Legendre and Chebyshev Polynomials on a Hyperelliptic Curve: an Elementary Account

1. Introduction

Let \(p(x) \in \C[x]\) satisfy \(p(0) \neq 0\), let \(A = \C[x^{\pm 1}, u \mid u^2 = p(x)]\) be the coordinate ring of the affine hyperelliptic curve \(u^2 = p(x)\) with the fibre over \(x=0\) removed, and let \(\pa = u\,\tfrac{d}{dx}\); when the roots of \(p\) are simple, \(\Der(A) = A\pa\) [8]. Two invariants of this data carry orthogonal polynomials. The first is the quotient \(A/\pa A\), whose reduction to a basis is performed by the three-term recurrence of the Legendre polynomials evaluated at the branch parameter; this is equation (2.22) of Cox and Zhao [10]. The second is the group of units \(A^\times\), determined for the quartic family in [8, Theorem 13] and described in [10, Theorem 5] by pairs \((u_n, v_{n-1})\) satisfying a polynomial Pell equation, of which the Chebyshev pairs \((T_n, U_{n-1})\) are the distinguished instance.

Both statements are proved in [10] inside a study of the universal central extension of \(\Der(A)\), for general branch data, and the mechanism producing them is not separable from that setting there. It is separable, and it is elementary. The purpose of this account is to carry it out.

We fix the three smallest families, \[ p(x) = x-a, \qquad p(x) = x^2-2ax+1, \qquad p(x) = x^4-2ax^2+1, \] and compute \(A/\pa A\) for each from the two formulas \(\pa(x^n) = n x^{n-1}u\) and \(\pa(x^n u) = n x^{n-1}p + \tfrac12 x^n p'\), which is the whole of the apparatus. The three dimensions themselves are instances of a closed formula: for \(p\) of degree \(n\) with simple nonzero roots, \(\dim A/\pa A = n+1\). On the differential side this is Bremner [3, Theorem 3.4], who exhibits the basis \(x^{-1}dx,\, x^{-1}u\,dx,\, \ldots,\, x^{-n}u\,dx\) of \(\Omega^1_A/dA\); on the side of \(A/\pa A\) it is [8, Theorem 9], quoted as [10, Theorem 2]. What the computation below supplies is the reduction that produces the number, class by class, and the recurrence that performs it, which for the quadratic family is Bonnet's. The dimension is then the number of classes the recurrence fails to determine from their neighbours: one for each integer at which the leading or the trailing coefficient vanishes and the reduction chain breaks, together with the free initial values of a chain that runs through all of \(\Z\) unbroken. Nothing below uses Lie theory, cohomology, or the theory of orthogonal polynomials beyond the definitions.

Sections ? to ? carry out the three computations, and §? treats the degenerate parameters \(a=\pm1\), at which \(p\) acquires a repeated root and \(A\) ceases to be a domain. One expects the dimension to drop there. It does not: the leading and trailing coefficients of the recurrence are \(n+1\) and \(n\), which vanish at the same two integers for every \(a\), so the chain breaks in the same places and \(\dim A/\pa A\) is \(3\) and \(5\) as in the smooth case (Propositions 6.2 and 6.3). We have not found this recorded. What does degenerate is the geometric side, and §? says where.

Section ? recovers the same three dimensions from the topology of the curve, through the classical count \(\dim \Omega^1_A/dA = 2g+r-1\), and exhibits the isomorphism \([f] \mapsto [f\,dx/u]\) responsible for the agreement. The puncture count is the step at which this calculation is usually got wrong, and Lemma 7.2 states it: \(\Spec A\) omits the fibre over \(x=\infty\) as well as the fibre over \(x=0\), because \(x\) itself lies in \(A\).

Sections ? and ? identify the two classical families. The second reads [10, Theorem 5] in the ring \(\C[a,w \mid w^2 = a^2-1]\) generated by the branch parameter, where the Pell equation is the classical one and the unit \(\varepsilon = a + w\) is fundamental. Placing the two families side by side puts one polynomial in two roles: the \(1-2az+z^2\) that carries the Legendre generating function as \((1-2az+z^2)^{-1/2}\) carries the Chebyshev one as \((1-2az+z^2)^{-1}\), so that \(U_n(a) = \sum_{k=0}^n P_k(a)P_{n-k}(a)\) (Proposition 9.3). That identity is classical. What we have not found stated is that its two sides come from the same curve, once through its defining equation and once through the norm form of the Pell equation for its parameter.

Section ? states a case we could not settle.

For the Lie-theoretic setting from which these quotients come we refer to Kassel and Loday [13] and Kassel [12] for the general theory of universal central extensions and Kähler differentials, to Krichever and Novikov [14] and Schlichenmaier [17] for the algebras attached to a curve, and to [5, 8, 9, 7] and [1, 16] for the hyperelliptic and superelliptic cases. For the classical polynomials we follow Szegő [18]; Chihara [4], Andrews, Askey and Roy [2] and Ismail [11] are standard alternatives.

2. The ring, the derivation, and the two formulas

Differentiating \(u^2 = p(x)\) gives \(2u\,\tfrac{du}{dx} = p'(x)\), whence \(\pa(u) = u\,\tfrac{du}{dx} = \tfrac12 p'(x)\), and with the Leibniz rule this yields the two formulas on which everything below rests: for every \(n \in \Z\), \begin{align} \pa(x^n) &= n\,x^{n-1}u, \label{eq:master1}\\ \pa(x^n u) &= n\,x^{n-1}p(x) + \tfrac{1}{2}\,x^n\,p'(x), \label{eq:master2} \end{align} the first at once and the second from \(\pa(x^n u) = n x^{n-1}u^2 + x^n\,\pa(u)\) together with \(u^2 = p(x)\).

The images (?) lie in \(\C[x^{\pm1}]u\) and the images (?) lie in \(\C[x^{\pm1}]\), so the quotient splits, \begin{equation}\label{eq:split} A/\pa A \;=\; \bigl(\C[x^{\pm 1}] \big/ \langle \pa(x^n u)\rangle\bigr) \;\oplus\; \bigl(\C[x^{\pm 1}]u \big/ \langle \pa(x^n)\rangle\bigr), \end{equation} and the second summand is independent of \(p\).

All the content is therefore in the first summand of (?), which depends on \(p\) through (?).

3. The linear case

For \(p(x) = x - a\) with \(a \neq 0\) we have \(p'(x)=1\), and (?) reads \begin{equation}\label{eq:rec_linear} \pa(x^n u) = \bigl(n+\tfrac{1}{2}\bigr)x^n - a\,n\,x^{n-1}. \end{equation} This is not a hyperelliptic curve, the genus being zero and \(p\) linear, and we treat it first because the mechanism is visible in two lines.

Two integers carry the dimension here, \(n=0\) in each of (?) and (?), so the count of breaks is the answer. That is an accident of the short recurrence. From §? on, the recurrence splits by parity, and one of the two halves runs through all of \(\Z\) without breaking anywhere and still contributes two classes. What carries over is the method: locate the breaks, then count what the chains leave behind.

4. The quadratic case

Let \(p(x) = x^2 - 2ax + 1\), of discriminant \(4(a^2-1)\), so that \(p\) has simple roots exactly when \(a \neq \pm 1\). Then \(p'(x) = 2x-2a\) and (?) gives \begin{equation}\label{eq:rec_quad} \pa(x^n u) = (n+1)\,x^{n+1} - a(2n+1)\,x^n + n\,x^{n-1}, \end{equation} which is the relation of [10, eq. (2.22)].

5. The quartic case

Let \(p(x) = x^4 - 2ax^2 + 1\), again with simple roots exactly for \(a \neq \pm 1\). Here \(p'(x) = 4x^3 - 4ax\) and (?) gives \begin{equation}\label{eq:rec_quart} \pa(x^n u) = (n+2)\,x^{n+3} - 2a(n+1)\,x^{n+1} + n\,x^{n-1}. \end{equation} The exponents differ by \(2\), so the even and the odd powers of \(x\) never interact and the first summand of (?) splits again, into an even sector and an odd sector.

6. The degenerate parameters

At \(a=1\) the quadratic \(p(x) = (x-1)^2\) has a repeated root, the relation \(u^2 = (x-1)^2\) factors as \((u-(x-1))(u+(x-1))=0\), and \(A\) is no longer a domain: the curve is two copies of \(\C^\times\) meeting at \((x,u)=(1,0)\).

The identification is a fibre product and not a direct product, which would describe two disjoint curves. The ring degenerates and the curve degenerates; the quotient does not.

The quotient \(A/\pa A\) is insensitive to the degeneration because the recurrence is a polynomial identity in \(a\) whose break structure is fixed by the integer coefficients alone. The geometry is not insensitive. At \(a=\pm1\) the curve acquires a node, the smooth-curve count of Theorem 7.1 below does not apply, and the isomorphism of Proposition 7.4 fails, since \(dx/u\) ceases to generate \(\Omega^1_A\) at the node. We claim nothing about \(\Omega^1_A/dA\) at the node; computing it directly would measure how far the two sides separate there.

One route to the two propositions above deserves a warning, because it is the natural one and gives the wrong answer. Instead of running the recurrence at \(a=1\) one may pass to the normalization, replace \(A\) by its branches and correct by the first Betti number of the dual graph. On a single branch \(\pa_1(x^n) = n(x^n - x^{n-1})\) identifies \([x^n]\) with \([x^{n-1}]\) only for \(n\neq 0\); the chain breaks at \(n=0\) as everywhere else, so \(\dim B/\pa_1 B = 2\) rather than \(1\). Two branches give \(4\). The fibre product condition then costs exactly one: it does not shrink \(\pa A\), because a constant may be added to either component without changing its image under \(\pa_1\), so \(\pa A\) is all of \(\pa_1 B \times \pa_1 B\), while \(A\) itself has codimension one in \(B \times B\). The answer is again \(3\). Reading the identification as unconditional loses that break and costs one dimension.

7. The same dimensions from the topology of the curve

Let \(\Omega^1_A\) be the \(A\)-module of Kähler differentials and \(dA\) its submodule of exact forms. The quotient \(\Omega^1_A/dA\) is computed without any recurrence. On a curve there are no forms of degree two, so it is the first algebraic de Rham cohomology of \(\Spec A\); it is also the first cyclic homology \(HC_1(A)\) in the sense of [15], and it is in that shape that it enters the theory of central extensions, as the centre of the universal one [13, 12]. The count itself is topological and classical.

Algebraic and topological de Rham cohomology agree on a smooth affine variety, and a genus-\(g\) surface with \(r\) punctures has the homotopy type of a wedge of \(2g+r-1\) circles. Applying the theorem requires counting those punctures.

The clause about \(x=\infty\) is the one that is missed by the argument that \(A\) inverts \(x\) and therefore deletes only the fibre over \(0\).

Proposition 7.4 identifies the two invariants once and for all, so the basis of \(\Omega^1_A/dA\) recorded in [3, Theorem 3.4] transports to \(A/\pa A\): the class \(x^{-1}dx\) corresponds to \([x^{-1}u]\) and \(x^{-i}u\,dx\) to \([x^{-i}p]\), for \(i = 1,\ldots,\deg p\).

8. The Legendre family

Recall Bonnet's recurrence for the Legendre polynomials, \((n+1)P_{n+1}(t) = (2n+1)\,t\,P_n(t) - n\,P_{n-1}(t)\), with \(P_0=1\) and \(P_1(t)=t\) [18, §4.7]. The following is [10, eq. (2.22)], in the notation of §?.

The numbers \(1,3,13,63\) of Example 4.2 are \(P_0(3),\dots,P_3(3)\).

The polynomial \(1-2az+z^2\) here is the reciprocal of the curve polynomial \(x^2-2ax+1\), and since that polynomial is palindromic it is its own reciprocal, so the two are the same polynomial in different variables. The curve chooses the generating function, and the generating function chooses the family.

9. The Chebyshev family and the Pell equation

Take the quartic in the normalisation \(p(t) = (t^4-2\beta t^2+1)/(\beta^2-1)\), the case of Date, Jimbo, Kashiwara and Miwa, whose algebras are studied in [6]. Its unit group was determined by Cox, Guo, Lu and Zhao [8, Theorem 13]: \(A^\times \cong \C^\times \times \Z^3\). Writing \(u_n + v_{n-1}\sqrt{p} = \lambda^n\) for a distinguished unit \(\lambda\), the pair \((u_n,v_{n-1})\) obeys \(y_{n+2} - 2q\,y_{n+1} + y_n = 0\) with \(q = (t^2-\beta)/ \sqrt{\beta^2-1}\), which is the Chebyshev recursion in \(q\); this is [10, Theorem 5].

The same statement in the ring generated by the branch parameter is the classical Pell equation, and it is worth seeing on its own, because the roots of \(x^2-2ax+1\) are \(a \pm \sqrt{a^2-1}\), of product \(1\). Set \[ R \coloneqq \C[a, w]\big/(w^2 - (a^2-1)), \] a ring of the shape of Definition 2.1 one degree lower, with the branch parameter as its variable. Both statements below are classical in this ring: the first is the polynomial Pell equation and the second its solution by Chebyshev pairs. We give the arguments because they are short and because they are what [10, Theorem 5] becomes here.

Theorem 9.2 is [10, Theorem 5] read in \(R\), where \(q\) becomes the variable \(a\) and the recursion becomes the classical one. Placing it beside Theorem 8.2 puts the two generating functions side by side.

The identity (?) between the two generating functions is classical and belongs to the theory of the Gegenbauer family, of which both are members. We record its source here: the same polynomial \(1-2az+z^2\) occurs as the defining equation of the curve, where its reciprocal square root reduces \(A/\pa A\), and as the norm form of the Pell equation for the branch parameter, where its reciprocal enumerates the units. The exponents \(-\tfrac12\) and \(-1\) are the square root in the curve equation and the norm in the Pell equation.

10. A case we could not settle

The next family is \(u^2 = x^N - 1\) with \(N\geq3\), whose branch points are the \(N\)-th roots of unity. Here \(p'(x) = Nx^{N-1}\) and (?) gives \[ \pa(x^n u) = \Bigl(n+\tfrac{N}{2}\Bigr)x^{n+N-1} - n\,x^{n-1}, \] a two-term relation identifying \([x^{n+N-1}]\) with a multiple of \([x^{n-1}]\), in steps of \(N\), except at \(n=0\) and, when \(N\) is even, at \(n=-N/2\). Being two-term, it admits no orthogonal family in the sense of Favard's theorem [4, Theorem 4.4], and the reduction coefficients are products of linear factors.

Cox and Zhao treat the general branch data in [10], where associated Legendre polynomials appear; the question here is which family survives the reciprocal symmetry \(p(x) = x^N p(1/x)\), which the deformed equation has exactly when \(N\) is even. For even \(N\) we expect a Chebyshev or Dickson family, on the grounds that §? produced one from the same shape of equation one degree lower; for odd \(N\) the deformation is not self-reciprocal and we have no expectation to offer. We have no proof in either case and state this as an expectation only.

Acknowledgements

The first author is supported by FAPESP grant 2024/14914-9.