Hopf Algebroid Twists for Deformation Quantization of Linear Poisson
Structures Hopf Algebroid Twists for Deformation Quantization
of Linear Poisson Structures

Stjepan MELJANAC † and Zoran ŠKODA ‡§ S. Meljanac and Z. Škoda

† Theoretical Physics Division, Institute Rudjer Bošković,
† Bijenička cesta 54, P.O. Box 180, HR-10002 Zagreb, Croatia<EMAIL_ADDRESS>‡
Faculty of Science, University of Hradec Králové,
‡ Rokitanského 62, Hradec Králové, Czech Republic<EMAIL_ADDRESS>§
University of Zadar, Department of Teachers’ Education,
§ Franje Tudjmana 24, 23000 Zadar, Croatia

Received May 24, 2017, in final form March 13, 2018; Published online March
25, 2018

In our earlier article [Lett. Math. Phys. 107 (2017), 475–503], we explicitly
described a topological Hopf algebroid playing the role of the noncommutative
phase space of Lie algebra type. Ping Xu has shown that every deformation
quantization leads to a Drinfeld twist of the associative bialgebroid of
$h$-adic series of differential operators on a fixed Poisson manifold. In the
case of linear Poisson structures, the twisted bialgebroid essentially
coincides with our construction. Using our explicit description of the Hopf
algebroid, we compute the corresponding Drinfeld twist explicitly as a product
of two exponential expressions.

deformation quantization; Hopf algebroid; noncommutative phase space; Drinfeld
twist; linear Poisson structure 53D55; 16S30; 16T05

##  1 Introduction

Given a possibly noncommutative associative $\mathbf{k}$-algebra $A$, a (left)
$A$-bialgebroid $H$ is an associative algebra with an additional $A$-bimodule
structure ${}_{A}H_{A}$, coproduct $\Delta\colon H\to H\otimes_{A}H$ and
counit $\epsilon\colon H\to A$ generalizing appropriately the structures of a
$\mathbf{k}$-vector space, coproduct and counit in the definition of a
$\mathbf{k}$-bialgebra. We refer to $A$ as the base algebra and $H$ as the
total algebra of the bialgebroid. For details see [4, 7] and Section 2. Lu
[13] introduced a class of $A$-bialgebroids meant to be noncommutative
analogues of transformation groupoids, hence a novel symmetry useful in
geometry and mathematical physics. These bialgebroids are smash products of a
Hopf algebra and a braided commutative algebra in its category of
Yetter–Drinfeld modules and nowadays they are often called scalar extension
bialgebroids. A particular case is the smash product of a Hopf algebra and its
dual (with its canonical Yetter–Drinfeld module structure); this smash product
is known under the name of Heisenberg double and may require some completions
in the infinite dimensional case. Our present focus is on the Lie algebra
noncommutative phase spaces whose Hopf algebroid structure is presented in
detail in [18], crudely shown to be a Heisenberg double of the universal
enveloping algebra $U(\mathfrak{g})$ in [22]. An improved variant of this
Heisenberg double and an entire class of generalizations are categorically
treated in [24], namely the Heisenberg doubles (and more generally, scalar
extension Hopf algebroids) in the internalized context of the symmetric
monoidal category of (countably cofinite strict) filtered-cofiltered vector
spaces, see [24]. They are shown to be internal Hopf algebroids in the sense
of Böhm [3]. It is compelling to interpret the Heisenberg double
$H_{\mathfrak{g}}$ of the universal enveloping algebra $U(\mathfrak{g})$ of a
Lie algebra $\mathfrak{g}$ as a noncommutative phase space of Lie algebra type
where the universal enveloping algebra $U(\mathfrak{g})$ is interpreted as its
coordinate sector (configuration space) and its (topological) Hopf dual as the
momentum sector. In the case of $\kappa$-deformed Minkowski space, one can
extend the Hopf algebroid adding full $\kappa$-deformed Poincaré symmetry into
the Heisenberg double [14, 15]. The physical discussion of the coproduct for
the momentum sector has been studied in many references including [1, 11].

Based on a work of Xu [25], this example can also be understood as a
deformation quantization of the phase space with the corresponding linear
Poisson structure. Xu defines an analogue of Drinfeld twist $\mathcal{F}\in
H\otimes_{A}H$ twisting the coproduct $\Delta$ to a new coproduct
$\Delta_{\mathcal{F}}$, by a recipe
$\Delta_{\mathcal{F}}(h)=\mathcal{F}^{-1}\Delta(h)\mathcal{F}$, for all $h\in
H$, where however the Hopf algebra $H$ from Drinfeld’s theory is replaced by
an $A$-bialgebroid $H$. On the other hand, if we consider the formal power
series in one variable $t$ with coefficients in the Poisson algebra of
functions, then the deformation quantization gives a recipe for an associative
star product $f\star g$ which is a formal power series in $t$ whose zero term
is the usual product of functions and the first correction gives the Poisson
bracket. One usually restricts to the case when there exist a series
$\mathcal{F}$ in $t$ whose coefficients are bidifferential operators such that
the star product is of the form $f\star g=(\mu\circ\mathcal{F})(f\otimes g)$.
Xu shows that such formal bidifferential operator $\mathcal{F}$ is in fact a
twist for the topological bialgebroid whose total algebra consists of formal
power series with coefficients in the original Poisson algebra of functions.

Excluding the bialgebroid twists induced by bialgebra twists [6], very few
examples where both the deformation twist and the twisted bialgebroid
structures are given by explicit formulas are known. The linear Poisson
structures are clearly among the most important classes to study and here we
find two explicit expressions $\mathcal{F}_{l}$ and $\mathcal{F}_{r}$ giving
the same twist for the left bialgebroid with completed Heisenberg–Weyl algebra
$\hat{A}_{n}$ to $H_{\mathfrak{g}}$ as the total algebra. Though
$\mathcal{F}_{l}=\mathcal{F}_{r}$ as elements in
$H_{\mathfrak{g}}\otimes_{U(\mathfrak{g})}H_{\mathfrak{g}}$ the corresponding
formulas come by projecting from two different elements
$\tilde{\mathcal{F}}_{l}$, $\tilde{\mathcal{F}}_{r}$ in the usual tensor
square $H_{\mathfrak{g}}\otimes H_{\mathfrak{g}}$. In [17] we exhibited
another formula $\mathcal{F}_{c}$ for this twist which is however far less
explicit and involves a series where each term involves both the original and
twisted coproduct, see (4.8).

Our main motivation is to find explicit examples of Xu’s construction, and is
reinforced by the recent interest in field theories on Lie algebra type
noncommutative spaces, including $\kappa$-deformations.

A Hopf algebroid is defined in Section 6 as a left bialgebroid with an
antihomomorphism $S\colon H\to H$ called the antipode and satisfying some
axioms. Unlike for the bialgebra twists, it is not known in general if the
bialgebroid twists of Hopf algebroids are sufficient to formulate a general
recipe for twisting the antipode as well. In our case, both the bialgebroid
before and the bialgebroid after the twist have an antipode. We give a
conjectural formula $S(h)=V_{\mathcal{F}}^{-1}S_{0}(h)V_{\mathcal{F}}$ for the
new antipode $S$ in terms of old antipode $S_{0}$ along with a partial
argument for it where
$V_{\mathcal{F}}=\mu(S_{0}\otimes\mathrm{id})(\mathcal{F})$. We can show that
$S(h)=V^{-1}S_{0}(h)V$ for some $V$, but we do not have the complete proof
that $V_{\mathcal{F}}=V$. Both can be expressed as exponentials in formal
power series of only the momentum variables $\partial^{\mu}$, starting with
free term $1$ and agreeing in the first few terms. Note that Xu [25] says Hopf
algebroid for the notion which does not involve an antipode and which is in
[7] shown to be equivalent to the left bialgebroid of other authors.

The representation of $H_{\mathfrak{g}}$ via a concrete twist, besides its
conceptual appeal, is useful to twist systematically many other constructions
(for example, basic constructions in differential geometry) from the
undeformed Heisenberg–Weyl algebra case to the case of phase space of Lie
algebra type. This is an additional tool for physical applications, e.g.,
development of field theories on the noncommutative spaces of Lie algebra
type. Such applications are under investigation.

To orient the reader within the subject, we note that some other physically
important examples of Hopf algebroids are built from the data of weak Hopf
algebras [4] (as those coming from the symmetries in low dimensional QFTs
[16]) and the study of the dynamical quantum Yang–Baxter equation [8, 25].

Conventions. In this article, all algebras are over a field $\mathbf{k}$ of
characteristic $0$, and the unadorned tensor product $\otimes$ is over the
ground field (in the deformation quantization and quantum gravity examples,
the field of real or complex numbers). We freely use Sweedler notation
$\Delta(h)=\sum h_{(1)}\otimes h_{(2)}$ for the coproducts with or without
explicit summation sign. If $A$ is an algebra, $A^{\mathrm{op}}$ denotes the
algebra with opposite multiplication. We use the Einstein summation
convention.

##  2 Bialgebroids

In this section, we define bialgebroids and some auxiliary constructions.

######  Definition 2.1 ([4, 7, 13]).

Given an associative algebra $(A,\mu_{A})$, which is in this context called
the base algebra, a left $A$-bialgebroid
$(H,\mu,\alpha,\beta,\Delta,\epsilon)$ consists of

  * •

an associative algebra $(H,\mu)$;

  * •

two algebra maps, the source map $\alpha\colon A\to H$ and the target map
$\beta\colon A^{\mathrm{op}}\to H$ such that $[\alpha(a),\beta(a^{\prime})]=0$
for all $a,a^{\prime}\in A$; assume in the following that on $H$ we fix the
$A$-bimodule structure given by $a.h.a^{\prime}=\alpha(a)\beta(a^{\prime})h$,
$a,a^{\prime}\in A$, $h\in H$;

  * •

$A$-bimodule map $\Delta\colon H\to H\otimes_{A}H$, called coproduct,
satisfying the coassociativity

$\displaystyle(\Delta\otimes_{A}\mathrm{id}_{H})\circ\Delta=(\mathrm{id}_{H}\otimes_{A}\Delta)\circ\Delta;$

  * •

$A$-bimodule map $\epsilon\colon H\to A$, called counit, satisfying
$\alpha(\epsilon(h_{(1)}))h_{(2)}=h=\beta(\epsilon(h_{(2)}))h_{(1)}$.

The following compatibilities are required for these data:

(i) Formula $\sum_{\lambda}h_{\lambda}\otimes
f_{\lambda}\mapsto\epsilon(\sum_{\lambda}h_{\lambda}\alpha(f_{\lambda}))$
defines an action $\blacktriangleright\colon H\otimes A\to A$ which extends
the left regular action $A\otimes A\to A$ along the inclusion $A\otimes
A\stackrel{{\scriptstyle\alpha\otimes A}}{{\longrightarrow}}H\otimes A$.

(ii) The $A$-subbimodule $H\times_{A}H\subset H\otimes_{A}H$ (called the
Takeuchi product [4]) defined by

$\displaystyle H\times_{A}H=\left\\{\sum_{i}b_{i}\otimes b^{\prime}_{i}\in
H\otimes_{A}H\,|\,\sum_{i}b_{i}\otimes
b^{\prime}_{i}\alpha(a)=\sum_{i}b_{i}\beta(a)\otimes b^{\prime}_{i},\
\forall\,a\in A\right\\}$

contains the image of $\Delta$ and the corestriction $\Delta|\colon H\to
H\times_{A}H$ is an algebra map with respect to the factorwise multiplication.

To see the meaning of (ii), notice that, unlike for the Takeuchi product
$H\times_{A}H$, the factorwise multiplication on $H\otimes H$ does not factor
to a well defined map on the vector space $H\otimes_{A}H$ in general.
Accordingly, it does not make sense to say that $\Delta\colon H\to
H\otimes_{A}H$ is an algebra map and the multiplicativity of $\Delta$ should
be interpreted differently, say within $H\times_{A}H$. Indeed, the kernel

$\displaystyle I_{A}=\operatorname{Ker}(H\otimes H\to H\otimes_{A}H)$ (2.1)

of the canonical projection of the $A$-bimodules is the right ideal in the
algebra $H\otimes H$ generated by $\beta(a)\otimes 1-1\otimes\alpha(a)$ for
all $a\in A$, but this kernel is not a two-sided ideal in general.

The compatibility (ii) in the definition of a bialgebroid holds iff the map
$H\otimes(H\otimes_{A}H)\to H$, $(g,\sum_{i}h_{i}\otimes
k_{i})\mapsto\Delta(g)(\sum_{i}h_{i}\otimes k_{i})$ (multiplied in each tensor
factor) is well defined (does not depend on the choice of the sum within the
equivalence class in $H\otimes_{A}H=H\otimes H/I_{A}$). In other words,
$\Delta(g)\cdot I_{A}\subset I_{A}$ for all $g\in H$. This last
characterization will be useful in the proof of Proposition 5.9.

######  Lemma 2.2.

The composition

$\displaystyle(H\otimes H)\otimes(A\otimes
A)\stackrel{{\scriptstyle\cong}}{{\longrightarrow}}(H\otimes
A)\otimes(H\otimes
A)\stackrel{{\scriptstyle\blacktriangleright\otimes\blacktriangleright}}{{\longrightarrow}}A\otimes
A\stackrel{{\scriptstyle\mu_{A}}}{{\longrightarrow}}A$ (2.2)

factors down along the projection $(H\otimes H)\otimes(A\otimes
A)\to(H\otimes_{A}H)\otimes(A\otimes A)$ to a $\mathbf{k}$-linear map

$\displaystyle\mu_{A}\circ(\blacktriangleright\otimes\blacktriangleright)\colon\
(H\otimes_{A}H)\otimes(A\otimes A)\longrightarrow A.$

By similar factorization, we get analogous maps from the multiple tensor
products $(H\otimes_{A}\cdots\otimes_{A}H)\otimes(A\otimes\cdots\otimes A)\to
A$. For elements we extend the operation syntax $h\blacktriangleright
a:=\blacktriangleright(h\otimes a)$ to products
$\blacktriangleright\otimes\blacktriangleright$ (and multiple analogues),
e.g., if $\mathcal{F}\in H\otimes_{A}H$ and $a,b\in A$ we write

$\displaystyle\mu_{A}\mathcal{F}(\blacktriangleright\otimes\blacktriangleright)(a\otimes
b):=(\mu_{A}\circ(\blacktriangleright\otimes\blacktriangleright))(\mathcal{F}\otimes(a\otimes
b)).$

###### Proof.

We need to show that the composition (2.2) gives zero on the subspace
$I_{A}\otimes(A\otimes A)$. We show this in terms of the generators. For
$a,b,c\in A$, $h,h^{\prime}\in H$ we need

$\displaystyle\mu_{A}(((\beta(c)h)\blacktriangleright
a)\otimes(h^{\prime}\blacktriangleright b))=\mu_{A}((h\blacktriangleright
a)\otimes((\alpha(c)h^{\prime})\blacktriangleright b)).$

By the definition of $\blacktriangleright$ (see (i) above), it is hence enough
to show

$\displaystyle\mu_{A}(\epsilon(\beta(c)h\alpha(a))\otimes\epsilon(h^{\prime}\alpha(b)))=\mu_{A}(\epsilon(h\alpha(a))\otimes\epsilon(\alpha(c)h^{\prime}\alpha(b))).$

Since $\epsilon$ is a map of $A$-bimodules, both sides yield
$\epsilon(h\alpha(a))\cdot c\cdot\epsilon(h^{\prime}\alpha(b))\in A$. ∎

##  3 Twists for bialgebroids

In this section, we explain how Ping Xu [25] generalized the Drinfeld twists
to bialgebroids (see also [8]), give some examples and focus on the case
relevant for deformation quantization. This involves formal completions and
completed tensor products and somewhat different formal completions in the
later part of the article. For that reason, in this section, we also include
an extensive review of completions adapted to our formalism (see also Remark
4.5). Unlike in some other publications, we here use Xu’s convention for
twists (elsewhere we use $\mathcal{F}$ for his $\mathcal{F}^{-1}$).

######  Definition 3.1 ([25]).

$\mathcal{F}\in H\otimes_{A}H$ is a Drinfeld twist for a left $A$-bialgebroid
$(H,\mu,\alpha,\beta,\Delta,\epsilon)$ if the 2-cocycle condition

$\displaystyle[(\Delta\otimes_{A}\mathrm{id})(\mathcal{F})](\mathcal{F}\otimes_{A}1)=[(\mathrm{id}\otimes_{A}\Delta)(\mathcal{F})](1\otimes_{A}\mathcal{F})$
(3.1)

and the counitality
$(\epsilon\otimes_{A}\mathrm{id})(\mathcal{F})=1_{H}=(\mathrm{id}\otimes_{A}\epsilon)(\mathcal{F})$
hold.

We use the Sweedler-like notation for the twist
$\mathcal{F}=\mathcal{F}^{(1)}\otimes\mathcal{F}^{(2)}$ (upper labels, while
we use lower for the coproduct) and the notation from Lemma 2.2.

######  Theorem 3.2 ([25]).

If $H$ is a left $A$-bialgebroid and $\mathcal{F}\in H\otimes_{A}H$ a Drinfeld
twist, then the formula

$\displaystyle a\star
b=\mu\mathcal{F}(\blacktriangleright\otimes\blacktriangleright)(f\otimes
g)=\big{(}\mathcal{F}^{(1)}\blacktriangleright
a\big{)}\big{(}\mathcal{F}^{(2)}\blacktriangleright b\big{)}$ (3.2)

defines an associative $\mathbf{k}$-algebra $A_{\star}=(A,\star)$ structure on
$A$ with the same unit; the formulas
$\alpha_{\mathcal{F}}(a)=\alpha\big{(}\mathcal{F}^{(1)}\blacktriangleright
a\big{)}\mathcal{F}^{(2)}$ and
$\beta_{\mathcal{F}}(a)=\beta\big{(}\mathcal{F}^{(2)}\blacktriangleright
a\big{)}\mathcal{F}^{(1)}$ define respectively an algebra homomorphism
$\alpha_{\mathcal{F}}\colon A_{\star}\to H$ and antihomomorphism
$\beta_{\mathcal{F}}\colon A_{\star}\to H$ with
$[\alpha_{\mathcal{F}}(a),\beta_{\mathcal{F}}(a)]=0$ for all $a\in A_{\star}$.
Given the $A_{\star}$-bimodule structure on $H$ by
$a.h.b:=\alpha_{\mathcal{F}}(a)\beta_{\mathcal{F}}(b)h$ and denoting by
$I_{A_{\star}}$ the kernel of the canonical projection $H\otimes H\to
H\otimes_{A_{\star}}H$, there is an inclusion $\mathcal{F}\cdot
I_{A_{\star}}\subset I_{A}\subset H\otimes H$ $($cf. (2.1)$)$. In other words,
$\mathcal{F}$ induces a linear map $\mathcal{F}^{\sharp}\colon
H\otimes_{A_{\star}}H\to H\otimes_{A}H$ given by the left multiplication of
classes in $H\otimes H$ by $\mathcal{F}$, that is by
$\mathcal{F}^{\sharp}(a\otimes_{A_{\star}}b)=f^{1}a\otimes_{A}f_{1}b$. Suppose
the map $\mathcal{F}^{\sharp}$ is invertible. Then the formula

$\displaystyle\Delta_{\mathcal{F}}(h)=\mathcal{F}^{\sharp-1}\Delta(h)\mathcal{F}$

defines a map $\Delta_{\mathcal{F}}\colon H\to H\otimes_{A_{\star}}H$ which is
coassociative and counital with the same counit. Moreover,
$H_{\mathcal{F}}=(H,\mu,\alpha_{\mathcal{F}},\beta_{\mathcal{F}},\Delta_{\mathcal{F}},\epsilon)$
is a left $A_{\star}$-bialgebroid.

######  Remark 3.3 (inverse cocycle).

For $\mathcal{F}^{\sharp-1}\colon H\otimes_{A}H\to H\otimes_{A_{\star}}H$ to
exist, it is sufficient that there is an element $\mathcal{F}^{-1}\in
H\otimes_{A_{\star}}H$ such that $\mathcal{F}^{-1}\cdot I_{A}\subset
I_{A_{\star}}$ and the identities $\mathcal{F}^{-1}\mathcal{F}=1\otimes
1+I_{A_{\star}}$ and $\mathcal{F}\mathcal{F}^{-1}=1\otimes 1+I_{A_{\star}}$
hold in $H\otimes H$. In that case,
$\mathcal{F}^{\sharp-1}=\mathcal{F}^{-1\sharp}$. This will be the situation
throughout the article, hence we do not need to distinguish
$\mathcal{F}^{\sharp-1}$ from the left multiplication with $\mathcal{F}^{-1}$.
In terms of $\mathcal{F}^{-1}$ and $A_{\star}$, the cocycle condition (3.1) is
equivalent to the condition

$\displaystyle\big{(}\mathcal{F}^{-1}\otimes_{A_{\star}}1\big{)}(\Delta\otimes_{A_{\star}}\mathrm{id})\big{(}\mathcal{F}^{-1}\big{)}=\big{(}1\otimes_{A_{\star}}\mathcal{F}^{-1}\big{)}(\mathrm{id}\otimes_{A_{\star}}\Delta)\big{(}\mathcal{F}^{-1}\big{)}.$

######  Remark 3.4.

The proof of the associativity of $\star$ in the theorem follows from
comparing $(a\star b)\star c$ and $a\star(b\star c)$, which are from the
definition (3.2) easily calculated to be

$\displaystyle(a\star b)\star
c=\mu(\mu\otimes\mathrm{id})[(\Delta\otimes\mathrm{id})\mathcal{F}](\mathcal{F}\otimes\mathrm{id})(\blacktriangleright\otimes\blacktriangleright\otimes\blacktriangleright)(a\otimes
b\otimes c),$ $\displaystyle a\star(b\star
c)=\mu(\mathrm{id}\otimes\mu)[(\mathrm{id}\otimes\Delta)\mathcal{F}](\mathrm{id}\otimes\mathcal{F})(\blacktriangleright\otimes\blacktriangleright\otimes\blacktriangleright)(a\otimes
b\otimes c).$

Thus the cocycle condition (3.1) implies the coassociativity. The converse
does not hold for arbitrary bialgebroids. Indeed, let
$\mu_{2}=\mu(\mu\otimes\mathrm{id})=\mu(\mathrm{id}\otimes\mu)$ be the second
iterate of the multiplication and $\mathcal{F}_{3a}$ and $\mathcal{F}_{3b}$
the left and the right-hand sides of the cocycle condition (3.1); we can see
only that the coassociativity implies
$\mu_{3}(\mathcal{F}_{3a}-\mathcal{F}_{3b})(\blacktriangleright\otimes\blacktriangleright\otimes\blacktriangleright)(a\otimes
b\otimes c)=0$ for all $a,b,c\in H$. This implies
$\mathcal{F}_{3a}-\mathcal{F}_{3b}=0$ in $H\otimes_{A}H\otimes_{A}H$ if
$\blacktriangleright\otimes\blacktriangleright\otimes\blacktriangleright$ is
nondegenerate in the first argument, which is not true for an arbitrary
bialgebroid. The nondegeneracy holds for the undeformed Heisenberg algebra and
by Theorem 4.3 more generally for the $U(\mathfrak{g})$-bialgebroid
$H_{\mathfrak{g}}$ studied in this paper – this is the content of Theorem 4.3
with the consequence for the cocycle condition, Corollary 4.4.

######  Example 3.5 (basic example of a noncommutative bialgebroid over a
commutative base).

Let $A=C^{\infty}(M)$, where $M$ is a smooth real manifold and let
$H=\mathcal{D}$ be the algebra of global differential operators with smooth
coefficients. The formula $\Delta(D)(f,g)=D(f\cdot g)$ defines a
bidifferential operator
$\Delta(D)\in\mathcal{D}\otimes_{C^{\infty}(M)}\mathcal{D}$; this defines a
cocommutative coproduct $\Delta$. The base $C^{\infty}(M)$ is commutative and
$\alpha=\beta$ is the canonical embedding of the algebra of functions into the
algebra of differential operators.

This example generalizes. The algebra of differential operators $\mathcal{D}$
is canonically isomorphism to the universal enveloping algebra of the Lie
algebroid given by the tangent bundle $V=TM$ where the anchor map $a$ is the
identity. Universal enveloping algebras of Lie algebroids are bialgebroids in
a canonical way. To explain this in few lines let us say that a Lie algebroid
$L=(V,a,[\,,\,])$ is a vector bundle $V$ over a manifold $M$ together with a
map $a\colon V\to TM$ of vector bundles (called the anchor) and an
antisymmetric $\mathbb{R}$-bilinear bracket $[\,,\,]$ on the
$C^{\infty}(M)$-module of global sections $\Gamma V$. It is required that the
map, $a_{*}\colon\Gamma V\to\Gamma TM$ is a map of $\mathbb{R}$-Lie algebras
(where the bracket on $\Gamma TM$ is the Lie bracket of vector fields) and is
satisfying the Leibniz rule $[X,fY]=f[X,Y]+a(X)(f)Y$. For any Lie algebroid
$V$ one defines the universal enveloping algebra $U(L)$ as the tensor algebra
of the space of sections $\Gamma V$ modulo the ideal generated by the
relations $[X,Y]-X\otimes Y+Y\otimes X$, and $XfY-a(X)(f)Y-fXY$, where $f\in
C^{\infty}(M)$, $X,Y\in\Gamma V$. It becomes a bialgebroid over the
commutative algebra $C^{\infty}(M)$ via the formula $\Delta(f)=f\otimes 1$ for
$f\in C^{\infty}(M)\subset U(V)$ and $\Delta(X)=X\otimes 1+1\otimes X$ for
$X\in\Gamma V\subset U(V)$.

Thesis [12] settled the question when bialgebroid structures on universal
enveloping algebras of Lie algebroids, and more generally of Lie–Rinehart
algebras, have an antipode (see Section 6), namely iff a flat connection in
the appropriate context exists. In [25] an erroneous argument (confusing the
role of the antipode with the duality for functional spaces) was given that
the bialgebroids of differential operators can not have an antipode. The
special case of the Heisenberg–Weyl algebra in this article of course has an
antipode.

######  Remark 3.6 (notation on completions).

Given a vector space $V$ an $\mathbb{N}$-cofiltration of $V$ is given by
specifying for all $n\in\mathbb{N}$ the quotient spaces $V_{n}$ called
cofiltering components with projections $\pi_{n}\colon V\to V_{n}$, connecting
quotient maps $\pi_{nm}\colon V_{m}\to V_{n}$ for all $n\leq m$, satisfying
$\pi_{nm}\circ\pi_{mk}=\pi_{nk}$ for all $n\leq m\leq k$, and commuting with
the projections, $\pi_{nm}\circ\pi_{m}=\pi_{n}$. One considers the completion
$\hat{V}$ as the inverse limit $\underset{\longleftarrow}{\lim}{}_{n}V_{n}$
which may be realized by threads, that is sequences $(v_{n})_{n\in\mathbb{N}}$
where $v_{n}\in V_{n}$ and $\pi_{nm}(v_{m})=v_{n}$ for all $n\leq m$. The
canonical map $V\to\hat{V}=\underset{\longleftarrow}{\lim}{}_{n}V_{n}$,
$v\mapsto(\pi_{n}(v))_{n}$ is injective iff for every $v\in V$ there is
$n\in\mathbb{N}$ such that $\pi_{n}(v)\neq 0$. In that case, we usually
identify $V$ with its image within $\hat{V}$. If the canonical map
$V\to\hat{V}$ is an isomorphism we say that $V$ is complete. A cofiltration is
often given by a decreasing filtration, that is a decreasing sequence of
subspaces $V^{\prime}_{n}$ with empty common intersection; the quotients are
given by $V_{n}=V/V^{\prime}_{n}$ and the inclusions
$V^{\prime}_{n}\hookrightarrow V^{\prime}_{n+1}$ induce the quotient maps
$V/V^{\prime}_{n+1}\to V/V^{\prime}_{n}$. If all $V^{\prime}_{n}$ are of
finite codimension in $V$, then the completion may be understood in the
topological sense (consisting of all equivalence classes of Cauchy sequences),
where $V$ is understood as a topological vector space with linear topology for
which the subspaces $V^{\prime}_{n}$ form a basis of neighborhoods of $0$.
Given two cofiltered spaces $V$ and $W$, there is a completed tensor product
also adorned with the hat symbol,
$V\hat{\otimes}W=\underset{\longleftarrow}{\lim}{}_{n,m}V_{n}\otimes W_{m}$,
where the inverse limit is over the directed set $\mathbb{N}\times\mathbb{N}$;
the spaces $V_{n}\otimes V_{m}$ together with the connecting morphisms form an
$\mathbb{N}\times\mathbb{N}$-cofiltration [24], which can be replaced, in the
generality of this paper, by an equivalent $\mathbb{N}$-cofiltration [18,
Appendix 2]. Given a cofiltered vector space $V$, an expression
$\sum_{\lambda\in\Lambda}v_{\lambda}$ where $v_{\lambda}\in V$ is a profinite
(or formal) sum [24] if for each $n$ there are only finitely many
$\lambda\in\Lambda$ such that $\pi_{n}(v_{\lambda})\neq 0$. Then the finite
sums $s_{n}=\sum_{\lambda,\pi_{n}(\lambda)\neq 0}\pi_{n}(v_{\lambda})\in
V_{n}$ define a thread $(s_{n})_{n\in\mathbb{N}}$ which is therefore an
element of the completion $\hat{V}$ called the value $v$ of the profinite sum
$\sum_{\lambda\in\Lambda}v_{\lambda}$ and we write
$v=\sum_{\lambda\in\Lambda}v_{\lambda}$. A linear map $f\colon V\to W$ between
cofiltered vector spaces distributes over profinite sums (equivalently: is a
cofiltered map) if for each profinite sum
$\sum_{\lambda\in\Lambda}v_{\lambda}$ in $V$ the expression
$\sum_{\lambda\in\Lambda}f(v_{\lambda})$ is profinite in $W$ and the equation
$f\big{(}\sum_{\lambda\in\Lambda}v_{\lambda}\big{)}=\sum_{\lambda\in\Lambda}f(v_{\lambda})$
holds for the values; cofiltered spaces (even if noncomplete) with cofiltered
maps form a monoidal category with the usual tensor product $\otimes$.
Similarly, the subcategory of complete cofiltered spaces is a monoidal
category with respect to the completed tensor product $\hat{\otimes}$. Each
cofiltered map $f\colon V\to W$ induces the morphisms among the limits of
cofiltrations, which can be interpreted as the completed cofiltered map
$\hat{f}\colon\hat{V}\to\hat{W}$, the completion of $f$, which commutes with
the canonical maps $i_{V}\colon V\to\hat{V}$, $i_{W}\colon W\to\hat{W}$ into
the completions, $\hat{f}\circ i_{V}=i_{W}\circ f$. It can be seen that a
subspace $Z$ of a complete cofiltered space
$V\cong\underset{\longleftarrow}{\lim}{}_{\lambda}V_{\lambda}$ is complete iff
for all profinite sums whose summands are in $Z$ their values are also in $Z$.
The cofiltered maps among the complete vector spaces for which the topological
interpretation holds, coincide with the continuous linear maps. Algebra in the
monoidal category of cofiltered vector spaces with the usual tensor product is
said to be a noncomplete cofiltered algebra. The modules and ideals of
complete algebras should be completed as well, the fact which we often pass
over silently. In particular, if $A$ is a noncomplete cofiltered algebra, $M$
a right $A$-module and $N$ a left $A$-module, such that the actions are
cofiltered maps, the kernel $I_{A}=\operatorname{ker}(M\otimes N\to
M\otimes_{A}N)$ is a subspace $I_{A}\subset M\otimes N\hookrightarrow
M\hat{\otimes}N$ and its completion is identified with a subspace
$\hat{I}_{A}\subset M\hat{\otimes}N$. If the actions on $M$ and $N$ are
cofiltered maps, then the vector space
$M\hat{\otimes}_{A}N=M\hat{\otimes}N/\hat{I}_{A}$ is cofiltered, and called
the completed tensor product of $M$ and $N$ over $A$, see [24]. The
multiplication of $A$ is cofiltered, hence it extends to a cofiltered map
$\hat{A}\otimes\hat{A}\to\hat{A}$ and even to a unique cofiltered map
$A\hat{\otimes}A\cong\hat{A}\hat{\otimes}\hat{A}\to\hat{A}$; thus $\hat{A}$
becomes an algebra in the category of complete cofiltered vector spaces with
the completed tensor product. In this situation, if $M$ is a right $A$-module
and the action is a cofiltered map $A\otimes M\to M$, then $\hat{M}$ is a
right complete cofiltered $\hat{A}$-module, in other words the action is a
cofiltered map $\hat{A}\hat{\otimes}\hat{M}\to\hat{M}$ (note the complete
tensor product), hence by restriction a fortiori an $A$-module. It is
straightforward to observe that
$\hat{M}\hat{\otimes}_{\hat{A}}\hat{N}\cong\hat{M}\hat{\otimes}_{A}\hat{N}\cong
M\hat{\otimes}_{A}N$ (indeed, $\hat{I}_{\hat{A}}=\hat{I}_{A}$ as a subset of
$M\hat{\otimes}N\cong\hat{M}\hat{\otimes}\hat{N}$). Therefore, if $U$ is the
(faithful) forgetful functor from the category of $\hat{A}$-bimodules with the
completed tensor product to the category of the usual $A$-bimodules, then
there is a natural isomorphism of functors $\hat{\otimes}_{A}\circ(U\times
U)\cong U\circ\hat{\otimes}_{\hat{A}}$. The following is a simple example of
our interest. If $t$ is a formal variable and $B$ an algebra then the
completion of the polynomial ring $B[t]$ with respect to the cofiltrations by
the finite dimensional subspaces $B[t]/(t^{n}B[t])$ of truncations modulo
$t^{n}$ is the formal series ring $B[[t]]$. Then, the restriction of the
(bi)functor $\otimes_{B[t]}$ to the subcategory of complete $B[t]$-modules
(with the usual $\otimes$) such that the actions are cofiltered, agrees in the
above sense with the (bi)functor $\otimes_{B[[t]]}$. In particular, we can
move the formal series from $B[[t]]$ across the tensor product as a limiting
case of moving its finite truncations in $B[t]/(t^{n}B[t])\cong
B[[t]]/(t^{n}B[[t]])$ at the level of cofiltered components. Note a minor
difference between the two (bi)functors, namely for $\otimes_{B[[t]]}$ we work
with the $B[[t]]$-bimodule structures in the monoidal category of complete
cofiltered spaces with the $\hat{\otimes}$-tensor product, rather than with
the usual $B[t]$-bimodule structures.

Unlike in [18], we do not pay full attention to completions.

######  Example 3.7 (an algebraic analogue of the basic Example 3.5 of a
noncommutative bialgebroid over commutative base).

Here $A=\mathcal{O}(M)$ is the algebra of regular functions on a smooth affine
variety over a field $\mathbf{k}$ of characteristic $0$ and $H=\mathcal{D}$ is
the algebra of regular differential operators. Again, $\Delta(D)(f,g)=D(f\cdot
g)$, the base $A$ is commutative and $\alpha=\beta$ is the canonical inclusion
of elements of $A$ as the operators of left multiplication. Here and in
Example 3.5 the counit is taking the constant term and $\blacktriangleright$
is the usual action of differential operators on functions (below denoted
$\triangleright$ or as evaluation $D(f)=D\triangleright f$). If $M$ is the
affine $n$-space, $H$ is the Weyl algebra $A_{n}$ and the corresponding
coproduct $\Delta_{0}$ is primitive on the generators
$\partial^{1},\dots,\partial^{n}$ and $\Delta_{0}(x_{\mu})=x_{\mu}$ on the
remaining generators $x_{1},\dots,x_{n}$ of the Weyl algebra. This
$\mathcal{O}(M)$-bialgebroid is in fact a Hopf algebroid with the antipode
$S(x_{\mu})=x_{\mu}$ and $S(\partial^{\nu})=-\partial^{\nu}$. We may complete
$A_{n}$ by the degree of a differential operator to the algebra $\hat{A}_{n}$
and the Hopf algebroid structure continuously extends ([18]) to a topological
coproduct on $\hat{A}_{n}$ with values in the completed tensor product
$\hat{A}_{n}\hat{\otimes}_{A}\hat{A}_{n}$ relative over
$A=\mathbf{k}[x_{1},\dots,x_{n}]$,

$\displaystyle\Delta_{0}\colon\
\hat{A}_{n}\to\hat{A}_{n}\hat{\otimes}_{\mathbf{k}[x_{1},\dots,x_{n}]}\hat{A}_{n},\qquad\Delta_{0}(x_{\mu})=x_{\mu}\otimes
1,\qquad\Delta_{0}(\partial^{\nu})=1\otimes\partial^{\nu}+\partial^{\nu}\otimes
1.$ (3.3)

By abuse of notation, we denote by $\Delta_{0}$ also the restriction of
$\Delta_{0}$ to the polynomial algebra of $\partial^{1},\dots,\partial^{n}$
(or to its completion $\mathbf{k}[[\partial^{1},\dots,\partial^{n}]]$, the
algebra of formal power series), which therefore becomes a Hopf
$\mathbf{k}$-algebra (respectively, a topological Hopf $\mathbf{k}$-algebra).

######  Example 3.8 (related to deformation quantization).

Ping Xu [25] extends the base algebra $A=C^{\infty}(M)$ from Example 3.5 or
$A=\mathbf{k}[x_{1},\dots,x_{n}]$ from Example 3.7 to the algebra of formal
power series $A[[h]]$ in a formal variable $h$ with coefficients in $A$. Then
$\mathcal{D}[[h]]$ carries a left $A[[h]]$-bialgebroid structure obtained from
$A$-bialgebroid $\mathcal{D}$ by extending the scalars from $A$ to $A[[h]]$
and one works with the $h$-adically completed tensor product
$\hat{\otimes}_{\mathbf{k}[[h]]}$.

######  Theorem 3.9 ([25]).

Suppose $M$ is a Poisson manifold and a formal bidifferential operator
$\mathcal{F}\in(\mathcal{D}\otimes_{C^{\infty}(M)}\mathcal{D})[[h]]$ is given.
If $\mathcal{F}$ defines a deformation quantization of $M$ via the star
product $\mu\mathcal{F}(f\otimes g)$ then $\mathcal{F}$ is a Drinfeld twist
for the left $C^{\infty}(M)[[h]]$-bialgebroid $\mathcal{D}[[h]]$ of formal
power series in regular differential operators. Consequently, by Theorem 3.2,
each deformation quantization of $M$ defines also a twist deformation of the
bialgebroid $\mathcal{D}[[h]]$, where the twisted base algebra is
$C^{\infty}(M)[[h]]$, equipped with the star product $($both in the sense of
constructed from a bialgebroid twist and in the sense of deformation
quantization$)$.

We are interested in using the bialgebroid techniques to find explicit
formulas for $\mathcal{F}$ and also to describe the Xu’s bialgebroid in detail
in special cases. There are rather few explicit formulas for bialgebroid
twists in the literature which are not induced from bialgebra twists by a
procedure studied in [6].

##  4 Phase spaces of Lie type as bialgebroids

In this section, we present the bialgebroid $H_{\mathfrak{g}}$ from our
earlier article [18]. This bialgebroid is in fact a Hopf algebroid, but the
discussion of the antipode is deferred to Section 6.

Throughout, $\mathfrak{g}$ is a fixed Lie algebra over $\mathbf{k}$ with basis
$\hat{x}_{1},\dots,\hat{x}_{n}$, $U(\mathfrak{g})$ is the universal enveloping
algebra and $S(\mathfrak{g})$ the symmetric algebra of $\mathfrak{g}$. The
generators of $U(\mathfrak{g})$ are also denoted
$\hat{x}_{1},\dots,\hat{x}_{n}$ (and viewed as noncommutative coordinates),
but the corresponding generators of $S(\mathfrak{g})$ are $x_{1},\dots,x_{n}$
(and viewed as coordinates on the undeformed commutative space). The structure
constants $C_{\mu\nu}^{\lambda}$ are given by

$\displaystyle[\hat{x}_{\mu},\hat{x}_{\nu}]=C_{\mu\nu}^{\lambda}\hat{x}_{\lambda}.$

Let $\partial^{1},\dots,\partial^{n}$ be the dual basis of $\mathfrak{g}^{*}$,
which are also (commuting) generators of $S(\mathfrak{g}^{*})$. Let
$\hat{S}(\mathfrak{g}^{*})$ be the formal completion of $S(\mathfrak{g}^{*})$
(which should be interpreted as the algebra of formal functions on the space
of deformed momenta). A basis of neighborhoods of $0$ is made out of powers of
the ideal $J=\mathfrak{g}\cdot\hat{S}(\mathfrak{g}^{*})$ generated by
$\mathfrak{g}$ ($J$-adic topology of the formal power series ring). The ring
$M_{n}(\hat{S}(\mathfrak{g}^{*}))$ of $n\times n$-matrices with entries in
$\hat{S}(\mathfrak{g}^{*})$ is rank $n^{2}$ free
$\hat{S}(\mathfrak{g}^{*})$-module and as such it inherits the topology of
componentwise convergence. By a straightforward check we can see that this
topology is equivalent to the $M_{n}(J)$-adic topology, where
$M_{n}(J)=M_{n}(\mathfrak{g}\cdot\hat{S}(\mathfrak{g}^{*}))$ consists of all
matrices with entries in $\mathfrak{g}\cdot\hat{S}(\mathfrak{g}^{*})$.
Therefore $M_{n}(\hat{S}(\mathfrak{g}^{*})$ is a linearly compact topological
ring as well. We introduce an auxiliary matrix $\mathcal{C}\in
M_{n}(\hat{S}(\mathfrak{g}^{*}))$ with entries

$\displaystyle\mathcal{C}^{\alpha}_{\beta}:=C^{\alpha}_{\beta\gamma}\partial^{\gamma}\in\hat{S}(\mathfrak{g}^{*}),$
(4.1)

with the summation over repeated Greek indices and where, for the purposes of
this article, we use the convention that $\alpha$ is the row and $\beta$ the
column index. In this notation, we introduce matrices

$\displaystyle\phi:=\frac{-\mathcal{C}}{e^{-\mathcal{C}}-1}=\sum_{N=0}^{\infty}\frac{(-1)^{N}B_{N}}{N!}\mathcal{C}^{N},\qquad\tilde{\phi}:=\frac{\mathcal{C}}{e^{\mathcal{C}}-1}=\sum_{N=0}^{\infty}\frac{B_{N}}{N!}\mathcal{C}^{N},$
(4.2)

where $B_{N}$ are the Bernoulli numbers. By a simple comparison of the
expressions (4.2) we obtain

$\displaystyle\tilde{\phi}_{\alpha}^{\beta}=\phi^{\alpha}_{\rho}\mathcal{O}^{-1\rho}_{\beta},\qquad\text{where}\qquad\mathcal{O}:=e^{\mathcal{C}}\in
M_{n}\big{(}\hat{S}(\mathfrak{g}^{*})\big{)}.$ (4.3)

By $\hat{A}_{n}$ denote the completion by the degree of differential operators
of the $n$-th Weyl algebra $A_{n}$ with generators $x_{1},\dots,x_{n}$,
$\partial^{1},\dots,\partial^{n}$ hence the underlying vector space of $A_{n}$
is $S(\mathfrak{g})\otimes S(\mathfrak{g}^{*})$ and $\hat{A}_{n}$ is some
completion of it. The symmetric algebra $S(\mathfrak{g})$ is a Hopf algebra
which acts on $\hat{S}(\mathfrak{g}^{*})$ by a unique action
${\boldsymbol{\delta}}\colon
S(\mathfrak{g})\to\operatorname{End}_{\mathbf{k}}(\hat{S}(\mathfrak{g}^{*}))$
which is a Hopf action, namely

$\displaystyle{\boldsymbol{\delta}}(f)(P\cdot
Q)={\boldsymbol{\delta}}(f_{(1)})(P){\boldsymbol{\delta}}(f_{(2)})(Q)\qquad\text{and}\qquad{\boldsymbol{\delta}}(f)(1)=\epsilon(f)1,$

and which satisfies
${\boldsymbol{\delta}}(x_{\nu})(\partial^{\nu})=\delta^{\nu}_{\mu}$ on the
generators. It is a useful point of view for the deformations below that the
product in $A_{n}$ is the multiplication of the smash product algebra given by
the formula

$\displaystyle(f\otimes P)(g\otimes Q)=\sum f\cdot
g_{(1)}\otimes{\boldsymbol{\delta}}(g_{(2)})(P)\cdot Q\qquad\text{for}\quad
f,g\in S(\mathfrak{g}),\quad P,Q\in\hat{S}(\mathfrak{g}^{*}).$

The algebra $\hat{A}_{n}$ acts on the symmetric algebra $S(\mathfrak{g})$ of
polynomials in $x_{1},\dots,x_{n}$, extending the Fock action of $A_{n}$ on
$S(\mathfrak{g})$ by differential operators $(P,f)\mapsto P(f)$.

Now define the elements $\hat{x}^{\phi},\hat{y}^{\phi}\in\hat{A}_{n}$

$\displaystyle\hat{x}^{\phi}_{\rho}:=\sum_{\tau}x_{\tau}\phi^{\tau}_{\rho},\qquad\hat{y}^{\phi}_{\rho}:=\sum_{\tau}x_{\tau}\tilde{\phi}^{\tau}_{\rho}.$
(4.4)

With our choice of the matrix ${\boldsymbol{\phi}}$, the right-hand side in
the equations (4.4) are dual, as explained and derived in [18], Section 2, to
the expressions for the left and right invariant vector fields respectively,
corresponding to the Lie algebra generators, in the chart given by the
exponential map. The expressions for $\phi^{\tau}_{\rho}$,
$\hat{x}^{\phi}_{\rho}$ and $\hat{y}^{\phi}_{\rho}$ were derived many times
historically, most notably by Berezin in the geometric context [2] and the
corresponding expression for the star product studied in $C^{\infty}$-context
has been introduced by Gutt [10]; the supersymmetric generalization of the
formula in the algebraic context and in the dual language of coderivations
(rather than derivations and vector fields here) is in [20]; finally the
article [9] gives three proofs of the formula valid over any ring containing
rationals: a direct calculation, a proof in formal geometry over arbitrary
rings and an algebraic proof close in spirit to the approach in [20]. The
expression for ${\boldsymbol{\phi}}$ corresponds to the part of
Baker–Hausdorff series linear in one of the variables. The same star product
is often defined for exponentials via the full Baker–Hausdorff series [1, 11,
19].

From (4.3) it follows immediately that

$\displaystyle\hat{y}^{\phi}_{\alpha}=\hat{x}^{\phi}_{\beta}\mathcal{O}^{-1\beta}_{\alpha},$
(4.5)

and one can also prove (see, e.g., [18, Appendix 1])

$\displaystyle[\hat{x}^{\phi}_{\alpha},\hat{y}^{\phi}_{\beta}]=0.$ (4.6)

In the geometric interpretation, equation (4.6) means that all left invariant
vector fields on a Lie group commute with all right invariant vector fields.
Then $\hat{x}_{\rho}\mapsto\hat{x}_{\rho}^{\phi}$ extends to a unique algebra
map $(-)^{\phi}\colon U(\mathfrak{g})\to\hat{A}_{n}$. This realization map
$(-)^{\phi}$ is related to the symmetrization (PBW) isomorphism

$\displaystyle\xi\colon\ S(\mathfrak{g})\cong U(\mathfrak{g}),\qquad
w_{1}\cdots w_{r}\mapsto\frac{1}{r!}\Sigma_{\sigma\in\Sigma(r)}\hat{w}_{\sigma
1}\cdots\hat{w}_{\sigma r},\qquad w_{i}\in\mathfrak{g}\hookrightarrow
U(\mathfrak{g}),$

where $\Sigma(r)$ is the symmetric group on $r$ letters, in the sense that
$\xi^{-1}(u)=u^{\phi}(1)$ ($\phi$-realization of $u$ acting on $1$) and more
generally $u\cdot\xi(f)=u^{\phi}(f)$ for all $u\in U(\mathfrak{g}),f\in
S(\mathfrak{g})$, hence the star product $f\star
g:=\xi^{-1}(\xi(f)\cdot_{U(\mathfrak{g})}\xi(g))$ on $\mathfrak{g}$ may be
written as $f\star g=\xi(f)^{\phi}(g)$. Thus, in physics literature, our
choice of $\phi$ is said to correspond to the symmetric ordering. Analogously,
other coalgebra isomorphisms $S(\mathfrak{g})\cong U(\mathfrak{g})$ identical
on $\mathfrak{g}$ correspond to different choices of $\phi$ or to different
orderings, see [17]. Our $\phi$ corresponds to the symmetric ordering [9]. The
formula ${\boldsymbol{\phi}}(\hat{x}_{\mu})(\partial^{\nu})=\phi^{\nu}_{\mu}$
defines a linear map
${\boldsymbol{\phi}}(\hat{x}_{\mu})\colon\mathfrak{g}^{*}\to\hat{S}(\mathfrak{g}^{*})$
which extends by the chain rule to a (equally denoted) unique continuous
derivation
${\boldsymbol{\phi}}(\hat{x}_{\mu})\colon\hat{S}(\mathfrak{g}^{*})\to\hat{S}(\mathfrak{g}^{*})$.
Denote by $\operatorname{Derc}(\hat{S}(\mathfrak{g}^{*}))$ the Lie algebra of
all continuous derivations of $\hat{S}(\mathfrak{g}^{*})$. A key property of
$\phi$ is that the linear map
${\boldsymbol{\phi}}\colon\mathfrak{g}\to\operatorname{Derc}(\hat{S}(\mathfrak{g}^{*}))$
extending the rule
$\hat{x}_{\mu}\mapsto{\boldsymbol{\phi}}(\hat{x}_{\mu})\in\operatorname{Derc}(\hat{S}(\mathfrak{g}^{*}))$
is a Lie algebra homomorphism, hence it further extends to a unique right Hopf
action

$\displaystyle{\boldsymbol{\phi}}\colon\
U(\mathfrak{g})\to\operatorname{End}^{\mathrm{op}}\big{(}\hat{S}(\mathfrak{g}^{*})\big{)},$

which we denoted by the same symbol. The property that this action
${\boldsymbol{\phi}}$ is Hopf ensures that we can define the smash product
algebra

$\displaystyle
H_{\mathfrak{g}}=U(\mathfrak{g})\sharp_{{\boldsymbol{\phi}}}\hat{S}(\mathfrak{g}^{*})$

in the usual way: it is the tensor product of vector spaces
$U(\mathfrak{g})\otimes\hat{S}(\mathfrak{g}^{*})$ with the multiplication

$\displaystyle(u\sharp f)(v\sharp g)=\sum u\cdot
v_{(1)}\sharp{\boldsymbol{\phi}}(v_{(2)})(f)\cdot g\qquad\text{for}\quad
u,v\in U(\mathfrak{g}),\quad f,g\in\hat{S}(\mathfrak{g}^{*}),$

where, for the emphasis, the expression $u\sharp f$ denotes $u\otimes f$ as an
element in the smash product (whenever the meaning is clear the simpler
product notation $uf$ may be used as well). One can check that the algebra map
$H_{\mathfrak{g}}\to\hat{A}_{n}$, which on elements of
$U(\mathfrak{g})\sharp\mathbf{k}\hookrightarrow H_{\mathfrak{g}}$ agrees with
the realization $\hat{x}_{\mu}\mapsto\hat{x}^{\phi}_{\mu}$ and sends
$1\sharp\partial^{\beta}\in\mathbf{k}\otimes\hat{S}(\mathfrak{g}^{*})\subset
H_{\mathfrak{g}}$ to $\partial^{\beta}\in A_{n}$, is an algebra isomorphism;
hence from now on this algebra map $H_{\mathfrak{g}}\to\hat{A}_{n}$ is viewed
as an identification and sometimes called realization as well. We thus
identify $\hat{x}_{\mu}\in U(\mathfrak{g})$ and
$\hat{x}^{\phi}_{\mu}\in\hat{A}_{n}$ etc.

The isomorphism of coalgebras $\xi\colon S(\mathfrak{g})\to U(\mathfrak{g})$
induces the dual isomorphism of algebras $\xi^{T}\colon U(\mathfrak{g})^{*}\to
S(\mathfrak{g})^{*}\cong\hat{S}(\mathfrak{g}^{*})$ hence the map $\mu^{T}$
dual to the multiplication $\mu\colon U(\mathfrak{g})\otimes
U(\mathfrak{g})\to U(\mathfrak{g})$ can be identified with a deformed
comultiplication
$\Delta_{\hat{S}(\mathfrak{g}^{*})}=\big{(}\xi^{T}\otimes\xi^{T}\big{)}\circ\mu^{T}\colon\hat{S}(\mathfrak{g}^{*})\to\hat{S}(\mathfrak{g}^{*})\hat{\otimes}\hat{S}(\mathfrak{g}^{*})$
(the completion on $\otimes$ comes from (co)filtrations, see [18]). The source
map $\alpha\colon U(\mathfrak{g})\to H_{\mathfrak{g}}$ is given by $u\mapsto
u\sharp 1$. Introduce

$\displaystyle\hat{y}_{\mu}:=\hat{x}_{\lambda}\sharp\mathcal{O}^{\lambda}_{\mu}\in
U(\mathfrak{g})\sharp_{\phi}\hat{S}(\mathfrak{g}^{*})=H_{\mathfrak{g}}.$

In the realization $H_{\mathfrak{g}}\cong\hat{A}_{n}$, this expression becomes
$\hat{y}_{\mu}^{\phi}$. The rule $\hat{x}_{\mu}\mapsto\hat{y}_{\mu}$ extends
to a unique algebra map $\beta\colon U(\mathfrak{g})^{\mathrm{op}}\to
H_{\mathfrak{g}}$, the target map. Equation (4.6) implies that
$[\alpha(u),\beta(v)]=0$ for all $u,v\in U(\mathfrak{g})$. We also often
identify $u$ with its image $\alpha(u)$ and $P\in\hat{S}(\mathfrak{g}^{*})$
with $1\sharp P\in H_{\mathfrak{g}}$. Thus, $H_{\mathfrak{g}}$ becomes a
$U(\mathfrak{g})$-bimodule via $u.h.v=\alpha(u)\beta(v)h$. It has been shown
in [18] that, with the appropriate completions, $H_{\mathfrak{g}}$ is a
formally completed version of a Hopf $U(\mathfrak{g})$-algebroid with the
coproduct

$\displaystyle\Delta_{H}\colon\ H_{\mathfrak{g}}\to
H_{\mathfrak{g}}\hat{\otimes}_{U(\mathfrak{g})}H_{\mathfrak{g}},\qquad u\sharp
P\stackrel{{\scriptstyle\Delta_{H}}}{{\mapsto}}u\sharp
P_{(1)}\otimes_{U(\mathfrak{g})}P_{(2)},$
$\displaystyle\Delta_{\hat{S}(\mathfrak{g}^{*})}(P)=P_{(1)}\otimes
P_{(2)},\qquad u\in U(\mathfrak{g}),\qquad P\in\hat{S}(\mathfrak{g}^{*}).$

Notice that in the undeformed case (that is, when $\mathfrak{g}$ is Abelian)
the (topological) bialgebroid $H_{\mathfrak{g}}$ is some completion of the
Weyl algebra $A_{n}$ of regular differential operators on $\mathbf{k}^{n}$
where $A_{n}$ is viewed as a bialgebroid over the commutative base algebra
consisting of polynomial functions on $\mathbf{k}^{n}$. Hence, the undeformed
$H_{\mathfrak{g}}$ is very close to the bialgebroid of differential operators
over a smooth manifold.

Notation. In the case of deformed bialgebroid $H=H_{\mathfrak{g}}$, the action
$\blacktriangleright$ will be denoted

$\displaystyle\blacktriangleright_{\mathfrak{g}}\colon\
H_{\mathfrak{g}}\otimes U(\mathfrak{g})\to U(\mathfrak{g}),$

and in the undeformed case $\triangleright$. The action $\triangleright\colon
H_{\mathfrak{g}}\otimes S(\mathfrak{g})\to S(\mathfrak{g})$ is the “Fock
action”of $\hat{A}_{n}$ by differential operators on polynomials precomposed
by the realization isomorphism $H_{\mathfrak{g}}\cong\hat{A}_{n}$.

######  Proposition 4.1.

Given $u,v\in U(\mathfrak{g})$, $f,g\in S(\mathfrak{g})$ the following
formulas hold: $\xi(f)\triangleright g=f\star g$, $f\triangleright g=fg$,
$\partial^{\mu}\triangleright f=\partial^{\mu}(f)$,
$f\blacktriangleright_{\mathfrak{g}}1=\xi(f)$, $u\triangleright
1=\xi^{-1}(u)$, $u\blacktriangleright_{\mathfrak{g}}v=uv$,
$\partial^{\mu}\blacktriangleright_{\mathfrak{g}}(\hat{x}_{\nu}v)=\delta^{\mu}_{\nu}v+\hat{x}_{\nu}(\partial^{\mu}\blacktriangleright_{\mathfrak{g}}v)$,
$\hat{y}_{\mu}\blacktriangleright_{\mathfrak{g}}u=u\hat{x}_{\mu}$ and more
generally $\beta(v)\blacktriangleright_{\mathfrak{g}}u=uv$. The last two
formulas imply

$\displaystyle\hat{y}_{\mu}\triangleright f=f\star
x_{\mu},\qquad\beta(\xi(g))\triangleright f=f\star g,\qquad f,g\in
S(\mathfrak{g}).$ (4.7)

###### Proof.

These properties are at length discussed and proved in our articles [17, 18,
22]. The actions $\blacktriangleright_{\mathfrak{g}}$ and $\triangleright$ are
there denoted $\blacktriangleright$ and $\triangleright$ respectively. ∎

######  Lemma 4.2.

The right ideal $I_{U(\mathfrak{g})}$ $($compare (2.1)$)$ is generated by
$\hat{x}_{\rho}\otimes 1-\mathcal{O}^{\tau}_{\rho}\otimes\hat{x}_{\tau}$,
$\rho=1,\dots,n$.

###### Proof.

$I_{U(\mathfrak{g})}$ is generated by $\beta(u)\otimes 1-1\otimes\alpha(u)$
where $u\in U(\mathfrak{g})$ and it is enough to confine to the generators
$u=\hat{x}_{\mu}$ of $U(\mathfrak{g})$. Then $\beta(\hat{x}_{\mu})\otimes
1-1\otimes\alpha(\hat{x}_{\mu})=\hat{y}_{\mu}\otimes
1-1\otimes\hat{x}_{\mu}=\hat{x}_{\sigma}\mathcal{O}^{-1\sigma}_{\mu}\otimes
1-1\otimes\hat{x}_{\mu}=(\hat{x}_{\sigma}\otimes
1-\mathcal{O}^{\rho}_{\sigma}\otimes\hat{x}_{\rho})(\mathcal{O}^{-1\sigma}_{\mu}\otimes
1)$. Therefore $\beta(\hat{x}_{\mu})\otimes 1-1\otimes\alpha(\hat{x}_{\mu})$
is in the right ideal generated by elements $\hat{x}_{\sigma}\otimes
1-\mathcal{O}^{\rho}_{\sigma}\otimes\hat{x}_{\rho}$. Conversely,
$\hat{x}_{\sigma}\otimes
1-\mathcal{O}^{\rho}_{\sigma}\otimes\hat{x}_{\rho}=(\hat{y}_{\mu}\otimes
1-1\otimes\hat{x}_{\mu})(\mathcal{O}^{\mu}_{\sigma}\otimes 1)\in
I_{U(\mathfrak{g})}$. ∎

######  Theorem 4.3 ([18, Theorem 4]).

For the topological Heisenberg double
$H=H_{\mathfrak{g}}=U(\mathfrak{g})\hat{\sharp}\hat{S}(\mathfrak{g})$ of the
universal enveloping algebra $U(\mathfrak{g})$, considered as a bialgebroid
over $A=U(\mathfrak{g})$, the right ideal $I^{(k)}$ which is the kernel of the
canonical projection $H\otimes\cdots\otimes H\to
H\otimes_{A}\cdots\cdots\otimes_{A}H$ $(k$ factors, all tensor products
properly completed$)$ coincides with the right ideal $I^{\prime(k)}$
consisting of all elements $r$ in $H\otimes\cdots\otimes H$ such that

$\displaystyle\mu_{k-1}\left(r(\blacktriangleright_{\mathfrak{g}}\otimes\cdots\otimes\blacktriangleright_{\mathfrak{g}})(a_{1}\otimes\cdots\otimes
a_{k})\right)=0\qquad\text{for all}\quad a_{1},\dots,a_{k}\in H.$

For an arbitrary bialgebroid one can easily show that $I^{(k)}\subset
I^{\prime(k)}$, but the converse inclusion is not present in general.

######  Corollary 4.4.

Let $H=H_{\mathfrak{g}}$. For any $\mathcal{F}\in
H\hat{\otimes}_{S(\mathfrak{g})}H$ satisfying
$\mathcal{F}I_{0}=I_{U(\mathfrak{g})}$ and having an inverse
$\mathcal{F}^{-1}\in H\hat{\otimes}_{U(\mathfrak{g})}H$, the satisfaction of
the identity
$(\Delta_{H})_{\mathcal{F}}(h)=\mathcal{F}^{-1}\Delta_{H}(h)\mathcal{F}+I_{U(\mathfrak{g})}$
for all $h\in H$ implies the cocycle condition (3.1). In particular, this
holds for the undeformed case, where $H=\hat{A}_{n}$ and the coproduct is
$\Delta_{0}$ as in (3.3).

###### Proof.

This follows from the coassociativity $\Delta_{H}$ by the reasoning at the end
of the Remark 3.4 using that the nondegeneracy Theorem 4.3 holds in the
deformed case. Of course, the nondegeneracy is well known in the undeformed
case. For example, if $k=2$, then up to the issue of completions, it boils
down to the basic fact that the Weyl algebra as defined by generators and
relations is faithfully represented on the space of polynomials. ∎

In this paper, we need the corollary in the undeformed case, as we study below
the twists $\mathcal{F}_{l}$, $\mathcal{F}_{r}$ on the undeformed
$H=\hat{A}_{n}$. If $\mathcal{F}$ is a candidate for a twist on undeformed $H$
then the cocycle condition for $\mathcal{F}^{-1}$ (as a twist on deformed
$H_{\mathfrak{g}}$) may be derived using Theorem 4.3 in the nontrivial
deformed case. This is useful as the cocycle condition for some examples of
$\mathcal{F}$ or $\mathcal{F}^{-1}$ is studied in the literature without
previous knowledge of the cocycle condition for $\mathcal{F}$.

In [17], we proved another formula for a twist deforming the coproduct
$\Delta_{0}$ on $\hat{A}_{n}$ into the coproduct for $H_{\mathfrak{g}}$
(although without bialgebroid formalism) which is however less explicit:

$\displaystyle\mathcal{F}_{c}=\sum_{i_{1},\dots,i_{n}=0}^{\infty}\frac{x_{1}^{i_{1}}\cdots
x_{n}^{i_{n}}\otimes 1}{i_{1}!\cdots
i_{n}!}\prod_{\nu=1}^{n}\big{[}\big{(}\Delta_{\hat{S}(\mathfrak{g})}-\Delta_{0}\big{)}(\partial^{\nu})\big{]}^{i_{\nu}}$
(4.8)

as it is a series, where each summand involves
$\Delta_{\hat{S}(\mathfrak{g})}$ which is itself a series and not very
explicit. Symbolically, using the normal ordering ($x$-s to the left,
$\partial$-s to the right) operation ${:}\ {:}$ one can write this formula as
the normally ordered exponential

$\displaystyle\mathcal{F}_{c}={:}\exp\left(\sum_{\mu=1}^{n}(x_{\mu}\otimes
1)\big{(}\Delta_{\hat{S}(\mathfrak{g}^{*})}-\Delta_{0}\big{)}(\partial^{\mu})\right){:}.$

######  Remark 4.5 (warning on variants of completions).

The completions in [18] are, roughly speaking, with respect to the
cofiltrations on $U(\mathfrak{g})^{*}$ and $\hat{S}(\mathfrak{g}^{*})$ induced
by duality from the standard filtrations on $U(\mathfrak{g})$ and
$S(\mathfrak{g})$. This is essentially different from using the additional
formal variable $h$ and $h$-adic completions as in Xu’s work [25]. Naively, to
fit with his work, the Lie algebra generators (or equivalently the structure
constants) should be simply rescaled by the formal variable $h$. For many
simple purposes this gives an equivalent treatment to ours. The set of series
which formally converge in two variants differs however. For the main results
in the present article this is important. Namely, the twists
$\mathcal{F}_{l}$, $\mathcal{F}_{r}$ in Section 5 exist in both completions,
the formulas as products of two exponentials make sense in our formalism, but
these individual exponential factors do not exist in the $h$-adic completion,
even after rescaling. Indeed, $\exp(\partial^{\alpha}\otimes x_{\alpha})$ does
not involve a small parameter (the reason is that $\partial^{\alpha}$ and
$x_{\alpha}$ are dual and no rescaling could make their tensor product small)
and is in fact related to an infinite-dimensional version of the canonical
element, while (due cancellations in the expansion) the entire twist
$\mathcal{F}_{l}$ equals $1$ plus a series of corrections involving the small
parameter. Thus, unlike the exponential factors, the final result
$\mathcal{F}_{l}$ does exist in both formalisms and hence can be interpreted
as defining a deformation quantization in the sense used in Theorem 3.9.

######  Remark 4.6.

The smash product algebra
$U(\mathfrak{g})\sharp_{{\boldsymbol{\phi}}}\hat{S}(\mathfrak{g}^{*})$ can be
equipped with another bialgebroid structure, namely over the commutative base
algebra $\hat{S}(\mathfrak{g}^{*})$ and with the comultiplication
$\Delta^{\prime}$ for which $\Delta^{\prime}(u\sharp P)=(u_{(1)}\sharp
1)\otimes_{\hat{S}(\mathfrak{g}^{*})}(u_{(2)}\sharp P)$ for $u\in
U(\mathfrak{g})$ and $P\in\hat{S}(\mathfrak{g}^{*})$, where $u\mapsto
u_{(1)}\otimes u_{(2)}$ denotes the (standard) cocommutative coproduct of
$U(\mathfrak{g})$. While our coproduct $\Delta$ takes values in a completed
product over the noncommutative base, the coproduct $\Delta^{\prime}$ is
algebraic, taking values in the ordinary tensor product over the commutative
base. Such scalar extensions over a commutative base were known much before
[21, pp. 117–118]. This bialgebroid does not fit the physical interpretation
which we intended. The coproduct on $\hat{S}(\mathfrak{g}^{*})$ in our
bialgebroid structure in this paper is dual to the algebra structure on
$U(\mathfrak{g})$ manifest in both algebroids and the coproduct on
$U(\mathfrak{g})$ in the other bialgebroid structure is dual to the algebra
structure on $\hat{S}(\mathfrak{g}^{*})$ manifest in both algebroids. The two
entire bialgebroids are mutually not dual in some sense standard for
bialgebroids, although their factors $U(\mathfrak{g})$ and
$\hat{S}(\mathfrak{g}^{*})$ are involved in (topological) Hopf algebra
dualities.

##  5 New twist

In this section, we show that the two expressions $\mathcal{F}_{l}$ and
$\mathcal{F}_{r}$ define (the same) twist of a completed Heisenberg double and
that the twisted bialgebroid is $H_{\mathfrak{g}}$ from the previous sections.

######  Theorem 5.1.

  1. $(i)$

In symmetric ordering, the deformed coproduct
$\Delta_{\hat{S}(\mathfrak{g}^{*})}$ on $\hat{S}(\mathfrak{g}^{*})\cong
U(\mathfrak{g})^{*}$ is given by

$\displaystyle\Delta_{\hat{S}(\mathfrak{g}^{*})}\partial^{\mu}=1\otimes\partial^{\mu}+\partial^{\alpha}\otimes[\partial^{\mu},\hat{x}_{\alpha}]+\tfrac{1}{2}\partial^{\alpha}\partial^{\beta}\otimes[[\partial^{\mu},\hat{x}_{\alpha}],\hat{x}_{\beta}]+\cdots,$

where the right-hand side, though an element of
$\hat{S}(\mathfrak{g}^{*})\hat{\otimes}\hat{S}(\mathfrak{g}^{*})$, is
calculated in the bigger algebra $H\hat{\otimes}H$ $($the completed tensor
product over the ground field$)$. The same equation written in a symbolic form
is

$\displaystyle\Delta_{\hat{S}(\mathfrak{g}^{*})}\partial^{\mu}=\exp(\partial^{\alpha}\otimes\operatorname{ad}(-\hat{x}_{\alpha}))(1\otimes\partial^{\mu})=\exp(\operatorname{ad}(-\partial^{\alpha}\otimes\hat{x}_{\alpha}))(1\otimes\partial^{\mu}).$
(5.1)

$($The rightmost equality follows by noting that $[\partial^{\alpha},1]=0$.$)$

  2. $(ii)$

More generally, for $P\in\hat{S}(\mathfrak{g}^{*})$ we have

$\displaystyle\Delta_{\hat{S}(\mathfrak{g}^{*})}(P)=\exp(\partial^{\alpha}\otimes\operatorname{ad}(-\hat{x}_{\alpha}))(1\otimes
P)=\exp(\operatorname{ad}(-\partial^{\alpha}\otimes\hat{x}_{\alpha}))(1\otimes
P).$

###### Proof.

Part (i) is proven in [17] (and used in [22]).

(ii) To extend the identity to the formal power series just notice that the
conjugation is a homomorphism of completed algebras
$\hat{H}\otimes\hat{H}\to\hat{H}$(the usual non-completed tensor product of
completions). More explicitly, the deformed coproduct of the $n$-th order
monomial in $\partial$-s has all summands in order $n$ or higher. If in a
formal series we replace each monomial with a formal series with nondecreased
minimal order then we obtain only finitely many summands of each finite order
hence again a formal series. ∎

######  Corollary 5.2.

In symmetric ordering, the deformed coproduct
$\Delta_{\hat{S}(\mathfrak{g}^{*})}$ on $\hat{S}(\mathfrak{g}^{*})\cong
U(\mathfrak{g})^{*}$ is also given by

$\displaystyle\Delta_{\hat{S}(\mathfrak{g}^{*})}(\partial^{\mu})=\exp(\mathrm{ad}(\hat{y}_{\alpha}\otimes\partial^{\alpha}))(\partial^{\mu}\otimes
1)$ (5.2)

and more generally
$\Delta_{\hat{S}(\mathfrak{g}^{*})}(P)=\exp(\mathrm{ad}(\hat{y}_{\alpha}\otimes\partial^{\alpha}))(P\otimes
1)$ for all $P\in\hat{S}(\mathfrak{g}^{*})$.

###### Proof.

It is known [17] that the deformed coproduct
$\Delta_{\hat{S}(\mathfrak{g}^{*})}(\partial^{\mu})\in\hat{S}(\mathfrak{g}^{*})\hat{\otimes}\hat{S}(\mathfrak{g}^{*})$
has the same symmetries as the Hausdorff formula $H(Z,W)=-H(-W,-Z)$; thus if
we interchange the tensor factors in the coproduct and multiply each partial
derivative with $-1$ then we get the same coproduct with the overall minus
sign. This rule of course applies only when the coproduct is written
manifestly in terms of tensor products of formal power series in partial
derivatives. Changing the sign of each partial derivative in the realizations
$\hat{x}_{\alpha}$ manifest in (5.1) is equivalent to the change of $\phi$ to
$\tilde{\phi}$ (because partial derivatives are contracted to the structure
constants so it is the same as change of the sign in the structure constants),
that is $\hat{x}_{\alpha}$ to $\hat{y}_{\alpha}$. However, the commutators
like $[\partial^{\mu},\hat{x}_{\alpha}]$ when calculated have a different
number of partials in each monomial than the raw expression. Indeed, one
partial derivative and one power of $x$ (recall that
$\hat{x}_{\rho}=x_{\sigma}\phi^{\sigma}_{\rho}$ where $\phi^{\sigma}_{\rho}$
involves partial derivatives) drop out when commuting; similarly in each of
the further commutators one power of $x$ and one of the partial derivatives
within the $\tilde{\phi}$-s drops out. As the number of commutators in each of
the left tensor factors in (5.1) is by one less than the number of partial
derivatives in the raw tensor product expression, together with overall minus
sign it amounts to no change in sign besides the change accounted in
$\hat{x}_{\mu}\mapsto\hat{y}_{\mu}$. Thus

$\displaystyle\Delta_{\hat{S}(\mathfrak{g}^{*})}\partial^{\mu}=\partial^{\mu}\otimes
1+[\partial^{\mu},\hat{y}_{\alpha}]\otimes\partial^{\alpha}+\frac{1}{2}[[\partial^{\mu},\hat{y}_{\alpha}],\hat{y}_{\beta}]\otimes\partial^{\alpha}\partial^{\beta}+\cdots.$

The proof that the result extends to general $P\in\hat{S}(\mathfrak{g}^{*})$
is completely analogous to the proof of part (ii) in Theorem 5.1. ∎

######  Definition 5.3.

Denote

$\displaystyle\mathcal{F}_{l}=\exp(-\partial^{\rho}\otimes
x_{\rho})\exp(\partial^{\sigma}\otimes\hat{x}_{\sigma}),$ (5.3)
$\displaystyle\mathcal{F}_{R}=\exp(-x_{\rho}\otimes\partial^{\rho})\exp(\hat{y}_{\sigma}\otimes\partial^{\sigma})$
(5.4)

understood as elements in $H\hat{\otimes}_{S(\mathfrak{g})}H$ with
$\mathcal{F}^{-1}_{L},\mathcal{F}^{-1}_{R}\in
H\hat{\otimes}_{U(\mathfrak{g})}H$. The right-hand sides of the formulas
(5.3), (5.4) understood as elements in $H\hat{\otimes}H$ define
$\tilde{\mathcal{F}}_{l},\tilde{\mathcal{F}}_{r}\in H\hat{\otimes}H$ with
explicit inverses
$\tilde{\mathcal{F}}_{l}^{-1}=\exp(-\partial^{\sigma}\otimes\hat{x}_{\sigma})\exp(\partial^{\rho}\otimes
x_{\rho})$,
$\tilde{\mathcal{F}}_{r}^{-1}=\exp(-\hat{y}_{\sigma}\otimes\partial^{\sigma})\exp(x_{\rho}\otimes\partial^{\rho})$.

Clearly, $\mathcal{F}_{l}=\tilde{\mathcal{F}}_{l}+I_{0}$,
$\mathcal{F}_{r}=\tilde{\mathcal{F}}_{r}+I_{0}$. We define also
$\mathcal{F}_{l}^{-1}:=\tilde{\mathcal{F}}_{l}^{-1}+I_{U(\mathfrak{g})}$,
$\mathcal{F}_{r}^{-1}:=\tilde{\mathcal{F}}_{r}^{-1}+I_{U(\mathfrak{g})}$ (see
(2.1)), but at this point we can not yet claim that they are twist inverses in
the sense of Remark (3.3) as the expressions
$\mathcal{F}_{l}\mathcal{F}_{l}^{-1}$ etc. should be well defined what is
proven only below.

######  Lemma 5.4.

$\Delta_{\hat{S}(\mathfrak{g}^{*})}\partial^{\mu}=\tilde{\mathcal{F}}_{l}^{-1}\Delta_{0}(\partial^{\mu})\tilde{\mathcal{F}}_{l}=\tilde{\mathcal{F}}_{r}^{-1}\Delta_{0}(\partial^{\mu})\tilde{\mathcal{F}}_{r}$
for $\mu=1,\dots,n$. More generally, for any $P\in\hat{S}(\mathfrak{g}^{*})$
$($formal power series in $\partial^{1},\dots,\partial^{n})$,
$\Delta_{\hat{S}}(\mathfrak{g}^{*})P=\tilde{\mathcal{F}}_{l}^{-1}\Delta_{0}(P)\tilde{\mathcal{F}}_{l}=\tilde{\mathcal{F}}_{r}^{-1}\Delta_{0}(P)\tilde{\mathcal{F}}_{r}$.

###### Proof.

By the Hadamard’s formula $\exp(A)B\exp(-A)=\exp(\operatorname{ad}A)(B)$,
(5.1) and (5.2) read

$\displaystyle\Delta_{\hat{S}(\mathfrak{g}^{*})}\partial^{\mu}=\exp(-\partial^{\rho}\otimes\hat{x}_{\rho})(1\otimes\partial^{\mu})\exp(\partial^{\sigma}\otimes\hat{x}_{\sigma}),$
(5.5)
$\displaystyle\Delta_{\hat{S}(\mathfrak{g}^{*})}\partial^{\mu}=\exp(-\hat{y}_{\rho}\otimes\partial^{\rho})(\partial^{\mu}\otimes
1)\exp(\hat{y}_{\sigma}\otimes\partial^{\sigma}).$ (5.6)

In particular, in the undeformed case when $C_{\mu\nu}^{\lambda}=0$ and
$\hat{x}_{\alpha}$, $x_{\alpha}$ and $\hat{y}_{\alpha}$ coincide, we obtain
the formulas for the undeformed coproduct $\Delta_{0}$ (which can also easily
be checked directly)

$\displaystyle\Delta_{0}\partial^{\mu}=\exp(-\partial^{\alpha}\otimes
x_{\alpha})(1\otimes\partial^{\mu})\exp(\partial^{\alpha}\otimes x_{\alpha}),$
(5.7)
$\displaystyle\Delta_{0}\partial^{\mu}=\exp(-x_{\alpha}\otimes\partial^{\alpha})(\partial^{\mu}\otimes
1)\exp(x_{\alpha}\otimes\partial^{\alpha}).$ (5.8)

Comparing the formulas for the deformed and for the undeformed case we obtain
new formulas relating $\Delta_{0}$ to
$\Delta_{\hat{S}(\mathfrak{g}^{*})}(\mathfrak{g}^{*})$. Indeed, comparing
(5.5) and (5.7) we obtain

$\displaystyle\Delta_{\hat{S}(\mathfrak{g}^{*})}(\partial^{\mu})=\tilde{\mathcal{F}}^{-1}_{L}\Delta_{0}(\partial^{\mu})\tilde{\mathcal{F}}_{l},$

and similarly comparing (5.6) to (5.8) we obtain

$\displaystyle\Delta_{\hat{S}(\mathfrak{g}^{*})}(\partial^{\mu})=\tilde{\mathcal{F}}^{-1}_{R}\Delta_{0}(\partial^{\mu})\tilde{\mathcal{F}}_{r}.$

To extend the identities to the formal power series proceed as in the proof of
part (ii) to Theorem 5.1.

We would like to have the same identities in
$H\hat{\otimes}_{U(\mathfrak{g})}H$, with $\mathcal{F}$ instead of
$\tilde{\mathcal{F}}$, but for this the calculation should not depend on a
representative, that is $\mathcal{F}^{-1}_{L}\cdot I_{0}\subset
I_{U(\mathfrak{g})}$, which is proven below in Proposition 5.8. ∎

######  Definition 5.5.

Define the map of left $U(\mathfrak{g})$-modules $\tilde{\Delta}\colon H\to
H\hat{\otimes}H$ (completed tensor product over the field) by
$\tilde{\Delta}(u\sharp P)=u\Delta_{\hat{S}(\mathfrak{g}^{*})}(P)$ (the
inclusion
$\hat{S}(\mathfrak{g}^{*})\hat{\otimes}\hat{S}(\mathfrak{g}^{*})\hookrightarrow
H\hat{\otimes}H$ understood).

######  Remark 5.6.

Clearly $\tilde{\Delta}(u\sharp P)+I_{U(\mathfrak{g})}=\Delta_{H}(u\sharp P)$,
but unlike $\Delta_{H}$ and $\Delta_{\hat{S}(\mathfrak{g}^{*})}$ the map
$\tilde{\Delta}$ is not multiplicative, that is
$\tilde{\Delta}(h)\tilde{\Delta}(h^{\prime})\neq\tilde{\Delta}(hh^{\prime})$
for general $h,h^{\prime}\in H$. We use this map to compute more precisely
with $\tilde{\mathcal{F}}_{l},\tilde{\mathcal{F}}_{r}$.

######  Lemma 5.7.

  1. $(i)$

For $\mu=1,\dots,n$ we have

$\displaystyle\tilde{\mathcal{F}}_{l}^{-1}(x_{\mu}\otimes
1)\tilde{\mathcal{F}}_{l}=\tilde{\Delta}(x_{\mu})+\hat{x}_{\tau}\big{(}\phi^{-1}\big{)}^{\tau}_{\mu}\otimes
1-\big{(}\tilde{\phi}^{-1}\big{)}^{\tau}_{\mu}\otimes\hat{x}_{\tau}$
$\displaystyle\hphantom{\tilde{\mathcal{F}}_{l}^{-1}(x_{\mu}\otimes
1)\tilde{\mathcal{F}}_{l}=}{}+\big{(}\mathcal{O}^{\sigma}_{\lambda}\otimes\hat{x}_{\sigma}-\hat{x}_{\lambda}\otimes
1\big{)}\tilde{\Delta}\big{(}\big{(}\phi^{-1}\big{)}^{\lambda}_{\mu}),$ (5.9)
$\displaystyle\tilde{\mathcal{F}}_{l}^{-1}(1\otimes
x_{\mu})\tilde{\mathcal{F}}_{l}=\tilde{\Delta}(x_{\mu})+\big{(}\mathcal{O}^{\sigma}_{\lambda}\otimes\hat{x}_{\sigma}-\hat{x}_{\lambda}\otimes
1\big{)}\tilde{\Delta}\big{(}\big{(}\phi^{-1}\big{)}^{\lambda}_{\mu}\big{)},$
(5.10) $\displaystyle\tilde{\mathcal{F}}_{r}^{-1}(1\otimes
x_{\mu})\tilde{\mathcal{F}}_{r}=\tilde{\Delta}(x_{\mu})+1\otimes\hat{x}_{\mu}-\hat{y}_{\tau}\otimes\big{(}\phi^{-1}\big{)}^{\tau}_{\mu},$
(5.11) $\displaystyle\tilde{\mathcal{F}}_{r}^{-1}(x_{\mu}\otimes
1)\tilde{\mathcal{F}}_{r}=\tilde{\Delta}(x_{\mu}).$ (5.12)

  2. $(ii)$

$x_{\mu}\otimes
1-\big{(}\tilde{\phi}^{-1}\big{)}^{\tau}_{\mu}\otimes\hat{x}_{\tau}\in
I_{U(\mathfrak{g})}$,
$\mathcal{O}^{\sigma}_{\lambda}\otimes\hat{x}_{\sigma}-\hat{x}_{\lambda}\otimes
1\in I_{U(\mathfrak{g})}$ and
$1\otimes\hat{x}_{\mu}-\hat{y}_{\tau}\otimes\big{(}\phi^{-1}\big{)}^{\tau}_{\mu}\in
I_{U(\mathfrak{g})}$.

  3. $(iii)$

$\Delta_{H}x_{\mu}=\tilde{\mathcal{F}}_{l}^{-1}(x_{\mu}\otimes
1)\tilde{\mathcal{F}}_{l}+I_{U(\mathfrak{g})}=\tilde{\mathcal{F}}_{r}^{-1}(1\otimes
x_{\mu})\tilde{\mathcal{F}}_{r}+I_{U(\mathfrak{g})}$, where
$I_{U(\mathfrak{g})}$ is the right ideal generated by $\beta(u)\otimes
1-1\otimes\alpha(u)$ for all $u\in U(\mathfrak{g})$.

###### Proof.

We first compute $\tilde{\mathcal{F}}^{-1}_{l}(x_{\mu}\otimes
1)\tilde{\mathcal{F}}_{l}$ and $\tilde{\mathcal{F}}^{-1}_{l}(1\otimes
x_{\mu})\tilde{\mathcal{F}}_{l}$. In order to conjugate with
$\tilde{\mathcal{F}}^{-1}_{L}$ which is a product of exponentials, we first
conjugate with the “inner” exponential

$\displaystyle\exp(\partial^{\rho}\otimes x_{\rho})(x_{\mu}\otimes
1)\exp(-\partial^{\sigma}\otimes x_{\sigma})=x_{\mu}\otimes 1+1\otimes
x_{\mu},$ (5.13) $\displaystyle\exp(\partial^{\rho}\otimes x_{\rho})(1\otimes
x_{\mu})\exp(-\partial^{\sigma}\otimes x_{\sigma})=1\otimes x_{\mu}.$ (5.14)

Now we need to apply outer exponentials.

By induction on $k=0,1,2,\ldots$ one checks that (in notation (4.1))

$\displaystyle\operatorname{ad}^{k}(\partial^{\rho}\otimes\hat{x}_{\rho})(1\otimes\hat{x}_{\mu})=[(-\mathcal{C})^{k}]^{\tau}_{\mu}\otimes\hat{x}_{\tau}.$
(5.15)

Hadamard’s formula, identity
$\operatorname{ad}(\partial^{\rho}\otimes\hat{x}_{\rho})(x_{\mu}\otimes
1)=1\otimes\hat{x}_{\mu}$ and equation (5.15) imply

$\displaystyle\exp(-\partial^{\sigma}\otimes\hat{x}_{\sigma})(x_{\mu}\otimes
1)\exp(\partial^{\rho}\otimes\hat{x}_{\rho})=x_{\mu}\otimes
1-\sum_{k=1}^{\infty}\frac{(\mathcal{C}^{k-1})^{\tau}_{\mu}}{k!}\otimes\hat{x}_{\tau}$
$\displaystyle\hphantom{\exp(-\partial^{\sigma}\otimes\hat{x}_{\sigma})(x_{\mu}\otimes
1)\exp(\partial^{\rho}\otimes\hat{x}_{\rho})}{}=x_{\mu}\otimes
1-\left(\frac{e^{\mathcal{C}}-1}{\mathcal{C}}\right)^{\tau}_{\mu}\otimes\hat{x}_{\tau},$

therefore by (4.2)

$\displaystyle\exp(-\partial^{\sigma}\otimes\hat{x}_{\sigma})(x_{\mu}\otimes
1)\exp(\partial^{\rho}\otimes\hat{x}_{\rho})=x_{\mu}\otimes
1-\big{(}\tilde{\phi}^{-1}\big{)}^{\tau}_{\mu}\otimes\hat{x}_{\tau}.$ (5.16)

Now conjugate $1\otimes
x_{\mu}=1\otimes\hat{y}_{\tau}\mathcal{O}^{\tau}_{\sigma}(\phi^{-1})^{\sigma}_{\mu}$
with $\exp(-\partial^{\sigma}\otimes\hat{x}_{\sigma})$, using

$\displaystyle
1\otimes\hat{y}_{\tau}\mathcal{O}^{\tau}_{\sigma}\big{(}\phi^{-1}\big{)}^{\sigma}_{\mu}=(1\otimes\hat{y}_{\tau})\exp(\partial^{\nu}\otimes\hat{x}_{\nu})\exp(-\partial^{\lambda}\otimes\hat{x}_{\lambda})\big{(}1\otimes\mathcal{O}^{\tau}_{\sigma}\big{(}\phi^{-1}\big{)}^{\sigma}_{\mu}\big{)}$

and $[\hat{x}_{\sigma},\hat{y}_{\tau}]=0$, therefore obtaining

$\displaystyle\exp(-\partial^{\sigma}\otimes\hat{x}_{\sigma})\big{(}1\otimes\hat{y}_{\tau}\mathcal{O}^{\tau}_{\lambda}\big{(}\phi^{-1}\big{)}^{\lambda}_{\mu}\big{)}\exp(\partial^{\rho}\otimes\hat{x}_{\rho})$
$\displaystyle\qquad{}=(1\otimes\hat{y}_{\tau})\Delta_{\hat{S}(\mathfrak{g}^{*})}(\mathcal{O}^{\tau}_{\lambda})\Delta_{\hat{S}(\mathfrak{g}^{*})}\big{(}\big{(}\phi^{-1}\big{)}^{\lambda}_{\mu}\big{)}=(\mathcal{O}^{\sigma}_{\lambda}\otimes\hat{x}_{\sigma})\Delta_{\hat{S}(\mathfrak{g}^{*})}\big{(}\big{(}\phi^{-1}\big{)}^{\lambda}_{\mu}\big{)},$
$\displaystyle\exp(-\partial^{\sigma}\otimes\hat{x}_{\sigma})\big{(}1\otimes\hat{y}_{\tau}\mathcal{O}^{\tau}_{\lambda}\big{(}\phi^{-1}\big{)}^{\lambda}_{\mu}\big{)}\exp(\partial^{\rho}\otimes\hat{x}_{\rho})$
$\displaystyle\qquad{}=\tilde{\Delta}(x_{\mu})+(\mathcal{O}^{\sigma}_{\lambda}\otimes\hat{x}_{\sigma}-\hat{x}_{\lambda}\otimes
1)\tilde{\Delta}\big{(}\big{(}\phi^{-1}\big{)}^{\lambda}_{\mu}\big{)},$ (5.17)

where we also used the known fact [18] that
$\Delta_{S(\mathfrak{g}^{*})}(\mathcal{O}^{\tau}_{\lambda})=\mathcal{O}_{\lambda}^{\sigma}\otimes\mathcal{O}^{\tau}_{\sigma}$,
Theorem 5.1(ii) for $P=\mathcal{O}^{\tau}_{\sigma}$ and
$P=(\phi^{-1})^{\sigma}_{\mu}$ and the fact that
$(\Delta_{H})|_{\hat{S}(\mathfrak{g}^{*})}$ agrees with (deformed)
$\Delta_{\hat{S}(\mathfrak{g}^{*})}$ followed by the inclusion into
$H\hat{\otimes}_{U(\mathfrak{g})}H$. According to (5.13) we obtain (5.9) by
adding (5.16) and (5) and similarly according to (5.14) we obtain (5.10) from
(5).

To compute $\tilde{\mathcal{F}}^{-1}_{r}(1\otimes
x_{\mu})\tilde{\mathcal{F}}_{r}$ and
$\tilde{\mathcal{F}}^{-1}_{r}(x_{\mu}\otimes 1)\tilde{\mathcal{F}}_{r}$ we
firstly conjugate with $\exp(x_{\rho}\otimes\partial^{\rho})$,

$\displaystyle\exp(x_{\rho}\otimes\partial^{\rho})(1\otimes
x_{\mu})\exp(-x_{\sigma}\otimes\partial^{\sigma})=1\otimes
x_{\mu}+x_{\mu}\otimes 1,$
$\displaystyle\exp(x_{\rho}\otimes\partial^{\rho})(x_{\mu}\otimes
1)\exp(-x_{\sigma}\otimes\partial^{\sigma})=x_{\mu}\otimes 1.$ (5.18)

Then we need to conjugate each summand at the right-hand side with the outer
exponential $\exp(-\hat{y}_{\sigma}\otimes\partial^{\sigma})$. By induction on
$k=1,2,\ldots$ one shows the analogue of (5.15) (or just notice that
$\hat{x}\mapsto\hat{y}$ by changing the sign of the structure constants and
interchange the tensor factors):

$\displaystyle\operatorname{ad}^{k}(\hat{y}_{\rho}\otimes\partial^{\rho})(\hat{y}_{\mu}\otimes
1)=\hat{y}_{\tau}\otimes\big{(}\mathcal{C}^{k}\big{)}^{\tau}_{\mu}.$

Along with Hadamard’s formula and
$\operatorname{ad}(\hat{y}_{\rho}\otimes\partial^{\rho})(1\otimes
x_{\mu})=1\otimes\hat{y}_{\mu}$ this implies

$\displaystyle\begin{split}&\exp(-\hat{y}_{\sigma}\otimes\partial^{\sigma})(1\otimes
x_{\mu})\exp(\hat{y}_{\rho}\otimes\partial^{\rho})=1\otimes
x_{\mu}-\sum_{k=1}^{\infty}\hat{y}_{\tau}\otimes\frac{\big{(}(-\mathcal{C})^{k-1}\big{)}^{\tau}_{\mu}}{k!}\otimes\hat{x}_{\tau}\\\
&\hphantom{\exp(-\hat{y}_{\sigma}\otimes\partial^{\sigma})(1\otimes
x_{\mu})\exp(\hat{y}_{\rho}\otimes\partial^{\rho})}{}=1\otimes
x_{\mu}-\hat{y}_{\tau}\otimes\left(\frac{e^{-\mathcal{C}}-1}{-\mathcal{C}}\right)^{\tau}_{\mu},\end{split}$

therefore by (4.2)

$\displaystyle\exp(-\hat{y}_{\sigma}\otimes\partial^{\sigma})(1\otimes
x_{\mu})\exp(\hat{y}_{\rho}\otimes\partial^{\rho})=1\otimes
x_{\mu}-\hat{y}_{\tau}\otimes\big{(}\phi^{-1}\big{)}^{\tau}_{\mu}.$ (5.19)

Now we conjugate the second summand $x_{\mu}\otimes
1=\hat{x}_{\lambda}(\phi^{-1})^{\lambda}_{\mu}\otimes 1$ on the right-hand
side of (5.18) with $\exp(-\hat{y}_{\sigma}\otimes\partial^{\sigma})$, using
$[\hat{x}_{\sigma},\hat{y}_{\tau}]=0$, therefore obtaining

$\displaystyle\exp(-\hat{y}_{\sigma}\otimes\partial^{\sigma})\big{(}\hat{x}_{\lambda}\big{(}\phi^{-1}\big{)}^{\lambda}_{\mu}\otimes
1\big{)}\exp(\hat{y}_{\rho}\otimes\partial^{\rho})=(\hat{x}_{\lambda}\otimes_{U(\mathfrak{g})}1)\Delta_{\hat{S}(\mathfrak{g}^{*})}\big{(}\big{(}\phi^{-1}\big{)}^{\lambda}_{\mu}\big{)}=\tilde{\Delta}(x_{\mu}),\\!\\!\\!$
(5.20)

where we also used Corollary 5.2 for
$P=\big{(}\phi^{-1}\big{)}^{\sigma}_{\mu}$. Adding (5.19) and (5.20) we obtain
(5.11) and similarly (5.20) alone gives (5.12).

(ii) The formula (4.4) gives
$x_{\mu}=\hat{x}_{\sigma}\big{(}\phi^{-1}\big{)}^{\sigma}_{\mu}$ and (4.5)
gives $\tilde{\phi}^{-1}=\mathcal{O}\phi^{-1}$, while the right ideal
$I_{U(\mathfrak{g})}$ is by Lemma 4.2 generated by elements of the form
$\hat{x}_{\rho}\otimes 1-\mathcal{O}^{\tau}_{\rho}\otimes\hat{x}_{\tau}$,
hence

$\displaystyle x_{\mu}\otimes
1-\big{(}\tilde{\phi}^{-1}\big{)}^{\tau}_{\mu}\otimes\hat{x}_{\tau}=(\hat{x}_{\rho}\otimes
1-\mathcal{O}^{\tau}_{\rho}\otimes\hat{x}_{\tau})\big{(}\big{(}\phi^{-1}\big{)}^{\rho}_{\mu}\otimes
1\big{)}\in I_{U(\mathfrak{g})}.$

Similarly, $1\otimes
x_{\mu}-\hat{y}_{\tau}\otimes\big{(}\phi^{-1}\big{)}^{\tau}_{\mu}=(1\otimes\hat{x}_{\tau}-\hat{y}_{\tau}\otimes
1)\big{(}1\otimes\big{(}\phi^{-1}\big{)}^{\tau}_{\mu}\big{)}\in
I_{U(\mathfrak{g})}$.

Part (iii) follows from (i) and (ii) (note Remark 5.6). ∎

Finally, one proves that the undeformed right ideal $I_{0}$ generated by
$x_{\mu}\otimes 1-1\otimes x_{\mu}$ after twist ends in the deformed right
ideal $I_{U(\mathfrak{g})}$. In fact,

######  Proposition 5.8.

$\displaystyle\tilde{\mathcal{F}}_{l}^{-1}(x_{\mu}\otimes 1-1\otimes
x_{\mu})\tilde{\mathcal{F}}_{l}=\hat{x}_{\tau}\big{(}\phi^{-1}\big{)}^{\tau}_{\mu}\otimes
1-\big{(}\tilde{\phi}^{-1}\big{)}^{\tau}_{\mu}\otimes\hat{x}_{\tau}\in
I_{U(\mathfrak{g})},$
$\displaystyle\tilde{\mathcal{F}}_{r}^{-1}(x_{\mu}\otimes 1-1\otimes
x_{\mu})\tilde{\mathcal{F}}_{r}=\hat{y}_{\tau}\otimes\big{(}\phi^{-1}\big{)}^{\tau}_{\mu}-1\otimes\hat{x}_{\mu}\in
I_{U(\mathfrak{g})},$
$\displaystyle\tilde{\mathcal{F}}_{l}^{-1}(x_{\mu}\otimes 1-1\otimes
x_{\mu})\in
I_{U(\mathfrak{g})},\qquad\tilde{\mathcal{F}}_{r}^{-1}(x_{\mu}\otimes
1-1\otimes x_{\mu})\in I_{U(\mathfrak{g})},$
$\displaystyle\mathcal{F}_{l}^{-1}(x_{\mu}\otimes 1-1\otimes
x_{\mu})=\mathcal{F}_{r}^{-1}(x_{\mu}\otimes 1-1\otimes
x_{\mu})=0+I_{U(\mathfrak{g})}\in H\hat{\otimes}_{U(\mathfrak{g})}H,$
$\displaystyle\tilde{\mathcal{F}}_{l}^{-1}\cdot
I_{0}=I_{U(\mathfrak{g})}=\tilde{\mathcal{F}}_{R}^{-1}\cdot
I_{0},\qquad\tilde{\mathcal{F}}_{l}\cdot
I_{U(\mathfrak{g})}=I_{0}=\tilde{\mathcal{F}}_{R}\cdot I_{U(\mathfrak{g})},$
$\displaystyle\mathcal{F}_{l}\mathcal{F}_{l}^{-1}=1\otimes
1+I_{0},\qquad\mathcal{F}_{l}^{-1}\mathcal{F}_{l}=1\otimes
1+I_{U(\mathfrak{g})},$
$\displaystyle\mathcal{F}_{r}\mathcal{F}_{r}^{-1}=1\otimes
1+I_{0},\qquad\mathcal{F}_{r}^{-1}\mathcal{F}_{r}=1\otimes
1+I_{U(\mathfrak{g})}.$

###### Proof.

The first line follows by subtracting (5.10) from (5.9) and the second line by
subtracting (5.11) from (5.12). The third line follows by multiplying the
first by $\tilde{\mathcal{F}}_{r}^{-1}$ and multiplying the second line by
$\tilde{\mathcal{F}}_{r}^{-1}$. The fourth line is clearly just the
restatement of the third in the quotient.

$\tilde{\mathcal{F}}_{l}^{-1}I_{0}\subset I_{U(\mathfrak{g})}$ follows from
the first line by noticing that the elements $x_{\mu}\otimes 1-1\otimes
x_{\mu}$, $\mu=1,\dots,n$ generate $I_{0}$ and similarly
$\tilde{\mathcal{F}}_{r}^{-1}I_{0}\subset I_{U(\mathfrak{g})}$ follows from
the second line.

To show $\tilde{\mathcal{F}}_{l}I_{U(\mathfrak{g})}\subset I_{0}$ multiply the
first line with $\tilde{\mathcal{F}_{l}}$ from the left to obtain

$\displaystyle\tilde{\mathcal{F}_{l}}\big{(}x_{\mu}\otimes
1-\big{(}\tilde{\phi}^{-1}\big{)}^{\tau}_{\mu}\otimes\hat{x}_{\tau}\big{)}=(x_{\mu}\otimes
1-1\otimes x_{\mu})\cdot\tilde{\mathcal{F}}_{L}\in I_{U(\mathfrak{g})},$

and note that this is sufficient because the elements of the form
$x_{\tau}\otimes 1-(\tilde{\phi}^{-1})^{\tau}_{\mu}\otimes\hat{x}_{\tau}$
generate $I_{U(\mathfrak{g})}$ because $(x_{\mu}\otimes
1-(\tilde{\phi}^{-1})^{\tau}_{\mu}\otimes\hat{x}_{\tau})(\tilde{\phi}^{\mu}_{\nu}\otimes
1)=\hat{y}_{\nu}\otimes 1-1\otimes\hat{x}_{\nu}$. Similarly we multiply the
second line with $\tilde{\mathcal{F}}_{R}$ from the left to obtain

$\displaystyle\tilde{\mathcal{F}_{R}}\big{(}\hat{y}_{\tau}\otimes\big{(}\phi^{-1}\big{)}^{\tau}_{\mu}-1\otimes\hat{x}_{\mu}\big{)}=(x_{\mu}\otimes
1-1\otimes x_{\mu})\cdot\tilde{\mathcal{F}}_{R}^{-1}\in I_{0}.$

This is sufficient to conclude
$\tilde{\mathcal{F}}_{R}I_{U(\mathfrak{g})}\subset I_{0}$ after observing that
the elements of the form
$\hat{y}_{\tau}\otimes\big{(}\phi^{-1}\big{)}^{\tau}_{\mu}-1\otimes\hat{x}_{\mu}$
generate $I_{U(\mathfrak{g})}$, indeed
$\big{(}\hat{y}_{\tau}\otimes\big{(}\phi^{-1}\big{)}^{\tau}_{\mu}-1\otimes\hat{x}_{\mu}\big{)}(1\otimes\phi^{\tau}_{\nu}\otimes
1)=\hat{y}_{\nu}\otimes 1-1\otimes\hat{x}_{\nu}$. Now
$\tilde{\mathcal{F}}^{-1}I_{0}\subset I_{U(\mathfrak{g})}$ and
$\tilde{\mathcal{F}}I_{U(\mathfrak{g})}\subset I_{0}$ together imply the
equality whenever $\tilde{\mathcal{F}}^{-1}$ and $\tilde{\mathcal{F}}$ are
strict inverses in $H\hat{\otimes}H$.

The assertions on $\mathcal{F}_{l}$, $\mathcal{F}_{r}$ are the direct
corollary: the products $\mathcal{F}_{l}\mathcal{F}_{l}^{-1}$,
$\mathcal{F}_{r}\mathcal{F}_{r}^{-1}$, $\mathcal{F}^{-1}_{l}\mathcal{F}_{l}$,
$\mathcal{F}^{-1}_{r}\mathcal{F}_{r}$ are well defined so we can compute the
representatives using $\tilde{\mathcal{F}}_{l}$, $\tilde{\mathcal{F}}_{r}$,
$\tilde{\mathcal{F}}^{-1}_{l}$, $\tilde{\mathcal{F}}_{r}^{-1}$. ∎

######  Proposition 5.9.

For every $h\in H=H_{\mathfrak{g}}$,

$\displaystyle\Delta_{H}(h)=\mathcal{F}_{l}^{-1}\Delta_{0}(h)\mathcal{F}_{l}=\mathcal{F}_{r}^{-1}\Delta_{0}(h)\mathcal{F}_{r}.$

###### Proof.

By Proposition 5.8 the expressions
$\mathcal{F}_{l}^{-1}\Delta_{0}(h)\mathcal{F}_{l}$ and
$\mathcal{F}_{r}^{-1}\Delta_{0}(h)\mathcal{F}_{r}$ are well defined (do not
depend on the representative of $\Delta_{0}(h)$). We can compute a
representative of the resulting class modulo $I_{U(\mathfrak{g})}$ as
$\tilde{\mathcal{F}}^{-1}\Delta_{0}^{\prime}(h)\tilde{\mathcal{F}}$ where
$\Delta_{0}^{\prime}(h)$ is any representative of $\Delta_{0}(h)$ and
$\mathcal{F}$ is $\mathcal{F}_{l}$ or $\mathcal{F}_{r}$. This way the
statement of the proposition for the generators $h=\partial^{\mu}$ follows by
Lemma 5.4 and for the rest of generators $\hat{x}_{\alpha}$ by Lemma 5.7. We
need to extend the statement for all $h\in H$ by linearity and some sort of
multiplicativity. Some care is however needed to achieve this.

Namely, $\sum_{\alpha}h_{\alpha}\otimes
h^{\prime}_{\alpha}\mapsto\tilde{\mathcal{F}}^{-1}\big{(}\sum_{\alpha}h_{\alpha}\otimes
h^{\prime}_{\alpha}\big{)}\tilde{\mathcal{F}}$ is a homomorphism of algebras
$H\hat{\otimes}H\to H\hat{\otimes}H$ and by $\mathcal{F}^{-1}\cdot
I_{0}=I_{U(\mathfrak{g})}$ it induces a well defined map of vector spaces
$\mathcal{F}^{-1}(-)\mathcal{F}\colon H\hat{\otimes}_{S(\mathfrak{g})}H\to
H\hat{\otimes}_{U(\mathfrak{g})}H$. Regarding that
$H\hat{\otimes}_{U(\mathfrak{g})}H$ is not an algebra, we can not say that
this map is a homomorphism of algebras. Suppose $h_{1},h_{2}\in H$ are such
that $\mathcal{F}^{-1}\Delta_{0}(h_{i})\mathcal{F}=\Delta_{H}(h_{i})$.
Regarding that the images of $\Delta_{H}$ and $\Delta_{0}$ are algebras, that
$\Delta_{H},\Delta_{0}$ viewed as corestrictions to the images are
multiplicative, and $\mathcal{F}\mathcal{F}^{-1}=1\otimes 1+I_{0}$, we
calculate

$\displaystyle\mathcal{F}^{-1}\Delta_{0}(h_{1}h_{2})\mathcal{F}=\mathcal{F}^{-1}\Delta_{0}(h_{1})\mathcal{F}\mathcal{F}^{-1}\Delta_{0}(h_{2})\mathcal{F}=\Delta_{H}(h_{1})\Delta_{H}(h_{2})=\Delta_{H}(h_{1}h_{2}),$

which would be sufficient to end the proof. However, we freely used
associativity and cancellations though the factors do not belong to an
associative algebra. Associativity holds for the representatives in
$H\hat{\otimes}H$, hence it is enough that all the products involved are well
defined up to an appropriate right ideal. In our case we inspect this
sequentially from the right to the left for the products involved, using
$\mathcal{F}I_{U(\mathfrak{g})}=I_{0}$,
$\mathcal{F}^{-1}I_{0}=I_{U(\mathfrak{g})}$, $\Delta_{0}(h)I_{0}\subset
I_{0}$, $\Delta_{H}(h)I_{U(\mathfrak{g})}\subset I_{0}$ and that $I_{0}$,
$I_{U(\mathfrak{g})}$ are right ideals. ∎

######  Lemma 5.10 ([17]).

In symmetric ordering,
$\exp\big{(}\sum_{\alpha}t_{\alpha}x_{\alpha}\big{)}\blacktriangleright_{\mathfrak{g}}1=\exp\big{(}\sum_{\alpha}t_{\alpha}\hat{x}_{\alpha}\big{)}$
for any formal variables $t_{\alpha}$ which commute with $x_{\beta}$,
$\hat{x}_{\beta}$. Conversely,
$\exp\big{(}\sum_{\alpha}t_{\alpha}\hat{x}_{\alpha}\big{)}\triangleright
1=\exp\big{(}\sum_{\alpha}t_{\alpha}x_{\alpha}\big{)}$. In particular,

$\displaystyle\exp\bigg{(}\sum_{\alpha}\partial^{\alpha}\otimes
x_{\alpha}\bigg{)}(\cdot\otimes\blacktriangleright_{\mathfrak{g}})(1\otimes
1)=\exp\bigg{(}\sum_{\alpha}\partial^{\alpha}\otimes\hat{x}_{\alpha}\bigg{)},$
$\displaystyle\exp\bigg{(}\sum_{\alpha}\partial^{\alpha}\otimes\hat{x}_{\alpha}\bigg{)}(\cdot\otimes\triangleright)(1\otimes
1)=\exp\bigg{(}\sum_{\alpha}\partial^{\alpha}\otimes x_{\alpha}\bigg{)}.$
(5.21)

######  Theorem 5.11.

$\mathcal{F}_{l}$ and $\mathcal{F}_{r}$ are counital Drinfeld twists for
$S(\mathfrak{g})$-bialgebroid on completed Weyl algebra $\hat{A}_{n}$ and by
twisting they yield the Heisenberg double $H_{\mathfrak{g}}$ of the
corresponding universal enveloping algebra $U(\mathfrak{g})$ with its
canonical $U(\mathfrak{g})$-bialgebroid structure.

###### Proof.

Both $\mathcal{F}_{l}$ and $\mathcal{F}_{r}$ are invertible. Proposition 5.9
and Corollary 4.4 together imply that the Drinfeld cocycle condition (3.1)
holds. We need to show the counitality. One has to be careful when checking
this, regarding that $\epsilon\colon H_{\mathfrak{g}}\to U(\mathfrak{g})$ is
not a homomorphism. However, for the symmetric ordering, checking this is
still not a difficult. Recall from the axioms of the bialgebroid that the
undeformed counit is given by $\epsilon(h)=h\triangleright 1$. Thus

$\displaystyle(\epsilon\otimes 1)\mathcal{F}_{l}=\exp(-\partial^{\rho}\otimes
x_{\rho})\exp(\partial^{\sigma}\otimes\hat{x}_{\sigma})(\triangleright
1\otimes\cdot 1)=1\otimes 1,$

because all the higher order terms have positive power of at least some
$\partial^{\alpha}$-s thus yielding zero when acting upon $1$. The second
counitality condition is a bit more involved; using the fact that
$\triangleright$ extends the regular action of $U(\mathfrak{g})$ on itself and
(5.21) we compute

$\displaystyle(1\otimes\epsilon)\mathcal{F}_{l}$
$\displaystyle=\exp(-\partial^{\rho}\otimes
x_{\rho})\exp(\partial^{\sigma}\otimes\hat{x}_{\sigma})(\cdot
1\otimes\triangleright 1)$ $\displaystyle=\exp(-\partial^{\rho}\otimes
x_{\rho})(\mathrm{id}\otimes\triangleright)\exp(\partial^{\sigma}\otimes
x_{\sigma})$ $\displaystyle=\exp(-\partial^{\rho}\otimes
x_{\rho})(\mathrm{id}\otimes\triangleright)\exp(\partial^{\sigma}\otimes
x_{\sigma})(\cdot 1\otimes\triangleright 1)$
$\displaystyle=(\exp(-\partial^{\rho}\otimes
x_{\rho})\exp(\partial^{\sigma}\otimes x_{\sigma}))(\cdot
1\otimes\triangleright 1)$ $\displaystyle=1\otimes 1.$

In the third line we used Lemma 5.10. Similarly, one shows that
$\mathcal{F}_{r}$ is counital.

The new base algebra is $S(\mathfrak{g})$ with the $\mathcal{F}_{l}$-twisted
(or $\mathcal{F}_{r}$-twisted) star product. We need to show that it is
canonically isomorphic to $U(\mathfrak{g})$ in the sense that

$\displaystyle\mu\mathcal{F}_{l}(\triangleright\otimes\triangleright)(g\otimes
f)=\mu\mathcal{F}_{r}(\triangleright\otimes\triangleright)(g\otimes
f)=\xi^{-1}(\xi(g)\cdot_{U(\mathfrak{g})}\xi(f)),\qquad g,f\in
S(\mathfrak{g}),$

where $\xi\colon
S(\mathfrak{g})\stackrel{{\scriptstyle\cong}}{{\to}}U(\mathfrak{g})$ is the
symmetrization map. Since we know that $\mathcal{F}_{l},\mathcal{F}_{r}$ are
Drinfeld cocycles, we know that in both cases the corresponding star product
is associative. Regarding that $\xi$ transports the product in
$U(\mathfrak{g})$ to the star product $g\star
f:=\xi^{-1}(\xi(g)\cdot_{U(\mathfrak{g})}\xi(f))=\xi(g)\triangleright f$ every
element in $S(\mathfrak{g})$ is a star polynomial in the generators in
$\mathfrak{g}$. Therefore the associativity and the star products of the form
$x_{\mu}\star f$ with general $f$ (or alternatively, all star products of the
form $g\star x_{\mu}$) are sufficient to determine the star product $g\star f$
for general $g$, $f$. Thus for $\mathcal{F}_{l}$ it is sufficient to check
that $\mu\mathcal{F}_{l}(\triangleright x_{\mu}\otimes\triangleright
f)=\hat{x}_{\mu}\triangleright f=\hat{x}_{\mu}^{\phi}(f)$ for all $f$ and
$\mu\mathcal{F}_{r}(\triangleright g\otimes\triangleright
x_{\mu})=\hat{y}_{\mu}\triangleright g=\hat{y}^{\phi}_{\mu}(g)$ for all $g$.
When acting by $\triangleright$ we take the realization in the Weyl algebra
and apply the corresponding differential operator. Thus all the higher
derivatives drop out when acting on $x_{\mu}$ and we obtain
$\mathcal{F}_{l}(\triangleright x_{\mu}\otimes\triangleright f)=x_{\mu}\otimes
f-1\otimes x_{\mu}f+1\otimes\hat{x}^{\phi}_{\mu}(f)$. After applying $\mu$ on
this equality, the first two summands cancel and we obtain
$\hat{x}^{\phi}_{\mu}(f)$. Similarly, $\mathcal{F}_{r}(\triangleright
g\otimes\triangleright x_{\mu})=\hat{y}_{\mu}(g)\otimes 1-g\otimes
x_{\mu}+x_{\mu}g\otimes 1$ and after applying $\mu$ the second summand cancels
with the third and we obtain $\mu\mathcal{F}_{r}(\triangleright
g\otimes\triangleright x_{\mu})=\hat{y}_{\mu}\triangleright
g\stackrel{{\scriptstyle(\ref{eq:betastar})}}{{=}}g\star x_{\mu}$ as required.

Finally, we need to compute the deformed source and target map from the twist
and compare them with the known source and target map in the deformed case.
Regarding that we already know that $\mathcal{F}_{l}$ and $\mathcal{F}_{r}$
are Drinfeld twists, Xu’s theorem tells us that $\alpha_{\mathcal{F}_{l}}$,
$\beta_{\mathcal{F}_{l}}$, $\alpha_{\mathcal{F}_{r}}$,
$\beta_{\mathcal{F}_{r}}$ are algebra maps by the construction, it is
sufficient to check the agreement with known deformed source and target maps
on the algebra generators $x_{\mu}$

$\displaystyle\alpha_{\mathcal{F}_{l}}(f)=\mu(\alpha_{0}\otimes\mathrm{id})\exp(-\partial^{\rho}\otimes
x_{\rho})\exp(\partial^{\sigma}\otimes\hat{x}_{\sigma})(\triangleright
f\otimes 1),$

where $\alpha_{0}=\beta_{0}$ is the source map in the undeformed case, which
is then equal to the undeformed target map. When $f=x_{\mu}$, after expanding
the exponentials only three summands survive, with at most one partial
derivative before applying $\triangleright$. Thus we obtain
$\alpha_{\mathcal{F}_{L}}(x_{\mu})=\alpha(x_{\mu})-x_{\mu}+\hat{x}_{\mu}=\hat{x}_{\mu}=\alpha(\xi(x_{\mu}))$
as required. For $\beta$ we first interchange the tensor factors in
$\mathcal{F}_{l}$,

$\displaystyle\beta_{\mathcal{F}_{l}}(f)=\mu(\beta_{0}\otimes\mathrm{id})\exp(-x_{\rho}\otimes\partial^{\rho})\exp(\hat{x}_{\sigma}\otimes\partial^{\sigma})(\triangleright
f\otimes 1).$

Regarding that the second tensor factor commutes, we can compute
$\exp(-x_{\rho}\otimes\partial^{\rho})\exp(\hat{x}_{\sigma}\otimes\partial^{\sigma})(\triangleright
x_{\mu}\otimes 1)$ as if $\partial$-s are independent formal variables. Now
$\exp(k^{\sigma}\hat{x}_{\sigma})\triangleright
x_{\mu}=\exp(k^{\sigma}x_{\sigma})\star
x_{\mu}=\hat{y}_{\mu}\triangleright\exp(k^{\sigma}x_{\sigma})=x_{\tau}\tilde{\phi}^{\tau}_{\mu}(k)\exp(k^{\sigma}x_{\sigma})$.
Then, the exponential factors cancel and we get
$\beta_{\mathcal{F}_{l}}(x_{\mu})=x_{\tau}\phi^{\tau}_{\mu}(\partial)=\hat{y}_{\mu}=\beta(\xi(x_{\mu}))$
as required.

$\displaystyle\alpha_{\mathcal{F}_{r}}(f)=\mu(\alpha_{0}\otimes\mathrm{id})\exp(-x_{\rho}\otimes\partial^{\rho})\exp(\hat{y}_{\sigma}\otimes\partial^{\sigma})(\triangleright
f\otimes 1).$

For $f=x_{\mu}$ calculate
$\exp(\hat{y}_{\sigma}\otimes\partial^{\sigma})(\triangleright x_{\mu}\otimes
1)=(x_{\mu}\otimes
1)(\star\otimes\cdot)\exp(x_{\sigma}\otimes\partial^{\sigma})=(\hat{x}_{\mu}\otimes
1)(\triangleright\otimes\cdot)\exp(x_{\sigma}\otimes\partial^{\sigma})=\exp(x_{\sigma}\otimes\partial^{\sigma})(x_{\tau}\otimes\phi^{\tau}_{\mu})$,
hence
$\alpha_{\mathcal{F}_{r}}(x_{\mu})=\alpha_{0}(x_{\tau})\phi^{\tau}_{\mu}=\hat{x}_{\mu}=\alpha(\xi(x_{\mu}))$
as required.

$\displaystyle\beta_{\mathcal{F}_{r}}(f)=\mu(\beta_{0}\otimes\mathrm{id})\exp(-\partial^{\rho}\otimes
x_{\rho})\exp(\partial^{\sigma}\otimes\hat{y}_{\sigma})(\triangleright
f\otimes 1).$

For $f=x_{\mu}$ only the three summands with partial derivatives up to the
first order survive after applying $\triangleright x_{\mu}$. Thus

$\displaystyle\beta_{\mathcal{F}_{r}}(x_{\mu})=\mu(\alpha_{0}\otimes\mathrm{id})(x_{\mu}\otimes
1-1\otimes x_{\mu}+1\otimes\hat{y}_{\mu})=\hat{y}_{\mu}=\beta(\xi(x_{\mu}))$

as required. ∎

######  Corollary 5.12.

$\tilde{\mathcal{F}}_{l}-\tilde{\mathcal{F}}_{r}\in I_{U(\mathfrak{g})}$,
$\mathcal{F}_{l}=\mathcal{F}_{r}$.

###### Proof.

In the proof of Theorem 5.11 we have shown
$\mu\tilde{\mathcal{F}}_{l}(\triangleright\otimes\triangleright)(f\otimes
g)=f\star
g=\mu\tilde{\mathcal{F}}_{r}(\triangleright\otimes\triangleright)(f\otimes g)$
for all $f,g\in\hat{S}(\mathfrak{g})$. Thus
$\tilde{\mathcal{F}}_{l}-\tilde{\mathcal{F}}_{r}\in I_{U(\mathfrak{g})}$ from
the undeformed case of the Theorem 4.3 on nondegeneracy. Therefore for the
cosets we conclude
$\mathcal{F}_{l}=\tilde{\mathcal{F}_{l}}+I_{U(\mathfrak{g})}=\tilde{\mathcal{F}_{r}}+I_{U(\mathfrak{g})}=\mathcal{F}_{r}$.
∎

##  6 Twisting the antipode

In this section, we discuss the twisting of the antipode. We first recall the
definition of Hopf algebroids as bialgebroids with an antipode and then we
recall from [18] the antipode for $H_{\mathfrak{g}}$.

Several nonequivalent versions of the axioms for the antipode are used in the
literature (see, e.g., [4, 5, 13, 23]). In [18] we checked for
$H_{\mathfrak{g}}$ the axioms of the symmetric Hopf algebroid, which involve
both a left and a right bialgebroid. Thus a reasonable formalism for the
twisting of symmetric Hopf algebroids is expected to provide a twist for the
left and another twist for the right bialgebroid, these twists satisfying some
compatibilities. Instead of taking this not so obvious path, we here work with
a version of the axioms for the antipode involving only the left bialgebroid.
We can do this because the antipode for $H_{\mathfrak{g}}$ is invertible both
in the undeformed and deformed case. Namely, if the antipode is invertible, as
proven by Böhm, the structure of a symmetric Hopf $A$-algebroid $H$ is
equivalent to a left $A$-bialgebroid together with an antipode $S$ which is an
algebra antihomomorphism $H\to H$ having an inverse $S^{-1}$ satisfying for
all $h\in H$ the relations

$\displaystyle S\circ\beta=\alpha,$
$\displaystyle\big{(}S^{-1}h_{(2)}\big{)}_{(1)}\otimes_{A}\big{(}S^{-1}h_{(2)}\big{)}_{(2)}h_{(1)}=S^{-1}h\otimes_{A}1_{H},$
$\displaystyle\big{(}Sh_{(1)}\big{)}_{(1)}h_{(2)}\otimes_{A}\big{(}Sh_{(1)}\big{)}_{(2)}=1_{H}\otimes_{A}Sh.$

If a bialgebroid twist
$\mathcal{F}=\mathcal{F}^{(1)}\otimes_{A}\mathcal{F}^{(2)}$ on a Hopf
algebroid $H$ has an inverse cocycle
$\mathcal{F}^{-1}=\overline{\mathcal{F}}^{(1)}\otimes_{A_{\star}}\overline{\mathcal{F}}^{(2)}$
in the sense of Remark 3.4, then
$V_{\mathcal{F}}=\big{(}S\mathcal{F}^{(1)}\big{)}\mathcal{F}^{(2)}$ is a well-
defined element in $H$. If $H$ is a Hopf algebra then a standard calculation
shows that
$\overline{\mathcal{F}}^{(1)}\big{(}S\overline{\mathcal{F}}^{(2)}\big{)}$ is
the two-sided inverse of $V$ with respect to the multiplication in $H$. The
calculation does not extend to Hopf algebroids, namely not only that
$\overline{\mathcal{F}}^{(1)}\big{(}S\overline{\mathcal{F}}^{(2)}\big{)}$ is
not the inverse of $V$, but worse,
$\overline{\mathcal{F}}^{(1)}\big{(}S\overline{\mathcal{F}}^{(2)}\big{)}$ is
not a well defined expression because $\mu(\mathrm{id}\otimes
S)I_{A_{\star}}\neq 0$. We do not know if the inverse of $V_{\mathcal{F}}$
exists in general.

######  Proposition 6.1.

Suppose that
$V_{\mathcal{F}}=\big{(}S\mathcal{F}^{(1)}\big{)}\mathcal{F}^{(2)}$ has an
inverse $V^{-1}_{\mathcal{F}}$ in $H$. Then define

$\displaystyle
S_{\mathcal{F}}h=V^{-1}_{\mathcal{F}}(Sh)V_{\mathcal{F}}=V^{-1}_{\mathcal{F}}(Sh)\big{(}S\mathcal{F}^{(1)}\big{)}\mathcal{F}^{(2)}.$
(6.1)

The formula $h\mapsto S_{\mathcal{F}}h$ then defines an antihomomorphism of
algebras $S_{\mathcal{F}}\colon H\to H$ and

$\displaystyle
V^{-1}_{\mathcal{F}}=\big{(}S_{\mathcal{F}}\overline{\mathcal{F}}^{(1)}\big{)}\overline{\mathcal{F}}^{(2)},$
(6.2)

where the right-hand side is well defined and in particular
$\mu(S_{\mathcal{F}}\otimes\mathrm{id})I_{A_{\star}}=0$.

###### Proof.

$S$ is an antihomomorphism hence it is clear that $S_{\mathcal{F}}$ given by
(6.1) is an antihomomorphism as well. It follows that
$\big{(}S\overline{\mathcal{F}}^{(1)}\big{)}\big{(}S\mathcal{F}^{(1)^{\prime}}\big{)}\mathcal{F}^{(2)^{\prime}}\overline{\mathcal{F}}^{(2)}=1$.
Here the primed Sweedler indices $(1)^{\prime}$, $(2)^{\prime}$ refer to
another copy of $\mathcal{F}$

$\displaystyle\big{(}S_{\mathcal{F}}\overline{\mathcal{F}}^{(1)}\big{)}\overline{\mathcal{F}}^{(2)}\big{(}S\mathcal{F}^{(1)}\big{)}\mathcal{F}^{(2)}$
$\displaystyle=V^{-1}_{\mathcal{F}}\big{(}S\overline{\mathcal{F}}^{(1)}\big{)}V_{\mathcal{F}}\overline{\mathcal{F}}^{(2)}\big{(}S\mathcal{F}^{(1)}\big{)}\mathcal{F}^{(2)}$
$\displaystyle=V^{-1}_{\mathcal{F}}\big{(}S\overline{\mathcal{F}}^{(1)}\big{)}\big{(}S\mathcal{F}^{(1)^{\prime}}\big{)}\mathcal{F}^{(2)^{\prime}}\overline{\mathcal{F}}^{(2)}\big{(}S\mathcal{F}^{(1)}\big{)}\mathcal{F}^{(2)}$
$\displaystyle=V^{-1}_{\mathcal{F}}(S\mathcal{F}^{(1)})\mathcal{F}^{(2)}$
$\displaystyle=1,$
$\displaystyle\big{(}S\mathcal{F}^{(1)}\big{)}\mathcal{F}^{(2)}\big{(}S_{\mathcal{F}}\overline{\mathcal{F}}^{(1)}\big{)}\overline{\mathcal{F}}^{(2)}$
$\displaystyle=\big{(}S\mathcal{F}^{(1)}\big{)}\mathcal{F}^{(2)}V^{-1}_{\mathcal{F}}\big{(}S\overline{\mathcal{F}}^{(1)}\big{)}V_{\mathcal{F}}\overline{\mathcal{F}}^{(2)}$
$\displaystyle=\big{(}S\mathcal{F}^{(1)}\big{)}\mathcal{F}^{(2)}V^{-1}_{\mathcal{F}}\big{(}S\overline{\mathcal{F}}^{(1)}\big{)}\big{(}S\mathcal{F}^{(1)^{\prime}}\big{)}\mathcal{F}^{(2)^{\prime}}\overline{\mathcal{F}}^{(2)}$
$\displaystyle=\big{(}S\mathcal{F}^{(1)}\big{)}\mathcal{F}^{(2)}V^{-1}_{\mathcal{F}}$
$\displaystyle=1.$

Regarding that the two-sided inverse in an associative algebra is unique, we
conclude (6.2) with the right-hand side in (6.2) well defined. ∎

One would like to conclude that $S_{\mathcal{F}}$ is an antipode for the
twisted bialgebroid. The standard proofs for the Hopf algebras do not seem to
generalize in straightforward manner.

However, in our case, for $H_{\mathfrak{g}}$, we know the deformed antipode
$S$, and one can say a bit more. For $H_{\mathfrak{g}}\cong\hat{A}_{n}$, the
antipode $S_{0}$ for the undeformed coproduct is a continuous antihomomorphism
$S_{0}\colon H_{\mathfrak{g}}\to H_{\mathfrak{g}}$ given on the generators
$x_{\mu},\partial^{\nu}$ of the dense subalgebra $A_{n}$ by
$S_{0}(x_{\mu})=x_{\mu}$, $S_{0}(\partial^{\nu})=-\partial^{\nu}$. Similarly,
the antipode $S\colon H_{\mathfrak{g}}\to H_{\mathfrak{g}}$ for the deformed
coproduct is determined by the formulas

$\displaystyle S\hat{y}_{\mu}=\hat{x}_{\mu},\qquad
S\partial^{\nu}=S_{0}\partial^{\nu}=-\partial^{\nu}.$

Therefore using $\hat{x}_{\mu}=x_{\rho}\phi(\partial)^{\rho}_{\mu}$ and
$\hat{y}_{\mu}=x_{\rho}\phi(-\partial)^{\rho}_{\mu}$

$\displaystyle
S_{0}\hat{y}_{\mu}=S_{0}(x_{\rho}\phi(-\partial)^{\rho}_{\mu})=\phi(\partial)^{\rho}_{\mu}x_{\rho}=x_{\rho}\phi(\partial)^{\rho}_{\mu}+\partial_{\rho}\phi(\partial)^{\rho}_{\mu},$

where we denoted $\partial_{\rho}=\frac{\partial}{\partial(\partial^{\rho})}$.

We seek for $V$ such that $S(h)=V^{-1}S_{0}(h)V$, in parallel to equation
(6.1) defining $S_{\mathcal{F}}$ in terms of $V_{\mathcal{F}}$. Regarding that
$S(\partial^{\mu})=S_{0}(\partial^{\mu})=-\partial^{\mu}$, this forces that
$[\partial^{\mu},V]=0$, hence
$V=V(\partial^{1},\dots,\partial^{n})\in\hat{S}(\mathfrak{g}^{*})$. Moreover,

$\displaystyle
S\hat{y}_{\mu}=x_{\rho}\phi^{\rho}_{\mu}=V^{-1}\big{(}x_{\rho}\phi(\partial)^{\rho}_{\mu}+\partial_{\rho}\phi(\partial)^{\rho}_{\mu}\big{)}V=V^{-1}x_{\rho}V\phi^{\rho}_{\mu}+\partial_{\rho}\phi^{\rho}_{\mu},$
$\displaystyle
x_{\mu}-\big{(}\partial_{\rho}\phi^{\rho}_{\gamma}\big{)}\phi^{-1\gamma}_{\mu}=V^{-1}x_{\mu}V,$
$\displaystyle
V^{-1}[V,x_{\mu}]=(\partial_{\rho}\phi^{\rho}_{\gamma})\phi^{-1\gamma}_{\mu},$
$\displaystyle\partial_{\mu}{\ln|V|}=V^{-1}\partial_{\mu}V=\phi^{-1\gamma}_{\mu}\partial_{\rho}\phi^{\rho}_{\gamma},$
$\displaystyle R:=\ln|V|,$
$\displaystyle\partial_{\mu}R=\phi^{-1\gamma}_{\mu}\partial_{\rho}\phi^{\rho}_{\gamma}.$
(6.3)

Setting $F_{\mu}=\phi^{-1\gamma}_{\mu}\partial_{\rho}\phi^{\rho}_{\gamma}$, we
rewrite (6.3) as the system of formal differential equations

$\displaystyle\partial_{\mu}R=F_{\mu}$ (6.4)

for an unknown formal series $R$. In our case, $F_{\mu}$ are analytic
functions, hence the solution exists if the integrability condition
$\partial_{\nu}F_{\mu}=\partial_{\mu}F_{\nu}$ holds, which boils down to

$\displaystyle\partial_{\nu}\big{(}\phi^{-1\gamma}_{\mu}\partial_{\rho}\phi^{\rho}_{\gamma}\big{)}=\partial_{\mu}\big{(}\phi^{-1\gamma}_{\nu}\partial_{\rho}\phi^{\rho}_{\gamma}\big{)}.$
(6.5)

Any solution for $V=\exp(\ln|V|)=\exp(R)$ is clearly invertible with an
inverse $V^{-1}=\exp(-R)$. The constant term $V(0)$ of $V$ which is viewed as
a formal power series is nonzero, hence $V(0)$ is also invertible. Therefore
we can write $V=V(0)\cdot V_{1}$ where $V_{1}=V/V(0)$. It is clear that
$V_{1}$ is also a solution and the identity $V_{1}(0)=1$ holds. We conclude
that without loss of generality (by passing from $V$ to $V_{1}$) we may assume
that $V(0)=1$ and such a solution for $V$ is unique. The condition $V(0)=1$ is
equivalent to the boundary condition $R(0)=0$ for the “potential” $R$ when
solving the exact first order differential equation (6.4). This boundary
condition guarantees the uniqueness of the solution for $R$. If, instead of
the abstract algebra elements $\partial^{\nu}$, we introduce the real
variables $\xi^{\nu}$, we can write the solution formally

$\displaystyle R\big{(}\zeta^{1},\dots,\zeta^{n}\big{)}=\int_{\Gamma}\sum
F_{\mu}\big{(}\xi^{1},\dots,\xi^{n}\big{)}{\rm d}\xi^{\mu},$

where the line integral is along any path $\Gamma$ from $(0,\dots,0)$ to
$(\zeta^{1},\dots,\zeta^{n})$. Due to the integrability (6.5), the line
integral does not depend on the path chosen.

We have some evidence that $S_{\mathcal{F}}=S$ or equivalently that
$V_{\mathcal{F}}$ satisfies the equations for $V$. First of all, Proposition
6.2 below says that the element $V_{\mathcal{F}}\in\hat{A}_{n}$ is in fact the
formal power series in $\partial^{1},\dots,\partial^{n}$ (that is
$x_{1},\dots,x_{n}$ do not appear) making sense of the equations for $V$.
$V_{\mathcal{F}}$ clearly satisfies the required boundary condition
$V_{\mathcal{F}}(0)=1$. It follows that $V_{\mathcal{F}}$ is invertible, as
required in Proposition 6.1, particularly in equation (6.1) defining
$S_{\mathcal{F}}$. Using the Baker–Hausdorff formula for computing
$R_{\mathcal{F}}=\ln|V_{\mathcal{F}}|$ directly from the definition
$V_{\mathcal{F}}=\mu(S_{0}\otimes\mathrm{id})\mathcal{F}$, we have
perturbatively checked up to the third order in $\partial$-s that
$R_{\mathcal{F}}$ satisfies the system (6.4) for $R$.

######  Proposition 6.2.

For $\mathcal{F}$ being $\mathcal{F}_{c}=\mathcal{F}_{l}=\mathcal{F}_{r}$, the
element $V_{\mathcal{F}}\in H_{\mathfrak{g}}$ is a formal power series in
$\partial^{1},\dots,\partial^{n}$. As a corollary,
$S_{\mathcal{F}}(\partial^{\nu})=S_{0}\partial^{\nu}=S\partial^{\nu}$.

The corollary part of the proposition follows by the defining formula (6.1)
for $S_{\mathcal{F}}$ recalling that $\partial^{1},\dots,\partial^{n}$
mutually commute. The first statement in Proposition 6.2 is the content of
parts (v) and (vi) of the following lemma restated.

######  Lemma 6.3.

  1. $(i)$

$(S_{0}\otimes\mathrm{id})\tilde{\mathcal{F}}_{l}\in H_{\mathfrak{g}}\otimes
H_{\mathfrak{g}}$ can be written as $\exp(W)$ where $W$ is a formal sum of
summands which are up to a rational factor equal to
$\partial^{\alpha_{1}}\cdots\partial^{\alpha_{s}}\otimes
L(w_{\alpha_{1}},\dots,w_{\alpha_{s}})$, where $s\geq 1$ and $w_{\alpha_{i}}$
is either $x_{i}$ or $\hat{x}_{i}$ $($the choice depending on $i$ and on the
summand$)$ and $L$ is a Lie monomial $(L$ differing from a summand to
summand$)$, more precisely an iterated Lie bracket involving each of the
variables precisely once.

  2. $(ii)$

For every of the summands in $(i)$ it holds that
$L(w_{\alpha_{1}},\dots,w_{\alpha_{s}})=\sum_{\rho}x_{\rho}\zeta^{\rho}_{\alpha_{1},\dots,\alpha_{s}}$
where
$\zeta^{\rho}_{\alpha_{1},\dots,\alpha_{s}}\in\hat{S}(\mathfrak{g}^{*})$.

  3. $(iii)$

$(\phi^{\mu}_{\nu}-\delta^{\mu}_{\nu})\partial^{\nu}=0$, therefore
$(\hat{x}_{\nu}-x_{\nu})\partial^{\nu}=0$. The summands in $W$ with $s=1$ add
up to $\partial^{\nu}\otimes(\hat{x}_{\nu}-x_{\nu})$.

  4. $(iv)$

For every of the summands in $(ii)$ with $s\geq 2$,
$\partial^{\alpha_{1}}\cdots\partial^{\alpha_{s}}\cdot\zeta^{\rho}_{\alpha_{1},\dots,\alpha_{s}}=0$.

  5. $(v)$

$V_{\mathcal{F}_{l}}=\mu(S_{0}\otimes\mathrm{id})\mathcal{F}_{l}$ belongs to
$\hat{S}(\mathfrak{g}^{*})$.

  6. $(vi)$

$V_{\mathcal{F}_{c}}=\mu(S_{0}\otimes\mathrm{id})\mathcal{F}_{c}$ $($see
(4.8)$)$ belongs to $\hat{S}(\mathfrak{g}^{*})$.

###### Proof.

(i) The statement is a straightforward application of the Baker–Hausdorff
series to $\tilde{\mathcal{F}}_{l}=\exp(-\partial^{\rho}\otimes
x_{\rho})\exp(\partial^{\sigma}\otimes\hat{x}_{\sigma})$ and using
$S_{0}(\partial^{\mu})=-\partial^{\mu}$.

(ii) Easily follows by induction, computing in $\hat{A}_{n}$.

(iii) Recall (4.2) that
$\phi=\sum\limits_{N=0}^{\infty}\frac{(-1)^{N}B_{N}}{N!}\mathcal{C}^{N}$. If
$N\geq 1$ then
$(\mathcal{C}^{N})^{\mu}_{\nu}\partial^{\nu}=(\mathcal{C}^{N-1})^{\mu}_{\rho}C^{\rho}_{\nu\sigma}\partial^{\nu}\partial^{\sigma}=0$
as it involves a contraction of an antisymmetric tensor $C^{\rho}_{\mu\nu}$ in
$\mu$, $\nu$ with a symmetric tensor $\partial^{\mu}\partial^{\nu}$.

(iv) Along with the statement to prove, we also claim that $\zeta$ is an
expression which is a contraction (always one lower and one upper index are
contracted) of the tensors of the form $\mathcal{C}^{\gamma}_{\delta}$ and
$C^{\gamma}_{\delta\lambda}$ where the contracted form is connected and the
only external indices which remain are the labels of $\zeta$. In particular,
each $\alpha_{i}$ is either within a factor of the form
$\mathcal{C}^{\gamma}_{\alpha_{i}}=C^{\gamma}_{\alpha_{i}\lambda}\partial^{\lambda}$,
which contracted with $\partial^{\alpha_{i}}$ is clearly zero (because
$C^{\gamma}_{\alpha_{i}\lambda}$ is antisymmetric in lower and
$\partial^{\alpha_{i}}\partial^{\lambda}$ in upper labels), or $\alpha_{i}$
appears in a factor of the form $C^{\gamma}_{\alpha_{i}\alpha_{j}}$ which is
zero when contracted with $\partial^{\alpha_{i}}\partial^{\alpha_{j}}$. This
description is an easy observation for $s=2$, and then we proceed by induction
on $s$. Note that for any $\zeta$, $\zeta^{\prime}$ tensors whose components
are formal series in $\partial^{1},\dots,\partial^{n}$ there are new tensors
$\zeta^{\prime\prime}$ defined by the commutators
$\big{[}\sum_{\rho}x_{\rho}\zeta^{\rho}_{\alpha_{1},\dots,\alpha_{s}},\sum_{\sigma}x_{\sigma}(\zeta^{\prime})^{\sigma}_{\beta_{1},\dots,\beta_{r}}\big{]}=\sum_{\tau}x_{\tau}(\zeta^{\prime\prime})^{\tau}_{\alpha_{1},\dots,\alpha_{s}\beta_{1},\dots,\beta_{r}}$.
Explicitly,

$\displaystyle(\zeta^{\prime\prime})^{\tau}_{\alpha_{1},\dots,\alpha_{s}\beta_{1},\dots,\beta_{r}}=\frac{\partial}{\partial(\partial^{\lambda})}\big{(}\zeta^{\tau}_{\alpha_{1},\dots,\alpha_{s}}\big{)}\cdot(\zeta^{\prime})^{\lambda}_{\beta_{1},\dots,\beta_{r}}-\frac{\partial}{\partial(\partial^{\lambda})}\big{(}(\zeta^{\prime})^{\lambda}_{\beta_{1},\dots,\beta_{r}}\big{)}\cdot\zeta^{\tau}_{\alpha_{1},\dots,\alpha_{s}}.$

For the induction step one checks that if (iv) holds for $\zeta$ and
$\zeta^{\prime}$ then it holds also for $\zeta^{\prime\prime}$.

(v) This is somewhat tricky point because the multiplication does not commute
with evaluating the exponential series.
$\mu(S_{0}\otimes\mathrm{id})\mathcal{F}_{l}$ will be a sum of terms which are
proportional to products of the summands from part (i), that is of the form

$\displaystyle\big{(}\partial^{\alpha_{1}}\cdots\partial^{\alpha_{s}}\big{)}\big{(}\partial^{\beta_{1}}\cdots\partial^{\beta_{r}}\big{)}\cdots\big{(}x_{\rho}\zeta^{\rho}_{\alpha_{1},\dots,\alpha_{s}}\big{)}\big{(}x_{\sigma}\zeta^{\sigma}_{\beta_{1},\dots,\beta_{r}}\big{)}\cdots.$

Moving $x$-s to the left by commuting we bring this expression to the normally
ordered form. By (iv), a nonzero contribution can possibly happen only from
those terms in which each bracket of partials on the left has lost at least
one power by yielding a Kronecker delta when commuted with some $x$. There is
only one $x$ per bracket on the right-hand side; because the $\partial$-groups
on the left and the $x\zeta$-groups on the right match, that means that all
$x$-s disappear in the normally ordered form.

(vi) While this is trivially equivalent to the statement in (v), there is an
independent but similar proof for $\mathcal{F}_{c}$, using the defining
formula (4.8) and the form of the tensors appearing in developing
$\tilde{\mathcal{F}}_{c}$, as described in [17]. In particular, one needs to
observe that for every $s\geq 1$ the expression
$\partial^{\mu_{1}}\cdots\partial^{\mu_{s}}\cdot[\cdots[\partial^{\lambda},\hat{x}_{\mu_{1}}],\ldots,\hat{x}_{\mu_{s}}]$,
which appears as a summand in
$\Delta_{\hat{S}(\mathfrak{g}^{*})}\partial^{\lambda}$, vanishes. To this end,
one uses the description in [17] of the tensors which appear in developing
$[\cdots[\partial^{\lambda},\hat{x}_{\mu_{1}}],\dots,\hat{x}_{\mu_{s}}]$. ∎

### Acknowledgements

S.M. has been supported by Croatian Science Foundation under the Project no.
IP-2014-09-9582 and the H2020 Twinning project no. 692194 “RBI-T-WINNING”.
Z.Š. has been partly supported by grant no. 18-00496S of the Czech Science
Foundation. We thank A. Borowiec for his remarks on the paper and M. Stojić
for remarks on Sections 1 and 2. We thank the referees for bringing to our
attention numerous constructive suggestions, which helped extending and
improving the article significantly.

## References

  * [1] Amelino-Camelia G., Arzano M., Coproduct and star product in field theories on Lie-algebra noncommutative space-times, Phys. Rev. D 65 (2002), 084044, 8 pages, hep-th/0105120.
  * [2] Berezin F.A., Some remarks about the associated envelope of a Lie algebra, Funct. Anal. Appl. 1 (1967), 91–102.
  * [3] Böhm G., Internal bialgebroids, entwining structures and corings, in Algebraic Structures and their Representations, Contemp. Math., Vol. 376, Amer. Math. Soc., Providence, RI, 2005, 207–226, math.QA/0311244.
  * [4] Böhm G., Hopf algebroids, in Handbook of Algebra, Handb. Algebr., Vol. 6, Elsevier/North-Holland, Amsterdam, 2009, 173–235, arXiv:0805.3806.
  * [5] Böhm G., Szlachányi K., Hopf algebroid symmetry of abstract Frobenius extensions of depth 2, Comm. Algebra 32 (2004), 4433–4464, math.QA/0305136.
  * [6] Borowiec A., Pachoł A., Twisted bialgebroids versus bialgebroids from a Drinfeld twist, J. Phys. A: Math. Theor. 50 (2017), 055205, 17 pages, arXiv:1603.09280.
  * [7] Brzeziński T., Militaru G., Bialgebroids, $\times_{A}$-bialgebras and duality, J. Algebra 251 (2002), 279–294, math.QA/0012164.
  * [8] Donin J., Mudrov A., Quantum groupoids and dynamical categories, J. Algebra 296 (2006), 348–384, math.QA/0311316.
  * [9] Durov N., Meljanac S., Samsarov A., Škoda Z., A universal formula for representing Lie algebra generators as formal power series with coefficients in the Weyl algebra, J. Algebra 309 (2007), 318–359, math.RT/0604096.
  * [10] Gutt S., An explicit ∗-product on the cotangent bundle of a Lie group, Lett. Math. Phys. 7 (1983), 249–258.
  * [11] Halliday S., Szabo R.J., Noncommutative field theory on homogeneous gravitational waves, J. Phys. A: Math. Gen. 39 (2006), 5189–5225, hep-th/0602036.
  * [12] Kowalzig N., Hopf algebroids and their cyclic theory, Ph.D. Thesis, Radboud University, Nijmegen, 2009.
  * [13] Lu J.H., Hopf algebroids and quantum groupoids, Internat. J. Math. 7 (1996), 47–70, q-alg/9505024.
  * [14] Lukierski J., Škoda Z., Woronowicz M., $\kappa$-deformed covariant quantum phase spaces as Hopf algebroids, Phys. Lett. B 750 (2015), 401–406, arXiv:1507.02612.
  * [15] Lukierski J., Škoda Z., Woronowicz M., On Hopf algebroid structure of $\kappa$-deformed Heisenberg algebra, Phys. Atomic Nuclei 80 (2017), 569–578, arXiv:1601.01590.
  * [16] Mack G., Schomerus V., Quasi Hopf quantum symmetry in quantum theory, Nuclear Phys. B 370 (1992), 185–230.
  * [17] Meljanac S., Škoda Z., Leibniz rules for enveloping algebras and a diagrammatic expansion, arXiv:0711.0149.
  * [18] Meljanac S., Škoda Z., Stojić M., Lie algebra type noncommutative phase spaces are Hopf algebroids, Lett. Math. Phys. 107 (2017), 475–503, arXiv:1409.8188.
  * [19] Meljanac S., Škoda Z., Svrtan D., Exponential formulas and Lie algebra type star products, SIGMA 8 (2012), 013, 15 pages, arXiv:1006.0478.
  * [20] Petracci E., Universal representations of Lie algebras by coderivations, Bull. Sci. Math. 127 (2003), 439–465, math.RT/0303020.
  * [21] Sweedler M.E., Groups of simple algebras, Inst. Hautes Études Sci. Publ. Math. (1974), 79–189.
  * [22] Škoda Z., Heisenberg double versus deformed derivatives, Internat. J. Modern Phys. A 26 (2011), 4845–4854, arXiv:0806.0978.
  * [23] Škoda Z., Stojić M., A two-sided ideal trick in Hopf algebroid axiomatics, arXiv:1610.03837.
  * [24] Stojić M., Completed Hopf algebroids, Ph.D. Thesis, University of Zagreb, 2017 (in Croatian).
  * [25] Xu P., Quantum groupoids, Comm. Math. Phys. 216 (2001), 539–581, math.QA/9905192.