Co-Axial Metrics on the Sphere and Algebraic Numbers

In this paper, we consider the following curvature equation $$\Delta u+{\rm e}^u=4\pi\biggl((\theta_0-1)\delta_0+(\theta_1-1)\delta_1 +\sum_{j=1}^{n+m}\bigl(\theta_j'-1\bigr)\delta_{t_j}\biggr)\qquad \text{in}\ \mathbb R^2,$$ $$u(x)=-2(1+\theta_\infty)\ln|x|+O(1)\qquad \text{as} \ |x|\to\infty,$$ where $\theta_0$, $\theta_1$, $\theta_\infty$, and $\theta_{j}'$ are positive non-integers for $1\le j\le n$, while $\theta_{j}'\in\mathbb{N}_{\geq 2}$ are integers for $n+1\le j\le n+m$. Geometrically, a solution $u$ gives rise to a conical metric ${\rm d}s^2=\frac12 {\rm e}^u|{\rm d}x|^2$ of curvature $1$ on the sphere, with conical singularities at $0$, $1$, $\infty$, and $t_j$, $1\le j\le n+m$, with angles $2\pi\theta_0$, $2\pi\theta_1$, $2\pi\theta_\infty$, and $2\pi\theta_{j}'$ at $0$, $1$, $\infty$, and $t_j$, respectively. The metric ${\rm d}s^2$ or the solution $u$ is called co-axial, which was introduced by Mondello and Panov, if there is a developing map $h(x)$ of $u$ such that the projective monodromy group is contained in the unit circle. The sufficient and necessary conditions in terms of angles for the existence of such metrics were obtained by Mondello-Panov (2016) and Eremenko (2020). In this paper, we fix the angles and study the locations of the singularities $t_1,\dots,t_{n+m}$. Let $A\subset\mathbb{C}^{n+m}$ be the set of those $(t_1,\dots,t_{n+m})$'s such that a co-axial metric exists, among other things we prove that (i) If $m=1$, i.e., there is only one integer $\theta_{n+1}'$ among $\theta_j'$, then $A$ is a finite set. Moreover, for the case $n=0$, we obtain a sharp bound of the cardinality of the set $A$. We apply a result due to Eremenko, Gabrielov, and Tarasov (2016) and the monodromy of hypergeometric equations to obtain such a bound. (ii) If $m\ge 2$, then $A$ is an algebraic set of dimension $\leq m-1$.


Introduction
Let S 2 , g 0 be the 2-dimensional sphere with the standard smooth metric g 0 .In this paper, we consider the following classical problem in conformal geometry: Given t 1 , . . ., t N ∈ S 2 and a set of positive real numbers θ t 1 , . . ., θ t N , is there a metric of constant curvature 1 conformal to g 0 such that the metric is smooth on S 2 \ {t 1 , . . ., t N } and has conical singularities at t j with the angle 2πθ t j ?This problem has been widely studied in the literature, and in particular, it was solved in [8,9,10] in the cases where there are at most 2 non-integer angles among {θ t 1 , . . ., θ t N } We simply say that the angle is non-integer if θ j / ∈ Z .In this paper, we consider the case that there are at least 3 non-integer angles among {θ t 1 , . . ., θ t N }.Without loss of generality, we may assume that three singularities with non-integer angles are 0, 1, and ∞.Then it is well known that this problem is equivalent to solving the following curvature equation ∆u + e u = 4π α 0 δ 0 + α where δ p is the Dirac measure at p, n ≥ 0, m ≥ 1, α 0 , α 1 , α ∞ , α t j ∈ R >−1 \Z for all 1 ≤ j ≤ n and α t j ∈ N for all n + 1 ≤ j ≤ m.Geometrically, a solution u(x) of (1.1) leads to a conformal metric ds 2 = 1 2 e u |dx| 2 of curvature 1 on S 2 with angles 2πθ p at conical singularities p ∈ {0, 1, ∞, t j }, where the angles are given by It is known that if (1.1) has a solution u(x), then there is a developing map h(x) such that h(x) is a multi-valued meromorphic function on C satisfying x ∈ Ċ := C \ {0, 1, t 1 , . . ., t m+n }.
This is known as the Liouville formula and this h(x) is multi-valued and has its projective monodromy group contained in PSU (2).More precisely, given any ℓ ∈ π 1 Ċ, q 0 where q 0 ∈ Ċ is base point, the analytic continuation of h(x), denoted by ℓ * h(x), is also a developing map of u(x) and so ℓ * h(x) = ah(x)+b ch(x)+d for some projective monodromy matrix a b c d ∈ PSU (2).The curvature equation (1.1) has been extensively studied in the past several decades.For the recent development of this subject and its application, we refer [2,3,4,5,6,8,9,10,11,13,15,16,17,18] to the interested reader.So far, the existence of solutions for (1.1) with general singular sources is challenging and seems to be out of reach.Mondello and Panov [17] studied a reduced problem: To describe possible angles (i.e., to describe the data {θ 0 , θ 1 , θ ∞ , θ t 1 , . . ., θ t n+m }) such that (1.1) has solutions for some singular set {t 1 , . . ., t n+m }.They introduced the measure where Z n+m+3 o is the subset of Z n+m+3 consisting of vectors with odd sums of coordinates, and d 1 is the ℓ 1 -distance.Then they proved the following interesting result.
1.If (1.1) has a solution u(x) for some singular set {t 1 , . . ., t n+m }, then then there is a developing map h(x) such that the projective monodromy group of h(x) is contained in the unit circle, i.e., any projective monodromy matrix of h(x) is diagonal.
then there exists some singular set {t 1 , . . ., t n+m } such that (1.1) has solutions.Definition 1.2.In [17], the corresponding metric 1  2 e u(x) |dx| 2 (if exists) is called co-axial if there is a developing map h(x) such that the projective monodromy group of h(x) is contained in the unit circle.In this paper, we also call this solution u(x) co-axial for convenience.
This paper aims to study this problem.Note that (1.4) implies m ≥ 1, this is the reason why we assume m ≥ 1 in this paper.
Let us consider the special case m = 1 first.Denote Furthermore, for each t = (t 1 , . . ., t n+1 ) ∈ A, all t j 's are algebraic over Q θ and the field where M is a constant depending on n and θ ′ n+1 .
In the special case when there are only three non-integer angles, i.e., when n = 0, we can get a much sharper bound for the cardinality of A. In the following, we will write t 1 and θ ′ 1 simply by t and θ.
Theorem 1.6.Assume that n = 0 and m = 1.Assume that ϵ 0 , ϵ 1 ∈ {±1} are signs such that counted with multiplicities (see Section 5 for an explanation of multiplicities).Moreover, if t ∈ A, then the degree of t over Q θ is at most We remark that the condition θ ≡ k mod 2 is necessary for A ̸ = ∅, by (1.4).Interestingly, a key ingredient in the proof of the theorem is the monodromy of the classical hypergeometric equations.
Remark 1.7.Theorem 1.6 yields a sharp non-existence result for (1.1) when n = 0 and m = 1.That is, when θ ≡ k mod 2, if t ∈ C is transcendental over Q θ or is algebraic of degree greater than θ 2 − k 2 /4 over Q θ , then (1.1) has no solutions (not just co-axial solutions).Indeed, suppose (1.1) has a solution u(z), since θ ≡ k mod 2 easily yields d 1 Z 4 o , θ − 1 = 1, it follows from Theorem 1.1 that u(z) is co-axial, i.e., t ∈ A, a contradiction with Theorem 1.6.It would be an interesting problem to obtain a similar result for the case n > 0.
For general cases m ≥ 2, we can not expect that A is a finite set.Let Q denote the algebraic closure of Q.
then for any t = (t 1 , . . ., t n+m ) ∈ A, the transcendence degree of One of the main ideas in our proofs of all the main results is the integrability of the curvature equation.This integrability property allows us to connect the partial differential equation with a second-order Fuchsian ordinary differential equation.In Section 2, we will derive those ODE's.Applying integrability in terms of the developing maps, we will prove an equivalent condition for the existence of co-axial solutions.Based on this equivalent condition, we will prove Theorems 1.5 and 1.10 in Section 3. In Section 4, we give a proof of Theorem 1.4.In the final section, we present a proof of Theorem 1.6 using hypergeometric equations.

The integrability of curvature equations and complex ODEs
In this section, we briefly review how to connect the integrability of the curvature equation (1.1) with a class of second order ODEs in complex variables; see, e.g., [2,9].Given a solution u(x) of (1.1), we define Then it is easy to see Q x(x) ≡ 0, so Q(x) is a meromorphic function in C with poles belonging to Consequently, Thus, where and d p 's are some constants satisfying Thus, d 0 and d 1 can be uniquely determined by while d could be considered as n + m unknown coefficients.
On the other hand, as mentioned in the introduction, the Liouville theorem (see, e.g., [2]) says that for a solution u(x) of (1.1), there is a developing map h(x) such that x ∈ Ċ := C \ I. (2.4) The developing map h(x) is multi-valued in C and due to (2.4), the projective monodromy group of h(x) is contained in PSU(2).More precisely, given any ℓ ∈ π 1 Ċ, q 0 where q 0 ∈ Ċ is a base point, the analytic continuation of h(x), denoted by ℓ * h(x), is also a developing map of u(x) and so ℓ * h(x) = ah(x)+b ch(x)+d for some projective monodromy matrix a b c d ∈ PSU(2).To connect Q(x) and h(x), we note that (2.4) implies where {h(x), x} is the Schwarz derivative.We refer the reader to [2] for details of computation.Consider the following second order linear ODE with the complex variable x: Then a classical result (see [19]) says that for any basis (ŷ 1 , ŷ2 ) of (2.6), the Schwarz derivative {ŷ 2 /ŷ 1 , x} always satisfies From this and (2.5), we conclude that if (2.6) is derived from a solution of (1.1), then there is a basis (y 1 , y 2 ) of (2.6) such that Furthermore, the projective monodromy group of (2.6) with respect to y 1 and y 2 is the same as that of h(x), i.e., is contained in PSU(2).Conversely, given Q(x) via (2.1)-(2.3),if the corresponding ODE (2.6) has a basis (y 1 , y 2 ) such that the projective monodromy group of the ratio h(x) = y 2 (x)/y 1 (x) is contained in PSU(2), then u(x) := ln 8|h ′ (x)| 2 (1+|h(x)| 2 ) 2 is a solution of (1.1).For the reader's convenience, we now briefly recall some basic notions of ODEs like (2.6).We refer the reader to [14,19] for a comprehensive introduction.Generally, (2.6) is called Fuchsian because the pole order of Q(x) is at most 2 at any singularities.Note that the Riemann scheme (see [14, p. 11] for the definition of the Riemann scheme) of (2.6) is given by Observe that the exponent difference at t j is an integer for n + 1 comes from a solution u(x) of (1.1), it is well known that (2.6) is apparent at t j i.e., no logarithmic singularity near t j and so the local monodromy matrix of (2.6) at t j is (−1) αt j I 2 for all n+1 ≤ j ≤ n+m.In this paper, we will also consider a Fuchsian differential equation (2.6) which may not come from a solution of (1.1).In this case, unless otherwise specified, we always assume that (2.6) with the Riemann scheme (2.8) is apparent at t j for all n + 1 ≤ j ≤ n + m.
It is well known that the necessary and sufficient condition of (2.6) being apparent at t j for all n + 1 ≤ j ≤ n + m can be derived from the Frobenius method.
Theorem 2.1.There are polynomials Furthermore, Pj (d, t) can be written as for some N j ∈ Z ≥0 and P j (d, t) ∈ Q θ [d, t] such that P j (d, t) and p∈I\{t j } (p − t j ) N j are coprime.Thus, (2.6) is apparent at t j for all n + 1 ≤ j ≤ n + m if and only if P j (d, t) = 0 for all j = n + 1, . . ., n + m.
Proof .Let us take t n+m for example and in the following proof, we write t n+m = t for convenience.By the Frobenius method, (2.6) is apparent at the singularity t if and only if it has a solution of the form Observe that where where (j Clearly, (2.10) holds automatically for j = 0, and (2.10) with j = 1 leads to c 1 = dt 1−θt .By an induction argument, for any 2 ≤ j ≤ θ t − 1, c j can be uniquely solved by (2.10) as with total degree j in d, where Consequently, (2.10) with j = θ t leads to the existence of r θt ∈ Q θ \ {0} and a polynomial with total degree θ t in d such that Then it is standard by the Frobenius theory that (2.6) is apparent at the singularity t if and only if P (d, t) = 0. Furthermore, the above argument also implies the existence of an integer 0 ≤ N ≤ θ t such that and P (d, t), p∈I\{t} (t − p) N are coprime.■ The case n = 0 is special and we plan to study the solvability of (1.1) for this case in another paper.In this case, the ODE (2.6) can be transformed into a Heun equation and this Heun equation has been well studied in [8,12].In Section 5, we will use some results from [8] and hypergeometric equations to prove Theorem 1.6.

Proofs of Theorems 1.5 and 1.10
This section is devoted to the proofs of Theorems 1.5 and 1.10.In the following we use notations

General setting
As mentioned in the introduction, following [17], we call the metric 1 2 e u(x) |dx| 2 and also the solution u(x) of (1.1) to be co-axial if there is a developing map h(x) of u(x) such that the projective monodromy group of h(x) is contained in the unit circle, that is, for any γ ∈ π 1 Ċ, q 0 , the analytic continuation of h(x) along γ, denoted by Clearly, this is equivalent to that the monodromy group of the associated ODE (2.6) with Q In Section 2, we have discussed the equivalence between the existence of a solution u(x) of (1.1) and a Fuchsian ODE (2.6) with Q(x) given by (2.1) satisfying P j (d, t) = 0 for all n + 1 ≤ j ≤ n + m (see Theorem 2.1) such that the monodromy group is conjugate to a subgroup of SU (2).This equivalence can be further strengthened as follows when the solution u(x) is co-axial.
In fact, given such an ODE (2.6), take the basis of local solutions near ∞ as follows Suppose that (2.6) comes from a co-axial metric.By (2.7), we have h(x) = y 2 (x)/y 1 (x) for some solutions y i (x) of (2.6).Consider a large circle |x| = R and let h e 2πi x be the analytic continuation of h(x) along the circle.Then by the co-axial condition, one has for some λ ̸ = 0. On the other hand, there is a matrix These two identities together imply that b = c = 0. Thus, after multiplying by some nonzero constants, we may assume that which satisfies Then under the basis (y + (x), y − (x)), we have M ∞ = e −πiα∞ 0 0 e πiα∞ with e −πiα∞ ̸ = e πiα∞ .Here for any p ∈ I ∪ {∞}, we use M p ∈ SL(2, C) to denote the monodromy matrix of ODE (2.6) with respect to a simple loop in π 1 Ċ, q 0 that encircles p once.Now suppose that the monodromy group of (2.6) is commutative (this holds if (2.6) comes from a co-axial solution u(x)), then it follows that all the monodromy matrices under the basis (y + (x), y − (x)) are diagonal, i.e., M p = e ±πiαp 0 0 e ∓πiαp for p ∈ I 1 = {0, 1, t 1 , . . ., t n }.This implies that after analytic continuation, αp 2 + 1 are the local exponents of (2.6) at p and c ± p ̸ = 0. Thus there exists ϵ p ∈ {±1} for p ∈ I 1 such that From here and (3.3), we see that h(x) can be written as where ĥ(x) is meromorphic in C and satisfies ĥ(x) = x −θ∞−ϵ 0 θ 0 −ϵ 1 θ 1 − n j=1 ϵt j θt j 1 + O x −1 as x → ∞, so ĥ(x) is a rational function. (3.5) Conversely, given an ODE (2.6) with (2.8) which might not come from a solution of (1.1), we consider h(x) = y + (x)/y − (x) as in (3.1) and (3.2).If there are ϵ p ∈ {±1} for p ∈ I 1 such that h(x) has the expression (3.4) with ĥ(x) being a rational function, then for any γ ∈ π 1 Ċ, q 0 , γ * h(x) = λ(γ)h(x) with |λ(γ)| = 1 (i.e., the projective monodromy group of h(x) is a subgroup of the unit circle, or equivalently the monodromy matrices of (2.6) are all diagonal under the basis (y + (x), y − (x))).Thus is well defined in C. Then a direct computation shows that u(x) is a co-axial solution of (1.1) with h(x) being a developing map.Therefore, the above argument proves the following result.
Lemma 3.1.The equation (1.1) has co-axial solutions if and only if there is an ODE (2.6) with the Riemann scheme (2.8) such that h(x) = y + (x)/y − (x) has the expression (3.4) for some ϵ p ∈ {±1} for p ∈ I 1 and some rational function ĥ(x).
Remark that h(x) or equivalently ĥ(x) might have zeros or poles at t n+j for some j ≥ 1.More precisely, the Riemann scheme (2.8) indicates that the exponent difference is α t j + 1 = θ t j at t j , so near t j with n + 1 ≤ j ≤ n + m, we have and so does ĥ(x).From here and (3.5), there is where a j 's and b k 's satisfy In other words, Then by h(x) = y + (x)/y − (x), we see that each a j (resp.b j ) is a zero of y + (x) (resp.y − (x)) and must be simple, namely each a j is a simple zero of h(x) and each b j is a simple pole of h(x), so a j ̸ = a k and b j ̸ = b k for any j ̸ = k.Therefore, Clearly, (3.3) implies By (3.6), we have where G(x) is a polynomial defined by where |J 1 | := #J 1 ≥ n + 2 and Note that the last identity is due to (3.7).
Definition 3.4.We say that a common zero (a, b, t) ∈ C m 1 +m 2 +n+m of the polynomials B j 's in (3.18) is admissible if any two elements of (a, b, t 1 , . . ., t n+m ) do not equal and none of them equals to 0, 1.
Then the above argument shows that once (2.6) comes from a co-axial solution u(x) of (1.1), then there exist I 1 ⊂ J 1 ⫋ I and ϵ p ∈ {±1} for p ∈ J 1 such that m 1 , m 2 defined by (3.14) and (3.15) are non-negative integers, (3.10) holds and the corresponding polynomials B j 's in (3.18) has an admissible zero (a, b, t).The following result shows that the converse statement also holds.Lemma 3.5.Suppose there exist I 1 ⊂ J 1 ⫋ I and ϵ p ∈ {±1} for p ∈ J 1 such that m 1 , m 2 defined by (3.14) and (3.15) are non-negative integers, i.e., (3.10) holds and the corresponding polynomials B j 's in (3.18) are well defined.If B j 's have an admissible zero (a, b, t), then (1.1) has a co-axial solution u(x) with its developing map given by (3.6).
From here and e −2πiα∞ ̸ = 1, we easily obtain b = c = 0 and so h(x) = ay + (x)/dy − (x) = y + (x)/y − (x), where a/d = 1 follows from (3.20) and (3.1).Then by Lemma 3.1, we conclude that (1.1) has a co-axial solution u(x) with h(x) being its developing map.■ Therefore, the problem turns to study the admissible zeros of the polynomials B j 's.The number of the polynomials is and the number of variables is m 1 + m 2 + n + m ≥ L. Thus there are two different cases: (i) For the case (i), we may expect that the number of common zeros of the polynomials B j 's is finite, but we can not expect the validity of this assertion for the case (ii).
Therefore, let us study the case m 1 + m 2 + n + m = L first.In this case, |J 1 | = |I| − 1 = n + m + 1, so I \ J 1 consists of a single point belonging to {t n+j | 1 ≤ j ≤ m}.Without loss of generality, we may assume Then t 1 = (t 1 , . . ., t n+m−1 ), (3.17) becomes and so the polynomials B j 's become . Furthermore, for each t = (t 1 , . . ., t n+m ) ∈ A, all t j 's are algebraic over Q θ , and the field where M is a constant depending on the integer angles θ t j 's.
Proof .Given any t 0 = (t 0,1 , . . ., t 0,n+m ) ∈ A. Then the polynomials B j 's in (3.27) have an admissible zero (a, b, t) such that t = t 0 .We claim that The polynomial system B j (a, b, t) = 0 with 1 ≤ j ≤ L has at most L! = (θ t n+m − 1)! solutions. (3.28) Once this claim is proved, we can conclude that A is a finite set with #A ≤ due to the fact that B j is invariant under any permutation of a 1 , . . ., a m 1 and any permutation of b 1 , . . ., b m 2 for all j.
To prove the claim (3.28), we consider the homogenization of this polynomial system We show that the solution of (3.29) with ε = 0 must be 0, i.e., (a, b, t) = 0. First of all, since 0, 1 ∈ J 1 , we see from (3.9) and (3.27) that and so Thus B L (a, b, t, 0) = 0 implies t n+m = 0.
To prove that (a, b, t 1 , . . ., t n+m−1 ) = 0, we define i.e., replace the non-homogenous factor x − 1 of h(x) in (3.6) with x.Then where the expression of G(x) is obtained from that of G(x) in (3.9) by replacing the term x − 1 with x.In other words, where R j (a, b, t 1 ) is the homogeneous part of degree j of R j (a, b, t), and so Thus B j (a, b, t 1 , 0, 0) = 0 for any j yields R j (a, b, t 1 ) = 0 for any j and so Suppose the number of those a j 's being 0 is n 1 , the number of those b j 's being 0 is n 2 and denote and so for some c, c ′ ̸ = 0. Therefore, Therefore, the homogenized system (3.29) has no solutions at infinity.Then by the Bezout theorem, the polynomial system B j (a, b, t) = 0 with 1 ≤ j ≤ L has exactly L! solutions by counting multiplicity.This proves the claim (3.28).Furthermore, it follows from Lemma 3.7 below that for any solution (a, b, t), every element of (a, b, t) is algebraic over Q θ , i.e., every element belongs to Q θ , where Q θ denotes the algebraic closure of Q θ .
Proof .Let I be the ideal of K[x 1 , . . ., x ℓ ] generated by the polynomials H j .For i = 1, . . ., ℓ, consider the elimination ideal Under the assumption that (3.30) has finitely many solutions, I i is generated by some polynomial f i (x i ) ∈ K[x i ] whose roots are precisely those t i such that (t 1 , . . ., t ℓ ) is a solution of (3.30) for some (t 1 , . . ., t i−1 , t i+1 , . . ., t ℓ ), by the closure theorem in the elimination theory (see [7,Theorem 3,Section 3.3]).Since all roots of f i are in K, this shows that the coordinates of any solution (t 1 , . . ., t ℓ ) of (3.30) are algebraic over Q(y 1 , . . ., y ℓ ).■

The case m = 1
This section is devoted to the proof of Theorem 1.5.

3.3
The general case m ≥ 2 and in particular, the case m 1 + m 2 + n + m > L appears generally, so we can not expect that the corresponding polynomial system has only finitely many solutions.This is the difference from the m = 1 case.Our key observation is following.
Lemma 3.8.The solution set W of the polynomial system Consequently, the dimension of the set of admissible zeros of the polynomials B j 's is ≤ m − 1.
Proof .Note that is an open subset of W . Fix any (n + m + 2 − |J 1 |)-tupe (i.e., the first component of t 2,0 is 1.) we claim that there are L! points (by counting multiplicities) in W satisfying t 2 = tt 2,0 for some t ∈ C.
Once this claim is proved, then the dimension of To prove this claim, we insert t 2 = tt 2,0 into (3.17),we obtain Then the polynomial system (3.31)becomes The advantage of this new polynomial system (3.33) is that the numbers of unknowns and equations are the same.Note that C L (t 2,0 ) = (−1) L t αt j j,0 ̸ = 0. Then the same proof as (3.29) shows that the homogenized system of (3.33) has no solutions at infinity, so it follows from the Bezout theorem that the polynomial system (3.33) has exactly L! solutions by counting multiplicity.This proves the claim (3.32).■ Proof of Theorem 1.10.Due to the finite choices of J 1 and (ϵ p ) p∈J 1 , the assertion that the dimension of A is ≤ m − 1 follows from Lemma 3.8.Now we further assume (1.9).Then Given any t 0 = t1 , . . ., tn+m ∈ A, i.e., (1.1) has a co-axial solution for some t 0 = t1 , . . ., tn+m satisfying tj ̸ = 0, 1 for any j and tj ̸ = tk for any j ̸ = k.Without loss of generality, we may assume the corresponding Then there is (a it follows from Lemma 3.7 that t 1,0 , tn+i+1 is algebraic over The proof is complete.■

Eremenko's theorem
Eremenko's Theorem 1.3 is a deep result, and in this section, we would like to make some discussions about it and prove Theorem 1.4.By using the notions in Section 3, we suppose that there exist I 1 ⊂ J 1 ⫋ I and ϵ p ∈ {±1} for p ∈ J := J 1 ∪ {∞} such that Set ).

Suppose that
Either c is incommensurable or there is η Then the proof of Theorem 1.3 in [9] actually yields the following result, a more precise form of Theorem 1.3.3), the polynomials B j 's in (3.18) always have an admissible zero.This is really a remarkable result.

Example 4.2. Consider the case {θ
small, then it is easy to calculate those t's such that (1.1) has co-axial solutions.For example, if θ t = 2, then t = 2 3 ; if θ t = 3, then t = 2±2i 3 .When θ t ≥ 4, the polynomial system B j = 0 becomes very complicated.In Section 5, we will give more examples.
2 .Now we turn back to the condition (1.5) in Theorem 1.3 if the vector c is commensurable.Under the conditions (1.3)-(1.4), it is not difficult to see that (1.5) holds automatically provided k ′ + k ′′ ≥ 1 and θ t j > 1 for all j.A more interesting thing is the assertion of Theorem 1.4, which says that (1.5) holds provided that Mondello-Panov's condition (1.2) holds.

Proof of Theorem 1.6
The goal of this section is to prove Theorem 1.6.Write t 1 and θ t 1 simply by t and θ.The function Q(x) in the differential equation (2.6) associated to the curvature equation (1.1) in the case under consideration is where β p = α p (α p + 2)/4 = θ 2 p − 1 /4 and d's satisfy (2.3), i.e., Before we proceed further, we note that the set {α p /2 + 1, −α p /2} is invariant under the substitution α p → −2 − α p = −θ p − 1.For simplicity of discussion later on, we assume that α 0 and α 1 satisfy where ϵ 0 , ϵ 1 are given in the assumption of Theorem 1.6.Thus, α 0 +α 1 +α ∞ = ϵ 0 θ 0 +ϵ 1 θ 1 +θ ∞ −3 is an integer.In view of (5.2), we regard d 0 and d 1 as functions of d := d t and let Q d,t (x) denote the rational function Q(t) in (5.1).By Theorem 2.1, there exists a polynomial P(d, t) ∈ Q θ [d, t] of degree θ in d such that the differential equation is apparent at t if and only if P(d, t) = 0, where The degree of P(d, t) in d is θ, but its total degree is in general strictly larger than θ (see (2.9)).In our proof of Theorem 1.6, we shall introduce another pair (λ, t) of unknowns in place of (d, t).Let where the second equality follows from (5.2).Indeed, the parameter λ is used as the accessary parameter of a certain Heun equation considered in [11], where for the simplicity of computation, the apparent singularity was put at x = 0 instead of x = t in this paper, and the condition for apparentness at x = 0 was shown to be equivalent to the vanishing of the characteristic polynomial of a finite Jacobi matrix (see [11, Proposition 2.4 and equation (2.8)]).For the convenience of the reader, we recall the equivalence, stated in our setting, as follows (see also [6,Lemma B.8]): (5.4) is apparent at t if and only if λ is an eigenvalue of the θ × θ matrix (undisplayed entries are all 0), where That is, (5.4) is apparent at t if and only if λ is an eigenvalue of M (t), i.e., a zero of the polynomial The polynomial P (λ, t) has the following properties.
(i) The total degree and the degree in λ of P (λ, t) are both θ.
The second property follows immediately from the fact that A j (0) = 0 and hence the roots of P (λ, 0) are simply B j (0).The third property is proved in [6,Theorem B.3].Since the conclusions are stated in different ways, here we provide a sketch of proof.
Let u = 1/x.We check directly that y(x) is a solution of (5.4) if and only if y(u) := uy(1/u) satisfies where Clearly, this is a Fuchsian differential equation with Riemann scheme Thus, for some complex numbers d 0 , d 1 , and d 1/t satisfying In fact, computing the partial fraction decomposition of Q(u), we find that We now apply parts (i) and (ii) to (5.8).Let By parts (i) and (ii), there is a polynomial P λ, 1/t such that (5.8) is apparent at 1/t if and only if P λ, 1/t = 0 and the roots of P λ, 0 are On the other hand, we see from (5.2) and (5.5) that so λ and λ are related by Letting t → ∞, we conclude that the roots of P ∞ (λ) are given by (5.7).■ Corollary 5.2.Let λ and λj be given by (5.5) and (5.7), respectively.Let y + (x; λ, t) and y − (x; λ, t) be solutions of (5.4) of the form Then as t → ∞ with λ/t → λj , y ± (x; λ, t) converge to solutions of the Fuchsian differential equation with Riemann scheme where α∞ = α ∞ + α t − 2j + 2.
For small θ, it is possible to completely write down the set A. We give some examples below.

PProposition 5 . 4 .
b (λ, t) := b(λ,t)≡0 on A j P j (λ, t), P c (λ, t) := c(λ,t)≡0 on A j P j (λ, t), then the cardinality of the set A is equal to the number of intersections of the two curves P b (λ, t) = 0 and P c (λ, t) = 0 in the affine plane C 2 , with the multiplicity of a point in A defined to be that of the corresponding intersection point of P b and P c .Furthermore, by part (iii) of Lemma 5.1, the two curves P b (λ, t) = 0 and P c (λ, t) = 0 do not intersect at infinity.Therefore, by Bezout's theorem, the number of intersections of the two curves in the affine plane is (deg P b )(deg P c ).Hence, to prove Theorem 1.6, we only need to determine the degrees of P b and P c .Let k := θ ∞ + ϵ 0 θ 0 + ϵ 1 θ 1 be given as in the statement of Theorem 1.6.If θ ≤ |k|, then deg P b = 0 and deg P c = θ, if k > 0, deg P b = θ and deg P c = 0, if k < 0. If θ > |k|, then deg P b = (θ − k)/2 and deg P c = (θ + k)/2.