HYPERGEOMETRIC EQUATION AND MONODROMY GROUPS
EDMUND Y.-M. CHIANG
Abstract. We shall introduce the basics of classical Gauss hypergeometric equation and to study its monodromy problem. Historically, Riemann was the first one who worked out the monodromy group of the DE in 1857.
1. Fuchsian-type Differential Equations
Theorem 1.1. Let a1, a2, ··· an and ∞ be the regular singular points of d2w dw (1.1) + p(z) + q(z)w = 0. dz2 dz Then we have n n X Ak X Bk Ck (1.2) p(z) = , q(z) = 2 + , z − ak (z − ak) z − ak k=1 k=1 and the indicial equation at each ak is given by 2 α + (Ak − 1)α + Bk = 0. 0 0 Suppose {αk, αk} and {α∞, α∞} denote, respectively, the exponents of the corre- sponding indicial equations at a1, a2, ··· an and ∞. Then we have n 0 X 0 (α∞ + α∞) + (αk + αk) = n − 1, k=1
Proof. Since p(z) has at most a first order pole at each of the points a1, a2, ··· an, so the function φ(z) defind in n X Ak p(z) = + ϕ(z) z − ak k=1 where ϕ(z) is an entire function in C, and where the Ak is the residue of p(z) at ak (k = 1, 2, ··· n). On the other hand, the ∞ is also a regular singular point of p(z), it follows from Proposition 3.2 in part I and which states ∞ is regular singular iff zp and z2q are analytic at ∞. That is, p(z) = O(1/z). So ϕ(z) → 0 and hence must be the constant zero by applying Liouville’s theorem from complex analysis. Hence n X Ak p(z) = . z − ak k=1
Lectures given at South China Normal University given during the period 22-24th, May 2017. This project was partially supported by Hong Kong Research Grant Council, #16300814. 1 2 EDMUND Y.-M. CHIANG
Similarly, we have n X Bk Ck q(z) = 2 + + ψ(z), (z − ak) z − ak k=1 where ψ(z) is entire function. But z2q(z) is analytic at ∞, so q(z) = O(1/z2) as z → ∞. This shows that ψ(z) ≡ 0. The Proposition 3.2 (in Part I) again implies that n X Ck = 0. k=1
Considering the Taylor expansions around the regular singular point ak ∞ ∞ X j 2 X j (z − ak)p(z) = pj · (z − ak) , (z − ak) q(z) = qj · (z − ak) j=0 j=0 and ∞ X α+j w(z) = aj(z − ak) j=0 yields 2 α + (Ak − 1)α + Bk = 0. for k = 1, 2, ··· , n. We now deal with the indicial equation at z = ∞. Since the ∞ is also a regular singular point, so the Proposition 3.2 (in Part I) implies that 2 1 1 lim t − p = 2 − lim zp(z) t→0 t t2 t z→∞ Qn n j=1 Ak(z − aj) X j6=k = 2 − lim z Qn z→∞ (z − aj) k=1 j=1 Pn A zn + O(zn−1) = 2 − lim k=1 k z→∞ zn + O(zn−1) n X (1.3) = 2 − Ak. k=1 Similarly, we have 1 1 lim t2 · q = lim z2q(z) t→0 t4 t z→∞ n 2 X Bk Ck = lim z 2 + z→∞ (z − ak) z − ak k=1 n n X X Ck = Bk + lim z z→∞ 1 − ak/z k=1 k=1 n n X X ak 1 = Bk + lim z Ck 1 + + O z→∞ z z2 k=1 k=1 n n X X (1.4) = Bk + 0 + akCk, k=1 k=1 HYPERGEOMETRIC EQUATION AND MONODROMY GROUPS 3
Pn because k=1 Ck = 0. The indicial equation for the point z = 0 is of the form 2 α + (p0 − 1)α + q0 = 0, 2 where p0 = limz→0 zp(z) and q0 = limz→0 z q(z). Thus, the point at infinity is given by n n 2 X X α + 1 − Ak α + (Bk + akCk) = 0. k=1 k=1 0 2 Let αk, αk be the exponents of the indicial equation α + (Ak − 1)α + Bk = 0 for 0 k = 1, 2, ··· , n, and α∞, α∞ be the exponents of the indicial equation at ∞. Thus we deduce n n n n X 0 X X X (αk + αk) = − (Ak − 1) = (1 − Ak) = n − Ak, k=1 k=1 k=1 k=1 0 Pn and α∞ + α∞ = 1 − k=1 Ak. Hence we deduce n X 0 0 (αk + αk) + (α∞ + α∞) = n − 1, k=1 as required.
Exercise 1.2 (Klein). We note that if ∞ is an ordinary point, then show that 2 p p(z) = + 2 + ··· , z z2 and q q q(z) = 4 + 5 + ··· . z4 z5 Suppose the ∞ in Theorem 5.1 is an ordinary point. Then show that the following identities hold. n X Ak = 2, k=1 n X Ck = 0, k=1 n X (Bk + akCk) = 0, k=1 n X 2 (2akBk + akCk) = 0. k=1
2. Differential Equations with Three Regular Singular Points We again consider differential equation in the form d2w dw (2.1) + p(z) + q(z)w = 0. dz2 dz
The following statement was first written down by Papperitz in 1885. See [9] 4 EDMUND Y.-M. CHIANG
Theorem 2.1. Let the differential equation (2.1) has regular singular points at ξ, η and ζ in C. Suppose the α, α0; β, β0; γ, γ0 are, respectively, the exponents at the regular singular points ξ, η and ζ. Then the differential equation must assume the form d2w n1 − α − α0 1 − β − β0 1 − γ − γ0 odw + + + − dz2 z − ξ z − η z − ζ dz n αα0 ββ0 γγ0 o − + + × (z − ξ)(η − ζ) (z − η)(ζ − ξ) (z − ζ)(ξ − η) (ξ − η)(η − ζ)(ζ − ξ) (2.2) × w = 0. (z − ξ)(z − η)(z − ζ)
Proof. The equation has regular singular points at ξ, η and ζ in C. Thus the functions P (z) = (z − ξ)(z − η)(z − ζ) p(z), and Q(z) = (z − ξ)2(z − η)2(z − ζ)2 q(z) are both entire functions (complex differentiable everywhere). Besides, since the point at “∞” is an analytic point for both p(z) and q(z) (since we have assumed that none of the three singular points {ξ, η, ζ} is at ∞), so we deduce from the Exercise 1.2 that the functions that P (z) and Q(z) have degree two. That is, P (z) A B C p(z) = = + + , (z − ξ)(z − η)(z − ζ) z − ξ z − η z − ζ where A + B + C = 2 (by Exercise 1.2 again). Similarly, Q(z) (z − ξ)(z − η)(z − ζ)q(z) = (z − ξ)(z − η)(z − ζ) D E F = + + . z − ξ z − η z − ζ P∞ α+k Substitute w(z) = k=0 wk(z − ξ) into the differential equation and note the coefficient q(z) assumes the form D E(z − ξ) F (z − ξ) (z − ξ)2q(z) = + + . (z − η)(z − ζ) (z − η)2(x − ζ) (z − η)(z − ζ)2 D So q0 = (ξ−η)(ξ−ζ) . Thus the indicial equation at z = ξ is given by D D α(α − 1) + Aα + = α2 + (A − 1)α + = 0, (ξ − η)(ξ − ζ) (ξ − η)(ξ − ζ) where α + α0 = −(A − 1) = 1 − A or A = 1 − α − α0, and D αα0 = , (ξ − η)(ξ − ζ) or D = (ξ − η)(ξ − ζ)αα0. Similarly, we have , at the regular singular points η, ζ that B = 1 − β − β0 and C = 1 − γ − γ0, and hence E = (η − ξ)(η − ζ)ββ0,F = (ζ − ξ)(ζ − η)γγ0. This yields the differential equation of the required form. HYPERGEOMETRIC EQUATION AND MONODROMY GROUPS 5
Remark 2.2. We note that the constraint A + B + C = 2 implies that α + α0 + β + β0 + γ + γ0 = 1.
Example 2.1. Suppose that a second order linear differential equation has only one regular singular point at z = 0 and infinity is an ordinary point. Show that the equation is of the form zw00 + 2w0 = 0.
Proof. Let the equation be in the form w00 + p(z)w0 + q(z)w = 0. Since z = 0 is a regular singular point. So P (z) = zp(z) is an entire function. But z = ∞ A p2 p2 is an ordinary point, so P (z) = zp(z) = z z + z2 + ··· = A + z + ··· as z → ∞ by Exercise 1.2. We conclude that P is the constant A by Liouville’s 2 p2 theorem. We also deduce that A = 2 since p(z) = z + z2 + ··· . Similarly, we have Q(z) = z2q(z) to be an entire function. But z = ∞ is an ordinary point, so 2 2 q4 q5 q4 q4 Q(z) = z q(z) = z z4 + z5 + ··· = z2 + z3 + · · · → 0 as z → ∞ by Exercise 1.2 again. Thus Q(z) ≡ 0 by Liouville’s theorem again. Hence q(z) ≡ 0. We conclude 00 2 0 that w + z w + 0 = 0.
Exercise 2.3. Suppose a second order linear differential equation has both the origin and the infinity to be regular singular points, and that their respective expo- 0 0 nents are {α, α } and {α∞, α∞}. Derive the exact form of the differential equation 0 0 0 0 and show that α + α + α∞ + α∞ = 0 and α · α = α∞ · α∞ hold.
If we now choose η = ∞ in the differential equation in Theorem 2.1, then we easily obtain d2w n1 − α − α0 1 − γ − γ0 odw + + + dz2 z − ξ z − ζ dz nαα0(ξ − ζ) γγ0(ζ − ξ)o w (2.3) + + ββ0 + × = 0. z − ξ z − ζ (z − ξ)(z − ζ) For convenience we could set ξ = 0 and ζ = 1. So this differential equation will be further reduced to the form d2w n1 − α − α0 1 − γ − γ0 odw + + + dz2 z z − 1 dz nαα0(−1) γγ0(1 − 0)o w (2.4) + + ββ0 + × = 0. z z − 1 z(z − 1) We note that the exponent pair at each of the singularities is exactly the same as before. Exercise 2.4. Show that the above equation can be written in the from d2w n1 − α − α0 1 − γ − γ0 odw + + + dz2 z z − 1 dz nαα0 γγ0 ββ0 − αα0 − γγ0 o (2.5) + + + w = 0. z2 (z − 1)2 z(z − 1)
Before we embark on more structural results, we need the following result. 6 EDMUND Y.-M. CHIANG
Proposition 2.1. Let d2w dw (2.6) + p(z) + q(z)w = 0 dz2 dz 0 to have a regular singular point at z0 = 0 with exponent {α, α }. Then the function f(z) defined by w(z) = zλf(z) satisfies the differential equation d2f 2λ df λp(z) λ(λ − 1) (2.7) + p(z) + + q(z) + + f = 0, dz2 z dz z z2 with regular singular point also at z = 0 but the exponent pair is {α − λ, α0 − λ}. If z = ∞, then the equation (2.7) has exponent pair {α + λ, α0 + λ} instead.
Proof. Let w(z) = zλf(z). We leave the verification of the equation (2.7) as an exercise. Substitute the series ∞ ∞ ∞ X k 2 X k X α+k zp(z) = pkz , z q(z) = qkz , f(z) = akz k=0 k=0 k=0 into the differential equation (2.7) and consider the indicial equation: α(α − 1) + p0 + 2λ α + q0 + λp0 + λ(λ − 1) = 0, or 2 α + p0 − 1 + 2λ α + q0 + λp0 + λ(λ − 1) = 0. Let {α,˜ α˜0} be the exponent pair for the equation (2.7) at z = 0. But thenα ˜ +α ˜0 = 0 1 − p0 − 2λ = α + α − 2λ, and 0 α˜α˜ = q0 + λp0 + λ(λ − 1) = αα0 − λ(α + α0) + λ2 = (α − λ)(α0 − λ). Solving the above two algebraic equations proves the first part of the proposition.
Exercise 2.5. Complete the proof of the above Proposition of the transformation to ∞ and the change of the exponent pair.
Remark 2.6. If we apply the transformation w(z) = (z − 1)λf(z) to the differential equation (2.6), then the resulting equation (2.7) would still have {0, 1, ∞} as reg- ular singular points, but the exponent pair at z = 1 is now {β − λ, β0 − λ} instead. The idea could be apply to a more general situation. Suppose the differential equa- tion (2.6) has z = ξ as a regular singular point with exponent pair {α, α0}, then the resulting differential equation after the transformation w(z) = (z − ξ)λf(z) would still has z = ξ as a regular singular point but now the corresponding exponent pair is {α − λ, α0 − λ}. HYPERGEOMETRIC EQUATION AND MONODROMY GROUPS 7
3. Mondromy We consider linear differential equation with finite number of singularities in the 1 one-dimensional projective space CP , that is C ∪ {∞}. Definition 3.1. Let D be simply connected. Let b ∈ D. (1) A loop γ in D γ : I = [0, 1] → D such that γ(0) = b = γ(1). The b is called the base point of γ. Two loops γ0, γ1 in D are homotopy (equivalent), denoted by γ0, ' γ1, if γ0 can be deformed continusely to γ1 and vice-versa (but keeping the base point fixed). (2) Let L(D, b) be the set of loops with a base point at b. Then ' defines an equivalence relation on L(D, b). (3) The set of all equivalence classes [γ] of loops defined above forms a group π1(D, b) = L(D, b)/ ', called the fundamental group.We define the product of two loops γ0 ·γ1 to be a loop such that γ1 follows γ0. For each loop γ one can define its inverse γ−1 by γ−1(t) = γ(1 − t) and an identity (constant) loop c : I → D such that c(t) ≡ b for all t. It is easy to verify that one carry this struture to their equivalence classes. Hence one can really make π1(D, b) = L(D, b)/ ' to be a group: with α = [γ] ∈ π1(D, b), α−1 = [γ−1], etc.
4. Riemann’s P −Scheme The following fomulation of solution of Gauss hypergeometric equation appeared to be first put forward by B. Riemann (1826–1866) in 1857 [7]. We shall follow closely Riemann’s exposition.
Definition 4.1. Following Riemann, we write the solution w of the equation (2.2) in Riemann’s P −symbol: ξ η ζ (4.1) w(z) = P α β γ ; z . α0 β0 γ0
Remark 4.1. We see immediately that the meaning of Riemann’s P −function re- mains unchanged if we permuate the first three columns or interchange any of the two exponents for each of the three points {ξ, η, ζ}. Thus a solution of equation (2.3) can be denoted by ξ ∞ ζ w(z) = P α β γ ; z , α0 β0 γ0 while that of (2.5) can be denoted by 0 ∞ 1 (4.2) w(z) = P α β γ ; z . α0 β0 γ0
It follows from Proposition 2.1 that 8 EDMUND Y.-M. CHIANG
Exercise 4.2. 0 ∞ 1 0 ∞ 1 (4.3) zp(z−1)qP α β γ ; z = P α + p β − p − q γ + q ; z . α0 β0 γ0 α0 + p β0 − p − q γ0 + q Similarly, Exercise 4.3. (4.4) 0 ∞ 1 0 ∞ 1 z−p(z − 1)−qP α β γ ; z = P α − p β + p + q γ − q ; z . α0 β0 γ0 α0 − p β0 + p + q γ0 − q
Exercise 4.4. Prove by transforming the differential equation that 0 ∞ 1 1 0 ∞ 1 (4.5) P α β γ ; z = P α β γ ; 0 0 0 1 − z α β γ α0 β0 γ0 In general, we have Exercise 4.5. ξ η ζ ξ η ζ z − ξ p z − ζ q P α β γ ; z = P α + p β − p − q γ + q ; z . z − η z − η α0 β0 γ0 α0 + p β0 − p − q γ0 + q
az+b Exercise 4.6. Let M(z) = cz+d , where ad − bc 6= 0, be a M¨obiustransformation. Let M(ξ) = ξ1,M(η) = η1,M(ζ) = ζ1. Show that ξ η ζ ξ1 η1 ζ1 P α β γ ; z = P α β γ ; z1 α0 β0 γ0 α0 β0 γ0 where z1 = M(z).
However, Riemann’s original exposition is different from the above line of devel- opment. In fact, he considered funcitons that satisfy the following properties: (1) Every branch of (2.2) is finite, single-valued and holomorphic (analytic) at every z except at three singularities a, b, c; (2) Any three branches are linearly connected, or linearly dependent that is, c0P 0 + c00P 00 + c000P 000 = 0; (3) At any of the three singularities, such as z = a for example, there are 0 two branches, denoted by {P (α),P (α )}, corresponds to the two exponents 0 0 {α, α0}, such that (z − ξ)αP (α) and (z − ξ)α P (α ) remain single-valued. 0 0 Similarly, we have {P (β),P (β )} for {β, β0} and finally {P (γ),P (γ )} for {γ, γ0}; (4) that the six exponents {α, α0, β, β0, γ, γ0} are such that none of the dif- ferences (α − α0), (β − β0), (γ − γ0) is an integer, and further that they all satisfy α + α0 + β + β0 + γ + γ0 = 1. HYPERGEOMETRIC EQUATION AND MONODROMY GROUPS 9
Thus the above exercises follow easily from these four requirements. Moreover, we automatically obtain that any two of the following six functions generated by (4.2) are equal to each other: 0 ∞ 1 0 ∞ 1 0 ∞ 1 1 1 P α β γ ; z ,P β γ α ; ,P γ α β; 1 − 0 0 0 1 − z z α β γ β0 γ0 α0 γ0 α0 β0
(4.6) 0 ∞ 1 0 ∞ 1 0 ∞ 1 z 1 P α γ β ; ,P β α γ ; ,P γ β α; 1 − z . z − 1 z 0 0 0 α0 γ0 β0 β0 α0 γ0 γ β α The six M¨obiustransformations permute the three singularities in six possible ways. 4.1. Some reduced forms. The Exercise 4.4 implies that by choosing p = α and q = γ that we would obtain 0 ∞ 1 0 ∞ 1 P α β γ ; z = zα(1 − z)γ P α − α α + β + γ γ − γ ; z α0 β0 γ0 α0 − α α + β0 + γ γ0 − γ 0 ∞ 1 (4.7) = zα(1 − z)γ P 0 α + β + γ 0 ; z . α0 − α α + β0 + γ γ0 − γ So one can ensure that exactly one series solution at each of the regular singularities z = 0 and z = 1 is analytic there. By exchanging the exponent pair α and α0 in the above derivation suggests that one can obtain two different representations. The new one being 0 ∞ 1 0 ∞ 1 0 P α β γ ; z = zα (1 − z)γ P α − α0 α0 + β + γ γ − γ ; z α0 β0 γ0 α0 − α0 α0 + β0 + γ γ0 − γ 0 ∞ 1 0 = zα (1 − z)γ P α − α0 α0 + β + γ 0 ; z 0 α0 + β0 + γ γ0 − γ 0 ∞ 1 0 (4.8) = zα (1 − z)γ P 0 α0 + β0 + γ γ0 − γ ; z . α − α0 α0 + β + γ 0 Interchanging the exponent pair γ and γ0 will therefore produce further two and hence a total of four different representations together α and α0. On the other hand, mixing each of these four with the six different M¨obiustransformation relations (4.6) would give a total of twenty four different representations. We shall return to them in a later section. Let us introduce exponent-differences λ := α0 − α, µ := β0 − β, ν := γ0 − γ. Then 1 α + β + γ = (1 − λ − µ − ν), 2 10 EDMUND Y.-M. CHIANG and 1 α + β0 + γ = (1 − λ + µ − ν). 2 Thus the (4.7) becomes 0 ∞ 1 0 ∞ 1 P α β γ ; z = zα(1 − z)γ P 0 α + β + γ 0 ; z α0 β0 γ0 α0 − α α + β0 + γ γ0 − γ 0 ∞ 1 α γ 1 (4.9) = z (1 − z) P 0 2 (1 − λ − µ − ν) 0 ; z . 1 λ 2 (1 − λ + µ − ν) ν Hence the corresponding equation becomes d2w 1 − λ 1 − ν dw h(1 − λ − µ − ν)(1 − λ + µ − ν)i (4.10) + + + w = 0. dz2 z z − 1 dz 4z(z − 1) A final reduced from is obtained by eliminating the first derivative term from (2.5). It is now regarded as in the normal form (or sometimes called one- dimensional Schr¨odingerform) by selecting α + α0 = 1 and γ + γ0 = 1 in (4.9). That is, 0 ∞ 1 1 1 2 (1−λ) 2 (1−ν) 1 z (z − 1) P 0 2 (1 − λ − µ − ν) 0 ; z 1 λ 2 (1 − λ + µ − ν) ν 0 ∞ 1 1 1 1 = P 2 (1 − λ) − 2 (1 + µ) 2 (1 − ν); z 1 1 1 2 (1 + λ) − 2 (1 − µ) 2 (1 + ν) So the corresponding (2.5) becomes d2w 11 − λ2 1 − ν2 λ2 − µ2 + ν2 − 1 (4.11) + + + w = 0. dz2 4 z2 (z − 1)2 z(z − 1)
5. Gauss Hypergeometric Equation Let us re-derive some of the calculations done in the last section (z − ξ)(η − ζ) t = M(z) = (z − η)(ξ − ζ) be a M¨obiustransformation. We deduce from Exercise 4.6 that ξ η ζ 0 ∞ 1 P α β γ ; z = P α β γ ; t α0 β0 γ0 α0 β0 γ0 0 ∞ 1 = tα(1 − t)γ P α − α α + β + γ γ − γ ; t α0 − α α + β0 + γ γ0 − γ 0 ∞ 1 = tα(1 − t)γ P 0 α + β + γ 0 ; t . α0 − α α + β0 + γ γ0 − γ HYPERGEOMETRIC EQUATION AND MONODROMY GROUPS 11
It turns out that it is more efficient to set the following parametrization of the exponent pairs α + β + γ = a, α + β0 + γ = b and 1 − c = α0 − α. Note that c − a − b = 1 − γ − γ0 − β − β0 − 2γ = γ + γ0 − 2γ = γ0 − γ. Then the last Riemann-Scheme becomes 0 ∞ 1 P 0 a 0 ; t 1 − c b c − a − b satisfies the differential equation
d2w 1 − (1 − c) 1 − (c − a − b)dw ab + + + w = 0, dt2 t t − 1 dt t(t − 1) or
(5.1) t(1 − t)w00 + {c − (a + b + 1)t}w0 − abw = 0, which is called the Gauss hypergeometric equation, where we have assumed that none of the exponent pairs {0, 1 − c}, {0, c − a − b} and {a, b} is an integer. Note again that the sum of all exponents is unity.
The important associated Legendre’s differential equation
d2w dw n µ2 o (1 − z2) − 2z + ν(ν + 1) − w = 0 dz2 dz 1 − z2 is a Gauss hypergeometric differential equation, where ν, µ are arbitrary complex constants, with singularities at ±1, ∞. In terms of Riemann’s P notation, we have
−1 ∞ 1 µ µ w(z) = P 2 ν + 1 2 ; t . µ µ − 2 ν − 2 The equation is called Legendre’s differential equation if m = 0. Its solutions are called spherical harmonics. This equation is important because it is often encountered in solving boundary value problems of potential theory for spherical regions, and hence the name of spherical harmonics.
5.1. Relation with confluent hypergeometric equation. A transformation of Y (z) = y(z/b) in the (5.1) yields
d2Y dY (5.2) z(1 − z/b) + [c − (a + b + 1)z/b] − a Y = 0 dz2 dz which clearly has regular singular points at 0, b, ∞. Letting b → ∞ in the equation (5.2) results in the
d2y dy (5.3) z + (c − z) − a y = 0, dz2 dz which is called the confluent hypergeometric equation. 12 EDMUND Y.-M. CHIANG
5.2. Relation with Bessel Equations. We define the Bessel function of first kind of order ν to be the complex function represented by the power series +∞ +∞ X (−1)k( 1 z)ν+2k X (−1)k( 1 )ν+2k (5.4) J (z) = 2 = zν 2 z2k. ν Γ(ν + k + 1) k! Γ(ν + k + 1) k! k=0 k=0 Here ν is an arbitrary complex constant.
1 Let us set a = ν + 2 , c = 2ν + 1 and replace z by 2iz in the Kummer series. That is, ∞ 1 X (ν + )k Φ(ν + 1/2, 2ν + 1; 2iz) = 2 (2iz)k (2ν + 1)k k! k=0 1 ν + 2 = 1F1 2iz 2ν + 1 iz − z 2 = e 0F1 − ν + 1 2 ∞ k 2k iz X (−1) z = e 2k (ν + 1)k k! 2 k=0 ∞ k X (−1) Γ(ν + 1) z 2k = eiz Γ(ν + k + 1) k! 2 k=0 ∞ k z −ν z ν X (−1) z 2k = eizΓ(ν + 1) · 2 2 Γ(ν + k + 1) k! 2 k=0 z −ν = eizΓ(ν + 1) · J (z), 2 ν where we have applied Kummer’s second transformation a 2x − 2 (5.5) 1F1 4x = e 0F1 1 x 2a a + 2 in the third step above, and the identity (ν)k = Γ(ν + k)/Γ(ν) in the fourth step.
Since the confluent hypergeometric function Φ(ν + 1/2, 2ν + 1; 2iz) satisfies the confluent hypergeometric equation d2y dy (5.6) z + (c − z) − a y = 0 dz2 dz iz z −ν with a = ν + 1/2 and c = 2ν + 1. Substituting e Γ(ν + 1) 2 · Jν (z) into the (5.6) and simplfying lead to the equation d2y dy (5.7) z2 + x + (z2 − ν2)y = 0. dz2 dz This is called the Bessel equation with parameter ν. It is clearly seen that 1 z = 0 is a regular singular point and that Z = z has an irregular singular point at 1 Z = 0 = ∞ for the transformed equation with Y (z) = y(1/z). The above derivation shows that the Bessel function of the first kind with order ν is a special case of the confluent hypergeometric functions with specifically chosen parameters. HYPERGEOMETRIC EQUATION AND MONODROMY GROUPS 13
5.3. Relation with Hermite equation. To be completed.
6. Gauss Hypergeometric series Let us consider power series solutions of the Gauss Hypergeometric equation (6.1) z(1 − z)w00 + {c − (a + b + 1)z}w0 − abw = 0, P∞ k+α at the regulary singular point z = 0. Substitute the power series w = k=0 akz into the equation (6.1) yields the indicial equation α(α + c − 1) = 0, so that the coefficient ak satisfies the recurrence relation
(α + k)(α + c + k − 1)ak = (α + a + k − 1)(α + b + k − 1)ak−1, or
(α + 1 + k − 1)(α + c + k − 1)ak = (α + a + k − 1)(α + b + k − 1)ak−1. This gives (α + a + k − 1)(α + b + k − 1) a = a k (α + 1 + k − 1)(α + c + k − 1) k−1 (α + a + k − 1)(α + b + k − 1) (α + a + k − 2)(α + b + k − 2) = a (α + 1 + k − 1)(α + c + k − 1) (α + 1 + k − 2)(α + c + k − 2) k−2 ······ (α + a + k − 1)(α + b + k − 1) (α + a + 0)(α + b + 0) = ··· a (α + 1 + k − 1)(α + c + k − 1) (α + 1 + 0)(α + c + 0) 0
(α + a)k(α + b)k = a0, (α + 1)k(α + c)k where a0 is an arbitrarily chosen constant and we recall the Pochhammer notation that (α)0 = 1, (α)k = α(α + 1)(α + 2) ··· (α + k − 1), for each integer k. So we have obtained, assuming that 1 − c is not an integer, two linearly independent power series solutions ∞ ∞ X (a)k(b)k k 1−c X (a − c + 1)k(b − c + 1)k k w1(z) = z , w2(z) = z z . k!(c)k (2 − c)kk! k=0 k=0
Definition 6.1. A Gauss hypergeometric series is a power series given in the form ∞ ∞ a, b X k X (a)k(b)k k F (a, b; c; z) = 2F1 z = ckz = z c (c)k k! k=0 k=0 ab a(a + 1)b(b + 1) = 1 + z + z2+ c c(c + 1) 1! a(a + 1)(a + 2)b(b + 1)(b + 2) (6.2) + z3 + ··· c(c + 1)(c + 2) 2! 14 EDMUND Y.-M. CHIANG
We note that the Gauss series above terminates into a polynomial when either a or b is zero or a negative integer. The second series terminates when either a − c + 1 or b − c + 1 is zero or a negative integer. These polynomials are called Jacobi polynomials. its limiting cases include (orthogonal) Laguerre polynomials and Hermite polynomials. See [1] We note that the hypergeometric function includes many important functions that we have encountered. For examples, we have
−a a (1 − z) = 1F0 − z ; −
z − e = z 0F0 z ; −
1, 1 log(1 + z) = z 2F1 − z ; 2 2 − z sin z = z 0F1 3 − ; 2 4 2 − z cos z = z 0F1 1 − ; 2 4 1 1 −1 2 , 2 2 sin z = z 2F1 3 z 2
The series derives its name because the ratio between two consecutive terms, a that is k+1 is a rational function of k. ak Theorem 6.1 (Chu (1303)-Vandermonde (1772)). Let n be a positive integer. Then
−n, b (c − b)n (6.3) 2F1 1 = . c (c)n
Remark 6.2. According to Askey’s account [1], the Chinese mathematician S.-S. Chu [2] had already discovered the essential form of the identity four centuries before the French mathematician Vandermonde. Askey described Chu’s result was “absolutely incredible” as it was discovered even before adequate mathematical notation was available [1, p. 60].
Theorem 6.3. The series 2F1(a, b; c; z) (1) converges absolutely in |z| < 1; (2) converges absolutely if <(c − a − b) > 0 if |z| = 1; (3) diverges if <(c − a − b) ≤ −1.
(a)k(b)k k Proof. (1) This first part is easy. For let uk = z . Then (c)kk!
uk+1 (a + k)(b + k) = z → |z| uk (c + k)(1 + k) as k → ∞. Thus the hypergeometric series converges (uniformly) within the unit disc, and it diverges in |z| > 1 by the ratio test. HYPERGEOMETRIC EQUATION AND MONODROMY GROUPS 15
(2) Recall that z−1 Γ(z) = lim k!k /(z)k. k→∞ But then the k−th coefficient of the power series behaves asymptotically like a−1 b−1 (a)k(b)k Γ(c) k!k · k!k Γ(c) a+b−c−1 ≈ c−1 = k (c)kk! Γ(a)Γ(b) k!k · k! Γ(a)Γ(b) Γ(c) 1 = , Γ(a)Γ(b) kc+1−a−b for all n ≥ N say. But then ∞ ∞ ∞ X (a)k(b)k Γ(c) X 1 Γ(c) X 1 zk ≤ = c+1−a−b <(c+1−a−b) (c)kk! Γ(a)Γ(b) |k | Γ(a)Γ(b) k N N N P 1 which is known to be convergent when <(c − a − b) > 0 since k1+δ converges when <δ > 0. Alternatively, we observe that
ck (k + 1)(c + k) 1 + c − a − b 1 = = 1 + + O 2 , ck+1 (a + k)(b + k) k k that is, c 1 + <(c − a − b) 1 k ≥ 1 + + O 2 . ck+1 k k Hence Raabe’s test (lim k(| ck | − 1) > 1 means convergent, and k→∞ ck+1 ck limk→∞ k(| | − 1) < 1 implies divergent) implies that the Gauss series ck+1 converges on |z| = 1, z 6= 1. (3) In this case, since the general term of the series does not tend to zero (please verify), so the series diverges.
We are ready for Theorem 6.4 (Gauss 1812). Suppose <(c − a − b) > 0. Then ∞ X (a)k(b)k a, b Γ(c)Γ(c − a − b) = 2F1 1 = (c)k k! c Γ(c − a)Γ(c − b) k=0 holds.
Proof. Suppose that <(c − a − b) > 0. Then Theorem 6.3 implies that F (a, b; c; 1) converges. Abel’s theorem asserts that F (a, b; c; 1) = lim F (a, b; c; x). x→1− Let us write, ∞ ∞ X k X k F (a, b; c; z) = Akz ,F (a, b; c + 1; z) = Bkz . k=0 k=0 Then it is routine to check that
(a)k(b)k h (c − a)(c − b)i c(c − a − b)Ak − (c − a)(c − b)Bk = c − a − b − k!(c + 1)k−1 c + k 16 EDMUND Y.-M. CHIANG and (a)k(b)k h (a + k)(b + k)i c(kAk − (k + 1)Ak+1) = k − k!(c + 1)k−1 c + k hold and that the right-hand sides of the above identities are equal. Thus, we deduce c(c − a − b)Ak = (c − a)(c − b)Bk + ckAk − c(k + 1)Ak+1. Hence n n X X c(c − a − b) Ak = (c − a)(c − b) Bk − c(n + 1)An+1. k=0 k=0 Now let n → ∞ to establish (c − a)(c − b) (6.4) F (a, b; c; 1) = F (a, b; c + 1; 1) c(c − a − b) since Γ(c) 1 (n + 1)A ≈ → 0 n+1 Γ(a)Γ(b) (n + 1)c−a−b and <(c − b − a) > 0. Iterating (6.4) k − 1 times yields (c − a) (c − b) (6.5) F (a, b; c; 1) = k k F (a, b; c + k; 1). (c)k(c − a − b)k Recall that Γ(a + k) = (a + k)Γ(a + k) = (a + k)(a + k − 1)Γ(a + k − 1) = (a + k)(a + k − 1) ··· (a + k − k) ··· Γ(a + k − k) = (a + k)(a + k − 1) ··· (a + 1)(a)Γ(a)
= (a)kΓ(a) or Γ(a + k) = (a)kΓ(a). Applying this to (6.5) yields Γ(c − a)Γ(c − b) Γ(c + k − a)Γ(c + k − b) F (a, b; c; 1) = F (a, b; c + k; 1). Γ(c)Γ(c − a − b) Γ(c + k)Γ(c + k − a − b) We deduce from Stirling’s formula √ −z z− 1 1 Γ(z) = 2π e z 2 1 + o( ) z that Γ(c + k − a)Γ(c + k − b) → 1, k → ∞. Γ(c + k)Γ(c + k − a − b) It remains to prove the limit Γ(c − a)Γ(c − b) (6.6) lim F (a, b; c; 1) = lim F (a, b; c + k; 1) = 1 k→∞ Γ(c)Γ(c − a − b) k→∞ under the assumption that <(c − b − a) > 0.
Exercise 6.5. Complete the proof for (6.6). See [3]. HYPERGEOMETRIC EQUATION AND MONODROMY GROUPS 17
6.1. Euler’s transformation. The above constructed a solution in |z| < 1 (and also |z| = 1 with <(c − a − b) > 0). The 2F1 is more ”global” than that. Theorem 6.6 (Euler). Let
7. Symmetry: Kummer’s 24 Solutions We now consider internal symmetry of the hypergeometric equation/function. Let us follow Riemann’s approach. We shall write down the pairs of power series of the fundamential sets of solutions with respect to the three regular singular points 0 0 0 {0, 1, ∞}. Riemann uses the notation {P (α),P (α )}, {P (β),P (β )}, {P (γ),P (γ )}, while we shall use {w1(z), w2(z)}, {w3(z), w4(z)}, {w5(z), w6(z)} instead. Recall that the standard hypergeometric equation (6.1) has two power series solutions whose characteristic exponents 0 and 1 − c are recorded in the Riemann- Scheme 0 ∞ 1 P 0 a 0 ; z 1 − c b c − a − b We have the following observation 0 ∞ 1 0 ∞ 1 P 0 a 0 ; z = z1−cP c − 1 a − c + 1 0 ; z 1 − c b c − a − b 0 b − c + 1 c − a − b 0 ∞ 1 = z1−cP 1 − c0 a0 0 ; z , 0 b0 c0 − a0 − b0 where a0 = a−c+1, b0 = b−c+1 and c0 = 2−c. The above transformation indicates that one of the original power series solution of equation (6.1) with characteristic exponent 1−c can be written in terms of the characteristic exponent 0 power series solution F (a0, b0; c0; z) represented by the above last Riemann-Scheme. Thus we have two solutions ∞ a, b X (a)k(b)k k w1(z) = 2F1 z = z , c k!(c)k k=0 ∞ 1−c a − c + 1, b − c + 1 1−c X (a − c + 1)k(b − c + 1)k k w2(z) = z 2F1 z = z z , 2 − c (2 − c)kk! k=0 which are precisely two linearly independent solutions found in Section 6. We recall that we have assumed that 1 − c is not an integer. Otherwise, there will be a logarithm term in the solution (see [9]). This gives two series solutions around the regular singular point z = 0. 18 EDMUND Y.-M. CHIANG
We now consider the regular singular point z = 1. We now consider the trans- formation z = 1 − t. This will transform the hypergeometric equation to the form d2w(t) dw(t) t(1 − t) + [(1 + a + b − c) − (a + b + 1)t] − abw(t) = 0, dt2 dt where w(t) = y(z) where t = 1 − z. That is, we have shown that the relation 0 ∞ 1 0 ∞ 1 (7.1) P 0 a 0 ; z = P 0 a 0 ; 1 − z 1 − c b c − a − b c − a − b b 1 − c holds. Observe that 0 ∞ 1 0 ∞ 1 P 0 a 0 ; z = P 0 a 0 ; 1 − z 1 − c b c − a − b c − a − b b 1 − c 0 ∞ 1 = P 0 a0 0 ; 1 − z 1 − c0 b0 c0 − a0 − b0 where a0 = a, b0 = b and c0 = 1 + a + b − c. We obtain
a, b w3(z) = 2F1 1 − z . 1 + a + b − c
Applying the transformation from w1 to w2 above to w3 and identifying t = 1 − z and the parameters a0, b0 and c0 yields the fourth solution
0 0 0 0 1−c0 a − c + 1, b − c + 1 w4(z) = t 2F1 t 2 − c0
c−a−b c − b, c − a = (1 − z) 2F1 1 − z . 1 + c − a − b
Similarly, one can obtain two solutions around the regular singular point ∞:
Exercise 7.1.
−a a, a − c + 1 1 w5(z) = (−z) 2F1 , a − b + 1 z where c − a − b is not an integer, and
−b b, b − c + 1 1 w6(z) = (−z) 2F1 , b − a + 1 z where a−b is not an integer. The negative signs are introduced for convinence sake (see later).
Hence we have obtained three pairs of six solutions w1(z), ··· , w6(z) in total. Let us now take w1(z), that is the standard solution, as an example on how to generate more solutions (where we have used underline on the appropriate exponent HYPERGEOMETRIC EQUATION AND MONODROMY GROUPS 19 to indicate on which solution that we are considering):
a, b w1(z) = 2F1 z c 0 ∞ 1 = P 0 a 0 ; z 1 − c b c − a − b 0 ∞ 1 = (1 − z)c−a−bP 0 c − b a + b − c ; z . 1 − c c − a 0
Thus
a, b c−a−b c − a, c − b 2F1 z = C(1 − z) 2F1 z , c c for some constant C. If we choose the branch of (1 − z)c−a−b such that it equals 1 when z = 0, then C = 1 since both sides of the equation are equal to 1. Thus, we have
a, b c−a−b c − a, c − b w1(z) = 2F1 z = (1 − z) 2F1 z , c c where | arg(1 − z)| < π. On the other hand, we have
0 ∞ 1 0 ∞ 1 z P 0 a 0 ; z = P 0 0 a ; z − 1 1 − c b c − a − b 1 − c c − a − b b 0 ∞ 1 z a z = 1 − P 0 a 0 ; z − 1 z − 1 1 − c c − b b − a 0 ∞ 1 z = (1 − z)−aP 0 a 0 ; . z − 1 1 − c c − b b − a
Thus, we have
a, b −a a, c − b z 2F1 z = (1 − z) 2F1 , c c z − 1 provided that | arg(1 − z)| < π. Interchanging the a and b in the above analysis yields another format:
a, b −b b, c − a z 2F1 z = (1 − z) 2F1 , c c z − 1 20 EDMUND Y.-M. CHIANG provided that | arg(1 − z)| < π. Putting these cases together, we have
a, b w1(z) = 2F1 z c
c−a−b c − a, c − b = (1 − z) 2F1 z c
−a a, c − b z = (1 − z) 2F1 c z − 1
−b b, c − a z = (1 − z) 2F1 . c z − 1
Similar formulae exist for the remaining w2(z), ··· , w6(z), where each solution has three additional variartions, thus forming a total of twenty four solutions, which are called Kummer’s 24 solutions (Crelle, issue XV (1836)). We list the remaining sets of these formulae below.
Exercise 7.2. 1−c w2(z) = z 2F1(a + 1 − c, b + 1 − c; 2 − c; z) 1−c c−a−b = z (1 − z) 2F1(1 − a, 1 − b; 2 − c; z) 1−c c−a−1 = z (1 − z) 2F1(a + 1 − c, 1 − b; 2 − c; z/(z − 1)) 1−c c−b−1 = z (1 − z) 2F1(b + 1 − c, 1 − a; 2 − c; z/(z − 1)), Exercise 7.3.
w3(z) =2F1(a, b; a + b − c + 1; 1 − z) 1−c = z 2F1(a − c + 1, b − c + 1; a + b − c + 1; 1 − z) −a = z 2F1(a, a − c + 1; a + b − c + 1; (z − 1)/z) −b = z 2F1(b, b − c + 1; a + b − c + 1; (z − 1)/z), Exercise 7.4. c−a−b w4(z) =(1 − z) 2F1(c − a, c − b; c − a − b + 1; 1 − z) 1−c c−a−b = z (1 − z) 2F1(1 − a, 1 − b; c − a − b + 1; 1 − z) −a c−a−b = z (1 − z) 2F1(1 − a, c − a; c − a − b + 1; (z − 1)/z) −b c−a−b = z (1 − z) 2F1(1 − b, c − b; c − a − b + 1; (z − 1)/z),
Exercise 7.5. −a w5(z) = (−z) 2F1(a, a − c + 1; a − b + 1; 1/z) −a c−a−b = (−z) (1 − 1/z) 2F1(1 − b, c − b; a + 1 − b; 1/z) −a −a = (−z) (1 − 1/z) 2F1(a, c − b; a + 1 − b; 1/(1 − z)) −a c−a−1 = (−z) (1 − 1/z) 2F1(a + c − 1, 1 − b; a + 1 − b; 1/(1 − z))
and HYPERGEOMETRIC EQUATION AND MONODROMY GROUPS 21
Exercise 7.6. −b w6(z) = (−z) 2F1(b, b − c + 1; b − a + 1; 1/z) −b c−a−b = (−z) (1 − 1/z) 2F1(1 − a, c − a; b + 1 − a; 1/z) −b −b = (−z) (1 − 1/z) 2F1(b, c − a; b + 1 − a; 1/(1 − z)) −b c−b−1 = (−z) (1 − 1/z) 2F1(b − c + 1, 1 − a; b + 1 − a; 1/(1 − z)).
One can also interpret the above twenty four solutions by considering that there are six M¨obiustransformations that permute the three regular singular points {0, 1, ∞}: z 1 1 1 z, 1 − z, , , , 1 − . z − 1 z z − 1 z This list gives all the possible permutations of the regular singularities. Notice the function ρ σ ∗ ∗ ∗ ∗ z (1 − z) 2F1(a , b ; c ; z ), where ρ and σ can be chosen so that exactly one of the exponents at z = 0 and z = 1 becomes zero. But we have two choices of exponents at either of the points (0 or 1) so that there are a total of 6 × 2 × 2 = 24 combinations. We refer to Bateman’s project Vol. I [4] for a complete list of these twenty four solutions. 22 EDMUND Y.-M. CHIANG
8. Riemann’s monodromy group Recall the notation Riemann used 0 0 0 P (α),P (α ); P (β),P (β ); P (γ),P (γ ) to denote two branches of the regular singularities at 0, 1, ∞ respectively. Then the following figure shows that three appropriate solutions from the Kummer list of 24 solutions would satisfy the following equations (α) (β) (β0) (γ) (γ0) (8.1) P = αβP + αβ0 P = αγ P + αγ0 P , and (α0) 0 (β) 0 (β0) 0 (γ) 0 (γ0) (8.2) P = αβP + αβ0 P = αγ P + αγ0 P in the region as shown below:
The problem that Riemann tackled was that he showed that the ratios 0 αβ αβ αγ αγ0 0 , 0 , 0 , 0 αβ αβ0 αγ αγ0 are invariants with respect to the constant multiplier for each branch chosen. We now take a simple closed curve with positive orientation enclosing the points x = 0 and x = 1 beginning and ending at a point in the upper half-plane. On the one hand, this is equivalent to a negatively oriented curve around the infinity ∞. One the other hand, the analytic continuation principle implies that the combined effect of the simple curve surrounding x = 0 and x = 1 is effected by the sequence (α) α2πi (α) 2πiα 2πiγ (γ) 2πiγ0 (γ0) P 7→ e P 7→ e αγ e P + αγ0 e P Combining with the simultaneous negatively oriented curve about ∞, we deduce −2πiβ (β) −2πiβ0 (β0) 2πiα 2πiγ (γ) 2πiγ0 (γ0) (8.3) e αβP + e αβ0 P = e αγ e P + αγ0 e P
0 −2πiβ (β) 0 −2πiβ0 (β0) 2πiα0 0 2πiγ (γ) 0 2πiγ0 (γ0) (8.4) αβe P + αβ0 e P = e αγ e P + αγ0 e P Substitute (8.1) (the second equality) into (8.3) yields −2πiβ (β) −2πiβ0 (γ) (γ0) (β) αβe P + e αγ P + αγ0 P − γβP (8.5) 2πiα 2πiγ (γ) 2πiγ0 (γ0) = e αγ e P + αγ0 e P HYPERGEOMETRIC EQUATION AND MONODROMY GROUPS 23
Collecting like terms yields (β) −2πiβ −2πiβ0 (α+γ)2πi −2πiβ0 (γ) αβP (e − e ) = αγ (e − e )P (8.6) (α+γ0)2πi −2πiβ0 (γ0) + αγ0 (e − e )P
0 Multiplying eπ(β+β ) on both sides of the above equation and simplifying yields 0 (β) iπ(α+β+γ) 0 (γ) αβ sin π(β − β)P = αγ e sin(α + β + γ)πP (8.7) iπ(α+β+γ0) 0 0 (γ0) + αγ0 e sin(α + β + γ )πP .
Repeating the above steps for the equations (8.2) and (8.4) yield a similar equa- tion 0 0 (β) 0 iπ(α0+β+γ) 0 0 (γ) αβ sin π(β − β)P = αγ e sin(α + β + γ)πP (8.8) 0 iπ(α0+β+γ0) 0 0 0 (γ0) + αγ0 e sin(α + β + γ )πP . An application of α + α0 + β + β0 + γ + γ0 = 1 yields iπα 0 iπα 0 0 αβ αγ e sin π(α + β + γ) αγ0 e sin π(α + β + γ ) 0 = 0 iπα0 0 0 = 0 iπα0 0 0 0 αβ αγ e sin π(α + β + γ) αγ0 e sin π(α + β + γ ) 0 We may repeat a similar procedure to obtain two equations for P (β ) as follow 0 0 (β0) iπ(α+β0+γ) (γ) αβ sin π(β − β )P = αγ e sin(α + β + γ)πP (8.9) iπ(α+β0+γ0) 0 (γ0) + αγ0 e sin(α + β + γ )πP . and 0 0 (β0) 0 iπ(α0+β0+γ) 0 (γ) αβ0 sin π(β − β )P = αγ e sin(α + β + γ)πP (8.10) 0 iπ(α0+β0+γ0) 0 0 (γ0) + αγ0 e sin(α + β + γ )πP from which we deduce iπα iπα 0 αβ0 αγ e sin π(α + β + γ) αγ0 e sin π(α + β + γ ) 0 = 0 iπα0 0 = 0 iπα0 0 0 . αβ0 αγ e sin π(α + β + γ) αγ0 e sin π(α + β + γ ) 8.1. Evaluation of the invariants. We draw a cut on the whole of R and assume the followings: 0 (1) P (α) → 1, P (α ) ∼ x1−c, as x → 0. 0 (2) P (β) ∼ (1 − x)−a, P (β ) ∼ (1 − x)−b, x → ∞ 0 (3) P (γ) → 1, P (γ ) ∼ (1 − x)c−a−b, as x → 1. (4) 0 < arg(x) < π, −π < arg(1 − x) ≤ 0. Recall the following evaluation of Gauss: Γ(c)Γ(c − a − b) F (a, b; c; 1) = := φ(x). 2 1 Γ(c − a)Γ(c − b)
Consider (α) (γ) (γ0) (α0) 0 (γ) 0 (γ0) (8.11) P = αγ P + αγ0 P ,P = αγ P + αγ0 P again. We assume <(c − a − b) > 0 and applying the first of Kummer’s 24 solutions the (α) P = w1 = 2F1(a, b; c; x), 24 EDMUND Y.-M. CHIANG
(α0) 1−c P = w2 = x 2F1(a − c + 1, b − c + 1; 2 − c; x), and (γ) P = w3(x) = 2F1(a, b; a + b − c + 1; 1 − x), (γ0) c−a−b P = w4(x) = (1 − x) 2F1(c − a, c − b; c − a − b + 1; 1 − x),
We now let x → 1 from the upper half-plane say. , Then the first equation above has 0 P (α) → φ(a, b, c),P (α ) → φ((a − c + 1, b − c + 1, 2 − c) (8.12) 0 P (γ) → 1,P (γ ) → 0. This establishes 0 αγ = φ(a, b, c), αγ = φ(a − c + 1, b − c + 1, 2 − c). We can then remove the restriction <(c − a − b) > 0 by analytic continuation of the parameters a, b, c. If we assume <(c − a − b) < 0, then choosing the second of w1 and w2 from Kummer’s list as (α) c−b−a P = (1 − x) 2F1(c − a, c − b; c; x),
(α0) 1−c c−a−b P = x (1 − x) 2F1(1 − a, 1 − b; 2 − c; x) and also the second of (γ) (1−c) P = x 2F1(a − c + 1, b − c + 1; a + b − c + 1; 1 − x), and (γ0) 1−c c−a−b P = x (1 − x) 2F1(1 − a, 1 − b; c − a − b + 1; 1 − x). Notice that as x → 1, we have P (α) ∼ (1 − x)c−b−aφ(c − a, c − b, c),
0 P (α ) ∼ (1 − x)c−a−bφ(1 − a, 1 − b, 2 − c),
0 P (γ) ∼ 1,P (γ ) ∼ (1 − x)c−a−b. Hence we further deduce 0 αγ0 = φ(c − a, c − b, c), αγ0 = φ(1 − a, 1 − b, 2 − c). We then remove the restriction <(c − a − b) < 0 by analytic continuation. 0 Similarly, we can compute the αβ, α , etc. We record
αγ αγ0 φ(a, b, c) φ(c − a, c − b, c) 0 0 = αγ αγ0 φ(a − c + 1, b − c + 1, 2 − c) φ(1 − a, 1 − b, 2 − c) and αβ αβ0 φ(a, c − b, c) φ(b, c − a, c) 0 0 = iπ(1−c) iπ(1−c) αβ αβ0 e φ(a − c + 1, 1 − b, 2 − c) e φ(b − c + 1, 1 − a, 2 − c) HYPERGEOMETRIC EQUATION AND MONODROMY GROUPS 25
References 1. R. Askey, “Orthogonal Polynomials and Special Functions”, SIAM, Philadelphia, 1975. 2. Chu, Shih-Chieh, “Ssu Yuan Y¨uChien (Precious Mirror of the Four Elements)”, 1303 (Clas- sical Chinese) 3. E. T. Copson,, “An Introduction to the Theory of Complex Functions of A Complex Variable”, Clarendon Press, Oxford, 1935. 4. Erd´elyi,A. “Higher Transendental Functions”, Vol. I, II, III, A. Erd´elyi,ed., McGraw-Hill. Reprinted by Krieger Publishing Co., Malabar, FL 1981. 5. E. E. Kummer, “Uber¨ die hypergeometrische Reihe, J. f¨urdie reine und angew Math, XV (1836), 39–?? 6. E. G. C. Poole, , “Introduction to the Theory of Linear Differential Equations”, Clarendon Press, Oxford, 1936. 7. Riemann, B., “Beitr¨agezur Theorie der durch Gauss’sche Reihe F (α, β, γ, x) darstellbaren Functionen”, K. Ges. Wiss. G¨ottingen, 7 (1857) 3-32 (in Werke 67-83) You can find the paper from D. R. Wilkins’s edited “Gesammelte Mathematische Werke”. 8. G. N. Watson, A Treatise on the Theory of Bessel Functions (2nd ed.), Cambridge University Press, Cambridge 1944 (reprinted in 1995). 9. Whittaker, E. T. and Watson, G. N., “A Course of Modern Analysis” (4th ed.), Cambridge University Press, 1927, (reprinted in 1992).
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