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Hypergeometric Equation and Monodromy Groups

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Hypergeometric Equation and Monodromy Groups 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 1 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 fαk; αkg and fα1; α1g denote, respectively, the exponents of the corre- sponding indicial equations at a1; a2; ··· an and 1. Then we have n 0 X 0 (α1 + α1) + (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) = + '(z) z − ak k=1 where '(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 1 is also a regular singular point of p(z), it follows from Proposition 3.2 in part I and which states 1 is regular singular iff zp and z2q are analytic at 1. That is, p(z) = O(1=z). So '(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 + + (z); (z − ak) z − ak k=1 where (z) is entire function. But z2q(z) is analytic at 1, so q(z) = O(1=z2) as z ! 1. This shows that (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 1 1 X j 2 X j (z − ak)p(z) = pj · (z − ak) ; (z − ak) q(z) = qj · (z − ak) j=0 j=0 and 1 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 = 1. Since the 1 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!1 8 Qn 9 n j=1 Ak(z − aj) <X j6=k = = 2 − lim z Qn z!1 (z − aj) :k=1 j=1 ; Pn A zn + O(zn−1) = 2 − lim k=1 k z!1 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!1 n 2 X Bk Ck = lim z 2 + z!1 (z − ak) z − ak k=1 n n X X Ck = Bk + lim z z!1 1 − ak=z k=1 k=1 n n X X ak 1 = Bk + lim z Ck 1 + + O z!1 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 α1; α1 be the exponents of the indicial equation at 1. 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 α1 + α1 = 1 − k=1 Ak. Hence we deduce n X 0 0 (αk + αk) + (α1 + α1) = n − 1; k=1 as required. Exercise 1.2 (Klein). We note that if 1 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 1 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 \1" is an analytic point for both p(z) and q(z) (since we have assumed that none of the three singular points fξ; η; ζg is at 1), 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 P1 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 = 1 A p2 p2 is an ordinary point, so P (z) = zp(z) = z z + z2 + ··· = A + z + ··· as z ! 1 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 = 1 is an ordinary point, so 2 2 q4 q5 q4 q4 Q(z) = z q(z) = z z4 + z5 + ··· = z2 + z3 + ···! 0 as z ! 1 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 fα; α g and fα1; α1g. Derive the exact form of the differential equation 0 0 0 0 and show that α + α + α1 + α1 = 0 and α · α = α1 · α1 hold. If we now choose η = 1 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.
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