# Short star-products on modules.

Daniil Klyuev  Daniil Klyuev

# Unitarizability of Harish-Chandra bimodules over generalized Weyl and
$q$-Weyl algebras

Daniil Klyuev  Daniil Klyuev

##  1 Introduction.

Let $\mathcal{A}$ be an algebra over $\mathbb{C}$, $M$ be an
$\mathcal{A}-\overline{\mathcal{A}}$-bimodule. Choose an automorphism of
$\mathcal{A}$. It gives an antilinear isomorphism of algebras
$\rho\colon\mathcal{A}\to\overline{\mathcal{A}}$. A Hermitian form
$(\cdot,\cdot)$ on $M$ is said to be $\rho$-invariant if $(am,n)=(m,n\rho(a))$
for all $m,n\in M$, $a\in\mathcal{A}$.

The notion of Harish-Chandra bimodule [Lo] is a generalization of a classical
notion of $(\mathfrak{g},K)$-module. For an algebra $\mathcal{A}$ and a
Harish-Chandra bimodule $M$ one can ask the following question: what are the
invariant positive definite forms on $M$? For example, let $\mathfrak{g}$ be a
complex simple Lie algebra, $G$ be the corresponding simply-connected group.
If we take $\mathcal{A}$ to be a central reduction of $U(\mathfrak{g})$, the
classification of irreducible Harish-Chandra bimodules $M$ that have an
invariant positive definite form is equivalent to the classification of
irreducible unitary representations of $G$ with this central character.

We will consider the following case. Let $A$ be an algebra of function of a
Kleinian singularity of type $A$, meaning
$A=\mathbb{C}[x,y]^{\mathbb{Z}/n}=\mathbb{C}[x^{n},y^{n},xy]$. We take
$\mathcal{A}$ to be a filtered deformation or a $q$-deformation of $A$.

Filtered deformations $\mathcal{A}$ of $\mathbb{C}[x,y]^{C_{n}}$ are
parametrized by monic polynomials $P$ of degree $n$: deformation
$\mathcal{A}_{P}$ is generated by $u,v,z$ with relations

$[z,u]=-u,\quad[z,v]=v,\quad uv=P(z+\tfrac{1}{2}),\quad vu=P(z-\tfrac{1}{2}).$

These algebras are also called generalized Weyl algebras: when $n=1$,
$P(x)=x$, $\mathcal{A}$ is the usual Weyl algebra with generators $u,v$ and
relation $[u,v]=1$.

Similarly, $q$-deformations $\mathcal{A}$ of $\mathbb{C}[x,y]^{C_{n}}$
correspond to Laurent polynomials $P$: deformation $\mathcal{A}_{P}$ is
generated by $u,v,Z$ with relations

$ZuZ^{-1}=q^{2}u,\quad ZvZ^{-1}=q^{-2}v,\quad uv=P(q^{-1}Z),\quad vu=P(qZ).$

These algebras are also called generalized $q$-Weyl algebras: when $P(x)=1$,
$\mathcal{A}$ is the $q$-Weyl algebra with generators $u^{\pm 1},Z^{\pm 1}$
and relation $Zu=q^{2}uZ$.

Papers [EKRS], [K] classify positive definite invariant forms in the case when
$\mathcal{A}$ is isomorphic to $\overline{\mathcal{A}}$ and $M$ is the regular
bimodule, meaning $M=\mathcal{A}$ with the natural action.

In this paper we will classify positive definite invariant forms on certain
bimodules over ($q$-)deformations of $\mathbb{C}[x,y]^{C_{n}}$ in the case of
generic parameter. In the case of a filtered deformation this means that all
roots of $P(x)$ are distinct and no two roots can differ by an integer. In
this case we have a construction of $\mathcal{A}$ as a certain subalgebra
inside $\mathbb{C}[x,x^{-1},\partial_{x}]$ and $M$ as a certain subspace of
$\mathbb{C}[x,x^{-1},\partial_{x}]$.

We also do the following. Paper [ES] defined the notion of a short star-
product. This notion can be extended to a not necessarily commutative algebras
$A$ such that $A_{0}$ is a semisimple algebra. We prove some properties of
short star-products in this case. In particular, each positive trace on $M$
gives a short star-product on the commutative algebra
$\begin{pmatrix}\mathop{\mathrm{gr}}\nolimits\mathcal{A}&\mathop{\mathrm{gr}}\nolimits
M\\\
\mathop{\mathrm{gr}}\nolimits\overline{M}&\mathop{\mathrm{gr}}\nolimits\overline{\mathcal{A}}\end{pmatrix}$.
Here both $\mathop{\mathrm{gr}}\nolimits\mathcal{A}$ and
$\mathop{\mathrm{gr}}\nolimits\overline{\mathcal{A}}$ are isomorphic to
$\mathbb{C}[x,y]^{C_{n}}$ and $\mathop{\mathrm{gr}}\nolimits M$ is isomorphic
to a $\mathbb{C}[x,y]^{C_{n}}$-submodule of $\mathbb{C}[x,x^{-1},y]$.
Similarly to [EKRS] and [K] we expect that these short star-products are
useful in 3-dimensional superconformal field theories [BPR] and in the study
of the Coulomb branch of 4-dimensional superconformal field theories [DG].

The organization of paper is as follows. In Section 2 we introduce the notion
of a short star-product in more general case. We have an analogue of Theorem
3.1 in [ES] in this case: there is a correspondence between short star-
products on $A$ and twisted traces on filtered deformations of $A$. For an
automorphism $g$ of $\mathcal{A}$ by $g$-twisted trace we mean a linear map
$T\colon A\to\mathbb{C}$ such that $T(ab)=T(bg(a))$ for all
$a,b\in\mathcal{A}$. We also prove that under certain condition twisted traces
on a matrix algebra $\begin{pmatrix}\mathcal{A}&M\\\
\overline{M}&\overline{\mathcal{A}}\end{pmatrix}$ correspond to invariant
sesquilinear forms on $M$. Combining these two results we obtain short star-
products from positive definite invariant forms.

In Section 3 we classify positive definite invariant forms in the case when
$\mathcal{A}$ is a filtered deformation of $\mathbb{C}[x,y]^{C_{n}}$. In order
to do this first we note that invariant forms on $M$ are in one-to-one
correspondence with twisted traces on
$L=M\otimes_{\overline{\mathcal{A}}}\overline{M}$, where twisted traces are
defined similarly to the above. Then we prove that $L$ is isomorphic to
$\mathcal{A}$ as an $\mathcal{A}$-bimodule. We know the classification of
twisted traces on $\mathcal{A}$ from [EKRS]. For generic parameter each trace
can be expressed as a certain contour integral. This and the construction of
$M$ as a subspace of $\mathbb{C}[x,x^{-1},\partial_{x}]$ allow us to compute
the set of traces that give positive definite form on $M$.

The answer in [EKRS] depends on the number of roots $\alpha$ with
$\lvert\operatorname{Re}\alpha\rvert<\tfrac{1}{2}$ and the number of roots
with $\lvert\operatorname{Re}\alpha\rvert=\tfrac{1}{2}$. In our situation the
module $M$ exists only when for each index $i$ from $1$ to $n$ there exists an
index $j$, possibly equal to $i$ such that $\alpha_{i}+\overline{\alpha_{j}}$
is an integer. We say that $i$ is good if the real number
$\alpha_{i}-\alpha_{j}$ satisfies $\lvert\alpha_{i}-\alpha_{j}\rvert<1$. Note
that in the case $M=\mathcal{A}$ we have $\alpha_{i}+\overline{\alpha_{j}}=0$,
so that
$\lvert\alpha_{i}-\alpha_{j}\rvert=2\lvert\operatorname{Re}\alpha_{i}\rvert$.
Hence the notion of a good root is an analogue of a root $\alpha$ with
$\lvert\operatorname{Re}\alpha\rvert<\tfrac{1}{2}$. Since our parameter is
generic we don’t have an analogue of a root $\alpha$ with
$\lvert\operatorname{Re}\alpha\rvert=\tfrac{1}{2}$. In the end we get the same
answer as in [EKRS]: invariant positive definite forms are a convex cone of
dimension equal to the number of good roots minus a constant from $0$ to $4$
that depends on $n$ and $\rho$.

In Section 4 we classify positive definite invariant forms in the case when
$\mathcal{A}$ is a $q$-deformation of $\mathbb{C}[x,y]^{C_{n}}$. The reasoning
here is similar to the reasoning in Section 3. The answer is always a convex
cone of dimension equal to the number of good roots.

In the case of filtered deformation and $n=2$ the algebra $\mathcal{A}$ is
isomorphic to a central reduction of $U(\mathfrak{sl}_{2})$. In the case of
$q$-deformation and $P(x)=ax+b+cx^{-1}$ the algebra $\mathcal{A}$ is
isomorphic to a central reduction of $U_{q}(\mathfrak{sl}_{2})$. We check that
our results in these cases recover the classical results on irreducible
unitary representations of $\mathop{\mathrm{SL}}\nolimits(2,\mathbb{C})$ and
$\mathop{\mathrm{SL}}\nolimits_{q}(2)$ [P].

##  2 Generalized star-products.

Let $A=\bigoplus\limits_{k\geq 0}A_{k}$ be a positively graded algebra over
$\mathbb{C}$, where all graded pieces $A_{k}$ are finite-dimensional and
$A_{0}$ is a semisimple algebra.

We assume that filtered algebras have a $\mathbb{Z}_{\geq 0}$ filtration.

######  Definition 2.1.

A map $*\colon A\times A\to A$ is called a star-product if $(A,*)$ is an
associative algebra and for all homogeneous $a,b\in A$ we have

$a*b=ab+\sum\limits_{k>0}C_{k}(a,b),$

where $\deg C_{k}(a,b)=\deg a+\deg b-2k$.

Similarly to [ES] star-products are in one-to-one correspondence with
$\mathbb{Z}/2\mathbb{Z}$-equivariant filtered quantizations equipped with a
quantization map.

######  Definition 2.2.

A filtered deformation of $A$ is a pair of a filtered algebra $\mathcal{A}$
and an isomorphism between $\mathop{\mathrm{gr}}\nolimits\mathcal{A}$ and $A$.

In particular, $\mathcal{A}_{\leqslant 0}$ is identified with $A_{0}$. Usually
we will not mention the isomorphism and write ’$\mathcal{A}$ is a filtered
deformation of $A$’.

######  Definition 2.3.

A $\mathbb{Z}/2\mathbb{Z}$-equivariant quantization of $A$ is a pair
$(\mathcal{A},s)$, where $\mathcal{A}$ is a filtered quantization of $A$ and
$s$ is an involution of $\mathcal{A}$ such that $\mathop{\mathrm{gr}}\nolimits
s=(-1)^{d}$, meaning it acts on $A_{k}$ as $(-1)^{k}$ for all $k$.

######  Definition 2.4.

A quantization map is a linear map $\phi\colon A\to\mathcal{A}$ such that
$\phi(A_{k})\subset\mathcal{A}_{\leqslant k}$ and
$\mathop{\mathrm{gr}}\nolimits\phi$ is the identity. We say that $\phi$ is
$\mathbb{Z}/2\mathbb{Z}$-equivariant if $s(\phi(a))=(-1)^{d}\phi(a)$ for all
homogeneous $a\in A$.

We have the following

######  Proposition 2.5.

Star-products on $A$ are in one-to-one correspondence with pairs
$(\mathcal{A},\phi)$, where $\mathcal{A}$ is a
$\mathbb{Z}/2\mathbb{Z}$-equivariant filtered deformation of $A$ and $\phi$ is
a $\mathbb{Z}/2\mathbb{Z}$-equivariant quantization map.

###### Proof.

If $*$ is a star-product on $A$ then we define $\mathcal{A}$ to be $(A,*)$,
$s=(-1)^{d}$ and $\phi$ to be the identity map. On the other hand if we have
$(\mathcal{A},\phi)$, we define $a*b:=\phi^{-1}(\phi(a)\phi(b))$. In both
cases all of the required conditions are satisfied. ∎

######  Definition 2.6.

We say that a star-product $*$ is short if for all homogeneous $a,b\in A$ and
$k>\min(\deg a,\deg b)$ we have $C_{k}(a,b)=0$.

We note that if $*$ is a short star-product on $A$ then for any $k\geq 0$,
$a\in A_{0},b\in A_{k}$ we have $a*b=ab$ and $b*a=ba$. Hence the structure of
an $A_{0}$-bimodule on $(A,*)$ coincides with the structure of an
$A_{0}$-bimodule on $A$.

######  Definition 2.7.

Let $*$ be a short star-product on $A$. It is called nondegenerate if for all
$k\geq 0$ the left kernel of the bilinear map $C_{k}\colon A_{k}\times
A_{k}\to A_{0}$ is zero.

It follows from the next proposition and its proof that if the left kernel of
$C_{k}$ if zero then the right kernel of $C_{k}$ is zero and vice versa.

######  Proposition 2.8.

Let $T_{0}\colon A_{0}\to\mathbb{C}$ be a nondegenerate trace:
$T_{0}(ab)=T_{0}(ba)$ for all $a,b$ and $(a,b)\mapsto T_{0}(ab)$ is a
symmetric nondegenerate bilinear pairing on $A_{0}\times A_{0}$. Then for all
$k\geq 0$ the map $T_{0}\circ C_{k}$ gives a nondegenerate bilinear form on
$A_{k}$.

###### Proof.

Let $V_{1}\ldots,V_{l}$ be all simple representations of $A_{0}$, so that
$A_{0}=\bigoplus\limits_{i=1}^{l}R_{i}$, where $R_{i}=V_{i}\otimes V_{i}^{*}$
are matrix algebras. Fix $k\geq 0$. Let
$A_{k}=\bigoplus\limits_{i=1}^{l}V_{i}^{d_{i}}$ as a left $A_{0}$-module and
$\bigoplus\limits_{i=1}^{l}(V_{i}^{*})^{e_{i}}$ as a right $A_{0}$-module.

Since $T_{0}$ is a trace, its restriction to $V_{i}\otimes V_{i}^{*}$ is given
by $T_{0}(u\otimes v)=\alpha_{i}v(u)$, where $\alpha_{i}$ is some complex
number. Since $T_{0}$ is nondegenerate, $\alpha_{i}$ is nonzero.

Let $a\in A_{0}$, $b,c\in A_{k}$. Then $(ab)*c=(a*b)*c=a*(b*c)=a(b*c)$.
Comparing $C_{k}$ we get $C_{k}(ab,c)=aC_{k}(b,c)$. Similarly
$C_{k}(b,ca)=C_{k}(b,c)a$. Define a structure of an $A_{0}$-bimodule on
$A_{k}\otimes_{\mathbb{C}}A_{k}$ as follows: $a(b\otimes c):=ab\otimes c$,
$(b\otimes c)a:=b\otimes ca$. If we regard $C_{k}$ as a linear map from
$A_{k}\otimes_{\mathbb{C}}A_{k}$ to $A_{0}$, it will be a map of
$C_{k}$-bimodules.

We have $A_{k}\otimes A_{k}=\bigoplus_{i,j=1}^{l}(V_{i}\otimes
V_{j}^{*})^{d_{i}e_{j}}$. Using Schur Lemma we see that $C_{k}$ restricted to
$V_{i}\otimes V_{j}^{*}$ is zero when $i\neq j$. When $i=j$, the map $C_{k}$
is defined by a matrix $M_{i}$ of size $d_{i}\times e_{i}$: $C_{k}(u\otimes
v)=(M_{i})_{ab}(u\otimes v)$ for $u$ in the $a$-th copy of $V_{i}$ and $v$ the
in $b$-th copy of $V_{i}^{*}$.

Since $C_{k}$ has no left kernel, it follows that $M_{i}$ has rank at least
$d_{i}$. In particular, $d_{i}\geq e_{i}$. Computing the dimension of $A_{k}$
in two different ways, we get $\sum\limits_{i=1}^{l}d_{i}\dim
V_{i}=\sum\limits_{i=1}^{l}e_{i}\dim V_{i}$. It follows that $d_{i}=e_{i}$. We
get that $M_{i}$ is a nondegenerate square matrix.

Hence $T_{0}\circ C_{k}$ is a nondegenerate bilinear form on $A_{k}$, as we
wanted. ∎

From now on we fix a nondegenerate trace $T_{0}$ on $A_{0}$.

Let $\langle a,b\rangle=T(a*b)=T_{0}(CT(a*b))$, where $CT$ is the constant
term map. The shortness of $*$ implies that $A_{k}$ is orthogonal to $A_{m}$
with respect to $\langle\cdot,\cdot\rangle$ when $k\neq m$. Hence if $*$ is
nondegenerate then $\langle\cdot,\cdot\rangle$ is a nondegenerate bilinear
form.

Similarly to Proposition 3.3 in [ES] we want a bijection between nondegenerate
short star-products and nondegenerate twisted traces.

######  Definition 2.9.

Let $\mathcal{A}$ be a filtered quantization of $A$ and $g$ be an automorphism
of $\mathcal{A}$ such that $g$ restricted to $\mathcal{A}_{\leqslant 0}$ is
identity. We say that a map $T\colon\mathcal{A}\to\mathbb{C}$ is a $g$-twisted
trace if $T(ab)=T(bg(a))$ for all $a,b\in\mathcal{A}$. We say that $T$ extends
$T_{0}$ if $T|_{\mathcal{A}_{0}}=T_{0}$.

######  Definition 2.10.

We say that $T$ is nondegenerate if $(a,b)=T(ab)$ is nondegenerate on each
$\mathcal{A}_{\leqslant i}$.

In [ES] the trace $T_{0}$ is defined by one number $T_{0}(1)=T(1)$, so it is
clear that different choices $T_{0}$ give essentially the same set of traces.
In our case after changing $T_{0}$ we should multiply both $T|_{A_{i}}$ and
$g|_{A_{ij}}$ by nonzero numbers.

We have the following analog of Proposition 3.3 in [ES]:

######  Proposition 2.11.

There is a one-to-one correspondence between nondegenerate star-products on
$A$ and triples $(\mathcal{A},g,T)$, where $\mathcal{A}$ is a filtered
deformation of $A$, $g$ is an automorphism of $\mathcal{A}$ equal to identity
on $\mathcal{A}_{\leqslant 0}$ and $T$ is an $s$-invariant $g$-twisted trace
that extends $T_{0}$.

###### Proof.

Let $*$ be a nondegenerate star-product on $A$, $\mathcal{A}=(A,*)$. Define
$T=T_{0}\circ CT$, where $CT$ is the constant term map. This map commutes with
$s=(-1)^{d}$. For any $k$, since $\langle\cdot,\cdot\rangle$ is nondegenerate
on $A_{k}$, there exists a linear automorphism $g_{k}$ such that $\langle
a,b\rangle=\langle g_{k}(b),a\rangle$ for all $a,b\in A_{k}$. This means
$T(ab)=T(bg_{k}(a))$ for all $a,b\in A_{k}$. We define $g$ so that
$g|_{A_{k}}=g_{k}$. In this way $T$ becomes a $g$-twisted trace. It remains to
prove that $g$ is an automorphism of $\mathcal{A}$.

For all $a,b,c\in\mathcal{A}$ we have $T(abc)=T(bcg(a))=T(cg(a)g(b))$ on one
hand and $T(abc)=T(cg(ab))$ on the other hand. Since $T$ has zero kernel, we
get $g(a)g(b)=g(ab)$. Hence $g$ is an automorphism of $\mathcal{A}$.

Since $T_{0}$ is a trace on $A_{0}$, the restriction of $g_{0}$ to $A_{0}$ is
the identity.

Since $\langle\cdot,\cdot\rangle$ is nondegenerate and $A_{i}$ is orthogonal
to $A_{j}$ when $i\neq j$, $\langle\cdot,\cdot\rangle$ is nondegenerate on
$A_{0}\oplus\cdots\oplus A_{k}=\mathcal{A}_{\leqslant k}$. This precisely
means that $T$ is nondegenerate.

On the other hand, suppose that we have $(\mathcal{A},g,T)$ as above.

Define $(a,b)=T(ab)$. For all $a,b\in\mathcal{A}$ we have $(a,b)=(b,g(a))$.
Using this for $a\in\mathcal{A}_{\leqslant k}$ we see that the left and the
right orthogonal complement of $\mathcal{A}_{\leqslant k}$ coincide. For
$k\geq 0$ define $\mathcal{A}_{k}=\mathcal{A}_{\leqslant
k-1}^{\perp}\cap\mathcal{A}_{\leqslant k}$.

Since $T$ is nondegenerate, $\mathcal{A}_{k}$ has trivial intersection with
$\mathcal{A}_{\leqslant k-1}$. Counting dimensions we have
$\dim\mathcal{A}_{k}=\dim\mathcal{A}_{\leqslant k}-\dim\mathcal{A}_{\leqslant
k-1}=\dim A_{k}$. It follows that $\mathcal{A}_{k}$ is in natural bijection
with $\mathcal{A}_{\leqslant k}/\mathcal{A}_{\leqslant k-1}=A_{k}$.

Define $\phi\colon A\to\mathcal{A}$ using these bijections. Define $*$ by
$a*b=\phi^{-1}(\phi(a)\phi(b))$. Since $(A,*)$ is isomorphic to $\mathcal{A}$,
$*$ is a star-product.

Let us prove that $\phi$ is $\mathbb{Z}/2\mathbb{Z}$-invariant. Let $a$ be an
element of $\mathcal{A}_{k}$. Note that $(sa,b)=T(sa\cdot b)=T(a\cdot
sb)=(a,sb)$ since $T$ is $s$-invariant. Since $s$ is filtration-preserving, we
get $sa\in\mathcal{A}_{\leqslant k}\cap\mathcal{A}_{\leqslant
k-1}^{\perp}=\mathcal{A}_{k}$. On the other hand, since
$\mathop{\mathrm{gr}}\nolimits s=(-1)^{d}$, $a+(-1)^{k+1}sa$ belongs to
$\mathcal{A}_{\leqslant k-1}$. We deduce that $sa=(-1)^{k}a$ when $a\in
A_{k}$. It follows that $\phi$ is $\mathbb{Z}/2\mathbb{Z}$-invariant.

It remains to check that $*$ is short and nondegenerate.

We start with shortness. Let $a\in A_{k},b\in A_{l}$. Suppose that $k>l$.

We note that for $r>0$ we have $\mathcal{A}_{\leqslant
r}^{\perp}=\bigoplus\limits_{t>r}\mathcal{A}_{t}$. So in order to check that
$a*b$ belongs to $\bigoplus\limits_{s\geq k-l}A_{s}$ it is enough to check
that $\phi(a*b)=\phi(a)\phi(b)$ belongs to $\mathcal{A}_{\leqslant
k-l-1}^{\perp}$.

Let $c\in\mathcal{A}_{\leqslant k-l-1}$. We want to check that
$T(\phi(a)\phi(b)c)=0$. We note that $\phi(b)c\in\mathcal{A}_{\leqslant k-1}$.
Using the definition of $\mathcal{A}_{k}$, we deduce that
$T(\phi(a)\phi(b)c)=0$.

It remains to prove that $*$ is nondegenerate. Since $T$ is nondegenerate,
$(\cdot,\cdot)$ is nondegenerate on each $\mathcal{A}_{\leqslant k}$. Using
the definition of $\mathcal{A}_{k}$, we see that $(\cdot,\cdot)$ is
nondegenerate on each $A_{k}$. It follows that $*$ is nondegenerate.

Finally, it is straightforward to check that the maps
$*\mapsto(\mathcal{A},g,T)$ and $(\mathcal{A},g,T)\mapsto*$ are inverse to
each other. ∎

###  2.1 Example: $2\times 2$ matrices.

Suppose that $A_{0}$ is either the algebra of $2\times 2$ matrices or diagonal
$2\times 2$ matrices. Let $e_{1},e_{2}$ be the diagonal matrix units. In this
case $A=A_{11}\oplus A_{12}\oplus A_{21}\oplus A_{22}$, where
$A_{ij}=e_{i}Ae_{j}$. We see that $A_{11}$ and $A_{22}$ are subalgebras,
$A_{12}$ is an $A_{11}-A_{22}$ bimodule, $A_{21}$ is an $A_{22}-A_{11}$
bimodule, and there are maps of bimodules $\phi\colon
A_{12}\otimes_{A_{22}}A_{21}\to A_{11}$, $\psi\colon
A_{21}\otimes_{A_{11}}A_{12}\to A_{22}$. In this case
$A_{11},A_{22},A_{21},A_{12}$ form a Morita context.

Let $\mathcal{A}$ be a filtered deformation of $A$. Then
$\mathcal{A}_{\leqslant 0}=A_{0}$, so we can similarly define
$\mathcal{A}_{11}$, $\mathcal{A}_{22}$, $\mathcal{A}_{12}$,
$\mathcal{A}_{21}$. Since $\mathcal{A}$ is a filtered deformation, for any
$e\in A_{0}$ and $a\in\mathcal{A}_{\leqslant k}$ we have
$e(a+\mathcal{A}_{\leqslant k-1})=ea+\mathcal{A}_{\leqslant k-1}$. It follows
that the isomorphism $\mathop{\mathrm{gr}}\nolimits\mathcal{A}\cong A$ sends
$\mathop{\mathrm{gr}}\nolimits\mathcal{A}_{ij}$ to $A_{ij}$.

Let $g$ be an automorphism of $\mathcal{A}$ that restricts to the identity on
$A_{0}$. From $g(e_{i}a)=e_{i}g(a)$ and $g(ae_{i})=g(a)e_{i}$ we deduce that
$g$ preserves $\mathcal{A}_{ij}$ for all pairs $(i,j)$.

Let $T_{0}$ be the standard matrix trace. Any twisted trace $T$ on
$\mathcal{A}$ satisfies $T(e_{i}a)=T(ae_{i})$. It follows that $T$ is zero on
$\mathcal{A}_{12}$ and $\mathcal{A}_{21}$. Hence for a fixed $g$ a $g$-twisted
trace $T$ is supported on $\mathcal{A}_{11}\cup\mathcal{A}_{22}$ and satisfies
$T(ab)=T(gb(a))$ for all pairs $i,j\in\\{1,2\\}$, $a\in A_{ij},b\in A_{ji}$.

The condition $T|_{A_{0}}=T_{0}$ is equivalent to $T(e_{1})=T(e_{2})=1$.

Using $T(ab)=T(ba)$ for $a\in A_{0}$, $b\in\mathcal{A}$ we can prove that
$\mathcal{A}_{k}=\mathcal{A}_{\leqslant k}\cap\mathcal{A}_{\leqslant
k-1}^{\perp}$ is an $A_{0}$-submodule of $\mathcal{A}$. It follows that the
quantization map $\phi$ is an isomorphism of $A_{0}$-modules. Hence
$A_{ij}*A_{kl}$ is a subset of $A_{il}$ when $j=k$ and zero otherwise. Note
that star-products in general do not have to satisfy this property if we take
$\phi$ that does not respect the $A_{0}$-module structure.

###  2.2 Conjugations, Hermitian star-products, positive traces.

The notions of conjugation-invariant and Hermitian star-products in [ES] can
be defined in our situation. Similarly to [ES] there is a connection between
Hermitian star-products and Hermitian forms; we will work it out in the
example below. This gives a connection between our unitarizability results
below and star-products: any unitarizable bimodule $M_{c,c^{\prime}}$ will
give a star-product on the algebra $\begin{pmatrix}A&I\\\ I&A\end{pmatrix}$,
where $A=\mathbb{C}[x,y]^{C_{n}}$ and $I$ is a
$\mathbb{C}[x,y]^{C_{n}}$-submodule of $\mathbb{C}[x,x^{-1},y]$.

######  Definition 2.12.

  1. 1.

A conjugation on $A$ is an antilinear graded automorphism $\rho\colon A\to A$.

  2. 2.

Let $\mathcal{A}$ be a $\mathbb{Z}/2$-invariant deformation of $A$. A
conjugation of $\mathcal{A}$ is a filtered antilinear automorphism $\rho$ that
commutes with $s$.

Recall that $A_{0}$ is a semisimple algebra. Hence $A_{0}$ is a direct product
of matrix algebras and any automorphism of $A_{0}$ is a composition of inner
automorphism and a permutation of matrix algebras of the same size. The
antilinear automorphism $\rho|_{A_{0}}$ is a composition of the standard
complex conjugation and an automorphism of $A_{0}$. Choose $T_{0}$ satisfying
the following two conditions: $T_{0}$ is invariant under all automorphisms of
$A_{0}$ and $T_{0}(\overline{a})=\overline{T_{0}(a)}$, where $\overline{a}$ is
the standard complex conjugation on matrices. For example, $T_{0}$ that is
equal to the standard matrix trace on each matrix algebra satisfies these
conditions. It follows that $T_{0}(\rho(a))=\overline{T_{0}(a)}$ for any
conjugation $\rho$ of $A$.

From now on we fix such $T_{0}$.

######  Definition 2.13.

  1. 1.

A star-product $*$ on $A$ is conjugation-invariant
if$\rho(a*b)=\rho(a)*\rho(b)$.

  2. 2.

A star-product $*$ on $A$ is Hermitian if
$T_{0}(CT(a*b))=T_{0}(CT(b*\rho^{2}(a)))$ for all $a,b\in A$.

We have a lemma similar to Lemma 3.20 in [ES].

######  Lemma 2.14.

Let $*$ be a conjugation-invariant nondegenerate star-product on $A$. Then $*$
is Hermitian if and only if the form $(a,b)\mapsto T_{0}\circ CT(a*\rho(b))$
is Hermitian.

###### Proof.

The form $(\cdot,\cdot)$ is Hermitian when $(a,b)=\overline{(b,a)}$ for all
$a,b\in A$. We know that $T_{0}(\rho(a))=\overline{(T_{0}(a))}$ for $a\in
A_{0}$. Since $\rho$ is graded, we have $CT(\rho(a))=\rho(CT(a))$ for any
$a\in A$. Using this we get

$\overline{(b,a)}=\overline{T_{0}\circ
CT(b*\rho(a))}=T_{0}\big{(}\rho(CT(b*\rho(a)))\big{)}=T_{0}(CT(\rho(b)*\rho^{2}(a))).$

So $(\cdot,\cdot)$ is Hermitian if and only if

$T_{0}(CT(a*\rho(b)))=T_{0}(CT(\rho(b)*\rho^{2}(a)))$

for all $a,b\in A$.

Changing $b$ to $\rho^{-1}(b)$ we see that $(\cdot,\cdot)$ is Hermitian if and
only if

$T_{0}(CT(a*b))=T_{0}(CT(b*\rho^{2}(a)))$

for all $a,b\in A$. This means that $*$ is Hermitian. ∎

We have a lemma similar to Lemma 3.23 from [ES].

######  Lemma 2.15.

  1. 1.

The star-product $*$ on $A$ is conjugation-invariant if and only if $\rho$ is
a conjugation on $\mathcal{A}$ that conjugates $T$.

  2. 2.

In this situation $*$ is Hermitian if and only if in addition $\rho^{2}=g$.

###### Proof.

  1. 1.

Suppose that $*$ is conjugation-invariant. Then $\rho$ is by definition a
conjugation on $\mathcal{A}$. We have

$T(\rho(a))=T_{0}(CT(\rho(a))=T_{0}(\rho(CT(a))=\overline{(T_{0}(CT(a))}=\overline{T(a)}.$

Suppose that $\rho$ is a conjugation on $\mathcal{A}$ that conjugates $T$. We
want to prove that the corresponding short star-product $*$ on $A$
satisfies$\rho_{1}(a*b)=\rho_{1}(a)*\rho_{1}(b)$, where
$\rho_{1}=\mathop{\mathrm{gr}}\nolimits\rho$. Let $\phi\colon A\to\mathcal{A}$
be the corresponding quantization map. Since $\phi$ is a linear isomorphism,
it is enough to prove that
$\phi(\rho_{1}(a*b))=\phi(\rho_{1}(a)*\rho_{1}(b))$.

In order to do this, it is enough to prove that $\phi\rho_{1}=\rho\phi$.
Indeed, we constructed $*$ so that

$\phi(\rho_{1}(a)*\rho_{1}(b))=\phi(\rho_{1}(a))\phi(\rho_{1}(b)).$

This is equal to

$\rho(\phi(a))\rho(\phi(b))=\rho(\phi(a)\phi(b))=\rho(\phi(a*b))=\phi(\rho_{1}(a*b)).$

Note that the corresponding bilinear form $(a,b)=T(ab)$ satisfies

$(\rho(a),\rho(b))=T(\rho(a)\rho(b))=T(\rho(ab))=\overline{T(ab)}=\overline{(a,b)}.$

Since $\rho$ preserves both $\mathcal{A}_{\leqslant n}$ and
$\mathcal{A}_{\leqslant n-1}$, it preserves
$\mathcal{A}_{n}:=(\mathcal{A}_{\leqslant
n-1})^{\perp}\cap\mathcal{A}_{\leqslant n}$.

By construction we have $\phi(a+\mathcal{A}_{\leqslant n-1})=a$ when
$a\in\mathcal{A}_{n}$. Since $\rho_{1}=\mathop{\mathrm{gr}}\nolimits\rho$ for
any $a\in\mathcal{A}_{n}$ we have $\rho(a)+\mathcal{A}_{\leqslant
n-1}=\rho_{1}(a+\mathcal{A}_{\leqslant n-1})$. It follows that
$\phi^{-1}(\rho(a))=\rho_{1}(\phi^{-1}(a))$, hence $\phi\rho_{1}=\rho\phi$ as
we wanted.

  2. 2.

Suppose that $*$ is Hermitian. By definition this means that $T$ is
$\rho^{2}$-twisted. Since $T$ has no kernel, we get $g=\rho^{2}$.

Suppose that $T$ is a $\rho^{2}$-twisted trace on $\mathcal{A}$. Note that
$T_{0}(CT(a))=T(\phi(a))$ for any $a\in A$. It follows that $T_{0}\circ CT$ is
$\rho_{1}^{2}$-twisted, as required.

∎

###  2.3 Bimodules and positive definite Hermitian forms.

As in subsection 2.1, let $A_{0}=\operatorname{Mat}_{2}(\mathbb{C})$ or
$\mathbb{C}\oplus\mathbb{C}$, so that $A$ and $\mathcal{A}$ correspond to
Morita contexts. Let us change notation: graded algebra is
$\begin{pmatrix}A&M\\\ N&B\end{pmatrix}$, filtered deformation is
$\begin{pmatrix}\mathcal{A}&\mathcal{M}\\\
\mathcal{N}&\mathcal{B}\end{pmatrix}$. Assume that there exists antilinear
isomorphisms $\rho_{1}\colon\mathcal{A}\to\mathcal{B}$,
$\rho_{2}\colon\mathcal{B}\to\mathcal{A}$,
$\phi\colon\mathcal{M}\to\mathcal{N}$, $\psi\colon\mathcal{N}\to\mathcal{M}$
such that

$\rho\begin{pmatrix}a&m\\\
n&b\end{pmatrix}=\begin{pmatrix}\rho_{2}(b)&\psi(n)\\\
\phi(m)&\rho_{1}(a)\end{pmatrix}$

is a conjugation. For $\mathcal{M},\mathcal{N}$ this means that
$\phi\colon\mathcal{M}\to\overline{\mathcal{N}}_{\rho}$ is an isomorphism of
bimodules, where we use $\rho_{1},\rho_{2}$ to define bimodule structure on
$\overline{\mathcal{N}}_{\rho}$: $a.n=\rho_{1}(a)n$, $n.b=n\rho_{2}(b)$. A
similar statement holds for $\psi$. Since $\rho^{2}$ is an automorphism of the
matrix algebra, we get that $\psi\phi$ is an isomorphism between $\mathcal{M}$
and $\mathcal{M}$ with bimodule structure twisted by
$\rho_{1}\rho_{2},\rho_{2}\rho_{1}$ respectively. Abusing notation denote the
restriction of $\rho$ to $A\oplus B$ by $\rho$.

Recall that a trace $T$ on the matrix algebra is a pair of traces $T_{1}$,
$T_{2}$ on $\mathcal{A}$,$\mathcal{B}$. A trace $T$ is $\rho^{2}$-twisted if
$T_{1}$, $T_{2}$ are $\rho^{2}$-twisted and

$T_{1}(mn)=T_{2}(n\psi\phi(m)),\quad T_{2}(nm)=T_{1}(m\phi\psi(n)).$

Consider the sesquilinear form $(k,l):=T_{1}(k\phi(l))$ for
$k,l\in\mathcal{M}$. Note that it is invariant in the following sense:

$(am_{1},m_{2})=T(am_{1}\psi(m_{2}))=T(m_{1}\psi(m_{2})\rho_{2}\rho_{1}(a))=\\\
T(m_{1}\psi(m_{2}\rho_{1}(a)))=(m_{1},m_{2}\rho_{1}(a)).$

When this form is positive definite we say that $T$ is positive for bimodules.

Hence from a matrix algebra and a trace we get an invariant sesquilinear form.
On the other hand, suppose that we have algebras $\mathcal{A},\mathcal{B}$, an
$\mathcal{A}-\mathcal{B}$-bimodule $\mathcal{M}$, an antilinear isomorphism
$\rho=\rho_{1}\oplus\rho_{2}\colon\mathcal{A}\oplus\mathcal{B}\to\mathcal{B}\oplus\mathcal{A}$,
an isomorphism $g\colon\mathcal{M}\to\mathcal{M}_{\rho^{2}}$ and a
sesquilinear form on $\mathcal{M}$ such that

$(am_{1},m_{2})=(m_{1},m_{2}\rho(a)),$ $(m_{1}\rho(a),m_{2})=(m_{1},am_{2}).$

We say that such forms are $\rho$-invariant. Note that, strictly speaking,
such forms are $\rho_{1}$-invariant, we do not use $\rho_{2}$. Below we will
get a $\rho_{2}$-invariant form on $\mathcal{N}$, so that the picture becomes
symmetric.

Define $\mathcal{N}$ to be $\overline{\mathcal{M}}_{\rho^{-1}}$. Then we can
define $\phi\colon\mathcal{M}\to\overline{\mathcal{N}}_{\rho}$ to be the
identity map on the underlying set and
$\psi\colon\mathcal{N}\to\overline{\mathcal{M}}_{\rho}$ to be $g$ on the
underlying set. We see that both $\phi,\psi$ are maps of bimodules. Hence we
recover the matrix algebra $\begin{pmatrix}\mathcal{A}&\mathcal{M}\\\
\mathcal{N}&\mathcal{B}\end{pmatrix}$ and a conjugation.

Recall that for a twisted bimodule
$\mathcal{N}=\overline{\mathcal{M}}_{\rho^{-1}}$ we use $b.n$ to mean the
action on $\mathcal{N}$ and $an$ to mean the action on $\mathcal{M}$, so that
$b.n=\rho^{-1}(a)n$.

For $n_{1},n_{2}\in\mathcal{N}=\overline{\mathcal{M}}_{\rho^{-1}}$ define
$(n_{1},n_{2})_{\mathcal{N}}=(n_{2},n_{1})_{\mathcal{M}}$. Then

$(b.n_{1},n_{2})_{\mathcal{N}}=(\rho^{-1}(b)n_{1},n_{2})_{\mathcal{N}}=(n_{2},\rho^{-1}(b)n_{1})_{\mathcal{M}}=\\\
(n_{2}b,n_{1})_{\mathcal{M}}=(n_{1},n_{2}b)_{\mathcal{N}}=(n_{1},n_{2}.\rho(b)),$

hence $(\cdot,\cdot)_{\mathcal{N}}$ is also $\rho$-invariant, but here we use
$\rho_{2}$, not $\rho_{1}$.

We turn to the traces:

######  Proposition 2.16.

  1. 1.

$\rho$-invariant sesquilinear forms on $\mathcal{M}$ are in one-to-one
correspondence with $\rho^{2}$-twisted traces
$\mathcal{M}\otimes_{\mathcal{B}}\mathcal{N}\to\mathbb{C}$. Each invariant
form also gives a $\rho^{2}$-twisted trace on
$\mathcal{N}\otimes_{\mathcal{A}}\mathcal{M}$.

  2. 2.

Suppose that $\mathcal{M}$ and $\mathcal{N}$ provide a Morita equivalence
between $\mathcal{A}$ and $\mathcal{B}$. Then there is one-to-one
correspondence between $\rho$-invariant sesquilinear forms on $\mathcal{M}$
and $\begin{pmatrix}\rho^{2}&g\\\ h&\rho^{2}\end{pmatrix}$-twisted traces on
$\begin{pmatrix}\mathcal{A}&\mathcal{M}\\\
\mathcal{N}&\mathcal{B}\end{pmatrix}$. Here $h$ is a unique isomorphism from
$\mathcal{N}$ to $\mathcal{N}_{\rho^{2}}$ such that $g\otimes h=\rho^{2}$.

###### Proof.

  1. 1.

Let $(\cdot,\cdot)$ be a $\rho$-twisted form. Consider the map
$T\colon\mathcal{M}\otimes_{\mathcal{B}}\mathcal{N}\to\mathbb{C}$ given by
$T(m\otimes n)=(m,n)$. We have

$T(mb\otimes n)=(mb,n)=(m,\rho^{-1}(b)n)=T(m\otimes b.n),$

hence $T$ is well-defined. We also have

$T(am_{1}\otimes m_{2})=(am_{1},m_{2})=(m_{1},m_{2}\rho(a))=\\\
(m_{1},m_{2}.\rho^{2}(a))=T((m_{1}\otimes m_{2})\rho^{2}(a))$

In other words, $T$ is a $\rho^{2}$-twisted trace.

In the other direction, if $T$ is a $\rho^{2}$-twisted trace we can define
$(\cdot,\cdot)$ using $(m,n)=T(m\otimes n)$. The same computations as above
show that $(\cdot,\cdot)$ is $\rho$-invariant.

If we use the same reasoning for $\mathcal{N}$ instead of $\mathcal{M}$ we
obtain a $\rho^{2}$-twisted trace on
$\mathcal{N}\otimes_{\mathcal{B}}\mathcal{N}_{\rho^{-1}}$. Note that
$\mathcal{N}_{\rho^{-1}}=\mathcal{M}_{\rho^{-2}}$. Applying
$\operatorname{id}_{\mathcal{N}}\otimes g$ we get
$\mathcal{N}\otimes\mathcal{N}_{\rho^{-1}}\cong\mathcal{N}\otimes\mathcal{M}$.
Hence the second trace $T^{\prime}$ is given by

$T^{\prime}(n\otimes m)=(n,g^{-1}m)_{\mathcal{N}}=(g^{-1}m,n).$

  2. 2.

Since $\mathcal{M}\otimes_{\mathcal{A}}\mathcal{N}$ is isomorphic to
$\mathcal{A}$, there is one-to-one correspondence between invariant forms on
$\mathcal{M}$ and twisted traces on $\mathcal{A}$, similarly for
$\mathcal{B}$. It remains to prove that every such pair of traces $T_{1}$,
$T_{2}$ on $\mathcal{A}$, $\mathcal{B}$, satisfies

$T_{1}(mn)=T_{2}(ng(m)),\quad T_{2}(nm)=T_{1}(mg(n)).$

Indeed, $T_{1}(mn)=(m,n)=T_{2}(ng(m))$ and

$T_{2}(nm)=(g^{-1}m,n)=T_{1}(g^{-1}(m)n)=T_{1}(mh(n)).$

On the last step we used that $g\otimes h=\rho^{2}$ and $T_{1}$ is
$\rho^{2}$-twisted.

∎

In the case when $\mathcal{A}=\mathcal{B}=\mathcal{M}$,
$\rho_{1}=\rho_{2}=\rho$ we have just one $\rho^{2}$-invariant trace $T$ on
$\mathcal{A}$. In this case $\mathcal{N}=\mathcal{A}$, the corresponding
isomorphism of bimodules sends $a\in\mathcal{N}$ to $\rho(a)\in\mathcal{A}$.
We have $\mathcal{M}\otimes_{\mathcal{B}}\mathcal{N}=\mathcal{A}$, so the
corresponding trace is a $\rho^{2}$-twisted trace on $\mathcal{A}$. The tensor
product $\mathcal{M}\otimes_{\mathcal{A}}\mathcal{M}_{\rho^{-1}}$ is
isomorphic to $\mathcal{A}$ via $a\otimes b\mapsto a\rho(b)$, so the
positivity condition is $T(a\rho(a))>0$. We recover the positivity condition
from [EKRS] in this case.

In the case we are interested in, namely $\mathcal{A},\mathcal{B}$ are
($q$-)deformations of Kleinian singulartieis, bimodule positivity implies that
the corresponding traces on $\mathcal{A}$, $\mathcal{B}$ are nondegenerate. It
follows that the trace on the matrix algebra is also nondegenerate. Using
Proposition 2.11 we obtain a star-product on the algebra
$\begin{pmatrix}A&M\\\ N&B\end{pmatrix}$.

##  3 Unitarizability in the case of generic parameter.

###  3.1 Construction of Hermitian forms on a bimodule.

Let $c=(c_{1},\ldots,c_{n})$ be a sequence of complex numbers. When none of
$c_{i}-c_{j}$ are integers, we have

######  Lemma 3.1 ([BEF],Proposition 2.11).

Consider the algebra $A_{c}$ of differential operators with a pole at only
possibly zero that preserve each set $x^{c_{i}}\mathbb{C}[x]$, $i=1,\ldots,n$.
This algebra is generated by $v=x$, $z=E=x\partial_{x}$ and
$u=x^{-1}(E-c_{1})\cdots(E-c_{n})$ and the defining set of relations is
$[z,u]=-u$, $[z,v]=v$, $uv=P(z+\tfrac{1}{2})$, $vu=P(z-\tfrac{1}{2})$, where
$P(t)=\prod_{i=1}^{n}(t-c_{i}+\tfrac{1}{2})$.

The proof of this proposition can be found in [BEF]. Also Lemma 4.1 will have
a similar proof.

We say that $c$ is a generic parameter if none of $c_{i}-c_{j}$ are integers.

Suppose that $c$ and $c^{\prime}$ are two generic parameters such that
$c_{i}-c^{\prime}_{i}$ are integers. Consider the
$A_{c}$-$A_{c^{\prime}}$-bimodule $M=M_{c,c^{\prime}}$ that consists of
differential operators with pole at only possibly zero that send each
$x^{c^{\prime}_{i}}\mathbb{C}[x]$ to $x^{c_{i}}\mathbb{C}[x]$.

######  Lemma 3.2.

  1. 1.

The module $M_{c,c^{\prime}}$ is the direct sum of its
$\operatorname{ad}z$-eigenspaces. The eigenspace of weight $j$ is
$x^{j}R_{j}(z)\mathbb{C}[z]$, where $R_{j}$ is a monic polynomial with simple
zeroes at $c^{\prime}_{i},\ldots,c_{i}-j-1$ for all $i$ such that
$c_{i}-c^{\prime}_{i}-j>0$.

  2. 2.

$M_{c,c^{\prime}}$ is a Harish-Chandra bimodule in the sense of [Lo]: there
exists an increasing filtration $M_{\leqslant j}$ with $\cup_{j}M_{\leqslant
j}=M_{c,c^{\prime}}$ compatible with filtrations on $A_{c},A_{c^{\prime}}$
with the following properties:

     * •

For each $j$ the subspace $M_{\leqslant j}$ is preserved by
$\operatorname{ad}z$ and $[u,M_{\leqslant j}]\subset M_{\leqslant n+j-2}$,
$[v,M_{\leqslant j}]\subset M_{\leqslant n+j-2}$.

     * •

The previous condition makes $\mathop{\mathrm{gr}}\nolimits M_{c,c^{\prime}}$
into $\mathop{\mathrm{gr}}\nolimits
A_{c}\cong\mathbb{C}[x,y]^{C_{n}}\cong\mathop{\mathrm{gr}}\nolimits
A_{c^{\prime}}$-bimodule such that the action on the left coincides with the
action on the right. Then $\mathop{\mathrm{gr}}\nolimits M_{c,c^{\prime}}$ is
a finitely generated $\mathop{\mathrm{gr}}\nolimits A_{c}$-module.

###### Proof.

  1. 1.

For $a,b\in\mathbb{C}$, $m=\sum
x^{i}S_{i}(z)\in\mathbb{C}[x,x^{-1},\partial_{x}]$ we have the following:
$mx^{a}$ belongs to $x^{b}\mathbb{C}[x]$ if and only if each
$x^{i}S_{i}(z)x^{a}$ belongs to $x^{b}\mathbb{C}[x]$. It follows that $m=\sum
x^{i}S_{i}(z)$ belongs to $M_{c,c^{\prime}}$ if and only if each
$x^{i}S_{i}(z)$ belongs to $M_{c,c^{\prime}}$. This proves the first
statement.

Consider $x^{j}R(z)\in M_{c,c^{\prime}}$. This element sends
$x^{c^{\prime}_{i}+k}$ to $x^{c^{\prime}_{i}+k+j}R(c^{\prime}_{i}+k)$. Hence
when $k\geq 0$ and $c^{\prime}_{i}+k+j<c_{i}$ we should have
$R(c^{\prime}_{i}+k)=0$. It follows that $R$ has roots $c^{\prime}_{i}+k$ for
all $k$ such that $0\leqslant k<c_{i}-c^{\prime}_{i}-j$. This proves the
second statement.

  2. 2.

Denote $M:=M_{c,c^{\prime}}$ and denote by $M_{j}$ is an $\operatorname{ad}z$
eigenspace of $M$ of weight $j$.

Let $M_{\leqslant k}$ consist of all $x^{j}R(z)$ such that $nj+2\deg
R\leqslant k$. Note that for negative $j$ with $\lvert j\rvert$ large enough
the minimal possible degree of $R$ is $-nj$ plus some constant. This proves
that $M_{\leqslant k}$ is finite-dimensional for all $k$ and empty for large
negative $k$. We see that this filtration is compatible with filtrations on
$A_{c},A_{c^{\prime}}$ and each $M_{\leqslant k}$ is preserved by
$\operatorname{ad}h$. In order to check the adjoint action let us take
$x^{j}R(z)\in M_{\leqslant k}$. Then

$[v,x^{j}R(z)]=x^{j+1}\big{(}R(z)-R(z+1)\big{)},$
$[u,x^{j}R(z)]=x^{j-1}\big{(}P(z+j)R(z)-P_{1}(z)R(z-1)\big{)},$

where $P_{1}$ is a polynomial for algebra $A_{c^{\prime}}$. Both $R(z)-R(z+1)$
and $P(z+j)R(z)-P_{1}(z)R(z-1)$ are differences of two polynomials with the
same leading coefficient, hence their degrees are at most $\deg R-1$, $\deg
R+\deg P-1=\deg R+n-1$ respectively. We see that both $[u,x^{j}R(z)]$ and
$[v,x^{j}R(z)]$ belong to $M_{\leqslant k+n-2}$.

Let us compute the action of $\mathop{\mathrm{gr}}\nolimits
A_{c}=\mathbb{C}[x,y]^{C_{n}}=\mathbb{C}[u,v,z]/(uv-z^{n})$ on
$\mathop{\mathrm{gr}}\nolimits M$. For $a\in M_{\leqslant k}\setminus
M_{\leqslant k-1}$ denote by $\overline{a}$ the corresponding element of
$(\mathop{\mathrm{gr}}\nolimits M)_{k}$. Take
$\overline{a}\in\mathop{\mathrm{gr}}\nolimits M$. We can assume that
$a=x^{j}z^{l}$ with $z+2l=k$. Then

$v\overline{a}=\overline{va}=\overline{x^{j+1}z^{l}},$
$z\overline{a}=\overline{za}=\overline{zx^{j}z^{l}}=\overline{(x+1)^{j}z^{l+1}}=\overline{x^{j}z^{l}},$
$u\overline{a}=\overline{ua}=\overline{x^{-1}P(z-\tfrac{1}{2})x^{j}z^{l}}=\overline{x^{j-1}P(z+j-\tfrac{1}{2})z^{l}}=\overline{x^{j-1}z^{n+l}}.$

This means that $\mathop{\mathrm{gr}}\nolimits M$ is
$\mathbb{C}[x,y]^{C_{n}}$-submodule of $\mathbb{C}[x,x^{-1},y]$ with
$\overline{x^{j}z^{l}}\in\mathop{\mathrm{gr}}\nolimits M$ corresponding to
$x^{nj+l}y^{l}\in\mathbb{C}[x,x^{-1},y]$.

Note that for large enough $j\geq j_{0}$ we have $R_{j}(z)=1$ and

$R_{-j-1}(z)=(z-c_{1}-j)(z-c_{2}-j)\cdots(z-c_{n}-j)R_{-j}(z).$

We see that $\mathop{\mathrm{gr}}\nolimits M$ is generated by
$\overline{x^{j}R_{j}(z)}$ for all $j$ with $\lvert j\rvert\leq j_{0}$.

∎

######  Remark 3.3.

In the case $n=2$ algebras $A_{c}$, $A_{c^{\prime}}$ are central reductions of
$U(\mathfrak{sl}_{2})$ and our definition of Harish-Chandra bimodule agrees
with the standard one: the adjoint action of $U(\mathfrak{sl}_{2})$ is locally
finite. In the case $n>2$ we use generators $u,v,z$ to write a definition
similar to the one that Losev gives in [Lo]: $\mathcal{M}$ is a filtered
bimodule over a filtered algebra $\mathcal{A}$, $d$ is a positive integer such
that $[\mathcal{A}_{\leqslant i},\mathcal{M}_{\leqslant
j}]\subset\mathcal{M}_{\leqslant i+j-d}$, module
$\mathop{\mathrm{gr}}\nolimits\mathcal{M}$ is finitely generated over
$\mathop{\mathrm{gr}}\nolimits\mathcal{A}$. Our definition is equivalent to
Losev’s if we take $\mathcal{A}$ to be the algebra with generators $u,v,z$ and
relations $[z,v]=v$, $[z,u]=-u$, so that it has both $A_{c}$ and
$A_{c^{\prime}}$ as its quotients, $\mathcal{M}=M_{c,c^{\prime}}$ and $d=2$.
Note that there is no relation on $[u,v]$ because $A_{c}$ and $A_{c^{\prime}}$
have different expressions of $[u,v]$ as a polynomial in $z$.

######  Lemma 3.4.

$M_{c,c^{\prime}}$ and $M_{c^{\prime},c}$ give a Morita equivalence between
$A_{c}$ and $A_{c^{\prime}}$.

###### Proof.

Let $U$ be any of $A_{c},A_{c^{\prime}},M_{c,c^{\prime}},M_{c^{\prime},c}$. By
definition $U\subset k[x,x^{-1},\partial_{x}]$. We also see that $U$ is a
$\mathbb{C}[v]=\mathbb{C}[x]$-module and for any
$p\in\mathbb{C}[x,x^{-1},\partial_{x}]/U$ there exists $k$ such that
$x^{k}p=0$. It follows that $k[x,x^{-1}]\otimes_{k[x]}U$ is isomorphic to
$k[x,x^{-1},\partial_{x}]$.

Consider the map $\phi\colon
M_{c,c^{\prime}}\otimes_{A_{c^{\prime}}}M_{c^{\prime},c}\to A_{c}$ that sends
$f\otimes g$ to $fg$. It can be proved that
$M_{c,c^{\prime}}\otimes_{A_{c^{\prime}}}M_{c^{\prime},c}$ does not have
$x$-torsion. Alternatively, we can use the fact that it is enough to prove
surjectivity in Morita context. Since $k[x,x^{-1}]$ is a flat $k[x]$-module it
is enough to prove that $\psi=\operatorname{id}_{k[x,x^{-1}]}\otimes\phi$ is
an isomorphism. After identifying $k[x,x^{-1}]\otimes_{k[x]}M_{c,c^{\prime}}$
and $k[x,x^{-1}]\otimes_{k[x]}A_{c}$ with $k[x,x^{-1},\partial_{x}]$ we get
$\psi\colon
k[x,x^{-1},\partial_{x}]\otimes_{A_{c^{\prime}}}M_{c^{\prime},c}\to
k[x,x^{-1},\partial_{x}]$ given by $f\otimes g\mapsto fg$.

We have $k[x,x^{-1},\partial_{x}]=k[x,x^{-1}]\otimes_{k[x]}A_{c^{\prime}}$.
Using this we get

$k[x,x^{-1},\partial_{x}]\otimes_{A_{c^{\prime}}}M_{c^{\prime},c}=k[x,x^{-1}]\otimes_{k[x]}M_{c^{\prime},c}=k[x,x^{-1},\partial_{x}]$

and $\psi$ becomes identity.

Hence $\phi$ is an isomorphism. We similarly prove the similar map from
$M_{c^{\prime},c}\otimes_{A_{c}}M_{c,c^{\prime}}$ to $A_{c^{\prime}}$ is an
isomorphism. The lemma follows. ∎

Now we assume that $A_{c^{\prime}}$ is isomorphic to $\overline{A_{c}}$ and
that both maps $A_{c}\to\overline{A_{c^{\prime}}}$,
$A_{c^{\prime}}\to\overline{A_{c}}$ are given by $v\mapsto au$, $u\mapsto bv$,
$z\mapsto-z$. Abusing notation we denote both maps by $\rho$.

It follows from discussion in Section 2.3 in [EKRS] that we may take $a,b$
such that $\lvert a\rvert=1$ and $ab=(-1)^{n}$. Hence $a=\pm i^{n}e^{-\pi
ic}$, $b=\pm i^{n}e^{\pi ic}$. These isomorphisms are well-defined when
$P_{c^{\prime}}(x)=(-1)^{n}\overline{P_{c}}(-x)$. Both sides have the same
leading coefficient, so this is equivalent to having the same set of roots. We
get the following condition: for any $i$ from $1$ to $n$ there exists $j$ such
that $c_{i}-\tfrac{1}{2}=\tfrac{1}{2}-\overline{c^{\prime}_{j}}$. The latter
is equivalent to $c_{i}+\overline{c^{\prime}_{j}}=1$.

Suppose that $j$ corresponds to $i$ and $k$ corresponds to $j$:
$c_{i}+\overline{c^{\prime}_{j}}=1$, $c_{j}+\overline{c^{\prime}_{k}}=1$.
Conjugating the second equation and subtracting we get

$c_{i}+\overline{c^{\prime}_{j}-c_{j}}-c^{\prime}_{k}=0,$

hence

$c_{i}-c_{k}=\overline{c_{j}-c^{\prime}_{j}}+c^{\prime}_{k}-c_{k}$

is a sum of two integers. This contradicts our assumption that
$\\{c_{1},\ldots,c_{n}\\}$ is a generic parameter.

Hence numbers from $1$ to $n$ are divided into pairs $(i,j)$ and singletons
$i=j$ such that
$c_{i}+\overline{c^{\prime}_{j}}=c_{j}+\overline{c^{\prime}_{i}}=1$.

Proposition 2.16 says that $\rho$-invariant forms on $M_{c,c^{\prime}}$ are in
one-to-one correspondence with $\rho^{2}$-twisted traces on
$M_{c,c^{\prime}}\otimes_{A_{c^{\prime}}}M_{c,c^{\prime},\rho^{-1}}$. Here for
an $A_{c}-A_{c^{\prime}}$-bimodule $M$ by $M_{\rho}$ we mean $\overline{M}$
with the action $b.m=\rho(b)m$, $m.a=m\rho(a)$ for $a\in A_{c}$,$b\in
A_{c^{\prime}}$, $m\in\overline{M}$. We want
$A_{c},A_{c^{\prime}},M_{c,c^{\prime}},M_{c^{\prime},c}$ to form Morita
context with conjugation as in example in Section 2.2. The two remaining
pieces are isomorphisms $M_{c,c^{\prime}}\cong M_{c^{\prime},c,\rho}$ and
$M_{c^{\prime},c}\cong M_{c,c^{\prime},\rho}$. We can interchange $c$ and
$c^{\prime}$, so it is enough to find just one of these two isomorphisms.

The bimodule $M_{c,c^{\prime}}$ is Harish-Chandra. This shows why we want the
action of $\rho$ on generators $u,v,z$ to be the same for $A_{c}$ and
$A_{c^{\prime}}$: if $M_{c^{\prime},c^{\prime},\rho}$ is isomorphic to
$M_{c,c^{\prime}}$, it is also Harish-Chandra. Hence, for example, the action
$m\mapsto\rho_{1}(v)m-m\rho_{2}(v)$ should send $M_{\leqslant k}$ to
$M_{\leqslant k+n-2}$. This is possible only when $\rho_{1}(v)$ and
$\rho_{2}(v)$ are the same multiple of $u$. Similarly, $\rho(z)$, $\rho(u)$
should be the same for $A_{c}$, $A_{c^{\prime}}$.

######  Lemma 3.5.

The map $\phi$ given by
$\phi(x^{j}R_{j}(z)R(z))=x^{-j}S_{-j}(z)\overline{R}(-h)$ is an isomorphism
from $M_{c,c^{\prime}}$ to $M_{c^{\prime},c,\rho}$. The map $\phi$ also gives
an isomorphism from $M_{c,c^{\prime},\rho^{-1}}$ to $M_{c^{\prime},c}$. Here
for an integer $j$ and all $i$ such that $c^{\prime}_{i}-j-1\geq c_{i}$ the
polynomial $S_{j}$ has roots $c_{i},\ldots,c^{\prime}_{i}-j-1$. The leading
coefficient of $S_{j}$ is $C_{\phi}(-1)^{\deg R_{-j}}a^{-j}$, where $C_{\phi}$
is a constant corresponding to a choice of $\phi$.

###### Proof.

The second statement follows from the first after twisting the action from
both sides by $\rho^{-1}$.

We want to construct a linear isomorphism $\phi\colon M_{c,c^{\prime}}\to
M_{c^{\prime},c}$ such that $\phi(em)=v.\phi(m)=au\phi(m)$,
$\phi(um)=u.\phi(m)=bv\phi(m)$, $\phi(zv)=z.\phi(v)=-z\phi(v)$, similarly for
the right multiplication.

We define $\phi(x^{j}R_{j}(z)R(z)):=x^{-j}S_{-j}(z)\overline{R}(-h)$. Here
$S_{-j}$ is a polynomial that satisfies a similar condition on roots as
$R_{j}$ but is not necessarily monic. Namely, $S_{j}$ has roots
$c_{i},\ldots,c^{\prime}_{i}-j-1$ for all $i$ such that
$c^{\prime}_{i}-j-1\geq c_{i}$.

We see that $\phi$ is linear and satisfies $\phi([z,m])=[\phi(m),z]$ and
$\phi(mz)=-\phi(m)z$ for all $m\in M_{c,c^{\prime},\rho}$. Hence
$\phi(zm)=-z\phi(m)$ for all $m\in M_{c,c^{\prime},\rho}$.

It is enough to check all other conditions for $m=x^{j}R_{j}(z)$.

We have

$\phi(vm)=\phi(x^{j+1}R_{j}(z))=\phi(x^{j+1}R_{j+1}(z)L_{j}(z))=x^{-j-1}S_{-j-1}(z)\overline{L_{j}}(-h).$
(3.1)

Here $L_{j}(t)=\frac{R_{j}(t)}{R_{j+1}(t)}$ is a monic polynomial with roots
$c_{i}-j-1$ for all $i$ such that $c_{i}-j-1\geq c^{\prime}_{i}$.

Let $\sigma$ denote the permutation such that
$\overline{c_{i}}+c^{\prime}_{\sigma(i)}=1$, $\sigma^{2}=1$. Suppose that
$c_{i}-j-1\geq c^{\prime}_{i}$. We get
$\overline{c_{i}}-j-1\geq\overline{c^{\prime}_{i}}$, hence
$1-c^{\prime}_{\sigma(i)}-j-1\geq 1-c_{\sigma(i)}$. It follows that
$c_{\sigma(i)}-j-1\geq c^{\prime}_{\sigma(i)}$. We see that roots of $L$ come
in pairs $c_{i}-j-1$, $c_{\sigma(i)}-j-1$.

The polynomial $\overline{L_{j}}(-t)$ has roots
$-(\overline{c_{i-j-1}})=j+1-\overline{c_{i}}=j+c^{\prime}_{\sigma(i)}$ for
all $i$ such that $c_{i}-j-1\geq c^{\prime}_{i}$. Taking $i$ instead of
$\sigma(i)$, the roots become $j+c^{\prime}_{i}$.

We have $S_{-j-1}(t)=S_{-j}(t)M_{-j-1}(t)$, where $M_{-j-1}$ has roots
$j+c^{\prime}_{i}$ for all $i$ such that $c^{\prime}_{i}+j\geq c_{i}$.
Similarly to the above, roots of $M_{-j-1}$ are in pairs $j+c^{\prime}_{i}$,
$j+c^{\prime}_{\sigma(i)}$.

For every $i$ either $c_{i}-j-1\geq c^{\prime}_{i}$ or $c^{\prime}_{i}+j\geq
c_{i}$ is true but not both. We deduce that the union of roots of $M_{-j-1}$
and $\overline{L_{j}}(-t)$ is disjoint and equal to
$\\{c^{\prime}_{1}+j,\ldots,c^{\prime}_{n}+j\\}$.

Using (3.1) we get

$\phi(vm)=x^{-j-1}S_{-j-1}(z)\overline{L_{j}}(-z)=x^{-j-1}S_{-j}(z)M_{-j-1}(z)\overline{L_{j}}(-z).$
(3.2)

From $M_{-j-1}=\frac{S_{-j-1}}{S_{-j}}$ we deduce that $M_{-j-1}$ has leading
coefficient

$\frac{(-1)^{\deg R_{j+1}}a^{j+1}}{(-1)^{\deg R_{j}}a^{j}}=(-1)^{\deg
R_{j+1}-\deg R_{j}}a=(-1)^{\deg L_{j}}a.$

Hence $M_{-j-1}(z)\overline{L_{j}}(-z)$ has leading coefficient $a$ and
$M_{-j-1}(z)\overline{L_{j}}(-z)=a(z-c^{\prime}_{1}-j)\cdots(z-c^{\prime}_{n}-j)=P_{1}(z-\tfrac{1}{2}-j)$.
We deduce from (3.2) that
$\phi(vm)=ax^{-j-1}S_{-j}(z)P_{1}(z-\tfrac{1}{2}-j)$.

We have

$au\phi(m)=ax^{-1}P_{1}(z-\tfrac{1}{2})x^{-j}S_{-j}(z)=ax^{-j-1}P_{1}(z-\tfrac{1}{2}-j)S_{-j}(z)=\phi(vm).$

It follows from the description of $M_{c,c^{\prime}}$ that it is a torsion-
free $A_{c}$ and $A_{c^{\prime}}$ module. Hence in order to check that
$\phi(um)=bv\phi(m)$, we can check that $u\phi(um)=buv\phi(m)$. This follows
from what we already proved:

$buv\phi(m)=bP_{1}(z+\tfrac{1}{2})\phi(m)=b\phi(\overline{P_{1}}(\tfrac{1}{2}-z)m)=\\\
(-1)^{n}b\phi(P(z-\tfrac{1}{2})m)=(-1)^{n}b\phi(vum)=(-1)^{n}abu\phi(um)=u\phi(um).$

We used that $\overline{P_{1}}(-t)=(-1)^{n}P(t)$, $ab=(-1)^{n}$ and
$\phi(vum)=au\phi(um)$.

Similarly the condition $\phi(mu)=\phi(m)bv$ will follow from the condition
$\phi(mv)=\phi(m)au$ for all $m$. We will prove it now. We have

$\phi(mv)=\phi(x^{j}R_{j}(z)x)=\phi(x^{j+1}R_{j}(z+1))=\\\
\phi(x^{j+1}R_{j+1}(z)A_{j}(z))=x^{-j-1}S_{-j-1}(z)\overline{A_{j}}(-z).$
(3.3)

Here $A_{j}$ has roots $c^{\prime}_{i}-1$ for all $i$ such that $c_{i}-j-1\geq
c^{\prime}_{i}$ and $\overline{A_{j}}(-z)$ has roots
$-(\overline{c^{\prime}_{i}}-1)=c_{\sigma(i)}$ for the same $i$. Reasoning as
with $L$ we see that the roots of $A_{j}$ come in pairs $c_{i},c_{\sigma(i)}$.

We have $S_{-j-1}(t)=S_{-j}(t-1)B_{-j-1}(t)$. The polynomial $B_{-j-1}$ has
roots $c_{i}$ for all $i$ such that $c^{\prime}_{i}+j\geq c_{i}$. Reasoning as
above, we deduce that the union of the roots of $A_{j}(t)$ and $B_{-j-1}(t)$
is disjoint and equals to $\\{c_{1},\ldots,c_{n}\\}$.

Similarly to the above, we deduce that
$B_{-j-1}(z)\overline{A_{j}}(-h)=aP(z-\tfrac{1}{2})$. The leading signs
coincide because the leading sign of $B_{-j-1}$ equals to the leading sign of
$M_{-j-1}$ and the degree of a monic polynomial $A_{j}$ equals to the degree
of a monic polynomial $L_{j}$.

Combining this and (3.3) we get
$\phi(mv)=ax^{-j-1}S_{-j}(z-1)P(z-\tfrac{1}{2})$.

We have

$\phi(m)au=ax^{-j}S_{-j}(z)x^{-1}P(z-\tfrac{1}{2})=ax^{-j-1}S_{-j}(z-1)P(z-\tfrac{1}{2}).$

The lemma follows. ∎

###  3.2 The positive forms.

Proposition 2.16 says that any $\rho$-invariant sesquilinear form
$(\cdot,\cdot)$ on $M$ is given by $(m,n)=T(m\phi(n))$ , where $T$ is a
$g_{t}=\rho^{2}$-twisted trace. Here $t=ba^{-1}$. For fixed $t$ there are two
conjugations $\rho$ with $\rho^{2}=g_{t}$. One of them is $\rho_{+}$, the
other is $\rho_{-}$, later we will specify which is which. The answer for
$\rho_{+}$ and $\rho_{-}$ is sometimes different, as it was in [EKRS].

We want to describe all traces in convenient form. Recall that $\lvert
t\rvert=1$. Let $t=e^{2\pi ic}$, where $c\in[0,1)$. We will need the following
definition:

######  Definition 3.6.

We say that a non-self-intersecting curve $C$ on a complex plane is an good
contour if the following holds:

  1. 1.

There exists $r>0$ such that $C\setminus B_{r}(0)$ coincides with
$(a+i\mathbb{R})\setminus B_{r}(0)$ for some $a\in\mathbb{R}$. This allows us
to define the notions ’’to the left of $C$’’ and ’’to the right of $C$’’.

  2. 2.

For every $i=1,\ldots,n$ the set $c_{i}-\mathbb{Z}_{>0}$ is to the left of $C$
and $c_{i}+\mathbb{Z}_{\geq 0}$ is to the right of $C$.

We note that for generic $c=\\{c_{1},\ldots,c_{n}\\}$ there exist good
contours.

Let ${\mathbb{P}}(x)=\prod_{i=1}^{n}(x-e^{2\pi ic_{i}})$.

Recall that $A_{c}$ is graded by the action of $\operatorname{ad}h$ and the
zeroth component is $\mathbb{C}[h]$. We have the following proposition,
similar to Proposition 3.1 from [EKRS].

######  Proposition 3.7.

Let $C$ be a good contour. Then any $g_{t}$-twisted trace $T$ on $A_{c}$ is
zero on $\operatorname{ad}z$ eigenspaces of nonzero weight and given on
$\mathbb{C}[z]$ by

$T(R(z))=\int_{C}R(x)w(x)dx,\qquad R\in\mathbb{C}[z],$

where $w$ is a weight function defined by the formula $w(t)=e^{2\pi
icx}\frac{G(e^{2\pi ix})}{{\mathbb{P}}(e^{2\pi ix})}$ and $G$ is a polynomial
of degree at most $n-1$ such that $G(0)=0$ if $c=0$.

###### Proof.

We note that $w(x+1)=tw(x)$. Conditions on $G$ imply that $w$ decays
exponentially when $|\mathop{\mathrm{Im}}\nolimits z|$ goes to infinity. Since
$C$ is good, the integral $\int_{C}R(z)w(z)dz$ is defined for all
$R\in\mathbb{C}[z]$.

Proposition 2.3 from [EKRS] says that $T$ is a trace if and only if $T$ is
supported on $\mathbb{C}[h]$ and
$T\big{(}S(z-\tfrac{1}{2})P(z-\tfrac{1}{2})\big{)}=tT(S(z+\tfrac{1}{2})P(z+\tfrac{1}{2})$.
Similarly to the proof of Proposition 3.1 from [EKRS] we have

$T\big{(}S(z-\tfrac{1}{2})P(z-\tfrac{1}{2})-tS(z+\tfrac{1}{2})P(z+\tfrac{1}{2})\big{)}=-\int_{\partial
U}S(x-\tfrac{1}{2})P(x-\tfrac{1}{2})w(x)dx,$

where $U$ is the region between $C$ and $C+1$, so that the boundary of $U$ is
$C+1$ in positive direction and $C$ in negative. It is enough to prove that
$P(x-\tfrac{1}{2})w(x)$ has no poles between $C$ and $C+1$. By definition of
$C$ the poles of $w$ between $C$ and $C+1$ are contained in
$\\{c_{1},\ldots,c_{n}\\}$. The roots of $P$ are
$c_{1}-\tfrac{1}{2},\ldots,c_{n}-\tfrac{1}{2}$, so the roots of
$P(z-\tfrac{1}{2})$ are $c_{1},\ldots,c_{n}$. It follows that
$P(x-\tfrac{1}{2})w(x)$ has no poles between $C$ and $C+1$.

We obtained the subspace of $g_{t}$-twisted traces of dimension $n$ if $t\neq
1$ and $n-1$ if $t=1$. This is exactly the dimension of the space of traces
from Corollary 2.4 in [EKRS]. ∎

From now on, we will not need $t$ in our computations, only $c$. Since we use
$x$ to express elements of $M_{c,c^{\prime}}$, we will use $t$ as an
integration variable.

Now we start computing the cone of positive definite Hermitian forms for fixed
$\rho$. Note that different $\operatorname{ad}z$ eigenspaces are orthogonal
with respect to $(\cdot,\cdot)$, hence it is enough to check the condition
$(m,m)>0$ for $m$ in some eigenspace of $\operatorname{ad}z$. Suppose that
$(\cdot,\cdot)$ is positive definite. When $m=x^{j}R_{j}(z)R(z)$, we have

$(m,m)=T(m\phi(m))=T(x^{j}R_{j}(z)R(z)x^{-j}S_{-j}(z)\overline{R}(-z))=\\\
T(R_{j}(z-j)R(z-j)S_{-j}(z)\overline{R}(-z))=T(R_{j}(z-j)S_{-j}(z)R(z-j)\overline{R}(-z))=\\\
\int_{C}R(x-j)\overline{R}(-x)R_{j}(x-j)S_{-j}(x)w(x)dx.$

We use the same strategy as in [EKRS]: we try to shift the contour $C$ to
$i\mathbb{R}+\tfrac{j}{2}$. If there are poles between $C$ and
$i\mathbb{R}+\tfrac{j}{2}$, we prove that this integral is negative for some
$R$, a contradiction. If there are no poles, we use that polynomials are dense
in $L^{2}(\mathbb{R},\omega)$ for an exponentially decaying weight $\omega$ to
get that $R_{j}(z-j)S_{-j}(z)w(z)$ should be positive on
$i\mathbb{R}+\tfrac{j}{2}$. This gives a condition on $G$ similar to the one
in [EKRS].

We say that index $i$ is bad if there exists $j$ such that the function

$f(x)=\frac{R_{j}(x-j)S_{-j}(x)}{(e^{2\pi ix}-e^{2\pi ic_{i}})}$

has poles in the closed region between $C$ and $\frac{j}{2}+i\mathbb{R}$. If
this holds, we say that $j$ is bad for $i$, in the other case we say that $j$
is good for $i$.

######  Lemma 3.8.

An index $i$ is bad if and only if
$\operatorname{Re}c_{i}+\operatorname{Re}c^{\prime}_{i}\leqslant 0$ or
$\operatorname{Re}c_{i}+\operatorname{Re}c^{\prime}_{i}\geqslant 2$.

###### Proof.

Recall that the intersection of roots of $R_{j}(z)$ with $c_{i}+\mathbb{Z}$ is
$\\{c^{\prime}_{i},c^{\prime}_{i}+1,\ldots,c_{i}-j-1\\}$ in the case when
$c_{i}-c^{\prime}_{i}-j>0$ and empty otherwise. For $R_{j}(z-j)$ this becomes
$c^{\prime}_{i}+j,\ldots,c_{i}-1$ or empty. For $S_{j}$ this intersection
equals to $\\{c_{i},c_{i}+1,\ldots,c^{\prime}_{i}+j-1\\}$ in the case when
$c^{\prime}_{i}-c_{i}+j>0$ and empty otherwise.

Suppose that $i$ is bad, take $j$ that is bad for this $i$. Denote by $L$ the
line $\frac{j}{2}+i\mathbb{R}$. Assume that $f$ has a pole to the left of $C$
and to the right of $L$. Recall that the set $c_{i}-\mathbb{Z}_{>0}$ is to the
left of $C$ and $c_{i}+\mathbb{Z}_{\geqslant 0}$ is to the right of $C$. It
follows that this pole of $f$ is $c_{i}-k$ for some $k>0$. This pole is to the
right of $L$, hence $\operatorname{Re}c_{i}-k\geqslant\frac{j}{2}$. All roots
of the denominator $e^{2\pi iz}-e^{2\pi ic_{i}}$ of $f$ are simple, hence
$c_{i}-k$ cannot be a root of $R_{j}(z-j)S_{-j}(z)$.

If $c^{\prime}_{i}+j>c_{i}-k$ then $c_{i}-k$ cannot be a root of
$R_{j}(z-j)S_{-j}(z)$. If $c^{\prime}_{i}+j\leqslant c_{i}-k$ then
$c_{i}-c^{\prime}_{i}-j\geqslant k>0$, so that $R_{j}(z-j)$ has roots
$c^{\prime}_{i}+j,\ldots,c_{i}-1$, hence it has root $c_{i}-k$.

So the condition we get is $c^{\prime}_{i}+j\geq c_{i}-k+1$. Hence there are
two conditions on $j$:

$\frac{j}{2}\leqslant\operatorname{Re}c_{i}-k,$ $j\geq
c_{i}-c^{\prime}_{i}-k+1.$

They can be satisfied by an integer $j$ if an only if

$c_{i}-c^{\prime}_{i}-k+1\leqslant\lfloor 2\operatorname{Re}c_{i}\rfloor-2k.$

There exists such integer $k>0$ if and only if $k=1$ works:

$c_{i}-c^{\prime}_{i}\leqslant\lfloor 2\operatorname{Re}c_{i}\rfloor-2.$

The left-hand side is an integer, so we don’t need to take floor function on
the right. We also have
$c_{i}-c^{\prime}_{i}=\operatorname{Re}c_{i}-\operatorname{Re}c^{\prime}_{i}$.
In the end we get $\operatorname{Re}c_{i}+\operatorname{Re}c^{\prime}_{i}\geq
2$.

If there exists a pole of $f$ to the left of $L$ and to the right of $C$, the
reasoning is similar: the poles is $c_{i}+k$ for some $k\geq 0$ and
$\operatorname{Re}c_{i}+k\leqslant\frac{j}{2}$. It works if $c_{i}+k$ is not a
root of $S_{-j}(z)$, this is equivalent to $c^{\prime}_{i}+j-1<c_{i}+k$. So
the two conditions are

$j\geq 2\operatorname{Re}c_{i}+2k,$ $j\leqslant c_{i}-c^{\prime}_{i}+k.$

Similarly to the above they can be satisfied when

$2\operatorname{Re}c_{i}+2k\leqslant c_{i}-c^{\prime}_{i}+k.$

It is enough to take $k=0$ here. In the end we get
$\operatorname{Re}c_{i}+\operatorname{Re}c^{\prime}_{i}\leqslant 0$. ∎

We claim that the cone of positive traces on $M_{c,c^{\prime}}$ is isomorphic
to the cone from [EKRS], where instead of bad and good roots we count bad and
good indices $i$.

######  Proposition 3.9.

Suppose that there exists a bad $i$ such that $w$ has a pole in $c_{i}$. Then
$w$ does not give a positive definite form on $M_{c,c^{\prime}}$.

###### Proof.

The proof of this proposition is very similar to subsections 4.3-4.4 of [EKRS]
with simplifications because $w$ has only simple poles.

Assume that $w$ gives a positive definite form.

We fix bad $i$ and $j$ that is bad for this $i$. We have

$(x^{j}R_{j}(h)R(h),x^{j}R_{j}(h)R(h))=\\\
T(R_{j}(z-j)S_{-j}(z)R(z-j)\overline{R}(-z))=\\\
\int\limits_{C}R_{j}(s-j)S_{-j}(s)R(s-j)\overline{R}(-s)w(s)ds$

Let $S$ be a polynomial such that $S(t)w(t)$ has no poles between $C$ and
$i\mathbb{R}+\frac{j}{2}$. Then for any $R\in\mathbb{C}[x]$ we have

$(x^{j}R_{j}(z)R(z)S(z+j),x^{j}R_{j}(z)R(z)S(z+j))=\\\
\int\limits_{C}R_{j}(t-j)S_{-j}(t)R(t-j)S(t)\overline{R}(-t)\overline{S}(j-t)dt=\\\
\int\limits_{i\mathbb{R}+\frac{j}{2}}R_{j}(t-j)S_{-j}(t)R(t-j)S(t)\overline{R}(-t)\overline{S}(j-t)dt$

We will use Lemma 4.2 from [EKRS]:

######  Lemma 3.10.

  1. Suppose that $w(x)\geq 0$ is a measurable function on the real line such that $w(x)<ce^{-b|x|}$ for some $c,b>0$, $1\leqslant p<\infty$.

  2. 1.

Suppose that $H$ is a continuous complex-valued function on $\mathbb{R}$ with
finitely many zeroes and at most polynomial growth at infinity. Then the set
$\\{H(x)S(x)\mid S(x)\in\mathbb{C}[x]\\}$ is dense in the space
$L^{p}(\mathbb{R},w)$.

  3. 2.

Suppose that $M(x)$ is a nonzero polynomial nonnegative on the real line. Then
the closure of the set $\\{M(x)S(x)\overline{S}(x)\mid
S(x)\in\mathbb{C}[x]\\}$ in $L^{p}(\mathbb{R},w)$ is the subset of almost
everywhere nonnegative functions.

We have

$\int\limits_{i\mathbb{R}+\frac{j}{2}}R_{j}(t-j)S_{-j}(t)R(t-j)S(t)\overline{R}(-t)\overline{S}(j-t)w(t)dt>0$

for all $R\in\mathbb{C}[x]$. Multiplying $w$ by $\pm i$, we can change $dt$ to
$\lvert dt\rvert$, a positive measure. Using Lemma 3.10(2) for

$w=R_{j}(t-j)S_{-j}(t)w(t)$

and

$M=S(t)\overline{S}(j-t)$

after the change of argument $t\mapsto it+\frac{j}{2}$ we deduce that

$R_{j}(t-j)S_{-j}(t)w(t)$

is nonnegative on the line $i\mathbb{R}+\frac{j}{2}$.

In particular, $R_{j}(t-j)S_{-j}(t)w(t)$ has poles of even order on the line
$i\mathbb{R}+\frac{j}{2}$. On the other hand all poles of $w$ are simple.
Therefore $R_{j}(t-j)S_{-j}(t)w(t)$ has no poles on the line
$i\mathbb{R}+\frac{j}{2}$.

Since $i$ is bad, we deduce that $R_{j}(t-j)S_{-j}(t)w(t)$ has poles strictly
between $C$ and $i\mathbb{R}+\frac{j}{2}$. We write

$T(R(z)R_{j}(z-j)S_{-j}(z))=\int\limits_{C}R(t)R_{j}(t-j)S_{-j}(t)w(t)dt=\\\
\int\limits_{i\mathbb{R}+\frac{j}{2}}R(t)R_{j}(t-j)S_{-j}(t)w(t)dt+\Phi(R),$

where $\Phi(R)$ is a nonzero linear functional of the form $\sum
a_{i}R(t_{i})$, $a_{i}\in\mathbb{C}$, $t_{i}$ are poles of
$R_{j}(t-j)S_{-j}(t)w(t)$ between $C$ and $i\mathbb{R}+\frac{j}{2}$. We get a
contradiction with the following lemma for $w=R_{j}(t-j)S_{-j}(t)w(t)$ after a
change of argument $t\mapsto it+\frac{j}{2}$.

######  Lemma 3.11.

Suppose that $w(t)$ is almost everywhere nonnegative function on the real line
such that $w(t)<be^{-c\lvert t\rvert}$ for some $b,c>0$, $\Phi$ is a nonzero
linear functional on $\mathbb{C}[t]$ of the form

$\Phi(R)=\sum_{i=1}^{l}a_{i}R(t_{i}),$

where $a_{i}\in\mathbb{C}$, $t_{i}\notin\mathbb{R}$,

$T(R)=\int\limits_{\mathbb{R}}w(t)R(t)dt+\Phi(R).$

Then there exists $R\in\mathbb{C}[t]$ such that
$T(R(t)\overline{R}(t))\notin\mathbb{R}_{\geq 0}$.

###### Proof.

Let $S$ be a polynomial such that $S=\overline{S}$ and
$S(t_{1})=\cdots=S(t_{l})=0$. It follows that

$\Phi(SP)=\Phi(\overline{S}P)=\\{0\\}$ (3.4)

for any polynomial $P$. Let $R$ be any polynomial. Using Lemma 3.10(1) for
$H=S$ we find a sequence of polynomials $M_{n}$ such that $SM_{n}$ tends to
$R$ in $L^{2}(\mathbb{R},w)$. Using (3.4) we have

$T((R-SM_{n})(\overline{R}-\overline{S}\overline{M_{n}}))=\|R-SM_{n}\|^{2}_{L^{2}(\mathbb{R},w)}+\Phi(R\overline{R}).$

Since $\|R-SM_{n}\|_{L^{2}(\mathbb{R},w)}$ tends to zero, it is enough to find
$R\in\mathbb{C}[t]$ such that $\Phi(R\overline{R})$ is not a nonnegative real
number.

Let $\Phi(R)=\sum_{i=1}^{k}a_{i}R(t_{i})$, where $a_{1}\neq 0$. Taking
$a_{2}=0$ if necessary, we can assume that $t_{2}=\overline{t_{1}}$. Let $p,q$
be complex numbers. Let $R$ be a polynomial such that $R(t_{1})=p$,
$R(t_{2})=q$, $R(t_{i})=0$ for $i>2$. Then
$\Phi(R\overline{R})=a_{1}p\overline{q}+a_{2}q\overline{p}$. We can find $p,q$
such that $a_{1}p\overline{q}+a_{2}q\overline{p}$ is not a nonnegative real
number. The lemma follows. ∎

∎

Now we assume that $w$ does not have poles at $c_{i}$ for all bad $i$. In this
case we can write $w(x)=e^{2\pi icx}\frac{G(e^{2\pi ix})}{{\mathbb{P}}(e^{2\pi
ix})}$, where the new ${\mathbb{P}}$ has roots at $e^{2\pi ic_{i}}$ for all
good $i$. We have the following

######  Proposition 3.12.

The form $(m,n)=T(m\phi(n))$ is positive definite if and only if
$R_{j}(t-j)S_{-j}(t)w(t)\geq 0$ for all $j$ and $t\in
i\mathbb{R}+\tfrac{j}{2}$.

###### Proof.

Recall that

$(x^{j}R_{j}(z)R(z),x^{j}R_{j}(z)R(z))=\int_{C}R_{j}(t-j)S_{-j}(t)R(t-j)\overline{R}(-t)w(t)dt.$

Since $w$ is good, we can take $i\mathbb{R}+\tfrac{j}{2}$ instead of $C$ in
this integral. We can also change $dt$ to $\lvert dt\rvert$ for convenience.
Hence $(\cdot,\cdot)$ is positive definite if and only if

$\int_{i\mathbb{R}+\tfrac{j}{2}}R_{j}(t-j)S_{-j}(t)R(t-j)\overline{R}(-t)w(t)\lvert
dt\rvert>0$

for all integer $j$ and nonzero polynomials $R$. Using Lemma 3.10 after the
change of argument $t\mapsto it+\tfrac{j}{2}$ with $M=1$, we see that
$R_{j}(t-j)S_{-j}(t)w(t)$ should be nonnegative on $i\mathbb{R}+\tfrac{j}{2}$.
∎

It remains to understand $R_{j}(t-j)S_{-j}(t)w(t)$ when
$\operatorname{Re}t=\tfrac{j}{2}$. We start with describing the behavior of
$R_{j}(t-j)S_{-j}(t)$ on $i\mathbb{R}+\tfrac{j}{2}$. Recall that $\phi\colon
M_{c,c^{\prime},\rho^{-1}}\to M_{c^{\prime},c}$ is defined up to a constant.
Since $w$ can also be multiplied by any constant, we can choose any $\phi$ we
like and the answer will be the same.

######  Lemma 3.13.

The set of roots of $R_{j}(t-j)S_{-j}(t)$ is
$\\{c_{i},\ldots,c^{\prime}_{i}+j-1\\}$,
$\\{c^{\prime}_{i}+j,\ldots,c_{i}-1\\}$ or empty depending on the sign of
$c^{\prime}_{i}-c_{i}+j$. We can choose $\phi$ such that for all $j$ the
polynomial $(ai^{n})^{-j}R_{j}(t-j)S_{-j}(t)$ is real on the line
$\operatorname{Re}t=\frac{j}{2}$ and positive when
$\operatorname{Re}t=\frac{j}{2}$ and $\mathop{\mathrm{Im}}\nolimits t$ is
large enough.

###### Proof.

Recall that the intersection of $c_{i}+\mathbb{Z}$ with the roots of
$R_{j}(t)$ equals to $\\{c^{\prime}_{i},\ldots,c_{i}-j-1\\}$, it has size
$\max(c_{i}-c^{\prime}_{i}-j,0)$. Lemma 3.5 says that the intersection of
$c_{i}+\mathbb{Z}$ with the set of roots of $S_{-j}$ equals to
$\\{c_{i},\ldots,c^{\prime}_{i}+j-1\\}$, and has size
$\max(c^{\prime}_{i}-c_{i}+j,0)$. It also says that the leading coefficient of
$S_{-j}$ equals to $C_{\phi}(-1)^{\deg R_{j}}a^{j}$, where $C_{\phi}$ is a
constant that depends only on $\phi$.

Hence one of the intersections of $c_{i}+\mathbb{Z}$ with roots is empty and
the other has size $\lvert c_{i}-c^{\prime}_{i}-j\rvert$ and may be also
empty.

Recall that there exists index $k$, possibly equal to $i$, such that
$c_{i}+\overline{c^{\prime}_{k}}=1$, $c_{k}+\overline{c^{\prime}_{i}}=1$. It
follows that $c_{i}-c^{\prime}_{i}=c_{k}-c^{\prime}_{k}$. Hence
$c_{i}-c^{\prime}_{i}-j=c_{k}-c^{\prime}_{k}-j$, in particular they have the
same sign.

Suppose that the intersection of $c_{i}+\mathbb{Z}$ with the set of roots of
$R_{j}(t-j)S_{-j}(t)$ is $\\{c^{\prime}_{i}+j,\ldots,c_{i}-1\\}$. In this case
the intersection of $c_{k}+\mathbb{Z}$ with the set of roots of
$R_{j}(t-j)S_{-j}(t)$ is $\\{c^{\prime}_{k}+j,\ldots,c_{k}-1\\}$. We have
$-\overline{c^{\prime}_{i}+j}=j-\overline{c^{\prime}_{i}}=j+c_{k}-1$. In this
case we see that the intersection of $\\{c_{i},c_{k}\\}+\mathbb{Z}$ with the
set of roots of $R_{j}(t-j)S_{-j}(t)$ is symmetric with respect to the line
$\operatorname{Re}t=\frac{j}{2}$. The other case is done similarly.

Hence the roots of $R_{j}(t-j)S_{-j}(t)$ are symmetric with respect to the
line $\operatorname{Re}t=\frac{j}{2}$.

Note that $R_{j}$ is monic and the leading coefficient of $S_{-j}$ is
$C_{\phi}(-1)^{\deg R_{j}}a^{j}$. It follows that the argument of
$R_{j}(t-j)S_{-j}(t)$ tends to the argument of

$i^{\deg R_{j}+\deg S_{-j}}C_{\phi}(-1)^{\deg R_{j}}a^{j}=C_{\phi}i^{\deg
S_{-j}-\deg R_{j}}a^{j}$

when $t$ tends to $i\infty$.

We can compute $\deg S_{-j}-\deg R_{j}$ as the number of roots of $S_{-j}$
minus the number of roots of $R_{j}$. We see from the description of roots
above that each $i=1,\ldots,n$ contributes $j-c^{\prime}_{i}-c_{i}$ to this
expression, so that $\deg S_{-j}-\deg R_{j}=nj-\sum c^{\prime}_{i}-\sum
c_{i}$. Hence the argument of $R_{J}(t-j)S_{-j}(t)$ tends to the argument

$(ai^{n})^{j}C^{\prime}_{\phi}$

when $t$ tends to $i\infty$, where $C^{\prime}_{\phi}=i^{-\sum
c^{\prime}_{i}-\sum c_{i}}C_{\phi}$. Choosing $C^{\prime}_{\phi}=1$ we get
that the argument of $(ai^{n})^{-j}R_{j}(t-j)S_{-j}(t)$ tends to zero when $t$
tends to $i\infty$. On the other hand, this polynomial does not change
argument when $\mathop{\mathrm{Im}}\nolimits t$ is large enough and
$\operatorname{Re}t=\frac{j}{2}$, hence it is positive. Since the roots of
$R_{j}(t-j)S_{-j}(t)$ are symmetric with respect to
$\operatorname{Re}t=\frac{j}{2}$, the polynomial
$(ai^{n})^{-j}R_{j}(t-j)S_{-j}(t)$ is real on this line. ∎

######  Proposition 3.14.

In this case $w$ gives a positive definite form on $M_{c,c^{\prime}}$ if and
only if $G$ has certain behavior on the real line: $G(x)$ is nonnegative for
$x\in\mathbb{R}$ in the case when $\rho=\rho_{+}$, $G(x)$ is nonnegative for
$x>0$ and nonpositive for $x<0$ in the case when $\rho=\rho_{-}$.

###### Proof.

The condition on $w$ implies that

$\big{(}x^{j}R_{j}(z)R(z),x^{j}R_{j}(z)R(z)\big{)}=\int\limits_{i\mathbb{R}+\frac{j}{2}}R_{j}(t-j)S_{-j}(t)R(t-j)\overline{R}(-t)w(t)\lvert
dt\rvert.$

Polynomial $R(t-j)\overline{R}(-t)$ is nonnegative on
$i\mathbb{R}+\frac{j}{2}$. It remains to check that $R_{j}(t-j)S_{-j}(t)w(t)$
is nonnegative on $i\mathbb{R}+\frac{j}{2}$. Lemma 3.13 says that
$(ai^{n})^{-j}R_{j}(t)S_{-j}(t)$ is positive when
$\operatorname{Re}t=\frac{j}{2}$ and $\mathop{\mathrm{Im}}\nolimits t$ is
large enough.

We have $a=\varepsilon e^{-\pi ic}i^{n}$, where $\rho=\rho_{\varepsilon}$. It
follows that

$\varepsilon^{j}e^{\pi icj}(-1)^{nj}R_{j}(t-j)S_{-j}(t)$

is positive when $\operatorname{Re}t=\frac{j}{2}$ and
$\mathop{\mathrm{Im}}\nolimits t$ is large enough.

We note that the zeroes of ${\mathbb{P}}(e^{2\pi it})$ on
$i\mathbb{R}+\frac{j}{2}$ are simple and in one-to-one correspondence with the
roots of $R_{j}(t-j)S_{-j}(t)$ on $i\mathbb{R}+\frac{j}{2}$, this follows from
the definition of a good index. Hence $w(t)R(t-j)S_{-j}(t)$ does not change
argument on $i\mathbb{R}+\frac{j}{2}$ if and only if $G$ does not have roots
on $(-1)^{j}\mathbb{R}_{>0}$.

Hence the necessary condition for positivity is that $G$ does not have roots
on $\mathbb{R}\setminus\\{0\\}$. If this condition holds then
$w(t)R(t-j)S_{-j}(t)$ does not change argument on $i\mathbb{R}+\frac{j}{2}$
for all $j$.

We have

$w(t)=e^{2\pi ict}\frac{G(e^{2\pi it})}{{\mathbb{P}}(e^{2\pi it})}.$

When $\operatorname{Re}t=\frac{j}{2}$, the function $e^{2\pi ict}$ has
argument $\pi cj$.

It remains to look at the behavior of $\frac{G(e^{2\pi
it})}{{\mathbb{P}}(e^{2\pi it})}$ when $\operatorname{Re}t=\frac{j}{2}$ and
$\mathop{\mathrm{Im}}\nolimits t$ tends to infinity. Suppose that the lowest
term in $G(x)$ is $sx^{k}$. Then $\frac{G(e^{2\pi it})}{{\mathbb{P}}(e^{2\pi
it})}$ has sign $\frac{(-1)^{kj}s}{{\mathbb{P}}(0)}$. We get the condition
$\varepsilon^{j}(-1)^{nj+kj}\frac{s}{{\mathbb{P}}(0)}$ should be positive for
all $j$. This means that $\varepsilon=(-1)^{n+k}$ and
$\frac{s}{{\mathbb{P}}(0)}$ is positive. The proposition follows. ∎

######  Remark 3.15.

For $\rho=\rho_{-}$ the sign of our polynomials is flipped compared to [EKRS].
This happened because in our case we also have a choice of $\phi$: if we take
$-\phi$ instead of $\phi$, we should take $-T$ instead of $T$.

So we have proved the following

######  Theorem 3.16.

Let $m$ be the number of good indices $i$. Then the dimension of the cone of
positive forms is the same as in [EKRS], namely:

  * •

$n-1$ for even $n$ and $n-2$ for odd $n$ if $\rho=\rho_{-}$;

  * •

$n-1$ for even $n$ and $n$ for odd $n$ if $c\neq 0$ and $\rho=\rho_{+}$;

  * •

$n-3$ for even $n$ and $n-2$ for odd $n$ if $c=0$ and $\rho=\rho_{+}$.

If the dimension is $\leq 0$, the cone is empty.

##  4 Unitarizability for $q$-deformation.

###  4.1 Construction of Hermitian form on a bimodule.

Let $q$ be a positive number. Consider the algebra $A$ that is generated by
$x,x^{-1},D,D^{-1}$ with relations $Dx=q^{2}xD$. $A$ acts on the algebra
$\mathbb{R}[\mathbb{C}]=\oplus_{\operatorname{Re}s\in[0,1)}x^{s}\mathbb{C}[x,x^{-1}]$
by $x.P(x)=xP(x)$, $DP(x)=P(q^{2}x)$. Since $q$ is positive, any complex power
of $q$ is well-defined.

Let $c_{1},\ldots,c_{n}$ be complex numbers such that none of $c_{i}-c_{j}$
belong to the lattice $\mathbb{Z}+\frac{\pi i}{\ln q}\mathbb{Z}$. We have the
following lemma:

######  Lemma 4.1.

Consider the subalgebra $A_{c}\subset A$ of operators that preserve
$x^{c_{i}}\mathbb{C}[x]$ for all $i=1,\ldots,n$. It is generated by $u=x$,
$Z=D$, $Z^{-1}$ and $v=x^{-1}(D-q^{2c_{1}})\cdots(D-q^{2c_{n}})$. The set of
defining relations is

$ZuZ^{-1}=q^{2}u,\quad ZvZ^{-1}=q^{-2}v,\quad uv=P(q^{-1}Z),\quad vu=P(qz),$

where $P(z)=q^{n}(z-q^{2c_{1}-1})\cdots(z-q^{2c_{n}-1})$.

###### Proof.

We see that $A_{c}$ contains $u,v,Z,Z^{-1}$.

Let $a$ be an element of $A_{c}$, $a=\sum x^{k}P_{k}(D)$, where $P_{k}$ are
Laurent polynomials. We have
$ax^{c_{i}+l}=\sum\limits_{k}x^{k+c_{i}+l}P_{k}(q^{2c_{i}+2l})$. Since
$ax^{c_{i}+l}$ belongs to $x^{c_{i}}\mathbb{C}[x]$, we see that
$P_{k}(q^{2c_{i}+2l})=0$ for all $k,l$ such that $k+l<0$. In particular for
$k<0$ the polynomial $P_{k}$ has at least $nk$ roots: $P_{k}(q^{2c_{i}+2l})=0$
for all $i=1,\ldots,n$ and $l=0,\ldots,-k-1$. By our assumption numbers
$q^{2c_{i}+2l}$ are all distinct.

Fix a negative $k$. By induction on $k$ we prove that

$v^{-k}=x^{k}(D-q^{2c_{1}})(D-q^{2c_{1}+2})\cdots(D-q^{2c_{1}+2k-2})\cdots(D-q^{2c_{n}+2k-2}).$

Hence there exists a polynomial $Q_{k}$ such that
$x^{k}P_{k}(D)=v^{-k}Q_{k}(D)$. For $k\geq 0$ we have
$x^{k}P_{k}(D)=u^{k}P_{k}(D)$. Therefore $a$ belongs to the subalgebra of $A$
generated by $u,v,D$.

The elements $u,v,Z$ satisfy the relationships of the lemma statement. Using
these relationships, we see that the elements $u^{k}Z^{l}$ and $v^{k}Z^{l}$
span $A_{c}$, where we take $k\geq 0$. Using the action of $A_{c}$ on
$\mathbb{R}[\mathbb{C}]$, we see that $u^{k}Z^{l}$ and $v^{k}Z^{l}$ are a
basis of $A_{c}$. The lemma follows. ∎

We will require that $n=2m$ be even. We multiply $v$ by $Z^{-m}$ from the
right so that $P$ becomes $P(Z)=q^{m}Z^{m}+\cdots+tZ^{-m}$ for some nonzero
complex $t$. More precisely, by Vieta’s formula,
$t=q^{m}\prod_{i=1}^{n}q^{\sum_{i=1}^{n}(2c_{i}-1)}$. Then we multiply $v$ by
a complex number so that the coefficient of $P$ on $Z^{m}$ becomes $q^{m-\sum
c_{i}}$. The coefficient of $P$ on $Z^{-m}$ becomes $q^{\sum c_{i}-m}$. We can
write this as

$P(z)=Z^{m}q^{m-\sum c_{i}}Z^{m}+\ldots+Z^{-m}q^{\sum
c_{i}-m}=\prod_{i=1}^{n}(\sqrt{z}q^{\tfrac{1}{2}-c_{i}}-\sqrt{z}^{-1}q^{c_{i}-\tfrac{1}{2}}).$

After doing this we define $P_{c}(z)=P(z)$. This will be convenient for our
computations: a Laurent polynomial of the form $aZ^{m}+\cdots+a^{-1}Z^{-m}$ is
defined by its nonzero roots up to a sign, and the product or quotient (when
polynomial) of two such polynomials again has this form.

Let $c=(c_{1},\ldots,c_{n}),c^{\prime}=(c^{\prime}_{1},\ldots,c^{\prime}_{n})$
be parameters such that $c_{i}-c^{\prime}_{i}$ are integers. Consider the
subset $M_{c,c^{\prime}}\subset A$ of operators that send
$x^{c^{\prime}_{i}}\mathbb{C}[x]$ to $x^{c_{i}}\mathbb{C}[x]$ for each
$i=1,\ldots,n$. This is naturally an $A_{c}$-$A_{c^{\prime}}$-bimodule.
Moreover, we have a natural map from
$M_{c,c^{\prime}}\otimes_{A_{c^{\prime}}}M_{c^{\prime},c}$ to $A_{c}$.

The proof of the following lemma is the same as the proof of Lemma 3.4:

######  Lemma 4.2.

The maps $\phi\colon
M_{c,c^{\prime}}\otimes_{A_{c^{\prime}}}M_{c^{\prime},c}\to A_{c}$ and
$\psi\colon M_{c^{\prime},c}\otimes_{A_{c}}M_{c,c^{\prime}}\to A_{c^{\prime}}$
give a Morita equivalence between $A_{c}$ and $A_{c^{\prime}}$.

######  Lemma 4.3.

We have
$M_{c,c^{\prime}}=\bigoplus\limits_{j\in\mathbb{Z}}x^{j}R_{j}(Z)\mathbb{C}[Z,Z^{-1}]$,
where $R_{j}(Z)$ is a monic polynomial with the following set of roots: we
start with an empty set and for all $i=1,\ldots,n$ such that
$c^{\prime}_{i}\leqslant c_{i}-j-1$ we add
$q^{2c^{\prime}_{i}},q^{2c^{\prime}_{i}+2},\ldots,q^{2c_{i}-2j-2}$.

###### Proof.

Let $m$ be an element of $M=M_{c,c^{\prime}}$, $m=\sum x^{k}P_{k}(D)$, where
$P_{k}$ are Laurent polynomials. For all triples $i,k$ and $l\geq 0$ such that
$c^{\prime}_{i}+k+l<c_{i}$, we have $P_{k}(q^{2c^{\prime}_{i}+2l})=0$. For
fixed $i,k$ this is equivalent to $0\leqslant l\leqslant
c_{i}-c^{\prime}_{i}-1-k$. This gives
$P_{k}(q^{2c^{\prime}_{i}})=\cdots=P_{k}(q^{2c_{i}-2k-2})=0$. The lemma
follows. ∎

Below we will write $q^{2c^{\prime}_{i}},\ldots,q^{2c_{i}-2j-2}$ to mean
$c_{i}-j-c^{\prime}_{i}$ numbers when $c_{i}-j>c^{\prime}_{i}$ and empty set
otherwise.

In [K] we obtained a good theory of unitarizable bimodules for conjugations
$\rho$ such that $\rho(Z)=Z^{-1}$. Here we will do the same.

Now we assume that $A_{c^{\prime}}$ is isomorphic to $\overline{A_{c}}$ and
both maps $\rho$ are given by $u\mapsto av$, $v\mapsto bu$, $Z\mapsto Z^{-1}$.

This isomorphism exists when

$abP_{c}(qZ)=abvu=\rho(uv)=\rho(P_{c^{\prime}}(q^{-1}Z))=\overline{P_{c^{\prime}}}(q^{-1}Z^{-1}).$
(4.1)

From this we deduce the following: $P_{c}(z_{0})=0$ if and only if
$P_{c^{\prime}}(\overline{z_{0}^{-1}})=0$. Hence for any $i$ from $1$ to $n$
there exists $j$ from $1$ to $n$ such that
$q^{2c_{i}-1}=\overline{q^{1-2c_{j}^{\prime}}}$. It follows that

$q^{2c_{i}+\overline{2c_{j}^{\prime}}-2}=e^{2\ln
q(c_{i}+\overline{c^{\prime}_{j}}-1)}=1,$

hence $c_{i}+\overline{c^{\prime}_{j}}-1$ is a multiple of $\frac{\pi i}{\ln
q}$. Shifting $c_{i}$ by $\frac{\pi i}{\ln q}$ does not change $2c_{i}$, so we
can assume that $c_{i}+\overline{c^{\prime}_{j}}-1=0$. We can do this for all
$i=1,\ldots,n$. Similarly to the previous section from
$c_{j}+\overline{c^{\prime}_{k}}-1=0$ we deduce that $k=i$. Hence there exists
an involution $\sigma$ such that $c_{i}+\overline{c^{\prime}_{\sigma(i)}}=1$
for all $i$.

Note that for any complex number $s$ with $\lvert s\rvert=1$ we have
$\rho(sZ)=\overline{s}\rho(Z)=s^{-1}Z^{-1}=(sZ)^{-1}$, hence we can change $Z$
to $sZ$ in both algebras $A_{c}$, $A_{c^{\prime}}$.

Taking $a=q^{2i\lambda}$ shifts all $c_{j}$ by $-i\lambda$. Hence we can
choose generator $Z$ such that $\sum c_{i}$ is real. Note that

$2\sum
c_{i}=2\sum\operatorname{Re}c_{i}=\sum\operatorname{Re}c_{i}+\operatorname{Re}c_{\sigma(i)}=\\\
\sum\operatorname{Re}(c_{i}-c_{i}^{\prime})+\operatorname{Re}c^{\prime}_{i}+\operatorname{Re}c_{\sigma(i)}=\sum(c_{i}-c_{i}^{\prime}+1)$
(4.2)

is an integer. We require that $\sum c_{i}$ is an integer, not half-integer,
so that we don’t obtain square root of $Z$ in the proof of Lemma 4.4 below.
This means that the number of singletons $i=\sigma(i)$ with
$\operatorname{Re}c_{i}$ half-integer is even.

If we interchange $c$ and $c^{\prime}$ in (4.2) and add (4.2) we get $2\sum
c_{i}+2\sum c^{\prime}_{i}=2n$, hence $\sum c_{i}+\sum c^{\prime}_{i}=n$.
Recall that $P_{c}(Z)=q^{m-\sum c_{i}}Z^{m}+\cdots+q^{\sum c_{i}-m}Z^{-m}$,
$P_{c^{\prime}}(Z)=q^{m-\sum c^{\prime}_{i}}Z^{n}+\cdots+q^{m-\sum
c^{\prime}_{i}}$, where $n=2m$. Note that all coefficients of $Z^{m}$ and
$Z^{-m}$ are real and $m-\sum c_{i}=\sum c^{\prime}_{i}-m$. It follows that
$P_{c}(Z)$ and $\overline{P_{c^{\prime}}}(Z^{-1})$ have the same leading
coefficient. Comparing this with (4.1) we deduce that $ab=1$. We want $u\in
A_{c},A_{c^{\prime}}$ to be exactly $x$, so we will leave it at that and allow
any $a,b$ such that $ab=1$. Then $a=\lvert a\rvert e^{2\pi is}$, $b=\lvert
a\rvert^{-1}e^{-2\pi is}$, where $s\in[0,1)$. It follows from (4.1) that
$P_{c}(z)=\overline{P_{c^{\prime}}}(z^{-1})$.

On the other hand, any $a,b$ with $ab=1$ and parameters $c,c^{\prime}$ such
that for some involution $\sigma$ we have
$c_{i}+\overline{c^{\prime}_{\sigma(i)}}=1$ will give an antilinear
isomorphism $\rho$ between $A_{c}$ and $A_{c^{\prime}}$ as above.

Recall that we have the notion of $\rho$-invariant sesquilinear form and
Proposition 2.16 says that sesquilinear $\rho$-invariant forms on $M$ are in
one-to-one correspondence with $\rho^{2}$-twisted traces on
$M\otimes_{A_{c^{\prime}}}M_{\rho^{-1}}$. We use the same strategy as in
Section 3 in order to describe $\rho$-invariant forms: prove that
$M_{c,c^{\prime},\rho^{-1}}$ is isomorphic to $M_{c^{\prime},c}$ to get
$\rho^{2}$-twisted traces on
$M_{c,c^{\prime}}\otimes_{A_{c^{\prime}}}M_{c^{\prime},c}\cong A_{c}$.

In order to check that two usual polynomials are equal to each other, it is
enough to check that they have the same roots and the same leading
coefficient. For Laurent polynomials we should also check that they have root
or pole of the same order at zero. In order to deal with this, we will
consider balanced Laurent polynomials, meaning they have the form
$ax^{N}+\cdots+bx^{-N}$. It is possible for $R_{j}$ to have odd number of
nonzero roots, so we allow $N$ to be a half-integer. We will construct an
isomorphism between $M_{c,c^{\prime}}[\sqrt{Z}]$ and
$M_{c^{\prime},c,\rho}[\sqrt{Z}]$ below and check that it gives an isomorphism
between $M_{c,c^{\prime}}$ and $M_{c^{\prime},c,\rho}$.

From now on we shift all $R_{j}$ by a power of $\sqrt{Z}$ so that they become
symmetric.

######  Lemma 4.4.

  1. 1.

We have $M_{c,c^{\prime}}\cong M_{c^{\prime},c,\rho}$. Denote this isomorphism
by $\phi$. We also get that $\phi$ is an isomorphism from
$M_{c,c^{\prime},\rho^{-1}}$ to $M_{c^{\prime},c}$.

  2. 2.

Let us write $\phi$ as
$\phi(x^{j}R_{j}(Z)R(Z))=x^{-j}S_{-j}(Z)\overline{R}(Z^{-1})$. Then
$\frac{a^{2}S_{j+2}(q^{2}z)}{S_{j}(z)}$ is positive for all $z$ with $\lvert
z\rvert=1$.

  3. 3.

We can choose $\phi$ so that the polynomials
$a^{-j}R_{j}(q^{-j}z)S_{-j}(q^{j}z)$ are real for all $j$.

###### Proof.

The second statement follows from the first after twisting the action from
both sides by $\rho^{-1}$.

We will use . to denote the action of $A_{c},A_{c^{\prime}}$ on
$M_{c^{\prime},c,\rho}$, so that $Z.m=Z^{-1}m,u.m=avm$ and so on.

We will construct the map $\phi$ similarly to the proof of Lemma 3.5. Let
$\phi(x^{j}R_{j}(Z)R(Z))=x^{-j}S_{-j}(Z)\overline{R}(Z^{-1})$, where $S_{-j}$
has symmetric degree and has roots similar to the roots of $R_{j}$ but is not
necessarily monic.

Note that $\phi$ is antilinear and satisfies $\phi(mZ)=\phi(m)Z^{-1}$. We also
have $\phi(ZmZ^{-1})=Z^{-1}\phi(m)Z$. It follows that
$\phi(Zm)=Z^{-1}\phi(m)$.

Since elements $x^{j}R_{j}(Z)$ form a $\mathbb{C}[Z,Z^{-1}]$ basis of
$M_{c,c^{\prime}}$ both for the left and for the right action of
$\mathbb{C}[Z,Z^{-1}]$, it is enough to prove $\phi(um)=u.\phi(m)$ for
$m=x^{j}R_{j}(Z)$ and similarly for the other conditions.

We want to prove that $\phi(um)=u.\phi(m)=avm$ for $m=x^{j}R_{j}(Z)$. We have
$um=xm=x^{j+1}R_{j}(Z)$. Denote $\frac{R_{j}(Z)}{R_{j+1}(Z)}$ by $L_{j}(Z)$.
We get $\phi(um)=x^{-j-1}S_{-j-1}(Z)\overline{L_{j}}(Z^{-1})$.

By definition

$av\phi(m)=ax^{-1}P_{c^{\prime}}(q^{-1}Z)\phi(m)=ax^{-1}P_{c^{\prime}}(q^{-1}Z)x^{-j}S_{-j}(Z)=\\\
ax^{-j-1}P_{c^{\prime}}(q^{-1-2j}Z)S_{-j}(Z).$

Denote $\frac{S_{-j-1}(Z)}{S_{-j}(Z)}$ by $M_{-j-1}(Z)$. It follows that

$\phi(um)=x^{-j-1}M_{-j-1}(Z)\overline{L_{j}}(Z^{-1})S_{-j}(Z).$

So it is enough to prove that

$aP_{c^{\prime}}(q^{-1-2j}Z)=M_{-j-1}(Z)\overline{L_{j}}(Z^{-1}).$ (4.3)

Note that $P_{c^{\prime}},M_{-j-1}$ and $L_{j}$ are all balanced. Hence it is
enough to check that both sides have the same roots and choose $S_{j}$ such
that both sides have the same leading coefficient.

Similarly to Lemma 3.5 we see that $L_{j}(Z)$ has roots $q^{2c_{i}-2j-2}$ for
all $i$ satisfying $c_{i}-j-1\geq c_{i}^{\prime}$. Hence
$\overline{L_{j}}(Z^{-1})$ has roots
$q^{2+2j-2\overline{c_{i}}}=q^{2j+2c_{\sigma}(i)^{\prime}}$, where $\sigma$ is
the permutation such that $\overline{c_{i}}+c^{\prime}_{\sigma(i)}=1$ for all
$i$. Similarly to the proof of Lemma 3.5 we see that roots of $L_{j}$ come in
pairs $q^{2c_{i}-2j-2}$, $q^{2c_{\sigma(i)}-2j-2}$.

Similarly to the above and to the proof of Lemma 3.5, we see that
$M_{-j-1}(Z)$ is balanced and has roots $q^{2c^{\prime}_{i}+2j}$ for all $i$
satisfying $c^{\prime}_{i}+j\geq c_{i}$, so that
$\overline{L_{j}}(Z^{-1})M_{-j-1}(Z)$ has roots
$q^{2c^{\prime}_{1}+2j},\ldots,q^{2c^{\prime}_{n}+2j}$. Polynomial
$P_{c^{\prime}}(q^{-1-2j}Z)$ also has roots
$q^{2c^{\prime}_{1}+2j},\ldots,q^{2c^{\prime}_{n}+2j}$. It follows that
$P_{c^{\prime}}(q^{-1-2j}Z)$ and $\overline{L_{j}}(Z^{-1})M_{-j-1}(Z)$ have
the same set of roots. Multiplying each $S_{k}$ by its own nonzero constant we
can make $aP_{c^{\prime}}(q^{-1-2j}Z)$ equal to
$\overline{L_{j}}(Z^{-1})M_{-j-1}(Z)$ for all $j$.

Hence $\phi$ satisfies $\phi(um)=u.\phi(m)$.

It remains to check that $\phi(mu)=\phi(m).u=a\phi(m)v$: we claim that
$\phi(vm)=v.\phi(m)$ and $\phi(mv)=\phi(m).v$ follow from all the other
conditions. Indeed, since $M_{c^{\prime},c}$ is a torsion-free
$A_{c^{\prime}}$-module, it is enough to check that $u.\phi(vm)=uv.\phi(m)$.
We have

$u.\phi(vm)=\phi(uvm)=\phi(P_{c}(Z)m)=P_{c}(Z).\phi(m)=uv.\phi(m).$

Similarly $\phi(mv)=\phi(m).v$ follows from $\phi(mv).u=\phi(mvu)$.

Hence it remains to prove that $\phi(mu)=a\phi(m)v$. As before we can check
this for $m=x^{j}R_{j}(Z)$. We have

$mu=x^{j}R_{j}(Z)x=x^{j+1}R_{j}(q^{2}Z)=x^{j+1}R_{j+1}(Z)A_{j}(Z).$

Here, similarly to Lemma 3.5 $A_{j}(Z)=\frac{R_{j}(q^{2}Z)}{R_{j+1}(Z)}$ is a
polynomial that has roots $q^{2c^{\prime}_{i}-2}$ for all $i$ such that
$c^{\prime}_{i}\leqslant c_{i}-j-1$.

We have $\phi(mu)=x^{-j-1}S_{-j-1}(Z)\overline{A_{j}}(Z^{-1})$. Here
$\overline{A_{j}}(Z^{-1})$ is a polynomial in $Z^{-1}$ that has roots
$Z=\overline{(q^{2c^{\prime}_{i}-2})^{-1}}=q^{2-2\overline{c^{\prime}_{i}}}=q^{2c_{\sigma(i)}}$.

We have

$a\phi(m)v=ax^{-j}S_{-j}(Z)x^{-1}P_{c}(q^{-1}Z)=ax^{-j-1}S_{-j}(q^{-2}Z)P_{c}(q^{-1}Z)=\\\
ax^{-j-1}S_{-j-1}(Z)\frac{P_{c}(q^{-1}Z)}{B_{-j-1}(Z)},$

where $B_{-j-1}(Z)=\frac{S_{-j-1}(Z)}{S_{-j}(q^{-2}Z)}$ has roots $q^{2c_{i}}$
for all $i$ such that $c_{i}\leqslant c^{\prime}_{i}+j$. Hence we should prove
that

$\overline{A_{j}}(Z^{-1})B_{-j-1}(Z)=aP_{c}(q^{-1}Z).$ (4.4)

Similarly to the proof of Lemma 3.5 and to the reasoning above we deduce that
$\overline{A_{j}}(Z^{-1})B_{-j-1}(Z)$ and $aP_{c}(q^{-1}Z)$ have the same set
of roots. Since $A_{j},B_{-j-1}$ and $P_{c}$ are all symmetric, it is enough
to check that they have the same leading coefficient. In other words,

$\lim_{Z\to\infty}\frac{\overline{A_{j}}(Z^{-1})B_{-j-1}(Z)}{aP_{c}(q^{-1}Z)}$

should be equal to one. We compute this limit as a product

$\lim_{Z\to
0}\frac{\overline{A_{j}}(Z)}{\overline{L_{j}}(Z)}\lim_{Z\to\infty}\frac{B_{-j-1}(Z)}{M_{-j-1}(Z)}\lim_{Z\to\infty}\frac{P_{c^{\prime}}(q^{-1-2j}Z)}{P_{c}(q^{-1}Z)},$

since $\overline{L_{j}}(Z^{-1})M_{-j-1}(Z)=aP_{c^{\prime}}(q^{-1-2j}Z)$.

We have $L_{j}(Z)=\frac{R_{j}(Z)}{R_{j+1}(Z)}$,
$A_{j}(z)=\frac{R_{j}(q^{2}Z)}{R_{j+1}(Z)}$. Hence

$\lim_{Z\to 0}\frac{\overline{A_{j}}(Z)}{\overline{L_{j}}(Z)}=q^{-\deg
R_{j}}.$

We have $M_{-j-1}(Z)=\frac{S_{-j-1}(Z)}{S_{-j}(Z)}$,
$B_{-j-1}(Z)=\frac{S_{-j-1}(Z)}{S_{-j}(q^{-2}Z)}$, hence the leading
coefficient of

$\lim_{Z\to\infty}\frac{B_{-j-1}(Z)}{M_{-j-1}(Z)}=q^{\deg S_{-j}}.$

Recall that

$P_{c}(Z)=q^{-\sum c_{i}+m}Z^{m}+\cdots+q^{\sum c_{i}-m}Z^{-m},$
$P_{c^{\prime}}(Z)=q^{-\sum c^{\prime}_{i}+m}Z^{-m}+\cdots+q^{\sum
c^{\prime}_{i}-m}Z^{-m},$

hence

$\lim_{Z\to\infty}\frac{P_{c^{\prime}}(q^{-1-2j}Z)}{P_{c}(q^{-1}Z)}=q^{-2mj+\sum
c_{i}-\sum c^{\prime}_{i}}.$

It follows that

$\lim_{Z\to
0}\frac{\overline{A_{j}}(Z)}{\overline{L_{j}}(Z)}\lim_{Z\to\infty}\frac{B_{-j-1}(Z)}{M_{-j-1}(Z)}\lim_{Z\to\infty}\frac{P_{c^{\prime}}(q^{-1-2j}Z)}{P_{c}(q^{-1}Z)}=q^{\deg
S_{-j}-\deg R_{j}-2mj+\sum c_{i}-\sum c^{\prime}_{i}}.$ (4.5)

Note that each $i$ adds exactly $c^{\prime}_{i}+j-c_{i}$ to $\deg S_{-j}-\deg
R_{j}$: either $S_{-j}$ has roots $c_{i},\ldots,c^{\prime}_{i}+j-1$ or $R_{j}$
has roots $c^{\prime}_{i},\ldots,c_{i}-j-1$. Hence $\deg S_{-j}-\deg
R_{j}=\sum c^{\prime}_{i}-\sum c_{i}+2mj$. It follows that the right-hand side
in (4.5) equals to one. We deduce that $\overline{A_{j}}(Z^{-1})B_{-j-1}(Z)$
and $aP_{c}(q^{-1}Z)$ have the same set of roots, hence they coincide.

It follows that $\phi$ is an isomorphism of $A_{c}-A_{c^{\prime}}$ bimodules.
It remains to check that it sends $M_{c,c^{\prime}}\subset
M_{c,c^{\prime}}[\sqrt{Z}]$ to $M_{c^{\prime},c,\rho}\subset
M_{c^{\prime},c,\rho}[\sqrt{Z}]$. For each $j$ the polynomial $R_{j}$ has
either all integer monomials or all half-integer ones and similarly for
$S_{j}$. So we have to check that either both $R_{j}$ and $S_{-j}$ have
integer monomials or both have half-integer monomials. This depends on the
number of roots: integer degrees for even number of roots, half-integer for
odd. Hence $R_{j}$ and $S_{-j}$ should have the same parity of number of
roots.

We will prove by downward induction on $j$ that $R_{j}$ and $S_{-j}$ have the
same parity of number of roots. The base case is for $j$ large enough. In this
case $R_{j}=1$ has no roots and $S_{-j}$ has roots
$c_{i},\ldots,c^{\prime}_{i}+j-1$ for all $i$. Hence there are $\sum
c^{\prime}_{i}-\sum c_{i}+nj$ roots of $S_{-j}$. It has the same parity as
$\sum c^{\prime}_{i}-\sum c_{i}$. It follows from (4.2) that $\sum
c^{\prime}_{i}-\sum c_{i}$ has the same parity as $2\sum c_{i}$. By our
assumption $2\sum c_{i}$ is even, hence both $R_{j}$ and $S_{-j}$ have even
number of roots.

Induction step is $j+1\to j$. We have $R_{j}(Z)=L_{j}(z)R_{j+1}(Z)$,
$S_{-j}(Z)=\frac{S_{-j-1}(Z)}{M_{-j-1}(Z)}$. We should prove that $L_{j}$ and
$M_{-j-1}$ have the same parity of number of roots. Recall that
$\overline{L_{j}}(Z^{-1})M_{-j-1}(Z)=P_{c^{\prime}}(q^{-1-2j}Z)$. The Laurent
polynomial $P_{c^{\prime}}(Z)$ has even number of roots, this proves the
induction step.

We turn to the second claim. We have

$\frac{R_{j+2}(q^{-j-2}z)S_{-j-2}(q^{j+2}z)}{R_{j}(q^{-j}z)S_{-j}(q^{j}z)}=\\\
\frac{R_{j+1}(q^{-j-2}z)S_{-j-1}(q^{j+2}z)}{R_{j}(q^{-j}z)S_{-j}(q^{j}z)}\frac{M_{-j-2}(q^{j+2}z)}{L_{j+1}(q^{-j-2}z)}=\\\
\frac{B_{-j-1}(q^{j+2}z)M_{-j-2}(q^{j+2}z)}{A_{j}(q^{-j-2}z)L_{j+1}(q^{-j-2}z)}=\\\
\frac{B_{-j-1}(q^{j+2}z)\overline{A_{j}}(q^{-j-2}z^{-1})M_{-j-2}(q^{j+2}z)\overline{L_{j+1}}(q^{-j-2}z^{-1})}{A_{j}(q^{-j-2}z)\overline{A_{j}}(q^{-j-2}z^{-1})L_{j+1}(q^{-j-2}z)\overline{L_{j+1}}(q^{-j-2}z^{-1})}$

The denominator of this fraction is nonnegative and it follows from (4.4) and
(4.3) that the numerator is $a^{2}P_{c^{\prime}}(q^{1-j}z)P_{c}(q^{j-1}z)$.
Since $P_{c^{\prime}}(z)=\overline{P_{c}}(z^{-1})$, the polynomial
$P_{c^{\prime}}(q^{1-j}z)P_{c}(q^{j-1}z)$ is nonnegative when $\lvert
z\rvert=1$.

It remains to prove the third claim. Similarly to the proof of Lemma 3.13 we
get that the roots of $R_{j}(q^{-j}z)S_{-j}(q^{j}z)$ are symmetric with
respect to $z\mapsto\overline{z}^{-1}$. Since
$a^{-j}R_{j}(q^{-j}z)S_{-j}(q^{j}z)$ is balanced it is enough to check that
its leading coefficient and its negative leading coefficient are complex
conjugate. Hence it is enough to check that the leading coefficient and
negative leading coefficient of $a^{-j}R_{j}(z)S_{-j}(z)$ have opposite
arguments. We can prove this using induction. The base case $j=0$ is a choice
of $\phi$. The induction step is

$\frac{a^{-j-1}R_{j+1}(z)S_{-j-1}(z)}{a^{-j}R_{j}(z)S_{-j}(z)}=a^{-1}\frac{M_{-j-1}(z)}{L_{j}(z)}=\\\
a^{-1}\frac{M_{-j-1}(z)\overline{L_{j}}(z^{-1})}{L_{j}(z)\overline{L_{j}}(z^{-1})}=\frac{P_{c^{\prime}}(q^{-1-2j}z)}{L_{j}(z)\overline{L_{j}}(z^{-1})}$

Here the numerator’s leading and negative leading coefficients are positive
and the denominator is positive when $\lvert z\rvert=1$, hence its leading and
negative leading coefficients are complex conjugate.

∎

###  4.2 Positivity condition for the Hermitian form.

The invariant Hermitian form $(\cdot,\cdot)$ is given by $(u,v)=T(u\phi(v))$,
where $T$ is $g_{t}=\rho^{2}$-twisted. Here $t=ba^{-1}$. For fixed $t$ there
are two conjugations $\rho$ with $\rho^{2}=g_{t}$. We denote one of them by
$\rho_{+}$ and another by $\rho_{-}$.

It is enough to check that $(\cdot,\cdot)$ is positive definite on each space
$Z^{j}\mathbb{C}[Z,Z^{-1}]$. Let $m=x^{j}R_{j}(Z)R(Z)$. Then

$(m,m)=T(x^{j}R_{j}(Z)R(Z)x^{-j}S_{-j}(Z)\overline{R}(Z^{-1}))=\\\
T(R_{j}(q^{-2j}Z)R(q^{-2j}Z)S_{-j}(Z)\overline{R}(Z^{-1})).$

Let $t=\lvert a\rvert e^{2\pi is}$. Similarly to the previous section, any
trace $T$ can be written as $T(R)=\int_{C}w(t)R(t)dt$, where $w$ is a certain
quasi-periodic function. More precisely, $w(q^{2}x)=tw(x)$ and $w(qx)P(x)$ has
no poles between $q^{-1}C$ and $qC$. In other words,
$q^{2c_{i}}q^{2\mathbb{Z}_{\geq 0}}$ is inside $C$ and
$q^{2c_{i}}q^{2\mathbb{Z}_{<0}}$ is outside of $C$.

We say that $i$ is bad if for any $w$ as above that has simple poles at
$q^{2c_{i}+1+2\mathbb{Z}}$ there exists $j$ such that
$R_{j}(q^{-2j}x)S_{-j}(x)w(x)$ has poles between $C$ and $q^{j}S^{1}$,
including the circle.

######  Lemma 4.5.

An index $i$ is bad if and only if
$\operatorname{Re}c_{i}+\operatorname{Re}c^{\prime}_{i}\leqslant 0$ or
$\operatorname{Re}c_{i}+\operatorname{Re}c^{\prime}_{i}\geq 2$.

###### Proof.

The proof is similar to the proof of Lemma 3.8. ∎

######  Proposition 4.6.

Suppose that there exists a bad $i$ such that $w$ has a pole at $q^{2c_{i}}$.
Then $w$ does not give a positive definite form on $M_{c,c^{\prime}}$.

###### Proof.

Suppose that $T$ defined by $w$ gives a positive definite form on
$M_{c,c^{\prime}}$. Let $j$ be a number that is bad for $i$.

Similarly to the proof of Proposition 3.9 we deduce that $w$ has no poles on
$q^{j}S^{1}$ and write

$T\big{(}R_{j}(q^{-2j}Z)R(q^{-2j}Z)S_{-j}(Z)\overline{R}(Z^{-1})\big{)}=\\\
\int_{q^{j}S^{1}}R(q^{-2j}z)\overline{R}(z^{-1})w(z)dz+\Phi\big{(}R(q^{-2j}z)\overline{R}(z^{-1})\big{)},$

where $\Phi(S)$ is a nonzero linear functional of the form $\Phi(S)=\sum
c_{i}S(z_{i})$. Here $z_{i}$ are the poles of $w$ between the contour $C$ and
the circle $q^{j}S^{1}$.

Let $L_{0}$ be a polynomial that has no zeroes on $S^{1}$ such that

$\Phi(\overline{L_{0}(z^{-1})}\mathbb{C}[z])=\Phi(L_{0}(q^{-2j}z)\mathbb{C}[z])=\\{0\\}.$

We find a polynomial $L_{1}$ such that
$\Phi(L_{1}(q^{-2j}z)\overline{L_{1}}(z^{-1}))$ does not belong to
$\mathbb{R}_{\geq 0}$. We use the Stone-Weierstrass theorem to find a
polynomial $S$ such that $L_{0}S+L_{1}$ is uniformly small on $q^{j}S^{1}$. It
follows that for $R=L_{0}S+L_{1}$ we get a contradiction with positivity of
$T$. ∎

######  Theorem 4.7.

Assume that there are $2k$ good indices. Fix any $k$ distinct pairs of numbers
$(z_{k},\overline{z_{k}}^{-1})$. Then the cone of positive traces is
isomorphic to the cone of elliptic functions with simple poles at
$z_{k},\overline{z_{k}}^{-1}$ that are positive on $q^{j}S^{1}$ for all
integers $j$. In particular, this cone has dimension $2k$, it does not depend
on $s$ and on the particular choice of $P$, only on the number of good
indices.

###### Proof.

We know that a function $w$ with poles corresponding to bad indices does not
give a positive definite form. Hence we can assume that $w$ does not have
poles at $q^{2c_{i}}$ for all bad $i$.

Hence for any $j$ we have

$T(R_{j}(q^{-2j}Z)R(q^{-2j}Z)S_{-j}(Z)\overline{R}(Z^{-1}))=\\\
\int\limits_{q^{j}S^{1}}R(q^{-2j}z)\overline{R}(z^{-1})R_{j}(q^{-2j}z)S_{-j}(z)w(z)\lvert
dz\rvert.$

Using the Stone-Weierstrass theorem we see that this quantity is positive for
all nonzero Laurent polynomials $R$ if and only if
$R_{j}(q^{-2j}z)S_{-j}(z)w(z)$ is nonnegative on $q^{j}S^{1}$.

This is equivalent to saying that $R_{j}(q^{-j}z)S_{-j}(q^{j}z)w(q^{j}z)$ is
nonnegative on $S^{1}$. We have

$\frac{w(q^{j+2}z)}{w(z)}=t=ba^{-1}=a^{-2}.$

Using the second statement of Lemma 4.4 we get that

$\frac{R_{j}(q^{-j}z)S_{-j}(q^{j}z)w(q^{j}z)}{R_{j+2}(q^{-j-2}z)S_{-j-2}(q^{j+2}z)w(q^{j+2}z)}$

is nonnegative on $S^{1}$. Hence it is enough to check this condition for
$j=0,1$.

The roots of $R_{j}(q^{-2j}z)S_{-j}(z))$ are in one-to-one correspondence with
good indices $i$ such that $2\operatorname{Re}c_{i}$ is an integer with the
same parity as $j$. This follows from the definition of a good index for $i,j$
and a quasi-periodic function $w_{0}$ that has a pole at $q^{c_{i}}$.

Let $C$ be the cone of quasi-periodic functions that give a positive trace. We
have $w\in C$ if and only if $R_{0}(z)S_{0}(z)w(z)$ and
$R_{1}(q^{-1}z)S_{-1}(qz)w(qz)$ are nonnegative on $S^{1}$. Using the third
statement of Lemma 4.4 we see that $P_{0}=R_{0}(z)S_{0}(z)$ and $P_{1}=e^{\pi
is}R_{1}(q^{-1}z)S_{-1}(qz)$ are real on $S^{1}$. It follows from the proof of
Theorem 3.7 in [K] that in this case the cone of $w$ such that $P_{0}(z)w(z)$
and $e^{-\pi is}P_{1}(z)w(qz)$ are nonnegative on $S^{1}$ has dimension $2k$
and does not depend on $s,P_{0},P_{1}$. In particular, taking $s=0$,
$P_{0}=P_{1}=1$ we get the cone in the statement of the theorem.

∎

#### Example: $n=1$.

Note that for $q$-deformations $n$ should be odd and $n=0$ gives no traces. In
the case $n=1$ the algebra $A_{c}=W$ is isomorphic to Weyl algebra $W$, it is
generated by $u,v$ with relation $[u,v]=1$. The parameter $c_{1}$ defines
inclusion of $W$ into $\mathbb{C}[x,x^{-1},\partial_{x}]\colon u\mapsto
v,v\mapsto x^{-1}(x\partial_{x}-c_{1})=\partial_{x}-c_{1}x^{-1}$. The bimodule
$M_{c,c^{\prime}}$ consists of differential operators with a possible pole at
zero that send $x^{c_{1}^{\prime}}\mathbb{C}[x]$ to $x^{c_{1}}\mathbb{C}[x]$.
Since $x^{c_{1}-c_{1}^{\prime}}$ provides a linear isomorphism between these
two spaces, we have
$M_{c,c^{\prime}}=A_{c}x^{c_{1}-c_{1}^{\prime}}=x^{c_{1}-c_{1}^{\prime}}A_{c^{\prime}}$,
a free module of rank one from each side. Note that a linear isomorphism
$A_{c}\cong M_{c,c^{\prime}}\cong A_{c^{\prime}}$ sends $a\in A_{c}$ to
$x^{c_{1}^{\prime}-c_{1}}ax^{c_{1}-c_{1}^{\prime}}$, hence it is an
isomorphism of algebras that sends $x$ to $x$ and $\partial_{x}-c_{1}x^{-1}$
to $\partial_{x}-c_{1}^{\prime}x^{-1}$.

Note that the conjugation $\rho\colon A_{c}\to A_{c^{\prime}}$ does not depend
on $c,c^{\prime}$: it sends $u$ to $bv$ and $v$ to $au$.

It follows that invariant sesquilinear forms on $M_{c,c^{\prime}}$ are in one-
to-one correspondence with invariant sesquilinear forms on $W$. In this case
the only root of $P(x)=x$ is good and using Theorem 3.16 we deduce that
positive traces exist only when $\rho=\rho_{+}$, $t\neq 1$ and in this case a
positive trace is unique up to scaling.

###  4.3 The case $n=2$, connection with unitary representations of
$\mathop{\mathrm{SL}}\nolimits(2)$ and $\mathop{\mathrm{SL}}\nolimits_{q}(2)$.

Note that for any half-integer $r$ we can change $h$ to $h+r$ in $A_{c}$ and
change $h$ to $h-r$ in $A_{c^{\prime}}$. In the case of $q$-deformations we
change $Z$ to $q^{\pm r}Z$. The parameters $c_{i}$ will shift by $r$ and the
parameters $c^{\prime}_{i}$ will shift by $-r$. The conjugation $\rho$ is
still defined, and we can still define $M_{c,c^{\prime}}$ for these new
parameters. We will denote this new bimodule by $M_{c,c^{\prime},r}$ and the
old one by $M_{c,c^{\prime}}$. We claim that $\phi(m)=x^{r}mx^{r}$ is an
isomorphism between $M_{c,c^{\prime},r}$ and $M_{c,c^{\prime}}$. It is enough
to prove that it is a homomorphism. Indeed,

Then

$x\phi(m)=\phi(xm),$ $h\phi(m)=hx^{r}mx^{r}=x^{r}(z+r)mx^{r}=\phi((z+r)m),$
$f\phi(m)=x^{-1}P(z-\tfrac{1}{2})x^{r}mx^{r}=x^{r-1}P(z+r-\tfrac{1}{2})mx^{r}=\phi(fm).$

Hence we can shift parameters by a half-integer without changing anything.

In the case $n=2$ there are parameters
$c_{1},c_{2},c^{\prime}_{1},c^{\prime}_{2}$ such that $c_{1}-c^{\prime}_{1}$,
$c_{2}-c^{\prime}_{2}$ are integers, $c_{1}-c_{2}$ is not an integer. For some
permutation $\sigma$ of $\\{1,2\\}$ we have
$c_{i}+\overline{c^{\prime}_{\sigma(i)}}=1$.

In the case when $\sigma$ is trivial we get
$c_{i}+\overline{c^{\prime}_{i}}=1$ for $i=1,2$. Since $c_{i}-c^{\prime}_{i}$
belongs to $\mathbb{Z}$ we deduce that $2\operatorname{Re}c_{i}$ is an
integer. Hence $\operatorname{Re}c_{i}$ and $\operatorname{Re}c^{\prime}_{i}$
are integers or half-integers. An index $i$ is good if and only if
$0<\operatorname{Re}c_{i}+\operatorname{Re}c^{\prime}_{i}<2$. This is
satisfied since $c_{i}+\overline{c^{\prime}_{i}}=1$.

Hence any $c_{1}$ and $c_{2}$ such that $2\operatorname{Re}c_{1}$,
$2\operatorname{Re}c_{2}$ are integers, $c_{1}+c_{2}$ is an integer,
$c_{2}-c_{1}$ is not an integer, give a unitarizable bimodule. This holds both
for $q$-deformation and for a usual deformation. Note that we can shift
$c_{1}$ and $c_{2}$ by $\frac{c_{1}+c_{2}-1}{2}$ and get the following:
$c_{1}=c+\tfrac{1}{2}$, $c_{2}=-c+\tfrac{1}{2}$, $2\operatorname{Re}c$ is an
integer, $c$ is not real. We expect that half-integer $c$ also gives a
unitarizable bimodule. Note that $\operatorname{Re}c=0$ is a situation of
regular bimodule $M_{c,c^{\prime}}=A_{c}=A_{c^{\prime}}$.

In the case when $\sigma$ is nontrivial we get
$c_{1}+\overline{c^{\prime}_{2}}=1$, $c_{2}+\overline{c^{\prime}_{1}}=1$.

From this we get that $\mathop{\mathrm{Im}}\nolimits
c_{1}=\mathop{\mathrm{Im}}\nolimits c_{2}$. On the other hand,
$c_{2}-c^{\prime}_{2}$ is an integer, so $\mathop{\mathrm{Im}}\nolimits
c_{2}=\mathop{\mathrm{Im}}\nolimits c^{\prime}_{2}$. We assumed that
$c_{1}+c_{2}$ is an integer, hence $c_{1},c_{1}^{\prime},c_{2},c_{2}^{\prime}$
are real numbers such that $c^{\prime}_{2}=1-c_{1}$, $c^{\prime}_{1}=1-c_{2}$.
Shifting $c_{1},c_{2}$ by $\frac{c_{1}+c_{2}-1}{2}$ we get
$c_{1}=c+\tfrac{1}{2}$, $c_{2}=-c+\tfrac{1}{2}$,
$c^{\prime}_{1}=c+\tfrac{1}{2}$, $c^{\prime}_{2}=c-\tfrac{1}{2}$. In this case
the bimodule $M_{c,c^{\prime}}$ is the regular bimodule
$A_{c}=A_{c^{\prime}}$.

An index $i$ is good if and only if
$0<\operatorname{Re}c_{i}+\operatorname{Re}c^{\prime}_{i}<2$. This is
equivalent to $\lvert 2c\rvert<1$, so that $\lvert c\rvert<\tfrac{1}{2}$. This
is the same answer we had in [EKRS] and [K]: the roots $\pm\alpha$ or
$q^{\pm\alpha}$ of $P$ should satisfy
$\lvert\operatorname{Re}\alpha\rvert<\tfrac{1}{2}$.

Let us compare these results with the classical results on irrreducible
unitary representations of $\mathop{\mathrm{SL}}\nolimits(2)$ and
$\mathop{\mathrm{SL}}\nolimits_{q}(2)$. The results for
$\mathop{\mathrm{SL}}\nolimits_{q}(2)$ are obtained in [P]. The same article
writes classification of unitary representations of
$\mathop{\mathrm{SL}}\nolimits(2)$ in terms of Casimir element, so we will
also use it as a reference. Let
$U_{1}(\mathfrak{sl}_{2})=U(\mathfrak{sl}_{2})$.

Similarly to Subsection 6.3 in [K] in the case of $q$-deformations we can take
the locally finite part of $M_{c,c^{\prime}}$ with respect to the adjoint
action of the Hopf algebra $U_{q}(\mathfrak{sl}_{2})$ and $U_{q}$-invariant
forms correspond to $g_{q^{-2}}$-twisted traces on $A_{c}$. Note that there
are two traces on $A_{c}$, but a computation in [K] shows that the trace
corresponding to the weight $w=x$ is zero on the locally finite part. Hence
there one trace on the locally finite part of $A_{c}$ up to a constant, giving
one positive definite form on the locally finite part of $M_{c,c^{\prime}}$ up
to a positive constant.

Now let us compute what finite-dimensional representations of
$U_{q}(\mathfrak{sl}_{2})$ are inside $M_{c,c^{\prime}}$ and what values of
the Casimir element correspond to unitarizable parameters $c$.

There is freedom in choosing $h$, $Z$ in our case, but for
$U_{q}(\mathfrak{sl}_{2})$ the element $h$ or $Z$ is fixed. In the case of
$q=1$ element $h$ is uniquely defined by the condition that $ef+fe$ is an even
polynomial in $h$. This means that $P(z+\tfrac{1}{2})+P(z-\tfrac{1}{2})$ is
even, which is equivalent to $P$ being even. Since $P$ has roots
$c_{1}-\tfrac{1}{2}$, $c_{2}-\tfrac{1}{2}$, this means $c_{1}+c_{2}=1$. Our
choice of $c_{1},c_{2}$ satisfies $c_{1}+c_{2}=1$. In the case $q\neq 1$ the
product of roots of $P$ should be one, this is also equivalent to
$c_{1}+c_{2}=1$.

The homogeneous elements in $M_{c,c^{\prime}}$ have form $x^{k}R(z)$ or
$x^{k}R(Z)$. They commute with $u=x$ in the case when $R=1$. Hence finite-
dimensional representations of $U_{q}(\mathfrak{sl}_{2})$ are in one-to-one
correspondence with nonnegative integers $k$ such that $x^{k}$ belongs to
$M_{c,c^{\prime}}$. This is equivalent to $c_{i}^{\prime}+k\geq c_{i}$ for
$i=1,2$. In the case of trivial $\sigma$ this means
$k\geq\max(c_{1}-c^{\prime}_{1},c_{2}-c^{\prime}_{2})=\lvert
2\operatorname{Re}c\rvert$. In the case of nontrivial $\sigma$ the minimal $k$
is zero. We see that the minimal $k$ for the regular bimodule is zero.
Conversely, when the minimal $k$ is zero, $M_{c,c^{\prime}}$ is a regular
bimodule.

Suppose that the minimal $k$ is nonzero. Then $c=\frac{k}{2}+i\alpha$ for some
real number $\alpha$. In the beginning of Section 6.1 of [K] we proved that
the central reduction $U_{q}(\mathfrak{sl}_{2})/(\Omega-X_{0})$ has parameter
$P(z)=-\frac{z+z^{-1}-2}{(q-q^{-1})^{2}}+X_{0}$.

The polynomial $P$ has roots $q^{2\pm c}$. Using Vieta’s formula we have
$q^{2c}+q^{-2c}=2+X_{0}(q-q^{-1})^{2}$. Since $X_{0}$ depends linearly on
$q^{2c}+q^{-2c}$, it is enough to describe the locus of $q^{2c}+q^{-2c}$. We
have $q^{2c}+q^{-2c}=q^{k+2i\alpha}+q^{-k-2i\alpha}$. Let $q^{k}=r$,
$q^{2i\alpha}=\cos\phi+i\sin\phi$. Then
$q^{2c}+q^{-2c}=(r+r^{-1})\cos\phi+i(r-r^{-1})\sin\phi$. This is precisely the
ellipse $\mathcal{E}_{p}$ in [P], equation (0.3), multiplied by
$\frac{\sqrt{1+q^{2}}}{q}$.

The set $\frac{\sqrt{1+q^{2}}}{q}\mathcal{E}_{0}$ is the closed interval
$[-q-q^{-1},q+q^{-1}]$ and the endpoints correspond to one-dimensional
representations. In the case of the regular bimodule the sum of roots in [K]
belongs to $(-q-q^{-1},q+q^{-1})$. Hence our answer coincides with [P] except
that Pusz also allows positive half-integer $p$. If we add the square root of
$Z$ we get exactly the same answer.

We already checked in [EKRS] that in the case of $q=1$ and the regular
bimodule we get the classical theory of spherical unitary representation of
$\mathop{\mathrm{SL}}\nolimits(2,\mathbb{C})$. In the other cases we have
$c=\frac{k}{2}+i\alpha$, where $k$ is a nonzero number. It is checked in
[EKRS], Example 2.1.2 that $P(x)=x^{2}-X_{0}$, where $X_{0}$ is the value of
Casimir element. By Vieta’s formula
$X_{0}=c^{2}=\frac{k^{2}}{4}-\alpha^{2}+ik\alpha$. With $k=2p$ and $\alpha=t$
this coincides with $\frac{1}{2}\mathcal{P}_{p}-1$, where $\mathcal{P}_{p}$
are parabolas in [P].

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