95} does not mention them explicitly, though it
lists Ref.~\cite{Kuiken79} in its bibliography. Recently, one of Kuiken's
transformations was rediscovered by Ivanov~\cite{Ivanov2001}.
Unfortunately, the theorem of Ref.~\cite{Kuiken79} is incomplete. The
theorem asserts that a transformation to the hypergeometric equation, by a
rational change of the independent variable, is possible only~if the
singular points of the Heun equation form a harmonic quadruple in the sense
of projective geometry; in~which case the change of variables must be
quadratic. We~shall show that there are several other possibilities.
A~transformation may also be possible if the singular points form an
equianharmonic quadruple, in~which case the change of variables must be
cubic. There are additional singular point configurations that permit
transformations of degrees 3,~4, 5, and~6. Our main theorem
(Theorem~\ref{thm:main}) and its corollaries classify all such
transformations, up~to linear automorphisms of the Heun and hypergeometric
equations.
We were led to this correction and expansion of the theorem of
Ref.~\cite{Kuiken79} by considering a discovery of Clarkson and
Olver~\cite{Clarkson96}: an~unexpected reduction of the Weierstrass form of
the equianharmonic Lam\'e equation to the hypergeometric equation. Their
reduction involves a cubic change of the independent variable, which, it
turns~out, is a special case of a general transformation.
In~\S\ref{sec:CO}, we~determine the extent to~which their result can be
generalized.
Our new transformations are similar to the polynomial transformations which
appear in the classical hypergeometric identities (i.e.,
hypergeometric-to-hypergeometric transformations) of Goursat.
(See~\cite{Erdelyi53},~Chap.~2; also~\cite{Askey94}.) Transforming the
Heun equation to the hypergeometric equation is more difficult: it~is
possible only~if the singular point location parameter and normalized
accessory parameter are restricted to take values in related discrete sets.
Actually, the Heun-to-hypergeometric transformations classified in this
part of the paper (Part~I) are of a restricted type: unlike most classical
hypergeometric transformations, they include no~linear change of the
dependent variable. A~classification of Heun-to-hypergeometric
transformations of the more general type is possible, but is best phrased
in geometric terms: it~relies on a classification of certain branched
covers of the Riemann sphere by itself. Such a classification will appear
in Part~II\null.
\section{Preliminaries}
\subsection{The Equations}
\label{sec:defs}
The hypergeometric equation may be written in the form
$$
%\begin{equation}
%\label{eq:hyper}
\frac{d^2 y}{dz^2} + \left(\frac{c}{z} + \frac{a+b-c+1}{z-1}\right)
\frac{dy}{dz} + \frac{ab}{z(z-1)}\,y = 0.
\leqno (\mathfrak{h})
%\end{equation}
$$
It~and its solution space are specified by the $P$-symbol
\begin{equation}
\label{eq:HeunP}
P\left\{
\begin{array}{cccc}
0&1&\infty& \\
0&0&a& ;z \\
1-c&c-a-b&b&
\end{array}
\right\},
\end{equation}
where each column, except the last, refers to a singular point. The first
entry is its location, and the final two are the characteristic exponents
of the solutions there. The exponents at each singular point are obtained
by solving the Frobenius indicial equation~\cite{Hille76}. In~general,
each finite singular point~$z_0$ has $\zeta$~as a characteristic exponent
if~and only~if the equation has a solution of the form $(z-z_0)^\zeta h(z)$
in a neighborhood of~$z=z_0$, where $h$~is analytic and nonzero at~$z=z_0$.
If~the exponents at~$z=z_0$ differ by an integer, this statement must be
modified: the solution corresponding to the smaller exponent may have a
logarithmic singularity at~$z=z_0$. The definition extends in a
straightforward way to~$z_0=\infty$, and to ordinary points, each of which
is said to have exponents~$0,1$.
There are $2\times3=6$ local solutions of the hypergeometric equation
in~all: two per singular point. If $c$~is not a nonpositive integer, the
local solution at~$z=0$ belonging to the exponent zero will be analytic.
In~that case, when normalized to unity at~$z=0$, it~will be the Gauss
hypergeometric function ${}_2F_1(a,b;c;z)$~\cite{Erdelyi53}. This is the
sum of a hypergeometric series, which converges in a neighborhood of~$z=0$.
In~general, ${}_2F_1(a,b;c;z)$ is not defined when $c$~is a nonpositive
integer.
The Heun equation is conventionally written in the form
$$
%\begin{equation}
%\label{eq:Heun}
\frac{d^2 u}{d t^2}
+ \left( \frac\gamma t + \frac\delta{t-1} + \frac\epsilon{t-d}
\right)\frac{du}{dt} + \frac{\alpha\beta t - q}{t(t-1)(t-d)}\,u = 0.
\leqno ({\mathfrak H})
%\end{equation}
$$
Here $d\in\mathbb{C}$, the location of the fourth regular singular point,
is a parameter~($d\neq0,1$), and the exponent parameters
$\alpha,\beta,\gamma,\delta,\epsilon\in\mathbb{C}$ are constrained to
satisfy
\begin{equation}
\label{eq:Pconstraint}
\epsilon=\alpha+\beta-\gamma-\delta+1.
\end{equation}
The Heun equation and its solutions are specified by the Riemann $P$-symbol
\begin{equation}
\label{eq:Psymbol}
P\left\{
\begin{array}{ccccc}
0&1&d&\infty& \\
0&0&0&\alpha& ;t \\
1-\gamma&1-\delta&1-\epsilon&\beta&
\end{array}
\right\}.
\end{equation}
The condition~(\ref{eq:Pconstraint}) constrains the sum of the
characteristic exponents to equal~$2$. The $P$-symbol does not fully
specify the Heun equation and its solutions, since it omits the accessory
parameter~$q\in\mathbb{C}$.
There are $2\times4=8$ local solutions of the Heun equation in~all: two per
singular point. If $\gamma$~is not a nonpositive integer, the local
solution at~$t=0$ belonging to the exponent zero will be analytic. In~that
case, when normalized to unity at~$t=0$, the solution is called the local
Heun function, and is denoted
$\Hl(d,q;\alpha,\beta,\gamma,\delta;t)$~\cite{Ronveaux95}. It~is the sum
of a Heun series, which converges in a neighborhood
of~$t=0$~\cite{Ronveaux95,Snow52}. In~general,
$\Hl(d,q;\alpha,\beta,\gamma,\delta;t)$ is not defined when $\gamma$~is a
nonpositive integer.
The solution spaces of the hypergeometric equation and Heun equation are
$2$-dimensional vector spaces of global analytic functions, i.e.,
$\mathbb{CP}^1$-valued functions on the multiply punctured Riemann sphere
$\mathbb{CP}^1\setminus\{0,1,\infty\}$ (resp.\
$\mathbb{CP}^1\setminus\{0,1,d,\infty\}$). In~general, the solutions are
multivalued. They may be viewed as single-valued analytic functions on a
parameter-dependent Riemann surface~$\cal S$ over~$\mathbb{CP}^1$, branched
at $z=0,1,\infty$ (resp.\ $t=0,1,d,\infty$). For certain parameter
choices, $\cal S$~may be compact, in which case the solutions, projected to
the multiply punctured Riemann sphere, will be finite-valued, i.e.,
algebraic. We~mention this only in passing: Heun-to-hypergeometric
transformations will turn~out to exist even in the absence of algebraicity.
If $\epsilon=0$ and $q=\alpha\beta d$, the Heun equation loses a singular
point and becomes the hypergeometric equation. Similar losses occur if
$\delta=0$, $q=\alpha\beta$, or $\gamma=0$,~$q=0$. Our results will
exclude the degenerate case when the Heun equation has fewer than four
singular points, since transforming the hypergeometric equation into itself
is a separate problem, leading to hypergeometric identities. Also, our
treatment will initially exclude the following case, in which
{\rm ($\mathfrak{H}$)}~can be solved by elementary means.
\begin{definition}
If $\alpha\beta=0$ and~$q=0$, the Heun equation~{\rm ($\mathfrak{H}$)} is
said to be trivial. Triviality implies one of the exponents at~$t=\infty$
is zero (i.e., $\alpha\beta=0$), and is implied by absence of the singular
point at~$t=\infty$ (i.e., $\alpha\beta=0$, $\alpha+\beta=1$, $q=0$).
\end{definition}
\smallskip
The transformation to the Heun equation or hypergeometric equation of a
linear second-order Fuchsian differential equation with singular points at
$t=0,1,d,\infty$ (resp.\ $z=0,1,\infty$), and with arbitrary characteristic
exponents, is accomplished by linear changes of the dependent variable,
called F-homotopic transformations. (See~\cite{Erdelyi53}
and~\cite{Ronveaux95}, \S{A}2 and Addendum,~\S1.8.) If~an equation with
singular points at~$t=0,1,d,\infty$ has dependent variable~$u$,
carrying~out the substitution $\tilde
u(t)=t^{-\rho}(t-1)^{-\sigma}(t-d)^{-\tau} u(t)$ will convert the equation
to a new one, with the exponents at~$t=0,1,d$ reduced by~$\rho,\sigma,\tau$
respectively, and those at~$t=\infty$ increased by $\rho+\sigma+\tau$.
By~this technique, one of the exponents at each finite singular point can
be set to zero, yielding the Heun equation. $\rho,\sigma,\tau$ are not
unique: in~general, there are two possibilities for each.
In fact, the Heun equation has a $({\mathbb Z}_2)^3$ group of F-homotopic
automorphisms, since at each of $t=0,1,d$, the exponents~$0,\zeta$ can be
shifted to~$-\zeta,0$, which is equivalent to~$0,-\zeta$. Similarly, the
hypergeometric equation has a $({\mathbb Z}_2)^2$ group of F-homotopic
automorphisms. These groups act on the 6-dimensional and 3-dimensional
parameter spaces, respectively. For example, the latter includes
$(a,b;c)\mapsto(c-a,c-b;c)$, which is obtained from an F-homotopic
transformation at~$z=1$. The identity
\begin{equation}
\label{eq:flip}
{}_2F_1(a,\,b;\,c;\,z)= (1-z)^{c-a-b}{}_2F_1(c-a,\,c-b;\,c;\,z)
\end{equation}
is satisfied by~${}_2F_1$, since ${}_2F_1$~is a local solution at~$z=0$,
rather than at~$z=1$.
If the regular singular points of the Fuchsian differential equation are
arbitrarily placed, transforming it to the hypergeometric or Heun equation
will require a M\"obius (i.e., projective linear or homographic)
transformation, which repositions the singular points to the standard
locations. A~unique M\"obius transformation maps any three distinct points
in~$\mathbb{CP}^1$ to any other three distinct points; but the same is not
true of four points, which is why the Heun equation has~$d$ as a free
parameter.
\subsection{The Cross-Ratio}
Our characterization of Heun equations that can be reduced to the
hypergeometric equation will employ the cross-ratio orbit of
$\{0,1,d,\infty\}$, defined thus. If $A,B,C,D$ are four distinct points
in~$\mathbb{CP}^1$, their cross-ratio is
\begin{equation}
(A,B;C,D)\defeq
\frac{(C-A)(D-B)}{(D-A)(C-B)}\in\mathbb{CP}^1\setminus\{0,1,\infty\}.
\end{equation}
The cross-ratio is is invariant under M\"obius transformations; in~fact,
$(A_1,B_1;C_1,D_1)$ can be mapped to $(A_2,B_2;C_2,D_2)$ iff their
cross-ratios are equal. When studying the mapping of unordered point sets,
it~is necessary to take the action of permutations into account. Permuting
$A,B,C,D$ transforms $(A,B;C,D)$ in a well-defined way, yielding an action
of the symmetric group~$S_4$ on~$\mathbb{CP}^1\setminus\{0,1,\infty\}$.
$(A,B;C,D)$ is invariant under interchange of the pairs $A,B$ and~$C,D$,
and also under simultaneous interchange of the two points in each pair. So
each orbit contains no~more than $4!/4=6$ cross-ratio values. By
examination, the six possible actions of~$S_4$
on~$s\in\mathbb{CP}^1\setminus\{0,1,\infty\}$ are generated by
$s\mapsto1-s$ and $s\mapsto 1/s$, and the orbit of~$s$ comprises
\begin{equation}
s,\quad 1-s,\quad 1/s,\quad 1/(1-s),\quad s/(s-1),\quad (s-1)/s,
\end{equation}
which may not be distinct. This is called the cross-ratio orbit of~$s$;
or, if~$s=(A,B;C,D)$, the cross-ratio orbit of the unordered set
$\{A,B,C,D\}\subset\mathbb{CP}^1$.
Generic cross-ratio orbits contain six values, but there are two
exceptions: an~orbit containing exactly three values and an orbit
containing exactly two. If $(A,B;C,D)$ equals~$-1$, then the cross-ratio
orbit of $\{A,B,C,D\}$ will clearly be $\{-1,1/2,2\}$. The value~$-1$ for
$(A,B;C,D)$ defines a so-called harmonic configuration: $A,B$ and $C,D$ are
said to be harmonic pairs. More generally, if $(A,B;C,D)$ equals any of
$-1$, $1/2$, or~$2$, i.e., if $\{-1,1/2,2\}$ is the cross-ratio orbit of
the unordered set $\{A,B,C,D\}$, then $\{A,B,C,D\}$ is said to be a
harmonic quadruple. It~is easy to see that if $C$~is at infinity and
$A,B,D$ are distinct finite points, then $A,B$ and $C,D$ will be harmonic
pairs iff $D$~is the midpoint of the line segment~$\overline{AB}$.
In~consequence, $\{A,B,C=\infty,D\}$ will be a harmonic quadruple iff
$\{A,B,D\}$ comprises three points which are collinear and equally spaced.
So, $\{A,B,C,D\}\subset\mathbb{CP}^1$ will be a harmonic quadruple iff it
can be mapped by a M\"obius transformation to a set consisting of three
equally spaced finite points and the point at infinity; equivalently, to
the vertices of a square in~$\mathbb{C}$.
The cross-ratio orbit containing exactly two values comprises the two
non-real cube roots of~$-1$, i.e., is $\{(1+i\sqrt3)/2,(1-i\sqrt3)/2\}$.
If~this is the cross-ratio orbit of~$\{A,B,C,D\}$, $\{A,B,C,D\}$ is said to
be an equianharmonic quadruple. If $C$~is at infinity, $\{A,B,C,D\}$ will
be an equianharmonic quadruple iff $A,B,D$ are the vertices of an
equilateral triangle in~$\mathbb{C}$\null. If $\mathbb{CP}^1$ is
interpreted as a sphere via the usual stereographic projection, then by a
linear transformation (a~special M\"obius transformation), this situation
reduces to the case when $A,B,C=\infty,D$ are the vertices of a regular
tetrahedron. So, $\{A,B,C,D\}\subset\mathbb{CP}^1$ will be an
equianharmonic quadruple iff it can be mapped by a M\"obius transformation
to the vertices of a regular tetrahedron.
Cross-ratio orbits are of two sorts: real orbits such as the harmonic
orbit, and non-real orbits such as the equianharmonic orbit. All values in
a real orbit are real, and in a non-real orbit, all have a nonzero
imaginary part. So, $\{A,B,C,D\}$ will have a specified real orbit as its
cross-ratio orbit iff it can be mapped by a M\"obius transformation to a
set consisting of three specified collinear points in~$\mathbb{C}$ and the
point at infinity; equivalently, to the vertices of a specified quadrangle
(generically irregular) in~$\mathbb{C}$\null. Similarly, it will have a
specified non-real orbit as its cross-ratio orbit iff it can be mapped to a
set consisting of a specified triangle in~$\mathbb{C}$ and the point at
infinity; equivalently, to the vertices of a specified tetrahedron
(generically irregular) in~$\mathbb{CP}^1$.
The cross-ratio orbit of $\{0,1,d,\infty\}$ will be the harmonic orbit iff
$d$~equals $-1$, $1/2$, or~$2$, and the equianharmonic orbit iff $d$~equals
$(1\pm i\sqrt{3})/2$. In~contrast, it will be a specified generic orbit
iff $d$~takes one of six orbit-specific values.
The cross-ratio orbit of $\{0,1,d,\infty\}$ being a specified orbit is
equivalent to its being the same as the cross-ratio orbit of some specified
quadruple of the form $\{0,1,D,\infty\}$, i.e., to there being a M\"obius
transformation that maps $\{0,1,d,\infty\}$ onto $\{0,1,D,\infty\}$.
By~examination, this occurs iff $\{0,1,d\}$ is mapped onto $\{0,1,D\}$ by
some {\em linear\/} transformation, i.e., iff the the triangle
$\triangle01d$ is similar to some specified triangle in~$\mathbb{C}$\null.
This gives a geometric significance to the allowed values of~$d$.
For the equianharmonic orbit, in which the six values degenerate to two,
the triangle may be chosen to be any equilateral triangle. For a real
orbit, all values on the cross-ratio orbit are real. But the statement
about the orbit being characterized by $\triangle01d$ being similar to some
triangle in~$\mathbb{C}$ is still true in a degenerate sense, in which the
vertices of the triangles are allowed to be collinear. For example, the
cross-ratio orbit of $\{0,1,d,\infty\}$ being the harmonic orbit is
equivalent to the `triangle' $\triangle01d$ being similar to the `triangle'
$\triangle012$, i.e., to the unordered set $\{0,1,d\}$ being a set
consisting of three equally spaced collinear points. This will hold iff
$d$~equals $-1$, $1/2$, or~$2$, in agreement with the definition of a
harmonic quadruple.
\subsection{Automorphisms}
\label{subsec:auto}
According to the theory of the Riemann $P$-function, a M\"obius
transformation~$M$ of the independent variable will preserve characteristic
exponents. For the hypergeometric equation~$(\mathfrak{h})$, this implies
that if $M$~is one of the $3!$~M\"obius transformations that permute the
singular points $z=0,1,\infty$ (i.e.,~elements of the symmetric
group~$S_3$), the exponents of the transformed
equation~$(\tilde{\mathfrak{h}})$ at its singular points
$M(0),M(1),M(\infty)$ will be those of~$(\mathfrak{h})$ at~$0,1,\infty$.
But if $M$~is not linear, i.e., $M(\infty)\neq\infty$, then
$(\tilde{\mathfrak{h}})$~will not in general be a hypergeometric equation,
since its exponents at~$M(\infty)$ may both be nonzero. To~convert
$(\tilde {\mathfrak{h}})$ to a hypergeometric equation, each permutation
in~$S_3$ must in~general be followed by an F-homotopic transformation of
the form $\tilde y(z)=[z-M(\infty)]^{-a} y(z)$ or $\tilde
y(z)=[z-M(\infty)]^{-b} y(z)$.
\begin{definition}
${\rm Aut}(\mathfrak{h})$, the automorphism group of the hypergeometric
equation~${\rm (\mathfrak{h})}$, is the group of changes of variable
(M\"obius of the independent variable, linear of the dependent variable)
which leave~${\rm (\mathfrak{h})}$ invariant, except for changes of
parameter. Similarly, ${\rm Aut}(\mathfrak{H})$ is the automorphism group
of the Heun equation~${\rm (\mathfrak{H})}$.
\end{definition}
${\rm Aut}({\mathfrak{h}})$ acts on the 3-dimensional parameter space
of~${\rm (\mathfrak{h})}$. It~contains the group~$S_3$ of permutations of
singular points as a subgroup, and the group $({\mathbb Z}_2)^2$ of
F-homotopic transformations as a normal subgroup. So ${\rm
Aut}({\mathfrak{h}})\simeq ({\mathbb Z}_2)^2 \ltimes S_3$, a~semidirect
product.
\begin{definition}
Within ${\rm Aut}(\mathfrak h)$, the M\"obius automorphism group is the
subgroup ${\cal M}(\mathfrak{h})\defeq\{1\}\times S_3$, which permutes the
singular points $z=0,1,\infty$. The linear automorphism group is ${\cal
L}(\mathfrak{h})\defeq\{1\}\times S_2$, which permutes the\/ {\em finite\/}
singular points $z=0,1$, and fixes~$z=\infty$. The F-homotopic
automorphism group is $({\mathbb Z}_2)^2\times\{1\}$.
\end{definition}
The action of ${\rm Aut}(\mathfrak h)$ on the $2\times3=6$ local solutions
is as~follows. $\left|{\rm Aut}(\mathfrak h)\right|=2^2\times3!=24$, and
applying the transformations in~${\rm Aut}(\mathfrak h)$ to any single
local solution yields $24$~solutions of~($\mathfrak{h}$). Applying them
to~${}_2F_1$, for instance, yields the $24$~well-known series solutions of
Kummer. However, the $24$~solutions split into six sets of four, since for
each singular point $z_0\in\{0,1,\infty\}$ there is a subgroup of~${\rm
Aut}(\mathfrak{h})$ of order~$2^1\times2!=4$, each element of~which
fixes~$z=z_0$ and performs no~F-homotopy there; so it leaves each local
solution at~$z=z_0$ invariant.
For example, the four transformations in the subgroup associated to~$z=0$
yield four equivalent expressions for ${}_2F_1(a,b;c;z)$; one of which is
${}_2F_1(a,b;c;z)$ itself, and another of which appears above
in~(\ref{eq:flip}). The remaining
%remaining equivalent
expressions for~${}_2F_1(a,b;c;z)$ are
$(1-z)^{-a}{}_2F_1\left(a,c-b;c;z/(z-1)\right)$ and
$(1-z)^{-b}{}_2F_1\left(b,c-a;c;z/(z-1)\right)$. The five additional sets
of four are expressions for the five additional local solutions. One which
will play a role is the local solution at~$z=0$ belonging to the exponent
$1-c$. One of the four expressions for~it, in~terms of~${}_2F_1$,
is~\cite{Erdelyi53}
\begin{equation}
\label{eq:tilde1}
\widetilde{{}_2F_1}(a,\,b;\,c;\,z)\defeq z^{1-c}{}_2F_1(a-c+1,\,b-c+1;\,2-c;\,z).
\end{equation}
The quantity $\widetilde{{}_2F_1}(a,b;c;z)$~is defined if
$c\neq2,3,4,\ldots$\ \ The second local solution at~$z=0$ must be specified
differently if $c=2,3,4,\ldots$\ \ (See~\cite{Abramowitz65}, \S15.5.) The
same is true if~$c=1$, since in that case, $\widetilde{{}_2F_1}$ reduces
to~${{}_2F_1}$. When $\widetilde{{}_2F_1}$ is defined, we define it
uniquely in a neighborhood of~$z=0$ by choosing the principal branch
of~$z^{1-c}$.
The automorphism group of the Heun equation is slightly more complicated to
describe. There are $4!$~M\"obius transformations~$M$ that map the
singular points $t=0,1,d,\infty$ onto $t=0,1,d',\infty$, for
some~$d'\in\mathbb{CP}^1\setminus\{0,1,\infty\}$. The possible~$d'$
constitute the cross-ratio orbit of~$\{0,1,d,\infty\}$. Of~these
$4!$~transformations, $3!$~fix $t=\infty$, i.e., are linear. All
values~$d'$ on the orbit are obtained via linear transformations, i.e., a
mapping is possible iff $\triangle01d$ is similar to~$\triangle01d'$.
If~$M$ is not linear, it must be followed by an F-homotopic transformation
of the form $\tilde u(t)=[t-M(\infty)]^{-\alpha} u(t)$ or $\tilde
u(t)=[t-M(\infty)]^{-\beta} u(t)$.
${\rm Aut}({\mathfrak{H}})$ acts on the 6-dimensional parameter space
of~${\rm (\mathfrak{H})}$. It~contains the group~$S_4$ of permutations of
singular points as a subgroup, and the group $({\mathbb Z}_2)^3$ of
F-homotopic transformations as a normal subgroup. So ${\rm
Aut}({\mathfrak{H}})\simeq ({\mathbb Z}_2)^3 \ltimes S_4$, a~semidirect
product.
\begin{definition}
Within ${\rm Aut}(\mathfrak H)$, the M\"obius automorphism group is the
subgroup ${\cal M}(\mathfrak{H})\defeq\{1\}\times S_4$, which maps between
sets of singular points of the form $\{0,1,d',\infty\}$. The linear
automorphism group is ${\cal L}(\mathfrak{H})\defeq\{1\}\times S_3$, which
maps between sets of\/ {\em finite\/} singular points of the form
$\{0,1,d'\}$, and fixes~$t=\infty$. The F-homotopic automorphism group is
$({\mathbb Z}_2)^3\times\{1\}$.
\end{definition}
The action of ${\rm Aut}(\mathfrak H)$ on the $2\times4=8$ local solutions
is as~follows. $\left|{\rm Aut}(\mathfrak H)\right|=2^3\times4!=192$, and
applying the transformations in~${\rm Aut}(\mathfrak H)$ to any single
local solution yields $192$~solutions of~($\mathfrak{H}$). However, the
$192$~solutions split into eight sets of~$24$, since for each singular
point $t_0\in\{0,1,d,\infty\}$ there is a subgroup of~${\rm
Aut}(\mathfrak{H})$ of order~$2^2\times3!=24$, each element of~which
fixes~$t=t_0$ and performs no~F-homotopy there; so it leaves each local
solution at~$t=t_0$ invariant. This statement must be interpreted with
care: in~the event that $t_0=d$, what is really being chosen is a
cross-ratio orbit, rather than a single point.
The case $t_0=0$ should serve as an example. The $24$~transformations in
the subgroup associated to~$t=0$ yield $24$~equivalent expressions for
$\Hl(d,q;\alpha,\beta,\gamma,\delta;t)$, one of which, the~only nontrivial
one with no F-homotopic factor, is~\cite{Ronveaux95,Snow52}
\begin{equation}
\label{eq:newguy1}
\quad\Hl\left(d,\,q;\,\alpha,\,\beta,\,\gamma,\,\delta;\,t\right)
=\Hl\left(1/d,\,q/d;\,\alpha,\,\beta,\,\gamma,\,\alpha+\beta-\gamma-\delta+1;\,t/d\right).
\end{equation}
(The two sides are defined if $\gamma$ is not a nonpositive integer.) The
additional seven sets of~24 are expressions for the additional seven local
solutions. One which will play a role is the solution at~$t=0$ belonging
to the exponent $1-\gamma$. One of the $24$~expressions for~it, in~terms
of~$\Hl$, is~\cite{Snow52}
\begin{equation}
\label{eq:newguy2}
\quad\widetilde\Hl(d,\,q;\,\alpha,\,\beta,\,\gamma,\,\delta;\,t)\defeq
t^{1-\gamma}
\Hl(d,\,q';\,\alpha-\gamma+1,\,\beta-\gamma+1,\,2-\gamma,\,\delta;\,t),
\end{equation}
where the transformed accessory parameter~$q'$ equals
$q+(1-\gamma)(\epsilon+d\delta)$. The quantity
$\widetilde{\Hl}(d,q;\alpha,\beta,\gamma,\delta;t)$ is defined if
$\gamma\neq2,3,4,\ldots$\ \ The second local solution at~$t=0$ must be
specified differently if $\gamma=2,3,4,\ldots$\ \ The same is true
if~$\gamma=1$, since in that case, $\widetilde{\Hl}$ reduces to~$\Hl$.
When $\widetilde\Hl$ is defined, we define it uniquely in a neighborhood
of~$t=0$ by choosing the principal branch of~$t^{1-\gamma}$.
In general, automorphisms of~${\rm (\mathfrak{H})}$ will alter not merely
$d$~and the exponent parameters, but also the accessory parameter~$q$.
This is illustrated by (\ref{eq:newguy1}) and~(\ref{eq:newguy2}).
\section{Polynomial Heun-to-Hypergeometric Transformations}
\label{sec:main}
We can now state, and prove, our corrected and expanded version of the
theorem of Ref.~\cite{Kuiken79}.
The theorem will characterize when a homomorphism of rational substitution
type from the Heun equation~($\mathfrak{H}$) to the hypergeometric
equation~($\mathfrak{h}$) exists. It~will list the possible substitutions,
up~to linear automorphisms of the two equations. It~is really a statement
about which ${\cal L}(\mathfrak{H})$-orbits may be mapped by homomorphisms
of this type to ${\cal L}(\mathfrak{h})$-orbits. The possible
substitutions, it turns~out, are all polynomial.
For ease of understanding, the characterization of the theorem will be
concrete: it~will require that the triangle $\triangle01d$ be similar to
one of a set of triangles of the form $\triangle01D$. Similarity occurs
iff $d$~belongs to the cross-ratio orbit of~$D$, i.e., iff $d$~can be
generated from~$D$ by repeated application of $d\mapsto 1-d$ and
$d\mapsto1/d$. It~is worth noting that if ${\rm Re}\, D=1/2$, the orbit
of~$D$ is closed under complex conjugation.
For each value of~$D$, the polynomial mapping from $t\in\mathbb{CP}^1$ to
$z\in\mathbb{CP}^1$, which we denote~$R$, will be given explicitly
when~$d=D$. When $d\neq D$, the mapping can be computed by composing with
the unique linear transformation $L_1$ of~$\mathbb{C}$ that maps
$\triangle01d$ to~$\triangle01D$. When $d\neq D$, statements about
necessary conditions regarding the singular points, characteristic
exponents, and accessory parameter must also be modified. For example,
case~1 of the theorem refers to a distinguished singular point~$d_0$, the
mandatory value of which is given when~$d=D$. When $d\neq D$, its
mandatory value can be computed as the preimage of that point under~$L_1$.
Similarly, a statement like ``the characteristic exponents of $t=0$ are
required to be~$0,1/2$'', valid when $d=D$, would be interpreted when
$d\neq D$ as ``the characteristic exponents of the preimage of $t=0$
under~$L_1$ are required to be~$0,1/2$''.
Note that in the statement of the theorem and what follows, $S\defeq1-R$.
\smallskip
\begin{theorem}
\label{thm:main}
A Heun equation of the form~{\rm ($\mathfrak{H}$)}, which has four singular
points and is nontrivial (i.e., $\alpha\beta\neq0$ or~$q\neq0$), can be
transformed to a hypergeometric equation of the form~{\rm ($\mathfrak{h}$)}
by a rational substitution $z=R(t)$ if and only~if $R$~is a polynomial,
$\alpha\beta\neq0$, and one of the following two conditions is satisfied.
\begin{remunerate}
\item $\triangle01d$ is similar to $\triangle01D$, for one of the
values of~$D$ listed in subcases {\rm 1(a)}--{\rm 1(c)}; each of which is
real, so the triangle must be degenerate. Also, the normalized accessory
parameter $q/\alpha\beta$ must equal one of\/ $0,1,d$, which may be
denoted~$d_0$. The description of each subcase lists the value of~$d_0$
when~$d=D$.
\item $\triangle01d$ is similar to $\triangle01D$, for one of the
values of~$D$ listed in subcases {\rm 2(a)}--{\rm 2(d)}; each of which is
non-real and has real part equal to~$1/2$, so the triangle must be
isosceles. The description of each subcase lists the value
of~$q/\alpha\beta$ when $d=D$.
\end{remunerate}
Besides specifying $D$ and the value of~$q/\alpha\beta$ when $d=D$, each
subcase imposes restrictions on the characteristic exponent parameters at
the singular points\/ $0,1,d$. The subcases of case~{\rm 1} are the
following.
{\rm 1(a).} [Harmonic (equally spaced collinear points) case.] $D=2$.
Suppose $d=D$. Then $d_0$~must equal\/~$1$, and $t=0,d$ must have the
same characteristic exponents, i.e., $\gamma=\epsilon$. In~general, either
$R$ or~$S$ will be the degree-2 polynomial $t(2-t)$, which maps $t=0,d$
to~$z=0$ and $t=1$ to~$z=1$ (with double multiplicity). There are special
circumstances in which $R$~may be quartic, which are listed separately, as
subcase~{\rm 1(c)}.
{\rm 1(b).} $D=4$. Suppose $d=D$. Then $d_0$~must equal\/~$1$, the
point $t=1$ must have characteristic exponents that are double those
of~$t=d$, i.e., $1-\delta=2(1-\epsilon)$, and $t=0$ must have exponents
$0,1/2$, i.e., $\gamma=1/2$. Either $R$ or~$S$ will be the degree-3
polynomial $(t-1)^2(1-t/4)$, which maps $t=0$ to~$z=1$ and $t=1,d$ to~$z=0$
(the former with double multiplicity).
{\rm 1(c).} [Special harmonic case.] $D=2$. Suppose $d=D$. Then
$d_0$~must equal~$1$, and $t=0,d$ must have the same characteristic
exponents, i.e., $\gamma=\epsilon$. Moreover, the exponents of $t=1$ must
be twice those of $t=0,d$, i.e., $1-\delta=2(1-\gamma)=2(1-\epsilon)$.
Either $R$ or~$S$ will be the degree-4 polynomial $4[t(2-t)-\frac12]^2$,
which maps $t=0,1,d$ to~$z=1$ ($t=1$~with double multiplicity).
The subcases of case~{\rm 2} are the following.
{\rm 2(a).} [Equianharmonic (equilateral triangle) case.] $D=\frac12+
i\sqrt{3}/2$. $q/\alpha\beta$~must equal the mean of\/~$0,1,d$, and
$t=0,1,d$ must have the same characteristic exponents, i.e.,
$\gamma=\delta=\epsilon$. Suppose $d=D$. Then $q/\alpha\beta$ must
equal $\frac12+i\sqrt3/6$. In~general, either $R$ or~$S$ will be the
degree-3 polynomial $\left[1-t/(\frac12+i\sqrt3/6)\right]^3$, which maps
$t=0,1,d$ to~$z=1$ and $t=q/\alpha\beta$ to~$z=0$ (with triple
multiplicity). There are special circumstances in which $R$~may be sextic,
which are listed separately, as subcase~{\rm 2(d)}.
{\rm 2(b).} $D=\frac12+i5\sqrt2/4$. Suppose $d=D$. Then
$q/\alpha\beta$ must equal $\frac12+i\sqrt2/4$, $t=d$ must have
characteristic exponents $0,1/3$, i.e., $\epsilon=2/3$, and $t=0,1$ must
have exponents $0,1/2$, i.e., $\gamma=\delta=1/2$. Either $R$ or~$S$ will
be the degree-4 polynomial
$\left[1-t/(\frac12+i5\sqrt2/4)\right]\left[1-t/(\frac12+i\sqrt2/4)\right]^3$,
which maps $t=d,q/\alpha\beta$ to~$z=0$ (the latter with triple
multiplicity) and $t=0,1$ to~$z=1$.
{\rm 2(c).} $D=\frac12+i11\sqrt{15}/90$. Suppose $d=D$. Then
$q/\alpha\beta$ must equal $\frac12+i\sqrt{15}/18$, $t=d$ must have
characteristic exponents $0,1/2$, i.e., $\epsilon=1/2$, and $t=0,1$ must
have exponents $0,1/3$, i.e., $\gamma=\delta=2/3$. Either $R$ or~$S$ will
be the degree-5 polynomial
$At(t-1)\left[t-(\frac12+i\sqrt{15}/18)\right]^3$, which maps
$t=0,1,q/\alpha\beta$ to~$z=0$ (the last with triple multiplicity). The
factor $A$ is chosen so that it maps $t=d$ to~$z=1$, as~well; explicitly,
$A=-i2025\sqrt{15}/64$.
{\rm 2(d).} [Special equianharmonic case.] $D=\frac12+ i\sqrt{3}/2$.
$q/\alpha\beta$~must equal the mean of\/~$0,1,d$, and $t=0,1,d$ must have
characteristic exponents $0,1/3$, i.e., $\gamma=\delta=\epsilon=2/3$.
Suppose $d=D$. Then $q/\alpha\beta$ must equal $\frac12+i\sqrt3/6$.
Either $R$ or~$S$ will be the degree-6 polynomial
$4\left\{[1-t/(\frac12+i\sqrt3/6)]^3-\frac12\right\}^2$, which maps
$t=0,1,d,q/\alpha\beta$ to~$z=1$ (the last with triple multiplicity).
\end{theorem}
{\em Remark\/} 1. The origin of the special harmonic and equianharmonic
subcases is easy to understand. In subcase~1(c), $t\mapsto R(t)$ or~$S(t)$
is the composition of the quadratic map of~1(a) with the map
$z\mapsto4(z-\frac12)^2$. In subcase~2(d), $t\mapsto R(t)$ or~$S(t)$ is
similarly the composition of the cubic map of~2(a) with
$z\mapsto4(z-\frac12)^2$. The extra restrictions on exponents make
feasible the additional quadratic transformation of~$z$, which is a
classical transformation of the hypergeometric equation into
itself~\cite{Erdelyi53}.
{\em Remark\/} 2. $R$ is determined uniquely by the choices enumerated in
the theorem. There is a choice of subcase, a choice of~$d$ from the
cross-ratio orbit of~$D$, and a binary choice between $R$ and~$S$. The
final two choices amount to choosing maps $L_1\in{\cal L}(\mathfrak{H})$
and~$L_2\in{\cal L}(\mathfrak{h})$, i.e., $L_2(z)=z$ or~$1-z$, which
precede and follow a canonical substitution.
In~the harmonic case~1(a), in which the ${\cal L}(\mathfrak{H})$-orbit
includes three values of~$d$, there are accordingly $3\times2=6$
possibilities for~$R$; namely,
\begin{equation}
R=t^2,\,1-t^2;\quad (2t-1)^2,\,1-(2t-1)^2;\quad t(2-t),\,1-t(2-t),
\end{equation}
corresponding to $d=-1,-1;1/2,1/2;2,2$, respectively. These are the
quadratic transformations of Kuiken~\cite{Kuiken79}. In~the equianharmonic
case~2(a), in which the orbit includes only two values of~$d$, there are
$2\times2=4$ possibilities; namely,
\begin{equation}
R=[1-t/(\tfrac12\pm i\sqrt3/6)]^3,
\qquad
1-[1-t/(\tfrac12\pm i\sqrt3/6)]^3,
\end{equation}
corresponding to $d=\frac12\pm i\sqrt{3}/2$. The remaining subcases, with
the exception of 1(c) and~2(d), correspond to generic cross-ratio orbits:
each value of~$D$ specifies six values of~$d$. In~each of those subcases,
there are $6\times2=12$ possibilities. So in~all, there are
56~possibilities for~$R$.
{\em Remark\/} 3. The characteristic exponents of the singular points
$z=0,1,\infty$ follow from those of the singular points $t=0,1,d,\infty$,
together with the formula for~$R$. The computation relies on
Proposition~\ref{thm:basicprop} below, which may be summarized thus. If
$t=t_0$ is not a critical point of the map $t\mapsto z=R(t)$, then the
exponents of $z=R(t_0)$ are the same as those of~$t_0$. If, on the other
hand, $t=t_0$~is mapped to $z=z_0\defeq R(t_0)$ with multiplicity~$k>1$,
i.e., $t=t_0$ is a $k-1$-fold critical point of~$R$ and $z=z_0$ is a
critical value, then the exponents of $z_0$ are $1/k$~times those of~$t_0$.
For example, in the harmonic case~1(a), the map $t\mapsto z$ coalesces two
of $t=0,1,d$ to either $z=0$ or~$z=1$, and by examination, the coalesced
point is not a critical value of the map; so the characteristic exponents
of those two points are preserved, and must therefore be the same, as
stated in the theorem. On~the other hand, the characteristic exponents of
the third point of the three, $t=d_0$, are necessarily halved when it is
mapped to $z=1$ or~$z=0$, since by examination, $R$~always has a simple
critical point at~$t=d_0$, i.e., $z\sim{\rm const} + C(t-d_0)^2$ for some
nonzero~$C$. (These statements follow by considering the canonical $d=D$
case.) So if $\delta_0$~denotes the parameter (out~of
$\gamma,\delta,\epsilon$) corresponding to~$t=d_0$, the characteristic
exponents of $z=1$ or~$z=0$ will be $0,(1-\delta_0)/2$. $R$,~being a
quadratic polynomial, also has a simple critical point at~$t=\infty$, so
the characteristic exponents of~$z=\infty$ are one-half those
of~$t=\infty$, i.e., $\alpha/2,\beta/2$. It~follows that in the harmonic
case, the Gauss parameters $(a,b;c)$ of the resulting hypergeometric
equation will be $(\alpha/2,\beta/2;(\delta_0+1)/2)$ or
$(\alpha/2,\beta/2;(\alpha+\beta-\delta_0+1)/2)$.
In~the equianharmonic case~2(a), the map $t\mapsto z$ coalesces $t=0,1,d$
to either $z=0$ or~$z=1$; and by examination, the coalesced point is not a
critical value of the map; so the characteristic exponents of those three
points are preserved, and must therefore be the same, as stated in the
theorem. On~the other hand, at~$t=q/\alpha\beta$, which is mapped to $z=1$
or~$z=0$, $R$~has, by examination, a double critical point, i.e.,
$z\sim{\rm const}+ C(t-q/\alpha\beta)^3$ for some nonzero~$C$. So~the
characteristic exponents of $z=1$ or~$z=0$, since $t=q/\alpha\beta$ is an
ordinary point of the Heun equation and effectively has characteristic
exponents~$0,1$, are~$0,1/3$. $R$,~being a cubic polynomial, also has a
double critical point at~$t=\infty$, so the characteristic exponents
of~$z=\infty$ are one-third those of~$t=\infty$, i.e., $\alpha/3,\beta/3$.
It~follows that in the equianharmonic case, the parameters $(a,b;c)$ of the
resulting hypergeometric equation will be $(\alpha/3,\beta/3;2/3)$ or
$\left(\alpha/3,\beta/3;(\alpha+\beta+1)/3\right)$.
\begin{definition}
A rational map $R:\mathbb{CP}^1\to\mathbb{CP}^1$ is said to map the
characteristic exponents of the Heun equation~{\rm ($\mathfrak{H}$)} to the
characteristic exponents of the hypergeometric equation~{\rm
($\mathfrak{h}$)} if, for~all $t_0\in\mathbb{CP}^1$, the exponents of
$t=t_0$ according to the Heun equation, divided by the multiplicity of
$t_0\mapsto z_0\defeq R(t_0)$, equal the exponents of~$z=z_0$ according to
the hypergeometric equation.
\end{definition}
For example, if $t_0$ and~$z_0$ are both finite, this says that if $z\sim
z_0+C(t-t_0)^k$ to leading order, for some nonzero~$C$, then the exponents
of~$z=z_0$ must be those of~$t=t_0$, divided by~$k$. If~$t=t_0$ is an
ordinary point of the Heun equation, then the exponents of~$z=z_0$ will
be~$0,1/k$. This implies that if~$k>1$, $z=z_0$ must be a singular point
of the hypergeometric equation, rather than an ordinary point.
\begin{proposition}
\label{thm:basicprop}
A Heun equation of the form~{\rm ($\mathfrak{H}$)} will be transformed to a
hypergeometric equation of the form~{\rm ($\mathfrak{h}$)} by a specified
rational substitution $z=R(t)$ of its independent variable only~if $R$~maps
exponents to exponents.
\end{proposition}
Proposition~\ref{thm:basicprop}, which was already used in Remark~3 above,
is an immediate consequence of the following lemma, which has an elementary
proof. But the proposition is best seen as a special case of a basic fact
in the theory of the Riemann $P$-function: if~a rational change of the
independent variable transforms one Fuchsian equation on the Riemann sphere
to another, then the characteristic exponents are transformed
multiplicatively. In~that context, it~would be proved by examining the
effects of the change of independent variable on each local (Frobenius)
solution.
The lemma begins the study of {\em sufficient\/} conditions for the
existence of a Heun-to-hypergeometric transformation. Finding them
requires care, since an accessory parameter is involved. It~is useful to
perform the substitution $z=R(t)$ explicitly. Substituting $z=R(t)$ into
the hypergeometric equation~{\rm ($\mathfrak{h}$)} transforms~it to
\begin{equation}
\label{eq:substituted}
\frac{d^2 y}{dt^2} + \left\{
-\frac{\ddot R}{\dot R} + \frac{\dot R}{R(1-R)}
[c-(a+b+1)R]\right\}
\frac{dy}{dt}
- \frac{ab\dot R^2}{R(1-R)}\,y = 0.
\end{equation}
\begin{lemma}
\label{thm:newlemma}
The coefficient of the $dy/dt$ term in the transformed hypergeometric
equation~(\ref{eq:substituted}), which may be denoted~$W(t)$, equals the
coefficient of the $du/dt$ term in the Heun equation~{\rm ($\mathfrak{H}$)}, i.e.,
$\gamma/t+\delta/(t-1)+\epsilon/(t-d)$, if~and only~if $R$~maps exponents
to exponents. That~is, the transformation at~least partly `works' if~and
only~if $R$~maps exponents to exponents.
\end{lemma}
{\em Proof.} Suppose $R$~maps $t=t_0$ to $z=z_0\defeq R(t_0)$ with
multiplicity~$k$, i.e., to~leading order $R(t)\sim z_0+C(t-t_0)^k$; if
$t_0$ and~$z_0$ are finite, that~is. By~direct computation, the leading
behavior of~$W$ at~$t=t_0$ is the following. In~the case when $t_0$~is
finite, $W(t)\sim (1-k)(t-t_0)^{-1}$ if $z_0\neq0,1,\infty$;
$[1-k(1-c)](t-t_0)^{-1}$ if $z_0=0$; $[1-k(c-a-b)](t-t_0)^{-1}$ if $z_0=1$,
and $[1-k(a+b)](t-t_0)^{-1}$ if $z_0=\infty$. In~the case when
$t_0=\infty$, $W(t)\sim (1+k)t^{-1}$ if $z_0\neq0,1,\infty$;
$[1+k(1-c)]t^{-1}$ if $z_0=0$; $[1+k(c-a-b)]t^{-1}$ if $z_0=1$, and
$[1+k(a+b)]t^{-1}$ if $z_0=\infty$.
This may be restated as~follows. At $t=t_0$, for finite~$t_0$, the leading
behavior of~$W$ is $W(t)\sim(1-k\eta)(t-t_0)^{-1}$, where $k$~is the
multiplicity of $t_0\mapsto z_0\defeq R(t_0)$ and $\eta$~is the sum of the
two characteristic exponents of the hypergeometric equation at~$z=z_0$;
if~the coefficient $1-k\eta$ equals zero then $W$~has no~pole at~$t=t_0$.
Likewise, the leading behavior of~$W$ at~$t=\infty$ is
$W(t)\sim(1+k\eta)t^{-1}$, where $k$~is the multiplicity of $\infty\mapsto
z_0\defeq R(\infty)$ and $\eta$ is the sum of the two exponents at~$z=z_0$;
if~the coefficient $1+k\eta$ equals zero then $W$~has a higher-order zero
at~$t=\infty$.
By the definition of `mapping exponents to exponents', it follows that the
leading behavior of~$W$ at $t=t_0$, for all $t_0$~finite, is of the form
$W(t)\sim(1-\eta')(t-t_0)^{-1}$, and also at $t_0=\infty$, is of the form
$W(t)\sim(1+\eta')t^{-1}$, where in both cases $\eta'$~is the sum of the
exponents of the Heun equation at~$t=t_0$, iff $R$~maps exponents to
exponents.
That is, the rational function~$W$ has leading behavior $\gamma t^{-1}$
at~$t=0$, $\delta(t-1)^{-1}$ at~$t=1$, $\epsilon(t-d)^{-1}$ at~$t=d$,
$(1+\alpha+\beta)t^{-1}=(\gamma+\delta+\epsilon)t^{-1}$ at~$t=\infty$, and
is regular at all~$t$ other than $0,1,d,\infty$, iff $R$~maps exponents to
exponents. \qquad \endproof
The following propositions characterize when the transformed hypergeometric
equation~(\ref{eq:substituted}) is, in~fact, the Heun
equation~{\rm ($\mathfrak{H}$)}.
\begin{proposition}
\label{thm:trivialprop}
A Heun equation of the form~{\rm ($\mathfrak{H}$)}, which is trivial (i.e.,
$\alpha\beta=0$ and~$q=0$) will be transformed to a hypergeometric equation
of the form~{\rm ($\mathfrak{h}$)} by a specified rational substitution $z=R(t)$
of its independent variable if and only~if $R$~maps exponents to exponents.
\end{proposition}
{\em Proof.} The `if' half is new, and requires proof. By
Lemma~\ref{thm:newlemma}, the coefficients of $dy/dt$ and $du/dt$ agree iff
$R$~maps exponents to exponents, so it suffices to determine whether the
coefficients of $y$ and~$u$ agree. But by triviality, the coefficient
of~$u$ in~{\rm ($\mathfrak{H}$)} is zero. Also, $t=\infty$ has zero as one
of its exponents, so all points $t\in\mathbb{CP}^1$ have zero as an
exponent. By~the mapping of exponents to exponents, $z=\infty$ must also
have zero as an exponent, i.e., $ab=0$. So~the coefficient of~$y$
in~(\ref{eq:substituted}) is also zero. \qquad\endproof
\begin{proposition}
\label{thm:mainproposition}
A Heun equation of the form~{\rm ($\mathfrak{H}$)}, which has four singular
points and is nontrivial (i.e., $\alpha\beta\neq0$ or~$q\neq0$), will be
transformed to a hypergeometric equation of the form~{\rm ($\mathfrak{h}$)}
by a specified rational substitution $z=R(t)$ of its independent variable
if~and only~if $R$~maps exponents to exponents, and moreover, $R$~is a
polynomial, $\alpha\beta\neq0$, and one of the following two conditions on
the normalized accessory parameter $p\equiv q/\alpha\beta$ is satisfied.
\begin{remunerate}
\item $p$~equals one of\/ $0,1,d$. Call this point~$d_0$, and the other
two singular points $d_1$~and~$d_2$. In this case, $d_0$~must be a double
zero of $R$ or~$S$, and each of $d_1,d_2$ must be a simple zero of $R$
or~$S$.
\item $p$~does not equal any of\/ $0,1,d$. In this case, each of\/ $0,1,d$
must be a simple zero of $R$ or~$S$, and $p$~must be a triple zero of
either $R$ or~$S$.
\end{remunerate}
In both cases, $R$~and~$S$ can have no~additional simple zeroes or zeroes
of order greater than two. Also, if\/ $1-c$ (the nonzero exponent at
$z=0$) does not equal~$1/2$, then $R$~can have no~additional double zeroes;
and if $c-a-b$ (the nonzero exponent at~$z=1$) does not equal~$1/2$, then
$S$~can have no~additional double zeroes.
In both cases, no~additional double zero, if~any, can be mapped by~$R$ to
the point (out~of $z=0,1$) to which $p$~is mapped. So additional double
zeroes, if~any, must all be zeroes of~$R$, or all be zeroes of~$S$.
\end{proposition}
{\em Proof.} Like Proposition~\ref{thm:trivialprop}, this follows by
comparing the transformed hypergeometric equation~(\ref{eq:substituted}) to
the Heun equation~{\rm ($\mathfrak{H}$)}. By Lemma~\ref{thm:newlemma}, the
coefficients of $dy/dt$ and $du/dt$ agree iff $R$~maps exponents to
exponents, so it suffices to characterize when the coefficients of $y$
and~$u$ agree.
The coefficient of~$y$ in~(\ref{eq:substituted}) is to equal the
coefficient of~$u$ in~{\rm ($\mathfrak{H}$)}. It~follows that $ab=0$ is
possible iff $\alpha\beta=0$ and~$q=0$, which is ruled~out by
nontriviality. So $ab\neq0$, and equality of the coefficients can hold iff
\begin{equation}
\label{eq:cond}
U\equiv\frac{\dot R}{R}\,\frac{\dot S}{S} =
\frac{-(\alpha\beta t - q)/ab}{t(t-1)(t-d)}
\equiv\frac{C_0}{t} + \frac{C_1}{t-1} + \frac{C_d}{t-d},
\end{equation}
where $S\equiv1-R$, and $C_0,C_1,C_d$ are certain complex numbers, at~least
two of~which are nonzero.
Both $\dot R/R$ and~$\dot S/S$ are sums of terms of the form
$n(t-\lambda)^{-1}$, where $n$~is a nonzero integer and $\lambda$~is a zero
or a pole of $R$ or~$S$. Poles are impossible, since $\lambda$~is a pole
of~$R$ iff $\lambda$~is a pole of~$S$, and there are no double poles on the
right-hand side of~(\ref{eq:cond}). So $R$ must be a polynomial.
By examining the definition of~$U$ in~terms of $R$ and~$S$, one sees the
following is true of any $\lambda\in\mathbb{C}$\null: if $R$ or~$S$ has a
simple zero at~$t=\lambda$, then $U$~will have a simple pole
at~$t=\lambda$; if $R$ or~$S$ has a double zero at~$t=\lambda$, then
$U$~will have an ordinary point (non-zero, non-pole) at~$t=\lambda$, and if
$R$ or~$S$ has a zero of order $k>2$ at~$t=\lambda$, then $U$~will have a
zero of order $k-2$ at~$t=\lambda$.
Most of what follows is devoted to proving the `only~if' half of the
proposition in the light of these facts, by examining the consequences of
the equality~(\ref{eq:cond}). In the final paragraph, the `if'~half will
be proved.
There are exactly three ways in which the equality~(\ref{eq:cond}) can
hold.
\begin{zeroremunerate}
\item $\alpha\beta=0$, but due to nontriviality, $q\neq0$. $U$~has
three simple poles on~$\mathbb{C}$, at $t=0,1,d$. It~has no other poles,
and no zeroes. So each of~$0,1,d$ must be a simple zero of either $R$
or~$S$; also, $R$~and~$S$ can have no~other simple zeroes, and no zeroes of
order~$k>2$. Except for possible double zeroes, the zeroes of $R$ and~$S$
are determined. The degree of~$R$ must equal the number of zeroes of~$R$,
and also equal the number of zeroes of~$S$, counting multiplicity. But
irrespective of how many double zeroes are assigned to $R$ or~$S$, either
$R$ or~$S$ will have an odd number of zeroes, and the other an even number,
counting multiplicity. So case~0 is absurd: necessarily
$\alpha\beta\neq0$.
\item $\alpha\beta\neq0$ and $\alpha\beta t-q$ is a nonzero multiple of
$t-d_0$, where $d_0=0$,~$1$, or~$d$, so that exactly one of $C_0,C_1,C_d$
is zero. $U$~has two simple poles on~$\mathbb{C}$, at $t=d_1,d_2$ (the~two
singular points other than~$d_0$); it~has no other poles, and no zeroes.
So each of~$d_1,d_2$ must be a simple zero of either $R$ or~$S$; also,
$R$~and~$S$ can have no~other simple zeroes, and no zeroes of order~$k>2$.
Since by assumption the Heun equation has four singular points, each of
$0,1,d$ must be a singular point, so the coefficient of $dy/dt$
in~(\ref{eq:substituted}) must have a pole at~$t=d_0$, which implies that
$R$ or~$S$ must have a zero at~$d_0$ of the only remaining type: a~double
zero.
\item $\alpha\beta\neq0$ but $\alpha\beta t-q$ is not a multiple of
$t$,~$t-1$, or~$t-d$, so that none of $C_0,C_1,C_d$ is zero. $U$~has three
poles on~$\mathbb{C}$, and exactly one zero, at $t=p\equiv q/\alpha\beta$,
which is simple. So each of~$0,1,d$ must be a simple zero of either $R$
or~$S$, and $q/\alpha\beta$~must be a triple zero of either $R$ or~$S$.
Also, $R$~and~$S$ can have no~other simple zeroes, and no other zeroes of
order~$k>2$.
\end{zeroremunerate}
In cases 1 and~2, what remain to be determined are the (additional) double
zeroes of $R$ and~$S$, if~any. That~is, it must be determined whether any
ordinary point of the Heun equation can be mapped to $z=0$ or~$z=1$ with
double multiplicity. But by Proposition~\ref{thm:basicprop}, $R$~can map
an ordinary point $t=t_0$ to~$z=0$ (resp.\ $z=1$) with double multiplicity
only~if the exponents of~$z=0$ (resp.\ $z=1$) are~$0,1/2$.
Suppose this occurs. In case~1, if the exponents of~$t=p=d_0$ are
denoted~$0,\gamma_0$, the exponents of~$R(p)$ will be~$0,\gamma_0/2$, since
$t=p$ will be mapped with double multiplicity to~$z=R(p)$. So if
$R(t_0)=R(p)$ then $\gamma_0$~must equal~$1$, which, since $q=\alpha\beta
d_0$, is ruled~out by the assumption that each of $0,1,d$, including~$d_0$,
is a genuine singular point. It~follows that in case~1, $R(t_0)\neq R(p)$.
A~related argument applies in case~2. In case~2, the point~$p$ is an
ordinary point of the Heun equation, and a double critical point of the
$t\mapsto z$ map. So as a singular point of the hypergeometric equation,
$R(p)$ must have exponents~$0,1/3$. It~follows that $R(t_0)=R(p)$ is
impossible.
The `only if' half of the proposition has now been proved; the `if' half
remains. Just as the equality~(\ref{eq:cond}) implies the stated
conditions on~$R$, so the stated conditions must be shown to imply the
equality~(\ref{eq:cond}). But the conditions on~$R$ are equivalent to the
left-hand side and right-hand side having the same poles and zeroes, i.e.,
to their being the same up~to a constant factor. To~show the constant is
unity, it~is enough to consider the limit $t\to\infty$. If $\deg R=n$,
then $\dot R/R\sim n/t$ and $\dot S/S\sim -n/t$, so~$U$, i.e., the
left-hand side, has asymptotic behavior~$-n^2/t^2$. This will be the same
as that of the right-hand side if $(\alpha\beta)/(ab)=n^2$. But
$a=\alpha/n$ and $b=\beta/n$ follow from the assumption that $R$~maps
exponents to exponents. \qquad \endproof
{\em Proof (of Theorem~\ref{thm:main})\/}. By
Proposition~\ref{thm:mainproposition}, the preimages of $z=0,1$ under the
polynomial~$R$ must include $t=0,1,d$, and in case~2 of the proposition,
$t=p\equiv q/\alpha\beta$. They may also include $l$~(additional) double
zeroes of $R$ or of~$S$, which will be denoted $t=a_1,\ldots,a_l$. Cases
1~and~2 of the theorem correspond to cases 1~and~2 of the proposition, and
the subcases of the theorem correspond to distinct choices of~$l$.
Necessarily $\deg R=|R^{-1}(0)|=|R^{-1}(1)|$, where the inverse images are
defined as multisets rather than sets, to take multiplicity into account.
This places tight constraints on~$l$, since each of $0,1,d$ (and~$p$, in
case~2) may be assigned to either $R^{-1}(0)$ or~$R^{-1}(1)$, but by the
proposition, all of $a_1,\ldots,a_l$ must be assigned, twice, to one or the
other. In case~1, one of $0,1,d$ (denoted~$d_0$ in the proposition) has
multiplicity~$2$, and the other two (denoted~$d_1,d_2$) have
multiplicity~$1$. It~follows that $0\leq l\leq 2$, with $\deg R=l+2$. In
case~2, each of $0,1,d$ has multiplicity~$1$, and $p$~has multiplicity~$3$.
It~follows that $0\leq l\leq 3$, with $\deg R=l+3$. Subcases are
as~follows.
1(a). Case~1, $l=0$, $\deg R=2$. Necessarily $R^{-1}(0)$ and~$R^{-1}(1)$
are $\{d_0,d_0\}$ and $\{d_1,d_2\}$, or vice versa. Without loss of
generality (WLOG), assume the latter, and assume $d_0=1$. Then
$R^{-1}(0)=\{0,d\}$ and $R^{-1}(1)=\{1,1\}$, i.e., $S^{-1}(0)=\{1,1\}$ and
$S^{-1}(1)=\{0,d\}$. Since $t=1$ is a double zero of~$S$, $S(t)=C(t-1)^2$
for some~$C$. But $S(0)=1$, which implies $C=1$, and $S(d)=1$, which
implies~$d=2$. So $S(t)=(t-1)^2$ and $R(t)=t(2-t)$. Since $t=0,d$ are
both mapped singly to the singular point~$z=0$ by~$R$, their exponents must
be the same as those of~$z=0$, and hence must be identical.
1(b). Case~1, $l=1$, $\deg R=3$. Necessarily $R^{-1}(0)$ and~$R^{-1}(1)$
are $\{d_0,d_0,d_1\}$ and~$\{d_2,a_1,a_1\}$, or vice versa. WLOG, assume
the former, and also assume $d_0,d_1,d_2$ equal $1,d,0$, respectively.
Then $R^{-1}(0)=\{1,1,d\}$ and~$R^{-1}(1)=\{0,a_1,a_1\}$. It~follows that
$R(t)=(t-1)^2(1-t/d)$, where $d$~is determined by the condition that the
critical point of~$R$ other than~$t=1$ (i.e., $t=a_1$) be mapped to~$1$.
Solving $\dot R(t)=0$ for $t=a_1$ yields $a_1=(2d+1)/3$, and substitution
into $R(a_1)-1=0$ yields $d=4$ or~$-1/2$. But the latter is ruled~out by
the fact that it would imply $a_1=0$, which is impossible. So $d=4$ and
$a_1=3$. Since $t=1,d$ are mapped to the singular point~$z=0$, doubly and
singly respectively, the exponents of $t=1$ must be twice those of~$z=0$,
and the exponents of $t=d$ must be the same as those of~$z=0$.
1(c). Case~1, $l=2$, $\deg R=4$. Necessarily $R^{-1}(0)$ and~$R^{-1}(1)$
are $\{d_0,d_0,d_1,d_2\}$ and~$\{a_1,a_1,a_2,a_2\}$, or vice versa. WLOG,
assume the latter, and assume $d_0=1$. Then
$R^{-1}(0)=\{a_1,a_1,a_2,a_2\}$ and $R^{-1}(1)=\{0,1,1,d\}$, i.e.,
$S^{-1}(0)=\{0,1,1,d\}$ and $S^{-1}(1)=\{a_1,a_1,a_2,a_2\}$. So
$S(t)=At(t-1)^2(t-d)$, where $d$~is determined by the condition that
$S$~must have two critical points other than $t=1$, i.e., $t=a_1,a_2$,
which are mapped by~$S$ to the same critical value (in~fact, to~$z=1$).
Computation yields $\dot S=A(t-1)\left[4t^2-(3d+2)t+d\right]$, so $a_1,a_2$
must be the roots of $4t^2-(3d+2)t+d$. If~the corresponding critical
values are $Aw_1,Aw_2$, then $w_1,w_2$ are the roots of the polynomial
in~$w$ obtained by eliminating~$t$ between $w-S(t)/A$ and $4t^2-(3d+2)t+d$.
Its discriminant turns~out to be proportional to $(d-2)^2(9d^2-4d+4)^3$, so
the criterion for equal values is that $d=2$ or $9d^2-4d+4=0$. But the
latter can be ruled~out, since by examination it would result in $a_1,a_2$
being equal. So~$d=2$; $a_1,a_2=1\pm\sqrt2/2$; and $S(t)=At(t-1)^2(t-2)$
with $A=-4$, so that $S(a_i)=1$. Hence
$R(t)=4\left[t(2-t)-\frac12\right]^2$. Since $t=0,d$ are mapped simply
to~$z=1$ and $t=1$~is mapped doubly, the exponents of~$t=0,d$ must be the
same, and double those of~$t=1$.
2(a). Case~2, $l=0$, $\deg R=3$. Necessarily $R^{-1}(0)$ and~$R^{-1}(1)$
are $\{p,p,p\}$ and $\{0,1,d\}$, or vice versa. WLOG, assume the former.
Then $R(t)=A(t-p)^3$ for some~$A$. Since $t=0,1,d$ are to be mapped singly
to~$1$, they must be the vertices of an equilateral triangle of which
$p$~is the mean, so $A=-1/p^3$ and $R=(1-t/p)^3$. WLOG, assume
$d=\frac12+i\sqrt3/2$, in which case $p=\frac12+i\sqrt3/6$. The exponents
at $t=0,1,d$ must be equal to one another, since they all equal the
exponents at~$z=1$.
2(b). Case~2, $l=1$, $\deg R=4$. Assume WLOG that $R^{-1}(0)=\{p,p,p,d\}$
and $R^{-1}(1)=\{0,1,a_1,a_1\}$, i.e., $S^{-1}(0)=\{0,1,a_1,a_1\}$ and
$S^{-1}(1)=\{p,p,p,d\}$. It~follows that $R(t)=(1-t/d)(1-t/p)^3$, but to
determine $d$ and~$p$, it~is best to focus on~$S$. Necessarily
$S(t)=At(t-1)(t-a_1)^2$, and $p$~can be a triple zero of~$R$ iff it is a
double critical point of~$S$ as~well as~$R$. The condition that $S$~have a
double critical point determines~$a_1$. $\dot
S=A(t-a_1)\left[4t^2-(3+2a_1)t+a_1\right]$, so the polynomial
$4t^2-(3+2a_1)t+a_1$ must have a double root. Its discriminant is
$4a_1^2-4a_1+9$, which will equal zero iff $a_1=\frac12\pm i\sqrt{2}$. The
corresponding value of the double root, i.e., the mandatory value of~$p$,
is $\frac12\pm i\sqrt{2}/4$. The requirement that $S$~map $p$ to~$1$
implies~$A=1/p(p-1)(p-a_1)^2$. $d$~is determined as the root of~$R=1-S$
other than~$p$\/; some computation yields $\frac12\pm i5\sqrt2/4$. WLOG
the `$\pm$' in the expressions for $p$ and~$d$ can be replaced by~`$+$'.
Since $t=p$ is an ordinary point and $R$~maps $t=p$ triply to~$z=0$, $z=0$
must have exponents~$0,1/3$. Since $R$~maps $t=d$ simply to~$z=0$,
$t=d$~must also have exponents~$0,1/3$. Similarly, since $R$~maps the
ordinary point $t=a_1$ doubly to~$z=1$, $z=1$~must have exponents~$0,1/2$;
so $t=0$ and~$t=1$, which are mapped simply to~$z=1$, must also.
2(c). Case~2, $l=2$, $\deg R=5$. Assume WLOG that
$R^{-1}(0)=\{p,p,p,0,1\}$ and $R^{-1}(1)=\{d,a_1,a_1,a_2,a_2\}$. Then
$R(t)=At(t-1)(t-p)^3$, where $p$~is determined by $R$~having two critical
points other than $t=p$, i.e., $t=a_1,a_2$, which are mapped to the same
critical value (in~fact, to~$z=1$). $\dot
R(t)=A(t-p)^2\left[5t^2-(2p+4)t+p\right]$, so $a_1,a_2$ must be the roots
of $5t^2-(2p+4)t+p$. If~the corresponding critical values are $Aw_1,Aw_2$,
then $w_1,w_2$ are the roots of the polynomial in~$w$ obtained by
eliminating~$t$ between $w-R(t)/A$ and $5t^2-(2p+4)t+p$. Its discriminant
turns~out to be proportional to $(p^2-p+4)^3(27p^2-27p+8)^2$, so the
criterion for equal values is that $p^2-p+4=0$ or $27p^2-27p+8=0$. But the
former can be ruled~out, since by examination it would result in $a_1,a_2$
being equal. The latter is true iff $p=\frac12\pm i\sqrt{15}/18$. WLOG
the plus sign may be used. This yields $a_1,a_2=\frac12 \pm 2\sqrt3/9 +
i\sqrt{15}/90$. From the condition $R(a_i)=1$, it~follows that
$A=-i2025\sqrt{15}/64$. $d$~is determined as the root of $R(t)-1$ other
than~$a_1,a_2$; computation yields $d=\frac12+11\sqrt{15}/90$. Since $t=p$
is an ordinary point mapped triply to~$z=0$, $z=0$~must have
exponents~$0,1/3$. Similarly, since $R$~maps the ordinary points~$t=a_i$
to~$z=1$, $z=1$~must have exponents~$0,1/2$, so~$t=d$, which is mapped
singly to~it, must also.
2(d). Case~2, $l=3$, $\deg R=6$. Necessarily $R^{-1}(0)$ and~$R^{-1}(1)$
are $\{p,p,p,0,1,d\}$ and $\{a_1,a_1,a_2,a_2,a_3,a_3\}$, or vice versa.
WLOG, assume the latter. Then $R(t)=A(t-a_1)^2(t-a_2)^2(t-a_3)^2$ and
$S(t)=Bt(t-1)(t-d)(t-p)^3$. Since $t=p$ is a triple zero of~$S$,
$R(t)\sim1-C(t-p)^3$ for some nonzero~$C$. So~$\sqrt{R(t)}$, defined to
equal~$+1$ at~$t=p$, will have a similar Taylor series:
$\sqrt{A}(t-a_1)(t-a_2)(t-a_3)\sim1-C(t-p)^3/2$. This is possible only~if
$a_1,a_2,a_3$ are the vertices of an equilateral triangle, and $p$~is their
mean. It~follows that the roots of~$S$ other than~$t=p$, i.e., $t=0,1,d$,
are also the vertices of an equilateral triangle centered on~$p$. WLOG,
choose $d=\frac12+i\sqrt3/2$ and~$p=\frac12+i\sqrt3/6$. With a bit of
algebra, $R$~can be rewritten in the form given in the theorem. Since
$t=p$ is an ordinary point and $R$~maps it triply to~$z=0$, $z=0$ must have
exponents~$0,1/3$. Since $R$~maps $t=0,1,d$ simply to~$z=0$,
$t=0,1,d$~must also have exponents~$0,1/3$.
\qquad\endproof
\begin{corollary}
\label{thm:useful0}
Suppose a Heun equation has four singular points and is nontrivial
($\alpha\beta\neq0$ or~$q\neq0$). Then the only reductions of the
corresponding local Heun function~$\Hl$ to the hypergeometric
function~${}_2F_1$ which can be performed by a rational transformation of
the independent variable involve, in~fact, polynomial transformations of
degrees 2,~3, 4, 5, and~6. Up~to the choice of~$d$ from the possible
cross-ratio orbits, the following canonical reductions are the only such
reductions. Here $\alpha,\beta,\gamma$ are free parameters.
\begin{subequations}
\begin{eqnarray}
\label{eq:generalharmonic}
\qquad&&\Hl\left(2,\,\alpha\beta;\,\alpha,\,\beta,\,\gamma,\,\alpha+\beta-2\gamma+1;\,t\right)\\
\qquad&&\qquad={}_2F_1\left(\alpha/2,\,\beta/2;\,\gamma;\,t(2-t)\right),
\nonumber\\
\qquad&&\Hl\left(4,\,\alpha\beta;\,\alpha,\,\beta,\,\tfrac12,\,2(\alpha+\beta)/3;\,t\right)\\
\qquad&&\qquad={}_2F_1\left(\alpha/3,\,\beta/3;\,\tfrac12;\,1-(t-1)^2(1-t/4)\right),
\nonumber\\
\label{eq:specialharmonic}
\qquad&&\Hl\left(2,\,\alpha\beta;\,\alpha,\,\beta,\,(\alpha+\beta+2)/4,\,(\alpha+\beta)/2;\,t\right)\\
\qquad&&\qquad={}_2F_1\left(\alpha/4,\,\beta/4;\,(\alpha+\beta+2)/4;\,1-4\bigl[t(2-t)-\tfrac12\bigr]^2\right),
\nonumber\\
\label{eq:generalequianharmonic}
\qquad&&\Hl\left(\tfrac12+i\sqrt3/2,\,\alpha\beta(\tfrac12+i\sqrt3/6);\,\alpha,\,\beta,\,(\alpha+\beta+1)/3,\,(\alpha+\beta+1)/3;\,t\right)\\
\qquad&&\qquad={}_2F_1\left(\alpha/3,\,\beta/3;\,(\alpha+\beta+1)/3;\,1-\bigl[1-t/(\tfrac12+i\sqrt3/6)\bigr]^3\right),
\nonumber\\
\qquad&&\Hl\left(\tfrac12+i5\sqrt2/4,\,\alpha({\tfrac23}-\alpha)(\tfrac12+i\sqrt2/4);\,\alpha,\,{\tfrac23}-\alpha,\,\tfrac12,\,\tfrac12;\,t\right)\\
\qquad&&\qquad={}_2F_1\Bigl(\alpha/4,\,{\tfrac16}-\alpha/4;\,\tfrac12;\,
\nonumber\\[-1.0ex]
\qquad&&\qquad\qquad\qquad 1-\bigl[1-t/(\tfrac12+i5\sqrt2/4)\bigr]\bigl[1-t/(\tfrac12+i\sqrt2/4)\bigr]^3\Bigr),
\nonumber\\
\label{eq:neednorm}
\qquad&&\Hl\left(\tfrac12+i11\sqrt{15}/90,\,
\alpha({\tfrac56}-\alpha)(\tfrac12+i\sqrt{15}/18);\,
\alpha,\,{\tfrac56}-\alpha,\,{\tfrac23},\,{\tfrac23},\,t\right)\\
\qquad&&\qquad={}_2F_1\Bigl(\alpha/5,\,{\tfrac16}-\alpha/5;\,{\tfrac23};\,
\nonumber\\[-1.0ex]
\qquad&&\qquad\qquad\qquad (-i2025\sqrt{15}/64)\,t(t-1)\bigl[t-(\tfrac12+i\sqrt{15}/18)\bigr]^3\Bigr),
\nonumber\\
\label{eq:specialequianharmonic}
\qquad&&\Hl\left(\tfrac12+i\sqrt3/2,\,\alpha(1-\alpha)(\tfrac12+i\sqrt3/6);\,\alpha,\,1-\alpha,\,{\tfrac23},\,{\tfrac23},\,t\right)\\
\qquad&&\qquad={}_2F_1\left(\alpha/6,\,\tfrac16-\alpha/6;\,{\tfrac23};\,1-4\left\{\bigl[1-t/(\tfrac12+i\sqrt3/6)\bigr]^3-\tfrac12\right\}^2\right).
\nonumber
\end{eqnarray}
\end{subequations}
Each of these equalities holds in a neighborhood of $t=0$ whenever the two
sides are defined, e.g., whenever the fifth argument of~$\Hl$ and the third
argument of~${}_2F_1$ are not equal to a nonpositive integer.
\end{corollary}
{\em Remarks.} The equalities of Corollary~\ref{thm:useful0} hold even if
the Heun equation has fewer than four singular points, or is trivial; but
in either of those cases, additional reductions are possible. For the
trivial case, see~\S\ref{sec:trivial}.
The special harmonic reduction~(\ref{eq:specialharmonic}) is composite:
it~can be obtained from the special case $\gamma=(\alpha+\beta+2)/4$ of the
harmonic reduction~(\ref{eq:generalharmonic}) by applying the standard
quadratic hypergeometric transformation~\cite{Erdelyi53}
\begin{equation}
\label{eq:apply}
\quad{}_2F_1\bigl(a,\,b;\,(a+b+1)/2;\,z\bigr) = {}_2F_1\bigl(a/2,\,b/2;\,(a+b+1)/2;\,1-4(z-\tfrac12)^2\bigr)
\end{equation}
to its right-hand side. The special equianharmonic
reduction~(\ref{eq:specialequianharmonic}) can be obtained in the same way
from the case $\beta=1-\alpha$ of the equianharmonic
reduction~(\ref{eq:generalequianharmonic}).
One might think that (\ref{eq:apply})~could be applied to the right-hand
sides of the remaining reductions in
(\ref{eq:generalharmonic})--(\ref{eq:specialequianharmonic}), to generate
additional composite reduction formul\ae. However, there are only a few
cases in which it can be applied; and it is easily checked that when
it~can, it~imposes conditions on the parameters of~$\Hl$ which require that
the Heun equation of~which $\Hl$~is a solution have fewer than four
singular points.
{\em Proof.} $\Hl$ and~${}_2F_1$ are the local solutions of their
respective equations which belong to the exponent zero at~$t=0$ (resp.\
$z=0$), and are regular and normalized to unity there. So the corollary
follows from Theorem~\ref{thm:main}:
(\ref{eq:generalharmonic})--(\ref{eq:specialharmonic}) correspond to
subcases 1(a)--1(c), and
(\ref{eq:generalequianharmonic})--(\ref{eq:specialequianharmonic}) to
subcases 2(a)--2(d). In~each subcase, the Gauss parameters $(a,b;c)$
of~${}_2F_1$ are computed by first calculating the exponents at
$z=0,1,\infty$, in the way explained in Remark~3. In~some subcases, the
polynomial map supplied in Theorem~\ref{thm:main} must be chosen to be
$S$~rather than~$R$, due~to the need to map $t=0$ to $z=0$, not~$z=1$,
so~that the transformation will reduce~$\Hl$ to~${}_2F_1$, rather than to
another local solution of the hypergeometric equation. \qquad \endproof
In~all, there are $28=56/2$ reductions of $\Hl$ to~${}_2F_1$ of this
non-F-homotopic sort, rather than~$56$, since the $R$-vs.-$S$ choice
mentioned in Remark~2 does not apply, as noted in the proof. Equivalently,
for each subcase of Theorem~\ref{thm:main}, there is one reduction of this
sort for each of the possible values of~$d$.
Each reduction listed in Corollary~\ref{thm:useful0} corresponds to
choosing $d=D$. Any other~$d$ on the cross-ratio orbit of~$D$ may be
chosen, but the orbit is defined by $\triangle01d$ being one of the
triangles (at~most six) similar to~$\triangle01D$, i.e., being obtained
from~$\triangle01D$ by a linear transformation~$L_1\in{\cal
L}(\mathfrak{H})$. So for any corollary subcase and choice of~$d$, the
corresponding reduction is $z=L_2(R_1(L_1(t)))$, where $L_1$~is constrained
to map $\triangle01d$ to~$\triangle01D$, $R_1$~is the canonical
transformation that appears in the subcase, and $L_2\in{\cal
L}(\mathfrak{h})$, i.e., $L_2(z)=z$ or~$1-z$, must be chosen so that $t=0$
is mapped to~$z=0$.
For example, in~(\ref{eq:generalharmonic}), which corresponds to the
harmonic subcase~1(a), the canonical transformation is $R_1=t(2-t)$, and the
$d=-1$ reduction of~$\Hl$ to~${}_2F_1$ is
\begin{equation}
\label{eq:extrathing}
\quad\Hl\bigl(-1,\,0;\,\alpha,\,\beta,\,\gamma,\,(\alpha+\beta-\gamma+1)/2;\,t\bigr)
={}_2F_1\left(\alpha/2,\,\beta/2;\,(\gamma+1)/2;\,t^2\right).
\end{equation}
This reduction is obtained by choosing $L_1(t)=t+1$ and~$L_2(z)=1-z$.
\smallskip
In applications, it is seldom the case that the four regular singular
points of an equation of Heun type are located at $0,1,d,\infty$. But
Theorem~\ref{thm:main} and its corollary may readily be generalized.
Consider the canonical situation when three of the four have zero as a
characteristic exponent, since this may always be arranged by an
F-homotopic transformation. There are two situations of interest: either
the singular points include the point at infinity, and each of the finite
singular points has zero as a characteristic exponent, or the location of
the singular points is unrestricted. The latter includes the former. They
have the respective $P$-symbols
\begin{subequations}
\label{subeq:generalizedP}
\begin{equation}
\label{eq:generalizedP1}
P\left\{
\begin{array}{ccccc}
d_1&d_2&d_3&\infty& \\
0&0&0&\alpha& ;s \\
1-\gamma&1-\delta&1-\epsilon&\beta&
\end{array}
\right\},
\end{equation}
\begin{equation}
\label{eq:generalizedP2}
P\left\{
\begin{array}{ccccc}
d_1&d_2&d_3&d_4& \\
0&0&0&\alpha& ;s \\
1-\gamma&1-\delta&1-\epsilon&\beta&
\end{array}
\right\}.
\end{equation}
\end{subequations}
In the nomenclature of Ref.~\cite{Ronveaux95}, they are the canonical
natural general-form Heun equation and the canonical general-form Heun
equation. They are transformed to the Heun equation~(\ref{eq:HeunP}) by
the linear and M\"obius transformations
\begin{equation}
\label{eq:respmaps}
t=\frac{s-d_1}{d_2-d_1},
\qquad
t=\frac{(s-d_1)(d_2-d_4)}{(d_2-d_1)(s-d_4)},
\end{equation}
respectively. Each of the $P$-symbols~(\ref{subeq:generalizedP}) is
accompanied by an accessory parameter. The equation specified
by~(\ref{eq:generalizedP1}) can be written as
\begin{equation}
\label{eq:genHeun}
\quad\frac{d^2 u}{d s^2}
+ \left( \frac\gamma{s-d_1} + \frac\delta{s-d_2} + \frac\epsilon{s-d_3}
\right)\frac{du}{ds} + \frac{\alpha\beta s - q'}{(s-d_1)(s-d_2)(s-d_3)}\,u = 0,
\end{equation}
where $q'$ is the accessory parameter~\cite{Ronveaux95}. The equation
specified by~(\ref{eq:generalizedP2}) with $d_4\neq\infty$ can be written
as
\begin{eqnarray}
\label{eq:verygenHeun}
&&\frac{d^2 u}{d s^2}
+ \left( \frac\gamma{s-d_1} + \frac\delta{s-d_2} + \frac\epsilon{s-d_3}
+ \frac{1-\alpha-\beta}{s-d_4}
\right)\frac{du}{ds} \nonumber\\[-1.5ex]
\label{eq:gen2Heun} \\[-1.5ex]
&&\qquad\qquad{}+ \frac{\alpha\beta\left.\left[{\displaystyle\prod_{i=1}^3(d_4-d_i)}\right]\right/(s-d_4) - q''}
{(s-d_1)(s-d_2)(s-d_3)(s-d_4)}\,u = 0\nonumber
\end{eqnarray}
where $q''$ is the accessory parameter~\cite{Ronveaux95}.
There is an impediment to the satisfactory generalization of
Theorem~\ref{thm:main} to these two equations, which is the rigorous
specification of which cases should be excluded on~account of their being
`trivial', or having fewer than four singular points. The excluded cases
should really be specified not in~terms of the {\em ad~hoc\/} accessory
parameters $q'$ and~$q''$, but rather in an invariant way, in~terms of an
accessory parameter defined so~as to be invariant under linear or M\"obius
transformations. (See Ref.~\cite{Ronveaux95}, Addendum,~\S2.2.) However,
it is clear that (\ref{eq:genHeun})~is trivial, i.e., can be transformed to
a trivial Heun equation by a linear transformation, iff
$\alpha\beta=0$,~$q'=0$. Also, it has fewer than four singular points if
$\gamma=0$,~$q'=0$; or $\delta=0$, $q'=\alpha\beta$; or $\epsilon=0$,
$q'=\alpha\beta d$. Likewise, it~is fairly clear that
(\ref{eq:gen2Heun})~will be trivial, i.e., can be transformed to a trivial
Heun equation by a M\"obius transformation, iff $\alpha\beta=0$,~$q''=0$.
The conditions on the parameters for there to be a full set of singular
points are, however, more complicated.
The first generalization of Theorem~\ref{thm:main} is
Corollary~\ref{thm:gen1}, which follows from Theorem~\ref{thm:main} by
applying the linear transformation~(\ref{eq:respmaps}a). It~mentions a
polynomial transformation, which is the composition of the $s\mapsto t$
linear transformation with the $t\mapsto z$ polynomial map of
Theorem~\ref{thm:main}. To~avoid repetition, Corollary~\ref{thm:gen1}
simply cites Theorem~\ref{thm:main} for the necessary and sufficient
conditions on the characteristic exponent parameters and the accessory
parameter.
\begin{corollary}
\label{thm:gen1}
A canonical natural general-form Heun equation of the form
(\ref{eq:genHeun}), which has four singular points and is nontrivial (i.e.,
$\alpha\beta\neq0$ or~$q'\neq0$), can be transformed to a hypergeometric
equation of the form~{\rm ($\mathfrak{h}$)} by a rational substitution $z=R(s)$
iff $\alpha\beta\neq0$, $R$~is a polynomial, and the Heun equation
satisfies the following conditions.
$\triangle d_1d_2d_3$ must be similar to $\triangle 01D$, with $D=2$ or
$\frac12+i\sqrt3/2$, or $D=4$, $\frac12+i5\sqrt2/4$,
or~$\frac12+i11\sqrt{15}/90$. That is, it must either be a degenerate
triangle consisting of three equally spaced collinear points (the harmonic
case), or be an equilateral triangle (the equianharmonic case), or be
similar to one of three other specified triangles, of which one is
degenerate and two are isosceles. The characteristic exponent parameters
$\gamma,\delta,\epsilon$ must satisfy conditions that follow from the
corresponding subcases of Theorem~\ref{thm:main}, and the accessory
parameter~$q'$ must take a value that can be computed uniquely from the
parameters $\gamma,\delta,\epsilon$ and the choice of subcase.
For example, in the harmonic case, the two endpoints of the degenerate
triangle of singular points $\triangle d_1d_2d_3$ must have equal exponent
parameters, and $q'$~must equal $\alpha\beta$ times the intermediate point.
In~this case, $R$~will in~general be a quadratic polynomial. There are two
possibilities: $R$~will map the two endpoints to $z=0$ and the intermediate
point to $z=1$, or vice versa. If~the characteristic exponents of the
intermediate point are twice those of the endpoints, then $R$~may also be
quartic: the composition of either possible quadratic polynomial with a
subsequent ${z\mapsto 4(z-\frac12)^2}$ or $z\mapsto 4z(1-z)$ map.
In~the equianharmonic case, all three exponent parameters
$\gamma,\delta,\epsilon$ must be equal, and the accessory parameter~$q'$
must equal $\alpha\beta$ times the mean of $d_1,d_2,d_3$. In~this case,
$R$~will in~general be a cubic polynomial. There are two possibilities:
$R$~will map $d_1,d_2,d_3$ to $z=0$ and their mean to $z=1$, or vice versa.
If~the exponent parameters $\gamma,\delta,\epsilon$ equal~$2/3$, then
$R$~may also be sextic: the composition of either possible cubic polynomial
with a subsequent $z\mapsto 4(z-\frac12)^2$ or $z\mapsto 4z(1-z)$ map.
\end{corollary}
The further generalization of Theorem~\ref{thm:main} is
Corollary~\ref{thm:gen2}, which follows from Theorem~\ref{thm:main} by
applying the M\"obius transformation~(\ref{eq:respmaps}b). It~mentions a
rational substitution, which is the composition of the $s\mapsto t$
M\"obius transformation with the $t\mapsto z$ polynomial map of
Theorem~\ref{thm:main}.
\begin{corollary}
\label{thm:gen2}
A canonical general-form Heun equation of the form~(\ref{eq:gen2Heun}),
which has four singular points and is nontrivial (i.e., $\alpha\beta\neq0$
or~$q''\neq0$), can be transformed to a hypergeometric equation of the
form~{\rm ($\mathfrak{h}$)} by a rational substitution $z=R(s)$ iff
$\alpha\beta\neq0$, and the Heun equation satisfies the following
condition.
The cross-ratio orbit of $d_1,d_2,d_3,d_4$ must be the same as that of\/
$0,1,D,\infty$, where $D$~takes one of the five values enumerated in
Theorem~\ref{thm:main}. That~is, the cross-ratio orbit must be the
harmonic orbit, the equianharmonic orbit, or one of three specified generic
orbits, one real and two non-real. The characteristic exponent parameters
$\gamma,\delta,\epsilon$ must satisfy conditions that follow from the
corresponding subcases of Theorem~\ref{thm:main}, and the accessory
parameter~$q''$ must take a value that can be computed uniquely from the
parameters $\gamma,\delta,\epsilon$ and the choice of subcase.
\end{corollary}
{\em Example\/}~1. Suppose $d_1,d_2,d_3,d_4$ form a harmonic quadruple,
i.e., they can be mapped by a M\"obius transformation to the vertices of a
square in~$\mathbb{C}$\null. Moreover, two of $d_1,d_2,d_3$ have the same
characteristic exponents, and are mapped to diagonally opposite vertices of
the square. That~is, of the three parameters $\gamma,\delta,\epsilon$, the
two corresponding to a diagonally opposite pair must be equal.
In this case, provided the accessory parameter takes a value that can be
computed from the other parameters, a substitution~$R$ exists. In~general,
it will be a degree-2 rational function, the only critical points of which
are the third singular point (out~of $d_1,d_2,d_3$) and~$d_4$. Either
$R$~will map the two distinguished singular points to $z=1$ and the third
singular point to~$z=0$, or vice versa; and $d_4$ to~$z=\infty$. In~the
special case when the characteristic exponents of the third point are twice
those of the two distinguished points, degree-4 rational substitutions are
also possible.
{\em Example\/}~2. Suppose $d_1,d_2,d_3,d_4$ form an equianharmonic
quadruple, i.e., they can be mapped by a M\"obius transformation to the
vertices of a regular tetrahedron in~$\mathbb{CP}^1$. Moreover,
$d_1,d_2,d_3$ have the same characteristic exponents, i.e.,
$\gamma=\delta=\epsilon$.
In this case, provided the accessory parameter takes a value uniquely
determined by the other parameters, a substitution~$R$ exists. In~general,
$R$ will be a degree-3 rational function, the only critical points of which
are the mean of $d_1,d_2,d_3$ with respect to~$d_4$, and~$d_4$. Either
$R$~will map $d_1,d_2,d_3$ to $z=1$ and the mean of $d_1,d_2,d_3$ with
respect to~$d_4$ to~$z=0$, or vice versa; and $d_4$ to~$z=\infty$. In~the
special case when the characteristic exponents of each of $d_1,d_2,d_3$
equal $0,1/3$, degree-6 rational substitutions are also possible.
{\em Remark.} In Example~2, the concept of the mean of three points
in~$\mathbb{CP}^1$ with respect to a distinct fourth point was used.
A~projectively invariant definition is the following. If $T$~is a M\"obius
transformation that takes $d_4$ ($\neq d_1,d_2,d_3$) to the point at
infinity, the mean of $d_1,d_2,d_3$ with respect to~$d_4$ is the point that
would be mapped to the mean of $Td_1,Td_2,Td_3$ by~$T$.
\section{The Clarkson--Olver Transformation}
\label{sec:CO}
The transformation discovered by Clarkson and Olver~\cite{Clarkson96},
which stimulated these investigations, turns~out to be a special case of
the equianharmonic transformation discovered in~\S\ref{sec:main}. Their
transformation was originally given in a rather complicated form, which we
shall simplify.
Recall that the Weierstrass function $\wp(u)\equiv\wp(u;g_2,g_3)$ with
invariants $g_2,g_3\in\mathbb{C}$ has a double pole at~$u=0$, and satisfies
\begin{eqnarray}
\wp'^2 &=& 4\wp^3 - g_2\wp - g_3\nonumber\\[-1.5ex]
\label{eq:Wode} \\[-1.5ex]
&=& 4(\wp - e_1)(\wp - e_2)(\wp - e_3),\nonumber
\end{eqnarray}
where $e_1,e_2,e_3$, the zeroes of the defining cubic polynomial, are the
finite critical values of~$\wp$. By~convention, the polynomial has no
quadratic term (i.e., $g_1=0$), so
\begin{equation}
\label{eq:zerosum}
e_1+e_2+e_3 = 0.
\end{equation}
In~general, $\wp$~is doubly periodic on~$\mathbb{C}$, with periods denoted
$2\omega,2\omega'$. So $\wp$~can be viewed as a function on the torus
$\mathbb{T}\defeq\mathbb{C}/{\cal L}$, where ${\cal
L}=\{2n\omega+2n'\omega'\mid n,n'\in\mathbb{Z}\}$ is the period lattice.
It turns~out that the half-lattice $\{0,
\omega,\omega',\omega+\omega'\}+{\cal L}$ comprises the critical points
of~$\wp$. The map $\wp:\mathbb{T}\to\mathbb{CP}^1$ is a branched double
cover of the Riemann sphere, but $\mathbb{T}$~is uniquely coordinatized by
the value of the pair $(\wp,\wp')$. The properties of~$\mathbb{T}$ are
determined by the invariant $\Delta\defeq g_2^3-27g_3^2$, which is the
discriminant of the defining polynomial. That~is, $\Delta$~equals
$16\prod_{\langle ij\rangle} (e_i-e_j)^2$.
If $\Delta>0$ (the so-called real rectangular case, which predominates in
applications), $\omega$~and~$\omega'$ can be taken to be real and
imaginary, respectively. If $\Delta<0$ (the less familiar real rhombic
case), this is not possible. However, it is possible to choose them to be
complex conjugates, so that the third basic critical point
$\omega_2\defeq\omega+\omega'$ is real. The distinction between $\Delta>0$
and $\Delta<0$ is important, since tori~$\mathbb{T}$ corresponding to
different values of~$\Delta$ are not, in~general, homeomorphic as complex
analytic manifolds.
Clarkson and Olver considered the Weierstrass-form Lam\'e equation
\begin{equation}
\label{eq:Lame}
\frac{d^2\psi}{du^2} - \left[j(j+1)\wp(u) + r\right]\psi = 0,
\end{equation}
which can be viewed as a Fuchsian equation on~$\mathbb{T}$, with exactly
one regular singular point (at~$(\wp,\wp')=(\infty,\infty)$) and a single
accessory parameter,~$r$. [We~have altered their characteristic exponent
parameter~$\sigma$ to $-j(j+1)/36$, to agree with the literature,
and have added the accessory parameter.] In~particular, they considered
the case $g_2=0$, $g_3\neq0$, $r=0$. They mapped $u\in\mathbb{T}$ to
$z\in\mathbb{CP}^1$ via the formal substitution
\begin{equation}
\label{eq:COsubst}
u = \frac{i}{(16g_3)^{1/6}}
\int^{(1-z)^{1/3}}\frac{d\tau}{\sqrt{1-\tau^3}},
\end{equation}
and showed that the Lam\'e equation is transformed to
\begin{equation}
\label{eq:hyperCO}
z(1-z)\frac{d^2\psi}{dz^2} + \left(\frac12 - \frac76 z\right)
\frac{d\psi}{dz}
+\frac{j(j+1)}{36}\,\psi = 0.
\end{equation}
This is a hypergeometric equation with
$(a,b;c)=\left(-j/6,(j+1)/6;1/2\right)$. It~has characteristic exponents
$0,1/2$ at~$z=0$; $0,1/3$ at~$z=1$; and $-j/6,(j+1)/6$ at~$z=\infty$.
In elliptic function theory the case $g_2=0$, $g_3\neq0$ is called the
equianharmonic case, since it yields a triple of critical values
$e_1,e_2,e_3$ that are the vertices of an equilateral triangle
in~$\mathbb{C}$\null. If, for example, $g_3$~is real, then $\Delta<0$; and
by convention, $e_1,e_2,e_3$ correspond to $\omega,\omega+\omega',\omega'$,
respectively. $e_1$~and~$e_3$ are complex conjugates, and $e_2$~is real.
The triangle $\triangle0\omega_2\omega'$ is also equilateral
(see~\cite{Abramowitz65}, \S18.13). It~is customary to standardize the
equianharmonic case by choosing $g_3=1$, so that $\Delta=-27$. This is not
a major matter, however: since $\wp$~satisfies
$\wp(z;0,g_3)=g_3^{1/3}\wp(zg_3^{1/6};0,1)$, all nonzero~$g_3$ are
equivalent. In~fact, tori~$\mathbb{T}$ defined by $g_2=0$, $g_3\neq0$ are
homeomorphic as complex analytic manifolds, even if they have different
values of $g_3$, and hence~$\Delta$.
So, what Clarkson and Olver considered was the {\em equianharmonic\/}
Lam\'e equation, the natural domain of definition of which is a
torus~$\mathbb T$ (i.e., an elliptic curve) with special symmetries. For
the Lam\'e equation to be viewed as a Heun equation on~$\mathbb{CP}^1$,
it~must be transformed to its algebraic form~\cite{Hille76}, by~$s=\wp(u)$.
The algebraic form is
\begin{equation}
\label{eq:algLame}
\quad\frac{d^2 \psi}{d s^2}
+ \left( \frac{1/2}{s-e_1} + \frac{1/2}{s-e_2} + \frac{1/2}{s-e_3}
\right)\frac{d\psi}{ds} + \frac{[-j(j+1)/4]s-r/4}{(s-e_1)(s-e_2)(s-e_3)}\,\psi = 0.
\end{equation}
If $\Delta\neq0$, the three critical values $e_1,e_2,e_3$ are distinct;
in~which case (\ref{eq:algLame})~is a special case of~(\ref{eq:genHeun}),
the canonical natural general form of the Heun equation, with the distinct
finite singular points $d_1,d_2,d_3 = e_1,e_2,e_3$. Also,
$\alpha,\beta=-j/2,(j+1)/2$, $\gamma=\delta=\epsilon=1/2$, and~$q'=r/4$.
It~has characteristic exponents $0,1/2$ at $s=e_1,e_2,e_3$, and
$-j/2,(j+1)/2$ at~$s=\infty$.
Applying Corollary~\ref{thm:gen1} to~(\ref{eq:algLame}) yields the
following.
\begin{theorem}
The algebraic-form Lam\'e equation~(\ref{eq:algLame}), in the
equianharmonic case $g_2=0$, $g_3\neq0$, can be transformed
when~$j(j+1)\neq0$ to a hypergeometric equation of the
form~{\rm ($\mathfrak{h}$)} by a rational transformation $z=R(s)$ iff the
accessory parameter~$r$ equals zero. If~this is the case, $R$~will
necessarily be a cubic polynomial; both of
\begin{equation}
z=4s^3/g_3,\qquad 1-4s^3/g_3
\end{equation}
will work, and they are the only possibilities.
\end{theorem}
{\em Proof.} If $j(j+1)\neq0$, (\ref{eq:algLame})~is a nontrivial Heun
equation. If $g_3\neq0$, then $\Delta\neq0$ and the singular points~$e_i$
of~(\ref{eq:algLame}) are distinct, so the Heun equation has four singular
points; by~(\ref{eq:Wode}), the~$e_i$ are the cube roots of~$g_3/4$, and
are the vertices of an equilateral triangle. Since
$\gamma=\delta=\epsilon$, the equianharmonic case of
Corollary~\ref{thm:gen1} applies, and no~other.
As~noted in~(\ref{eq:zerosum}), the sum and hence the mean of the~$e_i$ are
zero. So the polynomial $4s^3/g_3$ is the cubic polynomial that maps each
singular point to~$1$, and their mean to zero; $1-4s^3/g_3$ does the
reverse. These are the only possibilities for the map~$s\mapsto z$, since
the sextic polynomials mentioned in the equianharmonic case can be employed
only~if $\gamma,\delta,\epsilon$ equal $2/3$, which is not the case here.
\qquad \endproof
{\em Remark}. The equianharmonic case of Corollary~\ref{thm:gen1} also
applies to the equianharmonic algebraic-form Lam\'e equation in the case
$j(j+1)=0$, $r\neq0$, and guarantees it cannot be transformed to the
hypergeometric equation by any rational substitution. That is because in
the sense defined in~\S\ref{sec:defs}, this case too is nontrivial.
\begin{corollary}
\label{thm:cor}
The Weierstrass-form Lam\'e equation~(\ref{eq:Lame}), in the equianharmonic
case $g_2=0$, $g_3\neq0$, can be transformed when $j(j+1)\neq0$ to a
hypergeometric equation of the form~{\rm ($\mathfrak{h}$)} by a substitution of
the form $z=R\left(\wp(u)\right)$, where $R$~is rational, iff the accessory
parameter~$r$ equals zero. If~this is the case, the substitutions
\begin{equation}
\label{eq:Wsubsts}
z=4\wp(u)^3/g_3,\qquad 1-4\wp(u)^3/g_3
\end{equation}
will work, and they are the only such substitutions.
\end{corollary}
In~fact, applying the substitution $z=1-4\wp(u)^3/g_3$ to the Lam\'e
equation~(\ref{eq:Lame}) transforms it to the hypergeometric
equation~(\ref{eq:hyperCO}) derived by Clarkson and Olver, as is readily
verified. The alternative substitution yields a closely related
hypergeometric equation, with the singular points $z=0$ and~$z=1$
interchanged.
From this perspective, all that remains to be checked is the validity of
the original Clarkson--Olver substitution,~(\ref{eq:COsubst}). It~contains
a multivalued elliptic integral, which may be inverted, with the aid
of~(\ref{eq:Wode}), to yield $z=1-4\wp(u)^3/g_3$. Since this is listed in
Corollary~\ref{thm:cor}, the Clarkson--Olver transformation fits into the
framework of~\S\ref{sec:main}.
A natural question is whether their transformation can be generalized.
Corollary~\ref{thm:cor} does not offer much hope, other than allowing an
arbitrary nonzero value of~$g_3$ (which may even be non-real, so that
$\Delta$~may be non-real). Actually, the harmonic case as well as the
equianharmonic case of Corollary~\ref{thm:gen1} can be applied to the
algebraic-form Lam\'e equation. One of the resulting quadratic
transformations was recently discovered by Ivanov~\cite{Ivanov2001}. But
quadratic rather than cubic changes of the independent variable, and more
general transformations of the Lam\'e equation, will be treated elsewhere.
The most noteworthy feature of the Clarkson--Olver transformation is that
it can be performed irrespective of the choice of characteristic exponent
parameter~$j$. Only the accessory parameter~$r$ is restricted. As~they
remark in their paper, when $j=1$,~$1/2$, $1/4$, or~$1/10$, it~is a
classical result of Schwarz that all solutions of the hypergeometric
equation~(\ref{eq:hyperCO}) are necessarily algebraic
(\cite{Hille76},~\S10.3). This implies that if~$r=0$, the same is true of
all solutions of the algebraic Lam\'e equation~(\ref{eq:algLame}); which
had previously been proved by Baldassarri~\cite{Baldassarri81}, using
rather different techniques. But irrespective of the choice of~$j$, the
solutions of the $r=0$ Lam\'e equation reduce to solutions of the
hypergeometric equation. This is quite unlike the other known classes of
exact solutions of the Lam\'e equation, which restrict~$j$ to take values
in a discrete set~(\cite{Morales99},~\S2.8.4). But it is typical of
hypergeometric reductions of the Heun equation; of~the sort that we have
considered, at~least. As the theorems of~$\S\ref{sec:main}$ make clear,
in~general it is possible to alter characteristic exponent parameters
continuously, without affecting the existence of a transformation to the
hypergeometric equation.
\section{The Seemingly Trivial Case $\alpha\beta=0$, $q=0$}
\label{sec:trivial}
In the trivial case, the Heun equation~{\rm ($\mathfrak{H}$)} may be solved
by quadratures. Its solutions are
\begin{equation}
\quad u_1(t) = {\rm const},\quad\,\,
u_2(t) = \int^t\exp\left[-\int^v\left(\frac\gamma w+\frac\delta{w-1} + \frac\epsilon{w-d}\right)\,dw\right]\,dv.
\end{equation}
In the trivial limit, the local Heun function
$\Hl(d,q;\alpha,\beta,\gamma,\delta;t)$ degenerates to the former (in~fact,
to unity), and the solution belonging to the exponent $1-\gamma$ at~$t=0$,
denoted $\widetilde\Hl(d,q;\alpha,\beta,\gamma,\delta;t)$ here, to the
latter. In~applications, explicit solutions, if~any, are what matter.
It~is nonetheless interesting to examine under what circumstances a trivial
Heun equation can be transformed to a hypergeometric equation.
The canonical polynomial substitutions of~\S\ref{sec:main}, each of which
maps exponents to exponents, give rise to a large number of nonpolynomial
rational transformations of the trivial Heun equation to the hypergeometric
equation, via composition with certain M\"obius transformations.
To~understand why, recall that Theorem~\ref{thm:main} and
Corollary~\ref{thm:useful0} characterized, up~to linear automorphisms of
the two equations, the possible polynomial substitutions that may be
applied to a nontrivial Heun equation. If $t\mapsto R_1(t)$ denotes a
canonical polynomial transformation, the full set of polynomial reductions
derived from~it comprises $t\mapsto L_2\left(R_1(L_1(t))\right)$, where
$L_1\in{\cal L}({\mathfrak H})$~is a linear automorphism of the Heun
equation, which maps $\{0,1,d\}$ onto $\{0,1,D\}$, and $L_2\in{\cal
L}({\mathfrak h})$~is a linear automorphism of the hypergeometric equation,
which maps $\{0,1\}$ onto~$\{0,1\}$. (The only two possibilities for~$L_2$
are $L_2(z)=z$ and $L_2(z)=1-z$.)
In the context of {\em nontrivial\/} Heun equations, automorphisms that are
not linear could not be employed; essentially because, as discussed
in~\S\ref{subsec:auto}, moving the point at infinity will require a
compensating F-homotopic transformation. But in the trivial case, no~such
issue arises: by Proposition~\ref{thm:trivialprop}, the Heun equation is
transformed to a hypergeometric equation by a rational substitution of its
independent variable, $z=R(t)$, iff the substitution maps exponents to
exponents. And non-F-homotopic automorphisms, including M\"obius
transformations which are not linear, certainly preserve exponents. The
following theorem is a consequence.
\begin{theorem}
\label{thm:trivial}
A Heun equation of the form~{\rm ($\mathfrak{H}$)}, which has four singular
points and is trivial (i.e., $\alpha\beta=0$ and~$q=0$), can be transformed
to a hypergeometric equation of the form~{\rm ($\mathfrak{h}$)} by any
rational substitution of the form $z=M_2\left(R_1(M_1(t))\right)$, where
$z=R_1(t)$~is a polynomial that maps $\{0,1,D\}$ to $\{0,1\}$, listed
(along with~$D$) in one of the seven subcases of Theorem~\ref{thm:main},
$M_1\in{\cal M}(\mathfrak{H})$, and $M_1\in{\cal M}(\mathfrak{h})$.
That~is, $M_1$~is a M\"obius transformation that maps $\{0,1,d,\infty\}$
onto $\{0,1,D,\infty\}$, and $M_2$~is a M\"obius transformation that maps
$\{0,1,\infty\}$ onto $\{0,1,\infty\}$. The necessary conditions on
characteristic exponents stated in Theorem~\ref{thm:main} must be
satisfied: the conditions on exponents at specified values of~$t$ are to be
taken as applying to the exponents at their preimages under~$M_1$.
\end{theorem}
{\em Remark.} As in the derivation of the reductions listed in
Corollary~\ref{thm:useful0}, the Gauss parameters $(a,b;c)$ of the
resulting hypergeometric function are computed by first calculating the
exponents at~$z=0,1,\infty$, using the mapping of exponents to exponents.
\smallskip
The following example shows how such rational substitutions are
constructed. In subcase~1(a) of Theorem~\ref{thm:main}, i.e., the harmonic
case, $D=2$ and the polynomial tranformation is $t\mapsto z=R_1(t)=t(2-t)$;
the necessary condition on exponents is that $t=0,d$ have identical
exponents. Consider $d=-1$, which is on the cross-ratio orbit of~$D$.
$M_1(t)=(t-1)/t$ can be chosen; also, let $M_2(z)=1/z$. Then the
composition
\begin{equation}
z=R(t)\equiv M_2\left(R_1(M_1(t))\right)= t^2/(t^2-1)
\end{equation}
maps $t=0$ to~$z=0$ and $t=\infty$ to~$z=1$ (both with double
multiplicity), and $t=1,d$ to~$z=\infty$. This substitution may be applied
to any trivial Heun equation with $d=-1$, provided it has identical
exponents at $t=1,d$, i.e., provided $\delta=\epsilon$.
In~this example, ${M}_1,{M}_2$ were selected with foresight, to ensure that
$R$~maps $t=0$ to~$z=0$. This makes it possible to regard the substitution
as a reduction of $\Hl$ to~${}_2F_1$, or of $\widetilde\Hl$
to~$\widetilde{{}_2F_1}$. By computation of exponents, the reduction is
\begin{eqnarray}
\label{eq:explicitred}
&&\widetilde\Hl\left(-1,\,0;\,0,\,\beta,\,\gamma,(1+\beta+\gamma)/2;\,t\right)\\
&&\qquad=(-1)^{(\gamma-1)/2}\,
\widetilde{{}_2F_1}\left(0,\,(1-\beta+\gamma)/2;\,(1+\gamma)/2;\,t^2/(t^2-1)\right).\nonumber
\end{eqnarray}
Here $\widetilde\Hl,\widetilde{{}_2F_1}$ appear, since the corresponding
reduction of $\Hl$ to~${}_2F_1$ is trivially valid (both sides are constant
functions of~$t$, and equal unity). The normalization factor
$(-1)^{(\gamma-1)/2}$ is present because by convention
$\widetilde\Hl(t)\sim t^{1-\gamma}$ and $\widetilde{{}_2F_1}(z)\sim
z^{1-c}$ in a neighborhood of~$t=0$ (resp.\ $z=0$), where the principal
branches are meant.
Working out the number of rational substitutions $z=R(t)$ that may be
applied to trivial Heun equations, where $R(\cdot)$~is of the form
$M_2\left(R_1(M_1(\cdot))\right)$, is a useful exercise. There are seven
subcases of Theorem~\ref{thm:main}, i.e., choices for the polynomial~$R_1$.
Each subcase allows $d$~to be chosen from an orbit consisting of
$m$~cross-ratio values: $m=3$~in the harmonic subcases 1(a) and~1(c), $m=2$
in the equianharmonic subcases 2(a) and~2(d), and $m=6$ in the others.
In~any subcase, the $4!$~choices for~$M_1$ are divided equally among the
$m$~values of~$d$, and there are also $3!$~choices for~$M_2$. So each
subcase yields $(4!/m)3!$ rational substitutions for each value of~$d$, but
not all are distinct.
To count {\em distinct\/} rational substitutions for each value of~$d$,
note the following. $R$~will map $t=0,1,d,\infty$ to~$z=0,1,\infty$. Each
of the subcases of Theorem~\ref{thm:main} has a `signature', specifying the
cardinalities of the inverse images of the points in~$\{0,1,\infty\}$. For
example, case~1(a) has signature $2;1;1$, which means that of those three
points, one has two preimages and the other two have one. (Order is
irrelevant.) In~all, subcases 1(a),1(b),2(b),2(c) have signature $2;1;1$,
and the others have signature $3;1;0$. By~inspection, the number of
distinct mappings of $t=0,1,d,\infty$ to~$z=0,1,\infty$ consistent with the
signature $2;1;1$ is~$36$, and the number consistent with $3;1;0$ is~$18$.
Kuiken~\cite{Kuiken79} supplies a useful list of the $36$~rational
substitutions arising from the harmonic subcase~1(a), but states
incorrectly that they are the only rational substitutions that may be
applied to the trivial Heun equation. Actually, subcases 1(a)--1(c) and
2(a)--2(d) give rise to $36,36,18;18,36,36,18$ rational substitutions,
respectively. By dividing by~$m$, it~follows that for each subcase, the
number of distinct rational substitutions per value of~$d$ is
$12,6,6;9,6,6,9$. Of~these, exactly one-third map $t=0$ to~$z=0$, rather
than to $z=1$ or~$z=\infty$, and consequently yield reductions of $\Hl$
to~${}_2F_1$, or of $\widetilde\Hl$ to~$\widetilde{{}_2F_1}$. So for each
subcase, the number of such reductions per value of~$d$ is $4,2,2;3,2,2,3$.
For example, the four such reductions with $d=-1$ that arise from the
harmonic subcase~1(a) are
\begin{subequations}
\begin{eqnarray}
\label{eq:firstred}
&&\widetilde\Hl\left(-1,\,0;\,0,\,\beta,\,\gamma,\,(1+\beta-\gamma)/2;\,t\right)\\
&&\qquad=\widetilde{{}_2F_1}\left(0,\,\beta/2;\,(1+\gamma)/2;\,t^2\right)\nonumber\\
\label{eq:secondred}
&&\widetilde\Hl\left(-1,\,0;\,0,\,\beta,\,\gamma,(1+\beta+\gamma)/2;\,t\right)\\
&&\qquad=(-1)^{(\gamma-1)/2}\,\widetilde{{}_2F_1}\left(0,\,(1-\beta+\gamma)/2;\,(1+\gamma)/2;\,t^2/(t^2-1)\right)\nonumber\\
\label{eq:newred1}
&&\widetilde\Hl\left(-1,\,0;\,0,\,\beta,\,1-\beta,\,\delta;\,t\right)\\
&&\qquad=\widetilde{{}_2F_1}\left(0,\,(1-2\beta+\delta)/2;\,1-\beta;\,4t/(t+1)^2\right)\nonumber\\
\label{eq:newred2}
&&\widetilde\Hl\left(-1,\,0;\,0,\,\beta,\,1-\beta,\,\delta,\,t\right)\\
&&\qquad=(-4)^{-\beta}\,\widetilde{{}_2F_1}\left(0,\,(1-\delta)/2;\,1-\beta;\,-4t/(t-1)^2\right)\nonumber
\end{eqnarray}
\end{subequations}
The reduction~(\ref{eq:firstred}), which is the only one of the four in
which the degree-2 rational function $R$~is a polynomial, is simply the
trivial (i.e., $\alpha=0$) case of~(\ref{eq:extrathing}), a~quadratic
reduction that applies to nontrivial Heun equations with $d=-1$, as~well.
The reduction~(\ref{eq:secondred}) was derived above
as~(\ref{eq:explicitred}), but (\ref{eq:newred1}) and~(\ref{eq:newred2})
are new. By examination, they are related by composition with the M\"obius
transformation $z\mapsto z/(z-1)$, i.e., by the automorphism in~${\cal
M}(\mathfrak{h})$ that interchanges $z=1$ and~$z=\infty$.
Remarkably, many rational reductions of trivial Heun equations to the
hypergeometric equation are {\em not\/} derived from the polynomial
transformations of Theorem~\ref{thm:main}. The following striking degree-4
transformation is an example. The function
\begin{equation}
Q(t)=1-\left(\frac{t-1-i}{t-1+i}\right)^4
=\frac{8it(t-1)(t-2)}{(t-1+i)^4}
\end{equation}
maps $t=0,1,d\equiv2,\infty$ to~$z=0$; and $t=1\pm i$ to~$z=1,\infty$ (both
with quadruple multiplicity). By Proposition~\ref{thm:trivialprop}, a
trivial Heun equation with $d=2$ will be transformed by~$Q$ to a
hypergeometric equation iff $Q$~maps exponents to exponents. This
constrains the singular points $t=0,1,d,\infty$ to have the same exponents;
which is possible only if each has exponents~$0,1/2$, which must also be
the exponents of~$z=0$. Also, since $t=1\pm i$ are ordinary points of the
Heun equation, the exponents of the hypergeometric equation at $z=1,\infty$
must be~$0,1/4$. It~follows that on the level of solutions, the
transformation is
\begin{equation}
\label{eq:tiring}
\widetilde\Hl(2,\,0;\,0,\,\tfrac12,\,\tfrac12,\,\tfrac12;\,t)
=(i/4)^{1/2}\,\widetilde{{}_2F_1}\left(
0,\,{\tfrac14};\,\tfrac12;\,
\frac{8it(t-1)(t-2)}{(t-1+i)^4}\right),
\end{equation}
where the normalization factor $(i/4)^{1/2}$ follows from the known
behavior of the function values $\widetilde\Hl(t)$ and
$\widetilde{{}_2F_1}(z)$ as~$t\to0$ and $z\to0$. Using the definitions
(\ref{eq:tilde1}) and~(\ref{eq:newguy2}), the
transformation~(\ref{eq:tiring}) can be rewritten as
\begin{eqnarray}
\label{eq:finalreduction}
&&\quad\quad\Hl(2,\,\tfrac34;\,\tfrac12,\,1,\,\tfrac32,\,\tfrac12;\,t)\\
&&\quad\qquad=
(1-t)^{1/2}(1-t/2)^{1/2}\left[1-t/(1-i)\right]^{-2}
{}_2F_1\left(
\tfrac12,\,\tfrac34;\,\tfrac32;\,
\frac{8it(t-1)(t-2)}{(t-1+i)^4}
\right).\nonumber
\end{eqnarray}
The equality~(\ref{eq:finalreduction}) holds in a neighborhood of~$t=0$
(it~should be noted that both sides are real when $t$~is real and
sufficiently small). It~is a harmonic-case reduction of $\Hl$
to~${}_2F_1$, since it applies when $d=2$, but it is not a special case of
the general harmonic reduction (\ref{eq:generalharmonic}), even though it
is much more specialized, since it has no free parameters.
The transformation~(\ref{eq:finalreduction}) is of a more general type than
has been considered so~far. In~the abstract language
of~\S\ref{subsec:auto}, it specifies a homomorphism from an ${\rm
Aut}(\mathfrak{H})$-orbit to an ${\rm Aut}(\mathfrak{h})$-orbit. However,
it includes a linear change of the dependent variable, resembling a
complicated F-homotopy, in addition to a rational change of the independent
variable. It~is clear that (\ref{eq:finalreduction})~cannot be derived
from any of the canonical (polynomial) reductions of
Corollary~\ref{thm:useful0} by applying automorphisms of~{\rm
($\mathfrak{H}$)} and~{\rm ($\mathfrak{h}$)} to its left-hand and
right-hand sides. Tranformations of this more general type will be
classified in Part~II\null.
%\bibliographystyle{siam}
%\bibliography{general}
\small\def\em{\it} \newcommand{\noopsort}[1]{} \newcommand{\printfirst}[2]{#1}
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\end{document}