Twisted Traces and Positive Forms on Quantized Kleinian Singularities of Type A

Following [Beem C., Peelaers W., Rastelli L., Comm. Math. Phys. 354 (2017), 345-392, arXiv:1601.05378] and [Etingof P., Stryker D., SIGMA 16 (2020), 014, 28 pages, arXiv:1909.13588], we undertake a detailed study of twisted traces on quantizations of Kleinian singularities of type $A_{n-1}$. In particular, we give explicit integral formulas for these traces and use them to determine when a trace defines a positive Hermitian form on the corresponding algebra. This leads to a classification of unitary short star-products for such quantizations, a problem posed by Beem, Peelaers and Rastelli in connection with 3-dimensional superconformal field theory. In particular, we confirm their conjecture that for $n\le 4$ a unitary short star-product is unique and compute its parameter as a function of the quantization parameters, giving exact formulas for the numerical functions by Beem, Peelaers and Rastelli. If $n=2$, this, in particular, recovers the theory of unitary spherical Harish-Chandra bimodules for ${\mathfrak{sl}}_2$. Thus the results of this paper may be viewed as a starting point for a generalization of the theory of unitary Harish-Chandra bimodules over enveloping algebras of reductive Lie algebras [Vogan Jr. D.A., Annals of Mathematics Studies, Vol. 118, Princeton University Press, Princeton, NJ, 1987] to more general quantum algebras. Finally, we derive recurrences to compute the coefficients of short star-products corresponding to twisted traces, which are generalizations of discrete Painlev\'e systems.


Introduction
The notion of a short star-product for a filtered quantization A of a hyperKähler cone was introduced by Beem, Peelaers and Rastelli in [2] motivated by the needs of 3-dimensional superconformal field theory (under the name "star-product satisfying the truncation condition"); this is an algebraic incarnation of non-holomorphic SU(2)-symmetry of such cones. Roughly speaking, these are star-products which have fewer terms than expected (in fact, as few as possible). The most important short star-products are nondegenerate ones, i.e., those for which the constant term CT(a * b) of a * b defines a nondegenerate pairing on A = gr A. Moreover, physically the most interesting ones among them are those for which an appropriate Hermitian version of this pairing is positive definite; such star-products are called unitary. Namely, short This paper is a contribution to the Special Issue on Representation Theory and Integrable Systems in honor of Vitaly Tarasov on the 60th birthday and Alexander Varchenko on the 70th birthday. The full collection is available at https://www.emis.de/journals/SIGMA/Tarasov-Varchenko.html star-products arising in 3-dimensional SCFT happen to be unitary, which is a motivation to take a closer look at them.
In fact, in order to compute the parameters of short star-products arising from 3-dimensional SCFT, in [2] the authors attempted to classify unitary short star-products for even quantizations of Kleinian singularities of type A n−1 for n ≤ 4. Their low-degree computations suggested that in these cases a unitary short star-product should be unique for each quantization. While the A 1 case is easy (as it reduces to the representation theory of SL 2 ), in the A 2 case the situation is already quite interesting. Namely, in this case an even quantization depends on one parameter κ, and Beem, Peelaers and Rastelli showed that (at least in low degrees) short star-products for such a quantization are parametrized by another parameter α. Moreover, they computed numerically the function α(κ) expressing the parameter of the unique unitary short star-product on the parameter of quantization [2, Fig. 2], but a formula for this function (even conjectural) remained unknown.
These results were improved upon by Dedushenko, Pufu and Yacoby in [5], who computed the short star-products coming from 3-dimensional SCFT in a different way. This made the need to understand all nondegenerate short star-products and in particular unitary ones less pressing for physics, but it remained a very interesting mathematical problem.
Motivated by [2], the first and the last author studied this problem in [6]. There they developed a mathematical theory of nondegenerate short star-products and obtained their classification. As a result, they confirmed the conjecture of [2] that such star-products exist for a wide class of hyperKähler cones and are parametrized by finitely many parameters. The main tool in this paper is the observation, due to Kontsevich, that nondegenerate short star-products correspond to nondegenerate twisted traces on the quantized algebra A, up to scaling. The reason this idea is effective is that traces are much more familiar objects (representing classes in the zeroth Hochschild homology of A), and can be treated by standard techniques of representation theory and noncommutative geometry. However, the specific example of type A n−1 Kleinian singularities and in particular the classification of unitary short star-products was not addressed in detail in [6].
The goal of the present paper is to apply the results of [6] to this example, improving on the results of [2]. Namely, we give an explicit classification of nondegenerate short star-products for type A n−1 Kleinian singularities, expressing the corresponding traces of weight 0 elements (i.e., polynomials P (z) in the weight zero generator z) as integrals iR P (x)w(x)|dx| of P (x) against a certain weight function w(x). As a result, the corresponding quantization map sends monomials z k to p k (z), where p k (x) are monic orthogonal polynomials with weight w(x) which belong to the class of semiclassical orthogonal polynomials. If n = 1, or n = 2 with special parameters, they reduce to classical hypergeometric orthogonal polynomials, but in general they do not. We also determine which of these short star-products are unitary, confirming the conjecture of [2] that for even quantizations of A n−1 , n ≤ 4 a unitary star product is unique. Moreover, we find the exact formula for the function α(κ) whose graph is given in Fig. 2 of [2]: In particular, this recovers the value α − 1 4 = 1 4 − 2 π 2 predicted in [2] and confirmed in [4,5]. It would be very interesting to develop a similar theory of positive traces for higher-dimensional quantizations, based on the algebraic results of [6]. It would also be interesting to extend this analysis from the algebra A to bimodules over A (e.g., Harish-Chandra bimodules, cf. [11]). Finally, it would be interesting to develop a q-analogue of this theory. These topics are beyond the scope of this paper, however, and are subject of future research. For instance, the q-analogue of our results for Kleinian singularities of type A will be worked out by the second author in a forthcoming paper [9]. Remark 1.1. We show in Example 4.10 that for n = 2 the theory of positive traces developed here recovers the classification of irreducible unitary spherical representations of SL 2 (C) [15]. Moreover, this can be extended to the non-spherical case if we consider traces on Harish-Chandra bimodules over quantizations (with different parameters on the left and the right, in general) rather than just quantizations themselves. One could expect that a similar theory for higherdimensional quantizations, in the special case of quotients of U (g) by a central character (i.e., quantizations of the nilpotent cone) would recover the theory of unitary representations of the complex reductive group G with Lie algebra g. This suggests that the theory of positive traces on filtered quantizations of hyperKähler cones may be viewed as a generalization of the theory of unitary Harish-Chandra bimodules for simple Lie algebras. A peculiar but essential new feature of this generalization (which may scare away classical representation theorists), is that a given simple bimodule may have more than one Hermitian (and even more than one unitary) structure up to scaling (namely, unitary structures form a cone, often of dimension > 1), and that a bimodule which admits a unitary structure need not be semisimple.
Remark 1.2. The second author studied the existence of unitary star-products for type A n−1 Kleinian singularities in [8] and obtained a partial classification of quantizations that admit a unitary star-product. That paper also contains examples of non-semisimple unitarizable bimodules. The present paper has stronger results: it contains a complete description of the set of unitary star-products for any type A n−1 Kleinian singularity.
The organization of the paper is as follows. Section 2 is dedicated to outlining the algebraic theory of filtered quantizations and twisted traces for Kleinian singularities of type A, following [6]. In Section 3 we introduce our main analytic tools, representing twisted traces by contour integrals against a weight function. In this section we also use this weight function to study the orthogonal polynomials arising from twisted traces. In Section 4, using the analytic approach of Section 3, we determine which twisted traces are positive. In particular, we confirm the conjecture of [2] that a positive trace is unique up to scaling for n ≤ 4 (for the choice of conjugation as in [2]), and find the exact dependence of the parameter of the positive trace on the quantization parameters for n = 3 and n = 4, which was computed numerically in [2]. 1 Finally, in Section 5 we discuss the problem of explicit computation of the coefficients a k , b k of the 3-term recurrence for the orthogonal polynomials arising from twisted traces, which appear as coefficients of the corresponding short star-product. Since these orthogonal polynomials are semiclassical, these coefficients can be computed using non-linear recurrences which are generalizations of discrete Painlevé systems.

Filtered quantizations and twisted traces 2.1 Filtered quantizations
Let X n be the Kleinian singularity of type A n−1 . Recall that where Z/n acts by p → e 2πi/n p, q → e −2πi/n q. Thus This algebra has a grading defined by the formulas deg(p) = deg(q) = 1, thus The Poisson bracket is given by {p, q} = 1 n and on A takes the form Also recall that filtered quantizations A of A are generalized Weyl algebras [1] which look as follows. Let P ∈ C[x] be a monic polynomial of degree n. Then A = A P is the algebra generated by u, v, z with defining relations and filtration defined by (2.1). Thus we have i.e., the quasiclassical limit indeed recovers the algebra A with the above Poisson bracket. Note that we may consider the algebra A P for a polynomial P that is not necessarily monic. However, we can always reduce to the monic case by rescaling u and/or v. Also by transformations z → z + β we can make sure that the subleading term of P is zero, i.e., Thus the quantization A depends on n − 1 essential parameters (the roots of P , which add up to zero).
The algebra A decomposes as a direct sum of eigenspaces of ad z: If b ∈ A m , we will say that b has weight m. The weight decomposition of A can be viewed as a C × -action; namely, for t ∈ C × let be the automorphism of A given by

Even quantizations
Let s be the automorphism of A given by in other words, we have s = g (−1) n . Thus gr s : A → A equals (−1) d , where d is the degree operator. Recall [6,Section 2.3], that a filtered quantization A is called even if it is equipped with an antiautomorphism σ such that σ 2 = s and gr σ = i d , and that σ is unique if exists [6,Remark 2.10]. This means that σ(z) = −z, σ(u) = i n u, σ(v) = i n v. It is easy to see that σ exists if and only if i.e., P contains only terms x n−2i . Thus even quantizations of A are parametrized by [n/2] essential parameters, and all quantizations for n ≤ 2 are even.

Twisted traces
Let A = A P be a filtered quantization of A. Recall [6, Section 3.1] that a g t -twisted trace on A is a linear map T : A → C such that T (ab) = T (bg t (a)), where g t is given by (2.2). It is shown in [6, Section 3], that (s-invariant) nondegenerate twisted traces, up to scaling, correspond to (s-invariant) nondegenerate short star-products on A.
Let us classify g t -twisted traces 2 T on A. The answer is given by the following proposition.  (2) T S z − 1 2 P z − 1 In particular, any twisted trace is automatically s-invariant.
Proof . Suppose T satisfies (1), (2). It is enough to check that The equality T (zb) = T (bz) says that T (A j ) = 0 for j = 0, which is condition (1). By (1), it is enough to check the equality T (ub) = t −1 T (bu) for b ∈ A −1 . In this case b = vS z + 1 2 for some polynomial S. We have T (ub) = T uvS z + 1 , which yields the desired identity using (2).
Conversely, the same argument shows that if T is a g t -twisted trace then (1), (2) hold.

Thus we get
Corollary 2.4. The space of g t -twisted traces on A is naturally isomorphic to the space and has dimension n if t = 1 and dimension n − 1 if t = 1.

The formal Stieltjes transform
There is a useful characterization of the space g t -twisted traces in terms of generating functions. Given a linear functional T on C[z], its formal Stieltjes transform is the generating function Proposition 2.5. The formal Stieltjes transform of a g t -twisted trace on A satisfies and this establishes an isomorphism of the space of g t -twisted traces with the space of polynomials of degree ≤ n − 1 (for t = 1) or ≤ n − 2 (for t = 1).
Proof . We may write In the final expression, the second term is the image under T of a polynomial in z and x, while the first term expands as , this establishes an injective map from g t -twisted traces to polynomials of degree < deg(P ). This establishes the conclusion for t = 1, with surjectivity following by dimension count. Finally, for t = 1, we need simply observe that for any and thus the polynomial has degree < deg(P ) − 1, and surjectivity again follows from dimension count.
Remark 2.6. It is easy to see that the map F (x) → F x + 1 2 − tF x − 1 2 acts triangularly on x −1 C x −1 , of degree 0 (with nonzero leading coefficients) if t = 1 and degree −1 (ditto) if t = 1, letting one see directly that there is a unique solution of P (x) F x+ 1 2 −tF x− 1 2 = R(x) for any polynomial R satisfying the degree constraint.
Remark 2.7. A similar argument establishes an isomorphism between linear functionals satis- or, for t = 1, between linear functionals satisfying and formal series satisfying 3 An analytic construction of twisted traces 3.1 Construction of twisted traces when all roots of P (x) satisfy | Re α| < 1 When P (x) satisfies the equation P (−x) = (−1) n P (x) (the condition for existence of a conjugation ρ) the polynomial P(X) has real coefficients.
Proposition 3.1. Assume that every root α of P (x) satisfies | Re α| < 1 2 . Also suppose first that t does not belong to R >0 \ {1}, i.e., Re c ∈ (0, 1) or c = 0. Then every g t -twisted trace is given by 3 where w is the weight function defined by the formula where G is a polynomial of degree ≤ n − 1 and Proof . It is easy to see that the function w(x) enjoys the following properties: (2) |w(x)| decays exponentially and uniformly when Im x tends to ±∞; Indeed, (2) holds because the degree of G is strictly less than the degree of P and either Re(c) > 0 or G(0) = 0, and (3) holds because all roots of P are in the strip |Reα| < 1 2 . Let T (R(z)) := iR R(x)w(x)|dx|. We should check that Here |dx| denotes the Lebesgue measure on the imaginary axis.
We have But this integral vanishes by the Cauchy theorem since S(x)P (x)w x + 1 2 is holomorphic when | Re x| ≤ 1 2 and decays exponentially as Im x → ±i∞. By Corollary 2.4, the space of polynomials G(X) has the same dimension as the space of g t -twisted traces, and the map sending polynomials G to traces is clearly injective, so we have described all traces. Now consider the remaining case t ∈ R \ {1}, i.e., c ∈ iR \ {0}. In this case the function w(x) does not decay at +i∞, so the integral in Proposition 3.1 is not convergent. However, we can write the formula for T (R(z)) as follows, so that it makes sense in this case: Alternatively, one may say that T (R(z)) is the value of the Fourier transform of the distribution R(−iy)w(0, −iy) at the point ic (it is easy to see that this Fourier transform is given by an analytic function outside of the origin). We then have the following easy generalization of Proposition 3.1: With this modification, Proposition 3.1 is valid for all t.
Consider now the special case of even quantizations. Recall [6, Section 3.3] that nondegenerate even short star-products on A correspond to nondegenerate s-twisted σ-invariant traces T on various even quantizations A of A, up to scaling. So let us classify such traces. As shown above, s-twisted traces T correspond to w(x) such that w(x + 1) = (−1) n w(x). Also, it is easy to see that such T is σ-invariant if and only if T (R(z)) = T (R(−z)). We have So T is σ-invariant of and only if w(x) = w(−x). Thus we have the following proposition. Proposition 3.3. Suppose that A is an even quantization of A. Then s-twisted σ-invariant traces T are given by the formula

Relation to orthogonal polynomials
We continue to assume that all roots of P are in the strip |Reα| < 1 2 . Assume that the trace T is nondegenerate, i.e., the form (a, b) → T (ab) defines an inner product on A nondegenerate on each filtration piece. This holds for generic parameters, e.g., specifically if w(x) is nonnegative on iR. Let φ : A → A be the quantization map defined by T (see [6,Section 3]). Namely, the form (a, b) allows us to split the filtration, and φ is precisely the splitting map. Thus, φ(z k ) = p k (z), where p k are monic orthogonal polynomials for the inner product Recall [14] that these polynomials satisfy a 3-term recurrence Thus the corresponding short star-product z * z k has the form Thus the numbers a k , b k are the matrix elements of multiplication by z in weight 0 for the short star-product attached to T . More general matrix elements of multiplication by u, v, z for this short star-product are computed similarly. In other words, to compute the short star-product attached to T , we need to compute explicitly the coefficients a k , b k and their generalizations. This problem is addressed in Section 5.
It is more customary to consider orthogonal polynomials on the real (rather than imaginary) axis, so let us make a change of variable x = −iy. Then we see that the monic polynomials P k (y) := i k p k (−iy) are orthogonal under the inner product where w(y) := w(−iy). Then the 3-term recurrence looks like Example 3.4. Let n = 1, P (x) = x, so P(X) = X + 1. Then a nonzero twisted trace exists if and only if c = 0, in which case it is unique up to scaling, and the corresponding weight function is The corresponding orthogonal polynomials P k (y) are the (monic) Meixner-Pollaczek polynomials with parameters λ = 1 2 , φ = πc [10, Section 1.7].
Example 3.5. Let n = 2, P (x) = x 2 + β 2 , so The space of twisted traces is 1-dimensional if c = 0 and 2-dimensional if c = 0. So for c = 0 the traces up to scaling are defined by the weight function and the limiting cases α → ±∞ along the real axis, which yield These formulas for the plus sign also define the unique up to scaling trace for c = 0; i.e., .
which defines an s-twisted trace. The corresponding orthogonal polynomials are semiclassical but not hypergeometric for n ≥ 3.

Conjugation-equivariant traces
Let now ρ be a conjugation on A (Section 2.3). Let us determine which g t -twisted traces are ρ-equivariant (see [6,Section 3.6]. . This happens if and only if w(x) is real on iR. Since w is meromorphic this means that w(x) = w(−x). So we have the following proposition.
Proposition 3.9. Suppose that A is a quantization of A with conjugation ρ. Then ρ-equivariant g t -twisted traces T on A are given by where w is as in Proposition 3.1 and Moreover, if A is even then σ-invariant traces among them correspond to the functions w with w(x) = w(−x).

Construction of traces when all roots of
From now on we suppose that i n P (x) is real on iR (so that the conjugations ρ ± are well defined).
In particular, the roots of P (x) are symmetric with respect to iR. Suppose that for all roots α of P (x) we have | Re α| ≤ 1 2 , and let us give a formula for twisted traces in this case. There are unique monic polynomials P * (x), Q(x) such that P (x) = P * (x)Q x + 1 2 Q x − 1 2 , all roots of P * (x) belong to the strip | Re x| < 1 2 and all roots of Q(x) belong to iR. Suppose that α 1 , . . . , α k are the roots of P * (x) and α k+1 , . . . , α m are the roots of Q x + 1 2 . Note that deg Q = n − m. Let P * (X) = m j=1 (X + e 2πiα j ), w(x) = e 2πicx G(e 2πix ) P * (e 2πix ) , where G(X) is a polynomial of degree at most m − 1 and G(0) = 0 when t = 1. We have (1) w(x + 1) = tw(x); (2) w(x)Q(x) is bounded on iR and decays exponentially and uniformly when Im x tends to ±∞; where w(x) is as above and φ is any linear functional.
Proof . The space of polynomials G has dimension m−δ 0 c , while the space of linear functionals φ has dimension deg Q = n − m. So the space of such linear functionals T has dimension n − δ 0 c . The space of all g t -twisted traces has the same dimension, so it is enough to prove that all linear functionals T of this form are g t -twisted traces. In other words, we should prove that Since w x + 1 2 P (x) is holomorphic on | Re x| ≤ 1 2 , we deduce that this integral is zero similarly to the proof of Proposition 3.1

Twisted traces in the general case
Let m(α) be the multiplicity of α as a root of P (x). Any linear functional φ on the space C[x]/ P (x)C[x] can be written as φ(S) = α, 0≤i<m(α) C αi S (i) (α), where C αi ∈ C. Therefore any g t -twisted trace T is given by Let P (x) be the following polynomial: all roots of P (x) belong to the strip | Re x| ≤ 1 2 and the multiplicity of a root α equals to So P (x) has the same degree as P (x) and its roots are obtained from roots of P (x) by the minimal integer shift into the strip | Re x| ≤ 1 2 . In particular, the roots of P (x) are symmetric with respect to iR.
Suppose that α ∈ C has real part bigger than 1 2 , S(x) is an arbitrary polynomial, Let r be the smallest positive integer such that Re(α) − r ≤ 1 2 . Then We can write a similar equation for α ∈ C with real part smaller than − 1 2 . Therefore where Φ(R) := Re a =0,k≥0 c ak R (k) (a), c ak ∈ C, T is a g t -twisted trace for the quantization defined by the polynomial P (x). Below we will abbreviate this sentence to " T is a trace for P ".
Let P • be the following polynomial: all the roots of P • belong to strip | Re x| ≤ 1 2 and the multiplicity of α, | Re α| ≤ 1 2 in P • equals the multiplicity of α in P . Since φ i,α are linearly independent for different i, α, we deduce that Φ = 0 if and only if T is a trace for P • .
So we have proved the following proposition: Proposition 3.11. Suppose that P is any polynomial, P is obtained from P by the minimal integer shift of roots into the strip | Re x| ≤ 1 2 , and P • is obtained from P by throwing out roots not in the strip | Re x| ≤ 1 2 . Then any twisted trace T on A P can be represented as T = Φ + T , where and T is a trace for P . Furthermore, if Φ = 0 then T is a trace for P • .
Remark 3.12. We may think about Proposition 3.11 as follows. When the roots of P lie inside the strip | Re x| < 1 2 , the trace of R(z) is given by the integral of R against the weight function w along the imaginary axis. However, when we vary P , as soon as its roots leave the strip | Re x| < 1 2 , poles of w start crossing the contour of integration. So for the formula to remain valid, we need to add the residues resulting from this. These residues give rise to the linear functional Φ.

Positivity of twisted traces 4.1 Analytic lemmas
We will use the following classical result: Lemma 4.1. Suppose that w(x) ≥ 0 is a measurable function on the real line such that w(x) < ce −b|x| for some c, b > 0. We also assume that w > 0 almost everywhere. Then polynomials are dense in the space L p (R, w(x)dx) for all 1 ≤ p < ∞.
Proof . Changing x to bx we can asssume that b = 1.
Fix p. Let 1 p + 1 q = 1. Since L p (R, w) * = L q (R, w), it suffices to show that any function f ∈ L q (R, w) such that R f (x)x n w(x)dx = 0 for all nonnegative integers n must be zero. Choose 0 < a < 1 p . We have e a|x| ∈ L p (R, w). Therefore f (x)e a|x| w(x) ∈ L 1 (R). Denote f (x)w(x) by F (x). Let F be the Fourier transform of F . Since F (x)e a|x| ∈ L 1 (R), F extends to a holomorphic function in the strip | Im x| < a.
Since R f (x)x n w(x)dx = 0, we have R F (x)x n dx = 0, so F (n) (0) = 0. Since F is a holomorphic function and all derivatives of F at the origin are zero, we deduce that F = 0. Therefore F = 0, so f = 0 almost everywhere, as desired.
We get the following corollaries:  Proof . 1. The function w(x)|H(x)| p satisfies the assumptions of Lemma 4.1. Therefore polynomials are dense in the space L p (R, w|H| p ). The map g → gH is an isometry between L p (R, w|H| p ) and L p (R, w). The statement follows.
2. Suppose that f ∈ L p (R, w) is nonnegative almost everywhere. Then √ f is an element of L 2p (R, w). Using (1), we find a sequence S n ∈ C[x] such that √ M S n tends to √ f in L 2p (R, w).
We use the following corollary of Cauchy-Schwarz inequality: if a k , b k tend to a, b respectively in L 2p (R, w) then a k b k tends to ab in L p (R, w). Applying this to a = b = √ f , a n = √ M S n , b k = √ M S n we deduce that M S n S n tends to f in L p (R, w). The statement follows.

4.2
The case when all roots of P (x) satisfy | Re α| < 1 2 Let A be a filtered quantization of A with conjugations ρ ± such that ρ 2 ± = g t . We want to classify positive definite Hermitian ρ ± -invariant forms on A, i.e., positive definite Hermitian forms (·, ·) on A such that (aρ(y), b) = (a, yb) for all a, b, y ∈ A, where ρ = ρ ± .
In this subsection we will do the classification in the case when all roots α of P (x) satisfy | Re α| < 1 2 . We start with general results that are true for all parameters P . It is easy to see that Hermitian ρ-invariant forms are in one-to-one correspondence with g t -twisted ρ-invariant traces, i.e., g t -twisted traces T such that T (ρ(a)) = T (a). The correspondence is as follows: T (a) = (a, 1).
Therefore it is enough to classify g t -twisted traces T such that the Hermitian form (a, b) = T (aρ(b)) is positive definite. This means that T (aρ(a)) > 0 for all nonzero a ∈ A. Recall that ad z acts on A diagonalizably, A = ⊕ d∈Z A d . Thus it is enough to check the condition T (aρ(a)) > 0 for homogeneous a. Proof . 1. Suppose that T (aρ(a)) > 0 for all nonzero a ∈ A of weight 0 or 1. Let a be a nonzero homogeneous element of A with positive weight. There exists b of weight 0 or 1 and nonnegative integer k such that a = v k bv k . We have is a homogeneous element of weight 0 or 1. Suppose that a is a nonzero homogeneous element of A with negative weight. Then a = ρ(b), where b is a homogeneous element with positive weight. We get T (aρ(a)) = T (ρ(b)ρ 2 (b)) = T (ρ(b)g t (b)) = T (bρ(b)) > 0.
2. Suppose that a is an element of A 0 . Then a = R(z) for some R ∈ C[x]. We have T (aρ(a)) = T (R(z)R(−z)).
Suppose that a is an element of A 1 . Then a = R z − 1 2 v for some R ∈ C[x]. We have T (aρ(a)) = λT R z − 1 We have In the last equality we used the Cauchy theorem and the fact that the function P (x)w x + From now on we assume that all roots α of P (x) satisfy | Re α| < 1 2 . In this case every trace T can be represented as T (R(z)) = iR R(x)w(x)|dx|. Recall that w(x) = e 2πicx G(e 2πix ) P(e 2πix ) , where G is any polynomial with deg G ≤ deg P in the case when c = 0 and deg G < deg P in the case when c = 0. 1. If λ = −i −n e −πic (i.e., ρ = ρ − ) then w(x) and λP (x)w x+ 1 2 are nonnegative on iR if and only if G(X) is nonnegative when X > 0 and nonpositive when X < 0.
Proof . It is easy to see that P(X) is positive when X > 0. Therefore w(x) is nonnegative on iR if and only if G(X) is nonnegative when X > 0.
We have It is clear that i −n P (x) P(−e 2πix ) belongs to R when x ∈ iR and does not change sign on iR. When x tends to −i∞, the functions i −n P (x) and P(−e 2πix ) have sign (−1) n . Therefore i −n P (x) P(e −2πix ) is positive on iR. We deduce that ±G(X) should be nonnegative when X < 0. So there are two cases: 1. If λ = −i −n e −πic then G(X) should be nonnegative when X > 0 and nonpositive when X < 0.
2. If λ = +i −n e −πic then G(X) should be nonnegative for all X ∈ R.
This proves the proposition.
We deduce the following theorem from Propositions 4.4 and 4.5: Theorem 4.6. Suppose that A is a deformation of A = C[p, q] Z/n with conjugation ρ as above, ρ 2 = g t , t = exp(2πic). Let P (x) be the parameter of A, ε = i n e πic λ = ±1 (so ρ = ρ ε ). Then the cone C + of positive definite ρ-invariant forms on A is isomorphic to the cone of nonzero polynomials G(X) of degree ≤ n − 1 with G(0) = 0 if c = 0 such that 1. If ε = −1 then G(X) is nonnegative when X > 0 and nonpositive when X < 0.
2. If ε = 1 then G(X) is nonnegative for all X ∈ R.
Thus for ρ = ρ − , G(X) = XU (X) where U (X) ≥ 0 is a polynomial of degree ≤ n − 2, and for ρ = ρ + , G(X) ≥ 0 is a polynomial of degree ≤ n − 1 with G(0) = 0 if c = 0; in the latter case G(X) = X 2 U (X) where U (X) ≥ 0 is a polynomial of degree ≤ n − 3. Therefore, we get (Here if the dimension is < 0, the cone C + is empty.) Consider now the special case of even short star-products (i.e., quaternionic structures). Let A be an even quantization of A, and C even + the cone of positive σ-stable s-twisted traces (i.e., those defining even short star-products). Then we have Proposition 4.8. The dimensions of C even + modulo scaling in various cases are as follows: Proposition 4.8 shows that the only cases of a unique positive σ-stable s-twisted trace are ρ = ρ + for n = 1, 4 and ρ = ρ − for n = 2, 3.
The paper [2] considers the case ρ = ρ + if n = 0, 1 mod 4 and ρ = ρ − if n = 2, 3 mod 4; this is the canonical quaternionic structure of the hyperKähler cone (see [6,Section 3.8]), since it is obtained from ρ + on C[p, q] by restricting to Z/n-invariants. Thus for n ≤ 4 the unitary even star-product is unique, as conjectured in [2]. However, for n ≥ 5 this is no longer so. For example, for n = 5 (a case commented on at the end of section 6 of [2]) by Proposition 4.8 the cone C even + modulo scaling is 2-dimensional (which disproves the most optimistic conjecture of [2] that a unitary even star-product is always unique). 5 Example 4.9. Let n = 1, P (x) = x, so P(X) = X + 1. Then for ρ = ρ − there are no positive traces while for ρ = ρ + positive traces exist only if c = 0. In this case there is a unique positive trace up to scaling given by the weight function In particular, the only quaternionic case is ρ = ρ + , c = 1 2 , which gives w(x) = 1 2 cos πx .
Example 4.10. Let n = 2, P (x) = x 2 + β 2 , β 2 ∈ R so we have P(X) = (X + e 2πβ )(X + e −2πβ ). We assume that β 2 > − 1 4 so that all roots of P are in the strip | Re x| < 1 2 . Then ρ = ρ − gives a unique up to scaling positive trace defined by the weight function w(x) = e 2πicx 4 cos π(x − β) cos π(x + β) , and ρ = ρ + is possible if and only if c = 0 and gives a unique up to scaling positive trace defined by the weight function .
In particular, the only quaternionic case is ρ = ρ − , c = 0, with which corresponds to the SL 2 -invariant short star-product. There are two subcases: β 2 ≥ 0, which corresponds to the spherical unitary principal series for SL 2 (C), and − 1 4 < β 2 < 0, which corresponds to the spherical unitary complementary series for the same group (namely, the trace form is exactly the positive inner product on the underlying Harish-Chandra bimodule).
Note that together with the trivial representation corresponding to β 2 = − 1 4 , these representations are well known to exhaust irreducible spherical unitary representations of SL 2 (C) [15].
In this case c = 1 2 so the trace T , up to scaling, is given by with deg(G) ≤ 2. Moreover, because of evenness we must have w(x) = w(−x), so G(X) = X 2 G X −1 . Up to scaling, such polynomials G form a 1-parameter family, parametrized by α. Following [2, Section 6.3], let us equip the corresponding quantum algebra A = A P with the quaternionic structure ρ − given by 6 and let us determine which traces are unitary for this quaternionic structure. According to Theorem 4.6, there is a unique such trace (which is automatically σ-stable), corresponding to G(X) = X. Thus this trace is given by the weight function w(x) = 1 cos πx cos π(x − iβ) cos π(x + iβ) . Hence, in particular, T (z k ) = 0 if k is odd. For even k this integral can be computed using the residue formula. Namely, assume β = 0 and let us first compute T (1). Replacing the contour iR by 1 + iR and subtracting, we find using the residue formula: Thus Note that this function has a finite value at β = 0, which is the answer in that case. Now let us compute T z 2 . Again replacing the contour iR with 1 + iR and subtracting, we get Thus, This gives the equation of the curve in Fig. 2 in [2]. We also note that for κ = − 1 4 (i.e., β = 0) we get α = 1 4 − 2 π 2 . the value found in [2].
where β 2 , γ 2 ∈ R. This is an even quantization of A = C[X 4 ] discussed in [2, Section 6.4]. Thus even short star-products are still parametrized by a single parameter α; namely, the corresponding σ-invariant s-twisted trace such that T (1) = 1 is determined by the condition that T z 2 = −α.

The case of a closed strip
Suppose now that all roots α of P satisfy | Re α| ≤ 1 2 . Recall that we have P (x) = P * (x)Q x+ 1 2 ×Q x− 1 2 where all roots of P * (x) satisfy | Re x| < 1 2 and all roots of Q(x) belong to iR. For any By Proposition 3.10 each g t -twisted trace can be obtained as where w(x) = e 2πicx G(e 2πix ) P(e 2πix ) and φ is any linear functional.
Proposition 4.14. Suppose that T is a trace as above and w(x) has poles on iR. Then T does not give a positive definite form.
Then there exists a linear functional ψ such that for any Suppose that T gives a positive definite form. Then T (S(z)S(−z)) > 0 for all nonzero S ∈ C[x]. Taking S(x) = Q * (x)S 1 (x) and using Lemma 4.2, we deduce that Q 2 * (x)w(x), hence w(x), is nonnegative on iR. In particular, all poles of w(x) have order at least 2.
Without loss of generality assume that w(x) has a pole at zero. Let where b ∈ R. Suppose that F n is a sequence of polynomials that tends to the function f := χ (−ε,ε) (the characteristic function of the interval) in the space L 2 iR, (Q * + Q 2 * w). In particular, F n tends to f in the spaces L 2 (iR, Q * w) and L 2 iR, Q 2 * w . Then we deduce from the Cauchy-Schwartz inequality that F n F n tends to f 2 in the space L 1 iR, Q 2 * w , and F n and F n tend to f in L 1 (iR, Q * w).
We have Therefore, when n tends to infinity, We have φ b 2 = Cb 2 for some C ≥ 0. Suppose that w has a pole of order M ≥ 2 at 0 and Q * has a zero of order N > 0 at 0. Then Q * w has a zero of order N − M at zero and Q 2 * w has a zero of order 2N − M at zero. We deduce that where c 1 = c 1 (ε), c 2 = c 2 (ε) are functions having strictly positive limits at ε = 0 . Therefore This is a quadratic polynomial of b with discriminant Since M ≥2, for small ε this discriminant is positive. In particular, for some b, lim n→∞ T (R n (z))<0, so for this b and some n, T (R n (z)) < 0, a contradiction.
Now we are left with the case when w(x) has no poles on iR. In this case T (R(z)) = iR R(x)w(x)|dx| + η(R 0 ), where η is some linear functional.
Proposition 4.15. T gives a positive definite form only when η(R 0 ) = j c j R 0 (z j ), where c j ≥ 0 and z j ∈ iR are the roots of Q.
Proof . Suppose that this is not the case. Then it is easy to find a polynomial S such that η SS 0 < 0. Then using Lemma 4.2(2) for M = Q, we find F n such that F n Q + S tends to zero in L 2 (iR, w). We deduce that which gives a contradiction.
In the proof of Proposition 4.14 we got that w is nonnegative on iR. We also note that Q(z) divides P z − 1 2 , hence Using the proof of Proposition 4.4 we see that λP (z)w z + 1 2 is nonnegative on iR. Assume that Q z − 1 2 Q z + 1 2 is positive on R. Then this is equivalent to λP * (z)w z + 1 2 being nonnegative on iR. So we have proved the following theorem.

The general case
Let A is be a filtered quantization of A with conjugation ρ given by formula (2.3). Let P (x) be its parameter. Let P (x) be the polynomial defined in Section 3.5: it has the same degree as P (x) and its roots are obtained from the roots of P (x) by minimal integer shift into the strip | Re x| ≤ 1 2 . Also recall from Section 3.5 that P • denotes the following polynomial: all roots of P • belong to strip | Re x| ≤ 1 2 and the multiplicity of α, | Re α| ≤ 1 2 in P • equals to multiplicity of α in P . Let n • := deg(P • ).
Proposition 3.11 says that any trace T can be represented as T = Φ + T , where Φ is a linear functional such that ∈ iR, and T is a trace for P . Furthermore, if Φ = 0 then T is just a trace for P • . Proof . For big enough k we have Φ((x − z 1 ) k · · · (x − z m ) k C[x]) = 0. Recall that there exists polynomial Q * (x) nonnegative on iR such that for R = R 1 Q * + R 0 , deg R 0 < deg Q * , we have T (R) = iR R 1 Q * w|dx| + ψ(R 0 ), where ψ is some linear functional. Let U (x) be a polynomial divisible by Q * such that Φ(U (x)C[x]) = 0. Let L be any polynomial. Using Lemma 4.2 for M = U , we find a sequence G n = U S n that tends to L in the space L 2 (iR, Q * w|dx|). We deduce that H n (x) := (G n (x) − L(x))(G n (−x) − L(−x)) tends to zero in L 1 (iR, Q * w). We have T (H n (z)Q * (z)) = iR H n (x)Q * w|dx|. We conclude that T (H n (z)Q * (z)) = H n L 1 (iR,Q * w) tends to zero when n tends to infinity.
It follows that T (H n (z)) tends to Φ(Q * (x)H n (x)) = Φ Q * (x)L(x)L(−x) . Since H n is nonnegative on iR, we have T (H n (z)) > 0. Now we get a contradiction with Proof . Let r be the biggest number such that there exists j with c jr = 0. Let Here we omit x − z j * in the product, where j * = j is such that z j * = −z j . We note that where a := Q * (z j )F (r) (z j )F (−z j ). Pick a ∈ C so that c jr a + c j * r a = 2Re(c jr a) < 0, and choose G ∈ C[x] which gives this value of a (e.g., we can choose G to be linear). Then Φ Q * (x)F (x)F (−x) < 0, as desired.
If A P• is the quantization with parameter P • then there is a conjugation ρ • on A P• given by the formulas λ. Therefore we can consider the cone of positive definite forms for A P• with respect to ρ • . Proof . We deduce from Proposition 4.17 that each trace T that gives a positive definite form should have Φ = 0. By Proposition 3.11, in this case T is a trace for the polynomial P • (x). So there exists polynomial Q * such that for R = R 1 Q * + R 0 , deg R 0 < deg Q * , and Using Proposition 4.14 and its proof, we deduce that w has no poles and that w(x) and λw x + 1 2 P (x) are nonnegative on iR. Therefore where ψ is some linear functional. Using Proposition 4.15, we deduce that this trace is positive if and only if ψ(R 0 ) = j c j R 0 (z j ), where c j ≥ 0 and z j ∈ iR.
Since (−1) is positive on iR, we see that λw x + 1 2 P (x) is nonnegative on iR if and only if λ • w x + 1 2 P • (x) is nonnegative on iR. Using Theorem 4.16 we then deduce that T is positive for P (x) if and only if it is positive for P • (x).
So we have proved the following theorem.
Theorem 4.20. Let A = A P be a filtered quantization of A with parameter P equipped with a conjugation ρ such that ρ 2 = g t . Let be the number of roots α of P such that | Re α| < 1 2 counted with multiplicities and r be the number of distinct roots α of P with Re α = − 1 2 . Then the cone C + of ρ-equivariant positive definite traces on A is isomorphic to C 1 + × C 2 + , where C 2 + = R r ≥0 , and C 1 + is the cone of nonzero polynomials G such that (1) G has degree less than ; (2) G(0) = 0 if t = 1; (3) G(X) ≥ 0 when X > 0; (4) G(X) is either nonnegative or nonpositive when X < 0 depending on whether ρ = ρ + or ρ − .
The conditions are the same as in Theorem 4.6.
5 Explicit computation of the coefficients a k , b k of the 3-term recurrence for orthogonal polynomials and discrete Painlevé systems As noted in Section 3.2, to compute the short star-product associated to a trace T , one needs to compute the coefficients a k , b k of the 3-term recurrence for the corresponding orthogonal polynomials: Also recall [14] that a k = ν k ν k−1 , where ν k := (P k , P k ). Finally, recall that ν k = D k D k−1 , where D k is the Gram determinant for 1, x, . . . , x k−1 , i.e., where M r is the r-th moment of the weight function w(x), i.e., In the even case w(−x) = w(x) we have b k = 0, so and p k can be easily computed recursively from the sequence a k . If the polynomials p k are Proposition 5.4. The orthogonal polynomials satisfy the difference equation Note that it is not the mere existence of a difference equation with rational coefficients that is significant (indeed, any pair of polynomials satisfies such an equation!), rather it is the fact that (a) the poles are bounded independently of n, and (b) so is the asymptotic behavior at infinity.
If we consider (for t = 1) the family of matrices satisfying the above conditions; that is, P A n is polynomial, det(A n ) = t −1 , and x 2 , we find that the family is classified by a rational moduli space. To be precise, let f (x) := 1 − t −1 −1 P (x)A n (x) 21 , and let g(x) ∈ C[x]/(f (x)) be the reduction of P (x)A n (x) 11 modulo f (x). Then f and g both vary over affine spaces of dimension deg(q) − 1, and generically determine A n . Indeed, A n (x) 21 is clearly determined by f , and since A n (x) 11 P (x) is specified by the asymptotics up to an additive polynomial of degree deg(P ) − 2, it is determined by f and g. For generic f , g, this also determines A n (x) 22 , since the determinant condition implies that for any root α of f , A n (α) 11 A n (α) 22 = t −1 . Moreover, this constraint forces P (x) 2 A n (x) 11 A n (x) 22 − t −1 to be a multiple of f (x), and thus the unique value of A n (x) 12 compatible with the determinant condition gives a matrix satisfying the desired conditions.
Moreover, given such a matrix, the three-term recurrence for orthogonal polynomials tells us that the corresponding A n+1 is the unique matrix satisfying its asymptotic conditions and having the form It is straightforward to see that a n , b n are determined by the leading terms in the asymptotics of A n (x) 12 , and thus in particular are rational functions of the parameters. We thus find that the map from the space of matrices A n to the space of matrices A n+1 is a rational map, and by considering the inverse process, is in fact birational, corresponding to a sequence F n of birational automorphisms of A 2 deg(P )−2 . Note that the equation A 0 , though not of the standard form, is still enough to determine A 1 , and thus gives (rationally) a P deg(P )−1 worth of initial conditions corresponding to orthogonal polynomials. (There is a deg(P )-dimensional space of valid functions F , but rescaling F merely rescales the trace, and thus does not affect the orthogonal polynomials.) Example 5.5. As an example, consider the case P (x) = x 2 , corresponding, e.g., to w(y) = e 2πcy cosh 2 πy , with c ∈ (0, 1). In this case, deg(P ) = 2, so we get a 2-dimensional family of linear equations, and thus a second-order nonlinear recurrence, with a 1-parameter family of initial conditions corresponding to orthogonal polynomials. Since the monic polynomial f is linear, we may use its root as one parameter f n , and g n = A n (f n ) 11 as the other parameter. We thus find that where a n = t (t − 1) 2 n 2 g n − f 2 n (g n − 1) 2 g n (5. 2) and f n , g n are determined from the recurrence f n+1 = f n (f n (g n − 1) − ng n )(f n (g n − 1) − n) n 2 g n − f 2 n (g n − 1) 2 , (5.3) g n+1 = (f n (g n − 1) − ng n ) 2 tg n (f n (g n − 1) − n) 2 . (5.4) The three-term recurrence for the orthogonal polynomials is then where a n is as above and The initial condition is given by , g 0 = 1.
(Note that the resulting A 0 is not actually correct, but this induces the correct values for f 1 , g 1 , noting that the recurrence simplifies for n = 0 to f 1 = −f 0 , g 1 = 1/tg 0 .) It follows from the general theory of isomonodromy deformations [13] that this recurrence is a discrete Painlevé equation (This will also be shown by direct computation in forthcoming work by N. Witte.). We also note that the recurrence satisfies a sort of time-reversal symmetry: there is a natural isomorphism between the space of equations for t, n and the space for t −1 , −n, coming (up to a diagonal change of basis) from the duality A → A T −1 , and this symmetry preserves the recurrence. (This follows from the fact that if two equations are related by the three-term recurrence, then so are their duals, albeit in the other order.) Remark 5.6. The fact that A n (x) 12 has a nice expression in terms of a n and f n+1 follows more generally from the fact (via the three-term recurrence) that A n (x) 12 = −a n A n+1 (x) 21 .

One similarly has
so that in general f n+1 (x) ∝ P (x)A n (x) 12 and g n+1 (x) = P (x)A n (x) 22 mod f n+1 (x). In particular, applying this to n = 0 tells us that the orthogonal polynomial case corresponds to the initial condition f 1 (x) ∝ L(x), g 1 (x) = t −1 P (x) mod L(x).
The above construction fails for t = 1, because the constraint on the asymptotics of the off-diagonal coefficients of A n is stricter in that case: A n (x) 21 = 2n−1 Remark 5.8. One can perform a similar calculation for the case P (x) = x 4 − e 1 x 2 + e 2 with even trace; again, one obtains a second-order nonlinear recurrence, but the result is significantly more complicated, even for e 1 = e 2 = 0.
In each case, when the moduli space is 0-dimensional, so that the conditions uniquely determine the equation, we get an explicit formula for A n . This, of course, is precisely the case that the orthogonal polynomial is classical.