Lobachevskii Journal of Mathematics http://ljm.ksu.ru ISSN 1818-9962 Vol. 24, 2006, 13–42

H. L. Huru
QUANTIZATIONS OF BRAIDED DERIVATIONS.
1. MONOIDAL CATEGORIES
(submitted by V. V. Lychagin)

Abstract. For monoidal categories we describe braidings and quantizations. We use them to ﬁnd quantizations of braided symmetric algebras and modules, braided derivations, braided connections, curvatures and diﬀerential operators.

1 Introduction

We consider quantizations $q$ , braidings $σ$ and quantizations of braidings $σ_{q}$ of monoidal categories. We mainly work with braidings that are symmetries.

We consider $σ$ -symmetric algebras $A$ , modules, co- and bialgebras and internal homomorphisms and ﬁnd quantizations of these.

Internal homomorphisms of $σ$ -symmetric modules has a braided Lie structure with respect to the braided commutator. Quantizations of the internal homomorphisms has the quantized braided Lie structure and can be realized within the original braided Lie structure by what we call dequantization.

We investigate braided derivations of $σ$ -symmetric algebras and modules. The $σ$ -bracket of two braided derivations is a braided derivation. We show that there is a braided Lie structure on the braided derivations.

A quantizations the braided derivations provides an isomorphism of the modules of braided derivations and quantized braided derivations. We also show that the quantizations of braided derivations has the braided Lie structure with respect to the quantizations of the braiding which can be realized within the original braided Lie structure by dequantization.

We deﬁne braided connections in modules and braided curvatures. We prove that the braided curvature is $A$ -linear, skew $σ$ -symmetric and is an $A$ -module homomorphism.

We ﬁnd quantizations of braided connections and braided curvatures. The quantization of the braided curvature is $A$ -linear, skew $σ_{q}$ -symmetric and an $A$ -module homomorphism with respect to the quantized braiding.

Finally we consider braided diﬀerential operators. We show that there is a braided Lie structure on the braided diﬀerential operators. A quantization the braided diﬀerential operators provides an isomorphism of the original braided diﬀerential operators and quantized ones. The quantization of the braided Lie structure can be realized within the original one by dequantization.

This paper is the ﬁrst in a trilogy.

We have found explicit descriptions of all quantizations and braidings in the monoidal category of modules graded by a ﬁnite commutative monoid, [7]. We have proved the same as for any monoidal category for this category, but the picture is somewhat more visible in this case. That is, we have a complete description for braided derivations of graded algebras and graded modules, braided connections, braided curvature, quantizations and so on. This is to be found in the second paper Quantizations of braided derivations. 2. Graded modules, [8].

In [7] we showed that the Fourier transform establishes an isomorphism between the categories of $Ĝ$ -graded modules and $G$ -modules where $G$ is a ﬁnite abelian group and $Ĝ$ is the dual of $G$ . Using this we ﬁnd a description of all quantizations and braiding also for the monoidal category of modules with action by a ﬁnite abelian group $G$ . Again, we have a complete and explicit description for $σ$ -derivations of algebras and modules, braided connections, curvature, diﬀerential operators and quantizations of these structures. This is to be found in the third paper Quantizations of braided derivations. 3. Modules with action by a group, [9].

There are many interesting applications of these results. One of the more interesting applications is quantizations of braided Lie algebras. In the paper [10], which is to be published, we show quantizations of semisimple Lie algebras by quantizations of derivations, for example an alternative quantization of $𝔰 𝔩_{2} (ℂ)$ .

Note that in all three papers we assume that the associativity constraint is trivial.

2 Quantizations and $σ$ -commutativity

In this section we shall recall some results needed later. We have to deﬁne quantizations and $σ$ -commutativity of algebras, modules, co- and bialgebras and internal homomorphisms. Most of this was done by V. V. Lychagin and many results are found in [17].

2.1 Quantizations

A quantization [17] of a monoidal category $C$ is a natural isomorphism of the tensor bifunctor

\begin{array}{rcl} q & : & \otimes \to \otimes, \\ q_{X, Y} & : & X \otimes Y \to X \otimes Y, \end{array}

$X, Y \in O b (C)$ , which preserves the unit and associativity so that the following diagram

(1)

commutes for all $X, Y, Z \in O b (C)$ . We call this the coherence condition for quantizations.

The composition of two quantizations $q_{1}$ and $q_{2}$ is a quantization and the inverse $q^{- 1}$ of a quantization $q$ is a quantization.

2.2 Braidings

A braiding [18] of a monoidal category $C$ is a natural isomorphism

\begin{array}{rcl} σ & : & \otimes \to \otimes \circ τ \\ σ & = & σ_{X, Y} : X \otimes Y \to Y \otimes X, \end{array}

$X, Y \in O b (C)$ , which preserves the unit and associativity such that the following diagrams

X⊗ (Y ⊗ Z) --α--(X ⊗ Y )⊗ Z
| |
| |
1⊗ σ | σ
| |
X⊗ (Z ⊗ Y) Z ⊗ (X ⊗ Y)
| |
| |
α | α
| |
(X⊗ Z) ⊗ Y σ-⊗1-(Z ⊗ X) ⊗ Y

X ⊗ (Y ⊗ Z) -α---(X ⊗ Y) ⊗ Z
| |
| |
σ| σ ⊗ 1
| |
(Y ⊗ Z)⊗ X (Y ⊗ X) ⊗ Z
| |
(−1)| |(−1)
α | α
| |
Y ⊗ (Z ⊗ X) 1⊗-σ-Y ⊗ (X ⊗ Z)

(2)

commute. This is the coherence condition on braidings.

The braiding $σ$ is a symmetry if

σ_{Y, X} \circ σ_{Y, Z} = I d,

(3)

and a monoidal category equipped with such is called symmetric. We shall work only with symmetries.

When the associativity constraint is trivial, the coherence condition gives what we call the bihomomorphism conditions for any braiding $σ$

\begin{array}{rcl} (σ_{X, Z} \otimes 1) \circ (1 \otimes σ_{Y, Z}) & = & σ_{X \otimes Y, Z}, & (4) \\ (1 \otimes σ_{X, Z}) \circ (σ_{X, Y} \otimes 1) & = & σ_{X, Y \otimes Z}, & (5) \end{array}

$X, Y, Z \in O b j (C)$ .

The trivial braiding is the twist, $τ : X \otimes Y \to Y \otimes X$ .

Any braiding composed with the twist, $τ \circ σ$ , is a quantization since the coherence condition for quantizations then is satisﬁed.

Quantizations act on the set of braidings as follows

{(σ_{q})}_{X, Y} = q_{Y, X}^{- 1} \circ σ_{X, Y} \circ q_{X, Y},

and $σ_{q}$ is also a braiding.

2.3 Algebras

Let $A$ be an algebra in a monoidal category $C$ with multiplication

μ : A \otimes A \to A

and unit $η : e \to A$ .

Given a braiding $σ$ we say that $A$ is $σ$ -commutative or $σ$ -symmetric, [16], [12], if

μ = μ \circ σ .

Note that when the associativity constraint is trivial, the bihomomorphism conditions for the $σ$ -commutativity of algebras is

\begin{array}{rcl} μ \circ (μ \otimes 1) \circ (σ_{x, z} \otimes 1) \circ (1 \otimes σ_{y, z}) & = & μ \circ σ_{x y, z} \circ (μ \otimes 1), & (6) \\ μ \circ (1 \otimes μ) \circ (1 \otimes σ_{x, z}) \circ (σ_{x, y} \otimes 1) & = & μ \circ σ_{x, y z} \circ (1 \otimes μ), & (7) \end{array}

for $x, y, z \in A$ .

Given a quantization $q$ we deﬁne a quantization $A_{q}$ of the algebra $A$ as the same object $A$ equipped with a new multiplication

μ_{q} = μ \circ q_{A, A} : A \otimes A \to A .

$A_{q} = (A, μ_{q}, η)$ is an algebra.

If an algebra $A$ in a monoidal category is $σ$ -commutative, then it’s quantization $A_{q}$ is $σ_{q}$ -commutative [17].

2.4 Modules

Let $(A, μ)$ be an algebra in a monoidal category $C$ . Let $E$ be a left $A$ -module with action

ν^{l} : A \otimes E \to E

in a monoidal category. If not stated otherwise, assume that all modules are left modules.

By a quantization $E_{q}$ of the $A$ -module $E$ we mean the same object $E$ equipped with a new action

ν_{q}^{l} = ν^{l} \circ q_{A, E} : A \otimes E \to E .

$E_{q} = (E, ν_{q}^{l})$ is also a left $A$ -module in $C$ .

Let $A$ be a $σ$ -symmetric algebra and $E$ be an $A$ - $A$ -bimodule. Given a braiding $σ$ we say that $E$ is $σ$ -commutative if $ν^{l} = ν^{r} \circ σ_{A, E}$ and $ν^{r} = ν^{l} \circ σ_{E, A}$ [16], that is,

A⊗ E --νl-- X
| |
| |
σ| = |
| |
E⊗ A --νr-- X

(8)

and

r
E⊗ A --ν--- X|
| |
| |
σ| = |
| l
A⊗ E --ν--- X

(9)

commutes. If $σ$ is a symmetry then an $A$ -module $E$ is left $σ$ -commutative if and only if it is right $σ$ -commutative, that is (8) and (9) implies each other. This means that when $σ$ is a symmetry we need only to consider left modules, not bimodules.

When the associativity constraint is trivial, the bihomomorphism conditions for any the $σ$ -commutativity of modules is

\begin{array}{rcl} M \circ (M \otimes 1) \circ (σ \otimes 1) \circ (1 \otimes σ) & = & M \circ σ \circ (M \otimes 1), & (10) \\ M \circ (1 \otimes M) \circ (1 \otimes σ) \circ (σ \otimes 1) & = & M \circ σ \circ (1 \otimes M), & (11) \end{array}

where $M$ is $μ$ , $ν^{l}$ or $ν^{r}$ , depending on the triplet of $A$ and $E$ .

For any $A$ - $A$ -bimodule $E$ the $σ$ -symmetric part is

E_{σ} = \{x \in E ∣ ν^{l} (a \otimes x) = ν^{r} \circ σ_{A, E} (a \otimes x), ν^{r} (x \otimes a) = ν^{l} \circ σ_{E, A} (x \otimes a), \forall a \in A\} .

The quantization of a right $A$ -module $E$ with the action

ν^{r} : E \otimes A \to E,

is done by giving $E$ the new action

ν_{q}^{r} = ν^{r} \circ q_{E, A} : E \otimes A \to E .

A quantization of a $A$ - $A$ -bimodule is the same object $E$ equipped with two new actions $E_{q} = (E, ν_{q}^{l}, ν_{q}^{r})$ . $E_{q}$ is an $A_{q} - A_{q}$ -bimodule in $C$ .

$E$ is $σ$ -commutative, then $E_{q}$ is $σ_{q}$ -commutative;

\begin{array}{rcl} ν^{l} \circ q_{A, E} & = & ν_{q}^{l} \\ = & ν_{q}^{r} \circ {(σ_{q})}_{A, E} \end{array}

and similarly $ν_{q}^{r} = ν_{q}^{l} \circ {(σ_{q})}_{E, A}$ .

Let $E_{σ}$ be the symmetric part of the $A$ - $A$ -bimodule $E$ . Consider the quotient bimodule $E ∕ E_{σ} = E_{σ}^{(0)}$ and deﬁne $E_{σ}^{(1)}$ as the preimage of ${(E ∕ E_{σ})}_{σ}$ with respect to the canonical projection $E \to E ∕ E_{σ}$ . Proceeding, we get a ﬁltration of $E$ by bimodules $E_{σ}^{(i)}, i = - 1, 0, 1, . . .$ , $E_{σ}^{(- 1)} = 0$ . We call the bimodule

E_{σ}^{*} = \cup E_{σ}^{(i)}

a diﬀerential approximation of the $A$ - $A$ -bimodule $E$ .

2.5 Coalgebras

Let $A$ be a coalgebra in $C$ with comultiplication $Δ : A \to A \otimes A$ and counit $ɛ : A \to e$ .

$A$ is $σ$ -cocommutative if

Δ = σ^{- 1} \circ Δ .

Deﬁne a quantization $A_{q}$ of the coalgebra $A$ as the same object $A$ equipped with a new comultiplication deﬁned by

Δ_{q} = q_{A, A}^{- 1} \circ Δ : A \to A \otimes A .

$A_{q} = (A, Δ_{q}, ɛ)$ is a coalgebra.

2.6 Bialgebras

Let $A$ be an algebra in $C$ . Then the tensor square $A \otimes A$ can be considered as an algebra with multiplication

μ_{σ}^{\otimes 2} = (μ \otimes μ) \circ (1 \otimes σ \otimes 1) .

Let $A$ be a coalgebra in $C$ . Then the tensor square $A \otimes A$ can be considered as a coalgebra with comultiplication

Δ_{σ}^{\otimes 2} = (1 \otimes σ \otimes 1) \circ Δ \otimes Δ .

A $σ$ -bialgebra $(A, μ, Δ)$ in a monoidal category $C$ is an algebra $(A, μ)$ and a coalgebra $(A, Δ)$ such that the diagonal

Δ : (A, μ) \to (A \otimes A, μ_{σ}^{\otimes 2})

and counit are algebra morphisms and the multiplication

μ : (A \otimes A, Δ_{σ}^{\otimes 2}) \to (A, Δ)

and unit are coalgebra morphisms.

The quantization $A_{q} = (A, μ_{q}, Δ_{q})$ is a bialgebra in $C$ , [16].

2.7 Internal homomorphisms

For a closed monoidal category $C$ and any two objects $X$ and $Y$ in $C$ there is the internal homomorphism bifunctor $hom$ and the internal homomorphism object $hom (X, Y)$ ,

hom : X \otimes Y \to hom (X, Y)

together with the composition

μ^{h} : hom (Y, Z) \otimes hom (X, Y) \to hom (X, Z),

g \otimes f \mapsto g \circ f = g * f .

The collection of internal homomorphisms is an ”algebra” with respect to $*$ .

If $C$ is equipped with a braiding $σ$ , then the composition of internal homomorphisms is called $σ$ -symmetric if

μ^{h} = μ^{h} \circ σ .

The morphism

e v_{X, Y} : hom (X, Y) \otimes X \to Y

is called the evaluation map if any morphism $f : X \otimes Z \to Y$ can be represented by the composition

f = e v_{X, Y} \circ (\hat{f} \otimes i d_{X})

for a unique morphism $\hat{f} : Z \to hom (X, Y)$ . That is $M o r (Z, hom (X, Y)) ≅ M o r (Z \otimes X, Y)$ .

The evaluation map also can be considered as a result of the multiplication $μ^{h}$ ,

e v_{X, Y} (f \otimes x) = f (x) = f * x,

$x \in X$ , $f \in hom (X, Y)$ .

Given a quantization $q$ , deﬁne a quantization of the internal homomorphisms to be a quantization as an algebra and we equip the internal homomorphisms with a new multiplication. Namely a quantization is a natural isomorphism

Q_{q} : hom (X, Y) \to hom (X, Y)

deﬁned by

Q_{q} (f) * x = f *_{q} x

for all $f \in hom (X, Y)$ and $x \in hom (1, X)$ , where

f *_{q} g = μ^{h} (q_{hom (X, Y), hom (Y, Z)} (f \otimes g)) = μ_{q}^{h} (f \otimes g) .

For any quantization we have

\begin{array}{rcl} Q_{q} (f) * Q_{q} (f) & = & Q_{q} (f *_{q} g), & (12) \\ Q_{q} (x) & = & x . & (13) \end{array}

3 Braided Lie structure of internal homomorphisms

We shall describe the module structure and braided Lie algebra structure of the internal homomorphisms. We describe the same for the quantizations of the internal homomorphisms.

3.1 Module structure of $hom (E, E^{'})$

Let $σ$ be a braiding in a monoidal category $C$ .

Let $A$ be a $σ$ -commutative algebra, let $E$ and $E^{'}$ be left $A$ -modules and let $hom (E, E^{'})$ be the set of internal homomorphisms.

The set $hom (E, E^{'})$ , is an $A$ - $A$ -bimodule with the left and right multiplications deﬁned by

\begin{array}{rcl} ν_{a}^{l} (f) (x) & = & ν^{l} (a \otimes f) (x) = a f (x), \\ ν_{a}^{r} (f) (x) & = & ν^{r} (f \otimes a) (x) = f (a x), \end{array}

$a \in A$ , $x \in E$ .

Proposition 1 Let $σ$ be a symmetry and $E$ and $E^{'}$ be $σ$ -commutative $A$ - $A$ -modules. Then $hom (E, E^{'})$ is $σ$ -commutative as a module, that is, the diagrams

l
A⊗ hom(X, Y) ν- hom(X, Y )
| |
| |
σ | = |
| r |
hom(X, Y )⊗ A ν- hom(X, Y )

(14)

and

νr
hom(X, Y )⊗ A -- hom(X, Y )
| |
σ | = |
| |
νl
A⊗ hom(X, Y) -- hom(X, Y )

(15)

commute.

Proof. Write the actions as

\begin{array}{rcl} e v_{E, E^{'}} \circ (ν^{l} \otimes 1) & = & v_{E^{'}}^{l} \circ (1 \otimes e v_{E, E^{'}}) : A \otimes hom (E, E^{'}) \otimes E \to E^{'}, \\ e v_{E, E^{'}} \circ (ν^{r} \otimes 1) & = & e v_{E, E^{'}} \circ (1 \otimes v_{E}^{l}) : hom (E, E^{'}) \otimes A \otimes E \to E^{'} . \end{array}

Then

\begin{array}{rcl} ν^{l} (a \otimes f) (x) & = & ν_{E^{'}}^{l} \circ (1 \otimes e v_{E, E^{'}}) (a \otimes f \otimes x) \\ = & ν_{E^{'}}^{r} \circ σ \circ (1 \otimes e v_{E, E^{'}}) (a \otimes f \otimes x) \\ = & e v_{E, E^{'}} \circ (1 \otimes ν_{E}^{r}) \circ (1 \otimes σ) \circ (σ \otimes 1) (a \otimes f \otimes x) \\ = & e v_{E, E^{'}} \circ (1 \otimes ν_{E}^{l}) \circ (σ \otimes 1) (a \otimes f \otimes x) \\ = & ν^{r} \circ σ_{A, hom (E, E^{'})} (a \otimes f) (x), \end{array}

and

\begin{array}{rcl} ν^{r} (f \otimes a) (x) & = & e v_{E, E^{'}} \circ (1 \otimes v_{E}^{l}) (f \otimes a \otimes x) \\ = & e v_{E, E^{'}} \circ (1 \otimes v_{E}^{r}) \circ (1 \otimes σ) (f \otimes a \otimes x) \\ = & ν_{E^{'}}^{r} \circ (e v_{E, E^{'}} \otimes 1) \circ (1 \otimes σ) (f \otimes a \otimes x) \\ = & ν_{E^{'}}^{l} \circ σ \circ (e v_{E, E^{'}} \otimes 1) \circ (1 \otimes σ) (f \otimes a \otimes x) \\ = & ν_{E^{'}}^{l} \circ (1 \otimes e v_{E, E^{'}}) \circ (σ \otimes 1) \circ (1 \otimes σ) \circ (1 \otimes σ) (f \otimes a \otimes x) \\ = & ν_{E^{'}}^{l} \circ (1 \otimes e v_{E, E^{'}}) \circ (σ \otimes 1) (f \otimes a \otimes x) \\ = & ν^{l} \circ σ (f \otimes a) (x), \end{array}

$a \in A$ , $f \in hom (E, E^{'})$ , $x \in E .$ _

3.2 Braided commutators and Lie structure of $hom (E, E)$

Let $σ$ be a braiding, $A$ be an $σ$ -symmetric algebra and $E$ be a left $σ$ -symmetric $A$ -module. Consider $hom (E, E)$ .

Deﬁnition 2 Deﬁne the $σ$ -bracket or $σ$ -commutator of $hom (E, E)$

c_{σ} : hom (E, E) \otimes hom (E, E) \to hom (E, E)

c_{σ} = {[_,_]}^{σ} = μ^{h} - μ^{h} \circ σ .

Proposition 3 Let $σ$ be a symmetry. The $σ$ -bracket satisﬁes the conditions,

\begin{array}{rcl} c_{σ} \circ (ν^{l} \otimes 1) (a \otimes f \otimes g) & = & (ν^{l} \circ (1 \otimes c_{σ}) - ν^{l} \circ (e v_{E, E} \otimes 1) \circ σ) (a \otimes f \otimes g), \\ c_{σ} \circ (1 \otimes ν^{l}) (f \otimes a \otimes g) & = & (ν^{l} \circ (1 \otimes c_{σ}) \circ (σ \otimes 1) + ν^{l} \circ (e v_{E, E} \otimes 1)) (f \otimes a \otimes g), \end{array}

$a \in A$ , $f, g \in hom (E, E)$ .

Proof.

\begin{array}{rcl} c_{σ} \circ (ν^{l} \otimes 1) & = & μ^{h} \circ (ν^{l} \otimes 1) - μ^{h} \circ σ \circ (ν^{l} \otimes 1) \\ = & ν^{l} \circ (1 \otimes μ^{h}) - μ^{h} \circ (1 \otimes ν^{l}) \circ (σ \otimes 1) \circ (1 \otimes σ) \\ = & ν^{l} \circ (1 \otimes μ^{h}) - μ^{h} \circ (ν^{r} \otimes 1) \circ (σ \otimes 1) \circ (1 \otimes σ) \\ - ν^{l} \circ (1 \otimes μ^{h}) \circ (σ \otimes 1) \circ (σ \otimes 1) \circ (1 \otimes σ) \\ = & ν^{l} \circ (1 \otimes μ^{h}) - μ^{h} \circ σ \circ (ν^{l} \otimes 1) - ν^{l} \circ (1 \otimes μ^{h}) \circ (1 \otimes σ) \\ = & ν^{l} \circ (1 \otimes c_{σ}) - ν^{l} \circ (e v_{E, E} \otimes 1) \circ σ, \end{array}

and

\begin{array}{rcl} c_{σ} \circ (1 \otimes ν^{l}) & = & μ^{h} \circ (1 \otimes ν^{l}) - μ^{h} \circ σ \circ (1 \otimes ν^{l}) \\ = & μ^{h} \circ (ν^{r} \otimes 1) + ν^{l} \circ (1 \otimes μ^{h}) \circ (σ \otimes 1) \\ - ν^{l} \circ (1 \otimes μ^{h}) \circ (1 \otimes σ) \circ (σ \otimes 1) \\ = & ν^{l} \circ (e v_{E, E} \otimes 1) + ν^{l} \circ (1 \otimes c_{σ}) \circ (σ \otimes 1) . \end{array}

Proposition 4 The $σ$ -bracket is $σ$ -invariant,

μ^{h} \circ σ \circ (c_{σ} \otimes 1) = μ^{h} \circ (1 \otimes c_{σ}) \circ (σ \otimes 1) \circ (1 \otimes σ) .

Proof.

\begin{array}{rcl} μ^{h} \circ σ \circ (c_{σ} \otimes 1) & = & μ^{h} \circ σ \circ ((μ^{h} - μ^{h} \circ σ) \otimes 1) \\ = & μ^{h} \circ σ \circ (μ^{h} \otimes 1) - μ^{h} \circ σ \circ (μ^{h} \otimes 1) \circ (σ \otimes 1), \end{array}

and

\begin{array}{rcl} μ^{h} \circ (1 \otimes c_{σ}) \circ (σ \otimes 1) \circ (1 \otimes σ) \\ = & μ^{h} \circ (1 \otimes (μ^{h} - μ^{h} \circ σ)) \circ (σ \otimes 1) \circ (1 \otimes σ) \\ = & μ^{h} \circ (1 \otimes μ^{h}) \circ (σ \otimes 1) \circ (1 \otimes σ) - μ^{h} \circ (1 \otimes μ^{h}) \circ (1 \otimes σ) \circ (σ \otimes 1) \circ (1 \otimes σ) \\ = & μ^{h} \circ (μ^{h} \otimes 1) \circ (σ \otimes 1) \circ (1 \otimes σ) - μ^{h} \circ (μ^{h} \otimes 1) \circ (σ \otimes 1) \circ (1 \otimes σ) \\ = & μ^{h} \circ σ \circ (μ^{h} \otimes 1) - μ^{h} \circ σ \circ (μ^{h} \otimes 1) \circ (σ \otimes 1) . \end{array}

Proposition 5 Let $σ$ be a symmetry. The $σ$ -commutator satisﬁes skew $σ$ -symmetricity, that is, the diagram

-cσ-
hom(X, X) ⊗|hom(X, X) hom(X,X)
| |
σ | = |
| |
−-cσ
hom(X, X) ⊗ hom(X, X) hom(X,X)

(16)

commutes, equivalently

c_{σ} = - c_{σ} \circ σ .

Proof.

\begin{array}{rcl} c_{σ} \circ σ (f \otimes g) & = & μ^{h} \circ σ (f \otimes g) - μ^{h} \circ σ \circ σ (f \otimes g) \\ = & μ^{h} \circ σ (f \otimes g) - μ^{h} (f \otimes g) \\ = & - c_{σ} (f \otimes g), \end{array}

$f, g \in hom (E, E)$ . _

Proposition 6 Let $σ$ be a symmetry. Then $hom (E, E)$ equipped with the $σ$ -commutator satisﬁes the $σ$ -Jacobi identity,

c_{σ} \circ (1 \otimes c_{σ}) = c_{σ} \circ (c_{σ} \otimes 1) + c_{σ} \circ (1 \otimes c_{σ}) \circ (σ \otimes 1) .

Proof. The Jacobi identity applied to three elements in $hom (E, E)$ is satisﬁed when

\begin{array}{rcl} c_{σ} \circ (1 \otimes c_{σ}) \\ = & μ^{h} \circ (1 \otimes μ^{h}) & (17) \\ - μ^{h} \circ (1 \otimes μ^{h}) \circ (1 \otimes σ) & (18) \\ - μ^{h} \circ σ \circ (1 \otimes μ^{h}) & (19) \\ + μ^{h} \circ σ \circ (1 \otimes μ^{h}) \circ (1 \otimes σ) & (20) \end{array}

is equal to

\begin{array}{rcl} c_{σ} \circ (c_{σ} \otimes 1) + c_{σ} \circ (1 \otimes c_{σ}) \circ (σ \otimes 1) \\ = & μ^{h} \circ (μ^{h} \otimes 1) & (21) \\ - μ^{h} \circ (μ^{h} \otimes 1) \circ (σ \otimes 1) & (22) \\ - μ^{h} \circ σ \circ (μ^{h} \otimes 1) & (23) \\ + μ^{h} \circ σ \circ (μ^{h} \otimes 1) \circ (σ \otimes 1) & (24) \\ + μ^{h} \circ (1 \otimes μ^{h}) \circ (σ \otimes 1) & (25) \\ - μ^{h} \circ (1 \otimes μ^{h}) \circ (1 \otimes σ) \circ (σ \otimes 1) & (26) \\ - μ^{h} \circ σ \circ (1 \otimes μ^{h}) \circ (σ \otimes 1) & (27) \\ + μ^{h} \circ σ_{g, h f} \circ (1 \otimes μ^{h}) \circ (1 \otimes σ) \circ (σ \otimes 1), & (28) \end{array}

which is the case since

$(a) = (e);$

μ^{h} \circ (1 \otimes μ^{h}) = μ^{h} \circ (μ^{h} \otimes 1),

$(b) = (k);$

\begin{array}{rcl} μ^{h} \circ (1 \otimes μ^{h}) \circ (1 \otimes σ) & = & μ^{h} \circ σ \circ (1 \otimes μ^{h}) \circ (σ \otimes 1) \\ = & μ^{h} \circ (1 \otimes μ^{h}) \circ (1 \otimes σ) \circ (σ \otimes 1) \circ (σ \otimes 1), \end{array}

$(c) = (j);$

μ^{h} \circ σ \circ (1 \otimes μ^{h}) = μ^{h} \circ (1 \otimes μ^{h}) \circ (1 \otimes σ) \circ (σ \otimes 1),

$(d) = (h);$

\begin{array}{rcl} μ^{h} \circ σ \circ (1 \otimes μ^{h}) \circ (1 \otimes σ) & = & μ^{h} \circ σ \circ (μ^{h} \otimes 1) \circ (σ \otimes 1) \\ μ^{h} \circ (1 \otimes μ^{h}) \circ (1 \otimes σ) \circ (σ \otimes 1) \circ (1 \otimes σ) & = & μ^{h} \circ (μ^{h} \otimes 1) \circ (σ \otimes 1) \circ (1 \otimes σ) \circ (σ \otimes 1), \end{array}

$(f) + (i) = 0;$

μ^{h} \circ (μ^{h} \otimes 1) \circ (σ \otimes 1) = μ^{h} \circ (1 \otimes μ^{h}) \circ (σ \otimes 1),

and $(g) + (l) = 0;$

μ^{h} \circ σ \circ (μ^{h} \otimes 1) = μ^{h} \circ σ \circ (1 \otimes μ^{h}) \circ (1 \otimes σ) \circ (σ \otimes 1),

since

μ^{h} \circ σ \circ (μ^{h} \otimes 1) = μ^{h} \circ (μ^{h} \otimes 1) \circ (σ \otimes 1) \circ (1 \otimes σ),

and

μ^{h} \circ σ \circ (1 \otimes μ^{h}) \circ (1 \otimes σ) \circ (σ \otimes 1) = μ^{h} \circ (1 \otimes μ^{h}) \circ (1 \otimes σ) \circ (σ \otimes 1) \circ (1 \otimes σ) \circ (σ \otimes 1) .

This is shown by using the identities (6), (7) and the fact that, $σ$ satisﬁes the Yang-Baxter equation,

(1 \otimes σ) \circ (σ \otimes 1) \circ (1 \otimes σ) = (σ \otimes 1) \circ (1 \otimes σ) \circ (σ \otimes 1) .

Remark 7 In the proof of the propositions 5 and 6 we can assume that the internal homomorphisms are equipped with the $σ$ -commutative composition $μ^{h} = μ^{h} \circ σ$ instead of using the fact that $σ$ is a symmetry.

Deﬁnition 8 A $σ$ -Lie algebra is an algebra equipped with a skew $σ$ -symmetric $σ$ -commutator that is $σ$ -invariant and satisﬁes the $σ$ -Jacobi identity.

This is the deﬁnition of braided Lie algebras as introduced by D. Gurevich, [4]. By the propositions 4, 5 and 6 we have proved the following.

Theorem 9 Let $σ$ be a symmetry and an algebra $A$ and a left $A$ -module $E$ be $σ$ -symmetric. Then $hom (E, E)$ is a $σ$ -Lie algebra.

3.3 Quantization of $hom (E, E)$

We will deﬁne the quantization of the internal homomorphisms and describe the structure on the set of all such, also within the original internal homomorphisms.

Deﬁnition 10 Given a quantization $q$ and $f \in hom (E, E)$ , its quantization is deﬁned by

f_{q} (x) = Q_{q} (f) (x) \overset{d e f}{=} e v_{E, E} \circ q (f \otimes x),

(29)

$x \in E$ .

It is easy to see that $Q_{q} (f)$ is an internal homomorphism of the quantized $A_{q}$ -module $E_{q}$ .

Let $f$ and $g$ be internal homomorphisms. The quantization of composition of internal homomorphisms is

f *_{q} g = μ_{q}^{h} (f \otimes g) .

(30)

The collection of all $Q_{q} (f)$ , $f \in hom (E, E)$ , equipped with the quantization of the composition is denoted by ${hom}^{q} (E_{q}, E_{q})$ .

By proposition 1 is ${hom}^{q} (E_{q}, E_{q})$ a $σ_{q}$ -symmetric module if $E$ is a $σ$ -symmetric $A$ - $A$ -bimodule.

By theorem 9, if $σ$ is a symmetry is ${hom}^{q} (E_{q}, E_{q})$ a $σ_{q}$ -Lie algebra with respect to the $σ_{q} - q$ -bracket (or simply $σ_{q}$ -bracket when it is clear that the multiplication or composition is the quantized multiplication).

Let $γ$ be a braiding and $p$ any quantization. Deﬁne the $γ - p$ -bracket on internal homomorphisms,

c_{γ}^{p} = {[-, -]}_{p}^{γ} = μ_{p}^{h} - μ_{p}^{h} \circ γ .

(31)

Lemma 11 The $σ_{q} - q$ -bracket satisﬁes

c_{σ_{q}}^{q} = c_{σ} \circ q .

Proof.

\begin{array}{rcl} c_{σ_{q}}^{q} & = & μ_{q}^{h} - μ_{q}^{h} \circ σ_{q} \\ = & μ_{q}^{h} - μ^{h} \circ q \circ q^{- 1} \circ σ \circ q \\ = & μ^{h} \circ q - σ \circ μ^{h} \circ q \\ = & c_{σ} \circ q . \end{array}

The inverse of the quantization of $f \in {hom}^{q} (E_{q}, E_{q})$ is denoted by

Q_{q}^{- 1} (f) = f_{c},

and is called the dequantization of $f$ .

Lemma 12 The composition of internal homomorphisms satisﬁes

f_{c} *_{q} g_{c} = {(f \circ g)}_{c},

(32)

$f, g \in {hom}^{q} (E_{q}, E_{q})$ .

Proof.

\begin{array}{rcl} f_{c} *_{q} g_{c} (x) & = & Q_{q}^{- 1} (f) *_{q} Q_{q}^{- 1} (g) (x) \\ = & e v_{E, E} \circ q^{- 1} \circ (1 \otimes e v_{E, E}) \circ (1 \otimes q^{- 1}) \circ (q \otimes 1) (f \otimes g \otimes x) \\ = & e v_{E, E} \circ q^{- 1} \circ (μ^{h} \otimes 1) \circ (q^{- 1} \otimes 1) \circ (q \otimes 1) (f \otimes g \otimes x) \\ = & Q_{q}^{- 1} (f \circ g) (x) \\ = & {(f \circ g)}_{c}, \end{array}

Note,

a_{c} = a,

for $a \in A_{q}$ .

The set of all $Q_{q}^{- 1} (f)$ , $f \in {hom}^{q} (E_{q}, E_{q})$ , is equipped with the bracket $c_{σ_{q}}^{q}$ and the $A$ -module structure

a *_{q} f_{c} = ν^{l} \circ q (a \otimes f_{c}),

(33)

$a \in A$ .

The dequantization of a internal homomorphism operates on $A$ in the classical manner, but satisﬁes somewhat diﬀerent properties than the classical, as the following theorem states. (See also [14])

Theorem 13 Let $σ$ be a symmetry. The $σ_{q}$ -Lie algebra structure of ${hom}^{q} (E_{q}, E_{q})$ can be realized within the classical, $hom (E, E)$ , by dequantization as follows.

Let $f, g \in {hom}^{q} (E_{q}, E_{q})$ , $a \in A_{q}$ . Then linearity is

{(f + g)}_{c} = f_{c} + g_{c},

(i)

$A$ -module structure,

{(a \circ f)}_{c} = a *_{q} f_{c},

(ii)

and the braided commutator satisﬁes

Q_{q}^{- 1} \circ c_{σ_{q}} = c_{σ_{q}}^{q} \circ (Q_{q}^{- 1} \otimes Q_{q}^{- 1}) .

(iii)

Proof. $(i)$ :

Q_{q}^{- 1} (f + g) = Q_{q}^{- 1} (f) + Q_{q}^{- 1} (g) .

$(i i)$ : By lemma 12,

a_{c} *_{q} f_{c} = a *_{q} f_{c} = {(a \circ f)}_{c} .

$(i i i)$ :

\begin{array}{rcl} {[Q_{q}^{- 1} (f), Q_{q}^{- 1} (g)]}_{q}^{σ_{q}} (x) \\ = & e v_{E, E} \circ q^{- 1} \circ (1 \otimes e v_{E, E}) \circ (1 \otimes q^{- 1}) \circ (q \otimes 1) (f \otimes g \otimes x) \\ - e v_{E, E} \circ q^{- 1} \circ (1 \otimes e v_{E, E}) \circ (1 \otimes q^{- 1}) \circ (q \otimes 1) \circ (σ_{q} \otimes 1) (f \otimes g \otimes x) \\ = & e v_{E, E} \circ q^{- 1} \circ (μ^{h} \otimes 1) \circ (q^{- 1} \otimes 1) \circ (q \otimes 1) (f \otimes g \otimes x) \\ - e v_{E, E} \circ q^{- 1} \circ (μ^{h} \otimes 1) \circ (q^{- 1} \otimes 1) \circ (q \otimes 1) \circ (σ_{q} \otimes 1) (f \otimes g \otimes x) \\ = & Q_{q}^{- 1} ({[f, g]}^{σ_{q}}) (x), \end{array}

$x \in E$ , $f, g \in {hom}^{q} (E_{q}, E_{q})$ . _

4 Braided derivations in monoidal categories

Let $C$ be a monoidal category equipped with a braiding $σ$ .

Let $A$ be a $σ$ -symmetric algebra in $C$ with the multiplication $μ$ , and $E$ and $E^{'}$ be $A$ - $A$ -modules.

Denote by ${hom}_{σ} (E, E^{'}) = {(hom (E, E^{'}))}_{σ}$ the $σ$ -symmetric part of the $A - A$ -bimodule $hom (E, E^{'})$ . The modules

{(hom (E, E^{'}))}_{σ}^{(i)} = D i f f_{i}^{σ} (E, E^{'})

are called the braided or $σ$ -diﬀerential operators, see [16].

4.1 Braided derivations of algebras

With an additional condition we deﬁne $σ$ -derivations of algebras.

Deﬁnition 14 Deﬁne $σ$ -derivations or braided derivations of $A$ with values in a $A$ - $A$ -bimodule $E$ as

D e r^{σ} (E) = \{f \in D i f f_{1}^{σ} (A, E) ∣ f (1) = 0\} .

An internal homomorphism $f : A \to E$ is a $σ$ -derivation if and only if $f : A \to E_{σ} \subset E$ and $f$ satisﬁes the braided or $σ$ -Leibniz rule, [17],

\begin{array}{rcl} e v_{A, E} \circ (1 \otimes μ) (f \otimes a \otimes b) & (34) \\ = & (ν^{r} \circ (e v_{A, E} \otimes 1) + ν^{l} \circ (1 \otimes e v_{A, E}) \circ (σ \otimes 1)) (f \otimes a \otimes b), \end{array}

\begin{array}{rcl} e v_{A, E} \circ (1 \otimes μ) \circ (σ \otimes 1) (a \otimes f \otimes b) & (35) \\ = & (ν^{l} \circ (1 \otimes e v_{A, E}) + ν^{r} \circ (e v_{A, E} \otimes 1) \circ (σ \otimes 1)) (a \otimes f \otimes b) . \end{array}

If the braiding is a symmetry, then the two Leibniz rules implies each other.

From now on, assume that every braiding

σ

is a symmetry. We can then assume $E$ is a left $A$ -module.

Proposition 15 $D e r^{σ} (E)$ has a left $A$ -module structure deﬁned by

ν^{l} (a \otimes \partial) (b) \overset{d e f}{=} ν_{E}^{l} \circ (1 \otimes e v_{A, E}) (a \otimes \partial \otimes b) = a \partial (b),

(36)

$\partial \in D e r^{σ} (E)$ , $a, b \in A$ , $ν_{E}^{l}$ is the action of $A$ on $E$ .

Proof. We need to show that $ν_{a}^{l} (\partial)$ is a $σ$ -derivation, that is satisﬁes the $σ$ -Leibniz rule, and using the $σ$ -Leibniz rule for $\partial$ ,

a \partial (b c) = (\begin{matrix} ν_{E}^{l} \circ (1 \otimes ν_{E}^{l}) \circ (1 \otimes e v_{A, A} \otimes 1) \\ + ν_{E}^{l} \circ (1 \otimes ν_{E}^{l}) (1 \otimes 1 \otimes e v_{E, E}) \circ (1 \otimes σ \otimes 1) \end{matrix}) (a \otimes \partial \otimes b \otimes c),

$a, b, c \in A$ , $\partial \in D e r^{σ} (E)$ , and

\begin{array}{rcl} ν_{E}^{l} \circ (1 \otimes ν_{E}^{l}) \circ (1 \otimes e v_{A, A} \otimes 1) + ν_{E}^{l} \circ (1 \otimes ν_{E}^{l}) (1 \otimes 1 \otimes e v_{E, E}) \circ (1 \otimes σ \otimes 1) \\ = & ν_{E}^{l} \circ (ν_{E}^{l} \otimes 1) \circ (1 \otimes e v_{A, A} \otimes 1) \\ + ν_{E}^{l} \circ ((μ \circ σ) \otimes 1) \circ (1 \otimes 1 \otimes e v_{E, E}) \circ (1 \otimes σ \otimes 1) \\ = & ν_{E}^{l} \circ (e v_{A, A} \otimes 1) \circ (ν^{l} \otimes 1 \otimes 1) \\ + ν_{E}^{l} \circ (1 \otimes ν_{E}^{l}) \circ (1 \otimes 1 \otimes e v_{E, E}) \circ (σ \otimes 1 \otimes 1) \circ (1 \otimes σ \otimes 1) \\ = & ν_{E}^{l} \circ (e v_{A, A} \otimes 1) \circ (ν^{l} \otimes 1 \otimes 1) \\ + ν_{E}^{l} \circ (1 \otimes e v_{E, E}) \circ (1 \otimes ν^{l} \otimes 1) \circ (σ \otimes 1 \otimes 1) \circ (1 \otimes σ \otimes 1), \end{array}

and the condition,

ν^{l} (a \otimes \partial) (1) = a \partial (1) = 0,

is satisﬁed. _

If we consider $D i f f_{1}^{σ} (A, E)$ and not $D e r^{σ} (E)$ then there is a right $A$ -module structure.

Proposition 16 $D i f f_{1}^{σ} (A, E)$ has in addition to the left $A$ -module structure deﬁned by (36), a right $A$ -module structure deﬁned by

ν^{r} (\partial \otimes a) (b) \overset{d e f}{=} e v_{A, E} \circ (1 \otimes μ) (\partial \otimes a \otimes b) = \partial (a b),

$\partial \in D i f f_{1}^{σ} (A, E)$ , $a, b \in A$ .

Proof. The $σ$ -Leibniz rule for $\partial a$ is,

\partial a (b c) = (\begin{matrix} ν_{E}^{r} \circ (e v_{A, E} \otimes 1) \circ (ν^{r} \otimes 1 \otimes 1) \\ + ν_{E}^{l} \circ (1 \otimes e v_{A, E}) \circ (σ_{\partial a, b} \otimes 1) \circ (ν^{r} \otimes 1 \otimes 1) \end{matrix}) (\partial \otimes a \otimes b \otimes c) .

(37)

Using the $σ$ -Leibniz rule for $\partial$ ,

\begin{array}{rcl} \partial a (b c) & = & e v_{A, E} \circ (1 \otimes μ) \circ (1 \otimes 1 \otimes μ) (\partial \otimes a \otimes b \otimes c) \\ = & e v_{A, E} \circ (1 \otimes μ) \circ (1 \otimes μ \otimes 1) (\partial \otimes a \otimes b \otimes c) \\ = & (\begin{matrix} ν_{E}^{r} \circ (e v_{A, E} \otimes 1) \circ (1 \otimes μ \otimes 1) \\ + ν_{E}^{l} \circ (1 \otimes e v_{A, E}) \circ (σ \otimes 1) \circ (1 \otimes μ \otimes 1) \end{matrix}) (\partial \otimes a \otimes b \otimes c), \end{array}

$a, b, c \in A$ , $\partial \in D i f f_{1}^{σ} (A, E)$ , we see that (37) is satisﬁed as

\begin{array}{rcl} ν_{E}^{l} \circ (1 \otimes e v_{A, E}) \circ (σ \otimes 1) \circ (1 \otimes μ \otimes 1) \\ = & ν_{E}^{l} \circ (1 \otimes e v_{A, E}) \circ (1 \otimes ν^{l} \otimes 1) \circ (1 \otimes σ \otimes 1) \circ (σ \otimes 1 \otimes 1) \\ = & ν_{E}^{l} \circ (1 \otimes ν^{l}) \circ (1 \otimes 1 \otimes e v_{A, E}) \circ (1 \otimes σ \otimes 1) \circ (σ \otimes 1 \otimes 1) \\ = & ν_{E}^{l} \circ (μ \otimes 1) \circ (1 \otimes 1 \otimes e v_{A, E}) \circ (1 \otimes σ \otimes 1) \circ (σ \otimes 1 \otimes 1) \\ = & ν_{E}^{l} \circ (μ \otimes 1) \circ (σ \otimes 1) \circ (1 \otimes 1 \otimes e v_{A, E}) \circ (1 \otimes σ \otimes 1) \circ (σ \otimes 1 \otimes 1) \\ = & ν_{E}^{l} \circ (1 \otimes e v_{A, E}) \circ (μ \otimes 1 \otimes 1) \circ (σ \otimes 1) \circ (1 \otimes σ \otimes 1) \circ (σ \otimes 1 \otimes 1) \\ = & ν_{E}^{l} \circ (1 \otimes e v_{A, E}) \circ (σ \otimes 1) \circ (ν^{l} \otimes 1 \otimes 1) \circ (σ \otimes 1 \otimes 1) \\ = & ν_{E}^{l} \circ (1 \otimes e v_{A, E}) \circ (σ \otimes 1) \circ (ν^{r} \otimes 1 \otimes 1) \circ (σ \otimes 1 \otimes 1) \circ (σ \otimes 1 \otimes 1) \\ = & ν_{E}^{l} \circ (1 \otimes e v_{A, E}) \circ (σ \otimes 1) \circ (ν^{r} \otimes 1 \otimes 1) . \end{array}

For the rest of the paper, consider the $σ$ -derivations of a $σ$ -symmetric algebra $A$ with values in $A$ , denoted by

D e r^{σ} (A) .

Proposition 17 The $σ$ -commutator of two $σ$ -derivations of $A$ is a $σ$ -derivation of $A$ ,

c_{σ} : D e r^{σ} (A) \otimes D e r^{σ} (A) \to D e r^{σ} (A) .

Proof. The $σ$ -Leibniz rule is satisﬁed for the $σ$ -commutator of two $σ$ -derivations,

\begin{array}{rcl} c_{σ} (\partial_{1} \otimes \partial_{2}) (a b) & = & μ^{h} (\partial_{1} \otimes \partial_{2}) (a b) - μ^{h} \circ σ (\partial_{1} \otimes \partial_{2}) (a b) \\ = & c_{σ} (\partial_{1} \otimes \partial_{2}) (a) b + μ \circ (1 \otimes e v_{A, A}) (σ (c_{σ} (\partial_{1} \otimes \partial_{2}) \otimes a) \otimes b), \end{array}

$a, b \in A$ , $\partial_{1}, \partial_{2} \in D e r^{σ} (A)$ , as we see,

\begin{array}{rcl} e v_{A, A} \circ (1 \otimes e v_{A, A}) \circ (1 \otimes 1 \otimes μ) - e v_{A, A} \circ (1 \otimes e v_{A, A}) \circ (σ \otimes 1 \otimes 1) \circ (1 \otimes 1 \otimes μ) \\ = & e v_{A, A} \circ (1 \otimes μ) \circ (1 \otimes e v_{A, A} \otimes 1) + e v_{A, A} \circ (1 \otimes μ) \circ (1 \otimes 1 \otimes e v_{A, A}) \circ (1 \otimes σ \otimes 1) \\ - e v_{A, A} \circ (1 \otimes μ) \circ (1 \otimes e v_{A, A} \otimes 1) \circ (σ \otimes 1 \otimes 1) \\ - e v_{A, A} \circ (1 \otimes μ) \circ (1 \otimes 1 \otimes e v_{A, A}) \circ (1 \otimes σ \otimes 1) \circ (σ \otimes 1 \otimes 1) \\ = & μ \circ (e v_{A, A} \otimes 1) \circ (1 \otimes e v_{A, A} \otimes 1) + μ \circ (1 \otimes e v_{A, A}) \circ (σ \otimes 1) \circ (1 \otimes e v_{A, A} \otimes 1) \\ + μ \circ (e v_{A, A} \otimes 1) \circ (1 \otimes 1 \otimes e v_{A, A}) \circ (1 \otimes σ \otimes 1) \\ + μ \circ (1 \otimes e v_{A, A}) \circ (1 \otimes 1 \otimes e v_{A, A}) \circ (σ \otimes 1 \otimes 1) \circ (1 \otimes σ \otimes 1) \\ - μ \circ (e v_{A, A} \otimes 1) \circ (1 \otimes e v_{A, A} \otimes 1) \circ (σ \otimes 1 \otimes 1) \\ - μ \circ (1 \otimes e v_{A, A}) \circ (σ \otimes 1) \circ (1 \otimes e v_{A, A} \otimes 1) \circ (σ \otimes 1 \otimes 1) \\ - μ \circ (e v_{A, A} \otimes 1) \circ (1 \otimes 1 \otimes e v_{A, A}) \circ (1 \otimes σ \otimes 1) \circ (σ \otimes 1 \otimes 1) \\ - μ \circ (1 \otimes e v_{A, A}) \circ (1 \otimes 1 \otimes e v_{A, A}) \circ (σ \otimes 1 \otimes 1) \circ (1 \otimes σ \otimes 1) \circ (σ \otimes 1 \otimes 1) \\ = & μ \circ (e v_{A, A} \otimes 1) \circ (c_{σ} \otimes 1 \otimes 1) + μ \circ (1 \otimes e v_{A, A}) \circ (σ \otimes 1) \circ (μ^{h} \otimes 1 \otimes 1) \\ - μ \circ (1 \otimes e v_{A, A}) \circ (σ \otimes 1) \circ (μ^{h} \otimes 1 \otimes 1) \circ (σ \otimes 1 \otimes 1) \\ = & μ \circ (e v_{A, A} \otimes 1) \circ (c_{σ} \otimes 1 \otimes 1) + μ \circ (1 \otimes e v_{A, A}) \circ (σ \otimes 1) \circ (c_{σ} \otimes 1 \otimes 1) . \end{array}

Furthermore,

c_{σ} (\partial_{1} \otimes \partial_{2}) (1) = μ^{h} (\partial_{1} \otimes \partial_{2}) (1) - μ^{h} \circ σ (\partial_{1} \otimes \partial_{2}) (1) = 0 .

Corollary 18 Let the braiding $σ$ be a symmetry. Then $D e r^{σ} (A)$ equipped with the $σ$ -commutator is a $σ$ -Lie algebra.

4.2 Braided derivations of modules

Let $A$ be a $σ$ -symmetric algebra and let $E$ be a $σ$ -symmetric $A$ -module.

Deﬁnition 19 An operator $\partial : E \to E$ is said to be a $σ$ -derivation of $E$ over $\partial^{A} \in D e r^{σ} (A)$ if $\partial$ satisfy the $σ$ -Leibniz rule with respect to $\partial^{A}$

\partial (a x) = \partial^{A} (a) x + ν_{E}^{l} \circ (1 \otimes e v_{A, A}) \circ (σ \otimes 1) (\partial \otimes a \otimes x),

(38)

$x \in E$ , $a \in A$ and $ν_{E}^{l}$ is the action of $A$ on $E$ .

The pair $(\partial, \partial_{A})$ is called a $σ$ -derivation of $E$ over $A$ . (We could also call this pair a $σ$ - $𝔇$ -module, [13].)

The morphism $π : (\partial, \partial_{A}) \to \partial_{A}$ we call the projection from the $σ$ -derivations of $E$ over $A$ to the $σ$ -derivations of $A$ .

The set of $σ$ -derivations of $E$ over $A$ is denoted by $D e r^{(σ, A)} (E)$ .

Proposition 20 $D e r^{(σ, A)} (E)$ has a left $A$ -module structure deﬁned by

ν_{a}^{l} (\partial) (x) \overset{d e f}{=} ν_{E}^{l} \circ (1 \otimes e v_{E, E}) (a \otimes \partial \otimes x) = a \partial (x),

$\partial \in D e r^{(σ, A)} (E)$ , $a \in A$ , $x \in E$ .

Proof. Just repeat the proof of proposition 15. _

If we consider $D i f f_{1}^{σ} (E, E)$ and not $D e r^{(σ, A)} (E)$ then there is a right $A$ -module structure.

Proposition 21 $D i f f_{1}^{σ} (E, E)$ has in addition a right $A$ -module structure deﬁned by

ν_{a}^{r} (\partial) (x) \overset{d e f}{=} e v_{E, E} \circ (1 \otimes ν_{E}^{l}) (\partial \otimes a \otimes x) = \partial (a x),

$\partial \in D i f f_{1}^{σ} (E, E)$ , $a \in A$ , $x \in E$ .

Proof. Just repeat the proof of proposition 16. _

Proposition 22 The $σ$ -commutator of two $(σ, A)$ -derivations of $E$ is a $(σ, A)$ -derivation of $E$ ,

c_{σ} : D e r^{(σ, A)} (E) \otimes D e r^{(σ, A)} (E) \to D e r^{(σ, A)} (E) .

Proof. Simply repeat the proof of proposition 17. _

Corollary 23 Let the braiding $σ$ be a symmetry. Then the $(σ, A)$ -derivations of $E$ equipped with the $σ$ -commutator is a $σ$ -Lie algebra.

We get the following sequence of $σ$ -symmetric left $A$ -modules and $σ$ -Lie algebras

0 \to hom (E, E) \to D e r^{(σ, A)} (E) \overset{π}{\to} D e r^{σ} (A) .

4.3 Quantization of braided derivations of algebras

Consider derivations of a $σ$ -symmetric algebra $A$ .

Deﬁnition 24 Given a quantization $q$ and $\partial \in D e r^{σ} (A)$ deﬁne the quantization of $\partial$ by

Q_{q} (\partial) (a) = e v_{A, A} \circ q (\partial \otimes a),

$a \in A$ .

Sometimes we use the notation $\partial_{q} = Q_{q} (\partial)$ .

$Q_{q} (\partial)$ is an operator of the quantized algebra $A_{q}$ . Denote by $D e r^{σ_{q}} (A_{q})$ the set of all $Q_{q} (\partial)$ , $\partial \in D e r^{σ} (A)$ , equipped with the quantized composition.

Theorem 25 Given a braiding $σ$ , let $σ_{q}$ be the quantization of $σ$ . The operator

\begin{array}{rcl} Q_{q} & : & (D e r^{σ} (A), c_{σ}) \to (D e r^{σ_{q}} (A_{q}), c_{σ_{q}}^{q}), & (39) \\ \partial & \in & D e r^{σ} (A) \mapsto Q_{q} (\partial) \in D e r^{σ_{q}} (A_{q}), \end{array}

is an isomorphism of modules between the $σ$ -derivations of $A$ and the $σ_{q}$ -derivations of $A_{q}$

Proof. We need to show that the $σ_{q}$ -Leibniz rule is satisﬁed

Q_{q} (\partial) (a *_{q} b) = Q_{q} (\partial) (a) *_{q} b + μ_{A} \circ q \circ (1 \otimes e v_{A, A}) \circ (1 \otimes q) \circ (σ_{q} \otimes 1) (\partial \otimes a \otimes b)

since

\begin{array}{rcl} e v_{A, A} \circ (1 \otimes μ_{A}) \circ q \circ (1 \otimes q) \\ = & (μ_{A} \circ (e v_{A, A} \otimes 1) + μ_{A} \circ (1 \otimes e v_{A, A}) \circ (σ \otimes 1)) \circ q \circ (1 \otimes q) \\ = & (μ_{A} \circ (e v_{A, A} \otimes 1) + μ_{A} \circ (1 \otimes e v_{A, A}) \circ (σ \otimes 1)) \circ q \circ (q \otimes 1) \\ = & (μ_{A} \circ q \circ (e v_{A, A} \otimes 1) + μ_{A} \circ (1 \otimes e v_{A, A}) \circ q \circ (σ \otimes 1)) \circ (q \otimes 1) \\ = & (μ_{A} \circ q \circ (e v_{A, A} \otimes 1) + μ_{A} \circ q \circ (1 \otimes e v_{A, A}) \circ (1 \otimes q) \circ (q^{- 1} \otimes 1) \circ (σ \otimes 1)) \circ (q \otimes 1) \\ = & μ_{A} \circ q \circ (e v_{A, A} \otimes 1) \circ (q \otimes 1) + μ_{A} \circ q \circ (1 \otimes e v_{A, A}) \circ (1 \otimes q) \circ (σ_{q} \otimes 1) \end{array}

where

q_{A \otimes D e r^{σ} (A), B} \circ (σ_{D e r^{σ} (A), A} \otimes 1) = (σ_{D e r^{σ} (A), A} \otimes 1) \circ q_{D e r^{σ} (A) \otimes A, B}

by naturality.

Note that the $σ_{q}$ -Leibniz rule is satisﬁed for the $σ_{q}$ -commutator of two $σ_{q}$ -derivations which follows from proposition 22. _

By proposition 1 is $D e r^{σ_{q}} (A_{q})$ a $σ_{q}$ -symmetric module and by theorem 9 a $σ_{q}$ -Lie algebra with respect to the $σ_{q} - q$ -bracket.

By theorem 13 the $σ_{q}$ -Lie algebra structure of $(D e r^{σ_{q}} (A_{q}), c_{σ_{q}}^{q})$ can be realized within the classical one by dequantization.

4.4 Evaluations and commutators

For both $σ$ - and $σ_{q}$ -derivations, evaluating a derivation of some element corresponds to taking the braided bracket of the derivation and that element.

Proposition 26 Let $A$ be $σ$ -commutative algebra, $\partial \in D e r^{σ} (A)$ and $a \in A$ . Then

\partial (a) = c_{σ} (\partial \otimes a) .

Let

\partial_{q} = Q_{q} (\partial_{c}) \in D e r^{σ_{q}} (A_{q})

and $a \in A_{q}$ . Then

\partial_{q} (a) = c_{σ_{q}}^{q} (\partial_{q} \otimes a) .

Proof. Let $\partial \in D e r^{σ} (A)$ and $a \in A, b \in A$ . The $σ$ -Leibniz rule is

\partial (a b) = \partial (a) b + μ \circ (1 \otimes e v_{A, A}) \circ (σ \otimes 1) (\partial \otimes a \otimes b)

and since

\partial (a b) = e v_{A, A} \circ (ν^{r} \otimes 1) (\partial \otimes a \otimes b),

when we consider $\partial$ simply as an internal homomorphism, and

μ \circ (1 \otimes e v_{A, A}) \circ (σ \otimes 1) (\partial \otimes a \otimes b) = e v_{A, A} \circ (ν^{l} \otimes 1) \circ (σ \otimes 1) (\partial \otimes a \otimes b),

clearly, by rearranging the Leibniz rule we get

\partial (a) = c_{σ} (\partial \otimes a) .

Let $\partial \in D e r^{σ} (A)$ . By the $σ_{q}$ -Leibniz rule we have

e v_{A, A} \circ (1 \otimes μ) (Q_{q} (\partial) \otimes a \otimes b) = (μ \circ (e v_{A, A} \otimes 1) + μ \circ (1 \otimes e v_{A, A}) \circ (σ \otimes 1)) (\partial \otimes a \otimes b) .

Since

e v_{A, A} \circ (1 \otimes μ) (Q_{q} (\partial) \otimes a \otimes b) = e v_{A, A} \circ (ν^{r} \otimes 1) (Q_{q} (\partial) \otimes a \otimes b)

and

μ \circ (1 \otimes e v_{A, A}) \circ (σ \otimes 1) = e v_{A, A} \circ (ν^{l} \otimes 1) \circ (σ \otimes 1),

clearly, by rearranging, the evaluation of $Q_{q} (\partial)$ on $a \in A_{q}$ is

\begin{array}{rcl} e v_{A, A} (Q_{q} (\partial) \otimes a) & = & (ν^{r} - ν^{l} \circ σ) (Q_{q} (\partial) \otimes a) \\ = & c_{σ_{q}}^{q} (Q_{q} (\partial) \otimes a) . \end{array}

4.5 Quantization of braided derivations of modules

Consider derivations of a $σ$ -symmetric algebra $A$ and a $σ$ -symmetric $A$ -module $E$ . Let $(\partial, \partial_{A}) \in D e r^{(σ, A)} (E)$ .

Deﬁnition 27 Given a quantization $q$ , deﬁne the quantization of $(\partial, \partial_{A})$ by

Q_{q} (\partial) (x) = e v_{A, A} \circ q (\partial \otimes x),

$x \in E$ . If $x \in A$ , then this is the quantization of $\partial_{A}$ deﬁned in section 4.3.

$Q_{q} (\partial)$ is an operator of the quantized module $E_{q}$ . Denote by $D e r^{(σ_{q}, A_{q})} (E_{q})$ the set of all $Q_{q} (\partial)$ , $\partial \in D e r^{(σ, A)} (E)$ , equipped with the quantized composition.

Theorem 28 Given a braiding $σ$ , let $σ_{q}$ be the quantization of $σ$ . The operator

\begin{array}{rcl} Q_{q} & : & (D e r^{(σ, A)} (E), c_{σ}) \to (D e r^{(σ_{q}, A_{q})} (E_{q}), c_{σ_{q}}^{q}), & (40) \\ \partial & \in & D e r^{(σ, A)} (E) \mapsto Q_{q} (\partial) \in D e r^{(σ_{q}, A_{q})} (E_{q}), \end{array}

is an isomorphism of modules between the $σ$ -derivations of $E$ over $A$ and the $σ_{q}$ -derivations of $E_{q}$ over $A_{q}$ .

By proposition 1 is $D e r^{(σ_{q}, A_{q})} (E_{q})$ a $σ_{q}$ -symmetric module and by theorem 9 a $σ_{q}$ -Lie algebra with respect to the $σ_{q} - q$ -bracket.

By theorem 13 the $σ$ -Lie algebra of the structure of $(D e r^{(σ_{q}, A_{q})} (E_{q}), c_{σ_{q}}^{q})$ can be realized within the classical one by dequantization.

5 Braided connections and curvatures

Let $σ$ be a symmetry, $A$ be a $σ$ -symmetric algebra and $E$ a $σ$ -commutative $A$ -module.

Deﬁnition 29 A $σ$ -connection in $E$ is a $A$ -module homomorphism $\nabla$

\nabla : D e r^{σ} (A) \to D e r^{(σ, A)} (E),

such that

π \circ \nabla = I d .

Deﬁnition 30 A $σ$ -connection $\nabla$ is ﬂat if it is a $σ$ -Lie algebra homomorphism, that is,

\nabla \circ c_{σ} = c_{σ} \circ (\nabla \otimes \nabla) .

Deﬁnition 31 In general, deﬁne the $σ$ -curvature of $\nabla$ to be

K_{\nabla} : D e r^{σ} (A) \otimes D e r^{σ} (A) \to hom (E, E),

K_{\nabla} = c_{σ} \circ (\nabla \otimes \nabla) - \nabla \circ c_{σ} .

Theorem 32 The $σ$ -curvature $K_{\nabla}$ is a $σ$ -homomorphism of $E$ , that is,

e v_{E, E} \circ (K_{\nabla} \otimes ν^{l}) = ν^{l} \circ (1 \otimes e v_{E, E}) \circ (σ \otimes 1) \circ (K_{\nabla} \otimes 1 \otimes 1),

(i)

which maps $D e r^{σ} (A) \otimes D e r^{σ} (A) \otimes A \otimes E$ to $E$ , and it is skew $σ$ -symmetric,

K_{\nabla} = - K_{\nabla} \circ σ .

(ii)

Furthermore $K_{\nabla}$ satisﬁes

\begin{array}{rcl} K_{\nabla} \circ (ν^{l} \otimes 1) & = & ν^{l} \circ K_{\nabla}, & (41) \\ K_{\nabla} \circ (1 \otimes ν^{l}) & = & ν^{l} \circ K_{\nabla} \circ (σ \otimes 1) . & (42) \end{array}

Proof. (i): Note that

e v_{E, E} \circ (K_{\nabla} \otimes ν^{l}) = e v_{E, E} \circ (1 \otimes ν^{l}) \circ (K_{\nabla} \otimes 1 \otimes 1) .

Then

e v_{E, E} \circ (K_{\nabla} \otimes ν^{l}) = ν^{l} \circ (1 \otimes e v_{E, E}) \circ (σ \otimes 1) \circ (K_{\nabla} \otimes 1 \otimes 1)

if and only if

e v_{E, E} \circ (1 \otimes ν^{l}) = ν^{l} \circ (1 \otimes e v_{E, E}) \circ (σ \otimes 1) : hom (E, E) \otimes A \otimes E \to E .

By proposition 1 and by the symmetry of $σ$ ,

v_{E^{'}}^{l} \circ (1 \otimes e v_{E, E^{'}}) \circ (σ_{hom (E, E^{'}), A} \otimes 1) = e v_{E, E^{'}} \circ (1 \otimes v_{E}^{l}) : hom (E, E^{'}) \otimes A \otimes E \to E^{'}

for $σ$ -commutative $A$ -modules $E$ and $E^{'}$ .

(ii):

\begin{array}{rcl} - K_{\nabla} \circ σ & = & - c_{σ} \circ (\nabla \otimes \nabla) \circ σ + \nabla \circ c_{σ} \circ σ \\ = & - c_{σ} \circ σ \circ (\nabla \otimes \nabla) + \nabla \circ c_{σ} \circ σ \\ = & c_{σ} \circ (\nabla \otimes \nabla) - \nabla \circ c_{σ} \\ = & K_{\nabla}, \end{array}

by proposition 5.

(iii):

\begin{array}{rcl} K_{\nabla} \circ (ν^{l} \otimes 1) & = & c_{σ} \circ (\nabla \otimes \nabla) \circ (ν^{l} \otimes 1) - \nabla \circ c_{σ} \circ (ν^{l} \otimes 1) \\ = & c_{σ} \circ (ν^{l} \otimes 1) \circ (1 \otimes \nabla \otimes \nabla) - \nabla \circ c_{σ} \circ (ν^{l} \otimes 1) \\ = & ν^{l} \circ (1 \otimes c_{σ}) \circ (1 \otimes \nabla \otimes \nabla) - ν^{l} \circ (e v_{A, A} \otimes 1) \circ σ \circ (1 \otimes \nabla \otimes \nabla) \\ - \nabla \circ ν^{l} \circ (1 \otimes c_{σ}) + \nabla \circ ν^{l} \circ (e v_{A, A} \otimes 1) \circ σ \\ = & ν^{l} \circ (1 \otimes c_{σ}) \circ (1 \otimes \nabla \otimes \nabla) - ν^{l} \circ (1 \otimes \nabla) \circ (1 \otimes c_{σ}) \\ = & ν^{l} \circ K_{\nabla}, \end{array}

if and only if

ν^{l} \circ (e v_{A, A} \otimes 1) \circ σ \circ (1 \otimes \nabla \otimes \nabla) = \nabla \circ ν^{l} \circ (e v_{A, A} \otimes 1) \circ σ,

which clearly is satisﬁed.

(iv):

\begin{array}{rcl} K_{\nabla} \circ (1 \otimes ν^{l}) & = & c_{σ} \circ (\nabla \otimes \nabla) \circ (1 \otimes ν^{l}) - \nabla \circ c_{σ} \circ (1 \otimes ν^{l}) \\ = & c_{σ} \circ (1 \otimes ν^{l}) \circ (\nabla \otimes 1 \otimes \nabla) - \nabla \circ c_{σ} \circ (1 \otimes ν^{l}) \\ = & ν^{l} \circ (1 \otimes c_{σ}) \circ (σ \otimes 1) \circ (\nabla \otimes 1 \otimes \nabla) + ν^{l} \circ (e v_{A, A} \otimes 1) \circ (\nabla \otimes 1 \otimes \nabla) \\ - \nabla \circ ν^{l} \circ (1 \otimes c_{σ}) \circ (σ \otimes 1) - \nabla \circ ν^{l} \circ (e v_{A, A} \otimes 1) \\ = & ν^{l} \circ (1 \otimes c_{σ}) \circ (1 \otimes \nabla \otimes \nabla) \circ (σ \otimes 1) - ν^{l} \circ (1 \otimes \nabla) \circ (1 \otimes c_{σ}) \circ (σ \otimes 1) \\ = & ν^{l} \circ K_{\nabla} \circ (σ \otimes 1) . \end{array}

Note that

σ \circ (\nabla \otimes 1) = (1 \otimes \nabla) \circ σ

by the naturality of braidings. _

5.1 Quantization of braided connections and curvatures

Let $\nabla$ be a $σ$ -connection in $E$ . Then the quantization of $\nabla$ ,

Q_{q} (\nabla) = \nabla_{q} : D e r^{σ_{q}} (A_{q}) \to D e r^{(σ_{q}, A_{q})} (E_{q}) .

is deﬁned by

\nabla_{q} \overset{d e f}{=} Q_{q} \circ \nabla \circ Q_{q}^{- 1},

(43)

that is, the following diagram commutes

(σ,A) -∇--- σ
Der | (E) Der|(A)
| |
Qq| Qq|
| |
(σq,Aq) ∇q- σq
Der (Eq) Der (Aq)

Proposition 33 The quantization $\nabla_{q}$ is a $σ_{q}$ -connection in $E_{q}$ .

Proof. Clearly,

π_{q} \circ \nabla_{q} = I d,

π_{q} = Q_{q} \circ π \circ Q_{q}^{- 1} .

Let $\nabla_{q}$ be a $σ_{q}$ -connection in $E_{q}$ . Then the $σ_{q} - q$ -curvature of $\nabla_{q}$ (or simply $σ_{q}$ -curvature when it is clear that the multiplication or composition is the quantized multiplication),

K_{\nabla_{q}}^{q} : D e r^{σ_{q}} (A_{q}) \cdot_{σ_{q}} D e r^{σ_{q}} (A_{q}) \to E n d_{A_{q}} (E_{q}),

is deﬁned by

K_{\nabla_{q}}^{q} = c_{σ_{q}}^{q} \circ (\nabla_{q} \otimes \nabla_{q}) - \nabla_{q} \circ c_{σ_{q}}^{q} .

Let us state the properties of the $σ_{q} - q$ -curvature.

Theorem 34 The $σ_{q} - q$ -curvature is a $σ_{q}$ -homomorphism, that is,

e v_{E, E} \circ (K_{\nabla_{q}}^{q} \otimes ν_{q}^{l}) = ν_{q}^{l} \circ (1 \otimes e v_{E, E}) \circ (σ_{q} \otimes 1) \circ (K_{\nabla_{q}}^{q} \otimes 1 \otimes 1),

(i)

which maps $D e r^{σ_{q}} (A_{q}) \otimes D e r^{σ_{q}} (A_{q}) \otimes A_{q} \otimes E_{q}$ to $E_{q}$ , and is skew $σ_{q}$ -symmetric,

K_{\nabla_{q}}^{q} = - K_{\nabla_{q}}^{q} \circ σ_{q} .

(ii)

Furthermore $K_{\nabla_{q}}^{q}$ satisﬁes

\begin{array}{rcl} K_{\nabla_{q}}^{q} \circ (ν_{q}^{l} \otimes 1) & = & ν_{q}^{l} \circ (1 \otimes K_{\nabla_{q}}^{q}), & (44) \\ K_{\nabla_{q}}^{q} \circ (1 \otimes ν_{q}^{l}) & = & ν_{q}^{l} \circ K_{\nabla_{q}}^{q} \circ (σ_{q} \otimes 1), & (45) \end{array}

$\forall a \in A$ .

Proof. (i): See the proof of (i) of theorem 32.

(ii):

\begin{array}{rcl} K_{\nabla_{q}}^{q} & = & c_{σ_{q}}^{q} \circ (\nabla_{q} \otimes \nabla_{q}) - \nabla_{q} \circ c_{σ_{q}}^{q} \\ = & - c_{σ_{q}}^{q} \circ σ_{q} \circ (\nabla_{q} \otimes \nabla_{q}) + \nabla_{q} \circ c_{σ_{q}}^{q} \circ σ_{q} \\ = & - K_{\nabla_{q}} \circ σ_{q} . \end{array}

(iii):

\begin{array}{rcl} K_{\nabla_{q}}^{q} \circ (ν_{q}^{l} \otimes 1) \\ = & c_{σ_{q}}^{q} \circ (\nabla_{q} \otimes \nabla_{q}) \circ (ν_{q}^{l} \otimes 1) - \nabla_{q} \circ c_{σ_{q}}^{q} \circ (ν_{q}^{l} \otimes 1) \\ = & c_{σ_{q}}^{q} \circ (ν_{q}^{l} \otimes 1) \circ (1 \otimes \nabla_{q} \otimes \nabla_{q}) - \nabla_{q} \circ c_{σ_{q}}^{q} \circ (ν_{q}^{l} \otimes 1) \\ = & ν_{q}^{l} \circ (1 \otimes c_{σ_{q}}^{q}) \circ (1 \otimes \nabla_{q} \otimes \nabla_{q}) - ν_{q}^{l} \circ (e v_{A, A} \otimes 1) \circ σ_{q} \circ (1 \otimes \nabla_{q} \otimes \nabla_{q}) \\ - \nabla_{q} \circ ν_{q}^{l} \circ (1 \otimes c_{σ_{q}}^{q}) + \nabla_{q} \circ ν_{q}^{l} \circ (e v_{A, A} \otimes 1) \circ σ_{q} \\ = & ν_{q}^{l} \circ (1 \otimes c_{σ_{q}}^{q}) \circ (1 \otimes \nabla_{q} \otimes \nabla_{q}) - ν_{q}^{l} \circ (1 \otimes \nabla_{q}) \circ (1 \otimes c_{σ_{q}}^{q}) \\ = & ν_{q}^{l} \circ (1 \otimes K_{\nabla_{q}}^{q}) \end{array}

if and only if,

ν_{q}^{l} \circ (e v_{A, A} \otimes 1) \circ σ_{q} \circ (1 \otimes \nabla_{q} \otimes \nabla_{q}) = \nabla_{q} \circ ν_{q}^{l} \circ (e v_{A, A} \otimes 1) \circ σ_{q},

which clearly is satisﬁed.

(iv):

\begin{array}{rcl} K_{\nabla_{q}}^{q} \circ (1 \otimes ν_{q}^{l}) \\ = & c_{σ_{q}}^{q} \circ (\nabla_{q} \otimes \nabla_{q}) \circ (1 \otimes ν_{q}^{l}) - \nabla_{q} \circ c_{σ_{q}}^{q} \circ (1 \otimes ν_{q}^{l}) \\ = & ν_{q}^{l} \circ (1 \otimes c_{σ_{q}}^{q}) \circ (σ_{q} \otimes 1) \circ (1 \otimes \nabla_{q} \otimes \nabla_{q}) + ν_{q}^{l} \circ (e v_{A, A} \otimes 1) \circ (1 \otimes \nabla_{q} \otimes \nabla_{q}) \\ - \nabla_{q} \circ ν_{q}^{l} \circ (1 \otimes c_{σ_{q}}^{q}) \circ (σ_{q} \otimes 1) - \nabla_{q} \circ ν_{q}^{l} \circ (e v_{A, A} \otimes 1) \\ = & ν_{q}^{l} \circ (1 \otimes c_{σ_{q}}^{q}) \circ (σ_{q} \otimes 1) \circ (1 \otimes \nabla_{q} \otimes \nabla_{q}) - \nabla_{q} \circ ν_{q}^{l} \circ (1 \otimes c_{σ_{q}}^{q}) \circ (σ_{q} \otimes 1) \\ = & ν_{q}^{l} \circ K_{\nabla_{q}}^{q} \circ (σ_{q} \otimes 1) . \end{array}

We have the following condition for the braided curvature of dequantizations of braided derivations.

Theorem 35 The $σ_{q}$ -curvature of the connection $\nabla_{q}$ deﬁned by

K_{\nabla_{q}} = c_{σ_{q}} \circ (\nabla_{q} \otimes \nabla_{q}) - \nabla_{q} \circ c_{σ_{q}},

(46)

and the $σ_{q} - q$ -curvature $K_{\nabla}^{q}$ of the $σ$ -connection $\nabla$ of $E$ deﬁned by

K_{\nabla}^{q} = c_{σ_{q}}^{q} \circ (\nabla \otimes \nabla) - \nabla \circ c_{σ_{q}}^{q},

(47)

are related as follows,

Q_{q}^{- 1} \circ K_{\nabla_{q}} = K_{\nabla}^{q} \circ (Q_{q}^{- 1} \otimes Q_{q}^{- 1}) .

(48)

Proof. Proof of (48):

\begin{array}{rcl} K_{\nabla_{q}} & = & c_{σ_{q}} \circ (\nabla_{q} \otimes \nabla_{q}) - \nabla_{q} \circ c_{σ_{q}} \\ = & Q_{q} \circ c_{σ_{q}}^{q} \circ (Q_{q}^{- 1} \otimes Q_{q}^{- 1}) \circ ((Q_{q} \circ \nabla \circ Q_{q}^{- 1}) \otimes (Q_{q} \circ \nabla \circ Q_{q}^{- 1})) \\ - Q_{q} \circ \nabla \circ Q_{q}^{- 1} \circ Q_{q} \circ c_{σ_{q}}^{q} \circ (Q_{q}^{- 1} \otimes Q_{q}^{- 1}) \\ = & Q_{q} \circ c_{σ_{q}}^{q} \circ (\nabla \otimes \nabla) \circ (Q_{q}^{- 1} \otimes Q_{q}^{- 1}) - Q_{q} \circ \nabla \circ c_{σ_{q}}^{q} \circ (Q_{q}^{- 1} \otimes Q_{q}^{- 1}) \\ = & Q_{q} \circ K_{\nabla}^{q} \circ (Q_{q}^{- 1} \otimes Q_{q}^{- 1}) . \end{array}

6 Braided diﬀerential operators

We shall see how the picture is for braided diﬀerential operators.

6.1 Braided diﬀerential operators in algebras

Let $σ$ be a braiding and $A$ be a $σ$ -symmetric algebra.

Recall that the module

{(hom (A, A))}_{σ}^{(k)} = D i f f_{k}^{σ} (A, A)

is called the braided or $σ$ -diﬀerential operators in $A$ order at most $k$ .

An equivalent and more familiar way to deﬁne a $σ$ -diﬀerential operator of order at most $k$ is the linear map

f : A \to A,

such that

c_{σ} \circ (1 \otimes c_{σ}) \circ \dots \circ (\underset{k}{\underset{︸}{1 \otimes \dots \otimes 1}} \otimes c_{σ}) (a_{0} \otimes a_{1} \otimes \dots \otimes a_{k} \otimes f) = 0

(49)

$\forall a_{0}, \dots, a_{k} \in A$ . Denote by $D i f f_{k}^{σ} (A, A)$ the set of $σ$ -diﬀerential operators of order at most $k$ . Note,

f \in D i f f_{k}^{σ} (A, A) \Leftrightarrow c_{σ} (f \otimes a) \in D i f f_{k - 1}^{σ} (A, A), \forall a \in A .

Let $D i f f^{σ} (A, A) = \cup D i f f_{k}^{σ} (A, A)$ .

From [15] we have the following two results.

Proposition 36 The $σ$ -commutator of two $σ$ -diﬀerential operators $f \in D i f f_{i}^{σ} (A, A)$ and $g \in D i f f_{j}^{σ} (A, A)$ is a $σ$ -diﬀerential operator of order at most $i + j - 1$ ,

c_{σ} (f \otimes g) \in D i f f_{i + j - 1}^{σ} (A, A) .

The next result also follows from theorem 9.

Corollary 37 If $σ$ is a symmetry and an algebra $A$ is $σ$ -symmetric then $D i f f^{σ} (A, A)$ is a $σ$ -Lie algebra.

Proposition 38 There is an $A - A$ -module structure on $D i f f^{σ} (A, A)$ deﬁned by

\begin{array}{rcl} ν_{a}^{l} (f) (b) & = & ν^{l} (a \otimes f) (b) = a f (b), & (50) \\ ν_{a}^{r} (f) (b) & = & ν^{r} (f \otimes a) (b) = f (a b), & (51) \end{array}

$a, b \in A$ , $f \in D i f f^{σ} (A, A)$ , and $ν^{l} (a \otimes f)$ , $ν^{r} (f \otimes a) \in D i f f_{k}^{σ} (A, A)$ , for $f \in D i f f_{k}^{σ} (A, A)$ .

Proof. Let

σ^{i, i + 1} = (\underset{i - 1}{\underset{︸}{1 \otimes \dots \otimes 1}} \otimes σ \otimes \underset{(k + 3) - (i + 1)}{\underset{︸}{1 \otimes \dots \otimes 1}}),

than the left action on a braided diﬀerential operator again is a braided diﬀerential operator,

\begin{array}{rcl} c_{σ} \circ (1 \otimes c_{σ}) \circ \dots \circ (\underset{k}{\underset{︸}{1 \otimes \dots \otimes 1}} \otimes c_{σ}) \circ (\underset{k + 1}{\underset{︸}{1 \otimes \dots \otimes 1}} \otimes ν^{l}) \\ = & c_{σ} \circ \dots \circ (\underset{k - 1}{\underset{︸}{1 \otimes \dots \otimes 1}} \otimes c_{σ}) \circ (\underset{k}{\underset{︸}{1 \otimes \dots \otimes 1}} \otimes ((ν^{l} + e v_{A, A} \circ σ) \circ (1 \otimes ν^{l}))) \\ = & c_{σ} \circ \dots \circ (\underset{k}{\underset{︸}{1 \otimes \dots \otimes 1}} \otimes ((ν^{l} \circ (μ \otimes 1) + μ \circ (1 \otimes e v_{A, A}) \circ (1 \otimes σ)) \circ (σ \otimes 1))) \\ = & c_{σ} \circ \dots \circ (\underset{k}{\underset{︸}{1 \otimes \dots \otimes 1}} \otimes ν^{l}) \circ (\underset{k + 1}{\underset{︸}{1 \otimes \dots \otimes 1}} \otimes c_{σ}) \circ (\underset{k}{\underset{︸}{1 \otimes \dots \otimes 1}} \otimes σ \otimes 1) \\ ⋮ \\ = & ν^{l} \circ (1 \otimes c_{σ}) \circ \dots \circ (\underset{k + 1}{\underset{︸}{1 \otimes \dots \otimes 1}} \otimes c_{σ}) \circ σ^{1, 2} \circ \dots \circ σ^{k + 1, k + 2} \\ = & 0, \end{array}

when applied to $a_{0} \otimes a_{1} \otimes \dots \otimes a_{k} \otimes b \otimes f$ , and also the right action on a braided diﬀerential operator again is a braided diﬀerential operator,

\begin{array}{rcl} c_{σ} \circ (1 \otimes c_{σ}) \circ \dots \circ (\underset{k}{\underset{︸}{1 \otimes \dots \otimes 1}} \otimes c_{σ}) \circ (\underset{k + 1}{\underset{︸}{1 \otimes \dots \otimes 1}} \otimes ν^{r}) \\ = & c_{σ} \circ \dots \circ (\underset{k - 1}{\underset{︸}{1 \otimes \dots \otimes 1}} \otimes c_{σ}) \circ (\underset{k}{\underset{︸}{1 \otimes \dots \otimes 1}} \otimes ((ν^{l} + e v_{A, A} \circ σ) \circ (1 \otimes ν^{r}))) \\ = & c_{σ} \circ \dots \circ (\underset{k}{\underset{︸}{1 \otimes \dots \otimes 1}} \otimes (ν^{r} \circ (ν^{l} \otimes 1) + e v_{A, A} \circ (1 \otimes μ) \circ (1 \otimes σ) \circ (σ \otimes 1))) \\ = & c_{σ} \circ \dots \circ (\underset{k}{\underset{︸}{1 \otimes \dots \otimes 1}} \otimes ν^{r}) \circ (\underset{k}{\underset{︸}{1 \otimes \dots \otimes 1}} \otimes c_{σ} \otimes 1) \\ ⋮ \\ = & ν^{r} \circ (c_{σ} \otimes 1) \circ \dots \circ (\underset{k}{\underset{︸}{1 \otimes \dots \otimes 1}} \otimes c_{σ} \otimes 1) \\ = & 0, \end{array}

when applied to $a_{0} \otimes a_{1} \otimes \dots \otimes a_{k} \otimes f \otimes b$ , $a_{0}, \dots, a_{k}, b \in A$ , $f \in D i f f_{k}^{σ} (A, A)$ . _

Consider the symbol of the diﬀerential operators which is the leading part with respect to derivatives,

S m b l_{k}^{σ} (A, A) = D i f f_{k}^{σ} (A, A) ∕ D i f f_{k - 1}^{σ} (A, A),

then we have the $ℤ$ -graded object

S m b l^{σ} (A, A) = \sum S m b l_{k}^{σ} (A, A) .

The class of ${[f, g]}^{σ} \in D i f f_{i + j - 1}^{σ} (A, A)$ ,

\bar{{[f, g]}^{σ}} \in S m b l_{i + j - 1}^{σ} (A, A),

depends on the class of the two $σ$ -diﬀerential operators $f \in D i f f_{i}^{σ} (A, A)$ and $g \in D i f f_{j}^{σ} (A, A)$ , hence there is a $σ$ -Poisson structure on the braided symbol algebra, [15].

6.2 Braided diﬀerential operators in modules

Let $σ$ be a braiding, $A$ be a $σ$ -symmetric algebra and let $E$ be a $σ$ -symmetric $A$ -module.

Deﬁnition 39 The module

{(hom (E, E))}_{σ}^{(k)} = D i f f_{k}^{σ} (E, E)

is called the braided or $σ$ -diﬀerential operators in order at most $k$ of $E$ .

Denote the $σ$ -diﬀerential operators of order at most $k$ in $E$ by $D i f f_{k}^{(σ, A)} (E, E)$ and we consider $D i f f^{(σ, A)} (E, E) = \cup D i f f_{k}^{(σ, A)} (E, E)$ .

From [15] we have the following two results.

Proposition 40 The $σ$ -commutator of two $σ$ -diﬀerential operators $f \in D i f f_{i}^{(σ, A)} (E, E)$ and $g \in D i f f_{j}^{(σ, A)} (E, E)$ ,

{[f, g]}^{σ} \in D i f f_{i + j}^{(σ, A)} (E, E),

and so has order at most $i + j$ .

Corollary 41 If $σ$ is a symmetry and an algebra $A$ and a left $A$ -module $E$ are $σ$ -symmetric then $D i f f^{σ} (E, E)$ is a $σ$ -Lie algebra.

We consider the $A - A$ -module structure on $D i f f^{σ} (E, E)$ .

Proposition 42 There is an $A - A$ -module structure on $D i f f^{σ} (E, E)$ deﬁned by

\begin{array}{rcl} ν_{a}^{l} (f) (x) & = & ν^{l} (a \otimes f) (x) = a f (x), \\ ν_{a}^{r} (f) (x) & = & ν^{r} (f \otimes a) (x) = f (a x), \end{array}

$a \in A$ , $x \in E$ , $f \in D i f f^{σ} (E, E)$ .

Proof. The proof is the same as for proposition 38. _

Consider the symbol of the diﬀerential operators which is the leading part with respect to derivatives,

S m b l_{k}^{σ} (E, E) = D i f f_{k}^{σ} (E, E) ∕ D i f f_{k - 1}^{σ} (E, E),

then we have the $ℤ$ -graded object

S m b l^{σ} (E, E) = \sum S m b l_{k}^{σ} (E, E) .

The class of ${[f, g]}^{σ} \in D i f f_{i + j}^{σ} (E, E)$ ,

\bar{{[f, g]}^{σ}} \in S m b l_{i + j}^{σ} (E, E),

depends on the class of the two $σ$ -diﬀerential operators $f \in D i f f_{i}^{σ} (E, E)$ and $g \in D i f f_{j}^{σ} (E, E)$ , hence there is a $σ$ -Poisson structure on the braided symbol algebra.

6.3 Quantizations of braided diﬀerential operators in algebras

We can deﬁne quantization of $σ$ -diﬀerential operators in algebras. Let $A$ be a $σ$ -commutative algebra.

Deﬁnition 43 Given a quantization $q$ and $f \in D i f f^{σ} (A, A)$ deﬁne the quantization of $f$ by

Q_{q} (f) (a) = f_{q} (a) \overset{d e f}{=} e v_{A, A} \circ q (f \otimes a),

$a \in A$ .

$Q_{q} (f)$ is an operator of the quantized algebra $A_{q}$ . From [14] we have the following theorem.

Theorem 44 Given a braiding $σ$ , let $σ_{q}$ be the quantization of $σ$ . The operator

\begin{array}{rcl} Q_{q} & : & (D i f f^{σ} (A, A), c_{σ}) \to (D i f f^{σ_{q}} (A_{q}, A_{q}), c_{σ_{q}}^{q}), & (52) \\ f & \in & D i f f^{σ} (A, A) \mapsto Q_{q} (f) \in D i f f^{σ_{q}} (A_{q}, A_{q}), \end{array}

is an isomorphism of modules.

The symbol of $Q_{q}$ is an isomorphism of modules

\begin{array}{rcl} S m b l (Q_{q}) & : & (S m b l^{σ} (A, A), c_{σ}) \to (S m b l^{σ_{q}} (A_{q}, A_{q}), c_{σ_{q}}^{q}), & (53) \\ f & \in & S m b l^{σ} (A, A) \mapsto S m b l (Q_{q}) (f) \in S m b l^{σ_{q}} (A_{q}, A_{q}) . \end{array}

By proposition 38 is $D i f f^{σ_{q}} (A_{q}, A_{q})$ a $σ_{q}$ -symmetric module. By corollary 37, if $σ$ is a symmetry then $D i f f^{σ_{q}} (A_{q}, A_{q})$ is a $σ_{q}$ -Lie algebra with respect to the $σ_{q} - q$ -bracket and the quantized composition. Furthermore there is a $σ_{q}$ -Poisson structure on the quantized braided symbol algebra $S m b l^{σ_{q}} (A_{q}, A_{q})$ .

By theorem 13 the $σ_{q}$ -Lie algebra structure of $(D i f f^{σ_{q}} (A_{q}, A_{q}), c_{σ_{q}}^{q})$ can be realized within the classical one by dequantization.

6.4 Quantizations of braided diﬀerential operators in modules

Let $A$ be a $σ$ -symmetric algebra and let $E$ be a $σ$ -symmetric $A$ -module.

Deﬁnition 45 Given a quantization $q$ and $f \in D i f f^{(σ, A)} (E, E)$ deﬁne the quantization of $f$ by

Q_{q} (f) (x) = f_{q} (x) \overset{d e f}{=} e v_{E, E} \circ q (f \otimes x),

$x \in E$ .

$Q_{q} (f)$ is an operator of the quantized module $E_{q}$ .

Theorem 46 Given a braiding $σ$ , let $σ_{q}$ be the quantization of $σ$ . The operator

\begin{array}{rcl} Q_{q} & : & (D i f f^{(σ, A)} (E, E), c_{σ}) \to (D i f f^{(σ_{q}, A_{q})} (E_{q}, E_{q}), c_{σ_{q}}^{q}), & (54) \\ f & \in & D i f f^{(σ, A)} (E, E) \mapsto Q_{q} (f) \in D i f f^{(σ_{q}, A_{q})} (E_{q}, E_{q}), \end{array}

is an isomorphism of modules.

The symbol of $Q_{q}$ is an isomorphism of modules

\begin{array}{rcl} S m b l (Q_{q}) & : & (S m b l^{(σ, A)} (E, E), c_{σ}) \to (S m b l^{(σ_{q}, A_{q})} (E_{q}, E_{q}), c_{σ_{q}}^{q}), & (55) \\ f & \in & S m b l^{(σ, A)} (E, E) \mapsto S m b l (Q_{q}) (f) \in S m b l^{(σ_{q}, A_{q})} (E_{q}, E_{q}) . \end{array}

Proof. The isomorphism as $σ$ -diﬀerential operators and $σ$ -symbols is shown in [14]. _

By proposition 42 is $D i f f^{(σ_{q}, A_{q})} (E_{q}, E_{q})$ a $σ_{q}$ -symmetric module. By corollary 41, if $σ$ is a symmetry then $D i f f^{(σ_{q}, A_{q})} (E_{q}, E_{q})$ is a $σ_{q}$ -Lie algebra with respect to the $σ_{q} - q$ -bracket and the quantized composition. Furthermore there is a $σ_{q}$ -Poisson structure on the quantized braided symbol algebra, $S m b l^{(σ_{q}, A_{q})} (E_{q}, E_{q})$ .

By theorem 13 the $σ_{q}$ -Lie algebra structure of $(D i f f^{(σ_{q}, A_{q})} (E_{q}, E_{q}), c_{σ_{q}}^{q})$ can be realized within the classical one by dequantization.

References

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[5] D. Gurevich. Algebraic aspects of quantum Yang Baxter equation, Algebra and Analysis, 2, 4, 1990.

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[7] H. L. Huru. Associativity constraints, braidings and quantizations of modules with grading and action. Vol. 23, Lobachevskii Journal of Mathematics, http://ljm.ksu.ru/vol23/110.html, 2006.

[8] H. L. Huru. Quantization of braided algebras. 2. Graded Modules. Submitted to Lobachevskii Journal of Mathematics, ljm.ksu.ru, November 2006.

[9] H. L. Huru. Quantization of braided algebras. 3. Modules with action by a group. Submitted to Lobachevskii Journal of Mathematics, ljm.ksu.ru, November 2006.

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[13] Cathrine V. Jensen. Linear ordinary diﬀerential equations and $D$ -modules, solving and reduction methods, Dr.Scient. thesis, The University of Tromsø, Nov. 2004.

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Department of Mathematics, The University of Tromsoe, N-9037 Tromsoe, Norway

E-mail address: Hilja.Huru@matnat.uit.no

Received November 7, 2006

1 Introduction

2 Quantizations and σ-commutativity

2.1 Quantizations

2.2 Braidings

2.3 Algebras

2.4 Modules

2.5 Coalgebras

2.6 Bialgebras

2.7 Internal homomorphisms

3 Braided Lie structure of internal homomorphisms

3.1 Module structure of hom E,E′

3.2 Braided commutators and Lie structure of hom E,E

3.3 Quantization of hom E,E

4 Braided derivations in monoidal categories

4.1 Braided derivations of algebras

4.2 Braided derivations of modules

4.3 Quantization of braided derivations of algebras

4.4 Evaluations and commutators

4.5 Quantization of braided derivations of modules

5 Braided connections and curvatures

5.1 Quantization of braided connections and curvatures

6 Braided diﬀerential operators

6.1 Braided diﬀerential operators in algebras

6.2 Braided diﬀerential operators in modules

6.3 Quantizations of braided diﬀerential operators in algebras

6.4 Quantizations of braided diﬀerential operators in modules

References

2 Quantizations and $σ$ -commutativity

3.1 Module structure of $hom (E, E^{'})$

3.2 Braided commutators and Lie structure of $hom (E, E)$

3.3 Quantization of $hom (E, E)$