Graded $q$-Differential Polynomial Algebra of Connection Form

Abstract

Given a graded associative unital algebra we construct a graded $q$-differential algebra, where $q$ is a primitive $N$th root of unity and prove that the generalized cohomologies of the corresponding $N$-complex are trivial. We construct a graded $q$-differential algebra of polynomials and introduce a notion of connection form. We find explicit formula for the curvature of connection form and prove Bianchi identity.

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1 Introduction

An idea to generalize the concept of a differential module and to elaborate the corresponding algebraic structures by giving the basic property of differential $d^2=0$ a more general form $d^N=0,\,N\ge 2$ seems to be very natural. Taking the equation $d^N=0$ as a starting point one should choose a space where a calculus with $d^N=0$ will be constructed. As a calculus with $d^N=0$ may be considered as a generalization of $d^2=0$ and taking into account that there is an exterior calculus of differential forms with exterior differential $d^2=0$ on a smooth manifold one way to construct $d^N=0$ is to take a smooth manifold and to consider objects on this manifold more general than the differentials forms. Our approach is based on $q$-deformed structures such as graded $q$-Leibniz rule, graded $q$-commutator, graded inner $q$-derivation, where $q$ is a primitive $N$th root of unity [1–6].

A notion of graded $q$-differential algebra was introduced in [7] (see also in [8–10]) and it may be viewed as a generalization of a graded differential algebra. Let us mention that a concept of graded $q$-differential algebra is closely related to the monoidal structure introduced in [11] for the category of $N$-complexes and it is proved in [12] that the monoids of the category of $N$-complexes can be identified as the graded $q$-differential algebras. It is well known that a connection and its curvature are basic elements of the theory of fiber bundles and they play an important role not only in a modern differential geometry but also in theoretical physics namely in a gauge field theory. A basic algebraic structure used in the theory of connections on modules is a graded differential algebra. A graded differential algebra is an algebraic model for the de Rham algebra of differential forms on a smooth manifold. Consequently considering a concept of graded $q$-differential algebra which is more general structure than a graded differential algebra we can develop a generalization of the theory of connections on modules. One of the aims of this paper is to present and study algebraic structures based on the relation $d^N=0$ and to generalize a concept of connection and its curvature applying a concept of graded $q$-differential algebra to the theory of connections on modules.

In Sect. 2 we prove Theorem 2.1 which is very useful in the sense that we can construct various cochain $N$-complexes by means of this theorem. Theorem 2.1 asserts if their exist an element v of grading one of a graded associative unital algebra $\fancyscript{A}$ which satisfies ${{ v}^{N}}\in \fancyscript{Z}(\fancyscript{A}),$ where $\fancyscript{Z}(\fancyscript{A})$ is the graded center of $\fancyscript{A},$ then the inner graded $q$-derivation $\text{ ad }_{ v}^q$ is $N$-differential. Next we prove that the generalized cohomologies of cochain $N$-complex of Theorem 2.1 are trivial. In Sect. 3 we give the definition of a graded $q$-differential algebra. We introduce the algebra of polynomials and endow it with the structure of graded $q$-differential algebra. We introduce two operators $D,$ $\nabla$ and the polynomials $f_k,$ which are defined with the help of recurrent relation. We prove the Theorem 3.2 which give explicit power expansion formulae for the operator $D$ and the polynomials $f_k$.

2 $N$-Complexes and Cohomologies

A concept of cohomology of a differential module or of a cochain complex with coboundary operator $d$ is based on the quadratic nilpotency condition $d^2=0$. It is obvious that one can construct a generalization of a concept of cohomology of a cochain complex if the quadratic nilpotency $d^2=0$ is replaced by a more general nilpotency condition $d^N=0$, where $N$ is an integer satisfying $N\ge 2$.

Let $\fancyscript{A}=\oplus _{k\in {\mathbb Z}_N}{\fancyscript{A}}^k=\fancyscript{A}^0\oplus \fancyscript{A}^1\oplus \ldots \oplus \fancyscript{A}^{N-1}$ be a ${\mathbb Z}_N$-graded associative unital $\mathbb C$-algebra whose identity element is denoted by $1\!\!1$. The subspace $\fancyscript{A}^0\subset \fancyscript{A}$ of elements of grading zero is the subalgebra of an algebra $\fancyscript{A}$. Since this subalgebra plays an important role in several structures related to a graded algebra $\fancyscript{A}$ we will denote it by $\mathfrak {A}$, i.e. $\mathfrak {A}\equiv \fancyscript{A}^0$. It is easy to see that each subspace $\fancyscript{A}^i\subset \fancyscript{A}$ of homogeneous elements of grading $i$ is the $\mathfrak {A}$-bimodule. Hence in the case of a graded algebra $\fancyscript{A}$ we have the set of $\mathfrak {A}$-bimodules $\fancyscript{A}^0,\fancyscript{A}^1,\fancyscript{A}^2,\ldots ,\fancyscript{A}^{N-1}$. The graded subspace $\fancyscript{Z}(\fancyscript{A})\subset \fancyscript{A}$ generated by homogeneous elements $u\in \fancyscript{A}^k$, which for any ${ v}\in \fancyscript{A}^l$ satisfy $u{ v}=(-1)^{kl}{ v}u$, is called a graded center of a graded algebra $\fancyscript{A}$.

The derivation of degree $m$ induced by an element ${ v}\in \fancyscript{A}^m$ will be denoted by

$$\begin{aligned} \text{ ad }_{ v}(u)=[{ v},u]={ v}\,u-(-1)^{ml}u{ v}, \end{aligned}$$

(1)

where $u\in \fancyscript{A}^l$. The graded derivation $\text{ ad }_{ v}$ is called an inner graded derivation of an algebra $\fancyscript{A}$.

The notions of graded commutator and graded derivation of a graded algebra can be generalized within the framework of noncommutative geometry and the theory of quantum groups with the help of $q$-deformations. Let $q$ be a primitive $N$th root of unity. The graded q-commutator $[,]_q:\fancyscript{A}^k\otimes \fancyscript{A}^l\rightarrow \fancyscript{A}^{k+l}$ is defined by

$$\begin{aligned}{}[u,{ v}]_q=u{ v}-q^{kl}{ v}u. \end{aligned}$$

(2)

A graded q-derivation of degree m of a graded algebra $\fancyscript{A}$ is a linear mapping $\delta :\fancyscript{A}\rightarrow \fancyscript{A}$ of degree $m$ with respect to graded structure of $\fancyscript{A}$, i.e. $\delta :\fancyscript{A}^i\rightarrow \fancyscript{A}^{i+m}$, which satisfies the graded $q$-Leibniz rule

$$\begin{aligned} \delta (u{ v})=\delta (u)\,{ v}+q^{ml}u\,\delta ({ v}), \end{aligned}$$

(3)

where $u$ is a homogeneous element of grading $l$, i.e. $u\in \fancyscript{A}^l$. In analogy with an inner graded derivation one defines an inner graded $q$-derivation of degree $m$ of a graded algebra $\fancyscript{A}$ associated to an element ${ v}\in \fancyscript{A}^m$ by the formula

$$\begin{aligned} \text{ ad }^q_{ v}(u)=[{ v},u]_q={ v}u-q^{ml}u{ v}, \end{aligned}$$

(4)

where $u\in \fancyscript{A}^l$.

A left $K$-module $E$ is said to be an $N$-differential module if it is equipped with an endomorphism $d:E\rightarrow E$ which satisfies $d^N=0$. An $N$-differential module $E$ with $N$-differential $d$ is said to be a cochain N-complex of modules or simply N-complex if $E$ is a graded module $E=\oplus _{k\in {\mathbb Z}}E^k$ and its $N$-differential $d$ has degree 1 with respect to a graded structure of $E$, i.e. $d:E^k\rightarrow E^{k+1}$.

We prove the following theorem which can be used to construct a cochain $N$-complex for a certain class of graded associative unital algebras (see also [8], p. 394).

Theorem 2.1

Let $\fancyscript{A}=\oplus _{k\in {\mathbb Z}_N}{\fancyscript{A}}^k$ be a graded associative unital algebra and $q$ be a primitive $N$th root of unity. If there exists an element ${ v}\in \fancyscript{A}^1$ of grading one which satisfies the condition ${ v}^N\in \fancyscript{Z}(\fancyscript{A})$ then the inner graded $q$-derivation $d=\text{ ad }^q_{ v}$ of degree 1 is an $N$-differential and the sequence

$$\begin{aligned} \fancyscript{A}^0\mathop {\rightarrow }\limits ^{d}\fancyscript{A}^1\mathop {\rightarrow }\limits ^{d}\fancyscript{A}^2\mathop {\rightarrow }\limits ^{d}\cdots \mathop {\rightarrow }\limits ^{d}\fancyscript{A}^{N-1} \end{aligned}$$

(5)

is the cochain $N$-complex.

Proof

We begin the proof with a power expansion of $d^k$, where $1\le k\le N$. Let $u$ be a homogeneous element of an algebra $\fancyscript{A}$ whose grading will be denoted by $|u|$. For the first values of $k=1,2,3$ a straightforward computation gives

$$\begin{aligned} du&= [{ v},u]_q={ v}u-q^{|u|}u{ v},\nonumber \\ d^2 u&= [{ v},[{ v},u]_q]_q={ v}^2u-q^{|u|}[2]_q{ v}u{ v}+q^{2|u|+1}u{ v}^2,\nonumber \\ d^3 u&= { v}^3u-q^{|u|}[3]_q{ v}^2u{ v} + q^{2|u|+1}[3]_q{ v}u{ v}^2-q^{3|u|+3}u{ v}^3.\nonumber \end{aligned}$$

We state that for any $k\in \{1,2,\ldots ,N\}$ and any homogeneous $u\in \fancyscript{A}$ a power expansion of $d^k$ has the form

$$\begin{aligned} d^ku=\sum \limits _{i=0}^{k}(-1)^ip_i \left[ \begin{array}{c} k\\ i \end{array} \right] _q { v}^{k-i}\,u{ v}^{i}, \end{aligned}$$

(6)

where $p_i=q^{i|u|+\sigma (i)}$ and $\sigma (i)=\frac{i(i-1)}{2}$. We proof this statement by means of mathematical induction assuming that the above power expansion (6) for $d^k$ is true and then showing that it has the same form for $k+1$. Indeed we have

$$\begin{aligned} d^{k+1}u&= d(d^ku)=d\Bigl (\sum \limits _{i=0}^k(-1)^ip_i\left[ \begin{array}{c} k\\ i \end{array} \right] _q{ v}^{k-i}u{ v}^i\Bigr )\nonumber \\&= \sum \limits _{i=0}^k(-1)^ip_i\left[ \begin{array}{c} k\\ i \end{array} \right] _q({ v}^{k+1-i}u{ v}^i-q^{|u|+k}{ v}^{k-i}u{ v}^{i+1})\nonumber \\&= \sum \limits _{i=0}^k(-1)^ip_i\left[ \begin{array}{c} k\\ i \end{array} \right] _q{ v}^{k+1-i}u{ v}^i-\sum \limits _{i=0}^k(-1)^iq^{|u|+k}p_i\left[ \begin{array}{c} k\\ i \end{array} \right] _q{ v}^{k-i}u{ v}^{i+1}\nonumber \\&= { v}^{k+1}u+\sum \limits _{i=1}^k(-1)^ip_i\left[ \begin{array}{c} k\\ i \end{array} \right] _q{ v}^{k+1-i}u{ v}^i\nonumber \\&\quad -\sum \limits _{i=0}^{k-1}(-1)^iq^{|u|+k}p_i\left[ \begin{array}{c} k\\ i \end{array} \right] _q{ v}^{k-i}u{ v}^{i+1}-(-1)^kq^{|u|+k}p_k\,u{ v}^{k+1}\nonumber \\&= { v}^{k+1}u+\sum \limits _{i=1}^k(-1)^ip_i\left[ \begin{array}{c} k\\ i \end{array} \right] _q{ v}^{k+1-i}u{ v}^i\nonumber \\&\quad +\sum \limits _{i=1}^{k}(-1)^iq^{|u|+k}p_{i-1}\left[ \begin{array}{c} k\\ i-1 \end{array} \right] _q{ v}^{k-i}u{ v}^{i+1}+(-1)^{k+1}q^{|u|+k}p_k\,u{ v}^{k+1}\nonumber \\&= { v}^{k+1}u+\sum \limits _{i=1}^k(-1)^i\Bigl (p_i\left[ \begin{array}{c} k\\ i \end{array} \right] _q+q^{|u|+k}p_{i-1}\left[ \begin{array}{c} k\\ i-1 \end{array} \right] _q\Bigr ){ v}^{k+1-i}u{ v}^i\nonumber \\&\quad +(-1)^{k+1}q^{|u|+k}p_k\,u{ v}^{k+1}.\nonumber \end{aligned}$$

Now the coefficients in the last expansion we can write as follows

$$ p_i\left[ \begin{array}{c} k\\ i \end{array} \right] _q+q^{|u|+k}p_{i-1}\left[ \begin{array}{c} k\\ i-1 \end{array} \right] _q=p_i\Bigl (\left[ \begin{array}{c} k\\ i \end{array} \right] _q+q^{k+\sigma (i-1)-\sigma (i)}\left[ \begin{array}{c} k\\ i-1 \end{array} \right] _q\Bigr ), $$

and making use of

$$ \sigma (i-1)-\sigma (i)=\frac{(i-1)(i-2)}{2}-\frac{i(i-1)}{2}=1-i $$

and making use of well known recurrent relation for $q$-binomial coefficients we get

$$ \left[ \begin{array}{c} k\\ i \end{array} \right] _q+q^{k+1-i}\left[ \begin{array}{c} k\\ i-1 \end{array} \right] _q=\left[ \begin{array}{c} k+1\\ i \end{array} \right] _q. $$

As $p_{k+1}=q^{|u|+k}p_k$ we finally obtain

$$ d^{k+1}u=\sum \limits _{i=0}^{k+1}(-1)^ip_i\left[ \begin{array}{c} k+1\\ i \end{array} \right] _q{ v}^{k-i}u{ v}^i, $$

and this ends the proof of the formula for power expansion of $d^k$.

Now our aim is to show that the power expansion (6) implies $d^Nu=0$ for any $u\in \fancyscript{A}$. Indeed making use of (6) we can express the $N$th power of $d$ as follows

$$\begin{aligned} d^{N}u=\sum \limits _{i=0}^{N}(-1)^ip_i\left[ \begin{array}{c} N\\ i \end{array} \right] _q{ v}^{k-i}u{ v}^i. \end{aligned}$$

(7)

Taking into account that $q$ is a primitive $N$th root of unity we get

$$ \left[ \begin{array}{c} N\\ i \end{array} \right] _q=0, \quad i\in \{1,2,\ldots ,N-1\}. $$

Hence the terms in (7), which are numbered with $i=1,2,\ldots ,N-1$, vanish, and we are left with two terms

$$ d^Nu={ v}^{N}\,u +(-1)^Nq^{\sigma (N)}u{ v}^{N}. $$

As ${ v}^N$ is the element of grading zero (modulo $N$) of the graded center ${\fancyscript{Z}}(\fancyscript{A})$ we can rewrite the above formula as follows

$$ d^Nu=(1 +(-1)^Nq^{\sigma (N)})\,u{ v}^N,\quad \sigma (N)=\frac{N(N-1)}{2}. $$

In order to show that the multiplier in the above formula vanish for any $N\ge 2$ we consider separately two cases for $N$ to be an odd or even positive integer. If $N$ is an odd positive integer then the multiplier $1 +(-1)^Nq^{\sigma (N)}$ vanish because in this case

$$ 1 +(-1)^Nq^{\sigma (N)}=1-(q^{N})^{\frac{N-1}{2}}=0. $$

If $N$ is an even positive integer then

$$ 1 +(-1)^Nq^{\sigma (N)}=1+(q^{\frac{N}{2}})^{N-1}=1+(-1)^{N-1}=0. $$

Hence for any $N\ge 2$ we have $d^N=0$, and this ends the proof of the theorem. $\square$

Let us fix a positive integer $m\in \{1,2,\ldots ,N-1 \}$ and split up the $N$th power of $N$-differential $d$ as follows $d^N=d^m\circ d^{N-m}$. Then the nilpotency condition for $N$-differential can be written in the form $d^N=d^m\circ d^{N-m}=0$ and this leads to possible generalization of a concept of cohomology. For each integer $1\le m\le N-1$ one can define the submodules

$$\begin{aligned} Z_m(E)&= \{ x\in E: d^m x=0\}\subset E,\end{aligned}$$

(8)

$$\begin{aligned} B_m(E)&= \{x\in E: \exists y\in E,\,\, x=d^{N-m}y \}\subset E. \end{aligned}$$

(9)

From $d^N=0$ it follows that $B_{m}(E)\subset Z_m(E)$. For each $m\in \{1,2,\ldots ,N-1 \}$ the quotient module $H_m(E):=Z_m(E)/B_m(E)$ is said to be a generalized homology of order m of $N$-differential module $E$. The following lemma which is proved in [8] gives a very useful criteria for the triviality of the generalized cohomologies of an $N$-differential module.

Lemma 2.1

Let $E$ be an $N$-differential module over a ring $\mathbf k$, $N\ge 2$ be an integer and $q$ be an element of $\mathbf k$ satisfying the conditions $[N]_q=0$ and $[n]_q$ is invertible for any integer $1\le n\le N-1$. If there is a module-endomorphism $h:E\rightarrow E$ satisfying $h \circ d - q\,d\circ h=\text{ Id }_E$ then the generalized cohomologies of an $N$-differential module $E$ are trivial, i.e. for any integer $1\le n\le N-1$ it holds $H_n(E)=0$.

Based on this lemma we can prove that the generalized cohomologies of the cochain $N$-complex described in Theorem 2.1 are trivial. It is worth mentioning that the same argument is used in [10] to show that the generalized cohomologies of the $N$-differential module constructed by means of the algebra of $N\times N$-matrices $M_N(\mathbf{k})$ are trivial.

Theorem 2.2

Let $q$ be a primitive $N$th root of unity, $\fancyscript{A}=\oplus _{i\in \mathbb {Z}_N}\fancyscript{A}^i$ be a graded associative unital algebra with an element ${ v}\in \fancyscript{A}^1$ satisfying ${ v}^N=\lambda \,1\!\!1$, where $\lambda \ne 0.$ Then the generalized cohomologies $H_n(\fancyscript{A})$ of the cochain $N$-complex of Theorem 2.1

$$\begin{aligned} \fancyscript{A}^0\mathop {\rightarrow }\limits ^{d}\fancyscript{A}^1\mathop {\rightarrow }\limits ^{d}\fancyscript{A}^2\mathop {\rightarrow }\limits ^{d}\cdots \mathop {\rightarrow }\limits ^{d}\fancyscript{A}^{N-1} \end{aligned}$$

(10)

with $N$-differential $d=\text{ ad }^q_{ v}$, induced by an element ${ v}$, are trivial, i.e. for any $n\in \{1,2,\ldots ,N-1\}$ we have $H_n(\fancyscript{A})=0$.

Proof

Let us define the endomorphism $h$ of the vector space of $\fancyscript{A}$ as follows

$$ h(u)=\frac{1}{(1-q)\,\lambda }\,{ v}^{N-1}\,u, $$

where $u$ is an element of an algebra $\fancyscript{A}$. If $u$ is a homogeneous element of a graded algebra $\fancyscript{A}$ then $|h(u)|=|u|+N-1$, where $|u|$ is the grading of an element $u$. For any homogeneous $u\in \fancyscript{A}$ we have

$$\begin{aligned} (h \circ d - q\,d\circ h)(u)&= h(du)-q\,d(h(u))\nonumber \\&= h(\text{ ad }^q_{ v}(u))-\frac{q}{(1-q)\,\lambda }\,\text{ ad }^q_{ v}({ v}^{N-1}u)\nonumber \\&= h([{ v},u]_q)-\frac{q}{(1-q)\,\lambda }\,[{ v},{ v}^{N-1}]_q\nonumber \\&= h\,({ v}\,u-q^{|u|}u{ v})-\frac{q}{(1-q)\,\lambda }\,({ v}^Nu-q^{|u|+N-1}{ v}^{N-1}u\,{ v})\nonumber \\&= \frac{1}{(1-q)\,\lambda }\,{ v}^Nu-\frac{q^{|u|}}{(1-q)\,\lambda }{ v}^{N-1}u\,{ v}\nonumber \\&-\frac{q}{(1-q)\,\lambda }\,{ v}^Nu+\frac{q^{|u|}}{(1-q)\,\lambda }{ v}^{N-1}u\,{ v}\nonumber \\&= \frac{(1-q)\,\lambda }{(1-q)\,\lambda }\,u=\text{ Id }_\fancyscript{A}(u).\nonumber \end{aligned}$$

The endomorphism $h:\fancyscript{A}\rightarrow \fancyscript{A}$ of the vector space of an algebra $\fancyscript{A}$ satisfies $h\circ d-q\,d\circ h=\text{ Id }_{\fancyscript{A}}$ and it follows from Lemma 2.1 that the generalized cohomology of the cochain $N$-complex

$$ \fancyscript{A}^0\mathop {\rightarrow }\limits ^{d}\fancyscript{A}^1\mathop {\rightarrow }\limits ^{d}\fancyscript{A}^2\mathop {\rightarrow }\limits ^{d}\cdots \mathop {\rightarrow }\limits ^{d}\fancyscript{A}^{N-1} $$

are trivial.$\square$

3 Graded $q$-Differential Algebras

In this section we use the cochain $N$-complex described in the Theorem 2.1 to construct a graded $q$-differential algebra which can be viewed as a natural generalization of the notion of graded differential. Then we will describe a graded $q$-differential polynomial algebra which arises in relation with a connection form which can be viewed as analog of connection form in a graded differential algebra introduced by Quillen in [13].

Definition 3.1

A graded q-differential algebra is a graded associative unital algebra $\fancyscript{A}=\oplus _{k\in \mathbb {Z}}\fancyscript{A}^k$ endowed with a linear mapping $d$ of degree one such that the sequence

$$ \cdots \mathop {\rightarrow }\limits ^{d} \fancyscript{A}^{k-1}\mathop {\rightarrow }\limits ^{d}\fancyscript{A}^{k}\mathop {\rightarrow }\limits ^{d} \fancyscript{A}^{k+1}\mathop {\rightarrow }\limits ^{d}\cdots $$

is an $N$-complex with $N$-differential $d$ satisfying the graded $q$-Leibniz rule

$$\begin{aligned} d(u\,{ v})=d(u){ v}+q^{k}ud({ v}), \end{aligned}$$

(11)

where $u\in \fancyscript{A}^k, { v}\in \fancyscript{A}$.

It follows from Theorem 2.1

Theorem 3.1

Let $\fancyscript{A}$ be a graded associative unital algebra $\fancyscript{A}=\oplus _{k}{\fancyscript{A}}^k$, and $q$ be a primitive $N$th root of unity. If there exists an element of grading one ${ v}\in \fancyscript{A}^1$ which satisfies the condition ${ v}^N\in \fancyscript{Z}(\fancyscript{A})$, where $\fancyscript{Z}(\fancyscript{A})$ is the graded center of $\fancyscript{A}$, then the graded algebra $\fancyscript{A}$ endowed with the inner graded $q$-derivation $d=\text{ ad }^q_{ v}$ is a graded $q$-differential algebra ($d$ is its $N$-differential).

Indeed we can prove this theorem by taking into account that an inner graded $q$-derivation satisfies the graded $q$-Leibniz rule (3) and the inner graded $q$-derivation $d=\text{ ad }^q_{ v}$, induced by an element of grading one ${ v}\in \fancyscript{A}^1$ such that ${ v}^N\in \fancyscript{Z}(\fancyscript{A})$, is the $N$-differential of the cochain complex (Theorem 2.1)

$$ \cdots \mathop {\rightarrow }\limits ^{d} \fancyscript{A}^{k-1}\mathop {\rightarrow }\limits ^{d}\fancyscript{A}^{k}\mathop {\rightarrow }\limits ^{d} \fancyscript{A}^{k+1}\mathop {\rightarrow }\limits ^{d}\cdots . $$

Now we introduce a graded $q$-differential algebra of polynomials which arises in relation with an algebraic model of a connection form and this algebraic model is based on exterior calculus with differential satisfying $d^N=0$. This algebra will be used in the next section in order to calculate the curvature of a connection form.

Let $\mathbb {N}_1=\{i\in \mathbb {Z}: i\ge 1\}$ be the set of integers greater than or equal to one and $\{\mathfrak {d},a_i\}_{i\in {\mathbb N}_1}$ be a set of variables. We consider the algebra of noncommutative polynomials $\mathfrak {P}_q[\mathfrak {d},a]$ over $\mathbb C$ generated by the variables $\{\mathfrak {d},a_i\}_{i\in {\mathbb N}_1}$ which are subjected to the commutation relations

$$\begin{aligned} \mathfrak {d}a_i=q^i\,a_i\mathfrak {d}+a_{i+1},\qquad \forall i\in {\mathbb N}_1 \end{aligned}$$

(12)

where $q$ is any complex number different from zero. We denote the identity element of this algebra by $\nVdash$. Obviously we can split up the set of variables of the algebra $\mathfrak {P}_q[\mathfrak {d},a]$ into two subsets $\{\mathfrak {d}\}$, $\{a_i\}_{i\in {\mathbb N}_1}$ which generate respectively the subalgebras ${\mathfrak {P}}_q[\mathfrak {d}]\subset \mathfrak {P}_q[\mathfrak {d},a]$ and ${\mathfrak {P}}_q[a]\subset \mathfrak {P}_q[\mathfrak {d},a]$. Hence the subalgebra ${\mathfrak {P}}_q[\mathfrak {d}]$ is generated by a single variable $\mathfrak {d}$, and the subalgebra ${\mathfrak {P}}_q[a]$ is freely generated by the variables $\{a_i\}_{i\in {\mathbb N}_1}$ because we do not assume any relation between variables $a_i$.

Now our aim is to equip the algebra of polynomials $\mathfrak {P}_q[\mathfrak {d},a]$ with a graded structure so that $\mathfrak {P}_q[\mathfrak {d},a]$ will become a graded algebra. This can be done as follows: we assign grading zero to the identity element $1\!\!1$ of the algebra ${\mathfrak {P}}_q[\mathfrak {d},a]$, grading one to the generator $\mathfrak {d}$ and grading $i$ to a generator $a_i$, where $i\in {\mathbb N}_1$. Thus making use of previously defined notations we can describe the graded structure of generators of $\mathfrak {P}_q[\mathfrak {d},a]$ by the formulae

$$\begin{aligned} |1\!\!1|=0,\quad |\mathfrak {d}|=|a_1|=1,\quad |a_i|=i,\quad i\ge 2. \end{aligned}$$

(13)

As usual we extend this graded structure to the whole algebra $\mathfrak {P}_q[\mathfrak {d},a]$ by defining the grading of any product of variables $\{\mathfrak {d},a_i\}_{i\in {\mathbb N}_1}$ as the sum of gradings of its factors. It is easy to see that the algebra of polynomials $\mathfrak {P}_q[\mathfrak {d},a]$ becomes the positively graded algebra. Hence we can write

$$ {\mathfrak {P}}_q[\mathfrak {d},a]=\oplus _{k\in \mathbb {N}}{\mathfrak {P}}^k_q[\mathfrak {d},a], $$

where ${\mathfrak {P}}^k_q[\mathfrak {d},a]$ is the subspace of homogeneous polynomials of grading $k$. It should be mentioned that the graded structure of ${\mathfrak {P}}[\mathfrak {d},a]$ induces the graded structures of the subalgebras ${\mathfrak {P}}_q[\mathfrak {d}],{\mathfrak {P}}_q[a]$ which are positively graded algebras as well. Clearly the positively graded algebra $\mathfrak {P}_q[\mathfrak {d},a]$ becomes the $\mathbb {Z}_N$-graded algebra, where $N$ any integer greater than 1, if we slightly modify the above described gradation by taking all gradings modulo $N$. Let us denote by $\text{ Lin }\,\mathfrak {P}_q[a]$ the algebra of $\mathbb C$-endomorphisms of vector space of ${\mathfrak {P}}_q[a]$. Obviously $\text{ Lin }\,\mathfrak {P}_q[a]$ is a graded algebra with gradation induced by the gradation of ${\mathfrak {P}}_q[a]$. Having defined the positively graded structure of the algebra $\mathfrak {P}_q[\mathfrak {d},a]$ we can apply the notions of graded commutator and inner graded $q$-derivation described in the previous chapter to study the structure of $\mathfrak {P}_q[\mathfrak {d},a]$. First of all we observe that the commutation relations (12) can be written by means of graded commutator and inner graded $q$-derivation in the form

$$\begin{aligned}{}[\mathfrak {d},a_i]_q=a_{i+1},\qquad \text{ or }\qquad \text{ ad }^{q}_{\mathfrak {d}}(a_i)=a_{i+1}, \end{aligned}$$

(14)

where $i\in \mathbb {N}_1$. This form of commutation relations suggests us to consider the inner graded $q$-derivation $\text{ ad }^{q}_{\mathfrak {d}}$ of the algebra $\mathfrak {P}_q[\mathfrak {d},a]$ associated with a variable $\mathfrak {d}$. If we restrict $\text{ ad }^{q}_{\mathfrak {d}}$ to the subalgebra ${\mathfrak {P}}_q[a]$ we get the graded $q$-derivation of subalgebra ${\mathfrak {P}}_q[a]$ which we will denote by $d$, i.e.

$$\begin{aligned} d:=\text{ ad }^{q}_{\mathfrak {d}}\mid _{{\mathfrak {P}}_q[a]},\qquad d:{\mathfrak {P}}_q[a]\rightarrow {\mathfrak {P}}_q[a]. \end{aligned}$$

(15)

Obviously $d$ is a graded $q$-derivation of grading one of the $\mathbb {Z}_N$-graded algebra ${\mathfrak {P}}_q[a]$. From the commutation relations (14) it follows that

$$ d(1\!\!1)=0,\;\;\;d(a_i)=a_{i+1}, $$

for any $i\ge 1$. Let us define $D,\nabla \in \text{ Lin }\,\mathfrak {P}_q[a]$ of grading one and the polynomials $f_k\in {\mathfrak {P}}_q[a]$, where $k$ is an integer greater than or equal to zero, by the formulae

$$\begin{aligned} D(P)&= d(P)+a_1\,P, \end{aligned}$$

(16)

$$\begin{aligned} \nabla (P)&= d(P)+[a_1,P]_q,\\ f_0&= 1\!\!1,\nonumber \\ f_1&= a_1,\nonumber \end{aligned}$$

(17)

$$\begin{aligned} f_k&= D(f_{k-1}), \end{aligned}$$

(18)

where $P\in {\mathfrak {P}}_q[a]$ is a homogeneous polynomial. We can write the linear mapping $\nabla$ in the form $\nabla =\text{ ad }^q_{\mathfrak {d}+a_1}$ which clearly shows that $\nabla$ is a inner graded $q$-derivation of the algebra ${\mathfrak {P}}_q[a]$. Hence for any polynomials $P,Q\in {\mathfrak {P}}_q[a]$, where $P$ is homogeneous, it holds

$$\begin{aligned} D(PQ)&= D(P)\,Q+q^{|P|}P\,d(Q),\end{aligned}$$

(19)

$$\begin{aligned} \nabla (PQ)&= \nabla (P)\,+q^{|P|}P\,\nabla (Q). \end{aligned}$$

(20)

For the first values of $k$ we calculate by means of the recurrent relation (18)

$$\begin{aligned} f_2&= a_2+a_1^2, \nonumber \\ f_3&= a_3+a_2\,a_1+[2]_q\,a_1\,a_2+a_1^3, \nonumber \\ f_4&= a_4+a_3\,a_1+[3]_q\,a_1a_3+[3]_q\,a_2^2\nonumber \\&+a_2a_1^2+[3]_q\,a_1^2a_2+[2]_q\,a_1a_2a_1+a_1^4,\\ f_5&= a_5+a_4a_1+[4]_q\,a_1a_4+[4]_q\,a_3a_2\nonumber \\&+\left[ \begin{array}{c} 4 \\ 2 \\ \end{array} \right] _q\,a_2a_3+a_3a_1^2+[3]_q\,a_2^2a_1+[4]_q\,a_2a_1a_2\nonumber \\&+[2]_q\,[4]_q\,a_1a_2^2+\left[ \begin{array}{c} 4 \\ 2 \\ \end{array} \right] _q\,a_1^2a_3+[3]_q\,a_1a_3a_1\nonumber \\&+[2]_q\,a_1a_2a_1^2+[3]_q\,a_1^2a_2a_1+a_2a_1^3+[4]_q\,a_1^3a_2+a_1^5.\nonumber \end{aligned}$$

(21)

Getting a bit ahead we would like to point out that the polynomials $f_k$ may be interpreted as the curvature of a connection if we view the generator $a_1$ as an algebraic model for a connection one form. Let us remind that if $k$ is a positive integer then a composition of $k$ is a representation of $k$ as the sum of a sequence of strictly positive integers, and two sequences that differ in the order of their terms give different compositions of their sum while they define the same partition of $k$. For example if $k=3$ then there are 4 compositions

$$ 3=3,\;3=2+1,\;3=1+2,\;3=1+1+1. $$

Let $\varPsi _k$ be the set of all compositions of an integer $k$. We will write a composition of an integer $k$ in the form of a sequence of strictly positive integers $\sigma =(i_1,i_2,\ldots ,i_r)$, where $i_1+i_2+\cdots +i_r=k$. Let us denote

$$\begin{aligned} k_1&= i_1,\nonumber \\ k_2&= i_1+i_2,\nonumber \\ k_3&= i_1+i_2+i_3,\nonumber \\&\ldots&\nonumber \\ k_{r-1}&= i_1+i_2+\cdots + i_{r-1}.\nonumber \end{aligned}$$

It can be proved [14] that the number of all possible compositions of a positive integer $k$ is $2^{k-1}$, i.e. the set $\varPsi _k$ contains $2^{k-1}$ elements. The following theorem gives an explicit formula for the polynomials $f_k$:

Theorem 3.2

For any integer $k\ge 2$ we have the following expansion of power of the operator $D$ and the expansion of a polynomial $f_k$ in terms of generators $a_i$:

$$\begin{aligned} D^k&= \sum \limits _{i=0}^k \left[ \begin{array}{c} k \\ i \\ \end{array} \right] _q f_i\,d^{k-i},\nonumber \\ f_k&= \sum \limits _{\sigma \in \varPsi _k}\left[ \begin{array}{c} k_2-1 \\ k_1\\ \end{array} \right] _q \left[ \begin{array}{c} k_3-1 \\ k_2\\ \end{array} \right] _q \ldots \left[ \begin{array}{c} k-1 \\ k_{r-1} \\ \end{array} \right] _q\,a_{i_1}a_{i_2}\ldots a_{i_r},\nonumber \end{aligned}$$

where $\sigma =(i_1,i_2,\ldots ,i_r)$ is a composition of an integer $k$.

Proof

We will prove the expansion formulae of this theorem by the method of mathematical induction. In order to prove the expansion of power of the operator $D$ by means of mathematical induction we begin with the base case and show that this formula holds when $k$ is equal to 1. This is true because

$$ D=\left[ \begin{array}{c} 1 \\ 0 \\ \end{array} \right] _qf_0\,d+\left[ \begin{array}{c} 1 \\ 1 \\ \end{array} \right] _q\,f_1=d+a_1. $$

Next step in the proof is an inductive step, i.e. we assume that the expansion formula holds for some integer $k>1$ and show that it also holds when $k+1$ is substituted for $k$. Indeed we have

$$\begin{aligned} D^{k+1}&= D(D^k)=D\Bigl (\sum \limits _{i=0}^k \left[ \begin{array}{c} k \\ i \\ \end{array} \right] _q f_i\,d^{k-i}\Bigr )\nonumber \\&= \sum \limits _{i=0}^k \left[ \begin{array}{c} k \\ i \\ \end{array} \right] _q \Bigl (D(f_i)\,d^{k-i}+q^i\,f_id^{k+1-i}\Bigr )\nonumber \\&= \sum \limits _{i=0}^k \left[ \begin{array}{c} k \\ i \\ \end{array} \right] _q \Bigl (f_{i+1}\,d^{k-i}+q^i\,f_id^{k+1-i}\Bigr )\nonumber \\&= f_{k+1}+\sum \limits _{i=0}^{k-1} \left[ \begin{array}{c} k \\ i \\ \end{array} \right] _q f_{i+1}\,d^{k-i}\nonumber +q^i\sum \limits _{i=1}^k \left[ \begin{array}{c} k \\ i \\ \end{array} \right] _q f_{i}\,d^{k+1-i}+d^{k+1}\nonumber \\&= f_{k+1}+\sum \limits _{i=1}^{k} \left[ \begin{array}{c} k \\ i-1 \\ \end{array} \right] _q f_{i}\,d^{k+1-i}\nonumber +q^i\sum \limits _{i=1}^k \left[ \begin{array}{c} k \\ i \\ \end{array} \right] _q f_{i}\,d^{k+1-i}+d^{k+1}\nonumber \\&= f_{k+1}+\sum \limits _{i=1}^{k} \Bigl (\left[ \begin{array}{c} k \\ i-1 \\ \end{array} \right] _q + q^i\left[ \begin{array}{c} k \\ i \\ \end{array} \right] _q\Bigr ) f_{i}\,d^{k+1-i}+d^{k+1}\nonumber \\&= f_{k+1}+\sum \limits _{i=1}^{k} \left[ \begin{array}{c} k+1 \\ i \\ \end{array} \right] _q\, f_{i}\,d^{k+1-i}+d^{k+1}\nonumber \\&= \sum \limits _{i=0}^{k+1} \left[ \begin{array}{c} k+1 \\ i \\ \end{array} \right] _q\, f_{i}\,d^{k+1-i}.\nonumber \end{aligned}$$

Thus the expansion of power of the operator $D$ is proved. Now if we apply the both sides of the proved formula to $a_1$ we obtain

$$\begin{aligned} f_{k+1}=\sum \limits _{i=0}^k \left[ \begin{array}{c} k \\ i \\ \end{array} \right] _q f_i\,a_{k+1-i}, \end{aligned}$$

(22)

and this is the recurrent formula for the polynomials $f_k$ which we will use in the second part of the present proof in order to prove the expansion formula for $f_k$.

We start the proof of the expansion formula for a polynomial $f_k$ with the base case when $k=2$. In this case there are two compositions $2=2,\;2=1+1$. Hence we have

$$ f_2=\left[ \begin{array}{c} 1 \\ 0 \\ \end{array} \right] _q\,a_2+\left[ \begin{array}{c} 1 \\ 0 \\ \end{array} \right] _q\,\left[ \begin{array}{c} 1 \\ 1 \\ \end{array} \right] _q\,a_1^2=a_2+a_1^2. $$

Comparing this result with the first formula in (21) we see that in the case when $k=2$ the expansion formula for $f_k$ is correct. The next step is an inductive step, i.e. we assume that the expansion formula holds for some positive integer $k>2$ and show that it also holds when $k+1$ is substituted for $k$. Let us consider the sum

$$\begin{aligned} \sum \limits _{\sigma \in \varPsi _{k+1}}\left[ \begin{array}{c} k_2-1 \\ k_1\\ \end{array} \right] _q \left[ \begin{array}{c} k_3-1 \\ k_2\\ \end{array} \right] _q \ldots \left[ \begin{array}{c} k \\ k_{r} \\ \end{array} \right] _q\,a_{i_1}a_{i_2}\ldots a_{i_{r+1}}, \end{aligned}$$

(23)

where $\sigma =(i_1,i_2,\ldots ,i_{r},i_{r+1})$ is a composition of an integer $k+1$. Hence $i_1+\cdots +i_{r}+i_{r+1}=k+1$. Our aim is to show that this sum is equal to the polynomial $f_{k+1}$. Let us fix an integer $i\in \{0,1,\ldots ,k\}$ and a generator $a_{k+1-i}$. It is clear that if we select the compositions of an integer $k+1$ which have the form $(i_1,i_2,\ldots ,i_r,k+1-i)$, i.e. the last integer of each composition is previously fixed integer $k+1-i$, and we remove in each composition the last integer then the set of compositions $(i_1,i_2,\ldots ,i_r)$ is the set of all compositions of an integer $i$, i.e. $\{(i_1,i_2,\ldots ,i_r)\}=\varPsi _i$. Indeed we have

$$ i_1+i_2+\cdots +i_r+k+1-i=k+1, $$

which implies $i_1+i_2+\cdots +i_r=i$. Consequently if we select in the sum (23) all terms with $i_{r+1}=k+1-i$ (i.e. containing a generator $a_{k+1-i}$ at the end of a product of generators) then we get the sum

$$\begin{aligned} \sum \limits _{\sigma \in \varPsi _{k+1}}\left[ \begin{array}{c} k_2-1 \\ k_1\\ \end{array} \right] _q \left[ \begin{array}{c} k_3-1 \\ k_2\\ \end{array} \right] _q \ldots \left[ \begin{array}{c} k \\ i \\ \end{array} \right] _q\,a_{i_1}a_{i_2}\ldots a_{i_r}a_{k+1-i}, \end{aligned}$$

(24)

where the sum is taken over the compositions of integer $k+1$ which have the form $\sigma =(i_1,i_2,\ldots ,i_r,k+1-i)\in \varPsi _{k+1}$. We would like to point out that the product of binomial coefficients of each term in this sum contains the factor

$$ \left[ \begin{array}{c} k \\ i \\ \end{array} \right] _q. $$

Hence we can write the sum (24) as follows

$$\begin{aligned} \left[ \begin{array}{c} k \\ i \\ \end{array} \right] _q\Bigl (\sum \limits _{\tau \in \varPsi _{i}}\left[ \begin{array}{c} k_2-1 \\ k_1\\ \end{array} \right] _q \left[ \begin{array}{c} k_3-1 \\ k_2\\ \end{array} \right] _q\ldots \left[ \begin{array}{c} i-1 \\ k_{r-1}\\ \end{array} \right] _q a_{i_1}a_{i_2}\ldots a_{i_r}\Bigr )\,a_{k+1-i},\nonumber \end{aligned}$$

where $\tau =(i_1,i_2,\ldots ,i_r)\in \varPsi _i$ and the sum is taken over all compositions of integer $i$. Now we make use of the assumption of an inductive step that the expansion formula for a polynomial $f_m$ holds for each integer $m\in \{1,2,\ldots ,k\}$. Hence the sum in the previous formula is equal to $f_i$, i.e

$$ \sum \limits _{\tau \in \varPsi _{i}}\left[ \begin{array}{c} k_2-1 \\ k_1\\ \end{array} \right] _q \left[ \begin{array}{c} k_3-1 \\ k_2\\ \end{array} \right] _q\ldots \left[ \begin{array}{c} i-1 \\ k_{r-1}\\ \end{array} \right] _q a_{i_1}a_{i_2}\ldots a_{i_r}=f_i. $$

Thus the sum (24) is equal to

$$ \left[ \begin{array}{c} k \\ i \\ \end{array} \right] _q\,f_i\,a_{k+1-i}, $$

and summing up all these terms with respect to $i$ we get the sum (23). Consequently the sum (23) we started with is equal to the sum

$$ \sum \limits _{i=0}^k\left[ \begin{array}{c} k \\ i \\ \end{array} \right] _q\,f_i\,a_{k+1-i}, $$

which in turn is equal to $f_{k+1}$ (see the recurrent relation (22)). This ends the proof.$\square$

We remind a reader that the parameter $q$ which plays an important role in the structure of the algebra $\mathfrak {P}_q[\mathfrak {d},a]$ is any complex number different from zero. Now we will study the structure of the algebra of polynomials $\mathfrak {P}_q[\mathfrak {d},a]$ at a primitive $N$th root of unity, i.e. we assume $q$ to be a primitive $N$th root of unity. We may expect that in this case the infinite set of variables $\{\mathfrak {d},a_1,a_2,\ldots \}$ is “cut off” and we get an algebra whose vector space is finite dimensional. Indeed we can prove the following proposition:

Proposition 3.1

Let $\mathfrak {P}_q[\mathfrak {d},a]$ be the algebra of polynomials generated by the set of variables $\{\mathfrak {d},a_i\}_{i\in {\mathbb N}_1}$ which obey the commutation relations (12). If we assume that $q$ is a primitive $N$th root of unity and the variable $\mathfrak {d}$ is subjected to the additional relation $\mathfrak {d}^N=\lambda \cdot 1\!\!1$, where $\lambda$ is a complex number, then for any integer $k>N$ a variable $a_k$ vanishes, i.e. the algebra $\mathfrak {P}_q[\mathfrak {d},a]$ is generated by the finite set of variables $\{\mathfrak {d},a_k\}_{k=1}^{N}$ which obey the relations

$$\begin{aligned} \mathfrak {d}a_1&= q\,a_1\mathfrak {d}+a_2,\nonumber \\ \mathfrak {d}a_2&= q^2\,a_2\mathfrak {d}+a_3,\nonumber \\&\cdots&\\ \mathfrak {d}a_{N-1}&= q^{N-1}\,a_{N-1}\mathfrak {d}+a_{N},\nonumber \\ \mathfrak {d}a_{N}&= a_{N}\mathfrak {d},\nonumber \\ \mathfrak {d}^N&= \lambda \cdot 1\!\!1.\nonumber \end{aligned}$$

(25)

The graded $q$-derivation $d=\text{ ad }^q_{\mathfrak {d}}:{\mathfrak {P}}_q[a]\rightarrow {\mathfrak {P}}_q[a]$ associated to variable $\mathfrak {d}$ is an $N$-differential, i.e. $d^N=0$, and the sequence

$$ \cdots \mathop {\rightarrow }\limits ^{d}{{\mathfrak {P}}^{i-1}_q[a]}\mathop {\rightarrow }\limits ^{d}{{\mathfrak {P}}^{i}_q[a]}\mathop {\rightarrow }\limits ^{d}{{\mathfrak {P}}^{i+1}_q[a]}\mathop {\rightarrow }\limits ^{d}\cdots $$

is a cochain $N$-complex. The graded algebra ${\mathfrak {P}}_q[a]$ equipped with the $N$-differential $d$ is a graded $q$-differential algebra.

Proof

We suppose that the algebra of polynomials is equipped with the $\mathbb {Z}_N$-gradation as it was explained earlier (13). It easily follows from the commutation relations of the algebra $\mathfrak {P}_q[\mathfrak {d},a]$ that for any integer $k\ge 2$ we have

$$ a_{k+1}=d^k(a_1), $$

where $d=\text{ ad }^q_{\mathfrak {d}}$ is the graded $q$-derivation associated with a variable $\mathfrak {d}$. Making use of the expansion of power of graded $q$-derivation used in the proof of Theorem 2.1 we obtain

$$ a_{k+1}=d^k(a_1)=(\text{ ad }^q_{\mathfrak {d}})^k(a_1)=\sum \limits _{i=0}^k\,(-1)^i\,p_i\left[ \begin{array}{c} k\\ i \end{array} \right] _q \mathfrak {d}^{k-i}\,u\,\mathfrak {d}^{i}. $$

Consequently if $q$ is a primitive $N$th root of unity, $\mathfrak {d}$ satisfies $\mathfrak {d}^N=\lambda \cdot 1\!\!1$ and $k=N$ then making use of the same arguments as in the proof of Theorem 2.1) we conclude that all terms of the sum at the right-hand side of the above expansion formula vanish. Consequently we have $a_{N+1}=a_{N+2}=\ldots =0$ and this ends the proof.$\square$

It is well known that locally a connection of a vector bundle can be described with the help of matrix-valued 1-form. From an algebraic point of view this matrix-valued 1-form is an element of degree one of differential algebra of matrix-valued differential forms, where differential is identified with exterior differential and graduation is induced by degree of differential form. Hence an algebraic model for a connection can be constructed if we take a differential algebra ${\fancyscript{A}}$ (over $\mathbb C$) and consider an element of degree one of this algebra $A$ calling it connection form. Then a covariant differential induced by this connection form is the operator $\nabla =d+A$, and the curvature is the element of degree 2 given by $F=dA+A^2=dA+\frac{1}{2}[A,A]$, where $[,]$ is the graded commutator of $\fancyscript{A}$. This approach was proposed by Quillen in [13]. Following this approach we introduce a notion of $N$-connection form which particularly gives a connection form if $N=2$. Let us denote by $\mathfrak {P}_q[\mathfrak {d},a]$ the finite dimensional graded algebra generated by $\{\mathfrak {d},a_k\}_{k=1}^{N}$ which obey relations (26) and by $\mathfrak {P}_q[a]$ the graded $q$-differential algebra generated by $\{a_k\}_{k=1}^{N}$ with $N$-differential $d$. Now we give the following definition:

Definition 3.2

The generator $a_1$ of $\mathbb {Z}_N$-graded $q$-differential algebra $\mathfrak {P}_q[a]$ will be referred to as an $N$-connection form and the algebra $\mathfrak {P}_q[a]$ will be referred to as an algebra of $N$-connection form. The operator $D=d+a_1:\mathfrak {P}_q[a]\rightarrow \mathfrak {P}_q[a]$ will be called a covariant $N$-differential, and the polynomial $f_N$, whose explicit power expansion formula given in (3.2), will be called the curvature of $N$-connection form $a_1$.

Proposition 3.2

If $\mathfrak {P}_q[a]$ is the algebra of $N$-connection form and $d$ is its $N$-differential then the $N$th power of the covariant $N$-differential $D$ is the operator of multiplication by the curvature of $N$-connection form $f_N$.

Proof

The proof of this proposition is based on the first expansion formula proved in the Theorem 3.2. Indeed we can expand an $N$th power of the covariant $N$-differential $D$ into the sum of products of polynomials $f_i$ and the powers of the $N$-differential $d$ as follows

$$ D^N = \sum \limits _{i=0}^N \left[ \begin{array}{c} N \\ i \\ \end{array} \right] _q f_i\,d^{N-i}. $$

As $q$ is a primitive $N$th root of unity this expansion can be essentially simplified in the case $k=N$ if we take into account that all $q$-binomial coefficients with $i\in \{1,2,\ldots ,N-1\}$ vanish. The first term of this expansion also vanishes because $d$ is the $N$-differential. Hence for any polynomial $P\in \mathfrak {P}_q[a]$ we have

$$ D^N(P)=f_N\cdot P, $$

and this ends the proof.$\square$

Proposition 3.3

If $\mathfrak {P}_q[a]$ is the algebra of connection form and $f_N$ is the curvature of connection form then the curvature satisfies the identity

$$\begin{aligned} \nabla (f_N)=0. \end{aligned}$$

(26)

Proof

Let us remind a reader that $\nabla =d+\text{ ad }^q_{a_1}$. We prove this proposition by means of the recurrent relation for polynomials $f_k$

$$ f_{k+1}=\sum \limits _{i=0}^k \left[ \begin{array}{c} k \\ i \\ \end{array} \right] _q f_i\,a_{k+1-i}. $$

Substituting $N$ for $k$ in the above relation we obtain

$$\begin{aligned} f_{N+1}=\sum \limits _{i=0}^N \left[ \begin{array}{c} N \\ i \\ \end{array} \right] _q f_i\,a_{N+1-i}. \end{aligned}$$

(27)

As $q$ is a primitive $N$th root of unity we have

$$ \left[ \begin{array}{c} N \\ i \\ \end{array} \right] _q=0, $$

for any integer $i\in \{1,2,\ldots ,N-1\}$. Consequently there are only two terms with non-zero $q$-binomial coefficients (labeled by $i=0,N$) at the right-hand side of the relation (27) and

$$ f_{N+1}=f_0\,a_{N+1}+f_N\,a_1. $$

The first term at the right-hand side of the above formula is also zero because of $a_{N+1}=0$ (Proposition 3.1). Hence

$$\begin{aligned} 0&= f_{N+1}-f_N\,a_1=D(f_N)-f_N\,a_1\\&= d(f_N)+a_1\,f_N-f_N\,a_1=d(f_N)+[a_1,f_N]_q= (d+\text{ ad }^q_{a_1})(f_N)=\nabla (f_N). \end{aligned}$$

$\square$

The identity (26) is an analogue of Bianchi identity for the curvature of $N$-connection form. It is worth mentioning that we can write the Bianchi identity for the curvature of $N$-connection form (26) in a different way if we consider the covariant $N$-differential $D$ and the curvature $f_N$ as the linear operators $D,f_N:{\mathfrak {P}}_q[a]\rightarrow {\mathfrak {P}}_q[a]$, i.e. $D,f_N\in \text{ Lin }\,\mathfrak {P}_q[a]$, where $f_N$ is the operator of multiplication by $f_N$ (we denote it by the same symbol as the curvature $f_N$ in order not to make the notations very complicated). Then the Bianchi identity may be written in the form

$$ [D,f_N]_q=0. $$

Indeed

$$\begin{aligned}{}[D,f_N]_q&= D\circ f_N-f_N\circ D\nonumber \\&= d(f_N)+f_N\circ d+a_1\,f_N-f_N\circ d-f_N\,a_1\nonumber \\&= d(f_N)+[a_1,f_N]_q=\nabla (f_N)=0.\nonumber \end{aligned}$$