ELLIPTIC SELBERG INTEGRALS AND CONFORMAL BLOCKS. 0 j<k 1. Γ(1 + γ + jγ)γ(α + jγ)γ(β + jγ) Γ(1 + γ)γ(α + β + (p + j 1)γ).

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1 ELLIPTIC SELBERG INTEGRALS AND CONFORMAL BLOCKS G. FELDER, L. STEVENS, AND A. VARCHENKO, arxiv:math.qa/ v 3 Oct 2002 Dpartmnt Mathmatik, ETH-Zntrum, 8092 Zürich, Switzrland, fldr@math.thz.ch Dpartmnt of Mathmatics, Univrsity of North Carolina at Chapl Hill, Chapl Hill, NC , USA, lstvns@mail.unc.du, anv@mail.unc.du Octobr, 2002 Abstract. W prsnt an lliptic vrsion of Slbrg s intgral formula.. Introduction Th Slbrg intgral is th intgral B p α, β, γ = t α j t j β p j= 0 j<k t j t k 2γ dt...dt p, whr p = {t R p 0 t p t }. Th Slbrg intgral is a gnralization of th bta function. It can b calculatd xplicitly, B p α, β, γ = p! p Γ + γ + jγγα + jγγβ + jγ Γ + γγα + β + p + j γ. Th Slbrg intgral has many applications, s [A, A2, As, D, DF, DF2, M, S]. In this papr, w prsnt lliptic vrsions of th Slbrg intgral. 2. Conformal Blocks on th Torus Lt C b such that Im > 0. Lt κ and p b non-ngativ intgrs satisfying κ 2p + 2. Th KZB-hat quation is th partial diffrntial quation κ u λ, = 2 u λ 2λ, + pp + ρ λ, uλ,. Supportd in part by NSF grant DMS

2 2 G. FELDER, L. STEVENS, AND A. VARCHENKO Hr, th prim dnots th drivativ with rspct to th first argumnt, and ρ is dfind in trms of th first Jacobi thta function, ϑ λ, = 2q 8 sin πλ q j λ q j λ q j, ρλ, = ϑ λ, ϑ λ,, j= whr q =. Holomorphic solutions of th KZB-hat quation with th proprtis, i uλ + 2, = uλ,, ii uλ + 2, = κλ+ uλ,, iii u λ, = p+ uλ,, iv uλ, = Oλ m n p+ as λ m + n for any m, n Z ar calld conformal blocks or lliptic hyprgomtric functions associatd with th family of lliptic curvs C/Z + Z with th markd point z = 0 and th irrducibl sl 2 rprsntation of dimnsion 2p +. It is known that th spac of conformal blocks has dimnsion κ 2p. 3. Intgral Rprsntations of Conformal Blocks Introduc spcial functions Considr th thta functions σ λ t, = ϑ λ t, ϑ 0, ϑ λ, ϑ t,, Et, = ϑ t, ϑ 0,. θ κ,n λ, = j Z κj+ n 2κ 2 +κj+ n 2κ λ, n Z/2κZ. Thy form a basis of th spac of thta functions of lvl κ. Thy satisfy th quations θ κ,n λ +, = n θ κ,n λ,, θ κ,n λ +, = πiκλ+ 2 θ κ,n+κ λ, and hav th modular proprtis n2 πi θ κ,n λ, + = 2κ θκ,n λ,, λ θ κ,n, = i λ 2 2κ πiκ 2 2κ m=0 mn πi κ θκ,m λ,, whr arg i < π/2. Lt θ s κ,n dnot th symmtrization of θ κ,n with rspct to λ, θ s κ,n λ, = θ κ,nλ, + θ κ,n λ,. Dfin u κ,n by u κ,n λ, = u p,κ,n λ, = J p,κ,n λ, + p+ J p,κ,n λ,,

3 whr J p,κ,n λ, = ELLIPTIC SELBERG INTEGRALS AND CONFORMAL BLOCKS 3 p j= Et j, 2p κ j<k p Et j t k, 2 κ σ λ t j, θ κ,n λ + 2 κ j= p t j, dt...dt p. Th branch of th logarithm is chosn in such a way that arg Et, 0 as t 0 +, and th intgral is undrstood as th analytic continuation from th rgion whr all of th xponnts in th intgrand hav positiv ral parts. Thorm 3.. [FV] For all n, th intgrals u κ,n λ, ar solutions of th KZB-hat quation having th proprtis i-iv. Thorm 3.2. [FSV2] W hav a u κ,n = u κ,n+2κ and u κ,n = pn/κ u κ, n. b Th st {u κ,n λ, n = p+,..., κ p } is a basis for th spac of conformal blocks. Th intgrals u κ,n ar idntically zro for all othr valus of n in th intrval from 0 to κ. 4. Transformations acting on th spac of conformal blocks Introduc four transformations A, B, T, and S dfind by Auλ, = uλ +,, Tuλ, = uλ, +, whr w fix arg 0, π. j= Buλ, = πiκλ+ 2 uλ +,, λ2 πiκ Suλ, = 2 2 pp+ κ u λ, Proposition 4.. If uλ, is a solution of th KZB-hat quation, thn Auλ,, Buλ,, Tuλ,, and Suλ, ar solutions too. Morovr, th transformations A, B, T and S prsrv th proprtis i-iv. Th proofs that T and S prsrv th spac of conformal blocks ar givn in [EK]. Th proofs that A and B also prsrv this spac ar straightforward and follow from th quations ϑ λ +, = ϑ λ,, ϑ λ +, = πi2λ+ ϑ λ,. Lmma 4.2. Rstrictd to th spac of conformal blocks, th transformations A, B, T, and S satisfy th rlations A 2 = I, B 2 = I, S 2 = p pp+ πi i κ I, ST 3 = p pp+ πi i κ I, SAS = B, AB = κ BA, TB = i κ BAT, whr I dnots th idntity transformation.,

4 4 G. FELDER, L. STEVENS, AND A. VARCHENKO Lmma 4.3. W hav Au κ,n λ, = n u κ,n λ,, Bu κ,n λ, = pn κ uκ,κ n λ,. Lt t m,n and s m,n b th matrics of th transformations T and S, rspctivly, with rspct to th basis {u κ,n λ, p + n κ p }, namly, Tu κ,n = κ p m=p+ t m,nu κ,m, Su κ,n = κ p m=p+ s m,nu κ,m. In Thorm 4.4, w giv formulas for th matrics of T and S in trms of Macdonald polynomials of typ A. Th Macdonald polynomials [Ma] of typ A ar x-vn polynomials in trms of ǫ mx, whr m Z. Thy dpnd on two paramtrs k and n, whr k and n ar non-ngativ intgrs. Thy ar dfind by th conditions: P n k x = ǫ nx + ǫ nx + lowr ordr trms, xcpt for P k 0 x =, 2 P m k, P n k = 0 for m n, whr f, g = k fg 2 Const Trm ǫ 2j+x ǫ 2j x. Thorm 4.4. [FSV2] Lt ǫ = πi/κ. For p + m, n κ p, w hav s m,n = t m,n = ǫ n2 πi 4 pp+ pn m ǫ 2 ǫ m ǫ m 2κ whr δ mn = for m = n and 0 othrwis. 2 δmn, ǫ n+j ǫ n j j= 5. Intgral idntitis P p+ n p m, To formulat our main rsult, w nd functions η, φ, φ 2, and φ 3. Th Ddkind η- function is th function η = q /24 j= qj. W hav ϑ 0, = 2πη3. Considr th functions [W] φ = η2 = q 48 + q j 2, φ 2 = η 2 η2 η = q 48 q j 2, W hav η 2 j= φ 3 = 2 η2 η = 2q 24 + q j. φ = φ, φ 2 = φ 3, φ 3 = φ 2, φ + = πi 24 φ2, φ 2 + = πi 24 φ, φ 3 + = πi 2 φ3. j= j=

5 ELLIPTIC SELBERG INTEGRALS AND CONFORMAL BLOCKS 5 Lt c κ,n = c p,κ,n = 2π pp+ p3p p+ πi πi n + κ 2κ 2 Bp κ, 2p κ, κ j= n+j κ. Hr, B p α, β, γ is th Slbrg intgral. Thorm 5.. W hav tn sris of idntitis, u 2p+2,p+ λ, = c 2p+2,p+ ϑ λ, p+, u 2p+3,p+ λ, = c 2p+3,p+ η 3p+ 2p+3 ϑ p+ λ, θ,0 λ,, u 2p+3,p+2 λ, = c 2p+3,p+2 η 3p+ 2p+3 ϑ p+ λ, θ, λ,, u 2p+4,p+2 λ, = 2 2p+ 2p+4 c2p+4,p+2 φ3 η 4p+ 2p+4 ϑ p+ λ, θ2, s λ,, 6 u 2p+4,p+ λ, + p+ pp+ 2p+4 u2p+4,p+3 λ, = c 2p+4,p+ φ2 η 4p+ 2p+4 ϑ p+ λ, θ 2,0 λ, θ 2,2 λ,, 7 u 2p+4,p+ λ, + p pp+ 2p+4 u2p+4,p+3 λ, = c 2p+4,p+ φ η 4p+ 2p+4 ϑ p+ λ, θ 2,0 λ, + θ 2,2 λ,, 8 u 2p+6,p+ λ, + p+ pp+ 2p+6 u2p+6,p+5 λ, = 2 3p+ 2p+6 c2p+6,p+ φ 3 η 6p+ 2p+6 ϑ p+ λ, θ 4,0 λ, θ 4,4 λ,, 9 u 2p+6,p+2 λ, + p pp+2 2p+6 u2p+6,p+4 λ, = c 2p+6,p+2 φ 2 η 6p+ 2p+6 ϑ p+ λ, θ4, s λ, + θs 4,3 λ,, 0 u 2p+6,p+2 λ, + p+ pp+2 2p+6 u2p+6,p+4 λ, = c 2p+6,p+2 φ η 6p+ 2p+6 ϑ p+ λ, θ4, s λ, θs 4,3 λ,, u 2p+8,p+2 λ, + p+ pp+2 2p+8 u2p+8,p+6 λ, = c 2p+8,p+2 η 8p+ 2p+8 ϑ p+ λ, θ6, s λ, θs 6,5 λ,. Th intgrals in Thorm 5. ar appropriatly calld th lliptic Slbrg intgrals. Th idntity 2 appars in [FSV] and in [FV] for p =.

6 6 G. FELDER, L. STEVENS, AND A. VARCHENKO 6. Diffrntial quations In Lmmas , lt dnot th drivativ with rspct to λ, lt dnot th drivativ with rspct to, and lt vλ, = ϑ p+ λ, c j θ,j s λ,. Lmma 6.. Th function vλ, is a solution of th KZB-hat quation if and only if th diffrntial quation 2 κ p + d d c j θ,j s ϑ = 2p + 2 κ ϑ c j θ s,j 2 c j θs,j + ϑ c j θ πi ϑ,j s holds. Th proof of Lmma 6. uss th idntitis 3 4 2p + 2 λ, = ϑ p+ λ, + pp + ρ λ, ϑ p+ λ,, ϑ p+ κ θ s κ,m λ, = θs κ,m λ,. Proof of Lmma 6.. Applying th diffrntial oprator κ / to vλ, givs [ κ p + ϑ ϑ p c j θ s,j + ϑp+ d d c j θ,j s + ϑ p+ c j θs,j ]. Applying th diffrntial oprator 2 / λ 2 + pp + ρ λ, to vλ, givs ϑ p+ c j θ s,j + 2p + ϑ ϑ p c j θ s,j + ϑ p+ c j θ,j s + pp + ρ ϑ p+ c j θ,j s. Applying 3 and 4, w obtain th rsult.

7 ELLIPTIC SELBERG INTEGRALS AND CONFORMAL BLOCKS 7 Lmma 6.2. If vλ, is a solution of th KZB-hat quation, thn th functions c j satisfy th diffrntial quation κ d p + d c j θ,j 0, = d 2p + 2 κ d ln ϑ 0, c j θ,j 0, + 2κ 2p 3 c j d d θ,j0,. Proof. For any fixd λ, quation 2 givs a diffrntial quation for th functions c j. W tak th limit of that quation as λ 0. In th ratio ϑ /ϑ, both th numrator and dnominator tnd to zro, so th limit of this trm as λ 0 is qual to th limit of th ratio of th drivativs of th numrator and th dnominator. Th limit of th ratio c j θ,j s /ϑ is calculatd in th sam way, sinc ach θ,j s is a symmtric function and thrfor θ,j s 0, = 0. Thn th rsult follows from 4. Lmma 6.3. If vλ, is a solution of th KZB-hat quation, thn th functions c j satisfy th diffrntial quation κ d p + d c j θ, j 0, = d 2p + 2 κ d ln ϑ 0, c j θ, j 0, + 2κ 2p 3 c j d d θ, j0,. Proof. W tak th limit of 2 as λ. This limit is calculatd in trms of th limit λ 0 using th formulas z θ z, z=λ+ = λ πi θ λ, θ λ,, z θ λ +, z z= = λ πi πiθ λ, + θ λ, θ λ,, z θs κ,m z, z=λ+ = πiκλ πiκ 2 πiκθ s κ,κ m λ, + θκ,κ m s λ,, z θs κ,m λ +, z z= = πiκλ πiκ 2 πi κ 2 θsκ,κ m λ, θsκ,κ m λ, + θ sκ,κ m λ,.

8 8 G. FELDER, L. STEVENS, AND A. VARCHENKO It is straightforward to calculat th limit of th lft hand sid. Using th abov formulas, th limit as λ of th right hand sid is qual to th limit as λ 0 of th xprssion πi2p+2 κλ+ 2 ϑ 2p + 2 κ c j θ, j s ϑ 2 c j θs, j + ϑ c j θ, j. s πi ϑ This limit is calculatd using L Hôpital s rul. 7. Idntitis for thta functions In th nxt sction, w giv th proofs of th intgral idntitis in Thorm 5.. W will us th following rsults. Lmma 7.. W hav θ s 2,λ = ϑ λ + /2. Lmma 7. is provd by comparing th Fourir sris xpansions of th functions. Corollary 7.2. W hav 2θ 2, 0 = ηφ 3 2. Lmma 7.3. Lt f = θ 4,0 θ 4,3 0 η, f 2 = θ 4,0 + θ 4,3 0, f 3 = θ 4,00 θ 4,4 0. η 2η Thn f = φ, f 2 = φ 2, f 3 = φ 3. Th proof of Lmma 7.3 is basd on th following rsult. Lmma 7.4. [W] Suppos g, g 2, and g 3 ar holomorphic functions on th uppr half plan C + satisfying th following conditions. P. Th functions g, g 2, and g 3 can b writtn in th forms g = q a 48 a j q j 2, g2 = q a 48 j a j q j 2, g3 = q a 24 whr a is an intgr, a j, b j C, and a 0 =. P2. Th functions g, g 2, and g 3 hav th modular proprtis g = g, g 2 = g 3, g 3 = g 2. Thn g i = φ i a, for i =, 2, 3. b j q j,

9 ELLIPTIC SELBERG INTEGRALS AND CONFORMAL BLOCKS 9 Proof of Proposition 7.3. W hav θ 4, 0 = q 4j+ 8 2 = q 6 q 2 8j2 +2j, j Z j Z θ 4,0 0 = q 4j2 = q 2j2, j Z j Z θ 4,3 0 = q 4j = q 6 q 2 8j2 +6j+, j Z j Z θ 4,4 0 = j Z q 4j+ 2 2 = q 2j+2. j Z Hnc, th functions f = q 24 θ4, 0 θ 4,3 0, f2 = q 24 θ4, 0 + θ 4,3 0, f 3 = 2 2 q 24 θ4,0 0 θ 4,4 0 ar holomorphic functions on C + with th proprty P for a = and g i = f i. Lt yq = j= c jq j b dfind by th condition + yq = q /24 η. Thn for i =, 2, 3, f i = f i + yq = f i yq + yq ar holomorphic functions on C + with th proprty P for a = and g i = f i. On chcks that f, f 2 and f 3 hav th proprty P2 using th modular proprtis of θ 4,n 0 and η. Lmma 7.5. W hav θ 6, 0 θ 6,5 0 = η. Lmma 7.5 is provd by comparing th infinit sris xpansions of th functions. 8. Th proof of Thorm Proof of 2. For κ = 2p + 2, th spac of conformal blocks is on-dimnsional. Th right hand sid of 2 is a solution of with th proprtis i-iv [FV]. According to Thorm 3., th lft hand sid also has ths proprtis. Thus th two functions ar proportional. Th cofficint of proportionality is calculatd by comparing th lading trms of ϑ p+ and u κ,p+ in th limit as i. Th lading trm of ϑ p+ is i p+ q p+/8 πiλ πiλ p+. Lt dt = dt...dt p. Th lading trm of u κ,p+ is πit j πit j 2p 2p+2 πit j t k πit j t k 2 2p+2 p j= 2π πi 2 j<k p πiλ tj πiλ t j πiλ πiλ j= + p+ πiλ+tj πiλ+t j πiλ πiλ j= 2π πi 2 q p+ 2 42p+2 πip+λ+ 2 p 2p+2 j= t j q p+ 2 42p+2 πip+ λ+ 2 2p+2 p j= t j dt.

10 0 G. FELDER, L. STEVENS, AND A. VARCHENKO Th abov xprssion is qual to 2π πi pp+ 2 2p+2 +p 2p2 p πi 2p+2 +p q p+ 8 πiλ πiλ p I l πi2l+λ πi2l+λ, whr I l = f l t,...,t p p j= l=0 t j p+2 2p t j 2p 2p+2 j<k p t j t k 2 2p+2 dt, for som function f l, symmtric in th variabls t,..., t p. Comparing th cofficints of πip+λ in th lading trms, w find that u κ,p+ = i p+ 2π πi 2 pp+ 2p+2 +p 2p2 πi 2p+2 +p I p ϑ p+. To complt th proof, it rmains to comput I p. It is not difficult to show that f p t,...,t p = p j= πit j. Lt x j = t j. Lt p b th imag of p undr th map t j x j. W hav I p = p x p j= p+2 2p+2 j x j 2p 2p+2 j<k p Applying th Stoks thorm, w dform th contour p to gt I p = p j+p+ p+2 2p+2 2p+2 x x j 2p 2p+2 Obsrv that j= x p j= p+2 2p+2 j p j= j x j 2p 2p+2 j<k p x j x k 2 2p+2 dx. j<k p x j x k 2 2p+2 dx x j x k 2 2p+2 dx. is th Slbrg intgral B p p + 2/2p + 2, 2p/2p + 2, /2p + 2. This complts th proof Proof of 3 and 4. Lt κ = 2p + 3. Thn any solution of with th proprtis i-iv has th form vλ, = ϑ p+ λ, c 0 θ,0 λ, + c θ, λ,. Lt A b th transformation introducd in sction 4. By Proposition 4., Avλ, = p+ ϑ p+ λ, c 0 θ,0 λ, c θ, λ, is also a solution. Hnc, for j = 0 or, th function v j λ, = c j ϑ p+ λ, θ,j λ, givs a solution too. Morovr, Av j = p++j v j. By Thorm 3.2, th intgrals u κ,p+ and u κ,p+2 span th spac of conformal blocks. By Lmma 4.3, Au κ,p+ = p+ u κ,p+ and Au κ,p+2 = p u κ,p+2. So for j = 0 or, th intgral u κ,p++j is proportional to v j. By Lmma 6.2, c j must satisfy th diffrntial quation κ d d p + d c j θ,j 0, = d ln ϑ 0, c j θ,j 0,.

11 ELLIPTIC SELBERG INTEGRALS AND CONFORMAL BLOCKS Th function c j = 2π ϑ 0, p+/κ = η 3p+/κ is a solution of this quation for j = 0 and j =. Th cofficints of proportionality ar computd in th limit as i, cf. th proof of 2. This complts th proof Proof of 5, 6, and 7. Lt κ = 2p+4. Lt v j λ, = ϑ p+ λ, θ2,j s λ,, 0 j 2. Any solution of with th proprtis i-iv has th form vλ, = 2 c jv j λ,. Lt A b th transformation in sction 4. By Proposition 4., Av = 2 p+j+ c j v j is also a solution. Hnc th function c v givs a solution too. Morovr, it is an ignvctor of A with ignvalu p. By Thorm 3.2, th spac of conformal blocks is thr-dimnsional with spanning st {u κ,n p + n p + 3}. According to Lmma 4.3, th ignspac of A corrsponding to th ignvalu p is on-dimnsional and is spannd by u κ,p+2. It follows that u κ,p+2 is proportional to c v. By Lmma 6.2, c must satisfy th diffrntial quation κ d d p + d c θ 2, 0, = 2 d ln ϑ 0, c θ 2, 0, + 2c d d θ 2,0,. Th function c = 4πθ 2, 0, ϑ 0, 2p+/κ is a solution of th abov quation. By Corollary 7.2, c = φ 3 η 4p+/κ. Th cofficint of proportionality is computd in th limit i, cf. th proof of 2. This provs 5. To prov 6, w apply th transformation S to both sids of 5. To prov 7, w apply th transformation T to both sids of 6. This complts th proof Proof of 8, 9, and 0. Lt κ = 2p+6. Lt v j λ, = ϑ p+ λ, θ4,jλ, s, 0 j 4. Any solution of with th proprtis i-iv has th form vλ, = 4 c jv j λ,. Lt A and B b th transformations in sction 4. By Proposition 4., Av = 4 p+j+ c j v j is also a solution. Hnc th function c 0 v 0 + c 2 v 2 + c 4 v 4 givs a solution too. Morovr, Bc 0 v 0 + c 2 v 2 + c 4 v 4 = p+ c 4 v 0 + c 2 v 2 + c 0 v 4 is also a solution. So thr xists a solution of th form cv 0 v 4. It is an ignvctor of A with ignvalu p+ and an ignvctor of B with ignvalu p. W show that th subspac of conformal blocks with this proprty is on-dimnsional. By Thorm 3.2, th spac of conformal blocks is fiv-dimnsional with spanning st {u κ,n p + n p + 5}. By Lmma 4.3, th ignspac of A corrsponding to th ignvalu p+ is thr-dimnsional and is spannd by u κ,p+, u κ,p+3, and u κ,p+5. By Lmma 4.3, th transformation B prsrvs th subspac u κ,p+, u κ,p+3, u κ,p+5. Th matrix of B rstrictd to this subspac is p+ 0 κ 0 0 pp+ pp+5 κ Thus th rstriction of B to u κ,p+, u κ,p+3, u κ,p+5 has ignvalus p and p+ of multiplicitis and 2, rspctivly. Th ignspac corrsponding to th ignvalu.

12 2 G. FELDER, L. STEVENS, AND A. VARCHENKO p is spannd by th vctor u κ,p+ + p+ pp+/κ u κ,p+5. It follows that this vctor is proportional to cv 0 v 4. By Lmma 6.2, c must satisfy th diffrntial quation κ d p + d c d 4 d ln ϑ 0, θ 4,0 0, θ 4,4 0, = cθ 4,0 0, θ 4,4 0, + 6c d d θ 4,00, θ 4,4 0,. Th function c = 2π 2 θ 4,0 0, θ 4,4 0, 3 ϑ 0, 2 2p+/κ is a solution of th abov quation. By Lmma 7.3, w hav c = 2 3p+/κ φ 3 η 6p+/κ. Th cofficint of proportionality is computd as in th proof of 2. This provs 8. To prov 9 and 0, apply th transformations S and TS, rspctivly, to both sids of 8. This complts th proof Proof of. Lt κ = 2p + 8. Lt v j λ, = ϑ p+ λ, θ6,jλ, s, 0 j 6. Any solution of with th proprtis i-iv has th form vλ, = 6 c jv j λ,. Lt A and B b as in sction 4. By Proposition 4., Av = 6 p+j+ c j v j is also a solution. Hnc th function c v + c 3 v 3 + c 5 v 5 givs a solution too. Th function Bc v + c 3 v 3 + c 5 v 5 = p+ c 5 v + c 3 v 3 + c v 5 also givs a solution. So thr is a solution of th form cv v 5 which is an ignvctor of A and B with ignvalu p undr both transformations. W show that th subspac of conformal blocks with this proprty is on-dimnsional. By Thorm 3.2, th spac of conformal blocks is svn-dimnsional with spanning st {u κ,n p + n p + 7}. By Lmma 4.3, th thr-dimnsional ignspac of A corrsponding to th ignvalu p is spannd by u κ,p+2, u κ,p+4, and u κ,p+6. Th matrix of B rstrictd to u κ,p+2, u κ,p+4, u κ,p+6 is p+ 0 κ 0 0 pp+2 pp+6 κ Thus, B has ignvalus p and p+ of multiplicitis and 2, rspctivly. Th ignspac corrsponding to th ignvalu p is spannd by th vctor u κ,p+2 + p+ pp+2/κ u κ,p+6. So this vctor is proportional to cv v 5. By Lmma 6.2, c must b a solution of th diffrntial quation κ d p + d c d 6 d ln ϑ 0, θ 6, 0, θ 6,5 0, =. cθ 6, 0, θ 6,5 0, + 0c d d θ 6,0, θ 6,5 0,.

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