1 Introduction

The hypergeometric series

$$\begin{aligned} \phi _0(x)=\sum _{n=0}^{\infty } \frac{(5n)!}{(n!)^5} x^n=1+120x+113400x^2+\cdots \in \textbf{Z}[[x]] \end{aligned}$$

became famous after in the paper [20] its dual interpretation was discovered: on the one hand, the series encodes enumerative information on rational curves on the general quintic Calabi–Yau threefold in \(\textbf{P}^4\), and on the other hand, it can be identified as a normalised period of the mirror quintic Calabi–Yau threefold. The power series is the unique (up to scalar) series solution of the hypergeometric operator

$$\begin{aligned} L:=\theta ^4-5^5x\left( \theta +\frac{1}{5}\right) \left( \theta +\frac{2}{5}\right) \left( \theta +\frac{3}{5}\right) \left( \theta +\frac{4}{5}\right) ,\;\;\;\theta =x\frac{d}{dx}.\end{aligned}$$

The operator L is the first of the family of 14 hypergeometric fourth-order operators related to mirror symmetry for complete intersections in weighted projective spaces. These operators are all of the form

$$\begin{aligned} \theta ^4-Nx(\theta +\alpha _1)(\theta +\alpha _2)(\theta +\alpha _3)(\theta +\alpha _4), \end{aligned}$$

where the integer N is chosen as to make the coefficients of the normalised holomorphic solution integral in a minimal way. The properties of these operators have been well-studied from various points of view [8, 19, 26, 27, 34, 38, 39].

The operator L is also the first member of the ever growing list of Calabi–Yau operators [3, 41], a notion that was introduced by the first author and Zudilin [4] by abstracting the properties of the operator L. Calabi–Yau operators in this sense are essentially self-adjoint fourth-order Fuchsian operators with a point of maximal unipotent monodromy for which strong integrality properties are supposed to hold. In a sense, the term Calabi–Yau operator is somewhat of a misnomer, as there exist families of Calabi–Yau varieties, whose Picard–Fuchs operators have no MUM point and hence are not Calabi–Yau operators in the above sense of the word. For a recent account, see [23].

Let us consider a general Nth-order differential operator written in \(\theta \)-form:

$$\begin{aligned} L=P_0(\theta )+x P_1(\theta )+\cdots +\cdots +x^rP_r(\theta ),\;\;\;\theta := x\frac{d}{dx},\end{aligned}$$

where the \(P_i\) are polynomials of degree N. We call the largest r with \(P_r \ne 0\) the degree of the operator. The differential equation

$$\begin{aligned} L \phi =0,\;\;\; \phi (x)=\sum _{n=0}^{\infty } a_n x^n \end{aligned}$$

translates into the recursion relation

$$\begin{aligned} P_0(n)a_n+P_1(n-1)a_{n-1}+\cdots +P_r(n-r)a_{n-r}=0 \end{aligned}$$

on the coefficients \(a_n\) of the series \(\phi (x)\), so the degree of L is equal to the length of this recursion.

The roots of \(P_0(\theta )\) are the exponents of the operator at 0. By translation, one can define exponents of P at each point of the Riemann sphere \(\textbf{P}^1\). For the point \(\infty \) we have to apply inversion \(x\mapsto 1/x\); the exponents then are the roots of \(P_r(-\theta )\). The Riemann symbol is a table that contains for each singular point the corresponding exponents. These encode information about the local ramification of the solutions. Logarithmic terms may and usually do appear at points with exponents with integral difference. This leads to Jordan blocks in the local monodromy, but these are usually left implicit in the Riemann symbol.

We will usually assume that \(N=4\) and that all exponents at 0 vanish; \(P_0(\theta )=\theta ^4\). The most salient feature of this situation is that at 0 we have a canonical Frobenius basis of solution \(\phi _0, \phi _1, \phi _2, \phi _3\), where \(\phi _k\) contains terms with \(\log ^k(x)\). The local monodromy has a Jordan block of maximal size, so 0 is a point of maximal unipotent monodromy: MUM.

The 14 hypergeometric operators have degree 1 and have a Riemann symbol of the form

$$\begin{aligned}\left\{ \begin{array}{ccc} 0&{}\quad 1/N &{}\quad \infty \\ \hline 0&{}0&{}\alpha _1\\ 0&{}1&{}\alpha _2\\ 0&{}1&{}\alpha _3\\ 0&{}2&{}\alpha _4\\ \end{array} \right\} , \end{aligned}$$

exhibiting a conifold singularity at the point 1/N. At such a point the exponents are 0, 1, 1, 2 and the local monodromy is a symplectic reflection; there is a Jordan block of size two. The exponents at \(\infty \) are \(\alpha _1 \le \alpha _2 \le \alpha _3 \le \alpha _4\) and add up in pairs to one:

$$\begin{aligned} \alpha _1+\alpha _4=\alpha _2+\alpha _3=1. \end{aligned}$$

Currently, we know over 500 Calabi–Yau operators with degrees running up to 40, but in no way do we expect the current list to be complete. An update of the AESZ list from [3] is in preparation [2].

The first Calabi–Yau operator of degree \(>1\) is #15 in the AESZ table and appeared in [10]:

$$\begin{aligned} \theta ^4-3x(3\theta +1)(3\theta +2)(7\theta ^2+7\theta +2)-72x^2(3\theta +1)(3\theta +2)(3\theta +4)(3\theta +5). \end{aligned}$$

Its Riemann symbol is

$$\begin{aligned}\left\{ \begin{array}{cccc} 0&{}1/216&{}-1/27&{}\infty \\ \hline 0&{}0&{}0&{}\frac{1}{3}\\ 0&{}1&{}1&{}\frac{2}{3}\\ 0&{}1&{}1&{}\frac{4}{3}\\ 0&{}2&{}2&{}\frac{5}{3}\\ \end{array} \right\} . \end{aligned}$$

Another operator of this type is #25 of the AESZ list. It arose from the quantum cohomology of the Grassmanian G(2, 5). The complete intersection X(1, 2, 2) of hypersurfaces of degrees 1, 2, 2 in the Plücker embedding is a Calabi–Yau threefold with the following mirror Picard–Fuchs operator:

$$\begin{aligned} \theta ^4 -4x (2\theta + 1)^2(11 \theta ^2 + 11 \theta + 3) -16 x^2(2 \theta + 1)^2 (2 \theta + 3)^2. \end{aligned}$$

Its holomorphic solution is

$$\begin{aligned} \phi _0(x)=\sum _{n=0}^{\infty } { 2n \atopwithdelims ()n}^2A_n x^n, \end{aligned}$$

where

$$\begin{aligned} A_n=\sum _{k=0}^n {n \atopwithdelims ()k}^2{n+k \atopwithdelims ()k} \end{aligned}$$

are the small Apéry numbers that were used in Apéry’s proof of the irrationality of \(\zeta (2)\).

At present we know 69 essentially distinct Calabi–Yau operators of order two, to which all further such degree two operators can be related by simple transformations. The operators for which the instanton numbers are 0 are obtained as \(\hbox {Sym}^3\) of a second-order operator and do not count as a proper Calabi–Yau operator and do not appear in [3].

But contrary to the hypergeometric case, it appears that these degree two operators do not all fall in a single family. First of all, an operator of degree two can have four singularities, with a Riemann symbol of the form

$$\begin{aligned} \left\{ \begin{array}{cccc} 0&{}\quad a&{}\quad b&{}\quad \infty \\ \hline 0&{}0&{}0&{}\alpha _1\\ 0&{}1&{}1&{}\alpha _2\\ 0&{}1&{}1&{}\alpha _3\\ 0&{}2&{}2&{}\alpha _4\\ \end{array} \right\} , \end{aligned}$$

exhibiting two distinct conifold points, as in the previous two examples. But in many cases the points a and b coalesce, producing an operator with only three singular points.

As one can observe in the above two examples, the exponents at infinity are symmetrically centred around the value 1 and this seemed to be the case for all further examples we found. It was M. Bogner who first found an example of an Calabi–Yau operator for which this is not the case: it is the operator

$$\begin{aligned} {\theta }^{4}-x( 216\,{\theta }^{4}+396\,{\theta }^{3}+366\,{\theta }^{2}-168\,\theta +30)+ 36\,x^2\left( 3\,\theta +2 \right) ^{2} \left( 6\,\theta +7 \right) ^{2}. \end{aligned}$$

with Riemann symbol

$$\begin{aligned} \left\{ \begin{array}{cccc} 0&{}\quad 1/108&{}\quad \infty \\ \hline 0&{}0&{}2/3\\ 0&{}1/6&{}2/3\\ 0&{}1&{}7/6\\ 0&{}7/6&{}7/6\\ \end{array} \right\} , \end{aligned}$$

and later two more such operators were found. The existence of such operators was initially a surprise to us, as it appears that, within the group of operators with three singular points, further distinctions can be made. It is this circumstance that led to this paper.

The structure of the paper is as follows. In the first section, we recall some basic facts about the Calabi–Yau condition and formulate the main result of this paper. In the second section, we describe how the result can be obtained from a series of simple computations. In the third section we summarise the properties of the operators corresponding to each of the components that we find. In a final section, we review the list all degree two Calabi–Yau operators that are presently known to us. In two appendices we include some further monodromy and modular form information on these examples, together with some properties of the lower-order operators used in many constructions.

2 The Calabi–Yau condition

The adjoint \(L^*\) of a differential operator L of the form

$$\begin{aligned} L =\frac{\textrm{d}^n}{\textrm{d}x^n}+a_{n-1} \frac{\textrm{d}^{n-1}}{\textrm{d}x^{n-1}}+\cdots +a_{1}\frac{\textrm{d}}{\textrm{d}x}+a_0 \in \textbf{Q}(x)\left[ \frac{\textrm{d}}{\textrm{d}x}\right] \end{aligned}$$

is obtained by reading the operator backwards with alternating signs:

$$\begin{aligned} L^* =\frac{\textrm{d}^n}{\textrm{d}x^n}-\frac{\textrm{d}^{n-1}}{\textrm{d}x^{n-1}}a_{n-1}+\cdots +\pm \frac{\textrm{d}}{\textrm{d}x}a_1 +(-1)^n a_0 \in \textbf{Q}(x)\left[ \frac{\textrm{d}}{\textrm{d}x}\right] , \end{aligned}$$

so that

$$\begin{aligned} L^*(y) =y^{(n)}- (a_{n-1}y)^{(n-1)}+\cdots +(-1)^n a_0 y \end{aligned}$$

The operator \(\mathcal {P}\) is called essentially self-adjoint, if there exists a function \(\alpha \ne 0\) (in an extension field of \(\textbf{Q}(x)\)), such that

$$\begin{aligned} L \alpha =\alpha L ^{*}. \end{aligned}$$

It is easy to see that any such \(\alpha \) has to satisfy the first-order differential equation

$$\begin{aligned} \alpha ' = -\frac{2}{n} a_{n-1} \alpha . \end{aligned}$$

This essential self-adjointness is equivalent to the existence of an invariant pairing on the solution space, symmetric if n is odd and alternating if n is even, [16, 17]. The condition of essential self-adjointness can be expressed by the vanishing of certain differential polynomials in the \(a_i\), described in [1, 4, 16, 17], called the Calabi–Yau condition.

For an operator of order two, the essential self-adjointness does not impose any conditions, whereas for an differential equation of order three of the form

$$\begin{aligned} y'''+a_2 y''+a_1y'+a_0y, \end{aligned}$$

the Calabi–Yau condition comes down to the vanishing of the quantity

$$\begin{aligned} \mathcal {W}:=\frac{1}{3}a_2''+\frac{2}{3}a_2a_2'+\frac{4}{27}a_2^3+2a_0-\frac{2}{3}a_1a_2-a_1', \end{aligned}$$

whereas for a fourth-order operator

$$\begin{aligned} y^{(iv)}+a_3 y'''+a_2 y''+a_1 y' +a_0 y = 0, \end{aligned}$$

the quantity

$$\begin{aligned} \mathcal {Q}:= \frac{1}{2} a_2 a_3-a_1-\frac{1}{8}a_3^3+a_2'-\frac{3}{4}a_3 (a_3)'-\frac{1}{2} a_3'' \end{aligned}$$

has to vanish. In that case the associated differential equation satisfied by the \(2\times 2\) Wronskians reduces its order from 6 to 5, [4].

For operators of higher order, one finds that more than one differential polynomial condition has to be satisfied; here we will be concerned with operators of three and mainly order four.

It is rather easy to fulfil the condition \(\mathcal {Q}=0\)! For example, there exists one particularly nice family of operators of degree n that we call the main family.

Proposition 1

If A, B, C are polynomials of degree n, then the operator

$$\begin{aligned} L= & {} \theta ^2 A \theta ^2+\theta B \theta +C\\= & {} A\theta ^4+2 A'\theta ^3+(B+A'')\theta ^2+B'\theta +C \end{aligned}$$

has degree n and satisfies the Calabi–Yau condition. If the constant term of A is 1 and of B and B is zero, then the exponents of L at 0 are all zero. If the roots of A are all distinct, L has n singular points with exponents 0, 1, 1, 2 and a further singular point at infinity. (Here \(A':=\theta (A)=x\partial A/\partial x\), etc.)

The operator of the proposition can be shown to satisfy the Calabi–Yau condition \(Q=0\) by a direct, but tedious computation. But the form of the operator makes the self-adjointness manifest. The statements about the exponents require a further direct calculation.

In this paper we take a closer look at differential operators of the form

$$\begin{aligned} \theta ^4+x(a\theta ^4+b\theta ^3+c\theta ^2+d\theta +e)+fx^2(\theta +\alpha )(\theta +\beta )(\theta +\gamma )(\theta +\delta ). \end{aligned}$$

If \(f \ne 0\), then the numbers \(\alpha ,\beta ,\gamma ,\delta \) are then the characteristic exponents of the operator at \(\infty \).

For this operator, the quantity \(\mathcal {Q}\) is a rational function with

$$\begin{aligned} x^2(1+ax+fx^2)^3 \end{aligned}$$

as denominator; the singular points of the operator are the roots of this polynomial, together with the point \(\infty \).

The numerator is a polynomial Q of degree 5 in the variable x, so we can write

$$\begin{aligned} Q=Q_0+Q_1x+Q_2x^2+Q_3x^3+Q_4x^4+Q_5x^5, \end{aligned}$$

where the coefficients \(Q_i\) are rather complicated polynomials in the parameters a, b, c, d, e, f and the exponents \(\alpha ,\beta ,\gamma , \delta \):

$$\begin{aligned} Q_0,\;\;Q_1,\;\;Q_2,\;\;Q_3,\;\;Q_4,\;\;Q_5 \in \textbf{Q}[a,b,c,d,e,f,\alpha ,\beta ,\gamma ,\delta ]. \end{aligned}$$

So, the Calabi–Yau condition is equivalent to six polynomial equations

$$\begin{aligned} Q_0=Q_1=Q_2=Q_3=Q_4=Q_5=0, \end{aligned}$$

which define a certain affine algebraic set

$$\begin{aligned} X=V(Q_0,Q_1,\ldots ,Q_5) \subset \textbf{C}^{10}. \end{aligned}$$

We will give a complete description of the set X, in particular we determine the irreducible components of X. As we have six equations in 10 variables, a first trivial remark is that all irreducible components of X have dimension at least four. We will give explicit parametrisations of all the irreducible components. Note also that the polynomials \(Q_i\) do not depend on the accessory parameter e, so the solution set X contains the e-line as trivial factor.

Theorem 1

The algebraic set X defined by the condition \(Q(a,b,c,d,e,f,\alpha ,\beta ,\gamma ,\delta )=0\) is the union of ten irreducible components, of which there are seven with \(f\ne 0\).

Note that if \(f=0\) we are dealing with an operator of lower degree (i.e. hypergeometric), so will from now on assume that \(f \ne 0\) and we will ignore the three components with \(f=0\). The seven remaining components differ in the behaviour of the average exponents at infinity, i.e. the quantity:

$$\begin{aligned} \sigma :=2 \frac{\alpha +\beta +\gamma +\delta }{4}. \end{aligned}$$

There is a single component for which \(\sigma \) is not fixed, that we will call the transverse component. On each of the remaining six components, \(\sigma \) has a fixed value. The dimensions and values of \(\sigma \) that appear are summarised in the following diagram:

$$\begin{aligned} \begin{array}{|c||c|c|c|c|c|} \hline \sigma &{}\quad 0&{}\quad 1&{}\quad 2&{}\quad 3&{}\quad 4\\ \hline &{} &{} &{}6 &{} &{}\\ \dim &{} &{} 5 &{} &{} 5&{} \\ &{}4&{} &{}4 &{} &{}4\\ \hline \end{array} \end{aligned}$$

The two five-dimensional components are related by a simple transformation, as are the three four-dimensional components.

All Calabi–Yau operators known to us belong to the six-dimensional main component (M):

$$\begin{aligned}{} & {} (M): \theta ^4+x(a\theta ^4+2 a \theta ^3+(a+d)\theta ^2+d\theta +e)\\{} & {} \quad +fx^2(\theta +\alpha )(\theta +\beta )(\theta +2-\alpha )(\theta +2-\beta ) \end{aligned}$$

or the transverse component (T) that we describe later.

We remark furthermore that the ideal

$$\begin{aligned} I=(Q_0,Q_1,Q_2,Q_3,Q_4,Q_5) \subset \textbf{C}[a,b,c,d,e,f,\alpha ,\beta ,\gamma , \delta ] \end{aligned}$$

is, from an algebraic point of view, rather complicated. It is not radical and the components appear with non-trivial multiplicities in the primary decomposition.

For third-order operators of degree two

$$\begin{aligned} \theta ^3+x(a\theta ^3+b\theta ^2+c\theta +d)+f x^2(\theta +\alpha )(\theta +\beta )(\theta +\gamma ), \end{aligned}$$

the Calabi–Yau condition \(\mathcal {W}=0\) leads in a similar way to an algebraic set

$$\begin{aligned} Y \subset \textbf{C}^8, \end{aligned}$$

defined by a system of polynomial equations

$$\begin{aligned} W_0=W_1=W_2=W_3=W_4=W_5=0, \end{aligned}$$

where now the polynomials

$$\begin{aligned} W_0,\;\;W_1,\;\;W_2,\;\;W_3,\;\;W_4,\;\;W_5 \in \textbf{Q}[a,b,c,d,f,\alpha ,\beta ,\gamma ]. \end{aligned}$$

are in eight variables. The analysis of this ideal is very similar to the ideal for the fourth-order operators, but hardly simpler. The result is, somewhat surprisingly, also very similar to the case of order four:

Theorem 2

The algebraic set \(Y \subset \textbf{C}^8\) defined by the condition \(W(a,b,c,d,f,\alpha ,\beta ,\gamma )=0\) is the union of ten irreducible components, of which there are seven with \(f\ne 0\).

The seven remaining components differ in the behaviour of the (doubled) average exponent

$$\begin{aligned} \sigma :=2 \frac{\alpha +\beta +\gamma }{3}. \end{aligned}$$

Again there is a single transverse component on which the value of \(\sigma \) is not fixed. On each of the remaining six components, \(\sigma \) has a fixed value. The dimensions and values of \(\sigma \) that appear are summarised in the following diagram:

$$\begin{aligned} \begin{array}{|c||c|c|c|c|c|} \hline \sigma &{}\quad 0&{}\quad 1&{}\quad 2&{}\quad 3&{}\quad 4\\ \hline &{} &{} &{} 4 &{} &{}\\ \dim &{} &{} 3&{} &{} 3 &{}\\ &{}2 &{} &{} 2 &{} &{} 2\\ \hline \end{array} \end{aligned}$$

The main component (m) is the one with \(\sigma =2\), \(\dim =4\) and is given by

$$\begin{aligned} (m): \theta ^3+x(2\theta +1)((c-2d) \theta ^2+(c-2d)\theta + d)+f x^2(\theta +\alpha ) (\theta +1) (\theta +2-\alpha ), \end{aligned}$$

whereas on the transverse component (t) we have the operator

$$\begin{aligned}{} & {} (t): \theta ^3+ \frac{a}{4} x(4\theta ^3+3(\alpha +\beta )\theta ^2+(\alpha +\beta +2\alpha \beta )\theta +\alpha \beta )\\{} & {} \quad + \left( \frac{a}{4}\right) ^2 x^2 (\theta +\alpha )\left( \theta +\frac{\alpha +\beta }{2}\right) (\theta +\beta ). \end{aligned}$$

Again, the operators on the two three-dimensional components are related by a simple transformation, as are the three two-dimensional components.

One may speculate that the above patterns extends to higher-order operators, but we will not try to pursue it here.

3 Proof of Theorem 1

The proof of Theorem 1 is purely computational and consists of systematically solving of the equations, using a computer algebra system like Maple. The miracle here is that this procedure actually works. We will describe only the most important steps in the process.

3.1 Preliminary reductions \(Q_0=Q_5=0\)

The value of b: The polynomial \(Q_0\) is found to be

$$\begin{aligned} -4 b - 8 d + 8 c, \end{aligned}$$

and this allows us directly to eliminate the variable b and thus we will put from now on

$$\begin{aligned} b=2(c-d). \end{aligned}$$

Symmetry of the exponents: The polynomial \(Q_5\) factors nicely as

$$\begin{aligned} -f^3(\alpha +\beta -\delta -\epsilon )(\alpha +\delta -\beta -\epsilon )(\alpha +\epsilon -\beta -\delta ). \end{aligned}$$

As we assume that \( f \ne 0\), this only vanishes if one of the three other factors vanish, meaning that the exponents come in two pairs that add up to the same value, that we shall call \(\sigma \). So, without loss of generality we can assume to have the pairs \(\alpha , \sigma -\alpha \) and \(\beta , \sigma -\beta \) and we will take the operator to be of the form

$$\begin{aligned} \theta ^4+x(a\theta ^4+b\theta ^3+c\theta ^2+d\theta +e)+fx^2(\theta +\alpha )(\theta +\beta )(\theta +\sigma -\alpha ) (\theta +\sigma -\beta ), \end{aligned}$$

with \(b=2(c-d)\) as our starting point.

3.2 The remaining equations

After solving the equations \(\textbf{Q}_0=0,\textbf{Q}_5=0\), one is left to solve the polynomial equations \(Q_1=Q_2=Q_3=Q_4=0\). For sake of concreteness, we give the explicit form of the polynomials:

$$\begin{aligned} {Q_1}:= & {} 8\,f{{ \alpha }}^{2}{ \sigma }+16\,f{ \sigma }\,{ \beta }-8\,f{{\sigma }}^{2}{ \beta }-16\,{c}^{2}+40\,cd-24\,{d}^{2}-8\,f{ \alpha }\,{{ \sigma }}^{2}-16\,f{{ \alpha }}^{2}\\{} & {} +16\,f{{ \sigma }}^{2}-16\,f{{ \beta }}^{2}-32\,f{ \sigma }-24\,da+16\,ca+16\,f{ \alpha }\,{ \sigma }+8\,f{ \sigma }\,{ { \beta }}^{2},\\ {Q_2}:= & {} -8\,f{{ \alpha }}^{2}c-8\,f{{ \beta }}^{2}c+8\,f{{ \beta }}^{2}d+ 8\,f{{ \alpha }}^{2}d-8\,cad+8\,f{{ \sigma }}^{2}c+72\,f{ \sigma }\,d\\{} & {} -8\,f{{ \sigma }}^{2}d+24\,{c}^{2}d-24\,c{d}^{2}+8\,{c}^{2}a-64\,cf{ \sigma }+24\,f{ \alpha }\,{ \sigma }\,a +24\,f{ \sigma }\,{ \beta }\,a\\{} & {} -16\,f{{ \sigma }}^{2}{ \beta }\,a+16\,f{{ \alpha }}^{2}{ \sigma }\,a-16\,f{ \alpha }\,{{ \sigma }}^{2}a+16\,f{ \sigma }\,{{ \beta }}^{2}a\\{} & {} -8\,f{ \alpha }\,{ \sigma }\,d+8\,f{\alpha }\,{ \sigma }\,c-8\,{c}^{3}+8\,{d}^{3}+ 8\,f{ \sigma }\,{ \beta }\,c-8\,f{ \sigma }\,{ \beta }\,d\\{} & {} -64\,df-8\,d{a}^{2}-24\,f{ \sigma }\,a-24\,f{{\alpha }}^{2}a+48\,cf+24\,f{{ \sigma }}^{2}a-24\,f{{ \beta }}^{2}a,\\ {Q_3}= & {} 8\,f \left( 2\,f{ \sigma }\,{ \beta }+2\,f{ \alpha }\,{ \sigma }+f{ \sigma }\,{{ \beta }}^{2}-f{{ \sigma }}^{2}{ \beta }+f{{ \alpha }}^{2}{\sigma }-f{ \alpha }\,{{ \sigma }}^{2}-2\,f{{ \beta }}^{2}\right. \\{} & {} -3\,da+{{ \sigma }}^{2}{a}^{2}+f{{ \sigma }}^{3}-{{ \beta }}^{2}{a}^{2}-{ \sigma }\,{a}^{2}-{ \sigma }\,{ \beta }\,ad+{ \sigma }\,{ \beta }\,ac-{ \alpha }\,{ \sigma }\,ad+{ \alpha }\,{ \sigma }\,ac\\{} & {} -{{ \alpha }}^{2}{a}^{2}-2\,c{ \sigma }\,a-2\,f{{ \alpha }}^{2}-4\,f{{ \sigma }}^{2}+4\,f{ \sigma }+4\,{c}^{2}-7\,cd+3\,{d}^{2} -{ \alpha }\,{{ \sigma }}^{2}{a}^{2}\\{} & {} +{ \sigma }\,{{ \beta }}^{2}{a}^{2}+6\,{ \sigma }\,cd+3\,{ \sigma }\,ad-{{ \beta }}^{2}ac +{{ \beta }}^{2}ad-{{ \alpha }}^{2}ac+{{ \alpha }}^{2}ad+{{ \sigma }}^{2}ac\\{} & {} \left. -{{ \sigma }}^{2}ad-3\,{ \sigma }\,{d}^{2}-3\,{ \sigma }\,{c}^{2}+{ \alpha }\,{ \sigma }\,{a}^{2}+ { \sigma }\,{ \beta }\,{a}^{2}-{{ \sigma }}^{2}{ \beta }\,{a}^{2}+{{ \alpha }}^{2}{ \sigma }\,{a}^{2} \right) ,\\ { Q_4}= & {} 8\,{f}^{2} \left( -2\,c-{{ \alpha }}^{2}c-2\,{{ \sigma }}^{2}c-3\,{ \sigma }\,d+2\,{{ \sigma }}^{2}d+{{ \beta }}^{2}d-2\,{{ \sigma }}^{2}a-{ { \beta }}^{2}a-{ \alpha }\,{ \sigma }\,d\right. \\{} & {} \left. +{ \alpha }\,{ \sigma }\,c+{ \sigma }\,{ \beta }\,c-{ \sigma }\,{ \beta }\,d+{ \sigma }\,{ \beta }\,a-{ \alpha }\,{{ \sigma }}^{2}a-{{ \sigma }}^{2}{ \beta }\,a+{ \alpha }\,{ \sigma }\,a+{{ \alpha }}^{2}{ \sigma }\,a+{{ \beta }}^{2}{ \sigma }\,a\right. \\{} & {} \left. { \sigma }\,a-{{ \alpha }}^{2} a+{{ \sigma }}^{3}a+4\,c{ \sigma }-{{ \beta }}^{2}c+{{\alpha }}^{2}d \right) . \end{aligned}$$

Somewhat to our surprise, it turns out to be rather easy to give a complete solution to these equations. It will be useful to introduce the parameters

$$\begin{aligned} A:=\alpha (\sigma -\alpha ),\;\;B:=\beta (\sigma -\beta ), \end{aligned}$$

and use these instead of \(\alpha \) and \(\beta \). One has

$$\begin{aligned}{} & {} (\theta +\alpha )(\theta +\sigma -\alpha )(\theta +\beta )(\theta +\sigma -\beta )=(\theta ^2+\sigma \theta +A)(\theta ^2+\sigma \theta +B)\\{} & {} \quad =\theta ^4+2\sigma \theta ^3+(\sigma ^2+(A+B))\theta ^2+\sigma (A+B)\theta +AB. \end{aligned}$$

We will see that most expressions only depend on the quantity

$$\begin{aligned} \Delta :=-(A+B)=\alpha ^2+\beta ^2-\sigma (\alpha +\beta ). \end{aligned}$$

3.3 The equation \(Q_4=0\)

The polynomial \(Q_4\) is an expression that is \(f^2\) times an expression linear in the variable a. As we will only consider the cases with \(f \ne 0\), we can solve for a and find

$$\begin{aligned} a=\frac{(2(\sigma -1)^2+\Delta )c-(\sigma (2\sigma -3)+\Delta )d}{(\sigma -1)(\sigma ^2-\sigma +\Delta )}\;, \end{aligned}$$

Note that this value for a introduces the denominator

$$\begin{aligned} (\sigma -1)(\sigma (\sigma -1)+\Delta ). \end{aligned}$$

So, if we work further with this operator, we are implicitly assuming that these two factors do not vanish. If one of them does vanish, we should go back to the previous step, impose these conditions and compute further. For now, we will assume that these factors are nonzero and come back to these exceptional cases later.

3.4 The case \(\sigma =2\)

It appears that with this value of a the quantity \(\mathcal {Q}\) becomes divisible by \(\sigma -2\). So, in this remarkable \(\sigma =2\) case, we obtain an operator family with \(\mathcal {Q}=0\). For \(\sigma =2\) the value of a simplifies to

$$\begin{aligned} a=\frac{(2+\Delta )c-(2+\Delta )c}{1\cdot (2+\Delta )}=c-d, \end{aligned}$$

and one obtains what we call the main component (M):

$$\begin{aligned} P:= & {} \theta ^4+x((c-d)\theta ^4+2(c-d)\theta ^3+c\theta ^2+d\theta +e)\\{} & {} +fx^2(\theta +\alpha )(\theta +\beta )(\theta +2-\alpha ) (\theta +2-\beta ), \end{aligned}$$

depending c, d, e, f and the exponents \(\alpha \) and \(\beta \) as free parameters. Almost all operators of degree two from the AESZ list are of this type. In this way we obtain our first component of dimension 6.

3.5 The case \(\sigma \ne 2\)

If, however, \(\sigma \ne 2\), we have to do a further analysis. Still using the above value for a, the coefficient of x of \(\mathcal {Q}\) is a complicated expression in c, d, e, f and the exponents that, however, is factored by Maple instantly into four factors. First, there are the factors

$$\begin{aligned} (\sigma -1),\; (\sigma -2),\;(\sigma (\sigma -1)+\Delta ). \end{aligned}$$

We assume first that these are nonzero, so one is forced to put to zero the fourth factor, which leads to a specific value for f:

$$\begin{aligned} f=(2c-3d){\frac{ ((\sigma -1)^2+\Delta )c-(\sigma (\sigma -2)+\Delta )c}{ \left( \sigma -1 \right) \left( 2 \sigma +\Delta \right) \left( {{\sigma }}^{2}-{\sigma } +\Delta \right) }}. \end{aligned}$$

Here a new factor \(2\sigma +\Delta \) is introduced in the denominator, so we have to come back later to the case \(2\sigma +\Delta =0\).

3.5.1 The subcase \(\sigma =3\)

Using the above values for a and f, the quantity \(\mathcal {Q}\) only contains terms with \(x^2\) and \(x^3\). It factors out the factor \(\sigma -3\), which leads to another remarkable operator family with \(\mathcal {Q}=0\):

$$\begin{aligned} a =\frac{(8+\Delta )c-(9+\Delta )d}{2(6+\Delta )},\quad f = (2c-3d) \frac{(4+\Delta )c-(3+\Delta )d}{2(6+\Delta )^2}. \end{aligned}$$

It has c, d, e and \(\alpha ,\beta \) as free parameters. This makes up a second component, of dimension 5.

3.5.2 The subcase \(\sigma \ne 3\)

Still with the given values for a and f, if \(\sigma \ne 3\), one has to analyse the quantity \(\mathcal {Q}\) further. Maple manages to factor the coefficient of \(x^2\) into a product of eight factors, of only three were not supposed to be nonzero at this stage. Each of these factors leads to a linear dependence between c and d.

Case A:

$$\begin{aligned} c=(\Delta -\sigma )\frac{d}{\Delta }. \end{aligned}$$

Upon substitution, we obtain an operator that satisfies the condition \(\mathcal {Q}=0\). For this operator one has

$$\begin{aligned} a=-2 \frac{d}{\Delta },\;\;\;b=-2\sigma \frac{d}{\Delta },\;\;\;c=-(\sigma -\Delta ) \frac{d}{\Delta },\;\;\;f= \left( \frac{d}{\Delta }\right) ^2, \end{aligned}$$

and has d, e and the exponents \(\alpha , \beta ,\sigma \) as free parameters. So, we obtain a further irreducible component of dimension 5. The remarkable thing is that now we have only \(\Delta \) appearing in the denominator.

The operator of Bogner mentioned in the introduction is an instance of case A for the parameter choice:

$$\begin{aligned} \alpha =\beta =2/3,\;\;\sigma =11/6,\;\;d=-168,\;\; e=30. \end{aligned}$$

Case B:

$$\begin{aligned} c=d \frac{\sigma ^2-2\sigma +\Delta }{(\sigma -1)^2+\Delta }. \end{aligned}$$

In this case f reduces to 0, so we do not get an operator of degree two.

Case C:

$$\begin{aligned} c=d \frac{\sigma ^2-6\sigma +4+\Delta }{(\sigma -1)(\sigma -4)+\Delta } \end{aligned}$$

In this case the condition \(\mathcal {Q}=0\) reduces to special relations between the exponents. The exceptional cases are:

$$\begin{aligned} \sigma \in \{0,1,2,3,4\} \end{aligned}$$

or

$$\begin{aligned} \left( {{\sigma }}^{2}-2\sigma +4+\Delta \right) =0,\;\; \left( {{\sigma }}^{2}-3\sigma +4+\Delta \right) =0,\;\; \left( {{\sigma }}^{2}-5\sigma +4+\Delta \right) =0 \end{aligned}$$

This concludes the list of all possibilities. We still have to backtrack some of the cases.

3.6 Backtracking the remaining cases

Earlier, we use the value for a which involved the denominator

$$\begin{aligned} (\sigma -1)(\sigma ^2-\sigma +\Delta ). \end{aligned}$$

The introduction of f led to a further denominator

$$\begin{aligned} 2\sigma +\Delta . \end{aligned}$$

The analysis of Case A gave \(\Delta \) as denominator, and Case C gave further factor \((\sigma -1)(\sigma -4)+\Delta \) in the denominator. In each of the cases we have to backtrack and see if we find additional solutions.

3.6.1 The case \(\sigma =1\)

Looking at the operator with \(\sigma =1\), we find from the coefficient of x of \(\mathcal {Q}\) the value of f. Using this value, and factoring the coefficient of \(x^2\) of \(\mathcal {Q}\), we find there is a unique value of c:

$$\begin{aligned} c=d\frac{\Delta -1}{\Delta }. \end{aligned}$$

Here a denominator \(\Delta \) appears, so we assume this to be \(\ne 0\); but the case \(\Delta =0\) has to be analysed anyway.

3.6.2 The case \(\Delta =0\)

To analyse the further cases, it is useful to write the operator in terms of A and B instead of \(\alpha \) and \(\beta \), so that \(\Delta =C\) can be implemented by putting \(B=C-A\). We start with \(C=0\). The following seven components make up the intersection with \(\Delta =0\):

  1. Case 1

    \(\sigma =\Delta =0,\;\;b,\;c,\;d=0\):

    $$\begin{aligned} \theta ^4+x(a\theta ^4+e)+fx^2(\theta ^4+u). \end{aligned}$$
  2. Case 2

    \(\sigma =1,\;\;\Delta =0,\;\;d=0,\;\;f=(a-c)c\);

    $$\begin{aligned} \theta ^4+x(a\theta ^4+2c\theta ^3+c\theta ^2+e)+(a-c)c(\theta ^4+2\theta ^2+3+\theta ^2+u). \end{aligned}$$
  3. Case 3

    \(\sigma =2,\;\;\Delta =0,\;\;a=c-d\). This is inside the main component.

  4. Case 4

    \(\sigma =2,\;\; \Delta =0,\;\;a=d,\;\;c=2d.\;\;f=d^2/4\).

  5. Case 5

    \(\sigma =3,\;\; \Delta =0,\;\;a=8c-9d/12,\;\;f=(2c-3d)(4c-3d)/72\).

    This is inside the \(\sigma =3\) component.

  6. Case 6

    \(\sigma =4,\Delta =0,\;\;d=0,\;\;c=2a,\;\;f=a^2/4\).

  7. Case 7

    \(d=\Delta =0\), there is the family with parameters \(c,e,\sigma ,\alpha \)

    $$\begin{aligned} a=\frac{2c}{\sigma },\quad b=2c,\quad d=0,\quad f=\frac{a^2}{4}. \end{aligned}$$

3.6.3 Remaining cases

There are more cases to check and this could be done along the above sketched lines. Instead of this, we used the computer algebra system Singular. There is a library called primdec.lib containing the algorithms for doing primary decomposition. On our computer, Singular could not straight away compute the primary decomposition of the ideal \(\mathcal {I}\) defined by the vanishing of the x-coefficients of \(\mathcal {Q}\). But after taking the ideal quotient of \(\mathcal {I}\) by the ideals already found, it turned out to be able to find the last component. In total we thus found seven irreducible components of the variety defined by \(\mathcal {Q}=0\).

3.7 Summary of the seven components

In total we have found seven components. Membership to a component is defined by certain polynomial relations between the coefficients. Below we give the defining relations for each of the components, the operator with its coefficients, together with its generic Riemann symbol.

The big \(\sigma =2\) component, also called the main component is defined by the linear equations

$$\begin{aligned} \sigma -2=0,\;\;a-(c-d)=0. \end{aligned}$$

For a general member of this component, the Riemann symbol is:

$$\begin{aligned} \left\{ \begin{array}{cccc} 0&{}\quad *&{}\quad *&{}\quad \infty \\ \hline 0&{}0&{}0&{}\alpha \\ 0&{}1&{}1&{}\beta \\ 0&{}1&{}1&{}2-\beta \\ 0&{}2&{}2&{}2-\alpha \end{array} \right\} , \end{aligned}$$

where \(*\) are the two roots of \(1+ax+fx^2=0\). The operator depends on the six parameters \(a,d,e,f,\alpha ,\beta \). The parameters a, f determine the position of the singular fibres. The parameters \(\alpha ,\beta \) determine the four exponents at \(\infty \) which are symmetric around the value 2. The parameters d, e are accessory parameters.

However, if the two roots coincide, \(a^2=4f\), then we obtain an operator with three singular points.

The \(\sigma =1\) component is defined by the four conditions:

$$\begin{aligned}{} & {} \sigma -1=0,\;\;c\Delta +d(\Delta -1)=0,\;\;\\{} & {} d(a-(c-d))+f\Delta =0,\;\;c(a-(c-d))+f(\Delta -1)=0. \end{aligned}$$

The Riemann symbol of the operator is:

$$\begin{aligned} \left\{ \begin{array}{cccc} 0&{}\quad x&{}\quad y&{}\quad \infty \\ \hline 0&{}0&{}0&{}\alpha \\ 0&{}1&{}1&{}\beta \\ 0&{}1&{}2&{}1-\beta \\ 0&{}2&{}3&{}1-\alpha \\ \end{array} \right\} , \end{aligned}$$

where the special points x and y are located at:

$$\begin{aligned} x=\frac{\Delta }{d},\;\;\; y=\frac{\Delta }{d+a\Delta }. \end{aligned}$$

The point y is not really a singular point of the differential operator, as the exponents \(0,\;1,\;2,\;3\) are those of a regular point of the differential equation. Indeed, it can be checked that no logarithms occur.

The \(\sigma =3\) component is defined by the conditions:

$$\begin{aligned} \begin{array}{rcl} 0&{}=&{}\sigma -3,\\ 0&{}=&{}4a^2-4c^2-3ad+11cd-7d^2-3fA+20f,\\ 0&{}=&{}2aA-cA+dA-12a+8c-9d,\\ 0&{}=&{}2ac-2c^2-3a^2aA-cA+dA-12a+8c-9d,\\ 0&{}=&{}2c^2A-5cdA+3d^2A+2fA^2-8c^2+18cd-9d^2-24fA+72f.\\ \end{array} \end{aligned}$$

The Riemann symbol of the operator is:

$$\begin{aligned} \left\{ \begin{array}{cccc} 0&{}\quad x&{}\quad y&{}\quad \infty \\ \hline 0&{}0&{}-1&{}\alpha \\ 0&{}1&{}0&{}\beta \\ 0&{}1&{}1&{}3-\beta \\ 0&{}2&{}2&{}3-\alpha \\ \end{array} \right\} , \end{aligned}$$

where the singular points are located at:

$$\begin{aligned} x=\frac{6+\Delta }{2c-3d},\;\;y=\frac{-2(6+\Delta )}{4c-3d+(c-d)\Delta }. \end{aligned}$$

The operator has five parameters \(c,d,e,\alpha ,\beta \). \(\alpha , \beta \) are parameters that determine the exponents at infinity, symmetric around 3. The parameters c, d determine together with \(\alpha \) and \(\beta \) the position of the two singularities. e is a single accessory parameter.

The \(\sigma =0\) component is defined by the conditions:

$$\begin{aligned} \sigma =0,\;\;c=0,\;\;d=0,\;\;\Delta =0. \end{aligned}$$

The generic Riemann symbol for this operator is:

$$\begin{aligned} \left\{ \begin{array}{cccc} 0&{}\quad x&{}\quad y&{}\quad \infty \\ \hline 0&{}0&{}0&{}\alpha \\ 0&{}1&{}1&{}i\alpha \\ 0&{}2&{}2&{}-\alpha \\ 0&{}3&{}3&{}-i\alpha \\ \end{array} \right\} , \end{aligned}$$

where the special points x, y are located at the solutions of \(1+ax+fx^2=0\). The operator depends on four parameters \(a,e,f,\alpha \).

The small \(\sigma =2\) component is defined by the four conditions:

$$\begin{aligned} \sigma -2=0,\;\;4ad-d^2-16f=0,\;\;2c-3d=0,\;\;\Delta +2=0. \end{aligned}$$

The Riemann symbol for a general member of this component is:

$$\begin{aligned} \left\{ \begin{array}{cccc} 0&{}\quad x&{}\quad y&{}\quad \infty \\ \hline 0&{}-1&{}0&{}1+\alpha \\ 0&{} 0&{}1&{}1+i\alpha \\ 0&{} 1&{}2&{}1-\alpha \\ 0&{} 2&{}3&{}1-i\alpha \\ \end{array} \right\} , \end{aligned}$$

where the special points are

$$\begin{aligned} x=-\frac{4}{d},\;\;y=-\frac{4}{4a-d}. \end{aligned}$$

The \(\sigma =4\) component is defined by the four conditions:

$$\begin{aligned} \sigma -4=0,\;\;6a-c=0,\;\;2c-3d,\;\;\Delta +8=0. \end{aligned}$$

The Riemann symbol of this operator is:

$$\begin{aligned} \left\{ \begin{array}{cccc} 0&{}\quad x&{}\quad y&{}\quad \infty \\ \hline 0&{}-1&{}-1&{}2+\alpha \\ 0&{} 0&{}0&{}2+i\alpha \\ 0&{} 1&{}1&{}2-\alpha \\ 0&{} 2&{}2&{}2-i\alpha \\ \end{array} \right\} , \end{aligned}$$

where the singular points x, y are solutions to \(1+ax+fx^2=0\).

The transverse component finally is defined by the equations

$$\begin{aligned} \begin{array}{rcl} 0&{}=&{}a^2-4f,\\ 0&{}=&{}a\sigma -2(c-d),\\ 0&{}=&{}a\Delta +2d,\\ 0&{}=&{}-f\Delta ^2+d^2,\\ 0&{}=&{}ad+2f\Delta ,\\ 0&{}=&{}d\sigma -c\Delta +d\Delta ,\\ 0&{}=&{}f\sigma \Delta -f\Delta ^2+cd,\\ 0&{}=&{}-f\sigma ^2+f\sigma \Delta +c^2-cd,\\ 0&{}=&{}ac-2f\sigma +2f\Delta .\\ \end{array} \end{aligned}$$

Solving for a, b, c, f we find the family of operators

$$\begin{aligned}{} & {} T(d,e,\alpha ,\beta ,\sigma ):= \theta ^4\\{} & {} \quad -x \frac{1}{\Delta }\left( 2d\theta ^4+2d \sigma \theta ^3+(\sigma -\Delta )\theta )\right) +x(\theta +e)\\{} & {} \quad -\left( \frac{d x}{\Delta }\right) ^2(\theta +\alpha )(\theta +\beta )(\theta +\sigma -\beta )(\theta +\sigma -\beta ) \end{aligned}$$

The Riemann symbol of the generic member in this family is:

$$\begin{aligned} \left\{ \begin{array}{cccc} 0&{}\quad x&{}\quad \infty \\ \hline 0&{} 0&{}\alpha \\ 0&{} 1&{}\beta \\ 0&{} 2-\sigma &{}\sigma -\beta \\ 0&{} 3-\sigma &{}\sigma -\alpha \\ \end{array} \right\} , \end{aligned}$$

where \(x=\frac{\Delta }{d}\).

The big and small \(\sigma =2\) components do intersect:

$$\begin{aligned} \sigma =2,\;\;\Delta =-2,\;\;2a=d,\;2c=3d,\;f=d^2/16. \end{aligned}$$

The transverse component intersects all other components. No other pair of components can intersect, as they have a different \(\sigma \)-value.

3.8 The small components

It turns out that the components with \(\sigma =0,\;\;2,\;\;4\) are closely related to each other and the solutions of the corresponding differential equations can be related in a simple manner. The families of operators in question are:

$$\begin{aligned} \begin{array}{rcl} P_0(a,e,f,A)&{}:=&{}\theta ^4+x(a\theta ^4+e)+x^2f(\theta ^4-A^2)\\ P_2(a,d,e,A)&{}:=&{}\theta ^4+x(a\theta ^4+d\theta ^3+(3/2)d\theta ^2+d\theta +e)\\ &{}&{}+\frac{d(4a-d)}{16}x^2((\theta +1)^4-(A-1)^2)\\ P_4(a,e,f,A)&{}:=&{}\theta ^4+x(a\theta ^4+4a\theta ^3+6a\theta ^2+4a\theta +e)\\ &{}&{}+x^2 f(\theta ^4+8\theta ^3+24\theta ^2+32\theta +A(8-A))\\ \end{array} \end{aligned}$$

Their relation is most easily understood from the perspective of the operator \(P_2(a,d,e,A)\) and its exponents. The discriminant \(\Delta (x)\) of the operator \(P_2(a,d,e,A)\) factors as \((1+dx/4)(1+(4a-d)x/4)\). The exponents at the first factor are \(-1, 0, 1, 2\) and at the second factor 0, 1, 2, 3. The exponents at infinity are of the form

$$\begin{aligned} 1+\alpha ,\;\; 1-\alpha ,\;\; 1+i \alpha ,\;\;1-i \alpha . \end{aligned}$$

Multiplication of a solution to \(P_2(a,d,e,A)\) by \((1+dx/4)\) will shift the exponents at the first singular point by \(+1\) and the exponents at infinity by \(-1\), which then leads to an operator with two points with exponents 0, 1, 2, 3 and

$$\begin{aligned} \alpha , -\alpha , i \alpha ,-i \alpha \end{aligned}$$

as exponents at infinity, i.e. with \(\sigma =0\) and \(\Delta =0\). Similarly, division of a solution to \(P_2(a,d,e,A)\) by \((1+(4a-d)x/4)\) leads to a shift at the second singularity by \(-1\) and hence produces two singularities with exponents \(-1,0,1,2\). At infinity the exponents shift by \(+1\) and are of the form

$$\begin{aligned} 2+\alpha ,\;\; 2-\alpha ,\;\; 2+i \alpha ,\;\;2-i \alpha , \end{aligned}$$

i.e. with \(\sigma =4\) and \(\Delta =-8\).

We remark that the exponents at infinity can only be all real in the very special case that these all coincide.

A precise statement relating these operators is the following:

Proposition 2

$$\begin{aligned} \begin{array}{c} y(x)~{\textit{is a solution of}}~P_2(a,d,e,A)\\ \Longleftrightarrow \\ (1+dx/4).y_0(x)~{\textit{is a solution of}}\\ P_0(a,e-d/4,(4a-d)d/16,A-1)\\ \Longleftrightarrow \\ (1+(4a-d)x/4)^{-1}y(x)~{\textit{is a solution of}}\\ P_4(a,e+(4a-d)/4,(4a-d)d/16), A-3) \end{array} \end{aligned}$$

It is a remarkable fact that the local monodromies around the special points \(\ne 0\) and \(\ne \infty \) are trivial. The monodromy around 0 is MUM, so has infinite order and thus the monodoromy group is the non-reductive group \(\textbf{Z}\), which clearly prevents these operators from being Picard–Fuchs operators of a family of algebraic varieties.

Something similar happens for the \(\sigma =1\) and \(\gamma =3\) components, given by the operator families

$$\begin{aligned} \begin{array}{rcl} P_1(a,d,e,\alpha ,\beta )&{}:=&{}\theta ^4+x\left( a\theta ^4-\frac{2d}{\Delta }\theta ^3+\frac{\Delta -1}{\Delta }\theta ^2+d\theta +e\right) \\ &{}&{}-\frac{d(d+a\Delta )}{\Delta ^2}(\theta +\alpha )(\theta +\beta )(\theta +1-\alpha )(\theta +1-\beta )\\ \end{array}, \end{aligned}$$

where \(\Delta =\alpha ^2+\beta ^2-\alpha -\beta \), and

$$\begin{aligned} P_3(c,d,e,\alpha ,\beta ):= & {} \theta ^4 +x \left( \frac{(\Delta +8)c+(\Delta +9)d}{2(\Delta +6)}+2(c-d)\theta ^3+c\theta ^2+d\theta +e\right) \\{} & {} +(2c-3d)\frac{(\Delta +4)c-(\Delta +3)d)}{2(\Delta +6)^2}x^2(\theta +\alpha )\\{} & {} (\theta +\beta )(\theta +3+\alpha )(\theta +3-\beta ), \end{aligned}$$

with \(\Delta =\alpha ^2+\beta ^2-3\alpha -3\beta \).

One verifies directly that

Proposition 3

$$\begin{aligned} \begin{array}{c} y(x)~{\textit{is a solution of}}~P_1(a,d,e,\alpha ,\beta )\\ \Longleftrightarrow \\ (\Delta -(d+a\Delta )x)y_0(x)~{\textit{is a solution of}}\\ P_3(d (\Delta -1)/\Delta ,d,e,\alpha +1,\beta +1) \end{array} \end{aligned}$$

These operators are also remarkable. One of the roots of the polynomial \(1+ax+fx^2\) is in fact a regular point; the other is a conifold point, so the three non-trivial monodromies are precisely as those for the hypergeometric operators. In fact, if we multiply the hypergeometric operator

$$\begin{aligned} \theta ^4-Nx(\theta +\alpha )(\theta ) \end{aligned}$$

with the linear factor \(1+Mx\), the result is the operator

$$\begin{aligned} P_1(M+N,-N\Delta ,e,\alpha ,\beta ),\;\;\;(\Delta =\alpha ^2+\beta ^2-\alpha -\beta ) \end{aligned}$$

for the specific value

$$\begin{aligned} e=N\alpha \beta (1-\alpha )(1-\beta ) \end{aligned}$$

of the accessory parameter e. If e takes on another value, the operator is no longer equal to linear factor \(\times \) hypergeometric.

4 The known Calabi–Yau operators of degree 2

Below we will give an overview of the 69 Calabi–Yau operators of degree two that are currently known to us. We also include some further remarkable operators that are not strictly Calabi–Yau in the sense of [4]. Of course, this is just a small portion of the list of the still growing list Calabi–Yau operators that was started in [3] and now contains more than 500 members. An update of this list in preparation [2].

For operators of degree two, there are cases with three and with four singular points. Most of the known operators can be related to hypergeometric operators or operators that are convolutions, that is, can be obtained via Hadamard product of operators of lower order. Recall that the Hadamard product of two power series \(\phi (x):=\sum _n a_n x^n\) and \(\psi (x):=\sum _n b_n x^n\) is the series

$$\begin{aligned} \phi *\psi (x) =\sum _n a_n b_n x^n \end{aligned}$$

and by a classical theorem, if \(\phi \) and \(\psi \) satisfy a linear differential equations, then so does \(\phi *\psi \). The Hadamard product corresponds to the multiplicative convolution of local systems, so are under very good control, see [24, 25].

Frobenius basis: For a fourth-order operator P with at MUM point at 0, there is a canonical basis of solutions \(y_0,y_1,y_2,y_3\) to \(P y=0\) on a sufficiently small slit disc around the origin, called the Frobenius basis:

$$\begin{aligned} \begin{array}{rcl} y_0(x)&{}=&{}f_0(x)\\ &{}&{}\\ y_1(x)&{}=&{}\log (x) f_0(x)+f_1(x)\\ &{}&{}\\ y_2(x)&{}=&{}\frac{1}{2}\log (x)^2 f_0(x)+\log (x)f_1(x)+f_2(x)\\ &{}&{}\\ y_3(x)&{}=&{}\frac{1}{6}\log (t)^3f_0(x)+\frac{1}{2}\log (x)^2 f_1(x)+\log (x)f_2(x)+f_3(x)\\ \end{array} \end{aligned}$$

where \(f_0(x) \in \textbf{Q}[[x]]\), \(f_i(x) \in x \textbf{Q}[[x]], i=1,2,3\).

The Calabi–Yau operators from [4] are characterised by three integrality conditions:

  1. (1)

    integrality of the solution \(y_0\):

    $$\begin{aligned} y_0(x) \in \textbf{Z}[[x]]. \end{aligned}$$
  2. (2)

    integrality of the q-coordinate q(x):

    $$\begin{aligned} q(x):=\exp (y_1(x)/y_0(x))=x \exp (f_1(x)/f_0(x))=x+\cdots \in \textbf{Z}[[x]] \end{aligned}$$
  3. (3)

    integrality of the instanton numbers \(n_d\).

These can be computed in several different ways. The second logarithmic derivative of \(y_2/y_0\) expressed in the q-coordinate is called the Yakawa-coupling K(q) of the operator P:

$$\begin{aligned} K(q) = \left( q\frac{d}{dq}\right) ^2 \left( \frac{y_2}{y_0} \right) \end{aligned}$$

Now expand K(q) as a Lambert series

$$\begin{aligned} K(q)=1+\sum _{k=1}^{\infty } \frac{ k^3 n_k q^k}{1-q^k}. \end{aligned}$$

The numbers \(n_d\) are called the (normalised, \(n_0=1\)) instanton numbers of P and are required to be integral. It is also natural not to require strict integrality, but rather to allow small denominators in \(f_0\), q(x) and \(n_d\) to appear, that is, to require only N-integrality, for some denominator N.

4.1 Description of the operators

Many of the Calabi–Yau operators of degree 2 that we will describe involve special lower-order operators of Calabi–Yau type. These are introduced and discussed in “Appendix B”.

4.1.1 Operators with three singular points

I. The 14 tilde operators There are 14 exponents \((\alpha _1,\alpha _2,\alpha _3,\alpha _4)\) with

$$\begin{aligned} \alpha _1 \le \alpha _2 \le \alpha _3 \le \alpha _4,\;\;\alpha _1+\alpha _4=\alpha _2+\alpha _4=1,\; \end{aligned}$$

for which the hypergeometric operator, scaled by N,

$$\begin{aligned} \theta ^4-Nx(\theta +\alpha _1)(\theta +\alpha _2)(\theta +\alpha _3)(\theta +\alpha _4), \end{aligned}$$

is a Calabi–Yau operator, [8]. Corresponding to these, there are also 14 hypergeometric fifth-order Calabi–Yau operator

$$\begin{aligned} \theta ^5-4Nx(\theta +\alpha _1)(\theta +\alpha _2)(\theta +\frac{1}{2})(\theta +\alpha _3)(\theta +\alpha _4), \end{aligned}$$

with Riemann symbol

$$\begin{aligned} \left\{ \begin{array}{ccc} 0&{}\quad 1/4N&{}\quad \infty \\ \hline 0&{}0&{}\alpha _1\\ 0&{}1&{}\alpha _2\\ 0&{}3/2&{}1/2\\ 0&{}2&{}\alpha _3\\ 0&{}3&{}\alpha _4\\ \end{array} \right\} . \end{aligned}$$

These operators have a Yifan Yang pull-back to 14 special fourth-order operators, called the tilde-operators \(\widetilde{1},\widetilde{2},\ldots , \widetilde{14}\). These operators replace the more complicated hat-operators \(\hat{i},i=1,2,\ldots ,14\) that appeared in the list [3] into which they can be transformed. However, operator \(\widetilde{3}\) is not ’new’, as it can be reduced to operator 2.33.

II. The 16 operators of type \(H *\mu (H)\)

There are four special hypergeometric second-order operators called A, B, C, D. The Möbius transformation that interchanges the singularity with infinity leads to four Möbius transformed hypergeometric operators

$$\begin{aligned} \mu (A):=e,\;\;\; \mu (B)=h,\;\;\; \mu (C)=i,\;\;\; \mu (D)=j, \end{aligned}$$

where e, h, i, j refer to the names given in [4], see also “Appendix B”.

Thus, we can form the 16 Hadamard product \(A *\mu (B)\), etc., and obtain 16 fourth-order operators of degree two, with three singular points.

But there are two surprises: the operator \(C *\mu (A)\) can be reduced to the hypergeometric operator \(\hbox {AESZ}\,\#3=1.3\) and the operator \(C *\mu (B)\) has vanishing instanton numbers and in fact is the third symmetric power of the hypergeometric second-order operator

$$\begin{aligned} {\theta }^{2}-12\, x\left( 12\,\theta +7 \right) \left( 12\,\theta +1 \right) . \end{aligned}$$

So, from this we obtain only 14 Calabi–Yau operators of degree two that are ’new’.

III. The four Hadamard products \(I*\mu (H')\)

There are also four special hypergeometric third-order operators \(A',B',C',D'\), analogous to second-order operators A, B, C, D. Interchanging the singularity with the point at infinity, we obtain third-order operators

$$\begin{aligned} \mu (A')=\beta ,\;\;\;\mu (B')=\iota ,\;\;\;\mu (C')=\theta ,\;\;\;\mu (D')=\kappa \end{aligned}$$

of degree two, with three singular points. Taking Hadamard product with the central binomial coefficient produces four fourth-order operators of degree two. One of these, \(I *\beta \), can be transformed by a quadratic transformation to an operator \((I *\beta )^*\)

$$\begin{aligned} \theta ^4- 16(4\theta +1)(4\theta +3)(8\theta ^2+8\theta +3)+4096x^2(4\theta +1)(4\theta +3)(4\theta +5)(4\theta +7), \end{aligned}$$

which in turn can be transformed to the hypergeometric operator AESZ #3. The operator \(I *\theta \) can be transformed in 2.17, so only two of these give rise to ’new’ Calabi–Yau operators, [9].

IV. Four sporadic operators

The operator \(2.63=\#\hbox {AESZ} 84\) cannot be constructed in this way. Furthermore, there are three operators of Bogner-type that were obtained from transformations of the higher degree operators AESZ#245, AESZ#406 and AESZ #255. As these operators have unusual properties, they probably could better be understood in terms of their higher degree versions. We remark that the local monodromies around the two singularities \(\ne 0\) are of finite order.

4.1.2 Operators with four singular points

I. The 24 operators of type \(H *BZB\)

There are six special second-order operators of degree two with four singular fibres and Riemann symbol of the form

$$\begin{aligned} \left\{ \begin{array}{cccc}0 &{}\quad *&{}\quad *&{}\quad \infty \\ \hline 0&{}0&{}0&{}1\\ 0&{}0&{}0&{}1\\ \end{array}\right\} . \end{aligned}$$

So, we are dealing here with special Heun operators. They appear in the work of Beukers [13] and Zagier [44] and are related to the rational elliptic surfaces described by Beauville [12], so we call them the BZB operators. They were denoted a, b, c, d, f, g in [4].

Taking the Hadamard product with the hypergeometric operators A, B, C, D produces 24 fourth-order operators of degree two, with four singular points.

II. The six Hadamard products \(I *BZB'\)

The BZB operators have also analogues as third-order operators that we will call the BZB’-operators, that were called \(\alpha ,\gamma ,\delta ,\epsilon ,\zeta ,\eta \) in [4]. Taking \(I *\) with these operators produces six ’new’ fourth-order operators of degree two with four singularities.

III. Six Sporadic operators with four singularities

III.a. Three operators \(I *\) sporadic third order: There are three sporadic third-order operators of degree two, described in “Appendix B”. These lead to fourth-order operators of degree two by taking \(I *\) with them.

III.b: Three original operators: There are three operators that apparently cannot be constructed starting from simpler operators of lower order; we call them original operators.

The count

So, in total we have described

$$\begin{aligned} (14-1)+(16-2)+(4-2)+4=33 \end{aligned}$$

Calabi–Yau operators of degree two with three singular points and

$$\begin{aligned} 24+6+3+3=36 \end{aligned}$$

with four singular points, making up a total of 69. Because these all have different instanton numbers, these are essentially different and there are no algebraic transformations mapping one to any of the other operators.

Miscellaneous operators

There are two further types of operators of degree two that deserve to be mentioned at this place, as they challenge the precise definition of Calabi–Yau operator, as formulated in [4].

Reducible operators: A large scale Zagier-type search for degree two operators with integral solutions has been performed by Pavel Metelitsyn. Apart from the operators equivalent to the ones mentioned above, there are some interesting reducible fourth-order operators. With regard to the q-coordinate and instanton numbers these examples behave very much in the same way as the irreducible ones. An example is the operator

$$\begin{aligned}{} & {} {\theta }^{4}-12x\left( 6\theta +1 \right) \left( 6\theta +5 \right) \left( 2{\theta }^{2}+2\theta +1 \right) \\{} & {} \quad + 144 x^2 \left( 6\theta +1\right) \left( 6\theta +5 \right) \left( 6\theta +7\right) \left( 6\theta +11\right) . \end{aligned}$$

One finds:

$$\begin{aligned} \phi (x)= & {} 1 + 60 x + 13860 x^2 + 4084080 x^3 + 1338557220 x^4 \\{} & {} + 465817912560 x^5+\cdots \\ q(x)= & {} x + 312 x^2 + 107604 x^3 + 39073568 x^4 + 14645965026 x^5+\cdots \\ n_1= & {} -192,\;\;n_2=4182,\;\;n_3=-229568,\;\;n_4=19136058,\\ n_5= & {} -2006581440. \end{aligned}$$

But note that the above series \(\phi (x)\) is just the solution of the second-order operator D, and the above operator is just the compositional square of D. (Doing this with the hypergeometric second-order operators A, B, C produces periodic instanton numbers.)

A strange operator: Bogner [17] also found an example of an operator of degree two with an integer solution, integral q-coordinate, integral K(q)-series, but for which the instanton numbers are badly non-integral. We start from the third-order operator

$$\begin{aligned} L:=\theta ^{3}-4 x\left( 2\theta +1 \right) \left( 5\theta ^{2}+5 \theta +2\right) +48 x^2 \left( 3\theta +2 \right) \left( 3 \theta +4 \right) \left( \theta +1 \right) \end{aligned}$$

with integral solution

$$\begin{aligned} 1 + 8 x + 96 x^2 + 1280 x^3 + 17440 x^4 + 231168 x^5+\cdots \end{aligned}$$

The operator \(I *L\) is the fourth-order operator

$$\begin{aligned} \theta ^4-8 x (2\theta + 1)^2(5\theta ^2 + 5\theta +2) + 192 x^2(2\theta + 1)(3\theta + 2)(3\theta + 4)(2\theta + 3). \end{aligned}$$

We find the following nice integral series:

$$\begin{aligned} y_0(x)= & {} 1 + 16 x + 576 x^2 + 25600 x^3 + 1220800 x^4 + 58254336 x^5+\cdots \\ q(x):= & {} x + 40 x^2 + 1984 x^3 + 106496 x^4 + 5863168 x^5 + \cdots \\ K(q):= & {} 1 + 8 q^2 - 5632 q^3 - 456064 q^4 - 17708032 q^5+ \cdots \end{aligned}$$

However, if we expand K(q) as a Lambert series, we do not get integral instanton numbers:

$$\begin{aligned} n_1= & {} 8,n_2=-1,n_3=-{\frac{1880}{9}},n_4=-7126,n_5=-{\frac{3541608}{25}},\ldots ,\\ n_{17}= & {} \frac{2432475693294458880448632}{289},\ldots \end{aligned}$$

Experimentally, the denominator \(p^2\) appears in \(n_p\) for

$$\begin{aligned} p = 3,\;\; 5,\;\; 7,\;\; 11,\;\; 13,\;\; 17,\;\; 19,\ldots \end{aligned}$$

This is rather puzzling. Conjecturally, already the fact that \(\phi (x)\) is an integral series implies that the operator is a factor of a Picard–Fuchs equation of a family of varieties defined over \(\textbf{Q}\) and indeed our operator is of geometrical origin. It was conjectured in [4] that the integrality of the q-series implies the integrality of the instanton numbers, which in this example is not the case. The integrality of solution and mirror map clearly indicates that we have a rank four Calabi–Yau motive and one would expect the general arguments for the integrality of [43] to be applicable, but apparently they are not. There might exist a different scaling of the coordinate that repairs this defect, but up to now we have been unable to find it.