Orthogonal Hypergeometric Groups with a Maximally Unipotent Monodromy

Similar to the symplectic cases, there is a family of fourteen orthogonal hypergeometric groups with a maximally unipotent monodromy (cf. Table 1.1). We show that two of the fourteen orthogonal hypergeometric groups associated to the pairs of parameters $(0, 0, 0, 0, 0)$, $(\frac{1}{6}, \frac{1}{6}, \frac{5}{6}, \frac{5}{6}, \frac{1}{2})$; and $(0, 0, 0, 0, 0)$, $(\frac{1}{4}, \frac{1}{4}, \frac{3}{4}, \frac{3}{4}, \frac{1}{2})$ are arithmetic. We also give a table (cf. Table 2.1) which lists the quadratic forms $\mathrm{Q}$ preserved by these fourteen hypergeometric groups, and their two linearly independent $\mathrm{Q}$- orthogonal isotropic vectors in $\mathbb{Q}^5$; it shows in particular that the orthogonal groups of these quadratic forms have $\mathbb{Q}$- rank two.


Introduction
To explain the result of this paper, we first recall the definition of hypergeometric groups (which are monodromy groups of some hypergeometric differential equations).For, we denote θ = z d dz and write the differential operator: for α = (α 1 , . . ., α n ), β = (β 1 , . . ., β n ) ∈ C n ; and consider the hypergeometric differential equation D(α; β)w = 0 on P 1 (C) with regular singularities at the points 0, 1 and ∞, and regular elsewhere.
The hypergeometric groups ρ(π 1 ) are classified by a theorem of Levelt ( [9]; cf.[2, Theorem 3.5]): if α 1 , α 2 , . . ., α n , β 1 , β 2 , . . ., β n ∈ C such that α j − β k ∈ Z, for all j, k = 1, 2, . . ., n, then the hypergeometric group ρ(π 1 ) is (up to conjugation in GL n (C)) a subgroup of GL n (C) generated by the companion matrices A and B of f (X) = n j=1 (X − e 2πiα j ) and g(X) = n j=1 (X − e 2πiβ j ) resp., and the monodromy ρ is defined by We now denote the hypergeometric group ρ(π 1 ) by Γ(f, g) (which is a subgroup of GL n (C) generated by the companion matrices of f, g) and consider the cases where α, β ∈ Q n (that is, the roots of the polynomials f, g are the roots of unity) and the coefficients of f, g are integers (for example, one can take f, g as product of cyclotomic polynomials); in these cases, Γ(f, g) ⊂ GL n (Z).Note that the condition "α j − β k ∈ Z, for all j, k = 1, 2, . . ., n" means that f, g do not have any common root in C. We also assume that f, g form a primitive pair [2,Definition 5.1].
With the assumptions made in last paragraph, the Zariski closures G of the hypergeometric groups Γ(f, g) are completely determined by Beukers and Heckman in [2, Theorem 6.5]: if n is even and f (0) = g(0) = 1, then the hypergeometric group Γ(f, g) preserves a nondegenerate integral symplectic form Ω on Z n and Γ(f, g) Note that a hypergeometric group Γ(f, g) is called arithmetic, if it is of finite index in G(Z); and thin, if it has infinite index in G(Z) [11].In [11], Sarnak has asked a question to determine the cases where Γ(f, g) is arithmetic or thin.There have been some progress to answer the question of Sarnak.
For the symplectic cases: infinitely many arithmetic Γ(f, g) in Sp n (for any even n) are given by the author and Venkataramana in [13]; four other cases of arithmetic Γ(f, g) in Sp 4 are given by the author in [12]; seven thin Γ(f, g) in Sp 4 are given by Brav and Thomas in [3]; and in [6], Hofmann and van Straten have determined the index of Γ(f, g) in Sp 4 (Z) for some of the arithmetic cases of [12] and [13].
For the orthogonal cases: when the quadratic form Q has signature (n − 1, 1), Fuchs, Meiri and Sarnak give infinitely many thin Γ(f, g) in [8]; when the quadratic form Q has signature (p, q) with p, q ≥ 2, infinitely many arithmetic Γ(f, g) are given by Venkataramana in a very recent paper [16] (these are also the first examples of higher rank arithmetic orthogonal hypergeometric groups); and an example of thin Γ(f, g) in O(2, 2) is given by Fuchs in [7].
The question to determine the arithmeticity or thinness of these 14 symplectic hypergeometric groups is now completely solved (cf.Table 1.1): 7 of the 14 groups are arithmetic (by [12], [13]), and other 7 are thin (by [3]).
Note that, in this case we get f (0) g(0) = −1, therefore each of the 14 hypergeometric groups Γ(f, g) preserves some integral quadratic form Q on Z 5 and Γ(f, g) ⊂ O Q (Z) is Zariski dense.
Therefore similar to the 14 symplectic cases, we may ask the question to determine the arithmeticity or thinness of the 14 orthogonal cases; it is a special case of Sarnak's question [11].
Yes, [13] Yes, [12] Therefore 2 of the 14 orthogonal hypergeometric groups are arithmetic, and it will be very interesting to determine the dichotomy (similar to the symplectic cases) in the orthogonal cases also.
We now describe the proof of Theorem 1.2.By using [2, Theorem 4.5] (cf.Lemma 2.2), one can compute easily that for all the 14 orthogonal hypergeometric groups of Table 1.1, the quadratic form Q has signature (3,2) or (2,3), that is, the corresponding orthogonal group O Q has real rank 2 and Q-rank at least one (by the Hasse-Minkowski theorem).
For the pairs in Theorem 1.2, we explicitly compute, up to scalar multiples, the quadratic form Q on Q 5 and get a basis Note that the existence of such basis for (Q 5 , Q) ensures that the corresponding orthogonal group O Q has Q-rank two.
Also, with respect to the basis {ǫ 1 , ǫ 2 , u, ǫ * 2 , ǫ * 1 } of Q 5 , the group of diagonal matrices in SO Q form a torus T, the group of upper (resp.lower) triangular matrices in SO Q form a Borel subgroup B (resp.B − , opposite to B), and the group of unipotent upper (resp.lower) triangular matrices in SO Q form the unipotent radical U (resp.U − , opposed to U) of B (resp.B − ).
We prove the arithmeticity of the two orthogonal hypergeometric groups of Theorem 1.2, by showing that, with respect to the basis , and use a theorem of Tits [14] (cf.[15,Theorem 3.5]).
Also, to show that Γ(f, g) ∩ U(Z) is a subgroup of finite index in U(Z), it is enough to show that Γ(f, g) ∩ U(Z) is Zariski dense in U (since U is a nilpotent subgroup of GL 5 (R); cf.[10, Theorem 2.1]); and we show it, by showing that, Γ(f, g) ∩ U(Z) contains non-trivial unipotent elements corresponding to all positive roots of SO Q .
Let A and B be the companion matrices of f and g respectively.Let C = A −1 B. Then, one can check easily that .
Let e 1 , e 2 , e 3 , e 4 , e 5 be the standard basis vectors of Q 5 over Q, and v be the last column vector of C − I, where I is the identity matrix.Then, we get Ce 5 = v + e 5 , and hence Therefore, by using the invariance of Q under the action of C, we get that v is Q-orthogonal to the vectors e 1 , e 2 , e 3 , e 4 and Q(v, e 5 ) = 0 (since Q is non-degenerate).We may now assume that Q(v, e 5 ) = 1.
Proof.First, we note the following: Hence if we identity the vector space Q 5 as the quotient space Q[X] <f (X)> , where < f (X) > is the ideal generated by f (X) in the polynomial ring Q[X], and the action of <f (X)> .Now, suppose on contrary that the set {v, Av, A 2 v, A 3 v, A 4 v} is linearly dependent over Q. Therefore the set {Av, A 2 v, A 3 v, A 4 v, A 5 v} is also linearly dependent over Q, and hence there exists a non-zero polynomial h(X) of degree ≤ 4 such that That is, f (X) divides h(X).(f(X) − g(X)).Since f (X) and g(X) are co-prime, f (X) divides h(X) and hence h(X) = 0 (since the degree (≤ 4) of h(X) is less than the degree of f (X)), which is a contradiction to our assumption that h(X) = 0. Therefore the set {v, Av, A 2 v, A 3 v, A 4 v} is linearly independent over Q.
By using Lemma 2.1, to determine the quadratic form We prove further the following lemma, to determine the signature of a quadratic form preserved by the orthogonal hypergeometric groups of Table 1 Proof.By [2, Theorem 4.5], the signature (p, q) of the quadratic form Q preserved by a (orthogonal) hypergeometric group is given by the formula: where m j = #{k : β k < α j }, for all j = 1, 2, . . ., n.
By using both of the equations in p, q, we get that the quadratic forms Q preserved by the 14 orthogonal hypergeometric groups of Table 1.1 has signature (3, 2) or (2, 3); and hence the real rank of the orthogonal group O Q is 2 and the Q-rank of O Q is 1 or 2 (by the Hasse-Minkowski theorem).
Since Q is non-degenerate on Q 5 , the existence of two linearly independent isotropic vectors (which are not Q-orthogonal) for Q on Q 5 is now clear from Lemma 2.2.For the hypergeometric groups of Theorem 1.2, we show that there exist two linearly independent isotropic vectors (which are Q-orthogonal) for Q on Q 5 ; that is, the Q-rank of O Q is 2 for each of these two cases.
Note that we have used the invariance of Q, under the action of B, to write the above matrix form of Q.
If we denote by T , the matrix to change the basis {e 1 , e 2 , e 3 , e 4 , e 5 } to {v, Bv, B 2 v, B 3 v, B 4 v}, that is, then the matrix form of Q, with respect to the basis {e 1 , e 2 , e 3 , e 4 , e 5 }, where (T −1 ) t , A t , B t denote the transpose of the respective matrices.
It is now clear from the matrix form M Q of Q, that is an isotropic vector of Q; and by computation, we get 5 .We now by computation, get a basis {u 1 , u 2 , u 3 } of E ⊥ , where If we denote by L, the matrix to change the basis {e 1 , e 2 , e 3 , e 4 , e 5 } to {ǫ 1 , u 1 , u 2 , u 3 , ǫ * 1 }, that is, then the matrix form of Q, with respect to {ǫ 1 , u 1 , u 2 , u 3 , ǫ * 1 }, is: .
We now change the basis {u 1 , u 2 , u 3 } of E ⊥ to another basis {ǫ 2 , u, ǫ * 2 } with respect to which, the restriction of Q on E ⊥ is anti-diagonal, where If we denote by K, the matrix to change the basis {e 1 , e 2 , e 3 , e 4 , e 5 } to then the matrix form of Q, with respect to the basis {ǫ 1 , ǫ 2 , u, ǫ * 2 , ǫ * 1 }, is: Note that, the above matrix form of Q, with respect to the Q-basis Recall that, with respect to the basis {ǫ 1 , ǫ 2 , u, ǫ * 2 , ǫ * 1 } of Q 5 , the group of diagonal matrices in SO Q form a torus T, the group of upper (resp.lower) triangular matrices in SO Q form a Borel subgroup B (resp.B − , opposite to B), and the group of unipotent upper (resp.lower) triangular matrices in SO Q form the unipotent radical U (resp.U − , opposed to U) of B (resp.B − ).
Proof of the arithmeticity of Γ.To prove the arithmeticity of Γ, we show that with respect to the basis {ǫ , and use a theorem of Tits [14].Note that the proof also follows from a theorem of Venkataramana [15,Theorem 3.5].
Also, to show that Γ ∩ U − (Z) is a subgroup of finite index in U − (Z), it is enough to show that Γ ∩ U − (Z) is Zariski dense in U − (since U − is a nilpotent subgroup of GL 5 (R); cf.[10, Theorem 2.1]); and we show it, by showing that, Γ ∩ U − (Z) contains non-trivial unipotent elements corresponding to all negative roots of SO Q .
Let x, y denote the matrices A, B resp., with respect to the basis .
We now denote by c 1 , c 2 , . . ., c 19 , the following words in x and y: It is now clear from the above computations that the elements c 11 , c 14 , c 17 , c 19 ∈ Γ∩U − (Z), are non-trivial, and correspond to the negative root groups of SO Q , that is, Γ∩U − (Z) is Zariski dense in U − , and hence Γ ∩ U − (Z) is of finite index in U − (Z) (since U − is a nilpotent subgroup of GL 5 (R)).The proof now follows from [14] (cf.[15,Theorem 3.5]).
The invariant quadratic form.Let e 1 , e 2 , e 3 , e 4 , e 5 be the standard basis vectors of Q 5 over Q, and v = (C − I)e 5 .By doing similar computations as in Subsection 2.1, we find that the matrix form of Q, with respect to the basis {v, Bv, B 2 v, B 3 v, B 4 v} of Q 5 , is: If we denote by T , the matrix to change the basis {e 1 , e 2 , e 3 , e 4 , e 5 } to {v, Bv, B 2 v, B 3 v, B 4 v}, that is, If we denote by K, the matrix to change the basis {e 1 , e 2 , e 3 , e 4 , e 5 } to {ǫ 1 , ǫ 2 , u, ǫ * 2 , ǫ * 1 }, that is, then the matrix form of Q, with respect to the basis {ǫ 1 , ǫ 2 , u, ǫ * 2 , ǫ * 1 }, is: Note that, the above matrix form of Q, with respect to the Q-basis {ǫ 1 , ǫ 2 , u, ǫ * 2 , ǫ * 1 } of Q 5 , shows that the Q-rank of the orthogonal group O Q is two.
Proof of the arithmeticity of Γ.Let x, y denote the matrices A, B resp., with respect to the basis {ǫ 1 , ǫ 2 , u, ǫ * 2 , ǫ * 1 }, that is, x = K −1 AK and y = K −1 BK.Then It is now clear from the above computations that the elements c 2 , c 10 , c 11 , c 17 ∈ Γ ∩ U(Z), are non-trivial, and correspond to the positive root groups of SO Q , that is, Γ ∩ U(Z) is Zariski dense in U, and hence Γ ∩ U(Z) is of finite index in U(Z).The proof now follows from [14] (cf.[15,Theorem 3.5]).
be the subgroup of GL 5 (Z) generated by A and B. The invariant quadratic form.Note that A −1 B = B −1 A; and the statements of Lemma 2.1 and the succeeding paragraph, are unchanged, if we replace A by B. Therefore the vectors in the set {v, Bv, B 2 v, B 3 v, B 4 v} form a basis of Q 5 over Q. Recall that v is the vector (C − I)e 5 ; and e 1 , e 2 , e 3 , e 4 , e 5 are the standard basis vectors of Q 5 over Q.By computation, we get

. 1 :
Lemma 2.2.The quadratic forms Q preserved by the 14 orthogonal hypergeometric groups of Table1.1 has signature (3, 2) or (2, 3); that is, the real rank of the orthogonal group then the matrix form of Q, with respect to the basis {e 1 , e 2 , e 3 , e 4 , e 5 }, is N Q