[labelstyle=]

# Schwarz triangle functions and duality for certain parameters of the
generalised Chazy equation

Matthew Randall Department of Mathematics and Statistics
Faculty of Science, Masaryk University
Kotlářská 2, 611 37 Brno
Czech Republic<EMAIL_ADDRESS>
###### Abstract.

Schwarz triangle functions play a fundamental role in the solutions of the
generalised Chazy equation. Chazy has shown that for the parameters $k=2$ and
$3$, the equations can be linearised. We determine the Schwarz triangle
functions that represent the solutions in these cases. Some of these Schwarz
triangle functions also show up in the dual cases where $k=\frac{2}{3}$ and
$k=\frac{3}{2}$, which suggests intriguing connections between the solutions
for $k=2$ and $k=3$ with dihedral and tetrahedral symmetry respectively, and
$k=\frac{2}{3},\frac{3}{2}$ with $G_{2}$ symmetry. Integrating the solutions
when $k=\frac{2}{3}$, we obtain flat $(2,3,5)$-distributions parametrised
algebraically by the corresponding Schwarz triangle functions. The Legendre
dual of these curves are also algebraic, can be computed quite easily and are
related to the integral curves of the dual generalised Chazy equation with
parameter $\frac{3}{2}$.

This work is supported by the Grant agency of the Czech Republic P201/12/G028.

The generalised Chazy equation is a third order nonlinear autonomous ordinary
differential equation given by

(0.1)
$y^{\prime\prime\prime}-2y^{\prime\prime}y+3(y^{\prime})^{2}-\frac{4}{36-k^{2}}(6y^{\prime}-y^{2})^{2}=0$

for $k\neq 6$. The equation (0.1) was introduced by Jean Chazy in the papers
[6] and [7] in the context of investigating the Painlevé property for third
order ODEs.

The equation (0.1) can be solved by Schwarz triangle functions [1]. Schwarz
triangle functions determine through its inverse a map from the complex upper
half plane to an open triangular domain with boundary given by the edges of
the triangle. The angles of the triangle determined by the Schwarz functions
depend on the parameter $k$ in (0.1). When $k<6$, the image is a spherical
triangle. When $k=6$, the triangle is planar and when $k>6$, the triangle is
hyperbolic. In this article we present the spherical Schwarz triangle
functions corresponding to the solutions when $k$ is given by $\frac{2}{3}$,
$\frac{3}{2}$, $2$ and $3$ and determine them algebraically.

These four parameters are chosen because the equations also show up in the
context of the geometry of differential equations. The problem of determining
whether the solution set of a system of differential equation is equivalent to
another, via for instance point or contact preserving transformation, can be
solved using Cartan’s method of equivalence.

The equations for the parameters $k=2$ and $k=3$ are shown to be linearisable
by Chazy himself (see p. 346 of [7]). The equation when $k=2$ is linearisable
to the ODE $y^{\prime\prime\prime\prime}=0$. This is related to the third
order Riccati equation as observed in [16]. Applying Cartan’s method of
equivalence, this 4th order ODE has vanishing Wilczynski invariants in the
linear theory and also vanishing Bryant invariants in the non-linear theory
[4], [11], [14], [19]. We discuss the general solution when $k=2$ in Section
3.

The method of equivalence also applies to third order ODEs, as worked out by
Chern [8]. The equation for the parameter $k=3$ turns out to be the only
equation of the form (0.1) that has vanishing Wünschmann invariant. Third
order ODEs with vanishing Wünschmann invariant defines a conformal structure
of signature $(2,1)$ on the space of its solutions. The conformal metric is
obtained by quotienting out a degenerate split signature symmetric 2-tensor by
the vector field that annihilates the distribution encoding the ODE
$y^{\prime\prime\prime}=F(x,y,p,q)$. ODEs with vanishing Wünschmann invariant
and $F_{qqqq}=0$, satisfied for the $k=3$ equation, are contact equivalent to
the equation $y^{\prime\prime\prime}=0$ [13]. The equation for this parameter
$k=3$ is also linearisable and the general solution to this equation is
described in Section 5.

The equations for the parameters $k=\pm\frac{3}{2}$ and $k=\pm\frac{2}{3}$
show up in the local equivalence problem for generic maximally non-integrable
(or bracket generating) rank $2$ distributions on $5$-manifolds $M$ that
depend on a single function $F(x)$. Here the genericity condition implies that
$F^{\prime\prime}(x)\neq 0$. For such non-integrable distributions, the
bracket of the vector fields spanning the distribution $\mathcal{D}$
determines a filtration of the tangent bundle given by
$\mathcal{D}\subset[\mathcal{D},\mathcal{D}]\subset
TM=[[\mathcal{D},\mathcal{D}],\mathcal{D}]$ with the rank of $\mathcal{D}=2$
and the rank of $[\mathcal{D},\mathcal{D}]=3$. Such distributions are
therefore also known as $(2,3,5)$-distributions. Cartan solved the local
equivalence problem for such geometries in [5] and constructed the fundamental
curvature invariant. For distributions of the form $F(x)$ as described, the
curvature invariant vanishes when $F^{\prime\prime}(x)=e^{\frac{1}{2}\int
y(x){\rm d}x}$ where $y(x)$ is a solution to the generalised Chazy equation
(0.1) with parameter $k=\pm\frac{2}{3}$. In this case the distribution has
split $G_{2}$ as its local group of symmetries. Furthermore An and Nurowski
[3] showed that there is a duality that takes the solutions of this equation
to the solutions of the submaximal $7^{\rm th}$ order ODE

(0.2)
$10(y^{\prime\prime\prime})^{3}y^{(7)}-70(y^{\prime\prime\prime})^{2}y^{(4)}y^{(6)}-49(y^{\prime\prime\prime})^{2}(y^{(5)})^{2}+280y^{\prime\prime\prime}(y^{(4)})^{2}y^{(5)}-175(y^{(4)})^{4}=0$

studied in [20] (where it appears in equation (6.64)), [11] and [12].
Historically, this ODE appeared already in the thesis of Noth [18] in 1904.

This dual ODE (0.2) can also be reduced to a generalised Chazy equation but
now the Chazy parameter is given by $k=\pm\frac{3}{2}$. In this fashion, the
solutions with parameters $k=\frac{3}{2}$ and $k=\frac{2}{3}$ have vanishing
Cartan invariants. We discuss the solutions for both these equations in
Sections 4 and 6. We show in Appendix B that the Legendre duality property for
the generalized Chazy equations with parameters $k=\pm\frac{3}{2}$ and
$k=\pm\frac{2}{3}$ is unique only for these parameters.

Interestingly, one of the Schwarz triangle functions that solve the $k=2$
equation also show up in the solutions to the $k=\frac{2}{3}$ equation. Three
of the Schwarz triangle functions that solve the $k=3$ equation also show up
in the solutions to the $k=\frac{3}{2}$ equation. For the $k=2$ and
$k=\frac{2}{3}$ cases, the Schwarz triangle functions are pullbacks through
hypergeometric transformation of the Schwarz function
$s(\frac{1}{2},\frac{1}{3},\frac{1}{2},x)$ that appears in Schwarz’s list [22]
with dihedral symmetry. For the $k=3$ and $k=\frac{3}{2}$ cases, the functions
are pullbacks of the function $s(\frac{1}{2},\frac{1}{3},\frac{1}{3},x)$ that
appears in Schwarz’s list with tetrahedral symmetry. The result of Schwarz
[22] and Klein tells us that the Schwarz functions we obtain in these cases
are algebraic. The maps for the triangle functions that show up in the cases
$k=\frac{2}{3}$, $\frac{3}{2}$, $2$ and $3$ are given in the Diagrams 1, 2 and
3. They show the hypergeometric transformations that are given by quadratic,
cubic and quartic maps [15], [23].

Another motivation for the article is to work out the Schwarz functions that
solve the $k=\frac{2}{3}$ and $k=\frac{3}{2}$ equation and determine examples
of $(2,3,5)$-distributions with maximal symmetry group of split $G_{2}$. The
solutions to the $k=\frac{2}{3}$ equation has appeared in [21], but we present
them more explicitly here in the form of Table 2. We also present the
solutions when $k=\frac{3}{2}$ here. To work out the distributions $D_{F(x)}$
with vanishing Cartan curvature invariant, we have to determine
$F^{\prime\prime}(x)$ from the solutions of the $k=\frac{2}{3}$ Chazy equation
and integrate twice further. This gives an algebraic relation involving
$(x,F)$. The dual curve of this plane algebraic curve gives us integral curves
of equation (0.2).

The first two sections are background material. In Section 1 we set up the
preliminaries and consider $SL_{2}(\mathbb{C})$ equivalent classes of
solutions to (0.1). In Section 2 we review the definitions of Schwarz
functions and in Sections 3, 4, 5, 6 we present the Schwarz triangle functions
that appear in the solutions of the Chazy’s equation for $k=2$,
$k=\frac{2}{3}$, $k=3$ and $k=\frac{3}{2}$ respectively. The computations are
done through MAPLE 17.

###### Contents

  1. 1 $SL_{2}(\mathbb{C})$ action on the space of solutions
  2. 2 Schwarz functions and equivalent solutions under $SL_{2}(\mathbb{C})$
  3. 3 Generalised Chazy equation with $k=2$ and its Schwarz functions
  4. 4 Generalised Chazy equation with $k=\frac{2}{3}$ and its Schwarz functions
    1. 4.1 The Schwarz function $s(3,3,3,x)$
  5. 5 Generalised Chazy equation with $k=3$ and its Schwarz functions
  6. 6 Generalised Chazy equation with $k=\frac{3}{2}$ and its Schwarz functions
    1. 6.1 The Schwarz function $s(\frac{2}{3},\frac{1}{3},2,t)$
  7. A Solving quartic equation
  8. B Duality of generalised Chazy equations

##  1\. $SL_{2}(\mathbb{C})$ action on the space of solutions

The material in the first two sections is collated from [7], [9], [10], [2]
and [1]. We shall work over the complex field. Any element
$g={\bigl{(}\begin{smallmatrix}a&b\\\ c&d\end{smallmatrix}}\bigr{)}\in
SL_{2}(\mathbb{C})$ acts on $x$ by fractional linear transformations $g\cdot
x=\tilde{x}=\frac{ax+b}{cx+d}$.

######  Proposition 1.1.

(See also [9], [2]) Under this action of $SL_{2}(\mathbb{C})$, for any
solution $y(x)$ to the generalised Chazy equation (0.1) with $k\neq 6$, we
obtain new solutions to (0.1) by

(1.1)
$\displaystyle\tilde{y}(x)=\frac{1}{(cx+d)^{2}}y\left(\frac{ax+b}{cx+d}\right)-\frac{6c}{cx+d}.$

###### Proof.

The action of $SL_{2}$ gives the differential relation ${\rm
d}\tilde{x}=\frac{1}{(cx+d)^{2}}{\rm d}x$ and $\frac{{\rm d}}{{\rm
d}\tilde{x}}=(cx+d)^{2}\frac{{\rm d}}{{\rm d}x}$. Differentiating, we find

$\displaystyle(cx+d)^{4}(6\tilde{y}^{\prime}-\tilde{y}^{2})=6y^{\prime}-y^{2}$

and

$\displaystyle(cx+d)^{6}(9\tilde{y}^{\prime\prime}-9\tilde{y}\tilde{y}^{\prime}+\tilde{y}^{3})=9y^{\prime\prime}-9yy^{\prime}+y^{3},$

where prime on the left hand side denotes differentiation with respect to $x$
while prime on the right hand side denotes differentiation with respect to
$\tilde{x}.$ Differentiating once more gives

$\displaystyle(cx+d)^{8}\left(\tilde{y}^{\prime\prime\prime}-2\tilde{y}^{\prime\prime}\tilde{y}+3(\tilde{y}^{\prime})^{2}-\frac{4}{36-k^{2}}(6\tilde{y}^{\prime}-\tilde{y}^{2})^{2}\right)$
$\displaystyle=$ $\displaystyle
y^{\prime\prime\prime}-2y^{\prime\prime}y+3(y^{\prime})^{2}-\frac{4}{36-k^{2}}(6y^{\prime}-y^{2})^{2}.$

We see that $\tilde{y}(x)$ is a solution to (0.1) iff $y(\tilde{x})$ is a
solution as well. ∎

Let $f(x)=\exp(\frac{2}{k-6}\int y{\rm d}x)$. Chazy makes this substitution
and finds that $f$ satisfies the differential equation

(1.2) $\displaystyle
ff^{\prime\prime\prime\prime}-(k-2)f^{\prime}f^{\prime\prime\prime}+\frac{3k(k-2)}{2(k+6)}(f^{\prime\prime})^{2}=0.$

It is immediate from (1.2) that when $k=2$, the equation becomes linear, and
we shall discuss this further in Section 3. When we integrate (1.1), we obtain

$\displaystyle\int\tilde{y}(x){\rm d}x=$
$\displaystyle\int\frac{1}{(cx+d)^{2}}y(\tilde{x}){\rm
d}x-\int\frac{6c}{cx+d}{\rm d}x$ $\displaystyle=$ $\displaystyle\int
y(\tilde{x}){\rm d}\tilde{x}-6\log(cx+d)+c_{0}.$

We find that

$\displaystyle\tilde{f}(x)=\exp\left({\frac{2}{k-6}\int\tilde{y}(x){\rm
d}x}\right)=$
$\displaystyle\frac{\exp({\frac{2c_{0}}{k-6}})}{(cx+d)^{\frac{12}{k-6}}}\exp\left({\frac{2}{k-6}\int
y(\tilde{x}){\rm
d}\tilde{x}}\right)=\frac{\exp(\frac{2c_{0}}{k-6})}{(cx+d)^{\frac{12}{k-6}}}f(\tilde{x}).$

Absorbing constants (or normalising them so that $c_{0}=0$), we see that

$\displaystyle\tilde{f}(x)=\frac{1}{(cx+d)^{\frac{12}{k-6}}}f(\tilde{x}).$

This motivate the following definition.

######  Definition 1.1.

Suppose both functions $f(x)$ and
$\tilde{f}(x)=(cx+d)^{-\frac{12}{k-6}}f(\tilde{x})$ satisfy the same
differential equation (1.2). Then we say that the function $f(x)$ has weight
$\frac{12}{k-6}$ since $f(\tilde{x})=(cx+d)^{\frac{12}{k-6}}\tilde{f}(x)$
(following the convention in the literature about weights of modular forms).

Let us take $k=\frac{2}{3}$ and suppose
$f(x)=(F^{\prime\prime}(x))^{-\frac{3}{4}}$ for some $F(x)$. Then
$F^{\prime\prime}(x)$ has weight $3$ under the action of $SL_{2}(\mathbb{C})$
and we find that $F(x)$ satisfies the 6th order ODE

(1.3) $\displaystyle
10F^{(6)}(F^{\prime\prime})^{3}-80(F^{\prime\prime})^{2}F^{(3)}F^{(5)}-51(F^{\prime\prime})^{2}(F^{(4)})^{2}+336F^{\prime\prime}(F^{\prime\prime\prime})^{2}F^{(4)}-224(F^{\prime\prime\prime})^{4}=0$

in [3] upon substituting $f=(F^{\prime\prime})^{-\frac{3}{4}}$ into (1.2).

######  Proposition 1.2.

If $F(x)$ is a solution to the 6th order ODE (1.3), then so is

$\tilde{F}(x)=(cx+d)F\left(\frac{ax+b}{cx+d}\right)$

where $ad-bc=1$.

According to the definition given above, the function $F(x)$ has weight $-1$.
We have

######  Corollary 1.1.

The function $F(x)=x^{2}$ is a solution to the 6th order ODE (1.3), and
therefore so is

$\tilde{F}(x)=(cx+d)\left(\frac{ax+b}{cx+d}\right)^{2}=\frac{(ax+b)^{2}}{cx+d}$

where $ad-bc=1$.

Differentiating the above relation twice, we find that

$\displaystyle\tilde{F}^{\prime\prime}(x)=\frac{1}{(cx+d)^{3}}F^{\prime\prime}(\tilde{x}).$

Again the right hand side denotes differentiation with respect to $\tilde{x}$.
For any $F(x)$ with weight $-1$ satisfying the 6th order ODE (1.3), we
identify the solutions

$\left(x,\tilde{F}(x)\right)=\left(x,(cx+d)F(g\cdot x)\right)\sim\left(g\cdot
x,F(g\cdot x)\right).$

We define $SL_{2}(\mathbb{C})$ equivalent solutions to the generalised Chazy’s
equation in the following fashion (see also [9]).

######  Definition 1.2.

Two solutions $y(\tilde{x})$ and $\tilde{y}(x)$ to the generalised Chazy
equation are said to be equivalent if there exists an element $g$ of
$SL_{2}(\mathbb{C})$ such that $\tilde{x}=g\cdot x$ and (1.1) holds for
$y(g\cdot x)$ and $\tilde{y}(x)$.

From this we can identify the solutions

$\left(x,\tilde{y}(x)\right)\sim\left(\tilde{x},(cx+d)^{2}\tilde{y}(x)+6c(cx+d)\right)=(\tilde{x},y(\tilde{x})).$

It is well known that $F(x)=x^{m}$ for
$m\in\\{-1,\frac{1}{3},\frac{2}{3},2\\}$ solves equation (1.3) (see [5]). Let
us investigate how these solutions are equivalent solutions under
$SL_{2}(\mathbb{C})$. Using that $y(x)=2\frac{{\rm d}}{{\rm
d}x}\log(F^{\prime\prime}(x))$, this gives $y(x)=\frac{6}{x}$,
$y(x)=-\frac{10}{3x}$, $y(x)=-\frac{8}{3x}$ and $y(x)=0$ as solutions to the
$k=\frac{2}{3}$ equation. The solution $y=-\frac{8}{3(x+C)}-\frac{10}{3(x+B)}$
for constants $B$, $C$ was further obtained following [6] and [7]. This
corresponds to $F(x)=(x+B)^{\frac{1}{3}}(x+C)^{\frac{2}{3}}$.

######  Proposition 1.3.

The solutions to (0.1) for the parameter $k=\frac{2}{3}$ given by
$y=-\frac{10}{3x}$, $y=-\frac{8}{3x}$ and
$y=-\frac{8}{3(x+C)}-\frac{10}{3(x+B)}$ are equivalent in the sense of
Definition 1.2.

###### Proof.

Applying an arbitrary $g=\bigl{(}\begin{smallmatrix}a&b\\\
c&d\end{smallmatrix}\bigr{)}\in SL_{2}(\mathbb{C})$ to the solution given by
$y=-\frac{8}{3x}$, we obtain
$\tilde{y}=-\frac{8}{3(x+\frac{b}{a})}-\frac{10}{3(x+\frac{d}{c})}$. Applying
$g_{1}=\Bigl{(}\begin{smallmatrix}-ec&\frac{f}{c(e-f)}\\\
c&-\frac{1}{ec}(1+\frac{f}{e-f})\end{smallmatrix}\Bigr{)}\in
SL_{2}(\mathbb{C})$ to $\tilde{y}=-\frac{8}{3(x+e)}-\frac{10}{3(x+f)}$ with
$e\neq f$, we obtain $\tilde{\tilde{y}}=-\frac{10}{3x}$. To get back to
$y=-\frac{8}{3x}$, we use the transformation given by
$g_{2}=\bigl{(}\begin{smallmatrix}0&b\\\
-\frac{1}{b}&0\end{smallmatrix}\bigr{)}\in SL_{2}(\mathbb{C})$. ∎

As a consequence of Proposition 1.3, we see that the solutions given by
$F(x)=(x+B)^{\frac{1}{3}}(x+C)^{\frac{2}{3}}$, $x^{\frac{2}{3}}$ and
$x^{\frac{1}{3}}$ are equivalent to one another by this $SL_{2}$ action.

######  Proposition 1.4.

The solutions to (0.1) for the parameter $k=\frac{2}{3}$ given by $y=0$ and
$y=-\frac{6}{x}$ are equivalent in the sense of Definition 1.2.

###### Proof.

This is clear from applying $g=\bigl{(}\begin{smallmatrix}0&-1\\\
1&0\end{smallmatrix}\bigr{)}$ to the zero solution and its inverse
$g^{-1}=\bigl{(}\begin{smallmatrix}0&1\\\ -1&0\end{smallmatrix}\bigr{)}$ to
$y=-\frac{6}{x}$ to get the zero solution. ∎

##  2\. Schwarz functions and equivalent solutions under $SL_{2}(\mathbb{C})$

The solutions to equation (0.1) can be found using the techniques described in
[1] and [2]. They can be expressed in terms of logarithmic derivatives
involving Schwarz triangle functions. Chazy finds the solutions in [6] and [7]
by introducing the auxillary parameter $s$ and treating
$x=\frac{z_{2}(s)}{z_{1}(s)}$ and taking $y=6\frac{{\rm d}}{{\rm d}x}\log
z_{1}(s)$. The generalised equations are then satisfied if and only if
$z_{1}(s)$, $z_{2}(s)$ are linearly independent solutions to the
hypergeometric differential equation

(2.1) $\displaystyle s(1-s)z_{ss}+(c-(a+b+1)s)z_{s}-abz=0.$

Here the subscript denotes differentiation with respect to $s$ and $a$, $b$
and $c$ are constants that depend on $k$. The solutions can also be written as
a closed nonlinear system of first order autonomous differential equations,
called the generalised Darboux-Halphen system. From the first order system,
the equations can be transformed to a Schwarzian type equation with potential
term $V(s)$. The solutions are given precisely by Schwarz triangle functions.
Let prime denote differentiation with respect to $x$.

######  Definition 2.1.

A Schwarz triangle function $s(\alpha,\beta,\gamma,x)$ is a solution to the
following third order non-linear differential equation

$\\{s,x\\}+\frac{(s^{\prime})^{2}}{2}V(s)=0$

where $\\{s,x\\}=\frac{{\rm d}}{{\rm
d}x}\left(\frac{s^{\prime\prime}}{s^{\prime}}\right)-\frac{1}{2}\left(\frac{s^{\prime\prime}}{s^{\prime}}\right)^{2}$
is the Schwarzian derivative and

$V(s)=\frac{1-\beta^{2}}{s^{2}}+\frac{1-\gamma^{2}}{(s-1)^{2}}+\frac{\beta^{2}+\gamma^{2}-\alpha^{2}-1}{s(s-1)}$

is the potential.

A Schwarz triangle function determines through its inverse a mapping from the
complex upper half plane $\mathbb{H}=\\{z\in\mathbb{C}:\Im(z)>0\\}$ to a
spherical, planar or hyperbolic triangle $\Delta$ with angles between edges
given by $(\alpha\pi,\beta\pi,\gamma\pi)$. The edges of the triangle are given
by circular arcs. The inverse map $x:\mathbb{H}\to\Delta$ is single valued and
meromorphic given by

(2.2) $\displaystyle
x(s)=\frac{{}_{2}F_{1}(a-c+1,b-c+1;2-c;s)}{{}_{2}F_{1}(a,b;c;s)}s^{1-c}.$

The image of $0$, $\infty$ and $1$ under $s$ are the vertices of the triangle
with one vertex at the origin $x(0)=0$ and another $x(1)$ on the real line
(identifying the domain of the triangle as a subset of the complex plane).
Here we have

$\displaystyle a$ $\displaystyle=\frac{1}{2}(1-\alpha-\beta-\gamma),$
$\displaystyle b$ $\displaystyle=\frac{1}{2}(1+\alpha-\beta-\gamma),$
$\displaystyle c$ $\displaystyle=1-\beta.$

In the solutions to (0.1), $x$ is given by the quotient of linearly
independent solutions to (2.1). The general solution to (2.1) is given by
$\alpha z_{1}(s)+\beta z_{2}(s)$ where $z_{1}$, $z_{2}$ are linearly
independent. We form the quotient $x=\frac{z_{2}}{z_{1}}$. If we take a
different linear combination instead with

$\displaystyle\tilde{x}=\frac{\beta z_{1}-\delta z_{2}}{-\alpha z_{1}+\gamma
z_{2}}=\frac{\beta-\delta\frac{z_{2}}{z_{1}}}{-\alpha+\gamma\frac{z_{2}}{z_{1}}}=\frac{\beta-\delta
x}{-\alpha+\gamma x},$

then we find

$\displaystyle x=\frac{\alpha\tilde{x}+\beta}{\gamma\tilde{x}+\delta}.$

In other words, if we restrict to $\alpha$, $\beta$, $\gamma$, $\delta$ such
that $\alpha\delta-\beta\gamma=1$, then $x=g\cdot\tilde{x}$ and
$\tilde{x}=(g^{-1})\cdot x$ for $g\in SL_{2}$. Hence $SL_{2}$ equivalent
solutions to Chazy’s equation are determined by quotient
$\frac{z_{2}}{z_{1}}$, and thus are completely determined by the Schwarz
function $s$. Every distinct Schwarz function therefore gives rise to $SL_{2}$
equivalent solutions as in Definition 1.2. We will henceforth just consider
the quotient $x=\frac{z_{2}}{z_{1}}$ in our computations, modulo constants
that agree with the expression (2.2).

Our goal now is to present the various $(x(s),y(s))$, parametrised by the
distinct Schwarz functions $s$ that are found using the general method to
solve Chazy’s equation [1], [2]. Let us denote

$\displaystyle\Omega_{1}$ $\displaystyle=-\frac{1}{2}\frac{{\rm d}}{{\rm
d}x}\log\frac{s^{\prime}}{s(s-1)},$ $\displaystyle\Omega_{2}$
$\displaystyle=-\frac{1}{2}\frac{{\rm d}}{{\rm
d}x}\log\frac{s^{\prime}}{s-1},$ $\displaystyle\Omega_{3}$
$\displaystyle=-\frac{1}{2}\frac{{\rm d}}{{\rm d}x}\log\frac{s^{\prime}}{s}.$

Then $y=-2(\Omega_{1}+\Omega_{2}+\Omega_{3})$ solves (0.1) when
$(\alpha,\beta,\gamma)=(\frac{2}{k},\frac{2}{k},\frac{2}{k})$ or
$(\frac{2}{k},\frac{1}{3},\frac{1}{3})$ and its cyclic permutations.
Similarly, we find that $y=-\Omega_{1}-2\Omega_{2}-3\Omega_{3}$ solves (0.1)
when $(\alpha,\beta,\gamma)=(\frac{1}{k},\frac{1}{3},\frac{1}{2})$,
$(\frac{1}{k},\frac{2}{k},\frac{1}{2})$ or
$(\frac{1}{k},\frac{1}{3},\frac{3}{k})$. Also,
$y=-4\Omega_{1}-\Omega_{2}-\Omega_{3}$ solves (0.1) when
$(\alpha,\beta,\gamma)=(\frac{4}{k},\frac{1}{k},\frac{1}{k})$ or
$(\frac{2}{3},\frac{1}{k},\frac{1}{k})$. Each of these values of
$(\alpha,\beta,\gamma)$ determine the Schwarz triangle function and the values
of $(a,b,c)$ in (2.2). In the next four sections we compute these for $k=2$,
$k=\frac{2}{3}$, $k=3$ and $k=\frac{3}{2}$.

##  3\. Generalised Chazy equation with $k=2$ and its Schwarz functions

In this section we give the general solution to (0.1) for $k=2$ and the
Schwarz functions that solves the equation in Table 1.

######  Theorem 3.1.

The general solution to the generalized Chazy equation with $k=2$ over the
Riemann surface $\mathbb{P}^{1}=\mathbb{C}\cup\\{\infty\\}$ is given by

$\displaystyle
y(x)=-2\left(\frac{1}{x-x_{1}}+\frac{1}{x-x_{2}}+\frac{1}{x-x_{3}}\right)$

and depends on 3 arbitrary points $x_{1}$, $x_{2}$, $x_{3}$ on the Riemann
surface.

###### Proof.

We make the following observation over the complex plane $\mathbb{C}$. Under
the substitution $y=\frac{-2f^{\prime}}{f}$ for $f$ non-zero, the generalised
Chazy equation with $k=2$ is equivalent to the linear 4th order ODE
$f^{\prime\prime\prime\prime}=0$. The solution to
$f^{\prime\prime\prime\prime}=0$ is given by the cubic polynomial
$f=ax^{3}+3bx^{2}+6cx+d$ and therefore the general solution to the $k=2$ Chazy
equation over $\mathbb{C}$ is given by its logarithmic derivative

$\displaystyle y=-\frac{6ax^{2}+12bx+12c}{ax^{3}+3bx^{2}+6cx+d}$

where $a$, $b$, $c$, $d$ are constants of integration. For $a\neq 0$, we can
factorize $f=a(x-x_{1})(x-x_{2})(x-x_{3})$ over $\mathbb{C}$, and we obtain
$y=-2\left(\frac{1}{x-x_{1}}+\frac{1}{x-x_{2}}+\frac{1}{x-x_{3}}\right)$. The
general solution for $a\neq 0$ therefore depends on three generic points on
$\mathbb{C}$. If we include the point at infinity, we allow solutions with
$a=0$ of the form say $y=-2\left(\frac{1}{x-x_{1}}+\frac{1}{x-x_{2}}\right)$.
∎

The above fact has already been observed by Chazy in pages 346-347 of [7]. Let
us tabulate in Table 1 the Schwarz triangle functions that show up in the
solutions to the Chazy equation when $k=2$. In the first column, we give the
angles $(\alpha,\beta,\gamma)$ of the spherical triangle. This determines the
values
$(a,b,c)=(\frac{1}{2}(1-\alpha-\beta-\gamma),\frac{1}{2}(1+\alpha-\beta-\gamma),1-\beta)$
in the hypergeometric equation (2.1) and in the second column we give

$\displaystyle
x=\frac{{}_{2}F_{1}(a-c+1,b-c+1;2-c;s)}{{}_{2}F_{1}(a,b;c;s)}s^{1-c}.$

In the third column, we invert the second column to present the Schwarz
function, with a branch cut chosen for the functions in the last three rows.
The entry in the second column of the fifth row requires the solution of a
quartic polynomial determined by $x(r)$ on the second column. This same
Schwarz function will show up again in the solutions to the $k=\frac{2}{3}$
equation. Finally in the last column we present $y(x)$ as determined by the
formulas for $s(x)$ in the fourth column. Here $w=e^{\frac{2\pi i}{3}}$
denotes the cube root of unity. The formulas that give $y$ are discussed and
presented in [21]. The series expansion of $x(s)$ around a regular
neighbourhood can be computed and can be checked to see if it agrees with the
function presented in Table 1.

$(\alpha,\beta,\gamma)$ | $x(s)$ | $s(x)$ | $y(x)$
---|---|---|---
$(1,1,1)$ | $\frac{s}{1-s}$ | $\frac{x}{x+1}$ | $-\frac{2}{x}-\frac{2}{x+1}$
$(1,\frac{1}{3},\frac{1}{3})$ | $\left(\frac{s}{1-s}\right)^{\frac{1}{3}}$ | $\frac{x^{3}}{x^{3}+1}$ | $-\frac{2}{x+1}-\frac{2}{x+w}-\frac{2}{x+w^{2}}$
$\left(\frac{1}{2},\frac{1}{3},\frac{1}{2}\right)$ | $2^{\frac{2}{3}}\left(\frac{1-r}{1+r}\right)^{\frac{1}{3}}$ | $\frac{16x^{3}}{(x^{3}+4)^{2}}$ | $-\frac{6x^{2}}{x^{3}+4}$
|  and $s=1-r^{2}$ |  |
$\left(\frac{1}{2},1,\frac{1}{2}\right)$ | $\frac{2}{\sqrt{1-s}}-2$ | $1-\frac{4}{(x+2)^{2}}$ | $-\frac{2}{x}-\frac{2}{x+2}-\frac{2}{x+4}$
$\left(\frac{1}{2},\frac{1}{3},\frac{3}{2}\right)$ | $\frac{1}{2^{\frac{1}{3}}}\left(\frac{1-r}{1+r}\right)^{\frac{1}{3}}\left(\frac{3+r}{3-r}\right)$ | Roots of the quartic polynomial | $-\frac{3\cdot 2^{\frac{1}{3}}}{8}\left(\frac{r+1}{r-1}\right)^{\frac{1}{3}}\frac{r-3}{r^{3}}(r-1)(r+3)^{2}$
|  and $s=1-r^{2}$ | $\left(\frac{1}{2}-x^{3}\right)r^{4}+(8x^{3}+4)r^{3}$ | $=-\frac{12x^{2}}{2x^{3}-1}$
|  | $+(9-18x^{3})r^{2}+27x^{3}-\frac{27}{2}=0$ |
$\left(2,\frac{1}{2},\frac{1}{2}\right)$ | $\frac{\sqrt{s(1-s)}}{1-2s}$ | $\frac{1}{2}(1\pm\frac{1}{\sqrt{4x^{2}+1}})$ | $-\frac{2}{x}-\frac{16x}{4x^{2}+1}$
$\left(\frac{2}{3},\frac{1}{2},\frac{1}{2}\right)$ | $\frac{3}{2}\frac{\sin(r)}{\cos(r)}$ | $\frac{1}{2}\left(1+\frac{9(4x^{2}-3)}{(4x^{2}+9)^{\frac{3}{2}}}\right)$ | $-\frac{2}{x}-\frac{16x}{4x^{2}-27}$
|  and $s=\sin^{2}(\frac{3}{2}r)$ |  |
Table 1. Schwarz functions for $k=2$

Up to fractional linear transformations, the identity map $x=s$ is given by
$s(1,1,1,x)$. This function appears in the first row of Table 1. The upper
half plane is mapped to the hemisphere of $\mathbb{S}^{2}$ bound by a great
circle and the vertices of the triangle are three points lying on the great
circle. Geometrically speaking then the spherical triangles make sense only
when $(\alpha+\beta+\gamma)\pi\leq 3\pi$. The occurrence of angles adding up
to greater than $3\pi$ requires us to think of “triangles” overlapping onto
itself (or a branched cover or folded triangle) to realise such exaggeratedly
large angles.

The function appearing in the third row of Table 1 given by
$s(\frac{1}{2},\frac{1}{3},\frac{1}{2},x)$ appears in Schwarz’s list [22] and
has dihedral symmetry. Despite the complicated looking formula for $x$ in the
fifth row corresponding to $s(\frac{1}{2},\frac{1}{3},\frac{3}{2},x)$, it is
straightforward to determine $y$ in terms of $x$.

######  Proposition 3.1.

Let $s=s(\frac{1}{2},\frac{1}{3},\frac{3}{2},x)$ Then the solution to (0.1)
with $k=2$ is given by

$\displaystyle y=\frac{{\rm d}}{{\rm
d}x}\log\frac{(s^{\prime})^{3}}{s^{2}(s-1)^{\frac{3}{2}}}=-\frac{12x^{2}}{2x^{3}-1}.$

###### Proof.

From the parametrisation

$\displaystyle
x=\frac{1}{2^{\frac{1}{3}}}\left(\frac{1-r}{1+r}\right)^{\frac{1}{3}}\left(\frac{3+r}{3-r}\right)$

with $s=1-r^{2}$, we obtain

$\displaystyle{\rm
d}x=-\frac{8}{3}\frac{4^{\frac{1}{3}}r^{2}}{(r-3)^{2}(r-1)^{\frac{2}{3}}(r+1)^{\frac{4}{3}}}{\rm
d}r=\frac{1}{r^{\prime}}{\rm d}r.$

The corresponding formula for $y$ that solves (0.1) then gives

$\displaystyle y=\frac{{\rm d}}{{\rm
d}x}\log\frac{(s^{\prime})^{3}}{s^{2}(s-1)^{\frac{3}{2}}}=r^{\prime}\frac{{\rm
d}}{{\rm
d}r}\log\frac{(-2r^{\prime}r)^{3}}{(1-r^{2})^{2}(-r^{2})^{\frac{3}{2}}}$
$\displaystyle=-\frac{3\cdot
2^{\frac{1}{3}}}{8}\frac{1}{r^{3}}(r-3)(r+3)^{2}(r-1)^{\frac{2}{3}}(r+1)^{\frac{1}{3}}.$

From the formula for $x$, we find

$\displaystyle\frac{1}{2x^{3}-1}=\frac{(r-3)^{3}(r+1)}{16r^{3}},$

and therefore

$\displaystyle y=$
$\displaystyle-\frac{3}{4}\frac{1}{2^{\frac{2}{3}}}\left(\frac{1-r}{1+r}\right)^{\frac{2}{3}}\left(\frac{3+r}{3-r}\right)^{2}\frac{(r-3)^{3}(r+1)}{r^{3}}$
$\displaystyle=$
$\displaystyle-12\frac{1}{2^{\frac{2}{3}}}\left(\frac{1-r}{1+r}\right)^{\frac{2}{3}}\left(\frac{3+r}{3-r}\right)^{2}\frac{(r-3)^{3}(r+1)}{16r^{3}}$
$\displaystyle=$ $\displaystyle-\frac{12x^{2}}{2x^{3}-1}.$

∎

The maps between the Schwarz functions for the $k=2$ case are presented in
Figure 1. We use the notation adopted in [23], where the number over the arrow
denotes the degree of the algebraic transformation. These transformations are
classified in [23].

Figure 1. Mapping of Schwarz functions for $k=2$

##  4\. Generalised Chazy equation with $k=\frac{2}{3}$ and its Schwarz
functions

In this section we determine the Schwarz functions for the relevant angles
$(\alpha,\beta,\gamma)$ that show up in the solutions to the $k=\frac{2}{3}$
equation and present the flat or symmetric $(2,3,5)$-distribution that it
determines by computing the anti-derivative

$\displaystyle F(x)=\int\int e^{\frac{1}{2}\int y(x){\rm d}x}{\rm d}x{\rm
d}x.$

They are presented in Table 2. In the table we normalise the constants
appearing in $F(x)$ due to integration to be $1$.

$(\alpha,\beta,\gamma)$ |  $x(s)$ or $x(r)$ |  $F(x(s))$ or $F(x(r))$
---|---|---
$(3,3,3)$ | $\frac{1}{2}\frac{s-2}{2s-1}s^{3}$ | $s+\frac{1}{2(2s-1)}$
$(\frac{1}{3},\frac{1}{3},3)$ | $\frac{1}{2}\left(\frac{s+2}{2s+1}\right)s^{\frac{1}{3}}$ | $\frac{s^{\frac{2}{3}}}{2s+1}$
$\left(\frac{3}{2},\frac{1}{3},\frac{1}{2}\right)$ | $-\frac{1}{2^{\frac{1}{3}}}\left(\frac{1-r}{1+r}\right)^{\frac{1}{3}}\left(\frac{1+3r}{1-3r}\right)$ | $\frac{(r+1)^{\frac{1}{3}}(r-1)^{\frac{2}{3}}}{3r-1}$
|  and $s=1-r^{2}$ |
$\left(\frac{3}{2},3,\frac{1}{2}\right)$ | $-2\frac{s^{2}+4s-8}{\sqrt{1-s}}-16$ | $\frac{s-2}{\sqrt{s-1}}$
$\left(\frac{3}{2},\frac{1}{3},\frac{9}{2}\right)$ | $-2^{-\frac{4}{3}}\left(\frac{r+3}{r-3}\right)\left(\frac{1-r}{1+r}\right)^{\frac{1}{3}}\frac{3r^{4}-8r^{3}+54r^{2}-81}{3r^{4}+8r^{3}+54r^{2}-81}$ | $\left(\frac{r+1}{r-1}\right)^{\frac{1}{3}}\frac{(r+3)(r^{2}-9)(r-1)}{3r^{4}+8r^{3}+54r^{2}-81}$
|  and $s=1-r^{2}$ |
$\left(6,\frac{3}{2},\frac{3}{2}\right)$ | $\frac{(s(1-s))^{\frac{3}{2}}(2s-1)}{128s^{4}-256s^{3}+144s^{2}-16s-1}$ | $\frac{1}{128s^{4}-256s^{3}+144s^{2}-16s-1}$
$\left(\frac{2}{3},\frac{3}{2},\frac{3}{2}\right)$ | $-\frac{81}{64}\frac{\sin(8r)-2\sin(4r)}{\cos(8r)+2\cos(4r)}$ | $\frac{1}{\cos(8r)+2\cos(4r)}$
|  and $s=\sin^{2}(3r)$ |
Table 2. Parametrisations for $k=\frac{2}{3}$

Inverting the function for $s$ in terms of $x$ require the solution of a
quartic equation in $s$ except for two cases in the second and third last rows
of Table 2 where the degrees of the polynomials involved are larger. We
present $F(x)$ instead of the solutions $y(x)$ because the results take
precedence in the theory of $(2,3,5)$-distributions.

There is a Legendre duality discovered in [3] that takes equation (1.3) to
(0.2) given by the following:

$(x,F)\mapsto(t,H)=(F^{\prime},xF^{\prime}-F).$

We find that $H$ as a function of $t$ solves equation (0.2)

(4.1)
$10(H^{\prime\prime})^{3}H^{(6)}-70(H^{\prime\prime})^{2}H^{(3)}H^{(5)}-49(H^{\prime\prime})^{2}(H^{(4)})^{2}+280H^{\prime\prime}(H^{(3)})^{2}H^{(4)}-175(H^{(3)})^{4}=0$

iff equation (1.3) holds. The prime in equation (4.1) refers to
differentiation with respect to $t$. Equation (1.3) can be reduced to the
generalised Chazy equation with $k=\frac{2}{3}$ while equation (4.1) can be
reduced to the generalised Chazy equation with $k=\frac{3}{2}$. We now provide
an example of the computation of the entry in the first row of Table 2 with
the Schwarz function $s(3,3,3,x)$. The rest of the Schwarz functions that show
up in Table 2 can be determined similarly.

###  4.1. The Schwarz function $s(3,3,3,x)$

As an exercise let us determine the Schwarz triangle function $s(3,3,3,x)$ and
find the flat $(2,3,5)$-distribution that this function determines. See also
the first row of Table 1 in [21]. The solution to $u_{ss}+\frac{1}{4}V(s)u=0$
is

$\displaystyle u(s)=\alpha\frac{2s-1}{s(s-1)}+\gamma\frac{s^{2}(s-2)}{s-1}.$

We take $u_{1}=\frac{2s-1}{s(s-1)}$, $u_{2}=\frac{s^{2}(s-2)}{2(s-1)}$ and
obtain

(4.2) $\displaystyle x(s)=\frac{u_{2}}{u_{1}}=\frac{s^{3}(s-2)}{2(2s-1)}.$

This agrees with the expression given by (2.2). The Wronskian is given by
$c_{0}=W(u_{1},u_{2})=(u_{2})_{s}u_{1}-(u_{1})_{s}u_{2}=3$. We find

(4.3) $\displaystyle y(x(s))=\frac{u_{1}^{2}}{c_{0}}\frac{{\rm d}}{{\rm
d}s}\log\frac{u_{1}^{6}}{c_{0}^{3}s^{2}(s-1)^{2}}=-\frac{4(2s-1)(5s^{2}-5s+2)}{3s^{3}(s-1)^{3}}.$

Using (4.2) and (4.3), we can eliminate $s$ to obtain an equation involving
$x$ and $y$.

We can also express $y$ in terms of $x$ directly since (4.2) says that $s$ is
a root of the quartic polynomial

$\displaystyle s^{4}-2s^{3}-4xs+2x=0,$

which can always be solved by radicals. We solve this quartic polynomial in
Appendix A. Substituting this formula for $s=\xi+\frac{1}{2}$ into (4.3) gives
the solution to the generalised Chazy equation with $k=\frac{2}{3}$. The
formula for $y$ is really too long and too unsightly to reproduce here. As a
function of $\xi$, we have

$\displaystyle
y(\xi)=-\frac{128}{3}\frac{\xi(20\xi^{2}+3)}{(2\xi-1)^{3}(2\xi+1)^{3}}.$

Alternatively given the polynomial equation of degree 4

$\displaystyle s^{4}-2s^{3}-4xs+2x=0$

and the polynomial equation of degree 6

$\displaystyle
ys^{6}-3ys^{5}+3ys^{4}-(\frac{40}{3}-y)s^{3}-20s^{2}+12s-\frac{8}{3}=0$

that has to be satisfied because of (4.3), we can eliminate $s$ by forming the
resultant polynomial $P(x,y)$ of bidegree $(6,4)$:

$\displaystyle P(x,y)=$ $\displaystyle
y^{4}+\frac{8(4x+1)}{x(2x+1)}y^{3}+\frac{16}{3}\left(\frac{70x^{2}+35x+4}{x^{2}(2x+1)^{2}}\right)y^{2}$
$\displaystyle+\frac{256}{27}\left(\frac{(10x+1)(5x+2)(4x+1)}{x^{3}(2x+1)^{3}}\right)y+\frac{64}{9}\left(\frac{250x^{2}+125x+316}{x^{3}(2x+1)^{3}}\right)=0.$

This defines an implicit relation between $x$ and $y$, and the integral curves
are solutions to the generalised Chazy equation with parameter
$k=\pm\frac{2}{3}$.

To obtain the corresponding flat $(2,3,5)$-distribution $F(x)$, we have to
integrate further. Using (4.3) and the fact that

$\displaystyle{\rm d}x=3\frac{s^{2}(s-1)^{2}}{(2s-1)^{2}}{\rm d}s,$

we get

$\displaystyle F^{\prime\prime}(x(s))$
$\displaystyle=\frac{(2s-1)^{3}}{s^{4}(s-1)^{4}},$ $\displaystyle
F^{\prime}(x(s))$ $\displaystyle=3\left(\frac{1}{s}-\frac{1}{s-1}\right),$
$\displaystyle F(x(s))$ $\displaystyle=\int\int e^{\frac{1}{2}\int y{\rm
d}x}{\rm d}x{\rm d}x=-\frac{9}{4}s-\frac{9}{8(2s-1)}.$

Modulo constants of integration, the formula for $F(x(s))$ is just
$s+\frac{1}{2(2s-1)}$ as presented in the first row of Table 2. Eliminating
the parameter $s$ again gives a quartic algebraic curve $P_{1}(x,F)=0$. For
the family of quartic curves parametrised by

$\displaystyle(x,F)=\left(\frac{s^{3}(s-2)}{2(2s-1)},c\left(s+\frac{1}{2(2s-1)}\right)\right)$

where $c$ is constant, we can compute the Legendre dual $(x,F)\mapsto(t,H)$ by
determining $t=F^{\prime}$ and $H=xt-F$. This gives

$\displaystyle(t,H)=\left(\frac{4c}{3}\left(\frac{1}{s-1}-\frac{1}{s}\right),-c\left(\frac{1}{6}+\frac{2}{3}s+\frac{2}{3(s-1)}\right)\right).$

Each row of Table 2 gives a solution to equation (1.3). We have the following.

######  Theorem 4.1.

Each row of Table 2 determines a $(2,3,5)$-distribution $D_{F(x)}$ with split
$G_{2}$ symmetry.

We observe that the Schwarz triangle function
$s(\frac{1}{2},\frac{1}{3},\frac{3}{2},x)$ appears in both the solutions to
the $k=2$ Chazy equation and the $k=\frac{2}{3}$ equation. This gives an
intriguing relationship between the solutions to (0.1) for $k=\frac{2}{3}$ and
$k=2$ determined by this Schwarz function.

######  Proposition 4.1.

Let $s=s(\frac{3}{2},\frac{1}{3},\frac{1}{2},x)$ and suppose $s=1-r^{2}$ for
some $r(x)$. Take the parametrisation for $x(s)$ given in the third row of
Table 2. We find

$\displaystyle y_{2}=\frac{{\rm d}}{{\rm
d}x}\log\frac{(s^{\prime})^{3}}{(s-1)^{\frac{5}{2}}s^{2}}=-\frac{3}{4\cdot
2^{\frac{2}{3}}}\left(\frac{1-r}{1+r}\right)^{\frac{2}{3}}\frac{1}{r}(3r-1)(9r^{3}+15r^{2}+7r+1)$

is a solution to (0.1) with $k=2$ while

$\displaystyle y_{\frac{2}{3}}=\frac{{\rm d}}{{\rm
d}x}\log\frac{(s^{\prime})^{3}}{(s-1)^{\frac{3}{2}}s^{2}}=y_{2}+\frac{{\rm
d}}{{\rm
d}x}\log(s-1)=-\frac{3}{2^{\frac{2}{3}}}\left(\frac{1-r}{1+r}\right)^{\frac{2}{3}}(3r-1)(3r^{2}+5r+2)$

is a solution to (0.1) with $k=\frac{2}{3}$.

The Schwarz functions for the $k=\frac{2}{3}$ equation are determined by the
following pull-back maps in Figure 2, with the same notation from [23].

Figure 2. Mapping of Schwarz functions for $k=\frac{2}{3}$

The domain of this spherical triangle corresponding to the Schwarz function
$s(\frac{1}{2},\frac{1}{3},\frac{3}{2},x)$ has angles
$(\frac{1}{2}\pi,\frac{1}{3}\pi,\frac{3}{2}\pi)$. This triangle is the
complement of the triangle with angles
$(\frac{1}{2}\pi,\frac{2}{3}\pi,\frac{1}{2}\pi)$ in a hemisphere with the edge
of the hemisphere lying along the $\frac{1}{2}\pi$ and $\frac{2}{3}\pi$ edge
of the triangle. Reflecting along the $\frac{3}{2}\pi$, $\frac{1}{2}\pi$ edge
gives us the “triangle” with angles $(\frac{1}{3}\pi,\frac{1}{3}\pi,3\pi)$
branched over the vertex with angle $3\pi$. A reflection instead along the
equatorial $\frac{1}{3}\pi$, $\frac{1}{2}\pi$ edge gives the triangle with
angles $(\frac{2}{3}\pi,\frac{3}{2}\pi,\frac{3}{2}\pi)$, which is also the
complement of the $(\frac{4}{3}\pi,\frac{1}{2}\pi,\frac{1}{2}\pi)$ triangle in
the sphere.

##  5\. Generalised Chazy equation with $k=3$ and its Schwarz functions

We let $t$ denote now the independent variable. This is to avoid confusion
when we compute the Legendre dual curves later on which uses both independent
variables $x$ and $t$.

######  Theorem 5.1.

The general solution to the generalised Chazy equation with $k=3$ over the
Riemann surface $\mathbb{P}^{1}=\mathbb{C}\cup\\{\infty\\}$ is given by

(5.1) $\displaystyle
y(t)=-\frac{3}{2}\left(\frac{1}{t-t_{1}}+\frac{1}{t-t_{2}}+\frac{1}{t-t_{3}}+\frac{1}{t-t_{4}}\right)$

where the $4$ points $t_{1}$, $t_{2}$, $t_{3}$ $t_{4}$ on the Riemann surface
are subject to the constraint $\mathcal{Q}$ that $12ae-3bd+c^{2}=0$ where

$\displaystyle
a(t-t_{1})(t-t_{2})(t-t_{3})(t-t_{4})=at^{4}+bt^{3}+ct^{2}+dt+e.$

###### Proof.

Under the substitution $y=\frac{-3f^{\prime}}{2f}$ for $f$ non-zero, the
generalised Chazy equation with $k=3$ is equivalent to the nonlinear 4th order
ODE
$2f^{\prime\prime\prime\prime}f-2f^{\prime\prime\prime}f^{\prime}+(f^{\prime\prime})^{2}=0$.
Differentiating this ODE gives the linear ODE
$f^{\prime\prime\prime\prime\prime}=0$ whose solutions are given by
$f=at^{4}+bt^{3}+ct^{2}+dt+e$. Substituting this back into the 4th order ODE,
the coefficients are subject to the constraint $12ae-3bd+c^{2}=0$ and
therefore the general solution to the $k=3$ Chazy equation over $\mathbb{C}$
is given by (5.1). ∎

This general solution has been mentioned in [7] and [10]. For this parameter,
Chazy ([7] p. 347) makes the observation that assuming the four roots are
distinct, the roots $t_{1}$, $t_{2}$, $t_{3}$ and $t_{4}$ can be chosen to lie
on the sommets of the tétraèdre régulier. The condition $\mathcal{Q}$ defines
a projective variety in $\mathbb{P}^{3}$. When one of the points is
$t_{4}=\infty$, the condition on the remaining three roots to be a solution is
that they must lie on the conic
$\mathcal{C}=\\{t_{1},t_{2},t_{3}\in\mathbb{C}|t_{1}^{2}+t_{2}^{2}+t_{3}^{2}-t_{1}t_{2}-t_{1}t_{3}-t_{2}t_{3}=0\\}$.
In this situation,

$y=-2\left(\frac{1}{t-t_{1}}+\frac{1}{t-t_{2}}+\frac{1}{t-t_{3}}\right)$

satisfies the $k=2$ equation while

$y=-\frac{3}{2}\left(\frac{1}{t-t_{1}}+\frac{1}{t-t_{2}}+\frac{1}{t-t_{3}}\right)$

satisfies the $k=3$ equation. We have

######  Proposition 5.1.

The inclusion of varieties corresponds to the following isomorphism

$\left\\{\begin{tabular}[]{l}\mbox{Solutions to $k=3$ equation}\\\ \mbox{with
one pole at
$\\{\infty\\}$}\end{tabular}\right\\}\cong\left\\{\begin{tabular}[]{l}\mbox{Solutions
to $k=2$ equation}\\\ \mbox{with poles in
$\mathcal{C}$}\end{tabular}\right\\}$

where the isomorphism is given by a constant rescaling.

Determining the Schwarz functions when $k=3$ is a more complicated task than
the cases when $k=\frac{2}{3}$ and $k=2$. We present the functions in Table 3.
Observe that $s(1,\frac{1}{3},\frac{1}{3},t)$ appear both in the $k=2$ and
$k=3$ equations. Also observe that for
$s(\frac{2}{3},\frac{1}{3},\frac{1}{3},t)$ the formula for $y$ occurs twice,
with both

$\displaystyle y=\frac{{\rm d}}{{\rm
d}t}\log\frac{(s^{\prime})^{3}}{s^{2}(s-1)^{2}}\text{ and }y=\frac{{\rm
d}}{{\rm d}t}\log\frac{(s^{\prime})^{3}}{s^{\frac{5}{2}}(s-1)^{\frac{5}{2}}}\
$

satisfying the $k=3$ Chazy equation. The functions here that appear in
Schwarz’s list are $s(\frac{1}{3},\frac{1}{3},\frac{1}{2},t)$ and
$s(\frac{2}{3},\frac{1}{3},\frac{1}{3},t)$, both of which possess tetrahedral
symmetry. Again $w=e^{\frac{2\pi i}{3}}$ denotes the cube root of unity.

$(\alpha,\beta,\gamma)$ | $t(s)$ | $s(t)$ | $y(t)$
---|---|---|---
$(\frac{2}{3},\frac{2}{3},\frac{2}{3})$ | $\frac{{}_{2}F_{1}(\frac{1}{6},\frac{5}{6};\frac{5}{3};s)}{{}_{2}F_{1}(\frac{1}{6},-\frac{1}{2};\frac{1}{3};s)}s^{\frac{2}{3}}$ | $I(-2^{-\frac{2}{3}}w^{2}t)=\frac{s(t)^{2}}{4(s(t)-1)}$ |  Same as $y(t)$
|  |  where $I(t)=s(\frac{2}{3},\frac{1}{3},\frac{1}{2},t)$ |  for $s(\frac{2}{3},\frac{1}{3},\frac{1}{2},t)$
$(\frac{2}{3},\frac{1}{3},\frac{1}{3})$ | $\frac{{}_{2}F_{1}(\frac{1}{6},\frac{5}{6};\frac{4}{3};s)}{{}_{2}F_{1}(-\frac{1}{6},\frac{1}{2};\frac{2}{3};s)}s^{\frac{1}{3}}$ | $\frac{\sqrt{2}}{32}\frac{(\sqrt{t^{3}+2}+3\sqrt{2})^{3}(\sqrt{t^{3}+2}-\sqrt{2})}{(\sqrt{t^{3}+2})^{3}}$ | $-\frac{9}{2}\frac{t^{2}}{t^{3}+2}$
|  |  |  and $-\frac{6(t^{3}-4)}{t(t^{3}-16)}$
$\left(\frac{1}{3},\frac{1}{3},\frac{1}{2}\right)$ | $\frac{{}_{2}F_{1}(\frac{7}{12},\frac{1}{4};\frac{4}{3};s)}{{}_{2}F_{1}(-\frac{1}{12},\frac{1}{4};\frac{2}{3};s)}s^{\frac{1}{3}}$ | $-\frac{t^{3}(t^{3}-64)^{3}}{512(t^{3}+8)^{3}}$ | $-\frac{9}{2}\frac{t^{2}}{t^{3}+8}$
$\left(\frac{2}{3},\frac{1}{3},\frac{1}{2}\right)$ | $\frac{{}_{2}F_{1}(\frac{1}{12},\frac{3}{4};\frac{4}{3};s)}{{}_{2}F_{1}(\frac{5}{12},-\frac{1}{4};\frac{2}{3};s)}s^{\frac{1}{3}}$ | $\frac{s}{(4s-1)^{3}}=\frac{1}{512}\frac{t^{3}(27t^{3}+64)^{3}}{(27t^{3}-8)^{3}}$ | $-\frac{6}{t}\left(\frac{27t^{3}+16}{27t^{3}+64}\right)$
$\left(\frac{1}{3},\frac{1}{3},1\right)$ | $s^{\frac{1}{3}}$ | $t^{3}$ | $-\frac{9}{2}\frac{t^{2}}{t^{3}-1}$
$\left(\frac{4}{3},\frac{1}{3},\frac{1}{3}\right)$ | $\frac{{}_{2}F_{1}(\frac{7}{6},-\frac{1}{6};\frac{4}{3};s)}{{}_{2}F_{1}(-\frac{1}{2},\frac{5}{6};\frac{2}{3};s)}s^{\frac{1}{3}}$ | $I(2^{\frac{2}{3}}t)=4s(t)(1-s(t))$ |  Same as $y(t)$
|  |  where $I(t)=s(\frac{2}{3},\frac{1}{3},\frac{1}{2},t)$ |  for $s(\frac{2}{3},\frac{1}{3},\frac{1}{2},t)$
Table 3. Parametrisations for $k=3$

######  Proposition 5.2.

Let $s=s(\frac{2}{3},\frac{1}{3},\frac{1}{2},t)$. We have

(5.2)
$\displaystyle\frac{s}{(4s-1)^{3}}=\frac{1}{512}\frac{t^{3}(27t^{3}+64)^{3}}{(27t^{3}-8)^{3}}$

as written in the fourth row of Table 3.

###### Proof.

This is obtained from considering the pull back map
$(\frac{1}{3},\frac{1}{3},\frac{1}{2})\stackrel{{\scriptstyle
3}}{{\leftarrow}}(\frac{2}{3},\frac{1}{3},\frac{1}{2})$ in Figure 3. This
corresponds to the cubic transformation given in formula (121) of [15]. For
$\alpha=-\frac{1}{12}$, we have

$\displaystyle{}_{2}F_{1}\left(\frac{5}{12},-\frac{1}{4};\frac{2}{3};s\right)=(1-4s)^{\frac{1}{4}}{}_{2}F_{1}\left(-\frac{1}{12},\frac{1}{4};\frac{2}{3};\frac{27s}{(4s-1)^{3}}\right)$

while for $\alpha=\frac{1}{4}$, we have

$\displaystyle{}_{2}F_{1}\left(\frac{1}{12},\frac{3}{4};\frac{4}{3};s\right)=(1-4s)^{-\frac{3}{4}}{}_{2}F_{1}\left(\frac{7}{12},\frac{1}{4};\frac{4}{3};\frac{27s}{(4s-1)^{3}}\right).$

Together this gives

$\displaystyle\frac{{}_{2}F_{1}(\frac{1}{12},\frac{3}{4};\frac{4}{3};s)}{{}_{2}F_{1}(\frac{5}{12},-\frac{1}{4};\frac{2}{3};s)}s^{\frac{1}{3}}=-\frac{1}{3}\left(\frac{27s}{(4s-1)^{3}}\right)^{\frac{1}{3}}\frac{{}_{2}F_{1}(\frac{7}{12},\frac{1}{4};\frac{4}{3};\frac{27s}{(4s-1)^{3}})}{{}_{2}F_{1}(-\frac{1}{12},\frac{1}{4};\frac{2}{3};\frac{27s}{(4s-1)^{3}})}.$

Using the formula for the right hand side as given by the third row of Table 3
gives

$\displaystyle\frac{s}{(4s-1)^{3}}=\frac{1}{512}\frac{t^{3}(27t^{3}+64)^{3}}{(27t^{3}-8)^{3}}.$

∎

Let us now also explain how we determined the corresponding solution $y(t)$
for $s(\frac{2}{3},\frac{1}{3},\frac{1}{2},t)$, as this is not so
straightforward.

######  Proposition 5.3.

Let $s=s\left(\frac{2}{3},\frac{1}{3},\frac{1}{2},t\right)$. Then the solution
to (0.1) with $k=3$ is given by

$\displaystyle y=\frac{{\rm d}}{{\rm
d}t}\log\frac{(s^{\prime})^{3}}{s^{\frac{5}{2}}(s-1)^{\frac{3}{2}}}=-\frac{6}{t}\left(\frac{27t^{3}+16}{27t^{3}+64}\right).$

###### Proof.

From (5.2), we obtain

$\displaystyle\frac{s^{\frac{1}{3}}}{4s-1}=\frac{1}{8}\frac{t(27t^{3}+64)}{(27t^{3}-8)}.$

Let $Q$ denote the right hand side term of the above expression. We have

$\displaystyle-\frac{1}{3}\frac{8s+1}{s^{\frac{2}{3}}(4s-1)^{2}}{\rm
d}s=-\frac{1}{3}\frac{8s+1}{s(4s-1)}Q{\rm d}s=Q^{\prime}{\rm d}t,$

or alternatively,

$\displaystyle s^{\prime}=-3\frac{s(4s-1)}{8s+1}\frac{Q^{\prime}}{Q}.$

The formula for $y$ gives

$\displaystyle y=$ $\displaystyle\frac{{\rm d}}{{\rm
d}t}\log\frac{(s^{\prime})^{3}}{s^{\frac{5}{2}}(s-1)^{\frac{3}{2}}}$
$\displaystyle=$ $\displaystyle\frac{{\rm d}}{{\rm
d}t}\log\frac{\left(-3\frac{s(4s-1)}{8s+1}\frac{Q^{\prime}}{Q}\right)^{3}}{s^{\frac{5}{2}}(s-1)^{\frac{3}{2}}}$
$\displaystyle=$
$\displaystyle-3\frac{Q^{\prime}}{Q}\frac{s(4s-1)}{8s+1}\frac{{\rm d}}{{\rm
d}s}\log\frac{\left(-3\frac{s(4s-1)}{8s+1}\right)^{3}}{s^{\frac{5}{2}}(s-1)^{\frac{3}{2}}}+3\frac{{\rm
d}}{{\rm d}t}\log\frac{Q^{\prime}}{Q}$ $\displaystyle=$
$\displaystyle\frac{3}{2}\frac{Q^{\prime}}{Q}\frac{64s^{3}-48s^{2}+66s-1}{(8s+1)^{2}(s-1)}+3\frac{{\rm
d}}{{\rm d}t}\log\frac{Q^{\prime}}{Q}.$

Therefore we have

(5.3) $\displaystyle\frac{2Q}{3Q^{\prime}}\left(y-3\frac{{\rm d}}{{\rm
d}t}\log\frac{Q^{\prime}}{Q}\right)=\frac{64s^{3}-48s^{2}+66s-1}{(8s+1)^{2}(s-1)}.$

Using the fact that
$\frac{s}{(4s-1)^{3}}=\frac{1}{512}\frac{t^{3}(27t^{3}+64)^{3}}{(27t^{3}-8)^{3}}$,
we eliminate $s$ appearing in both this equation and (5.3) and get

$y=-\frac{6}{t}\left(\frac{27t^{3}+16}{27t^{3}+64}\right).$

∎

If we make the substitution $t=\frac{-4\sigma^{\frac{1}{3}}}{\sigma+2}$ and
$s=-\frac{1}{64}\frac{\sigma(\sigma+8)^{3}}{(1-\sigma)^{3}}$ into (5.2), we
get an identity. This alternative parametrisation comes from the mapping
$(\frac{2}{3},\frac{1}{3},\frac{1}{2})\stackrel{{\scriptstyle
4}}{{\leftarrow}}(\frac{2}{3},\frac{1}{3},2)$ in Figure 3 and $\sigma$ agrees
with $s(\frac{2}{3},\frac{1}{3},2,t)$ up to a constant reparametrisation of
$t$. From this, we find that $(t,y_{3})$ given by

$\displaystyle t$ $\displaystyle=\frac{\sigma^{\frac{1}{3}}}{\sigma+2}$
$\displaystyle y_{3}$
$\displaystyle=-\frac{3}{2}\frac{(\sigma+2)(\sigma^{3}+6\sigma^{2}-96\sigma+8)}{\sigma^{\frac{1}{3}}(\sigma+8)(\sigma-1)^{2}}$
$\displaystyle=-\frac{3}{2t}\left(1-\frac{8}{\sigma-1}-\frac{9}{(\sigma-1)^{2}}+\frac{8}{\sigma+8}\right)=-\frac{3}{2t}\left(\frac{1-108t^{3}}{1-27t^{3}}\right)$

solves the $k=3$ equation while $(t,y_{\frac{3}{2}})$ given by

$\displaystyle t$ $\displaystyle=\frac{\sigma^{\frac{1}{3}}}{\sigma+2}$
$\displaystyle y_{\frac{3}{2}}$
$\displaystyle=-\frac{9}{4}\frac{(\sigma+2)(\sigma-10)\sigma^{\frac{2}{3}}}{(\sigma-1)^{2}}$
$\displaystyle=-\frac{9}{4t}\left(1-\frac{8}{\sigma-1}-\frac{9}{(\sigma-1)^{2}}\right)=\frac{3}{2}y_{3}+\frac{9}{4t}\frac{8}{\sigma+8}$

solves the $k=\frac{3}{2}$ equation. The latter solution agrees with the
solution given in Section 6.1, which means that we have reparametrised the
solutions for $y(t)$ given by $s(\frac{2}{3},\frac{1}{3},\frac{1}{2},t)$ to
those given in the subsequent Section 6.1.

For brevity we denote $I(t)=s(\frac{2}{3},\frac{1}{3},\frac{1}{2},t)$,
$J(t)=s(\frac{4}{3},\frac{1}{3},\frac{1}{3},t)$ and
$K(t)=s(\frac{2}{3},\frac{2}{3},\frac{2}{3},t)$. We have found using quadratic
transformations (similar to those given in the next section) that
$I(2^{\frac{2}{3}}t)=4J(t)(1-J(t))$ and
$I(-2^{-\frac{2}{3}}w^{2}t)=\frac{K(t)^{2}}{4(K(t)-1)}$ where
$w=e^{\frac{2}{3}\pi i}$.

######  Proposition 5.4.

The functions $I$, $J$, $K$ determine the same solutions to the generalised
Chazy equation for $k=3$.

###### Proof.

We have $I(2^{\frac{2}{3}}t)=4J(t)(1-J(t))$ and
$I(-2^{-\frac{2}{3}}w^{2}t)=\frac{K(t)^{2}}{4(K(t)-1)}$. A computation shows
that

$\displaystyle\frac{(I^{\prime})^{3}}{I^{\frac{5}{2}}(I-1)^{\frac{3}{2}}}=c_{1}\frac{(J^{\prime})^{3}}{J^{\frac{5}{2}}(J-1)^{\frac{3}{2}}}=c_{2}\frac{(K^{\prime})^{3}}{K^{2}(K-1)^{2}}$

for some possible complex constants $c_{1}$ and $c_{2}$. By choosing the same
logarithmic branch, the logarithmic derivatives of these three functions give
the same solution to the $k=3$ Chazy equation. ∎

##  6\. Generalised Chazy equation with $k=\frac{3}{2}$ and its Schwarz
functions

In this section we present the Schwarz triangle functions that show up in the
solutions to the $k=\frac{3}{2}$ equation. The Schwarz triangle functions that
appear can be grouped into those that have already shown up in the the
solutions to the $k=3$ equation and those that have not. The additional ones
are presented in Table 4. The Schwarz triangle functions $I$, $J$ and $K$ have
already shown up in the solutions to the $k=3$ equation and they solve the
$k=\frac{3}{2}$ equation (see the previous section) by deforming the $k=3$
solutions along some function of $K$ (or $I$, $J$).

######  Proposition 6.1.

Let $K(t)=s(\frac{2}{3},\frac{2}{3},\frac{2}{3},t)$. We have

$\displaystyle\frac{{\rm d}}{{\rm
d}t}\log\frac{(K^{\prime})^{3}}{K^{\frac{5}{2}}(K-1)},\hskip 12.0pt\frac{{\rm
d}}{{\rm
d}t}\log\frac{(K^{\prime})^{3}}{K(K-1)^{\frac{5}{2}}}\mbox{~{}and~{}}\frac{{\rm
d}}{{\rm d}t}\log\frac{(K^{\prime})^{3}}{K^{\frac{5}{2}}(K-1)^{\frac{5}{2}}}$

all satisfying Chazy’s equation with $k=\frac{3}{2}$. This comes from
considering the different combinations $y=-4\Omega_{1}-\Omega_{2}-\Omega_{3}$,
$y=-\Omega_{1}-4\Omega_{2}-\Omega_{3}$ and
$y=-\Omega_{1}-\Omega_{2}-4\Omega_{3}$, discussed in [21]. In all three cases
the function $K$ remains the same since its triangular domain has the property
of being equilateral.

Let us denote $y_{3}=\frac{{\rm d}}{{\rm
d}t}\log\frac{(K^{\prime})^{3}}{K^{2}(K-1)^{2}}$. This is the solution to the
$k=3$ equation in the previous section.

######  Corollary 6.1.

The functions $y_{3}+\frac{{\rm d}}{{\rm d}t}\log\frac{K-1}{\sqrt{K}}$,
$y_{3}+\frac{{\rm d}}{{\rm d}t}\log\frac{K}{\sqrt{K-1}}$ and $y_{3}+\frac{{\rm
d}}{{\rm d}t}\log\frac{1}{\sqrt{K(K-1)}}$ are all solutions to Chazy’s
equations with $k=\frac{3}{2}$.

In the above expressions, $K$ can be substituted for $I$ or $J$ as well.

The remaining Schwarz functions that have not shown up in the $k=3$ case are
given by $s(\frac{4}{3},\frac{2}{3},\frac{1}{2},t)$,
$s(\frac{4}{3},\frac{4}{3},\frac{4}{3},t)$,
$s(\frac{8}{3},\frac{2}{3},\frac{2}{3},t)$ and
$s(\frac{2}{3},\frac{1}{3},2,t)$. We discuss $s(\frac{2}{3},\frac{1}{3},2,t)$
in Section 6.1. Again for brevity let us denote
$L(t)=s(\frac{4}{3},\frac{2}{3},\frac{1}{2},t)$,
$M(t)=s(\frac{4}{3},\frac{4}{3},\frac{4}{3},t)$ and
$N(t)=s(\frac{8}{3},\frac{2}{3},\frac{2}{3},t)$. The relationship between $L$,
$M$ and $N$ are presented in Table 4. We determine them in the following
fashion, using only quadratic transformations. The inversion formula for $L$
gives

$\displaystyle
t(L)=\frac{{}_{2}F_{1}(-\frac{1}{12},\frac{5}{4};\frac{5}{3};L)}{{}_{2}F_{1}(-\frac{3}{4},\frac{7}{12};\frac{1}{3};L)}L^{\frac{2}{3}}.$

We find

$\displaystyle\frac{{}_{2}F_{1}(-\frac{5}{6},\frac{11}{6};\frac{5}{3};s)}{{}_{2}F_{1}(-\frac{3}{2},\frac{7}{6};\frac{1}{3};s)}s^{\frac{2}{3}}=2^{-\frac{4}{3}}\frac{{}_{2}F_{1}(-\frac{1}{12},\frac{5}{4};\frac{5}{3};4s(1-s))}{{}_{2}F_{1}(-\frac{3}{4},\frac{7}{12};\frac{1}{3};4s(1-s))}(4s(1-s))^{\frac{2}{3}}.$

The left hand side is the inversion formula for $N(t)$, while the right hand
side is $2^{-\frac{4}{3}}t(4N(1-N))$ and so

$\displaystyle L(2^{\frac{4}{3}}t)=4N(t)(1-N(t)).$

Similarly, we find

$\displaystyle\frac{{}_{2}F_{1}(-\frac{1}{6},\frac{7}{6};\frac{7}{3};s)}{{}_{2}F_{1}(-\frac{1}{6},-\frac{3}{2};-\frac{1}{3};s)}s^{\frac{4}{3}}=2^{\frac{4}{3}}w^{2}\frac{{}_{2}F_{1}(-\frac{1}{12},\frac{5}{4};\frac{5}{3};\frac{s^{2}}{4(s-1)})}{{}_{2}F_{1}(-\frac{3}{4},\frac{7}{12};\frac{1}{3};\frac{s^{2}}{4(s-1)})}\left(\frac{s^{2}}{4(s-1)}\right)^{\frac{2}{3}},$

where $w=e^{\frac{2}{3}\pi i}$. The left hand side is the inversion formula
for $M(t)$, while the right hand side is
$2^{\frac{4}{3}}w^{2}t(\frac{M^{2}}{4(M-1)})$ and so

$\displaystyle L(2^{-\frac{4}{3}}wt)=\frac{M(t)^{2}}{4(M(t)-1)}.$

$(\alpha,\beta,\gamma)$ | $t(s)$ | $s(t)$
---|---|---
$(\frac{4}{3},\frac{4}{3},\frac{4}{3})$ | $\frac{{}_{2}F_{1}(-\frac{1}{6},\frac{7}{6};\frac{7}{3};s)}{{}_{2}F_{1}(-\frac{1}{6},-\frac{3}{2};-\frac{1}{3};s)}s^{\frac{4}{3}}$ | $L(2^{-\frac{4}{3}}wt)=\frac{s(t)^{2}}{4(s(t)-1)}$
$\left(\frac{4}{3},\frac{2}{3},\frac{1}{2}\right)$ | $\frac{{}_{2}F_{1}(-\frac{1}{12},\frac{5}{4};\frac{5}{3};s)}{{}_{2}F_{1}(-\frac{3}{4},\frac{7}{12};\frac{1}{3};s)}s^{\frac{2}{3}}$ | $L(t)$
$\left(\frac{8}{3},\frac{2}{3},\frac{2}{3}\right)$ | $\frac{{}_{2}F_{1}(-\frac{5}{6},\frac{11}{6};\frac{5}{3};s)}{{}_{2}F_{1}(-\frac{3}{2},\frac{7}{6};\frac{1}{3};s)}s^{\frac{2}{3}}$ | $L(2^{\frac{4}{3}}t)=4s(t)(1-s(t))$
$\left(\frac{2}{3},\frac{1}{3},2\right)$ | $\frac{2s^{\frac{1}{3}}}{s+2}$ | Roots of the cubic polynomial
|  | $s^{3}+6s^{2}+(12-\frac{8}{t^{3}})s+8=0$
Table 4. Parametrisations for $k=\frac{3}{2}$

######  Proposition 6.2.

The functions $L$, $M$, $N$ determine the same solutions to the generalised
Chazy equation for $k=\frac{3}{2}$.

###### Proof.

Similar to Proposition 5.4, a computation shows that

$\displaystyle\frac{(L^{\prime})^{3}}{L^{\frac{5}{2}}(L-1)^{\frac{3}{2}}}=c_{3}\frac{(M^{\prime})^{3}}{M^{2}(M-1)^{2}}=c_{4}\frac{(N^{\prime})^{3}}{N^{\frac{5}{2}}(N-1)^{\frac{5}{2}}}$

again for some possible complex constants $c_{3}$ and $c_{4}$. Again by
choosing the same logarithmic branch, the logarithmic derivatives of these
three functions give the same solution to the $k=\frac{3}{2}$ Chazy equation.
∎

We are now left with the problem of determining what

$\displaystyle y(t)=\frac{{\rm d}}{{\rm
d}t}\log\frac{(L^{\prime})^{3}}{L^{\frac{5}{2}}(L-1)^{\frac{3}{2}}}$

is. We shall show that this can be reparametrised to give the solution (6.2)
below.

######  Proposition 6.3.

Let $\tau=s(\frac{1}{3},\frac{2}{3},2,\frac{9t}{16})$ and
$L=s(\frac{4}{3},\frac{2}{3},\frac{1}{2},t)$. We have

(6.1)
$\displaystyle\frac{L^{\frac{1}{3}}}{4L-1}=\frac{4}{3}\tau^{\frac{1}{3}}\left(\frac{1-\tau}{1+8\tau}\right).$

###### Proof.

We first apply the cubic transformation
$(\frac{1}{3},\frac{2}{3},\frac{1}{2})\stackrel{{\scriptstyle
3}}{{\leftarrow}}(\frac{4}{3},\frac{2}{3},\frac{1}{2})$. Let
$v=\frac{27L}{(1+8L)^{2}(1-L)}$. We find
$\frac{v}{v-1}=\frac{27L}{(4L-1)^{3}}$. The formulas (121) of [15] give

$\displaystyle{}_{2}F_{1}\left(-\frac{3}{4},\frac{7}{12};\frac{1}{3};L\right)=(1-4L)^{\frac{3}{4}}{}_{2}F_{1}\left(-\frac{1}{4},\frac{1}{12};\frac{1}{3};\frac{27L}{(4L-1)^{3}}\right).$

for $\alpha=-\frac{1}{4}$ and

$\displaystyle{}_{2}F_{1}\left(-\frac{1}{12},\frac{5}{4};\frac{5}{3};L\right)=(1-4L)^{-\frac{5}{4}}{}_{2}F_{1}\left(\frac{5}{12},\frac{3}{4};\frac{5}{3};\frac{27L}{(4L-1)^{3}}\right).$

for $\alpha=\frac{5}{12}$. Together this gives the inversion formula for $L$

$\displaystyle
t=\frac{{}_{2}F_{1}(-\frac{1}{12},\frac{5}{4};\frac{5}{3};L)}{{}_{2}F_{1}(-\frac{3}{4},\frac{7}{12};\frac{1}{3};L)}L^{\frac{2}{3}}=$
$\displaystyle\frac{1}{9}\frac{{}_{2}F_{1}(\frac{5}{12},\frac{3}{4};\frac{5}{3};\frac{v}{v-1})}{{}_{2}F_{1}(-\frac{1}{4},\frac{1}{12};\frac{1}{3};\frac{v}{v-1})}\left(\frac{v}{1-v}\right)^{\frac{2}{3}}.$

Now we apply the degree $4$ transformation
$(\frac{1}{3},\frac{2}{3},\frac{1}{2})\stackrel{{\scriptstyle
4}}{{\leftarrow}}(\frac{1}{3},\frac{2}{3},2)$. Let
$w=\frac{64\tau(1-\tau)^{3}}{(1+8\tau)^{3}}$. We find using equation (127) of
[15] that the same parameter $\alpha=-\frac{1}{4}$ gives

$\displaystyle{}_{2}F_{1}\left(-\frac{1}{4},\frac{1}{12};\frac{1}{3};w\right)=(1+8\tau)^{-\frac{3}{4}}{}_{2}F_{1}\left(-1,-\frac{2}{3};\frac{1}{3};\tau\right)$

and $\alpha=\frac{5}{12}$ gives

$\displaystyle{}_{2}F_{1}\left(\frac{3}{4},\frac{5}{12};\frac{5}{3};w\right)=(1+8\tau)^{\frac{5}{4}}{}_{2}F_{1}\left(\frac{5}{3},2;\frac{5}{3};\tau\right).$

We therefore obtain

$\displaystyle\frac{{}_{2}F_{1}(\frac{3}{4},\frac{5}{12};\frac{5}{3};w)}{{}_{2}F_{1}(-\frac{1}{4},\frac{1}{12};\frac{1}{3};w)}w^{\frac{2}{3}}=$
$\displaystyle\frac{{}_{2}F_{1}(\frac{5}{3},2;\frac{5}{3};\tau)(1+8\tau)^{\frac{5}{4}}}{{}_{2}F_{1}(-1,-\frac{2}{3};\frac{1}{3};\tau)(1+8\tau)^{-\frac{3}{4}}}\left(\frac{4\tau^{\frac{1}{3}}(1-\tau)}{1+8\tau}\right)^{2}$
$\displaystyle=$ $\displaystyle\frac{16\tau^{\frac{2}{3}}}{1+2\tau}.$

Equating $w=\frac{v}{v-1}$, we obtain

$\displaystyle\frac{64\tau(1-\tau)^{3}}{(1+8\tau)^{3}}=\frac{27L}{(4L-1)^{3}}$

and the parameters are related by

$\displaystyle\frac{9t}{16}=\frac{\tau^{\frac{2}{3}}}{(1+2\tau)}$

where the right hand side is the inverse
$\tau=s(\frac{1}{3},\frac{2}{3},2,\frac{9t}{16})$. ∎

Let $R(\tau(t))$ denote the right hand side of (6.1). We have
$t=\frac{16\tau^{\frac{2}{3}}}{9(1+2\tau)}$ and

$\displaystyle L^{\prime}=-3\frac{L(4L-1)}{8L+1}\frac{R^{\prime}}{R}.$

Analogous to Proposition 5.3, we find that

$\displaystyle\frac{2R}{3R^{\prime}}\left(y-3\frac{{\rm d}}{{\rm
d}t}\log\frac{R^{\prime}}{R}\right)=\frac{64L^{3}-48L^{2}+66L-1}{(8L+1)^{2}(L-1)}.$

Eliminating $L$ between this equation and (6.1) give us the solution

(6.2)
$\displaystyle(t,y)=\left(\frac{16\tau^{\frac{2}{3}}}{9(1+2\tau)},\frac{81}{64}\frac{(1+2\tau)(10\tau-1)}{\tau^{\frac{2}{3}}(\tau-1)^{2}}\right).$

Using the formula $H=\int\int e^{\frac{2}{3}\int y{\rm d}t}{\rm d}t{\rm d}t$,
we obtain

$\displaystyle(t,H)=\left(\frac{16\tau^{\frac{2}{3}}}{9(1+2\tau)},\frac{1024}{81}\frac{\tau^{\frac{1}{3}}}{1+2\tau}\right).$

The dual curves are parametrised by

$\displaystyle(x,F)=\left(\frac{32(4\tau-1)}{9\tau^{\frac{1}{3}}(\tau-1)},\frac{512}{81}\frac{\tau^{\frac{1}{3}}}{\tau-1}\right).$

###  6.1. The Schwarz function $s(\frac{2}{3},\frac{1}{3},2,t)$

For the Schwarz function $s(\frac{2}{3},\frac{1}{3},2,t)$, the parametrisation
for $t$ is found to be given by

$t=\frac{2s^{\frac{1}{3}}}{(s+2)}$

while the formula for $H(t(s))$ is found to be given by

$H(t(s))=-\frac{4s^{\frac{2}{3}}}{s+2}.$

Let us explain how we determine this. For the values of
$(\alpha,\beta,\gamma)=(\frac{2}{3},\frac{1}{3},2)$, we find
$(a,b,c)=(-1,-\frac{1}{3},\frac{2}{3})$. The general solution to
$u_{ss}+\frac{1}{4}V(s)u=0$ with these corresponding values is given by
$u=\beta s^{\frac{1}{3}}+\alpha(s+2)$. Let $u_{1}=(s+2)$ and
$u_{2}=2s^{\frac{1}{3}}$. Then

$t=\frac{2s^{\frac{1}{3}}}{(s+2)}=\frac{s^{\frac{1}{3}}}{{}_{2}F_{1}(-1,-\frac{1}{3};\frac{2}{3};s)}$

agrees with the formula given by (2.2) and we find that

$\displaystyle y=\frac{{\rm d}}{{\rm
d}t}\log\frac{(s^{\prime})^{3}}{s^{2}(s-1)^{\frac{3}{2}}}=-\frac{9}{8}\frac{(s+2)(s-10)}{(s-1)^{2}}s^{\frac{2}{3}}.$

We thus find that $s(t)$ is given by the root of the cubic equation

$\displaystyle s^{3}+6s^{2}+\left(12-\frac{8}{t^{3}}\right)s+8=0.$

Eliminating the variable $s$, we obtain a polynomial algebraic curve $C(t,y)$
given by

$\displaystyle C(t,y)=$
$\displaystyle(2t-3)^{2}(4t^{2}+6t+9)^{2}y^{3}+18t(2t-3)(2t^{3}-27)(4t^{2}+6t+9)y^{2}$
$\displaystyle+324t^{2}(t^{6}-45t^{3}+243)y-1458t^{3}(5t^{3}-108)=0.$

We find that

$\displaystyle{\rm d}t=-\frac{4}{3}\frac{s-1}{(s+2)^{2}s^{\frac{2}{3}}}{\rm
d}s$

and therefore

$\displaystyle H=\int\int e^{\frac{2}{3}\int y{\rm d}t}{\rm d}t{\rm
d}t=-\frac{4s^{\frac{2}{3}}}{s+2}.$

We can eliminate $s$ to get

$\displaystyle H^{3}-16t^{3}-8Ht=0.$

Using the formula $(x,F)=(H^{\prime},tH^{\prime}-H)$, we find that the dual
curves are given by

$\displaystyle(x,F)=\left(-\frac{s-4}{s-1}s^{\frac{1}{3}},\frac{2s^{\frac{2}{3}}}{s-1}\right).$

To summarise the results of Section 6, we have the following

######  Theorem 6.1.

The functions $I$, $J$ $K$ give rise to the curve $(t,H)$ parametrised (up to
constants) by

(6.3)
$\displaystyle(t,H)=\left(\frac{2s^{\frac{1}{3}}}{s+2},-\frac{4s^{\frac{2}{3}}}{s+2}\right)$

where $s=s(\frac{2}{3},\frac{1}{3},2,t)$ and the dual curve

$\displaystyle(x,F)=\left(-\frac{s-4}{s-1}s^{\frac{1}{3}},\frac{2s^{\frac{2}{3}}}{s-1}\right)$

determines a flat $(2,3,5)$-distribution $\mathcal{D}_{F(x)}$. The functions
$L$, $M$ $N$ give rise to the curve $(t,H)$ parametrised (up to constants) by

$\displaystyle(t,H)=\left(\frac{2s^{\frac{2}{3}}}{1+2s},-\frac{4s^{\frac{1}{3}}}{1+2s}\right)$

where this is obtained by inverting $s\mapsto\frac{1}{s}$ in (6.3). The dual
curve given by

$\displaystyle(x,F)=\left(-\frac{4s-1}{s^{\frac{1}{3}}(s-1)},\frac{2s^{\frac{1}{3}}}{1-s}\right)$

also determines a flat $(2,3,5)$-distribution $\mathcal{D}_{F(x)}$.

Figure 3 shows the transformation maps between the Schwarz triangle functions
that show up in the $k=\frac{3}{2}$ and $k=3$ solutions. The Schwarz functions
$s(\frac{1}{2},\frac{1}{3},\frac{2}{3},x)$,
$s(\frac{2}{3},\frac{2}{3},\frac{2}{3},x)$ and
$s(\frac{4}{3},\frac{1}{3},\frac{1}{3},x)$ appear in the solutions to the
$k=3$ and $k=\frac{3}{2}$ equations. The three diagrams 1, 2 and 3 can be
combined at the nodes labelled by $(\frac{1}{3},\frac{1}{3},1)$ and
$\left(\frac{3}{2},\frac{1}{3},\frac{1}{2}\right)$.

Figure 3. Mapping of Schwarz functions for $k=\frac{3}{2}$ and $k=3$

We end by discussing the shapes of the spherical triangles that show up in the
$k=\frac{3}{2}$ and $k=3$ cases. The spherical triangle with angles
$(\frac{1}{2}\pi$, $\frac{1}{3}\pi$, $\frac{2}{3}\pi)$ corresponding to
$s(\frac{1}{2},\frac{1}{3},\frac{2}{3},x)$ is given by the following. Divide
the hemisphere equally into three lunes, with each end having angles
$\frac{\pi}{3}$. The domain for the
$(\frac{1}{2}\pi,\frac{1}{3}\pi,\frac{2}{3}\pi)$ triangle is the complement of
the fundamental domain of the $(\frac{1}{2}\pi,\frac{1}{3}\pi,\frac{1}{3}\pi)$
triangle with tetrahedral symmetry in this lune. Eight of these triangles tile
the whole sphere. The triangle with angles
$(\frac{2}{3}\pi,\frac{2}{3}\pi,\frac{2}{3}\pi)$ is generated by reflecting
this domain along the long edge meeting the right angle, while the triangle
with angles $(\frac{1}{3}\pi,\frac{1}{3}\pi,\frac{4}{3}\pi)$ is generated by
reflection along the short edge meeting the right angle.

When $k=\frac{3}{2}$, the triangle with angles $(\frac{1}{2}\pi$,
$\frac{2}{3}\pi,\frac{4}{3}\pi)$ also show up. This is the complement of the
$(\frac{1}{2}\pi$, $\frac{1}{3}\pi$, $\frac{2}{3}\pi)$ triangle in the
hemisphere with the edge between the $\frac{1}{2}\pi$ and $\frac{1}{3}\pi$
angles lying along the equator. The other triangle with angles
$(\frac{4}{3}\pi$, $\frac{4}{3}\pi,\frac{4}{3}\pi)$ is generated by reflecting
along the long edge meeting the right angle (or equivalently, the equator).
This is also the complement of the equilateral triangle with angles
$(\frac{2}{3}\pi$, $\frac{2}{3}\pi$, $\frac{2}{3}\pi)$ in the whole sphere.
Reflecting along the short edge meeting the right angle gives the “triangle”
with angles $(\frac{8}{3}\pi,\frac{2}{3}\pi,\frac{2}{3}\pi)$, which overlaps
at the vertex with angle $\frac{8}{3}\pi$.

##  Appendix A Solving quartic equation

The degree four polynomial equation

$s^{4}-2s^{3}-4xs+2x=0$

can be solved by radicals for $s$ in terms of $x$. Substituting
$s=\xi+\frac{1}{2}$, we bring it to the form

(A.1) $\xi^{4}-\frac{3}{2}\xi^{2}-(4x+1)\xi-\frac{3}{16}=0.$

From the coefficients of (A.1) we can form the resolvent cubic

(A.2) $\left(u+\frac{3}{2}\right)\left(u^{2}+\frac{3}{4}\right)=(4x+1)^{2}$

and solve for $u$ to obtain

$\displaystyle u=2(x(2x+1))^{\frac{1}{3}}-\frac{1}{2},\hskip
12.0pt\left(-\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)2(x(2x+1))^{\frac{1}{3}}-\frac{1}{2},\hskip
12.0pt\left(-\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)2(x(2x+1))^{\frac{1}{3}}-\frac{1}{2}$

as its roots. For $A=\sqrt{u+\frac{3}{2}}$ where $u$ is a solution to (A.2),
we form $P=\xi^{2}+\frac{1}{2}u$ and $Q=A\xi+\frac{4x+1}{2A}$. Then
$P^{2}=Q^{2}$ iff (A.1) holds, but now $P=\pm Q$ is quadratic in $\xi$ and so
the roots can be found using the quadratic formula. For

$u_{0}=(8x(2x+1))^{\frac{1}{3}}-\frac{1}{2},$

the roots for $\xi$ are given by:

$\displaystyle\xi=$
$\displaystyle\frac{1}{2\sqrt{(8x(2x+1))^{\frac{1}{3}}+1}}\left((8x(2x+1))^{\frac{1}{3}}+1\right)$
$\displaystyle\pm$
$\displaystyle\frac{\sqrt{-(8x(2x+1))^{\frac{2}{3}}+(8x(2x+1))^{\frac{1}{3}}+2(4x+1)\sqrt{(8x(2x+1))^{\frac{1}{3}}+1}+2}}{2\sqrt{(8x(2x+1))^{\frac{1}{3}}+1}},$
$\displaystyle\xi=$
$\displaystyle-\frac{1}{2\sqrt{(8x(2x+1))^{\frac{1}{3}}+1}}\left(8x(2x+1))^{\frac{1}{3}}+1\right)$
$\displaystyle\pm$
$\displaystyle\frac{\sqrt{-(8x(2x+1))^{\frac{2}{3}}+(8x(2x+1))^{\frac{1}{3}}-2(4x+1)\sqrt{(8x(2x+1))^{\frac{1}{3}}+1}+2}}{2\sqrt{(8x(2x+1))^{\frac{1}{3}}+1}}.$

##  Appendix B Duality of generalised Chazy equations

There is a Legendre duality [3] that takes the generalised Chazy equation with
parameters $k=\frac{2}{3}$ to its dual with parameters $k=\frac{3}{2}$. This
is explained in [3] and [21]. We show that the generalised Chazy equation that
is Legendre dual to another generalised Chazy equation has parameter $k$ given
either by $\pm\frac{2}{3}$ or $\pm\frac{3}{2}$.

######  Proposition B.1.

Let $m=\frac{4}{36-k^{2}}$ in equation (0.1). Any generalised Chazy equation
can be put into the form

(B.1) $\displaystyle
f^{3}f^{\prime\prime\prime\prime}-2(\ell+2)f^{2}f^{\prime}f^{\prime\prime\prime}+3(-12\ell
m+\ell-1)f^{2}(f^{\prime\prime})^{2}$ $\displaystyle+12(\ell^{2}m+6\ell
m+1)f(f^{\prime})^{2}f^{\prime\prime}-(\ell^{3}m+12\ell^{2}m+36\ell
m+\ell+6)(f^{\prime\prime})^{4}=0$

using the substitution $y=\ell\frac{f^{\prime}}{f}$ for $\ell$ non-zero.

To pass to the dual equation, we use the substitution $f=\frac{1}{h}$ and
$\frac{{\rm d}}{{\rm d}x}=\frac{1}{h}\frac{{\rm d}}{{\rm d}t}$ to determine
$f$, $f^{\prime}$, $f^{\prime\prime}$, $f^{\prime\prime\prime}$ and
$f^{\prime\prime\prime\prime}$ in terms of $h$ and its derivatives with
respect to $t$ and we obtain

$\displaystyle
h^{3}h^{\prime\prime\prime\prime}+(2\ell-11)h^{2}h^{\prime}h^{\prime\prime\prime}+(36\ell
m-3\ell-7)h^{2}(h^{\prime\prime})^{2}$ $\displaystyle+(12\ell^{2}m-144\ell
m-2\ell+59)h(h^{\prime})^{2}h^{\prime\prime}+(\ell^{3}m-24\ell^{2}m+144\ell
m+4\ell-48)(h^{\prime\prime})^{4}=0.$

Hence any equation of the form

$\displaystyle
h^{3}h^{\prime\prime\prime\prime}+(2j-11)h^{2}h^{\prime}h^{\prime\prime\prime}+(36jn-3j-7)h^{2}(h^{\prime\prime})^{2}$
$\displaystyle+(12j^{2}n-144jn-2j+59)h(h^{\prime})^{2}h^{\prime\prime}+(j^{3}n-24j^{2}n+144jn+4j-48)(h^{\prime\prime})^{4}=0$

is a Chazy equation only if the coefficients agree with that in equation
(B.1). In this case, $n$ and $m$ is determined completely and is given by
either $\frac{16}{135}$ or $\frac{9}{80}$. This gives $k=\pm\frac{2}{3}$ or
$k=\pm\frac{3}{2}$.

## Acknowledgements

The author would like to thank the anonymous reviewers for criticisms of the
earlier manuscript.

## References

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# Anomalous measurements:
recent results deepen flavour puzzle

Davide Lancierini1, Dan Moise2
1Physik-Institut, Universität Zürich, Zürich, Switzerland
2Imperial College London, London, United Kingdom

(21st June 2021)

###### Abstract

The crown jewel of particle physics, the Standard Model (SM), has withstood
numerous experimental trials. However, there are still some observations it
cannot explain. Examples such as dark matter and the matter-antimatter
imbalance in the Universe come to mind. The SM may be extended, by including
additional particles and interactions, so as to explain such phenomena. These
new particles and interactions are collectively referred to as “new physics”
(NP), and the results covered by this article provide a promising lead for
their discovery.

Invited contribution to the newsletter of the CERN EP Department, June 2021
(reformatted with minimal changes for submission to arXiv)
https://ep-news.web.cern.ch/node/3265

##  1 Flavour anomalies

In the past decade, a pattern has been emerging in the study of $b$-quark
decays with leptons in the final state. They are collectively referred to as
“flavour anomalies”, and they typically feature tensions at the level of
$2$–$3$ standard deviations between experimental results and SM predictions.
As such, individual anomalies are not sufficiently significant to claim
discovery of NP. However, the anomalies are often treated collectively in an
Effective Field Theory (EFT) framework, whereby short distance contributions
are separated from their long distance counterparts. This is similar to the
$4$-point approximation used to describe beta decay, where the process is
observed at long enough distances to regard the $W$ boson propagator as point-
like. The EFT approach leads to the formulation of an effective Lagrangian,
$\mathcal{L}_{\rm eff}=\sum_{i}C_{i}O_{i}$, where large energy-scale effects
are encoded in the so-called Wilson coefficients $C_{i}$, and small energy-
scale contributions are accounted for by the Wilson operators $O_{i}$. Given
the absence of direct discoveries at the LHC, NP is assumed to be
characterised by large (above TeV) energy scales. The anomalous flavour
observables could then be impacted indirectly, since NP particles may manifest
themselves virtually. From an EFT prespective, this would lead to values of
the $C_{i}$ coefficients that differ from the SM values.

Two examples of EFT analyses of flavour anomalies are shown in Figure 1 [1, 2,
3]. In the plot on the left-hand side, the $1\,\sigma$ and $2\,\sigma$
confidence levels for the best-fit point for the considered anomalies are
shown in red. On the right-hand side, the best-fit $1\,\sigma$, $2\,\sigma$,
and $3\,\sigma$ confidence levels are shown in green. The two analyses make
different assumptions on the presence of NP in the coefficients $C_{i}$ (as
shown by the axes), however they both find that the flavour anomalies prefer
$C_{i}$ values that are significantly different from the ones predicted by the
SM (the origin in each plot).

\begin{overpic}[width=160.43727pt]{Stangl-LaThuile2021-C9C10}
\put(70.5,37.7){$\star$ {\scriptsize SM}}
\end{overpic}\begin{overpic}[width=242.82613pt]{Kriewald-2020-C9univ-old}
\end{overpic} Figure 1: Examples of EFT fits to anomalous flavour observables
[1, 2, 3], under different NP scenarios. Blue confidence regions show values
preferred by flavour anomalies that are theoretically cleaner compared to
other anomalies, whose preferred values are depicted in orange. Confidence
regions preferred by all considered flavour anomalies are shown in red and
green.

##  2 Clean observables

There are several types of anomalous flavour observables, such as differential
branching fractions and angular coefficients. They are predicted in the SM
with various degrees of precision, and so particularly important are the
flavour anomalies that are theoretically clean. Recently, at the Electroweak
session of the Moriond 2021 conference, the LHCb collaboration has presented
the most precise individual measurements to date of two such theoretically-
clean anomalous observables. The first one is the ratio
${R_{K}}=\mathcal{B}(B^{\pm}\to K^{\pm}\mu^{+}\mu^{-})/\mathcal{B}(B^{\pm}\to
K^{\pm}e^{+}e^{-})$ [4]. The second one is the branching fraction of
$B_{s}^{0}\rightarrow\mu^{+}\mu^{-}$, including a search for
$B^{0}\rightarrow\mu^{+}\mu^{-}$ and
$B_{s}^{0}\rightarrow\mu^{+}\mu^{-}\gamma$ [5, 6]. The processes involved are
examples of Flavour Changing Neutral Currents (FCNCs), which are forbidden at
tree-level in the SM. As a result, the leading-order Feynman diagrams contain
loops, as exemplified by Figure 2. The loops introduce additional vertices
that make these processes rare. The branching fraction of
$B_{s}^{0}\rightarrow\mu^{+}\mu^{-}$ is predicted by the SM with sub-percent
precision, thanks to the purely-leptonic final state:
$\mathcal{B}(\mbox{$B_{s}^{0}\rightarrow\mu^{+}\mu^{-}$})=(3.66\pm 0.14)\times
10^{-9}$ [7]. The two $B^{\pm}\rightarrow K^{\pm}\ell^{+}\ell^{-}$ processes
upon which $R_{K}$ is built have branching fractions of $\mathcal{O}(10^{-6})$
[8]. They differ only by the leptons in the final state, which means $R_{K}$
can be predicted accurately by virtue of Lepton Flavour Universality (LFU).
This is an accidental symmetry of the SM, whereby the different lepton
flavours couple in the same way to vector bosons. As a result, $R_{K}$ is
predicted to be approximately $1$, with small deviations induced by phase
space differences and QED corrections [9, 10].

Figure 2: Examples of lowest-order Feynman diagrams allowed in the SM for
$B^{\pm}\rightarrow K^{\pm}\ell^{+}\ell^{-}$ (top) and
$B_{s}^{0}\rightarrow\mu^{+}\mu^{-}$ (bottom). Since FCNCs are forbidden at
tree-level, each of these diagrams contains a loop.

Despite the unambiguous SM predictions, experimental results on
$B_{(s)}^{0}\rightarrow\mu^{+}\mu^{-}$ and $R_{K}$ were exhibiting tensions
above $2\,\sigma$ before the Moriond 2021 conference, as shown in Figure 3.
The LHCb measurements shown here used $5\,{\rm fb}^{-1}$ of proton-proton
collision data, corresponding to approximately half of the currently available
dataset. Given the bigger picture in the context of the flavour anomalies,
updates to $B_{(s)}^{0}\rightarrow\mu^{+}\mu^{-}$ and $R_{K}$ based on the
full LHCb data set are crucial to better understanding whether there is NP in
rare $b$-decays.

\begin{overpic}[width=231.98828pt]{Bsdmumu-
Combo}\end{overpic}\begin{overpic}[width=197.29788pt]{RK2019}\end{overpic}
Figure 3: Experimental status of $B_{(s)}^{0}\rightarrow\mu^{+}\mu^{-}$ (left)
and $R_{K}$ (right) before the Moriond 2021 conference. Shown on the left are
$1$–$5\,\sigma$ confidence regions (shades of blue) for the combination [11]
of $B_{s}^{0}\rightarrow\mu^{+}\mu^{-}$ and $B^{0}\rightarrow\mu^{+}\mu^{-}$
measurements from ATLAS [12], CMS [13], and LHCb [14], alongside the SM
prediction (red). Shown on the right are results on $R_{K}$ from LHCb [15, 16]
(black and grey), BaBar [17] (green), and Belle [18] (blue); the latter has
since been updated [19]. The SM prediction is shown as a dashed line at
${R_{K}}=1$.

##  3 Measurement of $R_{K}$

Highly anticipated updates to the $R_{K}$ and
$B_{(s)}^{0}\rightarrow\mu^{+}\mu^{-}$ measurements, now including the full
dataset of $9\,{\rm fb}^{-1}$ of proton-proton collisions collected by LHCb,
were presented at the Moriond 2021 conference. At the core of the $R_{K}$
measurement is the comparison of selection efficiencies between electrons and
muons. The masses of the two leptons differ by two orders of mangitude, which
leads to them interacting differently with the detector. Muons traverse the
LHCb detector almost undisturbed, before finally being stopped by the muon
stations. However, electrons are subject to significant loss of energy to
bremsstrahlung radiation, leading to worse momentum and mass resolution
compared to muons. In order to suppress detection effects that are
systematically different between the two lepton flavours, $R_{K}$ is measured
as a double ratio:

$\displaystyle\begin{split}{R_{K}}&=\left.\frac{N(K^{\pm}\mu\mu)}{N(K^{\pm}ee)}\frac{\varepsilon(K^{\pm}ee)}{\varepsilon(K^{\pm}\mu\mu)}\middle/\underbrace{\frac{N(K^{\pm}J/\psi(\mu\mu))}{N(K^{\pm}J/\psi(ee))}\frac{\varepsilon(K^{\pm}J/\psi(ee))}{\varepsilon(K^{\pm}J/\psi(\mu\mu))}}_{r_{J/\psi}}\right.\,,\end{split}$
(1)

where $N(X)$ and $\varepsilon(X)$ represent respectively the yields and
efficiencies of selecting the decay of a $B^{+}$ meson into X. The $J/\psi$
modes upon which the single ratio $r_{{J/\psi}}$ is built are used to
calibrate efficiencies and conduct cross-checks. These modes are characterised
by high statistics, and are known to respect LFU [8]. In addition, the
$B^{\pm}\rightarrow K^{\pm}J/\psi(\ell^{+}\ell^{-})$ (control) channels are
kinematically similar to the $B^{\pm}\rightarrow K^{\pm}\ell^{+}\ell^{-}$
(rare) channels, and so the data are selected with identical requirements for
both. The only exceptions are the cut on the dilepton invariant mass squared,
$q^{2}$, and the reconstructed $B$ mass. The control data are selected using a
$q^{2}$ window around the $J/\psi$ resonance, whilst the $B^{\pm}\rightarrow
K^{\pm}\ell^{+}\ell^{-}$ candidates are required to have $q^{2}\in(1.1\,{\rm
GeV}^{2},\;6.0\,{\rm GeV}^{2})$. This $q^{2}$ window is chosen to prevent
contamination from resonances such as $\phi(1020)$ (at low $q^{2}$) and
$J/\psi$ (at high $q^{2}$).

Several cross-checks are performed to ensure the selection efficiencies are
well understood. Among these checks are the ratios $r_{{J/\psi}}$ and
$R_{{\psi(2S)}}$. Like $R_{K}$, the latter is a double ratio with respect to
$r_{{J/\psi}}$, the difference being that the numerator comes from the
$\psi(2S)$ resonance. The ratios are found to be ${r_{{J/\psi}}}=0.981\pm
0.020\,({\rm stat+syst})$, and ${R_{{\psi(2S)}}}=0.997\pm 0.011\,({\rm
stat+syst})$. Both are compatible with the LFU expectation of $1$. This
confirms that the efficiencies are valid across $q^{2}$, and that systematic
effects successfully cancel out in the double ratio.

The value of $R_{K}$ is extracted from an unbinned maximum likelihood fit to
the selected $B^{\pm}\rightarrow K^{\pm}e^{+}e^{-}$ and $B^{\pm}\rightarrow
K^{\pm}\mu^{+}\mu^{-}$ data. The different mass resolutions lead to different
background components in the fit model, as shown in Figure 4. The result for
$R_{K}$ [4] is:

$\displaystyle{R_{K}}=0.846\,^{+\,0.042}_{-\,0.039}~{}{\rm(stat.)}\,^{+\,0.013}_{-\,0.012}~{}{\rm(syst.)}\,.$
(2)

As expected, the uncertainty on the result is dominated by the statistical
component, rather than the systematic one. The dominant systematic effect is
the choice of the fit model, which contributes by around $1\%$. By comparison,
effects induced by the calculation of efficiencies are reduced to the permille
level by the double ratio.

Figure 4: Fit projections for $B^{\pm}\rightarrow K^{\pm}e^{+}e^{-}$ (left)
and $B^{\pm}\rightarrow K^{\pm}\mu^{+}\mu^{-}$ (right) candidates. The
resolutions and background components differ as a result of the different
behaviour of electrons and muons as they traverse the LHCb detector.

##  4 Branching fractions of $B_{(s)}^{0}\rightarrow\mu^{+}\mu^{-}$
$\\!(\gamma)$

Thanks to the excellent mass resolution of muons, the signature of
$B_{(s)}^{0}\rightarrow\mu^{+}\mu^{-}$ decays is very clean. It consists of a
pair of oppositely charged muons that have an invariant mass around mass of
the $B_{(s)}$, and that form a vertex that’s significantly displaced from the
interaction point. The branching fraction of the signal decays are extracted
from an unbinned maximum likelihood fit to the dimuon invariant mass and are
measured relatively to two normalisation channels
$B^{\pm}\rightarrow{J/\psi}K^{\pm}$ and $B^{0}\rightarrow K^{+}\pi^{-}$ which
both share similarities with the signal. The former is characterised by
similar particle identification and trigger performance, and the latter has
similar kinematics. The branching fraction is written as:

$\displaystyle\mathcal{B}(\mbox{$B_{(s)}^{0}\rightarrow\mu^{+}\mu^{-}$}(\gamma))=\mathcal{B}_{{\rm
norm}}\frac{f_{\rm sig}}{f_{\rm norm}}\frac{N_{\rm sig}}{N_{\rm
norm}}\frac{\varepsilon_{\rm sig}}{\varepsilon_{\rm norm}}\,,$ (3)

where $N_{\rm sig}$ and $N_{\rm norm}$ are the yields of the signal and
normalisation channels, respectively. The corresponding efficiencies and
fragmentation fractions are denoted by $\varepsilon_{\rm norm(sig)}$ and
$f_{\rm norm(sig)}$ respectively. LHCb provided a measurement for the latter
in [20]. In order to maximise sensitivity to the signal, a boosted decision
tree (BDT) is used to separate signal and background. The fit for the
$B_{(s)}^{0}\rightarrow\mu^{+}\mu^{-}$ yields is performed simultaneously on
subsets of the data separated by BDT output. The left-hand side plot of Figure
5 shows the expected relative yield of $B_{(s)}^{0}\rightarrow\mu^{+}\mu^{-}$,
in bins of BDT score, as estimated using two different methods. The red
distribution uses a BDT calibration based on the control channel
$B^{0}\rightarrow K^{+}\pi^{-}$, as done in the previous LHCb analysis [14].
The black histogram uses calibrated $B_{(s)}^{0}\rightarrow\mu^{+}\mu^{-}$
simulation. The two methods yield compatible results, and the new BDT
calibration method significantly improves the precision on the expected
relative yield. This is important because it directly improves the total
uncertainty of the branching fraction measurement.

Figure 5: Left: expected distribution of the relative signal yield, as a
function of the BDT score. Right: invariant-mass distribution of
$B_{(s)}^{0}\rightarrow\mu^{+}\mu^{-}$ candidates with BDT score above $0.5$.
The total fit model is shown in blue, alongside the individual components that
represent signal and background.

The branching fraction of $B_{s}^{0}\rightarrow\mu^{+}\mu^{-}$ is obtained
from a fit to the invariant mass of the dimuon system, which is shown on the
right-hand side of Figure 5. In addition to the
$B_{s}^{0}\rightarrow\mu^{+}\mu^{-}$ signal, the fit model contains
contributions from $B^{0}\rightarrow\mu^{+}\mu^{-}$ and
$B_{s}^{0}\rightarrow\mu^{+}\mu^{-}\gamma$. These two components are
compatible with the background-only hypothesis, and so upper limits are set on
their corresponding branching fractions. In summary, the results are:

$\displaystyle\begin{split}\mathcal{B}(\mbox{$B_{s}^{0}\rightarrow\mu^{+}\mu^{-}$})&=3.09\,^{+\,0.46}_{-\,0.43}~{}{\rm(stat.)}\,^{+\,0.15}_{-\,0.11}~{}{\rm(syst.)}\,,\\\
\mathcal{B}(\mbox{$B^{0}\rightarrow\mu^{+}\mu^{-}$})&<2.6\times
10^{-10}\textrm{ at }95\%\textrm{ C.L.}\,,\,\text{and}\\\
\mathcal{B}(\mbox{$B^{0}_{s}\rightarrow\mu^{+}\mu^{-}\gamma$})_{m_{\mu\mu}>4.9\,\textrm{GeV}}&<2.0\times
10^{-10}\textrm{ at }95\%\textrm{ C.L.}\,.\end{split}$ (4)

Like $R_{K}$, the measurement of
$\mathcal{B}(\mbox{$B_{s}^{0}\rightarrow\mu^{+}\mu^{-}$})$ is dominated by
statistics. The systematic uncertainty is largely due to the uncertainty on
$f_{s}/f_{d}$. The measured values are compatible with the previous results,
as well as with the SM predictions, at the $1\,\sigma$ level.

##  5 Status and prospects

The updaded results on $R_{K}$ and the $B_{(s)}^{0}\rightarrow\mu^{+}\mu^{-}$
branching fractions are summarised in Figure 6. While the SM prediction for
$\mathcal{B}(\mbox{$B_{(s)}^{0}\rightarrow\mu^{+}\mu^{-}$})$ still lies within
the $95\%\textrm{ C.L.}$ of the experimental result, the central value is
relatively unchanged with respect to the previous LHCb measurement. The same
is true for the central value of the $R_{K}$ result: thanks to the increase in
statistics with respect to the previous measurement, it now exhibits
${3.1}\,\sigma$ tension with respect to the value predicted by the SM [9, 10].
Therefore, this $R_{K}$ measurement represents the first evidence for the
violation of LFU in $B^{\pm}\rightarrow K^{\pm}\ell^{+}\ell^{-}$ decays.

\begin{overpic}[width=205.97214pt]{2DLbsmumu}\end{overpic}

\begin{overpic}[width=197.29788pt]{1DLRK}\end{overpic} Figure 6: Experimental
status of $B_{(s)}^{0}\rightarrow\mu^{+}\mu^{-}$ (left) and $R_{K}$ (right)
after the Moriond 2021 conference. Shown on the left are $1$–$3\,\sigma$
confidence regions (shades of blue) of $B_{s}^{0}\rightarrow\mu^{+}\mu^{-}$
and $B^{0}\rightarrow\mu^{+}\mu^{-}$ measurements from LHCb [6, 5], alongside
the SM prediction (red). Shown on the right is the likelihood function from
the fit to $B^{\pm}\rightarrow K^{\pm}\ell^{+}\ell^{-}$ candidates , profiled
in $R_{K}$. The extent of the dark, medium and light blue regions show the
values allowed for $R_{K}$ at the $1,\,3,$ and $5\,\sigma$ levels,
respectively. The SM prediction is shown as a solid line at ${R_{K}}=1$.

Since all presented measurements are dominated by the statistical uncertainty,
future updates are crucial to the better understanding of the flavour puzzle.
In Table 1 are summarised the projected improvements on the precision of the
measurements of $R_{K}$ and
$\mathcal{B}(\mbox{$B_{(s)}^{0}\rightarrow\mu^{+}\mu^{-}$})$ in the following
years of data taking. The limiting factor in the precision of $R_{K}$ is the
yield of the electron mode, which, assuming the same detector performance as
in Run 1, would increase more than $40$ times at the end of Run 4. This would
bring the statistical uncertainty on $R_{K}$ on par with the QED corrections.
The error projection on $B_{s}^{0}\rightarrow\mu^{+}\mu^{-}$ at future
upgrades depends on the assumptions made on the systematic uncertainties,
particularly $f_{s}/f_{d}$. A conservative estimate of $4\%$ would imply an
uncertainty of $\mathcal{B}(\mbox{$B_{s}^{0}\rightarrow\mu^{+}\mu^{-}$})$ to
be approximately $0.30\times 10^{-9}$ with $23\,{\rm fb}^{-1}$ and $0.16\times
10^{-9}$ with $300\,{\rm fb}^{-1}$ of data [21].

Table 1: Extrapolation, based on Run 1 results, of statistical uncertainties on $R_{K}$, $\mathcal{B}(\mbox{$B_{s}^{0}\rightarrow\mu^{+}\mu^{-}$})$ (expressed in units of their SM predictions), and the expected electron-mode yield at the end of future detector upgrades. The $b\overline{b}$ production cross-section is assumed to scale linearly with centre-of-mass energy, and the detector performance is assumed to be unchanged with respect to Run 1. Adapted from Ref. [21]. Stat. uncertainty | $9\,{\rm fb}^{-1}$ | $23\,{\rm fb}^{-1}$ | $300\,{\rm fb}^{-1}$
---|---|---|---
$R_{K}$ | $4.3\%$ | $2.5\%$ | $0.7\%$
$B_{s}^{0}\rightarrow\mu^{+}\mu^{-}$ | $12.5\%$ | $8.2\%$ | $4.4\%$
Yield | $9\,{\rm fb}^{-1}$ | $23\,{\rm fb}^{-1}$ | $300\,{\rm fb}^{-1}$
$B^{\pm}\rightarrow K^{\pm}e^{+}e^{-}$ | $1120$ | $3300$ | $46000$

The EFT interpretation in terms of shifts in Wilson Coefficients $\Delta
C^{\mu}_{i}$ from their SM value, in light of the new results of $R_{K}$ and
$B_{(s)}^{0}\rightarrow\mu^{+}\mu^{-}$, is updated in Figure 7 [22]. In the
plot on the left, the light orange horizontal band highlights the interval of
$\Delta C^{\mu}_{10}$ preferred by the combined ATLAS, CMS, and updated LHCb
measurements of $\mathcal{B}(\mbox{$B_{s}^{0}\rightarrow\mu^{+}\mu^{-}$})$,
while the elongated purple shape marks the regions of $\Delta C^{\mu}_{9}$ and
$\Delta C^{\mu}_{10}$ compatible with the measured values of $R_{K}$ and
$R_{K^{*}}$ [23]. Thanks to their different sensitivity to the $\Delta
C^{\mu}_{i}$ the combination of these two measurement forms the unfilled
contours, which are significantly displaced from the SM point. In the plot on
the right, the contribution from other $b\to s\ell^{+}\ell^{-}$ observables,
such as the angular ones in the $B\rightarrow K^{*}\mu^{+}\mu^{-}$ system [24]
is included in the $\Delta C_{i}$ fit. While the clean observables presented
in this article are insensitive to a flavour-universal redefinition of
$C^{\ell,{\rm SM}}_{9}$ in the SM, ($\Delta C^{U}_{9}$), the observables
characterised by less precise theoretical predictions are nonetheless able to
provide substantial sensitivity to a shift in $\Delta C^{\mu}_{9}$ and $\Delta
C^{\mu}_{10}$, when combined with $R_{K}$ and
$\mathcal{B}(\mbox{$B_{s}^{0}\rightarrow\mu^{+}\mu^{-}$})$.

\begin{overpic}[width=411.93767pt]{globalFitsNew_gino} \put(27.1,12.2){$\star$
{\scriptsize SM}} \put(77.4,31.8){$\star$ {\scriptsize SM}} \end{overpic}
Figure 7: EFT constraints from the $b\to s\ell^{+}\ell^{-}$ anomalies [22].
Left: Results of the two-dimensional fit to $\Delta C^{\mu}_{9}$ and $\Delta
C^{\mu}_{10}$ using clean observables only ($1\sigma$, $2\sigma$ and $3\sigma$
intervals). Also shown are the $1\sigma$ and $2\sigma$ intervals from $R_{K}$,
$R_{K^{*}}$, and $B_{s}^{0}\rightarrow\mu^{+}\mu^{-}$, the latter under the
hypothesis $\Delta C^{U}_{9}=0$. Right: Results of the two-dimensional fit
$\Delta C^{\mu}_{10}=-\Delta C^{\mu}_{9}$ vs. $\Delta C^{U}_{9}$ using all
$b\to s\ell^{+}\ell^{-}$ observables. The vertical band shows the result using
clean observables only ($1\sigma$ interval), while the ellipse denotes the
contribution of all the other observables, estimated using Flavio[25]
($1\sigma$ interval).

In light of the recent results from LHCb, the flavour puzzle has become even
more intriguing. LHCb will continue to improve the precision on its anomalous
flavour measurements, whilst also investigating related observables that could
provide complementary information. Verification from other experiments, such
as Belle II, is expected in the near future, so exciting times lay ahead of
the particle physics community!

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