Integrability of potentials of degree $k \neq \pm 2$. Second order variational equations between Kolchin solvability and Abelianity

In our previous paper: Integrability of Homogeneous potentials of degree $k = \pm 2$. An application of higher order variational equations, we tried to extract some particular structures of the higher variational equations (the ${VE}_p$ for $p>1 $), along particular solutions of some Hamiltonian systems. Then, we use them to get new Galois obstructions to the integrability of natural Hamiltonian with potential of degree $k = \pm 2$. In the present work, we apply the results of the previous paper, to the complementary cases, when the degrees of the potentials are relative integers $k$, with $|k|>2$. Since these cases are much more general and complicated, we reduce our study only to the second variational equation ${VE}_2$.


Presentation
This paper completes the previous one [4] by applying the same results and strategies to the complementary cases. That is, we still study the integrability of homogeneous potentials along Darboux points, but here we assume that the degree k is an arbitrary relative integer with |k| 3. As we shall see thing are technically much more complicated for the following reasons. At first, the assumption over the degree are more general, and secondly, here in contrast to the cases when |k| = 2, the Morales Ramis table (see Table 1) gives discreet obstructions at the level of the first order variational equation VE 1 . This will force the study to encompass a judge number of distinct cases. As a consequence, our major results, Proposition 5, Theorems 4 and 5 below, just concern the Galois group of the second variational equation VE 2 . They are not definitive results which guarantee that the Galois groups of the associated systems VE γ 2,α and EX γ 2,α,β are virtually Abelian in such and such cases, but they constitute effective algorithms to test such possibilities in particular cases. These results convert the virtual Abelianity of the Galois group to testing some Ostrowski relations between first level integrals Φ, Ψ α , Ψ γ etc, when they are so. The reader will see that in general, these integrals are very complicated. And the main, difficulty will be be the following : from the form of a given integrals, it will be in general very easy to predict that it can be algebraic if it is so. But in contrast when it does not have such specific form, it is very very difficult to decide when it is transcendental. Our main ingredient for this will be Remark 2 below.
In addition to the first paper, this one contains one additional idea which is interesting from a theoretical point of view. This is what we call the cohomological argument , which allow to test that a second level integral is indeed computable in closed form, that is to test if a Galois group is virtually Abelian, without computing effectively this second level integral in closed form explicitly. This procedure was discovered when dealing with the present second level integrals, but it can be applied in more general contexts.
Let us mention finally, that this approach is not isolated. Combot in [2], deal with the same problem for homogeneous potentials of degree k = −1, while in the same time Weil in [1], is studying exactly the same problem than us but from the complementary point of view of gaugetransformations, which has the advantage to convert, the original systems into new ones which are much more simple in the sens that the virtual Abelianity of the Galois group can be directly read in the form of the new system.
In order to helps the reader, into the reading of this paper, let's briefly present how it is organised. In Section 2, we present the systems VE γ 2,α and EX γ 2,α,β . Here the idea is that as for the study of VE 1 , these new systems are much more easy to study after the so called Yoshida-transformation, than in their original time-parametrisation. In Section 3, we present the associated second level integrals involved in those systems and their intrinsic hierarchy. Section 4, contains the technical ingredients both theoretical and practical which are going to be useful for the proofs of Theorems 4 and 5 (in Section 5), and for some effective remarks about VE γ 2,α and EX γ 2,α,β . These practical studies will be done in Sections 6 and 7 below. Moreover, we recommend the reader, to first take a look, to Section 6.4, where we present some experimental facts about the complexity of the law that govern the virtual Abelianity of VE γ 2,α . We hope that this will help him to understand how we are using the present criteria and why we where obliged to deal with such tremendous casuistic.

From VE 1 to VE through the Yoshida transformations
In all the previous works concerning the integrability of homogeneous potentials along Darboux points, the key result for the analysis of the first variational equation VE 1 , was the conversion of this differential system in time parametrisation to an equivalent one in new variable z. This is the Yoshida transformation given by This transformation convert the initial variational equations over a hyperelliptic curve into a Fuchsian equation over È 1 , with singularities at z ∈ {0; 1; ∞}.
In this Section we shall recall some results concerning this transformation, and we show how it applies to the study of the second variational equation VE 2 .

The subsystems of VE 2 to deal with
Here, we assume that V ′′ (c) is diagonalisable. Hence, VE 1 splits into a direct sum of equations which have the following formẍ where λ α are the eigenvalues of V ′′ (c), and ϕ = ϕ(t) is a particular solution defined by a Darboux point d. For each of these equations we denote by G 1 = G α = G(k, λ α ) its differential Galois group over the field K = (ϕ,φ). From Proposition 2.5 of [4], we know that the differential Galois group of VE 2 is virtually Abelian iff the same property hold true for the differential Galois groups of the systems VE γ 2,α and EX γ 2,α,β for all α; β; γ with α = β.

The Morales-Ramis table
The full list of all values of λ ∈ for which the differential Galois group of equation is virtually Abelian is given in the following table, where p denotes an integer.
λ arbitrary complex number 1 G a |k| 3 λ(k; p) = p + k 2 p(p − 1) 2 1 p  Table 1: Since we will now play with potential of degree k = ±2, the connected component of G 1 will be either G a or identity. See [5] and [3].

The Yoshida transformation of VE 1
Here we reproduce some of the computations which were more detailled in [3]. The Yoshida transformation consists of the change of the independent variable t −→ z = ϕ k (t), (2) in the considered equation. Thanks to the chain rule where . Now, after the classical Tschirnhaus change of dependent variable, equation (2.3) has the reduced form where r(z) := r λ (z) := r 0 (z) + λs(z) = Since the three above numbers are respectively the exponents differences at z = 0, z = 1, and z = ∞, of the reduced hypergeometric equation L 2 = x ′′ − r(z)x = 0; the solutions of L 2 = 0, belong to the Riemann scheme In Table 1, the group G(k, λ) appearing into the first column is precisely the differential Galois group of the equation L 2 = x ′′ − r(z)x = 0, with respect to the ground field (z).

2.4
Computation of the solution space of (1) when G(k, λ) • = G a .
We observed that after Yoshida transformation the new equation in z variable is L 2 = x ′′ − r λ (z)x = 0, and that the solutions of L 2 = 0 belong to the Riemann scheme where In Table 1, the group G(k, λ) appearing into the first column is precisely the differential Galois group of the equation L 2 = x ′′ − r(z)x = 0, with respect to the ground field (z). But its connected component coincide with the connected component of (1) over (ϕ,φ). Our Lemma 3.4 from [3] can be reformulated into the following way.
1. Up to a complex multiplicative constant, the algebraic solution x 1 is of the form and J(z) ∈ Ê[z] does not vanish at z ∈ {0, 1}.  From (9) and (8) we get , we get that From the four possibilities a ∈ { k−1 2k , k+1 2k }, b ∈ { 1 4 , 3 4 }, we get four possibles Riemann schemes. According to ([6] p. 95), the classical Jacobi polynomials J * (α,β) n (t)with parameters (α, β), and degree n ∈ AE, are defined by They belong to the Riemann scheme But here the singularities that we meet are {−1, 1, ∞} instead of {0, 1, ∞}. Hence we pass from the classical Jacobi polynomials to ours by putting t = 2z − 1, i.e., we set As a consequence, we get the following classification of the Jacobi polynomials J ∈ P 2 Here, n := deg(J(z)). To obtain this, we identified P 2 with P J . This gave α and β. To compute the degree, we observed that α + β ∈ . Hence, −n is the one of the two numbers ρ ∞ + a + b, or ρ ′ ∞ + a + b that belongs to . Thanks to the formula for J * = J * n (α,β) we get that up to a constant multiple Hence, according to the previous table and Lemma 1, the precise forms of the integrals I depends on the four cases and are given by the following Table 3: Remark 1 Compatibility between the above Table and Lemma is certainly true for |k| 3 but there is a problem for k = ±1. Indeed, thanks to ( [6] p.95), the differential equation for So by plugging z = 0, 1 in the later we get From Table 2, α = ±1/2 and β = ±1/k so n(α + β + n + 1) = 0. As a consequence, J(1) = 0. But when β = −1 that is when k = ±1, we get that J(0) = 0. Which is not compatible with the Lemma 1.
This is why, in this paper, we are going to work with the assumption that |k| 3, which is in fact complementary to what was done in [2].
With the above notations we get the following Proposition 1 By setting x = f (z)ζ, the differential equation in time t: is transformed into the differential equation in z variable Proof According to Yoshida, by dividing the original equation in time by ϕ k−2 and multiplying by −s(z) we get The field (z)[ω] is a finite Abelian extension of (z).

Corollary 1
The original systems in time VE γ 2,α and EX γ 2,α,β are equivalent to the following systems in z variable that we still denote by the same symbols: The coefficients of these systems are elements of K 0 := (z) [ω]. Here, in order to simplify notations we set: r α = r λα (z) = r 0 (z) + λ α s(z).
Proof In EX γ 2,α,β , let's perform the Tschirnhaus transformation by setting According to the above proposition, the system in time is therefore equivalent to Moreover direct computation gives s(z) Hence, the last equation of EX γ 2,α,β becomes Coming back to minuscule letter by setting we get the desired expression of EX γ 2,α,β into z variable. Similar computations hold for VE γ 2,α .
3 Second level integrals and hierarchy 3.1 The second level integral involved in VE γ 2,α and EX γ 2,α,β From now on, we will only work with differential equation in z variable, with coefficient in Proposition 2 Let F be a differential field containing the elements r, ω, b, and a basis of solutions {u 1 ; u 2 } of u ′′ = ru. Then, the field F contains all solutions of the differential equation iff it contains the two following integrals Proof Classical variation of constant.
Let us denote by F 1 /K 0 the Picard-Vessiot extension associated to the homogeneous part of one of the systems S := VE γ 2,α , or S := EX γ 2,α,β . Let also denote by F 2 /K 0 the Picard-Vessiot of S. From the above • For S = VE γ 2,α , F 2 /F 1 is generated by the four integrals Φ = ωyx 2 . That is, in term of basis {x 1 ; x 2 } and {y 1 ; y 2 } of the corresponding homogeneous equations, we have at most six generators, given by the integrals Φ := ωy i X, i = 1; 2.
These systems are complicated ones, nevertheless we get the following elimination result Theorem 1 If the groups G α , G β , G γ are finite then the Picard-Vessiot extensions of VE γ 2,α , and EX γ 2,α,β are virtually Abelian. Proof In these cases F 1 /K 0 is algebraic. It is also the algebraic closure of K in F 2 . Since F 2 /F 1 is generated by first level integral with respect to F 1 , Gal • (F 2 /K 0 ) is a vector group.
From now, since the casuistic is sufficiently tremendous, with those cases only, we shall assume up to the end of this paper that 3.2 Hierarchy of the integrals involved in VE γ 2,α and EX γ 2,α,β .
Since in the considered cases the group G λ of w ′′ = r λ w has the connected component G a , the equation always have an algebraic solution that we shall always denote by an index one, i.e., w 1 . The second solution is given by With this convention the second level integral involved in VE γ 2,α can be classify in following Observe that Φ is a first level integral with respect to the algebraic extension K of (z), with The remaining five integrals, are of second level with respect to K. Their complexity grow at each change of line in the diagram. For EX γ 2,α,β , we proceed similarly. Here, Φ = ωu 1 y 1 x 1 is first level with respect to K := (z)[ω; u 1 ; y 1 ; The diagram of complexity is now the following 4 Ingredients and tools 4

.1 A cohomological argument to decide between solvability and Abelianity
Here we recall one important result stated and proved in the first part of this paper (Theorem 3.1 of [4]). And we add to it some refinements which are going to be useful for effective testing of the virtual Abelianity.
Theorem 2 Let F 1 /K and F 2 /K be two Picard-Vessiot extensions with F 1 ⊂ F 2 and F 2 /F 1 generated by integral of second level. Then G 2 is virtually Abelian iff G 1 is virtually Abelian, and any second level integrals Φ ∈ F 2 can be expanded into the form where R 1 ∈ T (F 1 /K) and J ′ ∈K. HereK is the algebraic closure of K in F 2 . Moreover, for all Here we recall that T (F 1 /K) stands for the Picard-Vessiot ring of the extension F 1 /K. Let K ⊂ F 1 ⊂ F 2 be a tower a Picard-Vessiot extensions satisfying the assumptions of Theorem 2, with F 1 /K virtually Abelian. Let Φ ∈ F 2 be a second level integral. If G 2 is virtually Abelian, the mapping Now, let R ∈ M = T (F 1 /K), and h be an arbitrary mapping from G • 2 to the field of constats C, we say that the mapping is an extended coboundary from G • 2 with values in M . When moreover, h ∈ Hom(G • 2 , C), then we say that B is a coboundary. Direct computation shows that an extended coboundary B is a cocycle iff B is a coboundary.
In this language, Theorem 2, says that if G • 2 is Abelian, then any cocycle associated to a second level integral is a coboundary. We may wonder about a converse. Let Φ be a second level integral such that there exists R ∈ T (F 1 /K), and some function h : G • 2 → C, (here we do not assume a priory that h is additive), satisfying This relation yield the following implications Hence, there exists J such that J ′ is algebraic over K, and Φ = R + J. As a consequence, we proved that the cocycle associated to a second level integral Φ is a coboundary iff Φ can be computed in closed form.
Let us conclude this sub-section by explaining how we will use these observations. For any given tower of Picard-Vessiot extensions and an explicate second level integral Φ, the fact that the associated cocycle C(σ) = σ(Φ) − Φ belong to T (F 1 /K), will give us necessary conditions for the virtual Abelianity of G 2 . Next, thanks to the previous arguments, we will find sufficient conditions, showing that the cocycles are coboundaries.

Ostrowski relations and necessary conditions for virtual Abelianity
The following lemma was also stated into the first part of this paper as a consequence of Theorem 2, (see Lemma 3.3 of [4]). But we state it again in a more appropriated version to the present context. Moreover, we prove it again since its proof is better understandable into the framework of the cohomological arguments.
Lemma 2 Let K ⊂ F 1 ⊂ F 2 be a tower of Picard-Vessiot extensions of K, with the same field of constants C, and satisfying the following conditions , and I ′ ∈ K, • F 2 contains some second level integrals of the form Φ 1 := Φ ′ I, with Φ ′ ∈ K, and Φ ∈ F 2 .
If F 2 /K is virtually Abelian, then Φ and I satisfies Ostrowski relation of the form Conversely, if such a relation holds then the extension K(I, Φ 1 )/K is virtually Abelian.
Proof There exists c ∈ Hom(G • 2 , ) such that for all σ ∈ G • 2 , we have σ(I) = I + c(σ). Hence, , for some mapping d : G • 2 → . Now according to Theorem 2, the virtual Abelianity of F 2 /K implies that the cocycles σ( Let us choose σ = σ 0 such that c(σ 0 ) = 1. The last relation implies that Φ ∈ K[I]. Hence, the two primitive integrals over K, Φ and I must be dependant and we conclude thanks to Ostrowski.
For the converse, let us assume that Φ = dI + f for some f ∈ K. Integrating by part, we can compute Φ 1 in closed form thanks to first level integrals. Indeed we get and the claim follows since, f I ′ is a first level integral w.r.t K.
Alternatively, let us show on this example how the previous cohomological arguments are working here. If we assume that Φ = dI + f for some f ∈ K, we get a closed expression for the cocycle Let us show that this is an extended coboundary. If we set P (I) := dI 2 /2 + f I, we get σ(P (I)) − P (I) = P (I + c(σ)) − P (I) Hence, by comparing the above two formulae, we get where the function h : G • 2 → C, can be computed thanks to the formula given by h(σ) = −dc 2 (σ)/2 + d(σ). Since the cocycle: σ(Φ 1 ) − Φ 1 is an extended coboundary, and, in fact, is a coboundary, Φ 1 can be computed in closed form, and the result follows.
This proof explains why we introduced, the a priory artificial notion of an extended coboundary. Indeed, when looking to the complicated formula for h above, it not obvious that it is a group morphism from G • 2 to C. Nevertheless, what is really important for our purpose is that h takes constant values. In the more complicated cases that we shall meet below, we will not explicitly compute h, but we will only show its existence. Moreover, the advantage of this cohomological approach is that it shows that a second level integrals can be explicitly computed in closed form without having to make this computation explicitly. This will make things simpler in the more complicated cases below.

Testing Ostrowski relations thanks to characters
Let K/ (z) be an algebraic extension. Let Φ and I be two primitive integrals of elements belonging to K. In order to test if they satisfy an Ostrowski relation of the form for some d ∈ , we shall use the following observation by taking advantage that Φ and I are primitive of algebraic functions. Let σ be a Galois morphism fixing (z), for example a monodromy operator, and assume further that σ acts on Φ ′ and I ′ by characters according to the formulae By integrating we get relations of the form As a consequence, applying σ to (11), we get a system of two equations If χ(Φ) = χ(I) the matrix is invertible and (11) implies that Φ is algebraic. We have therefore proved the following criteria.

Lemma 3
1. Assume that Φ is not algebraic, and that there exists a monodromy operator σ acting by character on Φ ′ and I ′ . If χ(Φ) = χ(I), then (11) does not hold.

2.
Here is a generalisation: Let Φ, I 1 , . . . , I n be n + 1 integrals over K. Let σ be a Galois morphism acting by characters on the derivatives If, χ Φ ∈ {χ 1 , . . . , χ n }, then an arbitrary Ostrowski relation of the form The proof of the second point is similar to previous particular case. Indeed, by grouping together the I j corresponding to the same character, we are reduced to the case where all the χ j are distinct. Then, the action of the σ p for 0 p n lead to an invertible Vandermonde n × n system, which allows a similar conclusion as in point (1).
We are going to use this lemma in the proofs of Propositions 6, 7 and 8 below, by showing thanks to monodromies that some Ostrowski relations are impossible. This is one of the great advantage of Yoshida transformation in comparison to the time parametrisation, where the singularities of the corresponding complex functions are not well understood. Another advantage of Yoshida transformation is going to be shown right now.

To be or not to be an algebraic integral
Remark 2 Let F/K 0 be the Picard-Vessiot extension of either VE γ 2,α , or EX γ 2,α,β , over the field K 0 = (z)[ω]. The two above systems are Fuchsian with singularities at z ∈ {0, 1, ∞}. Indeed, according to Corollary 1 and Proposition 2, each solution of any of these systems is holomorphic in any simply connected domain of È 1 \{0; 1; ∞}, with at most exponential growth at the singularities.
As a consequence, the Schlesinger theorem implies that Gal(F/ (z)) is topologically generated by the two monodromies M 0 and M 1 .
This observation will have the following important consequence. Let Γ be a holonomic element of F fixed by M 0 , and having a finite orbit under < M 1 >. More generally, let us assume that Γ has a finite orbit under the monodromy group M ⊂ Gal(F/ (z)). Then necessarily, Γ is algebraic over (z). Indeed, since Γ is holonomic, there exists a -finite dimensional vector space V containing Γ on which Gal(F/ (z)) acts algebraically. As a consequence, the map is a morphism of algebraic variety. Since the image f (M) is finite, it is a Zariski closed subset of V , so f −1 (f (M)) is a closed subset of Gal(F/ (z)) containing M. Since M is dense in Gal(F/ (z)), we get that f −1 (f (M)) = Gal(F/ (z)) and the orbit of Γ under Gal(F/ (z)) is finite as has to be shown.
The most general integrals Φ and Ψ we shall meet below are Abelian integrals of the form where the exponents e 0 and e 1 are rational numbers > −1, with e 0 + e 1 ∈ .
Precisely, Γ will be an integral of the type Φ, for n = deg(J) = 0, and of the type Ψ otherwise. According to Remark 2, if one of the two exponents e 0 , or e 1 is an integer, then the corresponding integral Φ, or Ψ has finite orbit under the monodromy group, hence is algebraic. This is a very surprising fact especially for the integrals Ψ. Indeed, this shows that here, the P, Ω and J must be so specific that P Ω/J 2 does not has residues at none of the z i .
Away from those cases, we now have to investigate the integrals Γ, for which the two exponents are not integers in order to be able to test their eventual algebraicity.

Reduction of the integrals
When trying to compute Γ in closed form we get the following formula Viewed as a linear mapping of (z) to itself, T is injective. Indeed, , by computing the leading term of T (z r ), we see that the condition e 0 + e 1 ∈ implies that T increment the degree by n + 1. That is deg(T (R)) = deg(R) + n + 1. By counting dimensions we therefore get that for all N 0, we have the following direct sum decomposition Since it holds for all N 0, we get As a consequence, we can reduce any integral Γ by lowering the degree of the numerator in the following way : By multiplying this equality by Ω/J 2 , and integrating we get Since R Ω /J is algebraic, Γ(P ) is algebraic, iff Γ(Λ) is algebraic. For the study of this problem we get the following.
Theorem 3 Let P, Ω, J be as above with e 0 and e 1 in É\ .
1. The integral Γ(P ) is algebraic iff it belongs to the field (z) [Ω], that is iff there exists R ∈ (z) such that P = T (R).
4. If P is a non zero polynomial with deg(P ) n, then Γ(P ) is transcendental. Moreover. the n + 1 integrals z s Ω J 2 with 0 s n are algebraically independent. The ideas behind this result are very closed to what we did in [3]. Proof 1. Since the exponents are rational but not integers, there exist a minimal integral power d 2 such that Ω d ∈ (z). As a consequence, the field extension (z) . It can therefore be expanded into the form Taking derivative of the above equation, we obtain the following expression and the claim follows.
2. If P = T (R) with R ∈ (z), then the function is holomorphic in an arbitrary simply connected domain of \ {0, 1, z 1 , . . . , z n }. So, if R has got a pole, it must belong to {0, 1, z 1 , . . . , z n }. But for all p 1 , the leading term of T (1/z p ) is given by with h 1 holomorphic. Indeed, because Γ is algebraic Γ ′ does not have residue. So, with h 2 and h 3 holomorphic around z i . Since h 3 (z i ) = 0, R cannot have a pole at z i . Hence, R is polynomial. 3. and 4. follow from the following observation. Let Λ ∈ n [z] be such that Γ(Λ) is algebraic. Point 1 implies the existence of some R ∈ (z) with Λ = T (R). But according to point 2, R is a polynomial. If R = 0 then deg(Λ) = deg(R) + n + 1 > n which is contradictory. So Λ = T (0) = 0 and the integral is algebraic iff Λ = 0. The above result show that the algebraicity of an integral Γ(P ) reduces to the vanishing of the polynomial Λ appearing in equation (12). Although the correspondence P → Λ is linear, the decomposition given by equation (12), is very hard to perform effectively. Here, we are going to show that the vanishing of Λ can be controlled by the vanishing of some linear forms on P which can be directly computed thanks to some definite integrals.
In the most simple case, that is for J = 1, i.e,, when J(z) is a constant, this is achieved thanks to the following.
Proposition 3 Let e 0 and e 1 be two real numbers greater than −1, and belonging to É\ with e 0 + e 1 ∈ . Let us set Ω := z e0 (1 − z) e1 . Then we have 1. For any polynomial P , the primitive integral Φ := P Ω is algebraic iff where B(p; q) is the usual Euler Beta function and (x) n is the Pochammer symbol.
3. If P (z) = p n z n , then Φ is algebraic iff p n (e 0 + 1) n (e 0 + e 1 + 2) n = 0 The condition on the two exponents to be greater than -1, guaranties the convergence of the generalised integrals between 0 and 1. The first point shows that the set of polynomials for which Φ is algebraic is an hyperplane given by the kernel of the linear form µ.
Points 2 and 3, give explicit criterion on the coefficients of P to decide whether or not Φ is algebraic. Proof 1. Since J = 1, in the decomposition given by equation (12): P = T (R) + Λ, we have that Λ is a number. The corresponding relation (13), can be written Now let us compute Λ by evaluating the integrals between 0 and 1. Since e p + 1 > 0 for p ∈ {0; 1}, and R ∈ [z], we have

Is a direct consequence of the relation
3. Follows directly from the previous considerations.
On the basis of the same ideas we now treat the case when deg(J) = n 1. Let 0 < z 1 · · · z n < 1 be the roots of J. Let γ 0 be a half of the circle going counterclockwise from 0 to 1. Let γ i for 1 i n be some small trigonometric circles each enclosing z i and no other root z j . Let us consider the n + 1 linear forms on [z] given by Then we get the following Proposition 4 With the previous notations 1. If P = T (R) + Λ as in relation (12), then for all 0 i n, L i (P ) = L i (Λ).
2. The n + 1 linear forms L i are free and P Ω J 2 is algebraic iff for all 0 i n, L i (P ) = 0. Let us observe that this property is a generalisation of the previous one. Indeed, L 0 = µ, for deg(J) = 0, that is when J = 1. Here the problem is that we do not find comparable simple closed formulae for the linear forms L i . We mention this difficulty because the L i (P ) got the flavour of some periods on some Abelian variety. But we did not find this link precisely. This is probably the deep reason why things are so complicated in our context. Maybe we did not find the proper geometric space where the actual notions would get some more transparent meaning. Proof 1. This is a direct consequence of relation (13).
Now let us set P s := J 2 /(z − z s ), for 1 s n. The function P s Ω/J 2 = Ω/(z − z s ) has a non zero residue at z s , and zero residue elsewhere. So, L s (P s ) = 0 and L i (P s ) = 0 for i = s, and 1 i n. From this it follows immediately that the n + 1 linear forms L s are free on [z]. According to point 1, their restriction to n [z] form a basis of the dual space n [z] * . As a consequence, any Λ ∈ n [z] is zero iff it belongs to the common kernel of the linear forms. And we can therefore conclude that P Ω J 2 is algebraic iff for all 0 i n, L i (P ) = 0 according to Theorem 3.
5 Reducing the virtual Abelianity of VE γ 2,α and EX γ 2,α,β to Ostrowski relations In this section we exhibit the integrals and the Ostrowski relations which are going to govern the virtual Abelianity of the systems VE γ 2,α and EX γ 2,α,β . Next in the two sections that follow we will test effectively these results.

5.1
Getting obstruction thanks to the integrals Φ ν for ν ∈ {α, β, γ} Proposition 5 With the notation of Section 3.2, we get the following necessary conditions 1. If the differential Galois group of VE γ 2,α is virtually Abelian, then, we get two Ostrowski relations Φ + d γ I γ and Φ + d α I α are algebraic over (z) for some constants d γ and d α in . Here, Φ = ωy 1 x 2 1 .
Proof Let F/Kbe the Picard-Vessiot extension VE γ 2,α . According to Section 3.2, we get the following inclusion , and a similar one by changing γ to α. The fact that F/K is virtually Abelian implies the same property for F γ 2 /K, and F α 2 /K. Then we can conclude thanks to Lemma 2. Similar arguments hold when dealing with EX γ 2,α,β .

Strategy and Game
Two cases may a priory happen, when applying the above Proposition: Φ is either transcendental either algebraic over K or (z).

When Φ is transcendental
If this occurs, then the virtual Abelianity of the differential Galois group of S = VE γ 2,α , or S = EX γ 2,α,β implies that the corresponding constants d γ , d β and d α are non zero complex numbers. As a first consequence we must also get Ostrowski relations between the integrals I. For instance, d γ I γ − d α I α is algebraic...

When Φ is algebraic
In this situation, the Ostrowski relations of Proposition 5, hold with d γ = d β = d α = 0 and the proposition is helpless. Unfortunately, as we shall see in Section 5 and 6 below, Φ is algebraic very oftenly. This is the reason why we have to find new necessary conditions for the virtual Abelianity of differential Galois groups of VE γ 2,α and EX γ 2,α,β in this case. This will be the purpose of the next subsection.

Getting obstruction when Φ is algebraic
In this subsection, we give the two criteria such that groups of VE γ 2,α and EX γ 2,α,β are virtually Abelian when Φ is algebraic.

Integration by part and new second level integrals Ψ m associated to the Φ m
Here, we keep notations and formulae given in Section 3.2, and we assume that for each system S = VE γ 2,α or S = EX γ 2,α,β its corresponding Φ is algebraic, i.e., Φ ∈ K. We know that the Picard-Vessiot extension F/F 1 = PV(S)/F 1 is generated by second level integrals Φ ν , Φ i,j and Φ i,j,l which have, according to the given hierarchy, one, two or three indices. For each multi-index m ∈ {ν; (i, j); (i, j, l)}, let us symbolically write Φ m = Φ ′ I m . Integration by part gives Since Φ ∈ K, ΦI m ∈ T (F 1 /K) = K[I], and F/F 1 = PV(S)/F 1 is generated by the corresponding Ψ m . Our motivation to introduce these new integrals Ψ m is the fact that the computation of their associated cocycles σ(Ψ m ) − Ψ m is simpler than for the cocycles σ(Φ m ) − Φ m . Now, we give the precise formulae for the given Ψ m for one, two and three indices respectively. Next, we compute their cocycle, and give a general property about some specific cocycles.
Remark 3 Let us observe that the integrals are not well defined objects. This is because, as a primitive integral, Φ is only defined up to an additive constant. Therefore, two integrals Ψ 1 and Ψ 2 defined for the same "Φ" modulo constant terms, and the same I, are related by a relation of the form where (d, e) ∈ 2 . Hence, having an Ostrowski relation Ψ 1 + dI ∈ K is therefore equivalent to have a representative Ψ 2 = Ψ 1 + dI which is algebraic. We will therefore use both of the two expressions.
Let us observe also that point (3) of Theorem 5 below is coherent when the two representatives Ψ γ and Ψ α are defined with respect to the same Φ.
Observe also that the Ψ ν are integrals of first level with respect to K, since Φ, and the I ′ ν are in K.
For two indices For i = j, we have In the particular case where i = j = α, by simplicity we divide by two the original Ψ by setting For three indices We get General formulae for the cocycles C m (σ) := σ(Ψ m ) − Ψ m when σ ∈ G • Here G denotes the corresponding Galois group. In order to simplify notations we may some time write c ν instead of c ν (σ) in the relations σ(I ν ) = I ν + c ν (σ), for σ ∈ G • . We get the following formulae where the respective l m are function from G • to .
We make only the computation for C γ,2α (σ) since the others are similar.
The last relation has been obtained by integrating the previous one. As a consequence the function l γ,2α : G • → has constant values, but it is far to being a group morphism. This can be simply seen if we translate for l the cocycle relation for C.
Cocycles of degree in I and symmetric matrices When proving the two theorems below, we will get explicit polynomial expressions of the above cocycles, since a necessary condition for the virtual Abelianity of G is going to be In practice, these cocycles are going to be of degree one or two in I. Precisely, let us assume that F 1 /K = K(I 1 , . . . , I n ), is generated by n independent first level integrals. Let us denote by I T := (I 1 , . . . , I n ) and C T (σ) := (c 1 (σ), . . . , c n (σ)). We say that a cocycle C(σ) = σ(Ψ) − Ψ is of degree one in I, if it has the form where A ∈ M n (K) andÃ ∈ M n (K) are n × n matrices, F ∈ K n and, l is a constant valued function.
In the particular case where A andÃ are constant matrices (i.e., belong to M n ( )), the mapping σ → C(σ) TÃ C(σ) is a constant valued function. So, the general expression of the degree one cocycles with constant matrix can be more simply written With these notations, we get the following property which simplifies the proof of the theorems, enlightening a link between the abelianity of a group, and the general idea of symmetry which is realised here by symmetric matrices.
Lemma 4 Let us assume that F 1 /K = K(I 1 , . . . , I n ), is generated by n independent first level integrals over K.

Let
be a general cocycle of degree one in I. Then, C coincides with a coboundary iff A is a symmetric matrix, and A − 2Ã ∈ M n ( ).
2. Let C(σ) = σ(Ψ) − Ψ = C(σ) T AI + C(σ) T F + l(σ) be a degree one cocycle with constant matrices. Then C coincides with a coboundary iff A is symmetric. If it is the case, then the corresponding second level integral Ψ can be computed in a closed form thanks to a quadratic expression of the form 3. For n = 1 every degree one cocycle with constant matrix is a coboundary.
Proof (1) Since C is of degree one in I, if it coincides with a coboundary of the form ∆P + h(σ), then P (I) must be quadratic in I. It can therefore be written into the form P (I) = I T SI + B T I, for some symmetric matrix S ∈ M n (K), and B ∈ K n . The relation C(σ) = ∆P + h(σ) = P (I + c) − P (I) + h(σ), is therefore equivalent to having For a fixed value of σ, both side of this equation are affine linear forms in I with coefficients in K. Hence, we must have AC(σ) = 2SC(σ), for all σ ∈ G • . Since C(σ) span all n , when σ ∈ G • , we have that A = 2S is symmetric. Moreover, the previous equation is reduced to By derivating both sides of this equation, we obtain This is therefore equivalent toÃ ′ − S ′ = 0 and F ′ − B ′ = 0. That is to having A − 2Ã ∈ M n ( ). Conversely, if those two conditions are satisfied, we just have to choose B = F to get the desired coboundary.
(2) When C is of degree one with constant matrix, it coincides with a coboundary iff A is constant and symmetric, since in this case there is no condition onÃ. Conversely, if A is symmetric, the computation above withÃ = 0 shows that Ψ andΨ := 1 2 I T AI + F T I, have equal cocycles up to a constant valued function, therefore the difference Ψ −Ψ is a first level integral.
Let's observe that the computations of the second level integral Φ 1 appearing in Lemma 2, is a particular case of points (2) and (3) of the above lemma.

5.3.2
The criteria for the virtual Abelianity of EX γ 2,α,β when Φ is algebraic We decided to begin with EX γ 2,α,β because here, the role played by the integrals I ν is symmetric. This is not the case when dealing with VE γ 2,α . As a consequence, although they proceed with the same methods, the proof of Theorem 4 is more transparent than the proof of Theorem 5 below.
For the statement and proof of the following result, we use the notations introduced above. The first point of the theorem gives a necessary condition for the virtual Abelianity of EX γ 2,α,β . Having expressed this first necessary condition, the next three points will give sufficient conditions for the virtual Abelianity of EX γ 2,α,β . Each of them depends on the degree of dependence of the integrals I γ , I β and I α . Notice also that also that in the formulae G ⊂ K p , with p = 2 or 3 below, the letter G denotes a p-component vector which has nothing to do with the Galois group. Theorem 4 With the notation of Section 3.2, let us assume that Φ = ωu 1 y 1 x 1 is algebraic, i.e., Φ ∈ K.
1. Let us denote by Ψ = (Ψ α , Ψ β , Ψ γ ) T , and similar notations for the three component vector I. If EX γ 2,α,β has virtually Abelian differential Galois group, then there exists an Ostrowski relation between Ψ and I of the form : for some constant 3 × 3 matrix D and F ∈ K 3 .
2. If the integrals I α , I β , I γ are independent over K, then EX γ 2,α,β has a virtually Abelian differential Galois group iff the two following conditions are satisfied a) There exists a unique determination of Φ modulo constants such that each Ψ µ for µ ∈ {γ; β; α} is algebraic. This correspond to having D = 0 in point (1 3. If the integrals I α , I β , I γ form a system of rank one over K, that is if we have Ostrowski relations of the forms I β − θ β I α ∈ K and I γ − θ γ I α ∈ K, then EX γ 2,α,β has a virtually Abelian differential Galois group iff condition (1) holds and, for the first level integrals N i,j defined by equation (14) below, we get an Ostrowski relation of the form N β,γ + θ β N γ,α + θ γ N α,β = eI α + g, with e ∈ and g ∈ K.
4. If the integrals I α , I β , I γ form a system of rank two over K, that is if we get one Ostrowski relation of the form I γ − bI β − aI α ∈ K, and I α , I β are independent. Then, EX γ 2,α,β has a virtually Abelian differential Galois group iff the following conditions are satisfied a) There exists a choice of Φ modulo constants such that (1) can be written with x and y in . b) For the first level integrals N i,j defined by equation (15) below, we have a relation of the form where, E is a 2 × 2 constant symmetric matrix and G ∈ K 2 .
Proof (1) Let us set F/K = PV(EX γ 2,α,β )/K, and assume that G is virtually Abelian. According to Theorem 2, since each Ψ i,j ∈ F , the cocycles By considering all those possible relations with i = j, we get three relations which can be written into matrix form But det(C) = 2c α (σ)c β (σ)c γ (σ) = 0, for some σ ∈ G • . So, by inverting this system we get that all Ψ µ ∈ T (F 1 /K). So, Ψ µ , I α , I β , I γ are four dependant integrals of first level over K, from which we can deduce the desired Ostrowski relations.
(2) By plugging each Ψ i = µ d i,µ I µ +f µ into the above C i,j (σ), we get that each σ(Ψ i,j )−Ψ i,j is a cocycle of degree one with constant matrix. Indeed, let us do this for Ψ α,β : According to Theorem 2, the Abelianity of G • implies that the cocycle must be a coboundary. From Lemma 4 the equivalent condition is that A α,β is a symmetric matrix. As a consequence, we get that d β,γ = d α,γ = 0 and d α,α = d β,β .
Since the same arguments hold by considering C β,γ and C γ,α , we deduce that D = dI 3 , for some d ∈ . As a consequence, we can write for µ ∈ {α, β, γ}. Therefore, if we substitute, Φ − d to Φ in the definition of the corresponding Ψ µ , then they all are algebraic. This proves point (a). Let us assume from now that we are in this situation. Then, each M i,j = Ψ i I ′ j + Ψ j I ′ i is a first level integral. Since, Ψ α,β,γ = Ψ ′ γ I β I α + Ψ ′ β I γ I α + Ψ ′ α I γ I β is a second level integral that belongs to F its associated cocycles C α,β,γ (σ) ∈ T (F 1 /K) for all σ ∈ G • . But As a consequence, each Ψ i,j ∈ T (F 1 /K) since the vectors C(σ) = (c α (σ), c β (σ), c γ (σ)) T ranges 3 , when σ ranges G • . Therefore, each M i,j = Ψ i I j + Ψ j I i − Ψ i,j is a first level integral which belongs to T (F 1 /K). Hence, we get a matrix type Ostrowski relation of the form M = EI + G, where E is constant 3 × 3 matrix, and G ∈ K 3 . The previous relation implies that As a consequence, we get the following formulae for the cocycle Hence, C α,β,γ is a general coboundary of degree one. We get the formula of the first point of Lemma 4 by setting A := Y − EandÃ := Y /2.
Since Y is symmetric, C α,β,γ is a coboundary iff E is a symmetric matrix. This proves the claim.
As a consequence, the cocycles C i,j are of degree one with constant 1 × 1 matrices. Indeed we get Therefore, according to Lemma 4, those cocycles are coboundary and there exist some first level integrals N i,j such that Now, if we denote by the symbol the sum over the three cyclic permutations of the indices (α, β, γ), we get where we have set d := θ i θ j d h ∈ , f := θ h θ i f j ∈ K and N := θ h N i,j is a first level integral. Since C α,β,γ (σ) ∈ T (F 1 /K), for all σ ∈ G • , N also belongs to T (F 1 /K). As a consequence, we get an Ostrowski relation of the form This is precisely our additional necessary condition. Conversely, let's assume that (1) hold and N = eI + g. We already saw that all the Ψ i,j can be computed in closed form. We will conclude by showing that in fact C α,β,γ is a coboundary. By plugging N = eI +g inside the last expression of C α,β,γ , we see that the latter is of degree two in I. But we are going to decrease its degree thanks to the following trick So, C α,β,γ is a coboundary and G is virtually Abelian.
(4) Here we have c γ = bc β + ac α . If we set: I T := (I α , I β ) and C(σ) T := (c α , c β ) then the Ostrowski relation of point (1) can be written Again, the cocycles C i,j are going to be of degree one with constant matrices of size 2 × 2 this time. If we write C i,j (σ) = C(σ) T A i,j I + C(σ) T F i,j + l i,j (σ), the same computations as in point (3) give The three cocycles C i,j are coboundary iff the matrices A i,j are symmetric. This translates to having D of the form If we look at the first line of D this gives As a consequence, if we change Φ to Φ − d in the definition of the Ψ ν , the new corresponding matrix D will be simplified into the form and this relation is equivalent to condition (4.a).
When this condition is satisfied, the matrices A i,j are symmetric, hence by the second point of the lemma, we get explicit formulae of the form where the N i,j are first level integral. Precisely, by expressing the A i,j in terms of x, y, θ α , θ β , we get the following expressions Next, by plugging these expressions in closed form of Ψ i,j and Ψ ν into C α,β,γ we get a formula of degree two in I where A is the symmetric matrix given by A := af β f /2 f /2 bf α with f := f γ + bf β + af α : In the first line above we recognise an expression of the form ∆( ay 3 I 3 α + bx 3 I 3 β ) + l(σ). Moreover, since C α,β,γ (σ) ∈ T (F 1 /K)∀σ ∈ G • , the first level integrals N β,γ + aN α,β and N α,γ + bN α,β are in T (F 1 /K). So we get an Ostrowski relation of the form Hence, is a coboundary, iff E is a constant 2 × 2 symmetric matrix. This proves the claim.

5.3.3
The criteria for the virtual Abelianity of VE γ 2,α when Φ is algebraic Here, we also use the notations of Section 3.2. Again, the first three points of the theorem below, give necessary conditions for the virtual Abelianity of VE γ 2,α . Points 5 and 6, give sufficient conditions according to the dependence of the two integrals I α and I γ . Theorem 5 With the notation of Section 3.2, let us assume that Φ = ωy 1 x 2 1 is algebraic (i.e. Φ ∈ K). If VE γ 2,α is virtually Abelian, then we get the following 1. There exists an Ostrowski relation between Ψ α and I α : Ψ α − d α I α ∈ K.
2. There exists an Ostrowski relation between Ψ γ , I γ and I α : 3. If in the previous relations d γ = d α then there exists an Ostrowski relation between I γ and I α .
5. If Φ ∈ K and conditions (1), (2) and (3) hold true with I α and I γ independent. According to points (1) and (3), there is unique choice of Φ modulo constants such that Ψ α ∈ K and Ψ γ = dI α + f . Then VE γ 2,α is virtually Abelian iff the two following conditions are satisfied: • The integrals X and M have polynomial expressions of the form where M is defined in equation (16) below.
• Moreover, E is symmetric, that is we have a X = b M in equations (20) and (21) below.
(3) According to point (1) and Remark 3, we can choose a fixed representative of Φ such that Ψ α ∈ K. In other words we can assume that d α = 0 in (1). Let's compute Ψ γ,α thanks to the two previous Ostrowski relations modulo integral of first level and elements of T (F 1 /K). We get Since Ψ α ∈ K and Ψ γ can be written Ψ γ = d γ I γ + dI α + f with f ∈ K, the expression Ψ γ I α + Ψ α I γ ∈ T (F 1 /K) and M is a second level integral belonging to F . Moreover, If we set J 1 := f I ′ α + Ψ α I ′ γ ; this a first level integral over K and the last relation tells us that But F/K virtually Abelian implies that F (J 1 )/K is also virtually Abelian. Hence applying Lemma 2 again, we get an Ostrowski relation between I γ and I α if d γ = 0.
Before proving point (4), let us assume that Φ is algebraic and conditions (1), (2) and (3) of the theorem hold true. We have seen that these conditions can be restated into the following simpler form : Let us set X := Ψ α I ′ α and M := Ψ γ I ′ α + Ψ α I ′ γ as in (16). X is a first level integral. Here we are going to show in addition that the second level integrals M and Ψ γ,α can also be computed in closed form. Precisely we shall prove that any such integral coincides with a polynomial in I α , I γ with coefficients in K plus a first level integral. Proof of point (4). Since Ψ 2α = Ψ α I α − X, and X is a first level integral and Ψ 2α can be computed in closed form. Now, from point (3) we get two possibilities. If d γ = 0, then Hence, M can be computed in closed form since J 2 is a first level integral. For Ψ γ,α this follows from (16). As a consequence, according to Theorem 2, and Section 3.2, VE γ 2,α is virtually Abelian iff Φ γ,2α can be computed in closed form.
Computation of the cocycle C γ,2α (σ) = σ(Ψ γ,2α ) − Ψ γ,2α when (5.3.3) hold true By substituting the previous integrals by their expression in closed form into the formula C γ,2α (σ) = 2c α Ψ α,γ + 2c γ Ψ 2α + c 2 α Ψ γ + 2c α c γ Ψ α + l(σ), we get According to Theorem 2, G is virtually Abelian implies that Therefore, if we compute C γ,2α (σ) modulo T (F 1 /K) = K[I α ; I γ ], we find a new necessary condition for Abelianity: Two cases therefore happen First Case : If I γ and I α are independent over K, then (19) is equivalent to having both X and M in K[I α ; I γ ]. Since X is a first level integral this translates to having an Ostrowski relation of the form Since the independence of I γ and I α imposes to having d γ = 0 that is Ψ γ = dI α + f , and M = d 2 I 2 α + J 1 . Therefore, M ∈ K[I γ ; I α ] iff we get an Ostrowski relation of the form Now, if we substitute the actual expressions in closed polynomial form of Ψ γ , X and M into (18), and expand the result as a polynomial in {I α ; I γ } and {c α ; c γ } with coefficients in K, this gives the following formula for the cocycle: Again, as in point (4) of Theorem 4, this cocycle is of degree two, and by setting we get a formula But A − 2Ã is a constant matrix, and A is symmetric iff a X = b M . So, C γ,2α is a coboundary iff this latter condition is satisfied. This proves the criteria given in point (5).
Second case : Here, we assume that I γ and I α are dependant that is I γ = θI α + κ and c γ = θc α . For simplicity we will write I α = I and c α = c ⇒ c γ = θc. The Ostrowski relation for Ψ γ will be written Ψ γ = eI + f with e := θd γ + d. As a consequence the formula for M is now M = e 2 I 2 + J 2 (independently of the possible vanishing of d γ the important number with that respect is now e). Now (19) is equivalent to having Since, J 2 + θX is a first level integral this equivalent to having an expansion in closed polynomial form : Again, if we substitute the actual expressions of Ψ γ and M + θX in closed polynomial form in (18) with I γ = θI + κ and c γ = θc and expand the result as a polynomial in I and c, this gives the following formula for the cocycle There is no condition on the size, and we recognise a coboundary C γ,2α (σ) = ∆[ e 3 I 3 + (f + 2θΨ α − a)I 2 + 2(κΨ α − g)I] + l(σ).
and we do not get more constrain in this case, and point (6) follows.
6.1 Testing the algebraicity of Φ when |k| 3 According to Lemma 1, when |k| 3 and G • γ = G • α = G a , we can write y 1 and x 1 into the form , and Φ ′ = ωy 1 x 2 1 we get, As a consequence, Φ is an integral of the type described in the previous section, so according to Proposition 3, the algebraicity of Φ reduces to the vanishing of µ(P ) = µ(J γ J 2 α ). There are two ways of testing the vanishing of this number. One with the coefficients of P if they are explicitly known. And the other with the evaluation of the definite integral between 0 and 1.
About this second method it applies sometime efficiently in our context. Indeed, for any Jacobi polynomial J(z) = J (α,β) , we can associate a kernel Ω = Ω J = z β (1 − z) α , whose explicit formula is given thanks to Table 2. Moreover we know that J is orthogonal for the scalar product < P ; Q >= 1 0 P QΩ J , to any polynomial whose degree is smaller than deg(J). For examples : thanks to Tables 2 and 4, we have In case (γ, α) = (1; 1) then, Ω = Ω γ = Ω α therefore, Hence, 2 deg(J α ) < deg(J γ ) ⇒ µ(P ) = 0 and the corresponding Φ is algebraic. This of course give new cases when Φ is algebraic. Unfortunately, we have got no converse and Φ could be algebraic outside of these cases.
For the remaining four cases we have unfortunately no comparison of the expression of µ with one of the two scalar products. So in general we are not able to give any condition for the algebraicity ! Nevertheless, for applications in specific examples one is therefore reduced just to check the relation given by point 3 of Proposition 3.
6.2 Getting obstruction when Φ can be transcendental in the 6 possible cases of Table 4 Our main result is going to be the following Proposition 6 When |k| 3 and Φ is transcendental, then in the six cases of Table 4, VE γ 2,α is not virtually Abelian excepted maybe for |k| = 3 when (λ γ ; λ α ) ∈ Case 3 × {Case 3; Case 4}.
Proof Let's assume that VE γ 2,α is virtually Abelian and look to the restrictions imposed by this condition thanks to Proposition 5 and its consequences. According to Lemma 1, the action of the monodromy around z = 0, is given by characters thanks to the following formulae Since Φ is transcendental, the two constants d γ and d α in Proposition 5.1 both are non-zero. Therefore, according to the later, we get an Ostrowski between I γ and I α . As a consequence, thanks to Lemma 3, the Ostrowski relation between I γ and I α implies that But, by writing a = 1 2 + ε 2k , with ε = 1 in Cases 1 or 2, ε = −1 in Cases 3 or 4, we get that 2a γ − 2a α ∈ iff, ε γ = ε α . In other words we proved that an Ostrowski relation between I γ and I α implies that a γ = a α ⇔ (γ; α) ∈ {Cases 1 or 2} 2 ∪ {Cases 3 or 4} 2 .

Getting Obstruction when Φ is algebraic
Here, according to the second point of Theorem 5, we have to test an Ostrowski relation between Ψ α and I α . Again, if Ψ α is algebraic such a relation hold and we get nothing new. As a consequence our first task is going to be the study of this problem.

Testing the algebraicity of Ψ α and Ψ γ
Here, according to Lemma 1, , and similar formula with Ψ ′ γ = ΦI ′ γ = Φ/y 2 1 . According to Remark 3, we must compute the two Ψ ′ with the same Φ up to an additive constant. The precise forms of those expressions are respectively given for Ψ ′ α resp for Ψ ′ γ in the following two tables γ\α 1 2 3 4 1 According to Theorem 3 points 1 and 2, Φ is algebraic iff it can be computed in closed form .
As a consequence, when Φ is algebraic, we see form the above tables that Ψ α and Ψ γ are simultaneously algebraic very often. In fact in 10 cases over the 16. Since it is much more difficult to prove the transcendence than to show the algebraicity of an integral, we therefore cannot say something general in the six remaining cases, when for example λ γ ∈ Case 4.

Getting obstruction when Ψ α or Ψ γ is transcendental
Here our main result is going to be the following Proposition 7 If Φ is algebraic, and Ψ α is transcendental, then VE γ 2,α is not virtually Abelian except maybe for the two cases (λ γ ; λ α ) ∈ Case 4 × {Case 1 or 2}.
Proof From now we assume that VE γ 2,α is virtually Abelian. According to Theorem 5, we must get two Ostrowski relations: one between Ψ α and I α , and another one between Ψ γ , I γ and I α . Let's assume that we get such a relation between Ψ γ I γ and I α . The arguments for Ψ α being simpler. According to Lemma 3, if Ψ γ is transcendental, we must get the following relation between the exponents at z = 0 : Now a direct comparison, of Table 6, with Table 3, in the six cases where Ψ γ can be transcendental, gives that e 0 (Ψ ′ γ ) ∈ {e 0 (I ′ γ ); e 0 (I ′ α )} mod , when (λ γ ; λ α ) ∈ Case 4 × {Case 1 or 2} or when (λ γ ; λ α ) ∈ Case 2 × {Case 3 or 4}, if |k| = 3. Moreover, when (λ γ ; λ α ) ∈ Case 4 × {Case 1 or 2}, we get the same conclusion for the exponents at z = 1. This is the reason why, these two cases cannot be a-priory refined.
6.3.3 Some results when Φ, Ψ α and Ψ γ are algebraic As we said before those cases happen very often in 10 cases over the the 16. This the right moment to apply Theorem 5, with d α = d γ = e = 0. Here, X = Ψ α I ′ α and M = Ψ α I ′ γ + Ψ γ I ′ α are first level integrals and we first have to see if they can be computed in polynomial form. That is, if they satisfied Ostrowski relations with I α and I γ ...
Precisely, according to the theorem we will have two cases • When I α and I γ are independent then we must check the following simple form of the relation given in point 5 of the theorem where E is symmetric.
• When I α and I γ are dependant, that is I γ − θI α ∈ K, then we must check a relation of the form M + θX = aI + g.
Now, according to equation (23), we have a necessary condition for the independence of the two integrals I α and I γ , which we resume in the following new  So we see again, that in half of the cases the integrals are independent, and for the remaining cases we obviously do not know.
We have made explicit computations of the integrals X and M , they are first level integrals so they will obey the rule given by Remark 2. Nevertheless, they are integrals of the form As a consequence, they do not enter into the context of Theorem 3.
Here we are going to join the information given by Propositions 6 and 7, the previous tables and some experiments given by computers. This will give a picture of the behaviour of the Galois group of VE γ 2,α . For simplicity, we will assume that |k| 5.
Indeed, the Propositions above showed, that for |k| = 3 some exceptional behaviour occur. Experiments also show that for |k| = 4, we also get new exceptions. This is quite normal since these two cases correspond geometrically to the situation where the hyper-elliptic curve parametrised by t → (ϕ(t),φ(t)) is in fact an elliptic curve.
Step 1: when Φ is transcendental According to Section 6.1, we found that Φ can be transcendental in 6 over the 16 cases, with half of the possibilities when (γ, α) ∈ {(1, 1); (1, 2)}. It seems that experiments are showing that away from the predicted cases, Φ is transcendental. As a consequence, thanks to Proposition 6, the probability to get obstruction when Φ is transcendental is p Φ = 5/16.  The empty cases correspond to the four cases where Φ is transcendental. Again, for each letter X, we will denote by p X the probability to get obstruction in the context given by the corresponding letter. Here, the first thing to check is the possible transcendence of either Ψ α or Ψ γ in order to apply Proposition 7. If nothing subsequent occurs from this test, then we check one of the two relations (24) or (25) depending of the possible dependence of the integrals I α , I γ in order to apply Theorem 5.
In A and A ′ : According to Section 6.1, Φ is algebraic when 2 deg(J α ) < deg(J γ ) in A and, when 2 deg(J α ) + 1 < deg(J γ ) in A ′ . In both cases, according to Tables 5 and 6 Ψ α and Ψ γ are algebraic. Moreover, relation (24) hold for some symmetric matrix E. Observe that here, we do not need to check the dependence of the integrals I, since if this happens then we would get (24)⇒(25). As a consequence, there is no obstruction and For the remaining letters, Φ is always algebraic.
In B and B ′ : According to Tables 5 and 6, Ψ α and Ψ γ are algebraic. Moreover, experiments give that M and X are algebraic. As a consequence, relation (24) is trivially satisfied with E = 0. Hence, p B = p B ′ = 0.
In C and C ′ : According to Table 5 Ψ α is algebraic. Nevertheless, experience shows that Ψ γ is transcendental, hence according to Proposition 7, VE γ 2,α is not virtually Abelian and, In D and D ′ : According to Tables 5 and 6, Ψ α and Ψ γ are algebraic. Here, thanks to Table 7, the two integrals I α and I γ are independent. In both cases, experiments show that M is algebraic and X is transcendental. In D, we get an Ostrowski relation of the form X = aI α + g. Therefore, in (24) the matrix is not symmetric. Therefore, there is obstruction. In D ′ the situation is quite similar excepted that there is no Ostrowski relation between X, I α and I γ , so (24) is not satisfied. Hence, In E and E ′ : Here, we cannot apply directly Proposition 7. And in fact experiments show that we get two Ostrowski relations, one between Ψ α and I α and the other between Ψ γ and I γ . Moreover, Ψ α and Ψ γ are transcendental. But experiments also show that there is no possible Ostrowski relation between M , I α and I γ . Therefore, none of the equations (24) or (25) can be satisfied. So VE 2,α is not virtually Abelian and, In F and F ′ : Experiments show that: in F both Ψ α and Ψ γ are algebraic iff 2 deg(J α ) > deg(J γ ). Similarly, in F ′ , both Ψ α and Ψ γ are algebraic iff 2 deg(J α ) + 1 > deg(J γ ). Moreover, if these conditions on the degrees of the Jacobi polynomials are satisfied, then M and X are algebraic. Hence, we get half obstruction in each case and, As a consequence, the probability to get obstruction when Φ is algebraic is Hence, the total probability to get obstruction is p T = p Φ + p alg = 5/16 + 7/16 = 3/4.
To our point of view, there are two significant conclusions that can be derived, from this study : First, that there is still a lot a obstruction at the level of the second variational equation. Indeed, it seems that there is a quite big distance between solvable Galois groups and virtually Abelian ones. Secondly, although it is comparatively much more complicated to test, the most important obstruction to the virtual Abelianity of VE γ 2,α happens when Φ is algebraic.
7 Considerations about EX γ 2,α,β Here we follow the same strategy. But now the main technical difficulty comes in distinguishing the 64 possibilities for the cases satisfied by γ, β, α. We will therefore have to deal with spatial tables of 64 entries ! As a consequence, we will not give definitive results. Nevertheless, Theorem 4, can be used to deal with the complete study of specific potentials.
7.1 Counting the cases when Φ = ωu 1 y 1 x 1 is algebraic Since the most check able obstructions are going to happen when Φ is transcendental, at first glance, we count the possible numbers of such occurrences. Here, by using similar arguments as in Table 4, we are going to count 44, possibilities where Φ is algebraic. Therefore, we will be left with at most 20 cases where Φ can be transcendental ! Here, With P (z) = J γ J β J α and, Again, as in Section 4.4, Φ is going to be algebraic when at least one of the two exponents E 0 or E 1 is an integer distinct from -1.

Conclusion
Since we get 24 possibilities for E 0 ∈ , 32 possibilities for E 1 ∈ and 12 for both integral exponents. Φ is going to be algebraic in 44 = 24 + 32 − 12 cases. As a consequence, in the spatial table of 64 entries for (γ; β; α), there are at most 20 possibilities where Φ can be transcendental. 7.2 Counting when all the Ψ µ are algebraic in the 44 case where Φ is algebraic This is also made in order to find the cases where there is no obstruction.
In fact thanks to the consideration above we count 16 cases over 64 where everybody is certainly algebraic.

Getting obstruction with the assumption that Φ is transcendental
Here our main result is going to be the following one which is very similar in its statement and proof to Proposition 6.

Getting obstruction in the 44 cases when Φ is algebraic
In order to be able to exploit the Ostrowski relations given by Theorem 4, in these 44 cases, we first have to investigate the possible algebraicity of the Ψ µ , when µ ∈ {γ, β, α}.
2. For|k| = 3, we get the same conclusion if we assume that at least two of the three integrals Ψ µ are transcendental.
Proof Now we assume that EX γ 2,α,β is virtually Abelian. By symmetry let's assume that Ψ γ is transcendental. According to Theorem 4, we must have a non trivial Ostrowski relation The monodromy M 1 acts by characters on the derivatives of these four integrals. According to Lemma 3.2, the character of Ψ γ must be equal to the character of one of the I µ . But, as in the proof of Proposition 7, this condition is equivalent to having Now let's do the same job with M 0 . As in the proof of Proposition 7, the Ostrowski relation imposes that at least one of the three following numbers is an integer: Let us set ∆ β,α := E 0 − 2a β + 2a α . With this notation, we just have seen that Ψ γ transcendental implies that E 0 ∈ or∆ γ,β ∈ or∆ γ,α ∈ .
If E 0 ∈ , both exponent at z = 0 and z = 1 are integers, we are in the 12 cases over the 44 where Φ is algebraic and these arguments do not give any obstruction to the virtual Abelianity of EX γ 2,α,β . Now let's assume that E 0 ∈ . We are led to find obstructions from the conditions ∆ ∈ . But, by writing a = 1/2 + ε/2k, we get ∆ β,α = ε γ − ε β + 3ε α − 1 2k .
Its values depends on the eight possibilities given by ε µ = ±1. They are listed in the following table Here, we did not give the value of ∆ β,α for the three lines L 1 , L 2 , L 3 of Table 9, because they correspond to E 0 ∈ .
From this table, we get that ∆ ∈ except when |k| = 3, in the case of line L 6 . This prove the first point of the proposition.
So, the Ostrowski relation will not be satisfied for Ψ β . This prove the claim.