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] and [Etingof P., Stryker D., SIGMA 16 (2020), 014, 28 pages], we undertake a detailed study of twisted traces on quantizations of Kleinian singularities of type ₙ₋₁. In particular, we give explicit integral formula...
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| Цитувати: | Twisted Traces and Positive Forms on Quantized Kleinian Singularities of Type A. Pavel Etingof, Daniil Klyuev, Eric Rains and Douglas Stryker. SIGMA 17 (2021), 029, 31 pages |
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Digital Library of Periodicals of National Academy of Sciences of Ukraine| _version_ | 1859670977246396416 |
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| author | Etingof, Pavel Klyuev, Daniil Rains, Eric Stryker, Douglas |
| author_facet | Etingof, Pavel Klyuev, Daniil Rains, Eric Stryker, Douglas |
| citation_txt | Twisted Traces and Positive Forms on Quantized Kleinian Singularities of Type A. Pavel Etingof, Daniil Klyuev, Eric Rains and Douglas Stryker. SIGMA 17 (2021), 029, 31 pages |
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| container_title | Symmetry, Integrability and Geometry: Methods and Applications |
| description | Following [Beem C., Peelaers W., Rastelli L., Comm. Math. Phys. 354 (2017), 345-392] and [Etingof P., Stryker D., SIGMA 16 (2020), 014, 28 pages], we undertake a detailed study of twisted traces on quantizations of Kleinian singularities of type 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 ≤ 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 = 2, this, in particular, recovers the theory of unitary spherical Harish-Chandra bimodules for 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 Painlevé systems.
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Symmetry, Integrability and Geometry: Methods and Applications SIGMA 17 (2021), 029, 31 pages
Twisted Traces and Positive Forms
on Quantized Kleinian Singularities of Type A
Pavel ETINGOF a, Daniil KLYUEV a, Eric RAINS b and Douglas STRYKER a
a) Department of Mathematics, Massachusetts Institute of Technology, USA
E-mail: etingof@math.mit.edu, klyuev@mit.edu, stryker@mit.edu
b) Department of Mathematics, California Institute of Technology, Pasadena, CA 91125, USA
E-mail: rains@mit.edu
Received September 22, 2020, in final form March 08, 2021; Published online March 25, 2021
https://doi.org/10.3842/SIGMA.2021.029
Abstract. Following [Beem C., Peelaers W., Rastelli L., Comm. Math. Phys. 354 (2017),
345–392] and [Etingof P., Stryker D., SIGMA 16 (2020), 014, 28 pages], we undertake
a detailed study of twisted traces on quantizations of Kleinian singularities of type An−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 ≤ 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 sl2.
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 Painlevé systems.
Key words: star-product; orthogonal polynomial; quantization; trace
2020 Mathematics Subject Classification: 16W70; 33C47
To Vitaly Tarasov and Alexander Varchenko
with admiration
1 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 super-
conformal 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 = grA. 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
mailto:etingof@math.mit.edu
mailto:klyuev@mit.edu
mailto:stryker@mit.edu
mailto:rains@mit.edu
https://doi.org/10.3842/SIGMA.2021.029
https://www.emis.de/journals/SIGMA/Tarasov-Varchenko.html
2 P. Etingof, D. Klyuev, E. Rains and D. Stryker
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 An−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 A1 case is easy (as it reduces to the representation theory of SL2), in the A2 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 clas-
sification. 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 rea-
son 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 An−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 An−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 zk to pk(z), where pk(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 An−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]:
α(κ) =
1
4
−
κ+ 1
4
1− cos
(
π
√
κ+ 1
4
) .
In particular, this recovers the value α
(
−1
4
)
= 1
4 −
2
π2 predicted in [2] and confirmed in [4, 5].
It would be very interesting to develop a similar theory of positive traces for higher-dimen-
sional 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-
Twisted Traces and Positive Forms on Quantized Kleinian Singularities of Type A 3
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 SL2(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 higher-
dimensional 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 An−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 bimo-
dules. The present paper has stronger results: it contains a complete description of the set
of unitary star-products for any type An−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, follow-
ing [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 conjec-
ture 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 ak, bk 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.
2 Filtered quantizations and twisted traces
2.1 Filtered quantizations
Let Xn be the Kleinian singularity of type An−1. Recall that
A := C[Xn] = C[p, q]Z/n,
where Z/n acts by p 7→ e2πi/np, q 7→ e−2πi/nq. Thus
A = C[u, v, z]/(uv − zn),
1It is curious that, unlike classical representation theory, this dependence is given by a transcendental function.
4 P. Etingof, D. Klyuev, E. Rains and D. Stryker
where
u = pn, v = qn, z = pq.
This algebra has a grading defined by the formulas deg(p) = deg(q) = 1, thus
deg(u) = deg(v) = n, deg(z) = 2. (2.1)
The Poisson bracket is given by {p, q} = 1
n and on A takes the form
{z, u} = −u, {z, v} = v, {u, v} = nzn−1.
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 = AP is the algebra generated
by u, v, z with defining relations
[z, u] = −u, [z, v] = v, vu = P
(
z − 1
2
)
, uv = P
(
z + 1
2
)
and filtration defined by (2.1). Thus we have
[u, v] = P
(
z + 1
2
)
− P
(
z − 1
2
)
= nzn−1 + · · · ,
i.e., the quasiclassical limit indeed recovers the algebra A with the above Poisson bracket.
Note that we may consider the algebra AP 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 transforma-
tions z 7→ z + β we can make sure that the subleading term of P is zero, i.e.,
P (x) = xn + c2x
n−2 + · · ·+ cn.
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:
A = ⊕k∈ZAk.
If b ∈ Am, 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
gt = tad z : A → A (2.2)
be the automorphism of A given by
gt(v) = tv, gt(u) = t−1u, gt(z) = z.
Then gt(b) = tmb if b has weight m.
Example 2.1.
1. Let n = 1, P (x) = x. Then A is the Weyl algebra generated by u, v with [u, v] = 1, and
z = vu+ 1
2 = uv − 1
2 .
2. Let n = 2 and P (x) = x2 − C. Then setting e = v, f = −u, h = 2z, we get
[h, e] = 2e, [h, f ] = −2f, [e, f ] = h, fe = −
(
h+1
2
)2
+ C,
i.e., A is the quotient of the universal enveloping algebra U(sl2) by the relation fe +(
h+1
2
)2
= C, where fe+
(
h+1
2
)2
is the Casimir element.
Twisted Traces and Positive Forms on Quantized Kleinian Singularities of Type A 5
2.2 Even quantizations
Let s be the automorphism of A given by
s(u) = (−1)nu, s(v) = (−1)nv, s(z) = z,
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 σ2 = s and grσ = id, and that σ is unique if exists [6,
Remark 2.10]. This means that σ(z) = −z, σ(u) = inu, σ(v) = inv. It is easy to see that σ
exists if and only if
(−1)nP
(
z − 1
2
)
= (−1)nvu = σ(v)σ(u) = σ(uv) = σ
(
P
(
z + 1
2
))
= P
(
−z + 1
2
)
.
This is equivalent to
P (−x) = (−1)nP (x),
i.e., P contains only terms xn−2i. Thus even quantizations of A are parametrized by [n/2]
essential parameters, and all quantizations for n ≤ 2 are even.
2.3 Quantizations with a conjugation and a quaternionic structure
Recall [6, Section 3.6] that a conjugation on A is an antilinear filtration preserving automorphism
ρ : A → A that commutes with s. We will consider conjugations on A given by
ρ(v) = λu, ρ(u) = λ∗v, ρ(z) = −z, (2.3)
where λ, λ∗ ∈ C×; it easy to show that they are the only ones up to symmetry, using that
any two such conjugations differ by a filtration preserving automorphism commuting with s.
The automorphism u 7→ γ−1u, v 7→ γv rescales λ by |γ|−2 and λ∗ by |γ|2, so we may assume
that |λ| = 1, i.e.,
λ = ±i−ne−πic,
where c ∈ [0, 1). Then
P
(
−z + 1
2
)
= ρ
(
P
(
z + 1
2
))
= ρ(uv) = ρ(u)ρ(v) = λλ∗vu = λλ∗P
(
z − 1
2
)
,
i.e., P (−x) = λλ∗P (x). Thus λ∗ = (−1)nλ−1 = ±i−neπic (so |λ∗| = 1) and
P (−x) = (−1)nP (x),
i.e., inP is real on iR. We also have
ρ2(u) = λ∗λu, ρ2(v) = λλ∗v, ρ2(z) = z,
so ρ2 = gt, where gt is given by (2.2) and
t = (−1)nλλ−1 = (−1)nλ−2 = e2πic,
i.e., |t| = 1. Thus we see that for every t there are two non-equivalent conjugations, corresponding
to the two choices of sign for λ, which we denote by ρ+ and ρ−.
In particular, consider the special case t = (−1)n, i.e., gt = s. Then c = 1
2 for n odd and
c = 0 for n even. Thus λ = ±1, so the conjugation ρ on A is given by
ρ(v) = ±u, ρ(u) = ±(−1)nv, ρ(z) = −z.
Now assume in addition that A is even, i.e., P (−x) = (−1)nP (x). Then we have ρσ = σ−1ρ,
so ρ and σ give a quaternionic structure on A, cf. [6, Section 3.7]. So this quaternionic structure
exists if and only if P ∈ R[x], P (−x) = (−1)nP (x).
6 P. Etingof, D. Klyuev, E. Rains and D. Stryker
Example 2.2. Let n = 2, so A is the quotient of the enveloping algebra U(g), g = sl2, by the
relation fe+ (h+1)2
4 = C, where C ∈ R. Since
e = v, f = −u, h = 2z,
we have
ρ±(e) = ±f, ρ±(f) = ±e, ρ±(h) = −h.
So g+ := gρ+ has basis x = e+f
2 , y = i(e−f)
2 , z = ih
2 . Thus,
[x,y] = −z, [z,x] = y, [y, z] = x.
Hence, setting E := y − z, F := y + z, H := 2x, we have
[H,E] = 2E, [H,F ] = −2F, [E,F ] = H,
so g+ = sl2(R).
On the other hand, g− := gρ− has basis ix, iy, z, hence g− = so3(R) = su2.
So ρ+ and ρ− correspond to the split and compact form of g, respectively.
2.4 Twisted traces
Let A = AP be a filtered quantization of A. Recall [6, Section 3.1] that a gt-twisted trace on A
is a linear map T : A → C such that T (ab) = T (bgt(a)), where gt 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 gt-twisted traces2 T on A. The answer is given by the following proposition.
Proposition 2.3. T : A → C is a gt-twisted trace on A if and only if
(1) T (Aj) = 0 for j 6= 0;
(2) T
(
S
(
z − 1
2
)
P
(
z − 1
2
))
= tT
(
S
(
z + 1
2
)
P
(
z + 1
2
))
for all S ∈ C[x].
In particular, any twisted trace is automatically s-invariant.
Proof. Suppose T satisfies (1), (2). It is enough to check that
T (ub) = t−1T (bu), T (vb) = tT (bv), T (zb) = T (bz)
for b ∈ A.
The equality T (zb) = T (bz) says that T (Aj) = 0 for j 6= 0, which is condition (1).
By (1), it is enough to check the equality T (ub) = t−1T (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
2
))
= T
(
P
(
z + 1
2
)
S
(
z + 1
2
))
,
T (bu) = T
(
vS
(
z + 1
2
)
u
)
= T
(
vuS
(
z − 1
2
))
= T
(
P
(
z − 1
2
)
S
(
z − 1
2
))
,
which yields the desired identity using (2).
Similarly, it is enough to check the equality T (vb) = tT (bv) for b ∈ A1. In this case b =
uS
(
z − 1
2
)
. We have
T (vb) = T
(
vuS
(
z − 1
2
))
= T
(
P
(
z − 1
2
)
S
(
z − 1
2
))
,
T (bv) = T
(
uS
(
z − 1
2
)
v
)
= T
(
uvS
(
z + 1
2
))
= T
(
P
(
z + 1
2
)
S
(
z + 1
2
))
,
which again gives the desired identity using (2).
Conversely, the same argument shows that if T is a gt-twisted trace then (1), (2) hold. �
2One can show that for generic P and n > 2, the only possible filtration preserving automorphisms are gt.
Twisted Traces and Positive Forms on Quantized Kleinian Singularities of Type A 7
Thus we get
Corollary 2.4. The space of gt-twisted traces on A is naturally isomorphic to the space(
C[z]/
{
S
(
z − 1
2
)
P
(
z − 1
2
)
− tS
(
z + 1
2
)
P
(
z + 1
2
)
|S ∈ C[z]
})∗
and has dimension n if t 6= 1 and dimension n− 1 if t = 1.
2.5 The formal Stieltjes transform
There is a useful characterization of the space gt-twisted traces in terms of generating functions.
Given a linear functional T on C[z], its formal Stieltjes transform is the generating function
FT (x) :=
∑
n≥0
x−n−1T (zn) ∈ C
[[
x−1
]]
,
or equivalently FT (x) = T
(
(x − z)−1
)
, with (x − z)−1 itself expanded as a formal power series
in x−1.
Proposition 2.5. The formal Stieltjes transform of a gt-twisted trace on A satisfies
P (x)
(
FT
(
x+ 1
2
)
− tFT
(
x− 1
2
))
∈ C[x],
and this establishes an isomorphism of the space of gt-twisted traces with the space of polynomials
of degree ≤ n− 1 (for t 6= 1) or ≤ n− 2 (for t = 1).
Proof. We may write
P (x)
(
FT
(
x+ 1
2
)
− tFT
(
x− 1
2
))
= T
(
P (x)
x+ 1
2 − z
− t P (x)
x− 1
2 − z
)
= T
(
P
(
z − 1
2
)
x+ 1
2 − z
− t
P
(
z + 1
2
)
x− 1
2 − z
)
+ T
(
P (x)− P
(
z − 1
2
)
x−
(
z − 1
2
) − t
P (x)− P
(
z + 1
2
)
x−
(
z + 1
2
) )
.
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∑
n≥0
x−n−1T
(
P
(
z − 1
2
)(
z − 1
2
)n − tP (z + 1
2
)(
z + 1
2
)n)
= 0.
Since the map F 7→ F
(
x+ 1
2
)
−tF
(
x− 1
2
)
is injective on x−1C[[1/x]], this establishes an injective
map from gt-twisted traces to polynomials of degree < deg(P ). This establishes the conclusion
for t 6= 1, with surjectivity following by dimension count. Finally, for t = 1, we need simply
observe that for any F ∈ x−1C
[[
x−1
]]
, FT
(
x + 1
2
)
− FT
(
x − 1
2
)
∈ x−2C
[[
x−1
]]
, 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) 7→ F
(
x + 1
2
)
− tF
(
x − 1
2
)
acts triangularly
on x−1C
[[
x−1
]]
, of degree 0 (with nonzero leading coefficients) if t 6= 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.
8 P. Etingof, D. Klyuev, E. Rains and D. Stryker
Remark 2.7. A similar argument establishes an isomorphism between linear functionals satis-
fying T
(
P
(
q−
1
2 z
)
S
(
q−
1
2 z
)
− qtP
(
q
1
2 z
)
S
(
q
1
2 z
))
= 0 and elements F ∈ x−1C
[[
x−1
]]
such that
P (x)
(
FT
(
q
1
2x
)
− tFT
(
q−
1
2x
))
∈ C[x],
or, for t = 1, between linear functionals satisfying
T
(
z−1
(
P
(
q−
1
2 z
)
S
(
q−
1
2 z
)
− P
(
q
1
2 z
)
S
(
q
1
2 z
)))
= 0
and formal series satisfying
P (x)x−1
(
FT
(
q
1
2x
)
− FT
(
q−
1
2x
))
∈ C[x].
3 An analytic construction of twisted traces
3.1 Construction of twisted traces when all roots of P (x) satisfy |Reα| < 1
2
Let t = exp(2πic), where 0 ≤ Re c < 1 (clearly, such c exists and is unique).
Let P (x) =
∏n
j=1(x− αj). Define
P(X) :=
n∏
j=1
(
X + e2πiαj
)
.
When P (x) satisfies the equation P (−x) = (−1)nP (x) (the condition for existence of a conju-
gation ρ) the polynomial P(X) has real coefficients.
Proposition 3.1. Assume that every root α of P (x) satisfies |Reα| < 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 gt-twisted trace is
given by3
T (R(z)) =
∫
iR
R(x)w(x)|dx|, R ∈ C[x],
where w is the weight function defined by the formula
w(x) = w(c, x) := e2πicx
G(e2πix)
P(e2πix)
,
where G is a polynomial of degree ≤ n− 1 and G(0) = 0 if c = 0.
Proof. It is easy to see that the function w(x) enjoys the following properties:
(1) w(x+ 1) = tw(x);
(2) |w(x)| decays exponentially and uniformly when Imx tends to ±∞;
(3) w
(
x+ 1
2
)
P (x) is holomorphic when |Rex| ≤ 1
2 .
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)) :=
∫
iRR(x)w(x)|dx|. We should check that
T
(
tS
(
z + 1
2
)
P
(
z + 1
2
)
− S
(
z − 1
2
)
P
(
z − 1
2
))
= 0.
3Here |dx| denotes the Lebesgue measure on the imaginary axis.
Twisted Traces and Positive Forms on Quantized Kleinian Singularities of Type A 9
We have
T
(
tS
(
z + 1
2
)
P
(
z + 1
2
)
− S
(
z − 1
2
)
P
(
z − 1
2
))
=
∫
iR
tS
(
x+ 1
2
)
P
(
x+ 1
2
)
w(x)|dx| −
∫
iR
S
(
x− 1
2
)
P
(
x− 1
2
)
w(x)|dx|
=
∫
1
2
+iR
tS(x)P (x)w
(
x− 1
2
)
|dx| −
∫
− 1
2
+iR
S(x)P (x)w
(
x+ 1
2
)
|dx|
=
∫
1
2
+iR
S(x)P (x)w
(
x+ 1
2
)
|dx| −
∫
− 1
2
+iR
S(x)P (x)w
(
x+ 1
2
)
|dx|
=
1
i
∫
∂([− 1
2
, 1
2 ]×R)
S(x)P (x)w
(
x+ 1
2
)
dx.
But this integral vanishes by the Cauchy theorem since S(x)P (x)w
(
x+ 1
2
)
is holomorphic when
|Rex| ≤ 1
2 and decays exponentially as Imx→ ±i∞.
By Corollary 2.4, the space of polynomials G(X) has the same dimension as the space of
gt-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:
T (R(z)) = lim
δ→0+
∫
iR
R(x)w(c+ δ, x)|dx|.
Alternatively, one may say that T (R(z)) is the value of the Fourier transform of the distri-
bution 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:
Proposition 3.2. 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 nondegener-
ate 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)nw(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
T (R(−z)) =
∫
iR
R(−x)w(x)|dx| =
∫
iR
R(x)w(−x)|dx|.
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
T (R(z)) =
∫
iR
R(x)w(x)|dx|,
where w is as in Proposition 3.1 and w(x) = w(−x) = (−1)nw(x+ 1).
10 P. Etingof, D. Klyuev, E. Rains and D. Stryker
3.2 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) 7→ T (ab) defines an inner product onA 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(zk) = pk(z),
where pk are monic orthogonal polynomials for the inner product
(f1, f2)∗ :=
∫
iR
f1(x)f2(x)w(x)|dx|.
Recall [14] that these polynomials satisfy a 3-term recurrence
pk+1(x) = (x− bk)pk(x)− akpk−1(x),
for some numbers ak, bk, i.e.,
xpk(x) = pk+1(x) + bkpk(x) + akpk−1(x).
Thus the corresponding short star-product z ∗ zk has the form
z ∗ zk = φ−1
(
φ(z)φ
(
zk
))
= φ−1(zpk(z)) = φ−1(pk+1(z) + bkpk(z) + akpk−1(z))
= zk+1 + bkz
k + akz
k−1.
Thus the numbers ak, bk 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 ak, bk 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
Pk(y) := ikpk(−iy) are orthogonal under the inner product
(f1, f2) :=
∫ ∞
−∞
f1(y)f2(y)w(y)dy,
where w(y) := w(−iy). Then the 3-term recurrence looks like
Pk+1(y) = (y − ibk)Pk(y) + akPk−1(y)
(so for real parameters we’ll have ak ∈ R, bk ∈ iR).
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 6= 0, in which case it is unique up to scaling, and the corresponding weight
function is
w(x) =
e2πicx
e2πix + 1
=
e2πi(c−
1
2)x
2 cosπx
, w(y) =
e2π(c−
1
2)y
2 coshπy
.
The corresponding orthogonal polynomials Pk(y) are the (monic) Meixner–Pollaczek polynomials
with parameters λ = 1
2 , φ = πc [10, Section 1.7].
Twisted Traces and Positive Forms on Quantized Kleinian Singularities of Type A 11
Example 3.5. Let n = 2, P (x) = x2 + β2, so
P(X) =
(
X + e2πβ
)(
X + e−2πβ
)
.
The space of twisted traces is 1-dimensional if c = 0 and 2-dimensional if c 6= 0. So for c 6= 0
the traces up to scaling are defined by the weight function
w(x) =
e2πi(c−
1
2)x cosπ(x− iα)
2 cosπ(x− iβ) cosπ(x+ iβ)
, w(y) =
e2π(c−
1
2)y coshπ(y − α)
2 coshπ(y − β) coshπ(y + β)
,
and the limiting cases α→ ±∞ along the real axis, which yield
w(x) =
e2πi(c−
1
2
± 1
2)x
4 cosπ(x− iβ) cosπ(x+ iβ)
, w(y) =
e2π(c−
1
2
± 1
2)y
4 coshπ(y − β) coshπ(y + β)
.
These formulas for the plus sign also define the unique up to scaling trace for c = 0; i.e.,
w(x) =
1
4 cosπ(x− iβ) cosπ(x+ iβ)
, w(y) =
1
4 coshπ(y − β) coshπ(y + β)
.
In this case, the corresponding orthogonal polynomials Pk(y) are the continuous Hahn polyno-
mials with parameters 1
2 + iβ, 12 − iβ, 12 − iβ, 12 + iβ [10, Section 1.4].
Also for c = 1
2 , α = 0 we have
w(x) =
cosπx
2 cosπ(x+ iβ) coshπ(x− iβ)
, w(y) =
coshπy
2 coshπ(y + β) coshπ(y − β)
,
so Pk(y) are the continuous dual Hahn polynomials with a = 0, b = 1
2 − iβ, c = 1
2 + iβ [10,
Section 1.3].
Remark 3.6. In Example 3.4 (n = 1), the only even short star-product corresponds to w(y) =
1
2 coshπy . This is the Moyal–Weyl star-product. In Example 3.5 (n = 2), the only even short
star-product corresponds to w(y) = 1
4 coshπ(y−β) coshπ(y+β) . This is the unique SL2-invariant
star-product.
Example 3.7. Let t = (−1)n, G(X) = X [n/2]. Then
w(x) =
n∏
j=1
1
2 cosπ(x− iβj)
, w(y) =
n∏
j=1
1
2 coshπ(y + βj)
,
which defines an s-twisted trace. The corresponding orthogonal polynomials are semiclassical
but not hypergeometric for n ≥ 3.
Remark 3.8. The trace of Example 3.7 corresponds to the short star-product arising in the
3-d SCFT, as shown in [5, Section 8.1.2]. There the Kleinian singularity of type An−1 appears
as the Higgs branch, and the parameters βj are the FI parameters. The same trace also shows
up in [4, equation (5.27)], where the Kleinian singularity appears as the Coulomb branch, and
the parameters βj are the mass parameters.4
4We thank Mykola Dedushenko for this explanation.
12 P. Etingof, D. Klyuev, E. Rains and D. Stryker
3.3 Conjugation-equivariant traces
Let now ρ be a conjugation on A (Section 2.3). Let us determine which gt-twisted traces are
ρ-equivariant (see [6, Section 3.6]. A trace T is ρ-equivariant if T (R(z)) = T
(
R(−z)
)
, which
is equivalent to T being real on R[iz]. 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
gt-twisted traces T on A are given by
T (R(z)) =
∫
iR
R(x)w(x)|dx|,
where w is as in Proposition 3.1 and
w(x) = w(−x) = (−1)nw(x+ 1).
Moreover, if A is even then σ-invariant traces among them correspond to the functions w with
w(x) = w(−x).
3.4 Construction of traces when all roots of P (x) satisfy |Reα| ≤ 1
2
From now on we suppose that inP (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 |Reα| ≤ 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 |Rex| < 1
2 and all roots of Q(x)
belong to iR. Suppose that α1, . . . , αk are the roots of P∗(x) and αk+1, . . . , αm are the roots
of Q
(
x+ 1
2
)
. Note that degQ = n−m. Let P∗(X) =
∏m
j=1(X + e2πiαj ), w(x) = e2πicx G(e2πix)
P∗(e2πix)
,
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 Imx tends
to ±∞;
(3) w
(
x+ 1
2
)
P (x) is holomorphic on |Rex| ≤ 1
2 .
For any R ∈ C[x] let R(x) = R1(x)Q(x) +R0(x), where degR0 < degQ.
Proposition 3.10. A general gt-twisted trace on A has the form
T (R(z)) =
∫
iR
R1(x)Q(x)w(x)|dx|+ φ(R0),
where w(x) is as above and φ is any linear functional.
Proof. The space of polynomials G has dimension m−δ0c , while the space of linear functionals φ
has dimension degQ = n−m. So the space of such linear functionals T has dimension n− δ0c .
The space of all gt-twisted traces has the same dimension, so it is enough to prove that all
linear functionals T of this form are gt-twisted traces. In other words, we should prove that
T
(
S
(
z − 1
2
)
P
(
z − 1
2
)
− tS
(
z + 1
2
)
P
(
z + 1
2
))
= 0 for all S ∈ C[x].
We see that S
(
x− 1
2
)
P
(
x− 1
2
)
− tS
(
x+ 1
2
)
P
(
x+ 1
2
)
is divisible by Q(x), so
T
(
S
(
z − 1
2
)
P
(
z − 1
2
)
− tS
(
z + 1
2
)
P
(
z + 1
2
))
=
∫
iR
(
S
(
x− 1
2
)
P
(
x− 1
2
)
− tS
(
x+ 1
2
)
P
(
x+ 1
2
))
w(x)|dx|.
Since w
(
x+ 1
2
)
P (x) is holomorphic on |Rex| ≤ 1
2 , we deduce that this integral is zero similarly
to the proof of Proposition 3.1 �
Twisted Traces and Positive Forms on Quantized Kleinian Singularities of Type A 13
3.5 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)CαiS
(i)(α), where Cαi ∈ C. Therefore any
gt-twisted trace T is given by
T
(
S
(
z − 1
2
)
− tS
(
z + 1
2
))
=
∑
α, 0≤i<m(α)
CαiS
(i)(α).
Let P̃ (x) be the following polynomial: all roots of P̃ (x) belong to the strip |Rex| ≤ 1
2 and
the multiplicity of a root α equals to
�
∑
k∈Zm(α+ k) if |Reα| < 1
2 ;
�
∑
k≥0m(α+ k) if Reα = 1
2 ;
�
∑
k≤0m(α+ k) if Reα = −1
2 .
So P̃ (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 |Rex| ≤ 1
2 . In particular, the roots of P̃ (x) are symmetric
with respect to iR.
Suppose that α ∈ C has real part bigger than 1
2 , S(x) is an arbitrary polynomial, R(x) =
S
(
x − 1
2
)
− tS
(
x + 1
2
)
, i ≥ 0. Let r be the smallest positive integer such that Re(α) − r ≤ 1
2 .
Then
S(i)(α) =
r−1∑
k=0
(
t−kS(i)(α− k)− t−k−1S(i)(α− k − 1)
)
+ t−rS(i)(α− r)
=
r−1∑
k=0
t−k−1R(i)
(
α− k − 1
2
)
+ t−rS(i)(α− r) = φi,α(R) + t−rS(i)(α− r),
where
φi,α(R) :=
r−1∑
k=0
t−k−1R(i)
(
α− k − 1
2
)
.
We can write a similar equation for α ∈ C with real part smaller than −1
2 .
Therefore
T
(
S
(
z − 1
2
)
− tS
(
z + 1
2
))
=
∑
α, 0≤i<m(α)
CαiS
(i)(α)
=
∑
α, 0≤i<m(α)
Cαiφi,α(R) + t−r(α)CαiS
(i)(α− r) = Φ(R) + T̃ (R(z)),
where Φ(R) :=
∑
Re a6=0,k≥0 cakR
(k)(a), cak ∈ C, T̃ is a gt-twisted trace for the quantization
defined by the polynomial P̃ (x). Below we will abbreviate this sentence to “T̃ is a trace for P̃”.
Let P◦ be the following polynomial: all the roots of P◦ belong to strip |Rex| ≤ 1
2 and the
multiplicity of α, |Reα| ≤ 1
2 in P◦ equals the multiplicity of α in P .
Since φi,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:
14 P. Etingof, D. Klyuev, E. Rains and D. Stryker
Proposition 3.11. Suppose that P is any polynomial, P̃ is obtained from P by the minimal
integer shift of roots into the strip |Rex| ≤ 1
2 , and P◦ is obtained from P by throwing out roots
not in the strip |Rex| ≤ 1
2 . Then any twisted trace T on AP can be represented as T = Φ + T̃ ,
where
Φ(R) =
∑
a/∈iR, k≥0
cakR
(k)(a),
and T̃ is a trace for P̃ . 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 |Rex| < 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 |Rex| < 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.
4 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 Lp(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 Lp(R, w)∗ = Lq(R, w), it suffices to show that any function
f ∈ Lq(R, w) such that
∫
R f(x)xnw(x)dx = 0 for all nonnegative integers n must be zero.
Choose 0 < a < 1
p . We have ea|x| ∈ Lp(R, w). Therefore f(x)ea|x|w(x) ∈ L1(R). Denote
f(x)w(x) by F (x). Let F̂ be the Fourier transform of F . Since F (x)ea|x| ∈ L1(R), F̂ extends to
a holomorphic function in the strip | Imx| < a.
Since
∫
R f(x)xnw(x)dx = 0, we have
∫
R F (x)xndx = 0, so F̂ (n)(0) = 0. Since F̂ is a holomor-
phic function and all derivatives of F̂ at the origin are zero, we deduce that F̂ = 0. Therefore
F = 0, so f = 0 almost everywhere, as desired. �
We get the following corollaries:
Lemma 4.2. Let w satisfy the assumptions of Lemma 4.1.
1. Suppose that H(x) is a continuous complex-valued function on R with finitely many zeros
and at most polynomial growth at infinity. Then the set {H(x)S(x) |S(x) ∈ C[x]} is dense
in the space Lp(R, w).
2. Suppose that M(x) is a nonzero polynomial nonnegative on the real line. Then the closure
of the set {M(x)S(x)S(x) |S(x) ∈ C[x]} in Lp(R, w) is the subset of almost everywhere
nonnegative functions.
Proof. 1. The function w(x)|H(x)|p satisfies the assumptions of Lemma 4.1. Therefore poly-
nomials are dense in the space Lp(R, w|H|p). The map g 7→ gH is an isometry between
Lp(R, w|H|p) and Lp(R, w). The statement follows.
2. Suppose that f ∈ Lp(R, w) is nonnegative almost everywhere. Then
√
f is an element
of L2p(R, w). Using (1), we find a sequence Sn ∈ C[x] such that
√
MSn tends to
√
f in L2p(R, w).
Twisted Traces and Positive Forms on Quantized Kleinian Singularities of Type A 15
We use the following corollary of Cauchy–Schwarz inequality: if ak, bk tend to a, b respectively
in L2p(R, w) then akbk tends to ab in Lp(R, w). Applying this to a = b =
√
f , an =
√
MSn,
bk =
√
MSn we deduce that MSnSn tends to f in Lp(R, w). The statement follows. �
4.2 The case when all roots of P (x) satisfy |Reα| < 1
2
Let A be a filtered quantization of A with conjugations ρ± such that ρ2± = gt. 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
|Reα| < 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
gt-twisted ρ-invariant traces, i.e., gt-twisted traces T such that T (ρ(a)) = T (a). The correspon-
dence is as follows:
(a, b) = T (aρ(b)), T (a) = (a, 1).
Therefore it is enough to classify gt-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∈ZAd. Thus it is enough to check the condition
T (aρ(a)) > 0 for homogeneous a.
Lemma 4.3.
1. T gives a positive definite form if and only if one has T (aρ(a)) > 0 for all nonzero a ∈ A
of weight 0 or 1.
2. T gives a positive definite form if and only if
T
(
R(z)R(−z)
)
> 0 and λT
(
R
(
z − 1
2
)
R
(
1
2 − z
)
P
(
z − 1
2
))
> 0
for all nonzero R ∈ C[x].
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 = vkbvk. We have
T (aρ(a)) = λ2kT
(
vkbvkukρ(b)uk
)
= λ2kT
(
g−1t
(
uk
)
vkbvkukρ(b)
)
= λ2ktkT
(
ukvkbvkukρ(b)
)
= (−1)nkT
(
ukvkbvkukρ(b)
)
= T
(
ukvkbρ
(
ukvkb
))
> 0
since ukvkb 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)ρ2(b)) = T (ρ(b)gt(b)) = T (bρ(b)) > 0.
2. Suppose that a is an element of A0. 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 A1. Then a = R
(
z − 1
2
)
v for some R ∈ C[x]. We have
T (aρ(a)) = λT
(
R
(
z − 1
2
)
vR
(
− z − 1
2
)
u
)
= λT
(
R
(
z − 1
2
)
vuR
(
− z + 1
2
))
= λT
(
R
(
z − 1
2
)
R
(
1
2 − z
)
P
(
z − 1
2
))
.
The statement follows. �
16 P. Etingof, D. Klyuev, E. Rains and D. Stryker
Proposition 4.4. Suppose that T (R(z)) =
∫
iRR(x)w(x)|dx|. Then T gives positive definite
form if and only if w(x) and λw
(
x+ 1
2
)
P (x) are nonnegative on iR.
Proof. By Lemma 4.3 T gives positive definite form if and only if
T (R(z)R(−z)) > 0
and
λT
(
R
(
z − 1
2
)
R
(
1
2 − z
)
P
(
z − 1
2
))
> 0
for all nonzero R ∈ C[x]. A polynomial S ∈ C[x] can be represented as S(x) = R(x)R(−x)
if and only if S is nonnegative on iR. So we have T (R(z)R(−z)) > 0 for all nonzero R ∈ C[x]
if and only if∫
iR
S(x)w(x)|dx| > 0
for all nonzero S ∈ C[x] nonnegative on iR. Using Lemma 4.2(2) for M = 1, we see that this is
equivalent to w(x) being nonnegative on iR.
We have
T
(
R
(
z − 1
2
)
R
(
1
2 − z
)
P
(
z − 1
2
))
=
∫
iR
R
(
x− 1
2
)
R
(
1
2 − x
)
P
(
x− 1
2
)
w(x)|dx|
=
∫
1
2
+iR
R(x)R
(
− x
)
P (x)w
(
x+ 1
2
)
|dx|
=
∫
iR
R(x)R
(
− x
)
P (x)w
(
x+ 1
2
)
|dx|.
In the last equality we used the Cauchy theorem and the fact that the function P (x)w
(
x+ 1
2
)
is holomorphic when |Rex| ≤ 1
2 . Using Lemma 4.2(2) for M = 1 again, we see that λT
(
R
(
z − 1
2
)
×R
(
1
2 − z
)
P
(
z − 1
2
))
> 0 for all nonzero R ∈ C[x] if and only if λP (x)w
(
x+ 1
2
)
is nonnegative
on iR. �
From now on we assume that all roots α of P (x) satisfy |Reα| < 1
2 . In this case every trace T
can be represented as T (R(z)) =
∫
iRR(x)w(x)|dx|. Recall that w(x) = e2πicx G(e2πix)
P(e2πix)
, where G
is any polynomial with degG ≤ degP in the case when c 6= 0 and degG < degP in the case
when c = 0.
Proposition 4.5.
1. If λ = −i−ne−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.
2. If λ = +i−ne−πic (i.e., ρ = ρ+) then w(x) and λw
(
x+ 1
2
)
P (x) are nonnegative on iR if and
only if G(X) is nonnegative for all X ∈ R.
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
λP (x)w
(
x+
1
2
)
= ±i−nP (x)e2πicx
G(−e2πix)
P(−e2πix)
.
It is clear that i−nP (x)
P(−e2πix) belongs to R when x ∈ iR and does not change sign on iR. When x
tends to −i∞, the functions i−nP (x) and P(−e2πix) have sign (−1)n. Therefore i−nP (x)
P(e−2πix)
is
positive on iR. We deduce that ±G(X) should be nonnegative when X < 0. So there are two
cases:
Twisted Traces and Positive Forms on Quantized Kleinian Singularities of Type A 17
1. If λ = −i−ne−πic then G(X) should be nonnegative when X > 0 and nonpositive when
X < 0.
2. If λ = +i−ne−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,
ρ2 = gt, t = exp(2πic). Let P (x) be the parameter of A, ε = ineπicλ = ±1 (so ρ = ρε). 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) = X2U(X) where U(X) ≥ 0 is a polynomial of degree ≤ n− 3. Therefore, we get
Proposition 4.7. The dimension of C+ modulo scaling is
• n− 2 for even n and n− 3 for odd n if ρ = ρ−;
• n− 2 for even n and n− 1 for odd n if c 6= 0 and ρ = ρ+;
• n− 4 for even n and n− 3 for odd n if c = 0 and ρ = ρ+.
(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 Ceven+ 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 Ceven+ modulo scaling in various cases are as follows:
• n−3
2 if ρ = ρ−, n odd;
• n−1
2 if ρ = ρ+, n odd;
• n−2
2 if ρ = ρ−, n even;
• n−4
2 if ρ = ρ+, n even.
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 Ceven+ modulo scaling is 2-dimensional (which disproves the most optimistic conjecture
of [2] that a unitary even star-product is always unique).5
5It is curious that in the case considered in [2], the dimension of Ceven
+ modulo scaling is always even.
18 P. Etingof, D. Klyuev, E. Rains and D. Stryker
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 6= 0. In this case there is a unique positive
trace up to scaling given by the weight function
w(y) =
e2πi(c−
1
2)y
2 cosπy
.
In particular, the only quaternionic case is ρ = ρ+, c = 1
2 , which gives w(y) = 1
2 cosπy .
Example 4.10. Let n = 2, P (x) = x2 +β2, β2 ∈ R so we have P(X) = (X + e2πβ)(X + e−2πβ).
We assume that β2 > −1
4 so that all roots of P are in the strip |Rex| < 1
2 . Then ρ = ρ− gives
a unique up to scaling positive trace defined by the weight function
w(y) =
e2πcy
4 cosπ(y − β) cosπ(y + β)
,
and ρ = ρ+ is possible if and only if c 6= 0 and gives a unique up to scaling positive trace defined
by the weight function
w(y) =
e2π(c−1)y
4 cosπ(y − β) cosπ(y + β)
.
In particular, the only quaternionic case is ρ = ρ−, c = 0, with
w(y) =
1
4 cosπ(y − β) cosπ(y + β)
,
which corresponds to the SL2-invariant short star-product. There are two subcases: β2 ≥ 0,
which corresponds to the spherical unitary principal series for SL2(C), and −1
4 < β2 < 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 β2 = −1
4
)
, these repre-
sentations are well known to exhaust irreducible spherical unitary representations of SL2(C) [15].
Example 4.11. Let n = 3 and P (x) = x3 + β2x = x(x − iβ)(x + iβ), where β2 ∈ R. This
gives the algebra defined by formulas (6.17), (6.18) of [2], with ζ = 1; namely, the generators
X̂, Ŷ , Ẑ of [2] are v, u, z, respectively, and the parameter κ of [2] is κ = −β2 − 1
4 . This
is an even quantization of A = C[X3]. Thus even short star-products are 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 (z2) = −α (using the notation of [2]).
Assume that β2 > −1
4 (i.e., κ < 0), so that all the roots of P are in the strip |Rex| < 1
2 .
We have
P(X) = (X + 1)
(
X + e2πβ
)(
X + e−2πβ
)
.
In this case c = 1
2 so the trace T , up to scaling, is given by
T (R(z)) =
∫
iR
R(x)w(x)|dx|,
where
w(x) = eπix
G(e2πix)
(e2πix + 1)(e2πi(x−iβ) + 1)(e2πi(x+iβ) + 1)
,
Twisted Traces and Positive Forms on Quantized Kleinian Singularities of Type A 19
with deg(G) ≤ 2. Moreover, because of evenness we must have w(x) = w(−x), so G(X) =
X2G
(
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 = AP with the
quaternionic structure ρ− given by6
ρ−(v) = −u, ρ−(u) = v, ρ−(z) = −z,
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
cosπx cosπ(x− iβ) cosπ(x+ iβ)
.
Hence,
T (zk) =
∫
iR
xk|dx|
cosπx cosπ(x− iβ) cosπ(x+ iβ)
,
in particular, T (zk) = 0 if k is odd.
For even k this integral can be computed using the residue formula. Namely, assume β 6= 0
and let us first compute T (1). Replacing the contour iR by 1+iR and subtracting, we find using
the residue formula:
2T (1) = 2π
(
Res 1
2
w + Res 1
2
−iβw + Res 1
2
+iβw
)
.
Now,
Res 1
2
w =
1
π sinh2 πβ
,
while
Res 1
2
−iβw = Res 1
2
+iβw = − 1
π sinhπβ sinh 2πβ
.
Thus
T (1) =
1
sinh2 πβ
− 2
sinhπβ sinh 2πβ
=
1
sinh2 πβ
(
1− 1
coshπβ
)
=
1
2 cosh2(πβ2 ) coshπβ
.
Note that this function has a finite value at β = 0, which is the answer in that case.
Now let us compute T
(
z2
)
. Again replacing the contour iR with 1 + iR and subtracting,
we get
T (1) + 2T
(
z2
)
= T
(
z2
)
+ T
(
(z + 1)2
)
= 2π
(
Res 1
2
x2w + Res 1
2
−iβx
2w + Res 1
2
+iβx
2w
)
.
Now,
Res 1
2
x2w =
1
4π sinh2 πβ
,
while
Res 1
2
−iβx
2w + Res 1
2
+iβx
2w =
2β2 − 1
2
π sinhπβ sinh 2πβ
.
6Note that our ρ is ρ−1 in [2], so we use ρ− while [2] use ρ+ = ρ−1
− .
20 P. Etingof, D. Klyuev, E. Rains and D. Stryker
So
T
(
z2
)
= − 1
4 sinh2 πβ
+
2β2 + 1
2
sinhπβ sinh 2πβ
=
1
sinh2 πβ
(
− 1
4
+
β2 + 1
4
coshπβ
)
.
Thus,
α = −
T
(
z2
)
T (1)
=
1
4
+
β2
1− coshπβ
=
1
4
−
κ+ 1
4
1− cosπ
√
κ+ 1
4
.
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
π2 . the value found in [2].
Example 4.12. Let n = 4 and
P (x) =
(
x2 + β2
)(
x2 + γ2
)
= (x− iβ)(x+ iβ)(x− iγ)(x+ iγ),
where β2, γ2 ∈ R. This is an even quantization of A = C[X4] 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
(
z2
)
= −α.
Assume that β2, γ2 > −1
4 , so that all the roots of P are in the strip |Rex| < 1
2 . We have
P(X) =
(
X + e2πβ
)(
X + e−2πβ
)(
X + e2πγ
)(
X + e−2πγ
)
.
In this case c = 0 so the trace T , up to scaling, is given by
T (R(z)) =
∫
iR
R(x)w(x)|dx|,
where
w(x) =
G(e2πix)
(e2πi(x−iβ) + 1)(e2πi(x+iβ) + 1)(e2πi(x−iγ) + 1)(e2πi(x+iγ) + 1)
,
with deg(G) ≤ 3 and G(0) = 0. Moreover, because of evenness we must have w(x) = w(−x),
so G(X) = X4G
(
X−1
)
. Up to scaling, such polynomials G form a 1-parameter family, para-
metrized by α.
Let us equip the corresponding quantum algebra A = AP with the quaternionic structure ρ+
given by ρ+(v) = u, ρ+(u) = v, ρ+(z) = −z, 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) = X2. Thus this trace is given by the weight
function
w(x) =
1
cosπ(x− iβ) cosπ(x+ iβ) cosπ(x− iγ) cosπ(x+ iγ)
.
Thus,
T
(
zk
)
=
∫
iR
xk|dx|
cosπ(x− iβ) cosπ(x+ iβ) cosπ(x− iγ) cosπ(x+ iγ)
,
in particular, T
(
zk
)
= 0 if k is odd.
Twisted Traces and Positive Forms on Quantized Kleinian Singularities of Type A 21
As before, for even k this integral can be computed using the residue formula. Namely,
assume β 6= 0, γ 6= 0, β 6= ±γ, and let us first compute T (1). Replacing the contour iR by 1+iR
and subtracting, we find using the residue formula:
T
(
(z + 1)2
)
− T
(
z2
)
= T (1)
= −2π
(
Res 1
2
−iβx
2w + Res 1
2
+iβx
2w + Res 1
2
−iγx
2w + Res 1
2
+iγx
2w
)
.
Now,
Res 1
2
−iβx
2w + Res 1
2
+iβx
2w =
2β
π sinhπ(β + γ) sinhπ(γ − β) sinh 2πβ
.
Thus
T (1) =
1
sinhπ(β + γ) sinhπ(γ − β)
(
4β
sinh 2πβ
− 4γ
sinh 2πγ
)
.
Note that this function is regular when βγ(β − γ)(β + γ) = 0, and the corresponding limit is
the answer in that case.
We similarly have
T
(
(z + 1)4
)
− T
(
z4
)
= 6T
(
z2
)
+ T (1)
= −2π
(
Res 1
2
−iβx
4w + Res 1
2
+iβx
4w + Res 1
2
−iγx
4w + Res 1
2
+iγx
4w
)
,
and
Res 1
2
−iβx
4w + Res 1
2
+iβx
4w =
β − 4β3
π sinhπ(β + γ) sinhπ(γ − β) sinh 2πβ
.
Thus
6T
(
z2
)
+ T (1) =
2
sinhπ(β + γ) sinhπ(γ − β)
(
β − 4β3
sinh 2πβ
− γ − 4γ3
sinh 2πγ
)
.
Hence
T
(
z2
)
=
1
sinhπ(β + γ) sinhπ(γ − β)
(
γ + 4γ3
3 sinh 2πγ
− β + 4β3
3 sinh 2πβ
)
.
Thus
α = −
T
(
z2
)
T (1)
=
1
12
+
1
3
β3 sinh 2πγ − γ3 sinh 2πβ
β sinh 2πγ − γ sinh 2πβ
.
This is the equation (in appropriate coordinates) of the surface computed numerically in [2] and
shown in Fig. 4 of that paper. In particular, for β = γ = 0, we get
α =
1
12
− 1
2π2
.
Thus τ = 128α = 32(π2−6)
3π2 = 4.18211. . . is the number given by the complicated expression (B.16)
of [2] (as was pointed out in [5]).
Remark 4.13. Similar calculations can be found in [5, Section 8.1].
22 P. Etingof, D. Klyuev, E. Rains and D. Stryker
4.3 The case of a closed strip
Suppose now that all roots α of P satisfy |Reα| ≤ 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 |Rex| < 1
2 and all roots of Q(x) belong to iR. For any
R ∈ C[x] write R = R1Q+R0, where degR0 < degQ.
By Proposition 3.10 each gt-twisted trace can be obtained as
T (R(z)) =
∫
iR
R1(x)Q(x)w(x)|dx|+ φ(R0),
where w(x) = e2πicx G(e2πix)
P(e2πix)
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.
Proof. Let Q∗(x) = Q(x)Q(−x); note that Q∗(x) ≥ 0 for x ∈ iR. Then there exists a linear
functional ψ such that for any R = R1Q∗ +R0 with degR0 < degQ∗ we have
T (R(z)) =
∫
iR
R1(x)Q∗(x)w(x)|dx|+ ψ(R0).
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)S1(x) and using Lemma 4.2, we deduce that Q2
∗(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
Rn(x) := (FnQ∗ + b)
(
FnQ∗ + b
)
,
where b ∈ R. Suppose that Fn is a sequence of polynomials that tends to the function f := χ(−ε,ε)
(the characteristic function of the interval) in the space L2
(
iR, (Q∗ + Q2
∗
)
w). In particular,
Fn tends to f in the spaces L2(iR, Q∗w) and L2
(
iR, Q2
∗w
)
. Then we deduce from the Cauchy–
Schwartz inequality that FnFn tends to f2 in the space L1
(
iR, Q2
∗w
)
, and Fn and Fn tend to f
in L1(iR, Q∗w).
We have
T (Rn(z)) = T
((
FnFnQ∗ + Fnb+ Fnb
)
(z)Q∗(z) + b2
)
=
∫
iR
(
FnFnQ
2
∗ + FnbQ∗ + FnbQ∗
)
w|dx|+ φ
(
b2
)
.
Therefore, when n tends to infinity,
T (Rn(z))→
∫
iR
(
f2Q2
∗ + 2fbQ∗
)
w|dx|+ φ
(
b2
)
.
We have φ
(
b2
)
= Cb2 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 Q2
∗w has
a zero of order 2N −M at zero. We deduce that∫
iR
FnQ∗w|dx| → c1ε
N−M+1,
∫
iR
FnQ
2
∗w|dx| → c2ε
2N−M+1, n→∞,
where c1 = c1(ε), c2 = c2(ε) are functions having strictly positive limits at ε = 0 . Therefore
lim
n→∞
T (Rn(z)) = Cb2 + 2c1ε
N−M+1b+ c2ε
2N−M+1.
Twisted Traces and Positive Forms on Quantized Kleinian Singularities of Type A 23
This is a quadratic polynomial of b with discriminant
D = 4ε2N−2M+2
(
c21 − Cc2εM−1
)
.
Since M≥2, for small ε this discriminant is positive. In particular, for some b, lim
n→∞
T (Rn(z))<0,
so for this b and some n, T (Rn(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)) =∫
iRR(x)w(x)|dx|+ η(R0), where η is some linear functional.
Proposition 4.15. T gives a positive definite form only when η(R0) =
∑
j cjR0(zj), where
cj ≥ 0 and zj ∈ 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 Fn such that FnQ + S tends to
zero in L2(iR, w). We deduce that
T
(
(FnQ+ S)(z)(FnQ+ S)(z)
)
→ η
(
(SS)0
)
< 0,
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
T
(
R
(
z − 1
2
)
R
(
1
2 − z
)
P
(
z − 1
2
))
=
∫
iR
R
(
z − 1
2
)
R
(
1
2 − z
)
P
(
z − 1
2
)
w(z)|dz|.
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.
Theorem 4.16. Suppose that P (x) = P∗(x)Q
(
x− 1
2
)
Q
(
x+ 1
2
)
, where all roots of P∗ belong to
the set |Rex| < 1
2 and all roots of Q belong to iR. Suppose that α1, . . . , αk are all the different
roots of Q. Then positive traces T are in one-to-one correspondence with T̃ , c1, . . . , ck ≥ 0,
where T̃ is a positive trace for P∗; namely,
T (R(z)) = T̃ (R(z)) +
∑
ciR(αi).
4.4 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 P̃ (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
|Rex| ≤ 1
2 . Also recall from Section 3.5 that P◦ denotes the following polynomial: all roots
of P◦ belong to strip |Rex| ≤ 1
2 and the multiplicity of α, |Reα| ≤ 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 + T̃ , where Φ is a linear
functional such that
Φ(R) =
m∑
j=1
∑
k
cjkR
(k)(zj),
zj /∈ iR, and T̃ is a trace for P̃ . Furthermore, if Φ = 0 then T is just a trace for P◦.
24 P. Etingof, D. Klyuev, E. Rains and D. Stryker
Proposition 4.17. Let T be a trace such that Φ 6= 0. Then T does not give a positive definite
form.
Proof. For big enough k we have Φ((x − z1)k · · · (x − zm)kC[x]) = 0. Recall that there exists
polynomial Q∗(x) nonnegative on iR such that for R = R1Q∗ + R0, degR0 < degQ∗, we have
T̃ (R) =
∫
iRR1Q∗w|dx| + ψ(R0), 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 Gn = USn that
tends to L in the space L2(iR, Q∗w|dx|). We deduce that Hn(x) := (Gn(x) − L(x))(Gn(−x) −
L(−x)) tends to zero in L1(iR, Q∗w). We have T̃ (Hn(z)Q∗(z)) =
∫
iRHn(x)Q∗w|dx|. We con-
clude that T̃ (Hn(z)Q∗(z)) = ‖Hn‖L1(iR,Q∗w) tends to zero when n tends to infinity.
It follows that T (Hn(z)) tends to Φ(Q∗(x)Hn(x)) = Φ
(
Q∗(x)L(x)L(−x)
)
. Since Hn is non-
negative on iR, we have T (Hn(z)) > 0. Now we get a contradiction with
Lemma 4.18. There exists F (x) ∈ C[x] such that Φ
(
Q∗(x)F (x)F (−x)
)
< 0.
Proof. Let r be the biggest number such that there exists j with cjr 6= 0. Let
F (x) := G(x)(x− z1)r+1 · · · (x− zj)r · · · (x− zm)r+1.
Here we omit x − zj∗ in the product, where j∗ 6= j is such that zj∗ = −zj . We note that
cik
(
Q∗(x)F (x)F (−x)
)(k)
(zi) = 0 for all i, k except k = r and i = j or i = j∗. It follows that
Φ
(
Q∗(x)F (x)F (−x)
)
= cjr
(
Q∗(x)F (x)F (−x)
)(r)
(zj) + cj∗r
(
Q∗(x)F (x)F (−x)
)(r)
(zj∗)
= cjrQ∗(zj)F
(r)(zj)F (−zj) + (−1)rcj∗rQ∗(zj∗)F (zj∗)F
(r)
(−zj∗)
= cjra+ cj∗ra,
where a := Q∗(zj)F
(r)(zj)F (−zj). Pick a ∈ C so that
cjra+ cj∗ra = 2Re(cjra) < 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 AP◦ is the quantization with parameter P◦ then there is a conjugation ρ◦ on AP◦ given by
the formulas
ρ◦(v) = λ◦u, ρ◦(u) = (−1)nλ−1◦ v, ρ◦(z) = −z,
where λ◦ := (−1)
n−n◦
2 λ. Therefore we can consider the cone of positive definite forms for AP◦
with respect to ρ◦.
Corollary 4.19. The cone of positive definite forms on AP with respect to ρ coincides with the
cone of positive definite forms on AP◦ with respect to ρ◦. Namely, a trace T : C[x] → C for A
gives a positive definite form if and only if T is a trace for AP◦ that gives a positive definite
form on AP◦.
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 = R1Q∗ +R0, degR0 < degQ∗, and
T (R(z)) =
∫
iR
Q∗(x)R1(x)w(x)|dx|+ φ(R0).
Twisted Traces and Positive Forms on Quantized Kleinian Singularities of Type A 25
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
T (R(z)) =
∫
iR
R(x)w(x)|dx|+ ψ(R0),
where ψ is some linear functional. Using Proposition 4.15, we deduce that this trace is positive
if and only if ψ(R0) =
∑
j cjR0(zj), where cj ≥ 0 and zj ∈ iR.
Since (−1)
n−n◦
2
P (x)
P◦(x)
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 = AP be a filtered quantization of A with parameter P equipped with
a conjugation ρ such that ρ2 = gt. Let ` be the number of roots α of P such that |Reα| < 1
2
counted with multiplicities and r be the number of distinct roots α of P with Reα = −1
2 . Then
the cone C+ of ρ-equivariant positive definite traces on A is isomorphic to C1+ × C2+, where
C2+ = Rr≥0, and C1+ 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 ak, bk
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 ak, bk of the 3-term recurrence for the corresponding orthogonal
polynomials:
pk+1(x) = (x− bk)pk(x)− akpk−1(x).
Also recall [14] that ak = νk
νk−1
, where νk := (Pk, Pk). Finally, recall that νk = Dk
Dk−1
, where Dk
is the Gram determinant for 1, x, . . . , xk−1, i.e.,
Dk = det
0≤i, j≤k−1
(
xi, xj
)
= det
0≤i, j≤k−1
(Mi+j),
where Mr is the r-th moment of the weight function w(x), i.e.,
Mr =
∫
iR
xrw(x)|dx|.
In the even case w(−x) = w(x) we have bk = 0, so
pk+1(x) = xpk(x)− akpk−1(x),
26 P. Etingof, D. Klyuev, E. Rains and D. Stryker
and pk can be easily computed recursively from the sequence ak. If the polynomials pk are
q-hypergeometric (i.e., obtained by a limiting procedure from Askey–Wilson polynomials), then
Dk, νk, ak admit explicit product formulas, but in general they do not admit any closed expres-
sion and do not enjoy any nice algebraic properties beyond the above.
In our case, the hypergeometric case only arises for n = 1 or, in special cases, n = 2, but
the fact that the weight function for general n is essentially a higher complexity version of the
weight function for n = 1 suggests that there is still a weaker algebraic structure in the picture.
In fact, by [12] it follows immediately from the fact that the formal Stieltjes transform satisfies
an inhomogeneous first-order difference equation with rational coefficients that the corresponding
orthogonal polynomials pm(x) in the x-variable satisfy a family of difference equations(
pm
(
x+ 1
2
)
pm−1
(
x+ 1
2
)) = Am(x)
(
pm
(
x− 1
2
)
pm−1
(
x− 1
2
))
such that the matrix Am(x) has rational function coefficients of degree bounded by a linear
function of n alone. (Here we work with the “x” version of the polynomials, to avoid unnecessary
appearances of i.)
Since the results of [12] are stated in significantly more generality than we need, we sketch
how they apply in our special case. Let Y0 be the matrix
Y0(x) =
(
1 F (x)
0 1
)
,
where F is the formal Stieltjes transform of the given trace. Moreover, for each n, let qn(x)
pn(x)
be the
n-th Padé approximant to F (x) (with monic denominator), so that qn(x)
pn(x)
− F (x) = O
(
x−2n−1
)
.
If we define
Yn(x) :=
(
pn(x) −qn(x)
pn−1(x) −qn−1(x)
)
Y0(x)
for n > 0, then
Yn =
(
xn + o
(
xn
)
O
(
x−n−1
)
xn−1 + o
(
xn−1
)
O
(
x−n
) ) .
Lemma 5.1. The denominator pn of the n-th Padé approximant to F (x) is the degree n monic
orthogonal polynomial for the associated linear functional T .
Proof. If F = FT , then we find
pn(x)F (x) = T
(
pn(x)
x− z
)
= T
(
pn(x)− pn(z)
x− z
)
+ T
(
pn(z)
x− z
)
(where we evaluate T on functions of z, and x is a parameter). The two terms correspond to
the splitting of pn(x)F (x) into its polynomial part and its part vanishing at x =∞, so that
qn(x) = T
(
pn(x)− pn(z)
x− z
)
and
T
(
pn(z)
x− z
)
= pn(x)F (x)− qn(x) = O
(
x−n−1
)
.
Comparing coefficients of x−m−1 for 0 ≤ m < n implies that T (zmpn(z)) = 0 as required. �
Twisted Traces and Positive Forms on Quantized Kleinian Singularities of Type A 27
Remark 5.2.
1. It also follows that
Yn(x)12 = Nnx
−n−1 +O
(
x−n−2
)
, Yn(x)22 = Nn−1x
−n +O
(
x−n−1
)
.
2. Note that this is an algebraic/asymptotic version of the explicit solution of [3] to the
Riemann–Hilbert problem for orthogonal polynomials introduced in [7].
Lemma 5.3. We have det(Yn) = Nn−1 for all n > 0.
Proof. The definition of Yn implies that det(Yn) ∈ C[x], while the (formal) asymptotic behavior
implies that det(Yn) = Nn−1 +O
(
1
x
)
. �
The inhomogeneous difference equation satisfied by F trivially induces an inhomogeneous
difference equation satisfied by Y0:
Y0
(
x+ 1
2
)
=
(
1 t−1 L(x)P (x)
0 t−1
)
Y0
(
x− 1
2
)(1 0
0 t
)
,
where
L(x) = P (x)
(
F
(
x+ 1
2
)
− tF
(
x− 1
2
))
∈ C[x].
It follows immediately that Yn satisfies an analogous equation
Yn
(
x+ 1
2
)
= An(x)Yn
(
x− 1
2
)(1 0
0 t
)
,
where
An(x) =
(
pn
(
x+ 1
2
)
−qn
(
x+ 1
2
)
pn−1
(
x+ 1
2
)
−qn−1
(
x+ 1
2
))(1 t−1 L(x)P (x)
0 t−1
)(
pn
(
x− 1
2
)
−qn
(
x− 1
2
)
pn−1
(
x− 1
2
)
−qn−1
(
x− 1
2
))−1 .
Since det(Yn) = Nn−1, det(Y0) = 1, we can use the standard formula for the inverse of a 2 × 2
matrix to rewrite this as
An(x) = N−1n−1
(
pn
(
x+ 1
2
)
−qn
(
x+ 1
2
)
pn−1
(
x+ 1
2
)
−qn−1
(
x+ 1
2
))(1 t−1 L(x)P (x)
0 t−1
)(
−qn−1
(
x− 1
2
)
qn
(
x− 1
2
)
−pn−1
(
x− 1
2
)
pn
(
x− 1
2
)) .
It follows immediately that P (x)An(x) has polynomial coefficients. We can also compute the
asymptotic behavior of An(x) using the expression
An(x) = Yn
(
x+ 1
2
)(1 0
0 t−1
)
Yn
(
x− 1
2
)−1
to conclude that
An(x)11 = 1 + n
x +O
(
1
x2
)
, An(x)12 = − (1−t−1)an
x +O
(
1
x2
)
,
An(x)21 = 1−t−1
x +O
(
1
x2
)
, An(x)22 = t−1(1− n
x ) +O
(
1
x2
)
,
which when t = 1 refines to
An(x)11 = 1 + n
x +O
(
1
x2
)
, An(x)12 = − (2n+1)an
x2
+O
(
1
x3
)
,
An(x)21 = 2n−1
x2
+O
(
1
x3
)
, An(x)22 = 1− n
x +O
(
1
x2
)
.
Restricting to the first column of Yn(x) gives the following.
28 P. Etingof, D. Klyuev, E. Rains and D. Stryker
Proposition 5.4. The orthogonal polynomials satisfy the difference equation(
pn
(
x+ 1
2
)
pn−1
(
x+ 1
2
)) = An(x)
(
pn
(
x− 1
2
)
pn−1
(
x− 1
2
)) .
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 6= 1) the family of matrices satisfying the above conditions; that is,
PAn is polynomial, det(An) = t−1, and
An(x)11 = 1 + n
x +O
(
1
x2
)
, An(x)12 = O
(
1
x
)
,
An(x)21 = 1−t−1
x +O
(
1
x2
)
, An(x)22 = t−1
(
1− n
x
)
+O
(
1
x2
)
,
we find that the family is classified by a rational moduli space. To be precise, let f(x) :=(
1− t−1
)−1
P (x)An(x)21, and let g(x) ∈ C[x]/(f(x)) be the reduction of P (x)An(x)11 modu-
lo f(x). Then f and g both vary over affine spaces of dimension deg(q) − 1, and generically
determine An. Indeed, An(x)21 is clearly determined by f , and since An(x)11P (x) is speci-
fied 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 An(x)22, since the determinant condition
implies that for any root α of f , An(α)11An(α)22 = t−1. Moreover, this constraint forces
P (x)2
(
An(x)11An(x)22 − t−1
)
to be a multiple of f(x), and thus the unique value of An(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 An+1 is the unique matrix satisfying its asymptotic conditions and
having the form
An+1(x) =
(
x+ 1
2 − bn −an
1 0
)
An(x)
(
x− 1
2 − bn −an
1 0
)−1
.
It is straightforward to see that an, bn are determined by the leading terms in the asymptotics
of An(x)12, and thus in particular are rational functions of the parameters. We thus find that
the map from the space of matrices An to the space of matrices An+1 is a rational map, and
by considering the inverse process, is in fact birational, corresponding to a sequence Fn of bira-
tional automorphisms of A2 deg(P )−2. Note that the equation A0, though not of the standard
form, is still enough to determine A1, and thus gives (rationally) a Pdeg(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) = x2, corresponding, e.g., to
w(y) =
e2πcy
cosh2 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 fn, and gn = An(fn)11 as the other parameter. We thus find that
An(x) =
((
1− fn
x
)(
1 + fn+n
x
)
+ f2ngn
x2
−an 1−t−1
x
(
1− fn+1
x
)
1−t−1
x
(
1− fn
x
)
t−1
((
1− fn
x
)(
1 + fn−n
x
)
+ f2n
gnx2
)) , (5.1)
Twisted Traces and Positive Forms on Quantized Kleinian Singularities of Type A 29
where
an =
t
(t− 1)2
n2gn − f2n(gn − 1)2
gn
(5.2)
and fn, gn are determined from the recurrence
fn+1 =
fn(fn(gn − 1)− ngn)(fn(gn − 1)− n)
n2gn − f2n(gn − 1)2
, (5.3)
gn+1 =
(fn(gn − 1)− ngn)2
tgn(fn(gn − 1)− n)2
. (5.4)
The three-term recurrence for the orthogonal polynomials is then
pn+1(x) = (x− bn)pn(x)− anpn−1(x),
where an is as above and
bn = −fn+1 −
(t+ 1)
(
n+ 1
2
)
t− 1
.
The initial condition is given by
f0 = b0 +
t+ 1
2(t− 1)
, g0 = 1.
(Note that the resulting A0 is not actually correct, but this induces the correct values for f1, g1,
noting that the recurrence simplifies for n = 0 to f1 = −f0, g1 = 1/tg0.) 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 7→
(
AT
)−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 An(x)12 has a nice expression in terms of an and fn+1 follows more
generally from the fact (via the three-term recurrence) that
An(x)12 = −anAn+1(x)21.
One similarly has
An(x)22 = An+1(x)11 −
(
x+ 1
2 − bn
)
An+1(x)21,
so that in general fn+1(x) ∝ P (x)An(x)12 and gn+1(x) = P (x)An(x)22 mod fn+1(x). In par-
ticular, applying this to n = 0 tells us that the orthogonal polynomial case corresponds to the
initial condition f1(x) ∝ L(x), g1(x) = t−1P (x) mod L(x).
The above construction fails for t = 1, because the constraint on the asymptotics of the
off-diagonal coefficients of An is stricter in that case:
An(x)21 = 2n−1
x2
+O
(
1
x3
)
,
An(x)12 = O
(
1
x2
)
.
30 P. Etingof, D. Klyuev, E. Rains and D. Stryker
The moduli space is still rational, although the arguments is somewhat subtler. We can still
parametrize it by fn(x) := P (x)An(x)12 and gn(x) := P (x)An(x)11 mod fn(x) as above, which
is certainly enough to determine P (x)An(x)22 modulo fn(x). This still leaves two degrees of free-
dom in the diagonal coefficients, but det(An(x))+O
(
1
x4
)
depends only on the diagonal coefficients
and is linear in the remaining degrees of freedom, so we can solve for those. Once again, hav-
ing determined the coefficients on and below the diagonal, the 21 coefficient follows from the
determinant, and can be seen to have the correct poles and asymptotics. Note that now the
dimension of the moduli space is 2 deg(q) − 4; that the dimension is even in both cases follows
from the existence of a canonical symplectic structure on such moduli spaces, see [13].
There is a similar reduction in the number of parameters when the trace is even (forcing
t = (−1)n and P (x) = (−1)nP (−x)). The key observation in that case is that
Yn(−x) = (−1)n
(
1 0
0 −1
)
Yn(x)
(
1 0
0 −1
)
implying that An satisfies the symmetry
An(−x) =
(
1 0
0 −1
)
An(x)−1
(
1 0
0 −1
)
.
Since An is 2 × 2 and has determinant t−1 = (−1)n, this actually imposes linear constraints
on the coefficients of An:
An(−x)11 = (−1)nAn(x)22, An(−x)12 = (−1)nAn(x)12,
An(−x)21 = (−1)nAn(x)21, An(−x)22 = (−1)nAn(x)11.
In particular, An(x)21 has only about half the degrees of freedom one would otherwise expect,
and for any root of that polynomial, An(α)11An(−α)11 = 1, again halving the degrees of freedom
(and preserving rationality).
Example 5.7. Consider the case P (x) = x3 + β2x with t = −1 and even trace
(
e.g., for β = 0,
the weight function w(y) = 1
cosh3 πy
)
. Then An(x)21 has the form 2(x2−fn)
x3+β2x
, and An
(√
fn
)
11
is
of norm 1, which can be parametrized in the form
An
(√
fn
)
11
=
gn +
√
fn
gn −
√
fn
.
Applying this to both square roots gives two linear conditions on An(x)11, which suffices to
determine it, with An(x)22 following by symmetry and An(x)12 from the remaining determinant
conditions. We thus obtain
An(x) =
1+
n(x2−fn)
x(x2+β2)
+
2fn(fn+β2)(gn+x)
(g2n−fn)x(x2+β2)
−2an
x2 − fn+1
x(x2 + β2)
2
x2 − fn
x(x2 + β2)
−1+
n(x2−fn)
x(x2+β2)
+
2fn(fn+β2)(gn−x)
(g2n−fn)x(x2+β2)
,
where
an = −n
2
4
+
fn(fn + β2)
g2n − fn
and fn, gn are determined by the recurrence
gn+1 = −n
2
− 2gnan
ngn − 2fn
,
fn+1 = −
(ngn − 2fn)2g2n+1
4fnan
,
with initial condition f1 = −β2 − 1
4 − a1, g1 = 0.
Twisted Traces and Positive Forms on Quantized Kleinian Singularities of Type A 31
Remark 5.8. One can perform a similar calculation for the case P (x) = x4 − e1x2 + e2 with
even trace; again, one obtains a second-order nonlinear recurrence, but the result is significantly
more complicated, even for e1 = e2 = 0.
In each case, when the moduli space is 0-dimensional, so that the conditions uniquely deter-
mine the equation, we get an explicit formula for An. This, of course, is precisely the case that
the orthogonal polynomial is classical.
Acknowledgements
The work of P.E. was partially supported by the NSF grant DMS-1502244. P.E. is grateful to
Anton Kapustin for introducing him to the topic of this paper, and to Chris Beem, Mykola
Dedushenko and Leonardo Rastelli for useful discussions. E.R. would like to thank Nicholas
Witte for pointing out the reference [12].
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1 Introduction
2 Filtered quantizations and twisted traces
2.1 Filtered quantizations
2.2 Even quantizations
2.3 Quantizations with a conjugation and a quaternionic structure
2.4 Twisted traces
2.5 The formal Stieltjes transform
3 An analytic construction of twisted traces
3.1 Construction of twisted traces when all roots of P(x) satisfy |Re alpha|<1/2
3.2 Relation to orthogonal polynomials
3.3 Conjugation-equivariant traces
3.4 Construction of traces when all roots of P(x) satisfy |Re alpha| < 1/2
3.5 Twisted traces in the general case
4 Positivity of twisted traces
4.1 Analytic lemmas
4.2 The case when all roots of P(x) satisfy |Re alpha|<1/2
4.3 The case of a closed strip
4.4 The general case
5 Explicit computation of the coefficients ak, bk of the 3-term recurrence for orthogonal polynomials and discrete Painleve systems
References
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| id | nasplib_isofts_kiev_ua-123456789-211320 |
| institution | Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| issn | 1815-0659 |
| language | English |
| last_indexed | 2026-03-14T20:34:07Z |
| publishDate | 2021 |
| publisher | Інститут математики НАН України |
| record_format | dspace |
| spelling | Etingof, Pavel Klyuev, Daniil Rains, Eric Stryker, Douglas 2025-12-29T11:10:43Z 2021 Twisted Traces and Positive Forms on Quantized Kleinian Singularities of Type A. Pavel Etingof, Daniil Klyuev, Eric Rains and Douglas Stryker. SIGMA 17 (2021), 029, 31 pages 1815-0659 2020 Mathematics Subject Classification: 16W70; 33C47 arXiv:2009.09437 https://nasplib.isofts.kiev.ua/handle/123456789/211320 https://doi.org/10.3842/SIGMA.2021.029 Following [Beem C., Peelaers W., Rastelli L., Comm. Math. Phys. 354 (2017), 345-392] and [Etingof P., Stryker D., SIGMA 16 (2020), 014, 28 pages], we undertake a detailed study of twisted traces on quantizations of Kleinian singularities of type 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 ≤ 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 = 2, this, in particular, recovers the theory of unitary spherical Harish-Chandra bimodules for 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 Painlevé systems. The work of P.E. was partially supported by the NSF grant DMS-1502244. P.E. is grateful to Anton Kapustin for introducing him to the topic of this paper, and to Chris Beem, Mykola Dedushenko, and Leonardo Rastelli for useful discussions. E.R. would like to thank Nicholas Witte for pointing out the reference [12]. en Інститут математики НАН України Symmetry, Integrability and Geometry: Methods and Applications Twisted Traces and Positive Forms on Quantized Kleinian Singularities of Type A Article published earlier |
| spellingShingle | Twisted Traces and Positive Forms on Quantized Kleinian Singularities of Type A Etingof, Pavel Klyuev, Daniil Rains, Eric Stryker, Douglas |
| title | Twisted Traces and Positive Forms on Quantized Kleinian Singularities of Type A |
| title_full | Twisted Traces and Positive Forms on Quantized Kleinian Singularities of Type A |
| title_fullStr | Twisted Traces and Positive Forms on Quantized Kleinian Singularities of Type A |
| title_full_unstemmed | Twisted Traces and Positive Forms on Quantized Kleinian Singularities of Type A |
| title_short | Twisted Traces and Positive Forms on Quantized Kleinian Singularities of Type A |
| title_sort | twisted traces and positive forms on quantized kleinian singularities of type a |
| url | https://nasplib.isofts.kiev.ua/handle/123456789/211320 |
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