Ramanujan J https://doi.org/10.1007/s11139-017-9937-y

Certain families of polynomials arising in the study of hyperelliptic Lie algebras Ben Cox1

· Kaiming Zhao2,3

Received: 24 February 2016 / Accepted: 13 July 2017 © Springer Science+Business Media, LLC 2018

Abstract The commutative ring R(P(t)) = C[t ±1 , u | u 2 = P(t)], where P(t) = n n i i=0 ai t = k=1 (t − αi ) with αi ∈ C pairwise distinct, is the coordinate ring of a hyperelliptic curve when n > 4. The Lie algebra R(P(t)) = Der(R(P(t))) of derivations is called the hyperelliptic Lie algebra associated to P(t) and is a particular type of multipoint Krichever–Novikov algebra. In this paper, we describe the universal central extension of Der(R(P(t))) in terms of certain families of polynomials which in a particular case are associated Legendre polynomials. Moreover we describe certain families of polynomials that arise in the study of the group of units for the ring R(P(t)), where P(t) = t 4 − 2bt 2 + 1. In this study, pairs of Chebyshev √ polynomials (Un , Tn ) arise as particular cases of a pairs (rn , sn ) with rn + sn P(t) a unit in R(P(t)). We explicitly describe these polynomial pairs as coefficients of certain generating functions and show that certain of these polynomials satisfy particular second-order linear differential equations.

The first author is partially supported by a collaboration Grant from the Simons Foundation (#319261). The second author is partially supported by NSF of China (Grant 11471233) and NSERC.

B

Ben Cox [email protected] Kaiming Zhao [email protected]

1

Department of Mathematics, University of Charleston, South Carolina, 66 George Street, Charleston, SC 29424, USA

2

Department of Mathematics, Wilfrid Laurier University, Waterloo, ON N2L 3C5, Canada

3

College of Mathematics and Information Science, Hebei Normal (Teachers) University, Shijiazhuang 050016, Hebei, People’s Republic of China

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Keywords Krichever–Novikov algebras · Automorphism groups · Pell’s equation · Associated Legendre polynomials · Universal central extensions · Superelliptic Lie algebras · Superelliptic curves · DJKM algebras · Fáa di Bruno’s formula · Bell polynomials Mathematics Subject Classification 17B65 · 05E35 · 14H55 · 42C10

1 Introduction Throughout this paper we will take the set of natural numbers to be N = {1, 2, . . . }, the set of nonnegative integers will be denoted by Z+ = {0, 1, 2, 3, . . . }, and we will assume all vector spaces and algebras are defined over complex numbersC. n ±1 2 i nThe associative ring R(P(t)) = C[t , u | u = P(t)], where P(t) = i=0 ai t = k=1 (t − αi ) with αi ∈ C pairwise distinct, is the coordinate ring of a hyperelliptic curve (if deg(P) > 4). The Lie algebra R(P(t)) = Der(R(P(t))) of derivations is called the hyperelliptic Lie algebra associated to P(t) (even if deg(P) > 4 does not hold). In this paper we describe the universal central extension of Der(R(P(t))) in terms of Legendre polynomials, and we describe certain families of polynomials that arise in the study of the group of units for the ring R(P(t)), where P(t) = t 4 − 2bt 2 + 1 by using Chebyshev polynomials. We explicitly find the second-order linear differential equations these polynomial pairs satisfy. Let us give some background to this paper and more precise information of what is in the paper. The Laurent polynomial ring C[t, t −1 ] can be considered as the ring of rational functions on the Riemann sphere C ∪ {∞} with poles allowed only in {∞, 0}. From geometric point of view one can have a natural generalization of the loop algebra construction. Instead of the sphere with two punctures, one can use any complex algebraic curve X of genus g with a fixed subset P of n distinct points. This arrives at M. Schlichenmaier’s definition of multipoint algebras of Krichever–Novikov affine type if we replace C[t, t −1 ] with the ring R of meromorphic functions on X with poles allowed only in P in the construction of affine Kac–Moody algebras (see [13– 16]). The n-point affine Lie algebras which are a type of Krichever–Novikov algebra of genus zero also appeared in the work of Kazhdan and Lusztig [12, Sects. 4 & 7], [10, Chap. 12]. Krichever–Novikov algebras are used to construct analogues of important mathematical objects used in string theory and in the setting of a Riemann surface of arbitrary genus. Moreover Wess–Zumino–Witten–Novikov theory and analogues of the Knizhnik–Zamolodchikov equations are developed for analogues of the affine and Virasoro algebras (see the survey article [23], and for example [16,17,19,20,22]). In a recent paper [6], the n-point Virasoro algebras V˜ a were studied, which are natural generalizations of the classical Virasoro algebra and have as quotients multipoint genus zero Krichever–Novikov-type algebras. Necessary and sufficient conditions for the latter two such Lie algebras to be isomorphic, and their automorphisms, their derivation algebras were obtained, their universal central extensions and some other properties were also determined. Also a large class of modules which were called modules of densities were constructed, and necessary and sufficient conditions for them to be irreducible were obtained. The n-point Virasoro algebras all are coordinate

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Certain families of polynomials arising...

rings of the rings the Riemann sphere with n-points removed. See also [18] for more information on n-point Virasoro algebras. In another recent paper [7], mth superelliptic Lie algebras Rm (P) associated to P(t) ∈ C[t] were defined; the necessary and sufficient conditions for such Lie algebras to be simple were obtained; and their universal central extensions and their derivation algebras were determined. They are examples of genus greater than one Krichever– Novikov-type algebra. Sufficient conditions for Der(R) to be simple when R is a commutative algebra over C were obtained in [11,24]. The present paper is a sequel to [7]. The particular class we look at are hyperelliptic curves, i.e., m = 2. Let P(t) ∈ C[t]. Then we have the Riemann surfaces (commutative associative algebras) R(P) = C[t ±1 , u | u 2 = P(t)]. The Lie algebras R(P) = Der(R(P)) are called hyperelliptic Lie algebras due to the fact that u 2 = P(t) is a hyperelliptic curve when n > 4. Finally we should mention that other interesting families of nonclassical orthogonal polynomials appear in the description of the center of the universal enveloping algebra of other particular Krichever–Novikov algebras such as the DJKM algebra (see [5]). Other families such as the ultraspherical and Pollaczek polynomials come about from taking R = C[s, s −1 , (s − 1)−1 , (s − a)−1 ] ∼ = C[t, t −1 , u | u 2 = t 2 − 2bt + 1] and −1 2 3 2 R = C[t, t , u|u = t − 2bt + t], respectively (see [3] and [2]). Skryabin in 2004 (see [25]) gave a general description of the universal central extension for Der(R) where R is a commutative ring over C. In particular he showed that if L is the Lie algebra of vector fields on a Riemann surface  with a finite number of distinct poles at A = {a1 , . . . , a M } then H 2 (L , C) ∼ = Hd1R (\A) ∼ = H1 (\A, C). Explicit 2-cocycles in H 2 (L , C) can be derived from the formula 1 γC,R (e, f ) = 2π

  C

 1     (e f − e f ) − R(e f − e f ) dz, 2

where C is a cycle on the Riemann surface  with the points of A removed and R is a projective connection on . With this beautiful formula one still has to do a lot of computations. One of our goals is to give concrete and elementary formulas for universal central extensions. In the second section, we give an explicit basis for the universal central extension of the Lie algebras Rm (P) and in the particular case that 0 is not a root of P(t) explicitly describe a set of basis of two cocycles. We also explicitly give examples the value of any 2-cocycle on a basis of Rm (P). Our description uses Braá di Bruno’s formula and Bell polynomials to describe generating functions of families of polynomials appearing as coefficients of the basis elements. In the particular case of P(t) = t 2r − 2bt r + 1, r ∈ N, associated Legendre polynomials naturally arise as coefficients of a basis for the two cocycles. Having such an explicit description of the two cocycles will allow one to use conformal field theoretic tools to study free field type representations of these algebras.

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To study isomorphisms and automorphisms between the Lie algebras Rm (P), from [24], we know that it is equivalent to considering isomorphisms and automorphisms of the Riemann surfaces Rm (P). This is in general a very hard problem. In particular it is known that many sporadic simple groups can appear. For some results on higher genus Riemann surfaces, see [1,4] and the reference therein. In the third section, we describe the group of units of R2 (P) which will be used in the last section, in particular we explicitly describe this group in cases of P(t) = t 4 − 2bt + 1, b = ±1 which is the most interesting case studied by Date et al. [8]. In this later paper they investigated integrable systems arising from Landau–Lifshitz differential equation. When determining the unit group we find that it requires one find solutions of the polynomial Pell equation f 2 − g 2 P = 1 for a given P ∈ C[t] which is a very famous and very hard problem. See [9]. Then Theorem 4 describes the possible units in Rm (P) as Rm (P)∗ ∼ = C∗ × Z × Z × Z. One may want to compare this to the description of the automorphism group of a hyperellptic curve given in in terms of pairs of polynomials u n , [21]. The group of units √ of R2 (P) are described √ vn satisfying u n + vn P = (u 1 + v0 P)n and are similar in some sense to the Chebyshev polynomials of the first and second kinds. We explicitly describe these polynomial pairs as coefficients of generating function and show certain of these polynomials satisfy particular second-order linear differential equations.

2 The hyperelliptic Lie algebras R( P(t)) Let P(t) =

n+l 

bi t i = t l (t − a1 ) . . . (t − an )

i=l

for some l = 0 or 1, n ∈ N, and pairwise distinct nonzero a1 , . . . , an ∈ C. Note that bn+l = 1. From [7] we see that R(P) = C[t ±1 ] ⊕ C[t ±1 ]u and if we set  := ∂ + 2u ∂t∂ , where P  = ∂∂tP we have R(P) = R(P). One of the reasons we are P  ∂u interested in the Lie algebra R(P) := DerR(P(t)) is the following Theorem 1 [7, Theorem 4] The Lie algebra R(P(t)) is simple. The proof of this uses in an√ important fashion results found in [24]. Let R = R(P) and ∂ = P dtd . We know Theorem 2 [7, Theorem 9] Suppose that R = R(P) and P = t l (t −a1 ) . . . (t −an ) for l = 0 or 1, n ∈ N, and pairwise distinct nonzero a1 , . . . , an ∈ C. Then dim R/∂(R) = n + 1 and the universal central extension of R(P) is R(P) ⊕ R/∂(R) with brackets [ f ∂, g∂] = f ∂(g)∂ − g∂( f )∂ + ∂( f )∂(∂(g)), ∀ f, g ∈ R. The above theorems follow also from results in [25]. Note that R(P) is the classical centerless Virasoro algebra if n = 0, and R(P) with l > 1 is isomorphic to one of the Lie algebras in the above theorem. So our Lie algebra R(P) with the given restrictions on P cover all cases of P up to isomorphisms.

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Certain families of polynomials arising...

To study representations of the Lie algebras R(P)⊕ R/∂(R) with nontrivial central action, it is important to have a basis for its center R/∂(R) with concrete formulae for their brackets. This is the main purpose of this section, which turns out to be complicated. √ From the proof of [7, Theorem 9] we suppose a basis is given by ω0 := t −1 P, and  t −1 , . . . , t n−2 . In the rest of this section we will derive some formulae for the 2-cocycles of R(P) in terms of this precise basis. 2.1 The general case We can calculate any 2-cocycle as being a linear combination of the coefficients of ψ where ψ( f ∂, g∂) = ∂( f )∂ 2 (g). In particular  n   i ψ(t r ∂, t s ∂) = r s (2.1) s−1+ bi δr +s+i−2,0 ω0 2 i=0

(r s(r − s)b−r −s+2 ω0 )/2 if 0  −r − s + 2  n, = 0 otherwise. and

√ P∂, t s P∂) (2.2)     n n  j i r +i −2+ (r + i − 1)bi b j δr +s+i+ j−2,0 ω0 . r+ = 2 2

ψ(t r

√

i=0 j=0

√ For the cross term ψ(t r ∂, t s P∂), the formula is more complicated and we do not have a simple description of what it is but we have the following in the case l = 0 (for l = 1 a similar description can be made). For all r, s ∈ Z, we have ψ(t r ∂, t s

√

= r t r −1

P∂)

√

⎞  n   j s+ b j t s+ j−1 ⎠ P∂ ⎝ 2 ⎛

j=0

⎛ ⎞  n   √ √ j s+ (s + j − 1)b j t s+ j−2 P ⎠ = r t r −1 P ⎝ 2 j=0 ⎛ ⎞   n n   j ⎝ =r s+ (r + s + j − 2)b j ⎠ bi t r +s+i+ j−3 . 2 i=0

(2.3)

j=0

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We now consider the t s+i+ j−2 : As for any r ∈ Z we have ∂(t r

√

P) =

 n   i r+ bi t r +i−1 , 2 i=0

so in the quotient R/∂ R we have the recursion relation  n   i r+ bi t r +i−1 = 0. 2 i=0

As bn = 1, this proves that for r  0 t r +n−1

 n−1   i n −1  r+ bi t r +i−1 . =− r+ 2 2 i=0

This implies that {t k | k  n − 1} can be written as a linear combination of {t −1 , . . . , t n−2 }. More precisely we can find polynomials pr,i = pr,i (b0 , . . . , bn−1 ), −1,  i  n − 2, such that n−2  k t = pk,i t i (2.4) i=−1

for all k  n − 1. We will set pr,i = δr,i for −1  i, r  n − 2 so (2.4) holds for all k  −1. As b0 = 0 we have t r −1

 n  i 1  r+ bi t r +i−1 . =− r b0 2

(2.5)

i=1

This means that {t −k | k  2} can be written as a linear combination of {t −1 , . . . , t n−2 }. More precisely we can find rational functions qr,i = qr,i (b0 , . . . , bn−1 ), −1,  i  n − 2, such that n−2  t −k = qk,i t i (2.6) i=−1

for all k  1. As a consequence of (2.3), (2.4), and (2.6), we have ψ(t r ∂,t s

√

P∂)

=r

n−2  k=−1

123

⎛ ⎝

n 

i, j=0

(r + s + i + j − 2)

 × bi pr +s+i+ j−3,k (b) t k

   j (r + s + j − 2)b j s+ 2

Certain families of polynomials arising...

+r

n−2   n  k=−1

(−(r + s + i + j − 1))

i, j=0

    j (r + s + j − 2)b j bi q−(r +s+i+ j−3),k (b) t k × s+ 2 (2.7)

where

1 (k) = 0

if k  0, for k < 0,

is the Heaviside function. Because of this we then turn to giving very general formulae for the polynomials pk,i and qk,i . If we let pi,r = pr,i (b0 , . . . , bn−1 ) be polynomials in b0 , . . . , bn−1 , i = −1, 0, 1, 2 . . . , n − 2, r  −1 that satisfy the recursion relation  n   j r+ pr + j−1,i (b0 , . . . , bn−1 )b j = 0 2

for all r  n − 1,

(2.8)

j=0

with initial conditions pr,i = δi,r for −1  i, r  n − 2. Set Pi (b, z) :=



pk−1,i z k =

k0



pk,i z k+1

(2.9)

k−1

and ¯ P(z) :=

n 

b j z n− j ,

¯ Q(z) := z P¯  (z) − n P(z),

j=0

Ri (z) := z n

n 



(2l + j)b j pl+ j−1,i z l ,

j=0 − jl
¯ where −1  i  n − 2. Note that P(z), Q(z) and Ri (z) are all polynomials in z with coefficients polynomials in the bi . Then we have d ¯ Pi (b, z) + Q(z)Pi (b, z) = Ri (z), 2z P(z) dz

(2.10)

for i = 1, . . . , n. This has integrating factor  μ = exp

  n P¯  (z) −n/2 ¯ dz = P(z)z . − ¯ 2z 2 P(z)

(2.11)

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Here z −n/2 is only to be used formally in the following when integrating and is not to be thought of as a multivalued function. For i = −1, . . . , n − 2, we obtain generating functions as solutions to the family of n equations (2.10): z n/2 Pi (b, z) =  ¯ P(z)



Ri (z) dz  ¯ 2z P(z) n+2 2

(2.12)

and they are given by hyperelliptic integrals. In the above formula we expand 1 as a Laurent series in z, multiply by Ri (z)/2z (n+2)/2 , and then inte¯ 1/2 P(z) grate formally term by term. Let us explain this more precisely: If we expand ∞  1  h k (b0 , . . . , bn−1 )t k , then (2.12) becomes = ¯ P(t) k=0 Pi (b, z) =

∞ ∞ n 1  h k (b0 , . . . , bn−1 )(2l + j)b j pl+ j−1,i 2 k=0 m=0 j=0 − jl
∞

1 = 2 n



(2l + j)b j pl+ j−1,i h k (b0 , . . . , bn−1 )

k=0 m=0 j=0 − jl
× h m (b0 , . . . , bn−1 )C z k+(n/2) +

n ∞  ∞  



k=0 m=0 j=0 − jl
(2l + j)b j pl+ j−1,i h k (b0 , . . . , bn−1 ) n + 2(m + l)

× h m (b0 , . . . , bn−1 )z m+n+l+k ,

(2.13)

where C is a constant of integration (which is zero if n is odd as the left-hand power series has no fractional powers of z). Then (2.13) gives us a (rather complicated) description of pa,i as coefficient of z a in the above power series. So we are left with providing a description√of h k (b0 , . . . , bn−1 ) (see (2.15)). One can expand 1/ P(z) using Bell polynomials and Faà di Bruno’s formula as follows. Now the Bell polynomials in the variables z 1 , z 2 , z 3 , . . . are defined to be Bm,k (z 1 , . . . , z m−k+1 ) lm−k+1   z l1  z m−k+1 m! 1 ··· , := l1 !l2 ! · · · lm−k+1 ! 1! (m − k + 1)!! where the sum is over l1 + l2 + · · · = k and l1 + 2l2 + 3l3 + · · · = m. Now Faà di Bruno’s formula for the m-derivative of f (g(x)) is  dm f (g(x)) = f (l) (g(x))Bm,l (g  (x), g  (x), . . . , g (m−l+1) (x)). m dx m

l=0

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Certain families of polynomials arising...

√ ¯ Setting f (x) = 1/ x, g(x) = P(x) we get f (m) (x) =

(−1)m (2m − 1)!! 2m x (2m+1)/2

(2.14)

and P¯ (k) (0) = k!bn−k so that  (−1)l (2l − 1)!! dm f (g(x))| = x=0 (2l+1)/2 dx m 2l bn m

l=0

× Bm,l (bn−1 , 2bn−2 , . . . , (m − l + 1)!bn−m+l−1 ), √ where we set bk = 0 for k  −1 and (2k − 1)!! = (k + (1/2))2k / π . As a consequence ∞

 dk 1 f (g(t))|t=0 t k = dt k ¯ P(t) k=0   k ∞  1  (−1)l (2l − 1)!! = Bk,l (bn−1 , 2bn−2 , . . . , (k − l + 1)!bn−k+l−1 ) t k , (2l+1)/2 l k! 2 bn



k=0

l=0

and hence h k (b0 , . . . , bn−1 ) =

 k 1  (−1)l (2l − 1)!! (2l+1)/2 k! 2l bn l=0

 × Bk,l (bn−1 , 2bn−2 , . . . , (k − l + 1)!bn−k+l−1 ) .

(2.15)

Here √ (2k − 1)!! = (k + (1/2))2k / π and (z) is the Gamma function. Lastly we derive the formulae for the qk,i in (2.6) in terms of generating functions. For −1  i  n − 2 set Q i (b, z) :=



qk,i z k+n−2 =

k−n+2



qk−n+2,i z k ,

(2.16)

k0

where qk,i , k  −n + 2, are rational functions in bi and satisfy the recursion relation (t −1 )1−r = −

 n  i 1  bi t r +i−1 r+ r b0 2

(2.17)

i=1

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or (t −1 )k

 n   i 1 1−k+ bi (t −1 )k−i . =− (1 − k)b0 2

(2.18)

i=1

This gives us a recursion relation n 

(2k − ( j + 2)) b j qk− j,i = 0

(2.19)

j=0

for all k  2. These polynomials satisfy the initial condition qk,i = δk,−i for −n +2  i, k  1. Set P(z) =

n 

Q(z) := z P  (z) − 2(n − 1)P(z),

bi z i ,

(2.20)

i=0

Si (z) :=

n 



(2l − ( j + 2))b j ql− j,i z l+n−2 .

j=0 j−n+2l<2

Then 2z P(z)

d Q i (b, z) + Q(z)Q i (b, z) = Si (z), dz

which has integrating factor  μ = exp thus

 Q(z) dz = z −(n−1) P(z), 2z P(z)

z n−1 Q i (b, z) = √ P(z)



Si (z) dz. √ P(z)

2z n

(2.21)

As in the case of Pi (b, z), one can use Fáa de Bruno’s formula and Bell polynomials to find the Taylor series expansion of Q i (b, z) at z = 0. Next we illustrate the above generating functions in the examples below. First we can directly obtain Lemma 3 If P(t) = t 2 − 2bt with b = 0, then the center of the universal central extension of R(P) is two-dimensional and we can write any 2-cocycle ψ as   ψ(t r ∂, t s ∂) = r 3 δr +s,0 − br (r − 1)(2r − 1)δr +s−1,0 ω0 , √     ψ(t r ∂, t s P∂) = 3r br +s+1 r 4s 2 + 5s + 2 + 4s 3 + 10s 2 + 9s + 3 (2r + 2s − 3)!! × 1, (r + s + 1)!

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Certain families of polynomials arising...

ψ(t r

√

P∂, t s

√

P∂) = (r + 1)3 δr +s+2,0 ω0 − b(2r + 1)(2r 2 + 2r + 1)δr +s+1,0 ω0 + b2 (4r 2 − 1)r δr +s,0 ω0 ,

where by definition, (2r + 2s − 1)!! = (2r + 2s − 1) · (2r + 2s − 3) · · · 5 · 3 · 1 and ⎧ ⎪ ⎨0

i f r + s  −2, (2r + 2s − 3)!! 1 = 13 i f r + s = −1, ⎪ (r + s + 1)! ⎩ −1 i f r + s = 0. 2.2 The case R = C[t ±1 , u | u2 = t 2 − 2bt + c] with c = 0 Up to isomorphism we may assume that c = 1. In the case P(t) = t 2 − 2bt + 1, l = 0, the center of the universal√central extension is three-dimensional and in terms of the basis elements ω0 := t −1 P, t −1 , 1,  ψ(t r ∂, t s ∂) = r 3 δr +s,0 − br (r − 1)(2r − 1)δr +s−1,0  + r (r − 1)(r − 2)δr +s−2,0 ω0 ,   √ √ ψ(t r P∂, t s P∂) = (r + 1)3 δr +s+2,0 ω0 − b(2r + 1) 2r 2 + 2r + 1 δr +s+1,0 ω0      + b2 r 4r 2 − 1 + 2r r 2 + 1 δr +s,0 ω0 − 2b(r − 1)r (2r − 1)δr +s−1,0 ω0 + r (r − 1)(r − 2)δr +s−2,0 ω0 . In this case ∂(t k

√

P) = (k + 1)t k+1 − b (2k + 1) t k + kt k−1 ,

so in the quotient R/∂ R we have (k + 1)t k+1 − b (2k + 1) t k + kt k−1 = 0.

(2.22)

This is the recursion relation for the Legendre polynomials pk (b). Thus for k  0 we have t k = pk (b)1. For k  1 we have t −k = pk−1 (b)t −1 . Lastly we calculate ψ(t r ∂, t s

√

    P∂) = r s 2 − s t r +s−3 + r b s − 4s 2 t r +s−2   + r (4s 2 + 2s)b2 + 2s 2 + s + 1 t r +s−1   − r b 4s 2 + 5s + 2 t r +s + r (s + 1)2 t r +s+1 .

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When r + s  3 we get ψ(t r ∂, t s

√

    P∂) = r s 2 − s pr +s−3 (b)1 + r b s − 4s 2 pr +s−2 (b)1   + r (4s 2 + 2s)b2 + 2s 2 + s + 1 pr +s−1 (b)1   − r b 4s 2 + 5s + 2 pr +s (b)1 + r (s + 1)2 pr +s+1 (b)1.

If r + s = 2 we get ψ(t r ∂,t s

√

  1   P∂) = r (r − 1)(r − 2)t −1 + br b2 r 2 − 3r + 1 − 3r 2 + 9r − 5 1. 2

If r + s = 1 one has ψ(t r ∂, t s

√

  1   P∂) = −3b(r − 1)2 r t −1 + r b2 3r 2 − 6r + 2 + 3r 2 − 6r + 4 1. 2

When r + s = 0 we get ψ(t r ∂, t s

√

P∂) =

   1  r 3r (r − 1)b2 + 3r 2 − 3r + 2 t −1 − br 3r 2 − 3r + 1 1. 2

If r + s = −1 one obtains ψ(t r ∂, t s

√

P∂) =

  1  2 2 br b r − 1 − 3r 2 + 1 t −1 + r 3 1. 2

When r + s  −2 we get ψ(t r ∂, t s

√

    P∂) = r s 2 − s p−(r +s)+2 (b)t −1 + r b s − 4s 2 p−(r +s)+1 (b)t −1   + r (4s 2 + 2s)b2 + 2s 2 + s + 1 p−(r +s) (b)t −1   − r b 4s 2 + 5s + 2 p−(r +s)−1 (b)t −1 + r (s + 1)2 p−(r +s)−2 (b)t −1 .

2.3 The case R = C[t ±1 , u | u2 = t 3 + at + b] with b = 0 Up to isomorphism we may assume that b = 1. The center of the universal central √ extension is four-dimensional and in terms of the basis elements ω0 := t −1 P, t −1 , 1, t. We will express t r in terms of this basis. Here we have P(t) = t 3 + at + 1 and ∂(t k

123

√

P) =

√

P∂(t k

√

1 P) = kt k−1 P + t k P  2

Certain families of polynomials arising...

    3 k+2 1 t at k + kbt k−1 = k+ + k+ 2 2

(2.23)

so that a¯ t¯2 = − 1, 3 4 ¯ t 4 = − t¯ + 7 .. .. . . If we write

2¯ 3 t¯3 = − a t¯ − 1, 5 5 7 2 8a ¯ 5 2¯ ¯ a 1, t 5 = a t¯ + 1, 21 15 15

t r = pr,1 t + pr,0 1 + pr,−1 t −1

(2.24)

for some polynomials pr,i , i = −1, 0, 1 in a so that pk,i , i = −1, 0, 1 satisfy the same recurrence relation as t k but with initial conditions p−1,−1 = 0, p0,−1 = 0, p1,−1 = 0, p−1,0 = 0, p0,0 = 1, p1,0 = 0, p−1,1 = 0, p0,1 = 0, p1,1 = 1. So we get Q(z) = −3 − az 2 , R−1 (z) = 0, R0 (z) = z 3

(2.25) n 



(2l + j)b j pl+ j−1,0 z l = −z,

(2.26)

(2l + j)b j pl+ j−1,1 z l = z 2 ,

(2.27)

j=0 − jl0

R1 (z) = z 3

3   j=0 − jl0

and the generating series are given by (2.12) or rather (2.13) 

−z dz √ + 1 + az 2 + z 3   2z 4 5a 2 z 5 8 16 15a 3 7 az 3 − + + az 6 + − z + ··· , =z− 3 5 21 15 55 77  z 3/2 z2 P1 (z) = √ dz √ 1 + az 2 + z 3 2z 5/2 1 + az 2 + z 3 4z 5 7a 2 z 6 348 7 3az 4 − + + az + · · · = z2 − 5 7 15 385

P0 (z) = √

z 3/2

1 + az 2

z3

2z 5/2

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B. Cox, K. Zhao

Similarly one has from (2.5) a 1 t −2 = − t −1 + t, 2 2 2 3a 1 3a t −3 = t −1 − 1 − t, 8 4 8 15a 3 + 24b2 −1 5a 5a 2 t −4 = − t + 1+ t, 48 24 16 .. .. . . t −k = qk,−1 t −1 + qk,0 1 + qk,1 t. Then one needs to solve the differential equations 2z P(z)

d Q i (b, z) + Q(z)Q i (b, z) = Si (z), i = −1, 0, 1. dz

From (2.20), we have S−1 (z) = 0, S0 (z) = −2z − az 2 , S1 (z) = −4 − 3az, so that by (2.21) (and Fáa de Bruno’s formula and Bell polynomials) z2

Q −1 (z) = √

z 3 + az + 1 a 3a 2 4 15a 3 + 24 5 z − z + ··· , = z2 − z3 + 2 8 48 z2

Q 0 (z) = √



−2z − az 2 dz √ 2z 3 1 + az 2 + z 3

z 3 + az + 1 1 5a 5 = z − z4 + z + ··· , 4 24 

−4 − 3az dz √ + az + 1 1 + az 2 + z 3 1 3a 4 5a 2 5 z + z + ··· , = 1 + z3 − 2 8 16

Q 1 (z) = √

z2

z3

2z 3

where one needs to use a constant of integration 1, a/2, and −a 2 /8, respectively. Recall that (2.26) looks like the same recursion relation as (2.22) but allows one to solve for

123

Certain families of polynomials arising...

 the negative powers of t −k as a linear combination of t −1 , 1, t . Thus we needed to use a different set of initial conditions qk,i = δk,−i , −1  k  1, −1  i  1 than for the pk,i . Using (2.20) and (2.21) one is lead to the calculation given above for Q i , i = −1, 0, 1. 2.4 The case R = C[t ±1 , u | u2 = t 2r − 2bt r + 1], r ∈ N We assume r ∈ N is fixed throughout this subsection. The center of the universal central √ extension is 2r + 1-dimensional and in terms of the basis elements ω0 := t −1 P, t i for i = −1, 0, 1, · · · 2r − 2. We will explain how to express pk = t k in terms of this basis. In this case, we have ∂(t k

√

P) = (k + r )t 2r +k−1 − b (2k + r ) t r +k−1 + kt k−1 ,

which leads one to the recursion relation (k + r ) p2r +k−1 − b (2k + r ) pr +k−1 + kpk−1 = 0. Let now α, β ∈ C, c ∈ R, c > 0, and γ := α + β + 1. The associated Jacobi (α,β) polynomials ρk = Pk (x; c), k ∈ Z+ , ρ−1 = 0, ρ0 = 1 satisfy the recursion relation 2(k + c + 1)(k + c + γ )(2k + 2c + γ − 1)ρk+1  = (2k + 2c + γ ) (2k + 2c + γ − 1)(2k + 2c + γ + 1)x  + (γ − 1)(γ − 2β − 1) ρk − 2(k + c + γ − β − 1) × (k + c + β)(2k + 2c + γ + 1)ρk−1 for all k ∈ Z+ . Setting c = 1 gives the usual recursion relation for the Jacobi polynomials. Setting α = β = 0 into the associated Jacobi recursion relation above gives us (k + c + 1)ρk+1 = (2k + 2c + 1)xρk − (k + c)ρk−1 for k  0. The recursion relation we have is (k + r ) p2r +k−1 − b (2k + r ) pr +k−1 + kpk−1 = 0. Let k = r s + q where 0  q  r − 1. Then this becomes (r s + q + r ) p2r +r s+q−1 − b (2(r s + q) + r ) pr +r s+q−1 + (r s + q) pr s+q−1 = 0.

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B. Cox, K. Zhao q/r

For l ∈ Z with r and 0  q  r − 1 setting Pl

(b) := p(l+1)r +q−1 we get

    q   q q q/r q/r q/r + 1 Ps (b) + s Ps−1 (b) = 0. s + + 1 Ps+1 (b) − b 2s + 2 r r r q/r

Thus the Pl (b) and hence the p(l+1)r +q−1 are associated Legendre polynomials (or functions as the case may be). Similarly the q(l+1)r +q−1 are associated Legendre polynomials but with a different set of initial conditions. Then one can rewrite (2.7) in terms of the associate Legendre polynomials and the basis {t −1 , . . . , t n−2 }.

3 Chebyshev polynomials and some related polynomials We recall below our description of the group of units R ∗ (P) for the interesting case of P(t) studied by Date, Jimbo, Kashiwara, and Miwa [8] where they investigated integrable systems arising from Landau–Lifshitz differential equation. Let P(t) =

t 4 − 2βt 2 + 1 , β = ±1. β2 − 1

Observe that in this case P(t) = q(t)2 − 1 where q(t) = √t −β . 2 2

β −1

For convenience, let √ t2 − β λ0 =  + P, β2 − 1  β − 1√ t2 + 1 λ1 = √ P, + 2 2(β + 1)  β + 1√ t2 − 1 λ2 = √ P. + 2 2(β − 1) It is easy to verify that λ0 , λ1 , λ2 ∈ R ∗ (P). Actually, λ0 λ¯ 0 = 1, λ1 λ¯ 1 = t 2 , λ2 λ¯ 2 = t 2 , λ1 λ2 = t 2 λ0 . We have Theorem 4 [7, Theorem 13] Let P(t) be as above. (a) As a multiplicative group, R ∗ (P) is generated by C∗ , t, λ1 , λ2 . (b) R ∗ (P) C∗ × Z × Z × Z. Next we will study properties of elements in the unit group R ∗ (P). √ Theorem 5 Define polynomials u n , vn−1 by u n + vn−1 P = λn0 for n ∈ Z. Then u n+2 (t) − 2q(t)u n+1 (t) + u n (t) = 0, vn+2 (t) − 2q(t)vn+1 (t) + vn (t) = 0, and −P

123

 2  2   dy dq d2 y d q dq 2 dq + n + P − q y = 0, dx dx 2 dx 2 dx dx dx

(3.1)

Certain families of polynomials arising...

is the second-order differential equation satisfied by the u n . The vn (t) satisfies −P

 2    2 dy dq d2 y d q dq dq + P − 3q y = 0. + n(n + 2) 2 2 d dx dx dx dx dx

Proof We have 

u n (t)z n +

n0



  √ vn−1 (t) P(t)z n = (q(t) + P)n z n

n0

n0

√ 1 − q(t)z + P(t)z 1 = = . √ 1 − 2q(t)z + z 2 1 − (q(t) + P(t))z

We have the generating series for the set of polynomials as 

u n (t)z n =

n0

 1 − q(t)z 1 , vn−1 (t)z n = . 2 1 − 2q(t)z + z 1 − 2q(t)z + z 2 n0

The first equation gives us 

u n (t)z n −

n0



2q(t)u n (t)z n+1 +

n0



u n (t)z n+2 = 1 − q(t)z,

n0

or 

u n+2 (t)t n+2 −

n0



2q(t)u n+1 (t)z n+2 +

n0



u n (t)z n+2 = 0,

n0

which in turn gives us the recurrence relation u n+2 (t) − 2q(t)u n+1 (t) + u n (t) = 0,

(3.2)

with u 0 (t) = 1 and u 1 (t) = q(t). The second equation gives us  n0

vn−1 (t)z n −



2q(t)vn−1 (t)z n+1 +

n0



vn−1 (t)z n+2 = 1,

n0

or  n0

vn+1 (t)z n+2 −



2q(t)vn (t)z n+2 +

n0



vn−1 (t)z n+2 = 0,

n0

which in turn gives us the recurrence relation vn+2 (t) − 2q(t)vn+1 (t) + vn (t) = 0,

(3.3)

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B. Cox, K. Zhao

with v0 (t) = 1 and v1 (t) = 2q(t). The Chebyshev polynomials Tn (x) satisfy the recursion relation Tn+2 (x) − 2x Tn+1 (x) + Tn (t) = 0, with T0 (x) = 1 and T1 (x) = x. We note that this first family of Chebyshev polynomials satisfy the differential equation (1 − t 2 )y  (t) − t y  (t) + n 2 y = 0. Since this is similar to (3.2) we use the above differential equation to find one for the vn and u n as follows Now for q = q(x) by the chain rule dy(q) dq d y(q) = , dx dq dx   d2 d dy(q) dq dq y(q) = dx 2 dq dq dx dx

  d2 y(q) dq dq dy(q) d dq dq + d2 q dx dx dq dq dx dx  2 2 d y(q) dq dy(q) d2 q = + d2 q dx dq dx 2

=

so that 1 d dy(q) = dq y(q), dq dx dx  2  d2 y(q) d dy(q) d2 q 1 . y(q) − =  2 d2 q dx 2 dq dx 2 dq dx

Inserting these into the Chebyshev equation with respect to q, we get 1 − q2  2 dq dx



d2 dy(q) d2 q y(q) − 2 dx dq dx 2

 −

q d y(q) + n 2 y(q) = 0 dx

dq dx

or 

d2 dy(q) d2 q (1 − q ) y(q) − dx 2 dq dx 2 2

or



 2 dq d dq 2 −q y(q) + n y(q) =0 dx dx dx

 2  2   dy d q dq dq d2 y 2 dq +n + P 2 −q y = 0. −P dx dx 2 dx dx dx dx

This is the second-order differential equation satisfied by the u n .

123

(3.4)

Certain families of polynomials arising...

The second family of Chebyshev polynomials satisfy the differential equation (1 − t 2 )y  (t) − 3t y  (t) + n(n + 2)y = 0 so that the polynomials vn satisfy −P

 2  2   dy dq d q dq dq d2 y + n(n + 2) + P − 3q y = 0. dx dx 2 dx 2 dx dx dx

(3.5) 

√ Theorem 6 Define polynomials an , bn−1 by an + bn−1 P = λn2 for n ∈ N. Then an = b0−1 (bn − a1 bn−1 ), and bn (t) satisfies 0 = t 2 (t 2 − a12 )(a1 t − a1 )y      + − a1 t 3 (t 2 − a12 ) + 2(1 − n)t 3 − 3a1 a1 t 2 + a12 (2n + 1)t (a1 t − a1 ) y     + − a1 t 2 −nt 2 − a1 a1 (n + 2)t + 2a12 (n + 1)       + n(n − 1)+ n(a1 )2 − a1 a1 (n + 2) t 2 − na1 a1 (2n + 1) t (a1 t − a1 ) y. Proof The proof follows along the lines of the technique used in Arfken and Weber’s book “Mathematical Methods for Physicists” when deriving the second-order differential for the Legendre polynomials from the recurrence relation for these polynomials. We omit the details. Setting α = 1 we get the differential equation that the bn satisfy as follows: 0 = t 2 (t 2 − a12 )(a1 t − a1 )y      + − a1 t 3 (t 2 − a12 ) + 2(1 − n)t 3 − 3a1 a1 t 2 + a12 (2n + 1)t (a1 t − a1 ) y     + − a1 t 2 −nt 2 − a1 a1 (n + 2)t + 2a12 (n + 1)       + n(n − 1) + n(a1 )2 − a1 a1 (n + 2) t 2 − na1 a1 (2n + 1) t (a1 t − a1 ) y, (3.6) t2 − 1 p . . Note that t 2 − a12 = 2 β +1 2(β − 1) After dividing this by t and simplifying, this differential equation becomes

where a1 (t) := √

   0 = t t 2 + 1 t 4 − 2βt 2 + 1 y    − (2n − 3)t 6 + t 4 (−4βn + 2n − 5) + t 2 (4β − 4βn + 2n + 3) + 2n + 1 y    − 2 2nt 5 + nt 3 (β + (β + 1)n + 5) + nt (−β + (β + 1)n + 1) y. (3.7)

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Thus we have written the above differential equation in the form P 1 (t)y  + Q n (t)y  + Rn (t)y = 0, with    P 1 (t) = t t 2 + 1 t 4 − 2βt 2 + 1 ,   Q n (t) = − (2n − 3)t 6 + t 4 (−4βn + 2n − 5) + t 2 (4β − 4βn + 2n + 3) + 2n + 1    = −2(n − 1) t 2 + 1 t 4 − 2βt 2 + 1 + t 6 + (4β + 3)t 4 − 5t 2 − 3,   Rn (t) = −2n 2t 5 + t 3 (β + (β + 1)n + 5) + t (−β + (β + 1)n + 1) . Note Q n+1 (t) = Q n (t) − 2P 1 t −1 . The above derivation required that α ∈ / Z.



Proposition 7 The polynomials bn (t) satisfy (3.7) for all n  1. Proof The proof is by induction on n  0.

  In order to put this into Sturm–Liouville like form, we first divide by t t 2 + 1 4  t − 2βt 2 + 1 , and then multiply the differential equation above by the integrating factor 

 exp

3/2 4 t − 2βt 2 + 1 Q n (t)  dt =  2 . P 1 (t) t + 1 t 2n+1

We obtain d dt

 3/2   4 1/2 t 4 − 2βt 2 + 1 Rn (t) t − 2βt 2 + 1 dy   y = 0. + 2 2 2n+2 2 2n+1 dt (t + 1) t t +1 t

(3.8)

If we set 4 3/2 t − 2βt 2 + 1 d d  :=  2 , 2n+1 dτn dt t +1 t we get  τn (t) =

2  t + 1 t 2n+1  3/2 dt, t 4 − 2βt 2 + 1

which is an elliptic integral and can be expressed in the form of a series about t = 0. We will not write this out as it does not seem to simplify matters. Then (3.8) becomes 2  1/2 4 t + 1 t 2n+1 d2 y Rn (t) t − 2βt 2 + 1 + y = 0.  3/2 2 2 + 1)2 t 2n+2 4 2 dτ (t t − 2βt + 1 n

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Certain families of polynomials arising...

or

2 4 t (τn ) − 2βt 2 (τn ) + 1 Rn (t (τn )) d2 y y = 0. (3.9) + dτn2 (t 2 (τn ) + 1)3 t 4n+3 (τn ) √ Theorem 8 Define polynomials cn , dn−1 by cn + dn P = λn1 for n ∈ N. Then dn (t) satisfies P 2 (t)y  + Q 2n (t)y  + Rn2 (t)y = 0 with P 2 (t) = t (1 − t 2 )(t 4 − 2βt 2 + 1), Q 2n (t) = (2n − 3)t 6 − t 4 (4βn + 2n − 5) + t 2 (−4β + 4βn + 2n + 3) − 2n − 1,   Rn2 (t) = 2nt β − (β − 1)n + t 2 (β + (β − 1)n − 5) + 2t 4 + 1 .

Proof We next will derive second-order equations satisfied by cn √ linear differential √ and dn which are defined by cn + dn p(t) = (c1 + d1 p)n , for t2 + 1 c0 = 1, d0 = 0, c1 = √ d1 = 2(β + 1)



β −1 . 2

(3.10)

First we have    √ √ cn z n + dn z n p = (c1 + d1 p)n z n n0

n0

n0

=



n0

=



(−ı)n (λ1 (−β, ıt))n z n √ (−ı)n (an (−β, ıt) + bn (−β, ıt) p)z n

n0

by the following formula 

  −t 2 + 1 −β − 1 √ −ıλ1 (−β, ıt) = −ı √ p + 2 2(−β + 1)  t2 − 1 β + 1√ =√ p + 2 2(β − 1) = λ2 (β, t), where ı =

(3.11)

√ −1. Thus cn (β, t) = (−ı)n an (−β, ıt), dn (β, t) = (−ı)n bn (−β, ıt).

This finishes the proof.



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B. Cox, K. Zhao Acknowledgements The two authors would like to thank the Mittag-Leffler Institute in Djursholm, Sweden, for its hospitality and great working environment where part of this work was done.

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