Noncommutative symplectic manifolds for self-injective Nakayama algebras

Abstract

In the paper, we study noncommutative symplectic manifolds $(A,\llbracket \omega \rrbracket )$, where A is a self-injective Nakayama algebra. There is given a full description of exact symplectic manifolds for A being the trivial extension of an hereditary Nakayama algebra.

1 Introduction

This paper is devoted and motivated by noncommutative differential geometry, to be more precise, symplectic geometry. Noncommutative symplectic geometry was introduced by Kontsevich [9].

If we want to consider a structure of noncommutative symplectic manifold linked to an associative algebra A, we have to yield a differential calculus on the algebra A. Following Dubois-Violette [5], a differential calculus on the algebra A is a differential algebra $(\Omega ,d)$ such that $\Omega =\bigoplus ^{\infty }_{n=0}\Omega ^{n}$ is a graded algebra, $\Omega ^{0}=A$, and d is a super-derivation of degree 1 on $\Omega$ with $d^{2}=0$.

In the paper, we fix an algebraically closed field K whose characteristic is zero. By an algebra, we always mean an associative (in general noncommutative) unitary K-algebra A that is finite-dimensional. For such an algebra, we usually consider the Karoubi-de Rham complex $(DR^{\bullet }_{R}(A),d)$ for a fixed subalgebra R of A.

For a given algebra A, any element $\llbracket \omega \rrbracket \in DR^{n}_{R}(A)$ is called an R-relative differential n-form on A. If this form is non-degenerate and closed, then the pair $(A,\llbracket \omega \rrbracket )$ is said to be a symplectice manifold for the differential calculus $(DR^{\bullet }_{R}(A),d)$. If the algebra is noncommutative, then the symplectic manifold $(A,\llbracket \omega \rrbracket )$ is said to be noncommutative.

A particular class of finite-dimensional K-algebras form path algebras of quivers with relations, introduced by Gabriel in [6]. These algebras provide a wide class of examples to test many problems. For the Karoubi-de Rham differential calculus $(DR^{\bullet }_{R}(A),d)$, the path algebra $A=KQ$ of a quiver Q satisfies that there are no symplectic manifolds $(A,\llbracket \omega \rrbracket )$. That fact is shown in [2, 8, 12]. The shortest proof of this fact rests on computing the cohomology groups of the Karoubi-de Rham complex. The same holds for canonical algebras of Ringel (see [11]). The proof of this fact is obtained with a use of combinatorics. It seems to be possible to repeat the same arguments for any quiver algebra with relations, when the quiver has no oriented cycles.

On the contrary, there are symplectic manifolds $(K\bar{Q},\llbracket \omega \rrbracket )$ for the so-called preprojective algebras $K\bar{Q}$ of quivers Q (see [3]). These symplectic manifolds are not classified. Their existence is obtained with a help of deep tools that are developed in [3]. The preprojective algebras are self-injective. This is the reason why the self-injective Nakayama algebras are considered. The class of Nakayama algebras, studied in the paper, consists of symmetric algebras. The main aim for considering these algebras is to find direct methods to classify symplectic manifolds.

Our main goal is to give a full description of exact symplectic manifolds $(B_{r,l},\llbracket \omega \rrbracket )$ for a class of self-injective Nakayama algebras $B_{r,l}$, $r\ge 0$, $l=r+2$, (see [1, 10]). Exactness of the manifold means that the form $\llbracket \omega \rrbracket$ is exact, i.e., it is contained in the image of d. The considered class of algebras $B_{r,r+2}$ coincides to the class of the trivial extensions of the path algebras KQ of the quivers $Q=0{\mathop {\longrightarrow }\limits ^{\alpha _{0}}} 1{\mathop {\longrightarrow }\limits ^{\alpha _{1}}} \cdots {\mathop {\longrightarrow }\limits ^{\alpha _{r-1}}}r$.

The main results of the paper are the following.

Theorem 1.1

For any algebra $B_{r,r+2}$, $r\ge 0$, and the Karoubi-de Rham calculus $(DR^{\bullet }_{R}(B_{r,r+2}),d)$, the following conditions are satisfied:

• (1) There exist symplectic manifolds $(B_{r,r+2},\llbracket \omega \rrbracket )$.

• (2) A pair $(B_{r,r+2},\llbracket \omega \rrbracket )$ is an exact symplectic manifold if and only if $\llbracket \omega \rrbracket =\sum ^{r}_{i=0}k_{i}\llbracket e_{i}\otimes \overline{h_{i}}\otimes \overline{\alpha _{i}}\rrbracket \not =0$, and there is at most one index $i_{0}\in \{ 0,1,\ldots ,r\}$ such that $k_{i_{0}}=0$, where $k_{i}\in K$, $i=0,1,\ldots ,r.$

Theorem 1.2

For any algebra $B_{r,r+2}$, $r\ge 0$, every exact symplectic manifold $(B_{r,r+2},\llbracket \omega \rrbracket )$ is symplectomorphic to a manifold $(B_{r,r+2},\llbracket \omega _{0}\rrbracket )$ such that $\llbracket \omega _{0}\rrbracket = \sum ^{r}_{i=0}k_{i}\llbracket e_{i}\otimes \overline{h_{i}}\otimes \overline{\alpha _{i}}\rrbracket$ and $k_{0}\in \{0,1\}\subset K$ and $0\not =k_{i}\in K$, $i=1,2,\ldots ,r$.

In the sequel, there are explained all notations that have been used above. Furthermore, all information on path algebras of quivers can be found in [1].

2 Preliminaries

For a fixed K-algebra A that is unitary and finite-dimensional, let $\Omega ^{1}_{K}(A)$ be an A-bimodule of noncommutative differential 1-forms on A, i.e., $\Omega ^{1}_{K}(A)=\text{ ker }(m)$, where $m:A\otimes _{K}A\rightarrow A$ is an epimorphism given by $m(a\otimes b)=ab$.

Now, we denote by $\bar{A}$ the quotient linear space $\bar{A}=A/(K\cdot 1_{A})$ and let $A\rightarrow \bar{A}$ denote the projection. Then, the mapping $d_{A}:A\rightarrow \Omega ^{1}_{K}(A)$ given by $x\mapsto d_{A}x=x\otimes 1_{A}-1_{A}\otimes x$ is a derivation. Moreover, the mapping $A\otimes \bar{A} \rightarrow \Omega ^{1}_{K}(A)$ given by $x\otimes \bar{y}\mapsto x\otimes y -xy\otimes 1_{A} =xd_{A}y$ is well-defined and it is an A-bimodule isomorphism (comp. [7]; Proposition 10.1.3).

Following Ginzburg, we can proceed more generally (comp. [4]). For a given K-algebra A and its subalgebra R, we can consider

$$\begin{aligned} \Omega ^{n}_{R}(A)=A\otimes _{R}\underbrace{\bar{A}\otimes _{R}\cdots \otimes _{R}\bar{A}}_{n\times }, \end{aligned}$$

where now $\bar{A}=A/R$ as a linear K-space. Then,

$$\begin{aligned} \Omega ^{\bullet }_{R}(A)= \bigoplus ^{\infty }_{i=0}\Omega ^{i}_{R}(A). \end{aligned}$$

Further, we can define a differential $d_{n}:\Omega ^{n}_{R}(A) \rightarrow \Omega ^{n+1}_{R}(A)$ given by:

$$\begin{aligned} d_{n}(a_{0}\otimes \overline{a_{1}}\otimes \cdots \otimes \overline{a_{n}}) =1_{A}\otimes \overline{a_{0}}\otimes \overline{a_{1}}\otimes \cdots \otimes \overline{a_{n}}. \end{aligned}$$

In this manner, we obtain a complex $(\Omega ^{\bullet }_{R}(A),d)$. We define a product for this complex that is compatible with grading. We do it determining a map

$$\begin{aligned} \Omega ^{n}_{R}(A) \times \Omega ^{m-1-n}_{R}(A) \rightarrow \Omega ^{m}_{R}(A) \end{aligned}$$

as follows:

$$\begin{aligned} (a_{0}\otimes \overline{a_{1}}\otimes \cdots \otimes \overline{a_{n}})\cdot (a_{n+1} \otimes \overline{a_{n+2}} \otimes \cdots \otimes \overline{a_{m}}) = (-1)^{n}a_{0}a_{1}\otimes \overline{a_{2}}\otimes \cdots \otimes \overline{a_{m}} + \end{aligned}$$

$$\begin{aligned} \sum ^{n}_{i=1}(-1)^{n-i}a_{0} \otimes \overline{a_{1}} \otimes \cdots \otimes \overline{a_{i}a_{i+1}}\otimes \cdots \otimes \overline{a_{m}}. \end{aligned}$$

Now, let

$$\begin{aligned} DR^{\bullet }_{R}(A)=\Omega ^{\bullet }_{R}(A)/[\Omega ^{\bullet }_{R}(A),\Omega ^{\bullet }_{R}(A)]_{super} \end{aligned}$$

be the Karoubi-de Rham complex, where $[-,-]_{super}$ denotes the K-linear subspace spanned by the super-commutators. The differential $d:\Omega ^{\bullet }_{R}(A) \rightarrow \Omega ^{\bullet +1}_{R}(A)$ induces a well-defined differential $d:DR^{\bullet }_{R}(A)\rightarrow DR^{\bullet +1}_{R}(A)$.

Fixing an R-derivation $\theta :A\rightarrow A$, we obtain in the unique way a super-derivation of degree $(-1)$

$$\begin{aligned} i_{\theta }:DR^{\bullet }_{R}(A) \rightarrow DR^{\bullet -1}_{R}(A) \end{aligned}$$

of the algebra $DR^{\bullet }_{R}(A)$ that is given by the following formula:

$$\begin{aligned} i_{\theta }(\llbracket a_{0} \otimes \overline{a_{1}} \otimes \cdots \otimes \overline{a_{n}}\rrbracket ) =\sum ^{n}_{j=1}(-1)^{j-1}\llbracket a_{0}\otimes \overline{a_{1}}\otimes \cdots \otimes \overline{a_{j-1}\theta (a_{j})} \otimes \overline{a_{j+1}}\otimes \cdots \otimes \overline{a_{n}}\rrbracket , \end{aligned}$$

where $\llbracket a_{0}\otimes \overline{a_{1}}\otimes \cdots \otimes \overline{a_{n}}\rrbracket$ is the coset of the element $a_{0}\otimes \overline{a_{1}}\otimes \cdots \otimes \overline{a_{n}}\in \Omega ^{n}_{R}(A)$ modulo $[\Omega ^{n}_{R}(A),\Omega ^{n}_{R}(A)]_{super}$.

An element $\llbracket \omega \rrbracket \in DR^{n}_{R}(A)$ is said to be an R-relative differential n-form on A. For a fixed $\llbracket \omega \rrbracket \in DR^{2}_{R}(A)$, the map $i_{\llbracket \omega \rrbracket }:\text{ Der}_{R}(A)\rightarrow DR^{1}_{R}(A)$ is a K-linear transformation given by $\theta \mapsto i_{\theta }(\llbracket \omega \rrbracket )$, where $\text{ Der}_{R}(A)$ is the space of the R-derivations on A. An R-relative differential 2-form $\llbracket \omega \rrbracket$ is called non-degenerate provided that $i_{\llbracket \omega \rrbracket }$ is a K-linear isomorphism and is called degenerate otherwise. A 2-form $\llbracket \omega \rrbracket$ is exact if there is an R-relative differential 1-form $\llbracket \kappa \rrbracket$ such that $d(\llbracket \kappa \rrbracket )=\llbracket \omega \rrbracket$. Moreover, $\llbracket \omega \rrbracket$ is said to be closed provided that $d(\llbracket \omega \rrbracket )=0$ in $DR^{3}_{R}(A)$. Since $(DR^{\bullet }_{R}(A),d)$ is a complex, every exact 2-form is closed. Finally, a pair $(A,\llbracket \omega \rrbracket )$ is said to be a symplectic manifold if $\llbracket \omega \rrbracket \in DR^{2}_{R}(A)$ is non-degenerate and closed. This manifold is called exact if $\llbracket \omega \rrbracket$ is exact. For a symplectic manifold $(A,\llbracket \omega \rrbracket )$, we call the algebra A its support.

3 Nakayama algebras

A finite-dimensional associative unitary K-algebra A is called elementary (see [1]) if $A/\text{rad }(A)\cong K\times \cdots \times K$, where $\text{ rad }(A)$ is the Jacobson radical of the algebra A. P. Gabriel [6] attached a bound quiver $(Q_{A},I_{A})$ to every elementary K-algebra A in such a way that $A\cong KQ_{A}/I_{A}$, where $Q_{A}$ is a finite quiver, $KQ_{A}$ is the path algebra of the quiver $Q_{A}$, and $I_{A}$ is an admissible two-sided ideal in $KQ_{A}$. Now, many algebras are defined by their bound quivers. The same can be done for algebras introduced by Nakayama [10].

We shall consider finite-dimensional K-algebras of the form $B_{r,l}=KQ_{r}/I_{l}$, where $Q_{r}$ is a quiver of the form

$$\begin{aligned} \begin{array}{rcccl} 0 &{} {\mathop {\longrightarrow }\limits ^{\alpha _{0}}} &{} 1 &{} {\mathop {\longrightarrow }\limits ^{\alpha _{1}}} &{} 2 \\ \alpha _{r} \uparrow &{} &{} &{} &{} \downarrow \alpha _{2} \\ r &{} {\mathop {\longleftarrow }\limits ^{\alpha _{r-1}}} &{} &{} \cdots &{} \end{array} \end{aligned}$$

where $r=0,1,2,\ldots$ and $I_{l}$ is the two-sided ideal in the path algebra $KQ_{r}$ generated by the paths of length l, $l=2,3,\ldots$.

In the algebra $B_{r,l}$, every element $b\in B_{r,l}$ is said to be a path of length $d(b)=n\ge 1$ provided that $b=\beta _{n}\ldots \beta _{1}+I_{l}$, where $\beta _{j}$ is an arrow in the quiver $Q_{r}$, $j=1,2,\ldots ,n$, and the source $s(\beta _{j+1})$ of the arrow $\beta _{j+1}$ coincides to the sink $t(\beta _{j})$ of the arrow $\beta _{j}$ for each $j=1,2,\ldots ,n-1$ Further, $s(\beta _{1})$ is said to be the source of the path b and is denoted by s(b), and $t(\beta _{n})$ is said to be the sink of the path b and it is denoted by t(b). Moreover, each vertex $x\in \{0,1,\ldots ,r\}$ of the quiver $Q_{r}$ is identified to the trivial path $e_{x}+I_{l}$ of length 0. It is obvious that $s(e_{x})=t(e_{x})=x$. We shall usually write elements of an algebra $B_{r,l}$ as K-linear combinations of paths in the quiver $Q_{r}$, keeping in mind that if such a combination belongs to $I_{l}$, then we have a zero element of the algebra.

We shall denote by R the semi-simple subalgebra of $B_{r,l}$ that is generated by the trivial paths (orthogonal idempotente) $e_{i}$, $i=0,1,\ldots ,r$. Then, $R\cong K^{r+1}$.

For a fixed algebra $B_{r,l}$, $r\ge 0$, $l\ge 2$, an element $u_{0}\otimes \overline{u_{1}} \otimes \cdots \otimes \overline{u_{n}}\in \Omega ^{n}_{R}(B_{r,l})$ is defined to be a generalized path in $\Omega ^{n}_{R}(B_{r,l})$ if $u_{0},u_{1},\ldots ,u_{n}$ are paths in the quiver $Q_{r}$ and the following conditions are satisfied:

(1) $u_{0}\otimes \overline{u_{1}} \otimes \cdots \otimes \overline{u_{n}}\not =0$ in $\Omega ^{n}_{R}(B_{r,l})$.

(2) For each $i=0,1,\ldots ,n-1$, we have $s(u_{i})=t(u_{i+1})$ in $Q_{r}$.

A path u in the quiver $Q_{r}$ is called an oriented cycle provided that $s(u)=t(u)$ and u is non-trivial. That is why a generalized path in $\Omega ^{n}_{R}(B_{r,l})$ of the form $u_{0}\otimes \overline{u_{1}} \otimes \cdots \otimes \overline{u_{n}}$ is said to be a generalized cycle if $t(u_{0})=s(u_{n})$ in the quiver $Q_{r}$. Generalized cycles will play a prominent role in our further considerations.

For a fixed algebra $B_{r,l}$, $r\ge 0$, $l\ge 2$, and any integer $n\ge 1$ let $\mathscr {Y}^{(n)}(B_{r,l})= \{ u_{0}\otimes \overline{u_{1}} \otimes \cdots \otimes \overline{u_{n}} \in \Omega ^{n}_{R}(B_{r,l}); u_{0}\otimes \overline{u_{1}}\otimes \cdots \otimes \overline{u_{n}}$   is a generalized path that is not a generalized cycle}.

Lemma 3.1

For any algebra $B_{r,l}$, $r\ge 0$, $l\ge 2$, and any integer $n\ge 1$ we have $\mathscr {Y}^{(n)}(B_{r,l})\subset [B_{r,l},\Omega ^{n}_{R}(B_{r,l})]$.

Proof

We have that $[B_{r,l},\Omega ^{n}_{R}(B_{r,l})]$ is the K-linear subspace of $\Omega ^{n}_{R}(B_{r,l})$ spanned by the commutators [b, v] with $b\in B_{r,l}$ and $v\in \Omega ^{n}_{R}(B_{r,l})$. If $v_{0}\otimes \overline{v_{1}}\otimes \cdots \otimes \overline{v_{n}}\in \mathscr {Y}^{(n)}(B_{r,l})$, then $[e_{j},v_{0}\otimes \overline{v_{1}}\otimes \cdots \otimes \overline{v_{n}}] =e_{j}v_{0}\otimes \overline{v_{1}}\otimes \cdots \otimes \overline{v_{n}} - (-1)^{n} v_{0}v_{1} \otimes \overline{v_{2}} \otimes \cdots \otimes \overline{v_{n}} \otimes \overline{e_{j}} -$ $(-1)^{n-1} v_{0}\otimes \overline{v_{1}v_{2}}\otimes \overline{v_{3}}\otimes \cdots \otimes \overline{v_{n}}\otimes \overline{e_{j}} - \cdots - v_{0}\otimes \overline{v_{1}}\otimes \cdots \otimes \overline{v_{n}e_{j}}= v_{0}\otimes \overline{v_{1}}\otimes \cdots \otimes \overline{v_{n}}$ for $j=t(v_{0})$, because $w_{0}\otimes \overline{w_{1}}\otimes \cdots \otimes \overline{w_{n-1}}\otimes \overline{{e_{j}}}=0$ in $\Omega ^{n}_{R}(B_{r,l})$ and $v_{n}e_{j}=0$ since $t(v_{0})\not =s(v_{n})$. Therefore, the required condition holds. $\square$

Lemma 3.2

For any algebra $B_{r,l}$, $r\ge 0$, $l\ge 2$, and any integer $n\ge 1$, we have: if $u_{0}\otimes \overline{u_{1}} \otimes \cdots \otimes \overline{u_{n}}$ is a generalized path in $\Omega ^{n}_{R}(B_{r,l})$ and z is a nonzero path in $B_{r,l}$ such that $zu_{0}u_{1}\cdots u_{n}$ is an oriented cycle in the quiver $Q_{r}$, then the commutator $[z,u_{0}\otimes \overline{u_{1}} \otimes \cdots \otimes \overline{u_{n}}]$ is either a sum of generalized cycles or zero.

Proof

If $zu_{0}u_{1}\cdots u_{n}$ is an oriented cycle, then $[z,u_{0}\otimes \overline{u_{1}}\otimes \cdots \otimes \overline{u_{n}}]=zu_{0}\otimes \overline{u_{1}}\otimes \cdots \otimes \overline{u_{n}}$ $-(-1)^{n} u_{0}u_{1} \otimes \overline{u_{2}}\otimes \cdots \otimes \overline{u_{n}}\otimes {\overline{z}} -\cdots - u_{0}\otimes \overline{u_{1}}\otimes \cdots \otimes \overline{u_{n}z}$, where each of the summands can be zero or a generalized cycle. Thus, the required condition holds. $\square$

Lemma 3.3

For any algebra $B_{r,l}$, $r\ge 0$, $l\ge 2$, and any integer $n\ge 1$ we have: if $u_{0}\otimes \overline{u_{1}}\otimes \cdots \otimes \overline{u_{n}}$ is a generalized path in $\Omega ^{n}_{R}(B_{r,l})$ and z is a nonzero path in $B_{r,l}$ such that $zu_{0}u_{1}\cdots u_{n}$ is not an oriented cycle in $Q_{r}$, then the commutator $[z,u_{0}\otimes \overline{u_{1}}\otimes \cdots \otimes \overline{u_{n}}]$ is either zero or a sum of generalized paths that are not generalized cycles.

Proof

We have $[z,u_{0}\otimes \overline{u_{1}}\otimes \cdots \otimes \overline{u_{n}}]= zu_{0}\otimes \overline{u_{1}}\otimes \cdots \otimes \overline{u_{n}} -$ $(-1)^{n}u_{0}u_{1}\otimes \overline{u_{2}}\otimes \cdots \otimes \overline{u_{n}}\otimes {\overline{z}} - \cdots - u_{0}\otimes \overline{u_{1}} \otimes \cdots \otimes \overline{u_{n}z}$. Notice that if $zu_{0}u_{1}\cdots u_{n}$ is not an oriented cycle in $Q_{r}$, then $u_{0}u_{1}\cdots u_{n}z$ is not an oriented cycle, either. Thus, the above sum is either zero or is a sum of generalized paths that are not generalized cycles. $\square$

Lemma 3.4

For any algebra $B_{r,l}$, $r\ge 0$, $l\ge 2$, we have $[B_{r,l},B_{r,l}]$ is spanned by the noncyclic paths and differences of cyclic permutations of the cyclic paths.

Proof

We have by definition that $[B_{r,l},B_{r,l}]$ is the K-linear subspace of $B_{r,l}$ that is spanned by the elements of the form $[u,v]=uv-(-1)^{0\cdot 0}vu$, $u,v\in B_{r,l}$. It is clear that if $w\in I_{l}$, then $w=0$ in $B_{r,l}$ and the required condition holds.

Further notice that for any non-trivial path w in the quiver $Q_{r}$, we have $e_{t(w)}w-we_{t(w)}=e_{t(w)}w=w$ when $t(w)\not =s(w)$. Thus, for any non-trivial path w that is not an oriented cycle we obtain that $w\in [B_{r,l},B_{r,l}]$.

If c is an oriented cycle, then $c=vu$ and we have $[u,v]=uv-vu=c-vu$, and vu is a cyclic permutation of c. $\square$

4 R-derivations of the algebras $B_{r,l}$

First, we introduce a notion that will be used in the sequel. We shall denote by $c_{i}$, $i=0,1,\ldots ,r$, the oriented cycle $c_{i}=\alpha _{i-1}\cdots \alpha _{1}\alpha _{0} \cdots \alpha _{i}$ in the quiver $Q_{r}$.

We shall denote by $\text{ Der}_{R}(B_{r,l})$ the K-linear subspace of $\text{ Der}_{K}(B_{r,l})$ spanned by the derivations of $B_{r,l}$ vanishing on the subalgebra R.

Lemma 4.1

If $\theta \in \text{ Der}_{R}(B_{r,l})$, $r\ge 0$, $l\ge 2$, then for any arrow $\alpha _{i}$ in $Q_{r}$, $i=0,1,\ldots ,r$, we have $\theta (\alpha _{i})\in e_{i+1}B_{r,l}e_{i}$, where $i=s(\alpha _{i})$, $i+1=t(\alpha _{i})$.

Proof

We have the following series of equalities: $\theta (\alpha _{i})= \theta (e_{i+1}(\alpha _{i} e_{i}))= \theta (e_{i+1})\alpha _{i}e_{i} + e_{i+1}\theta (\alpha _{i}e_{i})= e_{i+1}\theta (\alpha _{i})e_{i} + e_{i+1}\alpha _{i}\theta (e_{i}) = e_{i+1}\theta (\alpha _{i})e_{i}\in e_{i+1}B_{r,l}e_{i}$. $\square$

Lemma 4.2

For any algebra $B_{r,l}$, $r\ge 0$, $l\ge 2$, and any derivation $\theta \in \text{ Der}_{R}(B_{r,l})$, the following conditions are satisfied:

• (1) If $l\le r+1$, then for any arrow $\alpha _{i}$, $i=0,1,\ldots ,r$, there exists $k_{i}\in K$ such that $\theta (\alpha _{i})=k_{i}\alpha _{i}$.

• (2) If $l\ge r+2$, then for any arrow $\alpha _{i}$, $i=0,1,\ldots ,r$, there exist $k_{i,0}, k_{i,1},\ldots ,k_{i,s}\in K$ such that $\theta (\alpha _{i})= \sum ^{s}_{j=0}k_{i,j}\alpha _{i}c^{j}_{i}$, where $l=s(r+1)+l_{0}$ and $1\le l_{0}\le r+1$.

Proof

The required conditions can be obtained straightforwardly by Lemma 4.1, because the system $\{ \alpha _{i}, \alpha _{i}c_{i}, \ldots , \alpha _{i}c^{t}_{i}\}$ forms a basis of the K-linear space $e_{i+1}B_{r,l}e_{i}$, where $t=\left\{ \begin{array}{ll} 0 &{} \text{ if } ~~ l\le r+1 \\ s &{} \text{ if }~~ l=s(r+1)+l_{0}~~ \text{ and }~~ 1\le l_{0}\le r+1. \end{array} \right.$ $\square$

Lemma 4.3

For any algebra $B_{r,l}$, $r\ge 0$, $2\le l\le r+2$, and any derivation $\theta \in \text{ Der}_{R}(B_{r,l})$, it holds: if $w=\beta _{z}\ldots \beta _{1}\not \in I_{l}$, $\beta _{j}$ is an arrow for $j=1,\ldots ,z$, then $\theta (w)=\sum ^{z}_{j=1}k_{j}w$, where $\theta (\beta _{j})=k_{j}\beta _{j}$, $j=1,\ldots ,z$.

Proof

We shall proceed by induction on z. If $z=1$, then the required condition is clear by Lemma 4.2. Assume that the required condition holds for the paths of length $\le z_{0}$ and consider a path $w^{\prime }=\beta _{z_{0}+1}w\not \in I_{l}$. Then, we have $\theta (w^{\prime })=\theta (\beta _{z_{0}+1})w + \beta _{z_{0}+1}\theta (w)=k_{z_{0}+1}w^{\prime }+\sum ^{z_{0}}_{j=1}k_{j}w^{\prime }$. $\square$

Now, we define a family of derivations in $\text{ Der}_{R}(B_{r,l})$ for a fixed algebra $B_{r,l}$. We assume that $l=s(r+1)+l_{0}$, $s\ge 0$, $1\le l_{0}\le r+1$. Let $\theta ^{j}_{(\alpha _{i})}\in \text{ Der}_{R}(B_{r,l})$, $i=0,1,\ldots ,r$, $j=0,1,\ldots ,s$, be as follows:

$$\begin{aligned} \theta ^{j}_{(\alpha _{i})}(\alpha _{n})=\left\{ \begin{array}{ll} \alpha _{i}c_{i}^{j} &{} \text{ if }~~ i=n \\ 0 &{} \text{ otherwise }.\end{array}\right. \end{aligned}$$

Lemma 4.4

For any derivation $\theta ^{0}_{(\alpha _{i})}$, $i=0,1,\ldots ,r$, and any path w in $Q_{r}$ that does not belong to $I_{l}$, we have

$$\begin{aligned} \theta ^{0}_{(\alpha _{i})}(w)=\left\{ \begin{array}{ll} pw_{1}\alpha _{i}w_{2} &{} \text{ if } ~~ w=w_{1}\alpha _{i}w_{2} \\ 0 &{} \text{ otherwise }, \end{array} \right. \end{aligned}$$

where p is the multiplicity of appearance of the arrow $\alpha _{i}$ in the path w.

Proof

We shall prove the lemma inductively on the length n of the path w. If $n=1$, then $w=\alpha _{t}$ and the required condition holds for each derivation $\theta ^{0}_{(\alpha _{i})}$, $i=0,1,\ldots ,r$.

Assume that for the paths w of length $\le n_{0}$ the required condition holds for each derivation $\theta ^{0}_{(\alpha _{i})}$, $i=0,1,\ldots ,r$. Now, consider a path $w=\beta _{n_{0}+1}\beta _{n_{0}}\cdots \beta _{1}$, $\beta _{j}$ is an arrow for each $j=1,2,\ldots , n_{0}+1$. Then, for a fixed $i_{0}\in \{ 0,1,\ldots ,r\}$, we have $\theta ^{0}_{(\alpha _{i_{0}})}(w)= \theta ^{0}_{(\alpha _{i_{0}})}(\beta _{n_{0}+1}\beta _{n_{0}}\cdots \beta _{1}) =\theta ^{0}_{(\alpha _{i_{0}})}(\beta _{n_{0}+1})\beta _{n_{0}}\cdots \beta _{1} + \beta _{n_{0}+1}\beta ^{n_{0}}\cdots \beta _{1}) =$ $\left\{ \begin{array}{ll} \beta _{n_{0}+1}\beta _{n_{0}}\cdots \beta _{1} +p\beta _{n_{0}+1}\beta _{n_{0}}\cdots \beta _{1} &{} \text{ if }~~ \alpha _{i_{0}}=\beta _{n_{0}+1} \\ p\beta _{n_{0}+1}\beta _{n_{0}}\cdots \beta _{1} &{} \text{ if }~~ \alpha _{i_{0}}\not =\beta _{n_{0}+1}, \end{array}\right.$

where the arrow $\alpha _{i_{0}}$ appears p times in the path $\beta _{n_{0}}\cdots \beta _{1}$. Hence, the required condition holds. $\square$

Proposition 4.5

For any algebra $B_{r,l}$, $r\ge 0$, $l\ge 2$, $l=s(r+1)+l_{0}$, $1\le l_{0}\le r+1$, the system $\{ \theta ^{j}_{(\alpha _{i})}\}_{i=0,1,\ldots , r;j=0,1,\ldots ,s}$ is a basis of the K-linear space $\text{ Der}_{R}(B_{r,l})$.

Proof

We start our proof with checking whether the system $\{\theta ^{j}_{(\alpha _{i})}\}_{i=0,1,\ldots ,r ;j= 0,1,\ldots ,s}$ is linearly independent. In order to do this, we consider a linear combination $\sum ^{s}_{j=0}\!\sum ^{r}_{i=0} k_{j,i}\theta ^{j}_{(\alpha _{i})}\!=\!0$. Then, for a fixed arrow $\alpha _{i_{0}}$ we have $\sum ^{s}_{j=0}\!\sum ^{r}_{i=0}k_{j,i}\theta ^{j}_{(\alpha _{i})}(\alpha _{i_{0}})= \sum ^{s}_{j=0}k_{j,i_{0}}\theta ^{j}_{(\alpha _{i_{0}})}(\alpha _{i_{0}})=$ $k_{0,i_{0}}\alpha _{i_{0}} +k_{1,i_{0}}\alpha _{i_{0}}c_{i_{0}} + \cdots + k_{s,i_{0}}\alpha _{i_{0}}c^{s}_{i_{0}}=0$. Hence, $k_{j,i_{0}}=0$ for $j=0,1,\ldots , s$. Since $i_{0}$ can be arbitrary, we obtain that all $k_{j,i}=0$, $i=0,1,\ldots ,r$, $j=0,1,\ldots ,s$. Therefore, the considered system is linearly independent.

Now, consider a derivation $\theta \in \text{ Der}_{R}(B_{r,l})$. We infer by Lemma 4.2 that for every arrow $\alpha _{i}$, $i=0,1,\ldots ,r$, there is $k_{i}\in K$, $i=0,1,\ldots ,r$ such that $\theta (\alpha _{i})= k_{i}\alpha _{i}$ or there are $k_{i,0},k_{i,1},\ldots ,k_{i,s}\in K$ such that $\theta (\alpha _{i})=\sum ^{s}_{j=0}k_{i,j}\alpha _{i}c^{j}_{i}$. Then, we get $\theta =\sum ^{r}_{i=0}k_{i}\theta ^{0}_{(\alpha _{i})}$ or $\theta =\sum ^{s}_{j=0}\sum ^{r}_{i=0}k_{i,j}\theta ^{j}_{(\alpha _{i})}$, respectively. Indeed, for any arrow $\alpha _{i_{0}}$ we have $\theta (\alpha _{i_{0}})=k_{i_{0}}\alpha _{i_{0}}$ and $\sum ^{r}_{i=0}k_{i}\theta ^{0}_{(\alpha _{i})}(\alpha _{i_{0}})=k_{i_{0}}\alpha _{i_{0}}$ in the first case. In the second case, similarly we have $\theta (\alpha _{i_{0}})=\sum ^{s}_{j=0}k_{i_{0},j} \alpha _{i_{0}}c^{j}_{i_{0}}$ and $\sum ^{s}_{j=0}\sum ^{r}_{i=0}k_{i,j}\theta ^{j}_{(\alpha _{i})}(\alpha _{i_{0}})= \sum ^{s}_{j=0}k_{i_{0},j}\theta ^{j}_{(\alpha _{i_{0}})}(\alpha _{i_{0}})= \sum ^{s}_{j=0} k_{i_{0},j}\alpha _{i_{0}}c^{j}_{i_{0}}$. Then, $\theta$ has the required form. $\square$

Corollary 4.6

For any algebra $B_{r,l}$, $r\ge 0$, $l\ge 2$, and $l=s(r+1) +l_{0}$, $1\le l_{0}\le r+1$, we have $\text{ dim}_{K}\text{ Der}_{R}(B_{r,l})=s(r+1)$.

Proof

Our corollary is an obvious consequence of Proposition 4.5.

5 Description of $DR^{0}_{R}(B_{r,l})$

In this section, we shall prove that $DR^{0}_{R}(B_{r,l})$ is always nonzero and indicate its basis.

Lemma 5.1

For any algebra $B_{r,l}$, $r\ge 0$, $l\ge 2$, the following conditions are satisfied:

• (1) Any trivial path $e_{i}$, $i=0,1,\ldots ,r$, does not belong to $[B_{r,l},B_{r,l}]$.

• (2) Any nonzero oriented cycle c does not belong to $[B_{r,l},B_{r,l}]$.

Proof

In order to prove condition (1), it is enough to check whether $e_{i}=[u,v]$ for some $u,v\in B_{r,l}$. Notice that if $u=e_{i_{1}}$, $v=e_{i_{2}}$, then $[e_{i_{1}},e_{i_{2}}]=0$. If $u=e_{i_{1}}$ and v is a non-trivial path, then $[e_{i_{1}},v]\in \{v,-v,0\}$. Hence, it cannot happen $e_{i}=[e_{i_{1}},v]$.

Now, consider the case when u, v are both non-trivial paths. Then, the commutator [u, v] is zero or a non-trivial path or else a nonzero difference of non-trivial paths. Therefore, it cannot be $e_{i}=[u,v]$ in the case, either. Consequently, condition (1) follows.

For a proof of condition (2) notice that any cycle $c_{i}$, $i=0,1,\ldots , r$, cannot be obtained as a commutator [u, v] for some paths u, v that satisfy $s(u)\not = t(v)$ or $s(v)\not = t(u)$. Hence, we have to consider only paths u, v such that $s(u)=t(v)$ and $s(v)=t(u)$. Then, we have $[u,v]=c_{s(v)}-c_{s(u)}$ and $[u,v]=0$ if $s(v)=s(u)$, and $[u,v]=c_{s(v)}-c_{s(u)}\not =0$ when $s(u)\not =s(v)$.

Now, suppose to the contrary that, say $c_{0}$, belongs to $[B_{r,l},B_{r,l}]$. Then, we infer by the above considerations that $c_{0}= \sum _{j}k^{\prime }_{j}[u_{j},v_{j}]=\sum ^{r}_{s,i}k_{i,s}(c_{i}-c_{j})$, where $k_{i,s}\in K$ and $k_{i,i}=0$, $i=0,1,\ldots ,r$. Then, we have $c_{0}=\sum ^{r}_{i=0}(k_{0,i}-k_{i,0})c_{0} +\sum ^{r}_{i=0}(k_{1,i}-k_{i,1})c_{1} + \cdots +\sum ^{r}_{i=0}(k_{i,r}-k_{r,i})c_{r}$. This equality implies the following system of equalities in the field K: $\sum ^{r}_{i=0}k_{0,i}-\sum ^{r}_{i=0}k_{i,0}=1$, $\sum ^{r}_{i=0}k_{1,i} -\sum ^{r}_{i=0}k_{i,1}=0$, $\sum ^{r}_{i=0}k_{2,i} -\sum ^{r}_{i=0}k_{i,2}=0$, $\ldots$, $\sum ^{r}_{i=0}k_{r,i}-\sum ^{r}_{i=0}k_{i,r}=0$. Further, we have that the equality $\sum ^{r}_{i=0}k_{1,i} -\sum ^{r}_{i=0}k_{i,1}=0$ implies that $k_{0,1}-k_{1,0} =(k_{1,2}-k_{2,1})+(k_{1,3}-k_{3,1}) +\cdots +(k_{1,r}-k_{r,1})$. The equality $\sum ^{r}_{i=0}k_{2,i}-\sum ^{r}_{i=0}k_{i,2}=0$ implies that $k_{0,2}-k_{2,0}=(k_{2,1}-k_{1,2}) +(k_{2,3}-k_{3,2}) +\cdots +(k_{2,r}-k_{r,2})$.

Generally, the equality $\sum ^{r}_{i=0}k_{j,i} -\sum ^{r}_{i=0}k_{i,j}=0$, $j=1,2,\ldots ,r$ implies that $k_{0,j}-k_{j,0}=(k_{j,1}-k_{1,j})+(k_{j,2}-k_{2,j})+\cdots +(k_{j,r}-k_{r,j})$.

Now, notice that the difference $k_{i,j}-k_{j,i}$ appears in the above presentation of $k_{0,i}-k_{i,0}$ if and only if the difference $k_{j,i}-k_{i,j}$ appears in the above presentation of $k_{0,j}-k_{j,0}$. Therefore, $\sum ^{r}_{i=0}k_{0,i}-\sum ^{r}_{i=0}k_{i,0}=\sum ^{r}_{i=0}(k_{0,i}-k_{i,0})=0$ which contradicts to the equality $\sum ^{r}_{i=0}k_{0,i}-\sum ^{r}_{i=0}k_{i,0}=1$, because $\text{ char }(K)=0$.

Finally, we obtain that $c_{0}\not \in [B_{r,l},B_{r,l}]$ and we infer by Lemma 3.4 that every oriented cycle $c_{i}\not \in [B_{r,l},B_{r,l}]$, $i=0,1,\ldots ,r$.

Since every nonzero oriented cycle c in $B_{r,l}$ is of the form $c^{j}_{i}$ for some $i=0,1,\ldots ,r$, and some integer $j\ge 1$, repeating the above arguments for $c^{j}_{0}$ we obtain similarly that $c^{j}_{i}\not \in [B_{r,l},B_{r,l}]$, which shows condition (2). $\square$

Lemma 5.2

For any algebra $B_{r,l}$, $r\ge 0$, $l\ge 2$, the K-linear space $B_{r,l}/[B_{r,l},B_{r,l}]$ has a basis of the form $\{\llbracket e_{i}\rrbracket \}_{i=0,1,\ldots ,r} \cup \{\llbracket c^{s}_{0}\rrbracket \}_{s=1,2,\ldots ,n}$, where $l=n(r+1) +l_{0}$, $1\le l_{0}\le r+1$ and $\llbracket b\rrbracket$ denotes the coset of $b\in B_{r,l}$ modulo $[B_{r,l},B_{r,l}]$.

Proof

First notice that the elements of the considered system are nonzero in view of Lemma 5.1. Further, we have that for any $j\in \{1,2,\ldots ,r\}$ there is a path $\alpha _{j-1}\cdots \alpha _{0}=u$ and a path $\alpha _{r}\cdots \alpha _{j}=v$. If $c_{0}$ is a nonzero oriented cycle, then the paths u, v are nonzero and $[u,v]=c_{j}-c_{0}$. Hence, $\llbracket c_{j}\rrbracket =\llbracket c_{0}\rrbracket$. Similarly, $\llbracket c^{s}_{j}\rrbracket = \llbracket c^{s}_{0}\rrbracket$ for $s=2,\ldots ,n$. Therefore, we infer by Lemma 3.4 that the system $\{\llbracket e_{i}\rrbracket \}_{i=0,1,\ldots ,r}\cup \{\llbracket c^{s}_{0}\rrbracket \}_{s=1,2,\ldots ,n}$ is indeed a set of generators of the K-linear space $B_{r,l}/[B_{r,l},B_{r,l}]$.

Now, consider a linear combination $\sum ^{r}_{i=0}k_{i}\llbracket e_{i}\rrbracket +\sum ^{n}_{s=1}k^{\prime }_{s}\llbracket c^{s}_{0}\rrbracket =0$ for some $k_{0},k_{1},\ldots ,k_{r}\in K$ and some $k_{1}^{\prime },k^{\prime }_{2},\ldots , k^{\prime }_{n}\in K$. Then, we have $\sum ^{r}_{i=0}k_{i}e_{i}+\sum ^{n}_{s=1}k^{\prime }_{s}c^{s}_{0}\in [B_{r,l},B_{r,l}]$. Repeating the arguments from the proof of Lemma 5.1, we obtain that $\sum ^{r}_{i=0}k_{i}e_{i} +\sum ^{n}_{s=1}k^{\prime }_{s}c^{s}_{0}= \sum ^{n}_{s=1}\sum ^{r}_{a,b=0}k_{a,b,s}(c^{s}_{a}-c^{s}_{b})$. Thus, it is clear that $k_{i}=0$ for $i=0,1,\ldots ,r$

If we have $k^{\prime }_{s_{0}}\not =0$ for some $s_{0}\in \{1,\ldots ,n\}$, then $\sum ^{r}_{a,b=0}k_{a,b,s_{0}}(c^{s_{0}}_{a}-c^{s_{0}}_{b})=k^{\prime }_{s_{0}}c^{s_{0}}_{0}$ and we obtain a contradiction as in the proof of Lemma 5.1. Therefore, all $k^{\prime }_{s}=0$, $s=1,2,\ldots ,n$. Consequently, the considered system is linearly independent and forms a basis of $B_{r,l}/[B_{r,l},B_{r,l}]$. $\square$

6 Description of $DR^{1}_{R}(B_{r,l})$

Our main aim of this section is a computation of the dimension of $DR^{1}_{R}(B_{r,l})$ for some $B_{r,l}$.

Lemma 6.1

For any algebra $B_{r,l}$, $r\ge 0$, $l\ge 2$, the following conditions are satisfied:

• (1) For every nonzero oriented cycle $c^{n+m}_{i}$, $i=0,1,\ldots ,r$; $n,m\ge 1$, we have $\llbracket c^{n}_{i}\otimes \overline{c^{m}_{i}}\rrbracket + \llbracket c^{m}_{i}\otimes \overline{c^{n}_{i}}\rrbracket = \llbracket e_{i}\otimes \overline{c^{n+m}_{i}}\rrbracket$ in $DR^{1}_{R}(B_{r,l})$.

• (2 ) If $c^{s}_{i}\notin I_{l}$ for some integer $s\ge 1$ and $vu=c^{s}_{i}$ in $Q_{r}$ for non-trivial paths u, v, then it holds $\llbracket u\otimes {\overline{v}}\rrbracket + \llbracket v\otimes {\overline{u}}\rrbracket = \llbracket e_{i}\otimes \overline{c^{s}_{i}}\rrbracket$ in $DR^{1}_{R}(B_{r,l})$.

Proof

In order to prove condition (1) notice that for a nonzero oriented cycle $c^{n+m}_{i}$, we have $[c^{n}_{i},e_{i}\otimes \overline{v^{m}_{i}}]=c^{n}_{i}(e_{i}\otimes \overline{c^{m}_{i}})- (e_{i}\otimes \overline{c^{m}_{i}})c^{n}_{i}= c^{n}_{i}\otimes \overline{c^{m}_{i}} -(-c^{m}_{i}\otimes \overline{c^{n}_{i}} + e_{i}\otimes \overline{c^{n+m}_{i}})=$ $c^{n}_{i}\otimes \overline{c^{m}_{i}} + c^{m}_{i}\otimes \overline{c^{n}_{i}} -e_{i}\otimes \overline{c^{n+m}_{i}}$. Hence, $\llbracket c^{n}_{i}\otimes \overline{c^{m}_{i}}\rrbracket + \llbracket c^{m}_{i}\otimes \overline{c^{n}_{i}}\rrbracket = \llbracket e_{i}\otimes \overline{c^{n+m}_{i}}\rrbracket$ in $DR^{1}_{R}(B_{r,l})$.

In order to prove condition (2), we have $[u, e_{i}\otimes {\overline{v}}]=u(e_{i}\otimes {\overline{v}})-(e_{i}\otimes {\overline{v}})u = u\otimes {\overline{v}} - (-v\otimes {\overline{u}}+ e_{i}\otimes {\overline{vu}})= u\otimes {\overline{v}} + v\otimes {\overline{u}} - e_{i}\otimes \overline{c^{s}_{i}}$, which shows (2). $\square$

Lemma 6.2

For any algebra $B_{r,l}$, $r\ge 0$, $l\ge 2$, and any nonzero oriented cycle $uv\beta \in B_{r,l}$, where $\beta$ is an arrow and u, v are non-trivial paths, the following equality $\llbracket u\otimes \overline{v\beta } \rrbracket = \llbracket \beta u\otimes {\overline{v}}\rrbracket + \llbracket uv\otimes {\overline{\beta }}\rrbracket$ holds in $DR^{1}_{R}(B_{r,l})$.

Proof

Notice that $[\beta , u\otimes {\overline{v}}]=\beta (u\otimes {\overline{v}})-(u\otimes {\overline{v}})\beta =\beta u\otimes {\overline{v}} +uv\otimes {\overline{\beta }} - u\otimes \overline{v\beta }$. Therefore, we have $\llbracket u\otimes \overline{v\beta }\rrbracket = \llbracket \beta u\otimes {\overline{v}}\rrbracket +\llbracket uv\otimes {\overline{\beta }}\rrbracket$ in $DR^{1}_{R}(B_{r,l})$. $\square$

Lemma 6.3

For any algebra $B_{r,l}$, $r\ge 0$, $l\ge 2$, and any oriented cycle $u\beta _{n}\cdots \beta _{1}$ in $Q_{r}$ that is nonzero in $B_{r,l}$ and $\beta _{1},\ldots ,\beta _{n}$ are arrows, $n\ge 2$, u is a non-trivial path, the following equality

$$\begin{aligned} \llbracket u\otimes \overline{\beta _{n}\cdots \beta _{1}}\rrbracket = \llbracket u\beta _{n}\cdots \beta _{2}\otimes \overline{\beta _{1}}\rrbracket +\llbracket \beta _{1} u\beta _{n}\cdots \beta _{3}\otimes \overline{\beta _{2}}\rrbracket + \cdots + \llbracket \beta _{n-1}\cdots \beta _{1}u\otimes \overline{\beta _{n}}\rrbracket \end{aligned}$$

holds in $DR^{1}_{R}(B_{r,l})$.

Proof

We know from Lemma 6.2 that $\llbracket u\otimes \overline{\beta _{n}\cdots \beta _{1}}\rrbracket = \llbracket \beta _{1}u\otimes \overline{\beta _{n}\cdots \beta _{2}}\rrbracket +$ $\llbracket u\beta _{n}\cdots \beta _{2}\otimes \overline{\beta _{1}}\rrbracket$. Further, applying Lemma 6.2, we obtain $\llbracket \beta _{1}u\otimes \overline{\beta _{n}\cdots \beta _{2}}\rrbracket =$ $\llbracket \beta _{2}\beta _{1} u\otimes \overline{\beta _{n}\cdots \beta _{3}}\rrbracket + \llbracket \beta _{1}u\beta _{n}\cdots \beta _{3}\otimes \overline{\beta _{2}}\rrbracket$. Hence, we get $\llbracket u\otimes \overline{\beta _{n}\cdots \beta _{1}}\rrbracket =$ $\llbracket \beta _{2}\beta _{1}u\otimes \overline{\beta _{n}\cdots \beta _{3}}\rrbracket +$ $\llbracket \beta _{1}u\beta _{n}\cdots \beta _{3}\otimes \overline{\beta _{2}}\rrbracket +$ $\llbracket u\beta _{n}\cdots \beta _{2}\otimes \overline{\beta _{1}}\rrbracket$. Continuing the above decompositions, we obtain the required equality. $\square$

Lemma 6.4

For any algebra $B_{r,l}$, $r\ge 0$, $l\ge 2$, and any generalized path $v_{0}\otimes \overline{v_{1}}\in \Omega ^{1}_{R}(B_{r,l})$ and a nonzero path u in $B_{r,l}$ such that $uv_{0}, v_{1}u\not \in I_{l}$, the following equivalence holds: $uv_{0}\otimes \overline{v_{1}}$ is a generalized cycle if and only if $v_{0}\otimes \overline{v_{1}u}$ is a generalized cycle.

Proof

Notice that if $uv_{0}\otimes \overline{v_{1}}$ is a generalized cycle in $\Omega ^{1}_{R}(B_{r,l})$, then $s(v_{1})=t(u)$, $s(u)=t(v_{0})$, $s(v_{0})=t(v_{1})$. Since $uv_{0}\otimes \overline{v_{1}}\not =0$ and $v_{0}\otimes \overline{v_{1}u}\not =0$, $v_{0}\otimes \overline{v_{1}u}$ is a generalized cycle.

Similar reasoning shows the opposite implication. $\square$

Proposition 6.5

For any algebra $B_{r,l}$, $r\ge 0$, $l=n(r+1)+l_{0}$, $1\le l_{0}\le r+1$, $n\ge 1$, the following conditions are satisfied:

• (1) For any generalized cycle $e_{i}\otimes \overline{c^{s}_{i}}$, $i=0,1,\ldots ,r$; $s=1,2,\ldots ,n$, we have $\llbracket e_{i}\otimes \overline{c^{s}_{i}}\rrbracket \not =0$ in $DR^{1}_{R}(B_{r,l})$.

• (2) For any generalized cycle $u_{i}\otimes \overline{v_{i}}$, $i=0,1,\ldots ,r$ such that $u_{i}v_{i}=c^{s}_{i}$ for some $s=1,2,\ldots ,n$ and $u_{i}$, $v_{i}$ are non-trivial paths in $Q_{r}$, we have $\llbracket u_{i}\otimes \overline{v_{i}}\rrbracket \not =0$ in $DR^{1}_{R}(B_{r,l})$.

Proof

We know from Section 2 that for any derivation $\theta \in \text{ Der}_{R}(B_{r,l})$ there is a linear transformation of contraction $i_{\theta }:DR^{1}_{R}(B_{r,l}) \rightarrow DR^{0}_{R}(B_{r,l})$ (see also [3, 7]) such that $i_{\theta }(\llbracket u\otimes {\overline{v}}\rrbracket )=\llbracket u\cdot \theta (v)\rrbracket$.

For a proof of condition (1) notice that for the derivation $\theta ^{0}_{(\alpha _{0})}$, we have $i_{\theta ^{0}_{(\alpha _{0})}}(\llbracket e_{i}\otimes \overline{c^{s}_{i}}\rrbracket )=s\llbracket c^{s}_{i}\rrbracket$ by Lemma 4.4. Further, we infer by Lemma 5.1(2) that $\llbracket c^{s}_{i}\rrbracket \not =0$ in $DR^{0}_{R}(B_{r,l})$. Therefore, $\llbracket e_{i}\otimes \overline{c^{s}_{i}}\rrbracket \not =0$ in $DR^{1}_{R}(B_{r,l})$, which shows (1).

In order to prove condition (2), it is enough to consider an arrow $\alpha _{i_{0}}$ that appears in the path $v_{i}$. The contraction $i_{\theta ^{0}_{(\alpha _{i_{0}})}}$ satisfies $i_{\theta ^{0}_{(\alpha _{i_{0}})}}(\llbracket u_{i}\otimes \overline{v_{i}}\rrbracket ) =p\llbracket u_{i}v_{i}\rrbracket =p\llbracket c^{s}_{i}\rrbracket$ for some $s=1,2,\ldots ,n$, where p is the multiplicity of appearance of the arrow $\alpha _{i_{0}}$ in the path $v_{i}$. Hence, again $\llbracket u_{i}\otimes \overline{v_{i}}\rrbracket \not =0$ in $DR^{1}_{R}(B_{r,l})$, which shows (2). $\square$

Corollary 6.6

For any algebra $B_{r,l}$, $r\ge 0$, $l\ge r+2$, the K-linear space $DR^{1}_{R}(B_{r,l})$ is nonzero.

Proof

We know from Proposition 6.5 that $\llbracket e_{i}\otimes \overline{c_{i}}\rrbracket \not =0$ in $DR^{1}_{R}(B_{r,l})$. $\square$

Lemma 6.7

For any algebra $B_{r,l}$, $r\ge 0$, $l\ge r+2$, we have $\llbracket e_{i}\otimes \overline{c^{s}_{i}}\rrbracket = \llbracket e_{j}\otimes \overline{c^{s}_{j}}\rrbracket$ in $DR^{1}_{R}(B_{r,l})$ for any $i,j\in \{0,1,\ldots ,r\}$ and any $s=1,2,\ldots ,n$, where $l=n(r+1)+l_{0}$, $1\le l_{0}\le r+1$, $n\ge 1$.

Proof

We infer by Lemma 6.1 that $\llbracket u_{0}\otimes \overline{u_{1}}\rrbracket + \llbracket u_{1}\otimes \overline{u_{0}}\rrbracket =\llbracket e_{j}\otimes \overline{c^{s}_{j}}\rrbracket$ in $DR^{1}_{R}(B_{r,l})$ for $c^{s}_{j}=u_{1}u_{0}$. In the same way, we have $\llbracket u_{1}\otimes \overline{u_{0}}\rrbracket +\llbracket u_{0}\otimes \overline{u_{1}}\rrbracket = \llbracket e_{i}\otimes \overline{c^{s}_{i}}\rrbracket$ for $c^{s}_{i}=u_{0}u_{1}$. Therefore, $\llbracket e_{j}\otimes \overline{c^{s}_{j}}\rrbracket = \llbracket e_{i}\otimes \overline{c^{s}_{i}}\rrbracket$ for $i\not = j$. If $i=j$, then the required condition is obvious. $\square$

For any arrow $\alpha _{i}$ in the quiver $Q_{r}$, $i=0,1,\ldots ,r$, we denote by $h_{i}$ the path in $Q_{r}$ that satisfies $h_{i}\alpha _{i}=c_{i}$ for the oriented cycle $c_{i}$

Proposition 6.8

For any algebra $B_{r,r+2}$, $r\ge 0$, the system $\{ \llbracket h_{i}\otimes \overline{\alpha _{i}}\rrbracket \}_{i=0,1,\ldots ,r}$ forms a basis of the K-linear space $DR^{1}_{R}(B_{r,r+2})$.

Proof

We deduce from Lemma 6.1 and Lemma 6.3 that the considered system generates the K-linear space $DR^{1}_{R}(B_{r,r+2})$.

In order to finish our proof, we have to check whether the system is linearly independent. Consider a linear combination $\sum ^{r}_{i=0}k_{i}\llbracket h_{i}\otimes \overline{\alpha _{i}}\rrbracket =0$, where $k_{i}\in K$, $i=0,1,\ldots ,r$. For any arrow $\alpha _{i}$, $i=0,1,\ldots ,r$, we have the derivation $\theta ^{0}_{(\alpha _{i})}$ in $\text{ Der}_{R}(B_{r,r+2})$ and the linear contraction $i_{\theta ^{0}_{(\alpha _{i})}}: DR^{1}_{R}(B_{r,r+2}) \rightarrow DR^{0}_{R}(B_{r,r+2})$. Notice that for a fixed arrow $\alpha _{i_{0}}$ we have $i_{\theta ^{0}_{(\alpha _{i_{0}})}}(\sum ^{r}_{i=0}\llbracket h_{i}\otimes \overline{\alpha _{i}}\rrbracket ) =\sum ^{r}_{i=0}k_{i}\theta ^{0}_{(\alpha _{i_{0}})}(\llbracket h_{i}\otimes \overline{\alpha _{i}}\rrbracket ) =k_{i_{0}}\llbracket c_{i_{0}}\rrbracket$. Hence, $k_{i_{0}}=0$, because $\llbracket c_{i_{0}}\rrbracket \not =0$ in $DR^{0}_{R}(B_{r,r+2})$ by Lemma 5.2. Considering that $\alpha _{i_{0}}$ is arbitrary, we obtain that $k_{0}=k_{1}=\cdots =k_{r}=0$. Therefore, the considered system is linearly independent. $\square$

Corollary 6.9

For any algebra $B_{r,r+2}$, $r\ge 0$, we have $\text{ dim}_{K} DR^{1}_{R}(B_{r,r+2})=r+1$.

Proof

It is an obvious consequence of Proposition 6.8. $\square$

7 Description of $DR^{2}_{R}(B_{r,l})$

The main aim of this section is to show that $DR^{2}_{R}(B_{r,r+2})\not =0$. Furthermore, we indicate some relations between elements of $DR^{2}_{R}(B_{r,r+2})$.

Lemma 7.1

For any algebra $B_{r,l}$, $r\ge 0$, $l\ge r+2$, the following condition holds: for each oriented cycle $c_{i}$, $i=0,1,\ldots ,r$, if $c^{n}_{i}$, $c^{m}_{i}$ are nonzero for some integers $n,m\ge 1$ then $\llbracket e_{i}\otimes \overline{c^{n}_{i}}\otimes \overline{c^{m}_{i}}\rrbracket = -\llbracket e_{i}\otimes \overline{c^{m}_{i}}\otimes \overline{c^{n}_{i}}\rrbracket$ in $DR^{2}_{R}(B_{r,l})$.

In particular, $\llbracket e_{i}\otimes \overline{c^{n}_{i}}\otimes \overline{c^{n}_{i}}\rrbracket =0$ for any $n\ge 1$.

Proof

We have $[e_{i}\otimes \overline{c^{n}_{i}},e_{i}\otimes \overline{c^{m}_{i}}]= (e_{i}\otimes \overline{c^{n}_{i}})(e_{i}\otimes \overline{c^{m}_{i}}) - (-1)^{1\cdot 1}(e_{i}\otimes \overline{c^{m}_{i}})(e_{i}\otimes \overline{c^{n}_{i}})=$ $(e_{i}\otimes \overline{c^{n}_{i}})(e_{i}\otimes \overline{c^{m}_{i}}) + (e_{i}\otimes \overline{c^{m}_{i}})(e_{i} \otimes \overline{c^{n}_{i}})=$ $-c^{n}_{i}\otimes \overline{e_{i}}\otimes \overline{c^{m}_{i}} + e_{i}\otimes \overline{c^{n}_{i}}\otimes \overline{c^{m}_{i}} - c^{m}_{i} \otimes \overline{e_{i}}\otimes \overline{c^{n}_{i}} + e_{i}\otimes \overline{c^{m}_{i}}\otimes \overline{c^{n}_{i}}=$ $e_{i}\otimes \overline{c^{n}_{i}}\otimes \overline{c^{m}_{i}} + e_{i}\otimes \overline{c^{m}_{i}}\otimes \overline{c^{n}_{i}}$. Hence, we obtain that $\llbracket e_{i}\otimes \overline{c^{n}_{i}}\otimes \overline{c^{m}_{i}}\rrbracket = -\llbracket e_{i}\otimes \overline{c^{m}_{i}}\otimes \overline{c^{n}_{i}}\rrbracket$ in $DR^{2}_{R}(B_{r,l})$ and $\llbracket e_{i}\otimes \overline{c^{n}_{i}}\otimes \overline{c^{n}_{i}}\rrbracket =0$ for $n\ge 1$, because $\text{ char }(K)=0$. $\square$

Proposition 7.2

For any algebra $B_{r,l}$, $r\ge 0$, $l\ge r+2$, the following conditions are satisfied:

• (1) For any generalized cycle $e_{i}\otimes \overline{w_{i}}\otimes \overline{v_{i}}$ in $\Omega ^{2}_{R}(B_{r,l})$, $i=0,1,\ldots ,r$, such that $w_{i}v_{i}=c_{i}$ and $w_{i}$, $v_{i}$ are non-trivial, we have $\llbracket e_{i}\otimes \overline{w_{i}}\otimes \overline{v_{i}}\rrbracket \not =0$ in $DR^{2}_{R}(B_{r,l})$.

• (2) For any generalized cycle $u_{i}\otimes \overline{w_{i}}\otimes \overline{v_{i}}$ in $\Omega ^{2}_{R}(B_{r,l})$, $i=0,1,\ldots ,r$, such that $u_{i}w_{i}v_{i}=c_{i}$ in $B_{r,l}$ and $u_{i}$, $w_{i}$, $v_{i}$ are non-trivial paths, we have $\llbracket u_{i}\otimes \overline{w_{i}}\otimes \overline{v_{i}}\rrbracket \not =0$ in $DR^{2}_{R}(B_{r,l})$.

Proof

For any derivation $\theta \in \text{ Der}_{R}(B_{r,l})$, there is a linear transformation of contraction $i_{\theta }:DR^{2}_{R}(B_{r,l}) \rightarrow DR^{1}_{R}(B_{r,l})$ (see [3, 7]) such that $i_{\theta }(\llbracket u\otimes {\overline{w}}\otimes {\overline{v}}\rrbracket )=\llbracket u\theta (w)\otimes {\overline{v}}\rrbracket -$ $\llbracket u\otimes \overline{w\theta (v)}\rrbracket$.

For a proof of condition (1) notice that if we take into account the derivation $\theta ^{0}_{(\alpha )}$ for an arrow $\alpha$ appearing in the path $v_{i}$, then $i_{\theta ^{0}_{(\alpha )}}(\llbracket e_{i}\otimes \overline{w_{i}}\otimes \overline{v_{i}}\rrbracket ) =\llbracket e_{i}\theta ^{0}_{(\alpha )}(w_{i})\otimes \overline{v_{i}}\rrbracket -$ $\llbracket e_{i} \otimes \overline{w_{i}\theta ^{0}_{(\alpha )}(v_{i})}\rrbracket = -\llbracket e_{i}\otimes \overline{w_{i}\theta ^{0}_{(\alpha )}(v_{i})}\rrbracket \not =0$ in $DR^{1}_{R}(B_{r,l})$ by Proposition 6.5. Therefore, $\llbracket e_{i}\otimes \overline{w_{i}}\otimes \overline{v_{i}}\rrbracket \not =0$ in $DR^{2}_{R}(B_{r,l})$, which shows condition (1).

A similar reasoning shows condition (2), we omit the details. $\square$

Corollary 7.3

For any algebra $B_{r,l}$, $r\ge 0$, $l\ge r+2$, the K-linear space $DR^{2}_{R}(B_{r,l})$ is nonzero.

Proof

We deduce from Proposition 7.2 that $\llbracket e_{i}\otimes \overline{w_{i}}\otimes \overline{v_{i}}\rrbracket \not =0$ in $DR^{2}_{R}(B_{r,l})$ for any generalized cycle $e_{i}\otimes \overline{w_{i}}\otimes \overline{v_{i}}$ in $\Omega ^{2}_{R}(B_{r,l})$ with $w_{i}v_{i}=c_{i}$ in $B_{r,l}$. $\square$

Proposition 7.4

For any algebra $B_{r,l}$, $r\ge 0$, $l\ge 2$, assume that $u\otimes {\overline{v}}\otimes \overline{\beta _{n}\cdots \beta _{1}}$ is a generalized cycle in $\Omega ^{2}_{R}(B_{r,l})$ such that u, v are non-trivial paths, $\beta _{j}$ is an arrow, $j=1,\ldots ,n$, $n\ge 1$. Then, the element $\llbracket u\otimes {\overline{v}}\otimes \overline{\beta _{n}\cdots \beta _{1}}\rrbracket$ in $DR^{2}_{R}(B_{r,l})$ is a linear combination of the elements $\llbracket uv\otimes \overline{\beta _{n}\cdots \beta _{2}}\otimes \overline{\beta _{1}}\rrbracket , \llbracket u\otimes \overline{v\beta _{n}\cdots \beta _{2}}\otimes \overline{\beta _{1}}\rrbracket ,$ $\llbracket \beta _{1}u\otimes \overline{v\beta _{n}\cdots \beta _{3}}\otimes \overline{\beta _{2}}\rrbracket , \ldots , \llbracket \beta _{n-2}\cdots \beta _{1}uv\otimes \overline{\beta _{n}}\otimes \overline{\beta _{n-1}}\rrbracket , \llbracket \beta _{n-2}\cdots \beta _{1}u\otimes \overline{v\beta _{n}}\otimes \overline{\beta _{n-1}}\rrbracket$.

Proof

We prove the proposition by induction on n. If $n=1$, then the required condition holds obviously.

Now, assume that for $n\le m$ the required condition holds. Consider a generalized cycle in $\Omega ^{2}_{R}(B_{r,l})$ of the form $u\otimes {\overline{v}}\otimes \overline{\beta _{m+1}\cdots \beta _{1}}$. Then, we have $[\beta _{1},u\otimes {\overline{v}}\otimes \overline{\beta _{m+1}\cdots \beta _{2}}] = \beta _{1}u\otimes {\overline{v}}\otimes \overline{\beta _{m+1}\cdots \beta _{2}} -(uv\otimes \overline{\beta _{m+1}\cdots \beta _{2}}\otimes \overline{\beta _{1}} - u\otimes \overline{v\beta _{m+1}\cdots \beta _{2}}\otimes \overline{\beta _{1}} + u\otimes {\overline{v}}\otimes \overline{\beta _{m+1}\cdots \beta _{1}})$. Hence, we have that$\llbracket u\otimes {\overline{v}}\otimes \overline{\beta _{m+1}\cdots \beta _{1}}\rrbracket = \llbracket \beta _{1}u \otimes {\overline{v}}\otimes \overline{\beta _{m+1}\cdots \beta _{2}}\rrbracket -$ $\llbracket uv\otimes \overline{\beta _{m+1}\cdots \beta _{2}}\otimes \overline{\beta _{1}}\rrbracket + \llbracket u\otimes \overline{v\beta _{m+1}\cdots \beta _{2}} \otimes \overline{\beta _{1}}\rrbracket$ in $DR^{2}_{R}(B_{r,l})$.

Applying the inductive assumption to $\llbracket \beta _{1}u\otimes {\overline{v}}\otimes \overline{\beta _{m+1}\cdots \beta _{2}}\rrbracket$, we obtain the required condition for $\llbracket u\otimes {\overline{v}}\otimes \overline{\beta _{m+1}\cdots \beta _{1}}\rrbracket$. $\square$

Proposition 7.5

For any algebra $B_{r,l}$, $r\ge 0$, $l\ge 2$, assume that $u\otimes {\overline{v}}\otimes {\overline{w}}$ is a generalized cycle in $\Omega ^{2}_{R}(B_{r,l})$ such that u, v, w are non-trivial paths in $Q_{r}$. Then, the following conditions are satisfied in $DR^{2}_{R}(B_{r,l})$:

• (1) $\llbracket u\otimes {\overline{v}}\otimes {\overline{w}}\rrbracket - \llbracket v\otimes {\overline{w}}\otimes {\overline{u}}\rrbracket = \llbracket e_{t(u)}\otimes {\overline{uv}}\otimes {\overline{w}}\rrbracket$.

• (2) $\llbracket u\otimes {\overline{v}}\otimes {\overline{w}}\rrbracket - \llbracket w\otimes {\overline{u}}\otimes {\overline{v}}\rrbracket = \llbracket e_{t(u)}\otimes {\overline{uv}}\otimes {\overline{w}}\rrbracket + \llbracket e_{t(v)}\otimes {\overline{vw}}\otimes {\overline{u}}\rrbracket$.

• (3) $\llbracket e_{t(u)}\otimes {\overline{uv}}\otimes {\overline{w}}\rrbracket + \llbracket e_{t(v)}\otimes {\overline{vw}}\otimes {\overline{u}}\rrbracket + \llbracket e_{t(w)}\otimes {\overline{wu}}\otimes {\overline{v}}\rrbracket =0$.

Proof

Consider the commutator $[e_{t(u)}\otimes {\overline{u}}, v\otimes {\overline{w}}]= (e_{t(u)}\otimes {\overline{u}})(v\otimes {\overline{w}}) -$ $(-1)^{1}(v\otimes {\overline{w}})(e_{t(u)}\otimes {\overline{u}})= -u\otimes {\overline{v}}\otimes {\overline{w}} + e_{t(u)}\otimes {\overline{uv}}\otimes {\overline{w}} - vw\otimes \overline{e_{t(u)}}\otimes {\overline{u}} + v\otimes {\overline{w}}\otimes {\overline{u}} =$ $-u\otimes {\overline{v}}\otimes {\overline{w}} + v\otimes {\overline{w}}\otimes {\overline{u}} + e_{t(u)}\otimes {\overline{uv}}\otimes {\overline{w}}$. Therefore, we obtain the equality in condition (1).

In order to prove condition (2) one applies condition (1) two times, and for a proof of condition (3) one applies condition (1) three times. $\square$

8 R-relative differential 2-forms

We are going to show in this section how important are exact R-relative differential 2-forms.

For a fixed $\llbracket \omega \rrbracket \in DR^{2}_{R}(B_{r,l})$, we have a K-linear transformation $i_{\llbracket \omega \rrbracket }: \text{ Der}_{R}(B_{r,l}) \rightarrow DR^{1}_{R}(B_{r,l})$ that is given by $\theta \mapsto i_{\theta }(\llbracket \omega \rrbracket )$, where $i_{\theta }:DR^{\bullet }_{R}(B_{r,l})\rightarrow DR^{\bullet -1}_{R}(B_{r,l})$ is a super-derivation of degree $(-1)$.

Lemma 8.1

For any algebra $B_{r,r+2}$, $r\ge 0$, assume that $\omega =\alpha _{i-1}\cdots \alpha _{j}\otimes \overline{\alpha _{j-1}\cdots \alpha _{t}}\otimes \overline{\alpha _{t-1}\cdots \alpha _{i}}$ is a generalized cycle in $\Omega ^{2}_{r}(B_{r,r+2})$ and $\alpha _{i-1}\cdots \alpha _{j}$ is a non-trivial path. Then, the R-relative differential 2-form $\llbracket \alpha _{i-1}\cdots \alpha _{j}\otimes \overline{\alpha _{j-1}\cdots \alpha _{t}}\otimes \overline{\alpha _{t-1}\cdots \alpha _{i}}\rrbracket$ is degenerate.

Proof

In order to have that the form $\llbracket \omega \rrbracket$ is non-degenerate, the transformation $i_{\llbracket \omega \rrbracket }:\text{ Der}_{R}(B_{r,r+2}) \rightarrow DR^{1}_{R}(B_{r,r+2})$ has to be an isomorphism. But for any derivation $\theta \in \text{ Der}_{R}(B_{r,r+2})$, we have $i_{\llbracket \omega \rrbracket }(\theta )= \alpha _{i-1}\cdots \alpha _{j}\theta (\alpha _{j-1}\cdots \alpha _{t})\otimes \overline{\alpha _{t-1}\cdots \alpha _{i}} -$ $\alpha _{i-1}\cdots \alpha _{j}\otimes \overline{\alpha _{j-1}\cdots \alpha _{t}\theta (\alpha _{t-1}\cdots \alpha _{i})}$. Thus, for each arrow $\alpha _{s}$ with $s\in \{i-1,\ldots ,j\}$ we have $i_{\llbracket \omega \rrbracket }(\theta ^{0}_{(\alpha _{s})})=0$. Therefore, $i_{\llbracket \omega \rrbracket }$ is not an isomorphism. Consequently, the R-relative differential 2-form $\llbracket \omega \rrbracket$ is degenerate. $\square$

Lemma 8.2

For any algebra $B_{r,r+2}$, $r\ge 0$, assume that $u^{\prime }_{0}\alpha u^{\prime \prime }_{0}\otimes \overline{u_{1}}\otimes \overline{u_{2}}$ is a generalized cycle in $\Omega ^{2}_{R}(B_{r,r+2})$ such that $\alpha$ is an arrow in $Q_{r}$ and $u^{\prime }_{0}$ is a non-trivial path. Then, $\llbracket u^{\prime }_{0}\alpha u^{\prime \prime }_{0}\otimes \overline{u_{1}}\otimes \overline{u_{2}}\rrbracket = \llbracket \alpha u^{\prime \prime }_{0}u_{1}\otimes \overline{u_{2}}\otimes \overline{u^{\prime }_{0}}\rrbracket - \llbracket \alpha u^{\prime \prime }_{0}\otimes \overline{u_{1}u_{2}}\otimes \overline{u^{\prime }_{0}}\rrbracket + \llbracket \alpha u^{\prime \prime }_{0}\otimes \overline{u_{1}}\otimes \overline{u_{2}u^{\prime }_{0}}\rrbracket$ in $DR^{2}_{R}(B_{r,r+2})$.

Proof

In order to prove the lemma, it is enough to consider the commutator $[u^{\prime }_{0},\alpha u^{\prime \prime }_{0}\otimes \overline{u_{1}}\otimes \overline{u_{2}}]$. We omit the details. $\square$

Proposition 8.3

For any algebra $B_{r,r+2}$, $r\ge 0$, assume that $\omega _{s}=\alpha _{i_{s}-1}\cdots \alpha _{j_{s}} \otimes \overline{\alpha _{j_{s}-1}\cdots \alpha _{t_{s}}}\otimes \overline{\alpha _{t_{s}-1}\cdots \alpha _{i_{s}}}$, $s=1,\ldots ,n$, are generalized cycles in $\Omega ^{2}_{R}(B_{r,r+2})$. Then, the following conditions are satisfied in $DR^{2}_{R}(B_{r,r+2})$:

• (1) If there is an arrow $\alpha$ in $Q_{r}$ such that it appears in every path $\alpha _{i_{s}-1}\cdots \alpha _{j_{s}}$ for $s=1,\ldots ,n$, then an R-relative differential 2-form $\llbracket \omega \rrbracket = \sum ^{n}_{s=1} k_{s}\llbracket \omega _{s}\rrbracket$ satisfies that $i_{\llbracket \omega \rrbracket }$ is not a monomorphism, where $k_{1},\ldots ,k_{n}\in K$.

• (2) If an R-relative differential 2-form $\llbracket \omega \rrbracket =\sum ^{n}_{s=1}k_{s}\llbracket \omega _{s}\rrbracket$, $k_{1},\ldots ,k_{n}\in K$, satisfies that $i_{\llbracket \omega \rrbracket }$ is a monomorphism, then there is a presentation $\llbracket \omega \rrbracket =$ $\sum ^{m}_{a=1}k^{\prime }_{a}\llbracket \omega ^{\prime }_{a}\rrbracket + \sum ^{d}_{b=1} k^{\prime \prime }_{b}\llbracket u_{b}\rrbracket$ such that $\omega ^{\prime }_{a}=\alpha \omega ^{\prime \prime }_{a}$ for some arrow $\alpha$, $a=1,2,\ldots ,m$, and for each $b=1,2,\ldots ,d$, it holds $u_{b}=e_{b}\otimes \overline{v_{b,0}}\otimes \overline{v_{b,1}}$, where $e_{b}=e_{j(b)}$ with $j(b)\in \{0,1,\ldots ,r\}$.

Proof

For a proof of condition (1) notice that if there is an arrow $\alpha$ in $Q_{r}$ that appears in each path $\alpha _{i_{s}-1}\cdots \alpha _{j_{s}}$, $s=1,2,\ldots ,n$, then $i_{\sum ^{n}_{s=1}k_{s}\llbracket \omega _{s}\rrbracket }(\theta ^{0}_{(\alpha )})=0$. Hence, $i_{\sum ^{n}_{s=1}k_{s}\llbracket \omega _{s}\rrbracket }$ is not a monomorphism for any $k_{1},\ldots ,k_{n}\in K$.

In order to prove condition (2) suppose that an R-relative differential 2-form $\llbracket \omega \rrbracket =\sum ^{n}_{s=1}k_{s}\llbracket \omega _{s}\rrbracket$ satisfies that $i_{\llbracket \omega \rrbracket }$ is a monomorphism. Then, we know from (1) that there is not an arrow $\alpha$ that appears in every path $\alpha _{i_{s}-1}\cdots \alpha _{j_{s}}$, $s=1,2,\ldots ,n$. Applying Lemma 8.2, we may assume that $\omega _{1},\ldots ,\omega _{s_{1}-1}$ have a form $\alpha _{i_{1}-1}\cdots \alpha _{j_{1}} \otimes \overline{\alpha _{j_{1}-1}\cdots \alpha _{t_{1}}}\otimes \overline{\alpha _{t_{1}-1}\cdots \alpha _{i_{1}}}$, $\alpha _{i_{1}-1}\cdots \alpha _{j_{2}}\otimes \overline{\alpha _{j_{2}-1}\cdots \alpha _{t_{2}}}\otimes \overline{\alpha _{t_{2}-1}\cdots \alpha _{i_{1}}}$, $\ldots$, $\alpha _{i_{1}-1}\cdots \alpha _{j_{s_{1}-1}}\otimes \overline{\alpha _{j_{s_{1}-1}-1} \cdots \alpha _{t_{s_{1}-1}}}\otimes \overline{\alpha _{t_{s_{1}-1}-1} \cdots \alpha _{i_{1}}}$.

Suppose that $\omega _{s_{1}}=\alpha _{i_{s_{1}}-1} \cdots \alpha _{j_{s_{1}}} \otimes \overline{\alpha _{j_{s_{1}}-1}\cdots \alpha _{t_{s_{1}}}} \otimes \overline{\alpha _{t_{s_{1}}-1} \cdots \alpha _{i{s_{1}}}}$ and $\alpha _{i_{s_{1}}-1}\not =\alpha _{i_{1}-1}$. If $\alpha _{i_{s_{1}}-1}\cdots \alpha _{j_{s_{1}}} =\alpha _{i_{s_{1}}-1} \cdots \alpha _{i_{1}}\alpha _{i_{1}-1} \cdots \alpha _{j_{s_{1}}}$, then we infer by Lemma 8.2 that we may change the form $\llbracket \omega _{s_{1}}\rrbracket$ onto a sum of three forms with $\alpha _{i_{1}-1}$ on the left hand side.

If the arrow $\alpha _{i_{1}-1}$ does not appear in the path $\alpha _{i_{s_{1}}-1} \cdots \alpha _{j_{s_{1}}}$, then $\omega _{s_{1}}=$ $\alpha _{i_{s_{1}}-1}\cdots \alpha _{j_{s_{1}}}\otimes \overline{\alpha _{j_{s_{1}}-1}\cdots \alpha _{i_{1}-1}\cdots \alpha _{t_{s_{1}}}} \otimes \overline{\alpha _{t_{s_{1}}-1} \cdots \alpha _{i_{s_{1}}}}$ or $\omega _{s_{1}}=$ $\alpha _{i_{s_{1}}-1} \cdots \alpha _{j_{s_{1}}}\otimes \overline{\alpha _{j_{s_{1}}-1}\cdots \alpha _{t_{s_{1}}}} \otimes \overline{\alpha _{t_{s_{1}}-1}\cdots \alpha _{i_{1}-1}\cdots \alpha _{i_{s_{1}}}}$.

In the first case, we have $[\alpha _{i_{s_{1}}-1}\cdots \alpha _{j_{s_{1}}}\otimes \overline{\alpha _{j_{s_{1}}-1}\cdots \alpha _{i_{1}}}, \alpha _{i_{1}-1}\cdots \alpha _{t_{s_{1}}}\otimes \overline{\alpha _{t_{s_{1}}-1}\cdots \alpha _{i_{s_{1}}}}]=$ $-\alpha _{i_{s_{1}}-1}\cdots \alpha _{j_{s_{1}}}\alpha _{j_{s_{1}}-1}\cdots \alpha _{i_{1}} \otimes \overline{\alpha _{i_{1}-1}\cdots \alpha _{t_{s_{1}}}} \otimes \overline{\alpha _{t_{s_{1}}-1}\cdots \alpha _{i_{s_{1}}}} +$ $\alpha _{i_{s_{1}}-1} \!\cdots \alpha _{j_{s_{1}}}\otimes \,\overline{\alpha _{j_{s_{1}}-1}\!\cdots \alpha _{i_{1}}\alpha _{i_{1}-1} \!\cdots \alpha _{t_{s_{1}}}}\,\otimes \,\overline{\alpha _{t_{s_{1}}-1} \!\cdots \alpha _{i_{s_{1}}}} -$ $\alpha _{i_{1}-1}\!\cdots \alpha _{t_{s_{1}}}\alpha _{t_{s_{1}}-1}\cdots \alpha _{i_{s_{1}}}\otimes \overline{\alpha _{i_{s_{1}}-1}\cdots \alpha _{j_{s_{1}}}} \otimes \overline{\alpha _{j_{s_{1}}-1} \cdots \alpha _{i_{1}}} +$ $\alpha _{i_{1}-1}\cdots \alpha _{t_{s_{1}}} \otimes \overline{\alpha _{t_{s_{1}}-1}\cdots \alpha _{i_{s_{1}}}\alpha _{i_{s_{1}}-1}\cdots \alpha _{j_{s_{1}}}} \otimes \overline{\alpha _{j_{s_{1}}-1}\cdots \alpha _{i_{1}}}$. Therefore, we deduce that $\llbracket \alpha _{i_{s_{1}}-1}\cdots \alpha _{j_{s_{1}}}\otimes \overline{\alpha _{j_{s_{1}}-1}\cdots \alpha _{i_{1}}\alpha _{i_{1}-1}\cdots \alpha _{t_{s_{1}}}} \otimes \overline{\alpha _{t_{s_{1}}-1} \cdots \alpha _{i_{s_{1}}}}\rrbracket =$ $\llbracket \alpha _{i_{1}-1}\cdots \alpha _{t_{s_{1}}}\alpha _{t_{s_{1}}-1}\cdots \alpha _{i_{s_{1}}}\otimes \overline{\alpha _{i_{s_{1}}-1}\cdots \alpha _{j_{s_{1}}}}\otimes \overline{\alpha _{j_{s_{1}}-1}\cdots \alpha _{i_{1}}} \,\rrbracket -$ $\llbracket \alpha _{i_{s_{1}}-1}\cdots \alpha _{i_{1}}\otimes \overline{\alpha _{i_{1}-1}\cdots \alpha _{t_{s_{1}}}} \otimes \overline{\alpha _{t_{s_{1}}-1}\cdots \alpha _{i_{s_{1}}}}\rrbracket +$ $\llbracket \alpha _{i_{1}-1}\cdots \alpha _{t_{s_{1}}} \otimes \overline{\alpha _{t_{s_{1}}-1}\cdots \alpha _{i_{s_{1}}}\alpha _{i_{s_{1}}-1}\cdots \alpha _{j_{s_{1}}}}\otimes \overline{\alpha _{j_{s_{1}}-1}\cdots \alpha _{i_{1}}}\rrbracket$. Further, using Proposition 7.5(1) we obtain that $\llbracket \alpha _{i_{s_{1}}\!-\!1}\cdots \alpha _{j_{s_{1}}}\alpha _{j_{s_{1}}-1}\cdots \alpha _{i_{1}}\otimes \overline{\alpha _{i_{1}\!-\!1}\cdots \alpha _{t_{s_{1}}}}\otimes \overline{\alpha _{t_{s_{1}}\!-\!1}\cdots \alpha _{i_{s_{1}}}}\rrbracket =$ $\llbracket \alpha _{i_{1}-1}\cdots \alpha _{t_{s_{1}}}\otimes \overline{\alpha _{t_{s_{1}}-1}\cdots \alpha _{i_{s_{1}}}} \otimes \overline{\alpha _{i_{s_{1}}-1}\cdots \alpha _{j_{s_{1}}}\alpha _{j_{s_{1}}-1}\cdots \alpha _{i_{1}}}\rrbracket +$ $\llbracket e_{i_{s_{1}}}\otimes \overline{\alpha _{i_{s_{1}}-1}\cdots \alpha _{j_{s_{1}}}\alpha _{j_{s_{1}}-1}\cdots \alpha _{i_{1}}\alpha _{i_{1}-1}\cdots \alpha _{t_{s_{1}}}}\otimes \overline{\alpha _{t_{s_{1}}-1}\cdots \alpha _{i_{s_{1}}}}\rrbracket$. Hence, we obtain that $\llbracket \alpha _{i_{s_{1}}-1}\cdots \alpha _{j_{s_{1}}}\otimes \overline{\alpha _{j_{s_{1}}-1}\cdots \alpha _{i_{1}}\cdots \alpha _{t_{s_{1}}}} \otimes \overline{\alpha _{t_{s_{1}}-1} \cdots \alpha _{i_{s_{1}}}}\rrbracket =$ $\llbracket \alpha _{i_{1}-1}\cdots \alpha _{t_{s_{1}}}\alpha _{t_{s_{1}}-1}\cdots \alpha _{i_{s_{1}}} \otimes \overline{\alpha _{i_{s_{1}}-1}\cdots \alpha _{j_{s{1}}}} \otimes \overline{\alpha _{j_{s_{1}}-1}\cdots \alpha _{i_{1}}}\rrbracket +$ $\llbracket \alpha _{i_{1}-1}\cdots \alpha _{t_{s_{1}}}\otimes \overline{\alpha _{t_{s_{1}}-1}\cdots \alpha _{i_{s_{1}}}\alpha _{i_{s_{1}}-1}\cdots \alpha _{j_{s_{1}}}} \otimes \overline{\alpha _{j_{s_{1}}-1}\cdots \alpha _{i_{1}}}\rrbracket -$ $\llbracket \alpha _{i_{1}-1} \cdots \alpha _{t_{s_{1}}} \otimes \overline{\alpha _{t_{s_{1}}-1}\cdots \alpha _{i_{s_{1}}}} \otimes \overline{\alpha _{i_{s_{1}}-1}\cdots \alpha _{j_{s_{1}}}\alpha _{j_{s_{1}}-1}\cdots \alpha _{i_{1}}}\rrbracket -$ $\llbracket e_{i_{s_{1}}}\otimes \overline{\alpha _{i_{s_{1}}-1}\cdots \alpha _{j_{s_{1}}}\alpha _{j_{s_{1}}-1}\cdots \alpha _{i_{s_{1}}}\alpha _{i_{s_{1}}-1}\cdots \alpha _{t_{s_{1}}}} \otimes \overline{\alpha _{t_{s_{1}}-1}\cdots \alpha _{i_{s_{1}}}}\rrbracket$. Replacing $\llbracket \omega _{s_{1}}\rrbracket$ by the above sum, we obtain $\llbracket \omega \rrbracket =$ $\sum ^{m_{1}}_{a_{1}=1}k^{(1)}_{a_{1}}\llbracket \omega ^{(1)}_{a_{1}} \rrbracket -k_{s_{1}}\llbracket e_{i_{s_{1}}}\otimes$ $\overline{\alpha _{i_{s_{1}}-1}\cdots \alpha _{j_{s_{1}}}\alpha _{j_{s_{1}}-1}\cdots \alpha _{i_{1}}\alpha _{i_{1}-1}\cdots \alpha _{t_{s_{1}}}}\otimes \overline{\alpha _{t_{s_{1}}-1}\cdots \alpha _{i_{s_{1}}}}\rrbracket$, where $\omega ^{(1)}_{a_{1}}=\omega _{a_{1}}$ for $a_{1}<s_{1}$, the next summands $\omega ^{(1)}_{s_{1}}$, $\omega ^{(1)}_{s_{1}+1}$, $\omega ^{(1)}_{s_{1}+2}$ are the first three summands in the above presentation of $\omega _{s_{1}}$. Further, $\omega ^{(1)}_{a_{1}}=\omega _{a_{1}-2}$ for $a_{1}\ge s_{1}+3$. Moreover,

$$\begin{aligned} k^{(1)}_{a_{1}}=\left\{ \begin{array}{ll} k_{a_{1}} &{} \text{ if }~~ a_{1}\le s_{1} \\ k_{s_{1}} &{} \text{ if } ~~ a_{1}=s_{1}+1 \\ -k_{s_{1}} &{} \text{ if }~~ a_{1}=s_{1}+2 \\ k_{a_{1}-2} &{} \text{ if }~~ a_{1}\ge s_{1}+3 .\end{array}\right. \end{aligned}$$

If $\omega _{s_{1}}=\alpha _{i_{s_{1}}-1}\cdots \alpha _{j_{s_{1}}}\otimes \overline{\alpha _{j_{s_{1}}-1}\cdots \alpha _{t_{s_{1}}}}\otimes \overline{\alpha _{t_{s_{1}}-1}\cdots \alpha _{i_{1}-1}\alpha _{i_{1}}\cdots \alpha _{i_{s_{1}}}}$, then we replace $\omega _{s_{1}}$ applying Proposition 7.5(1),(2) and again, we obtain a new presentation $\llbracket \omega \rrbracket =\sum ^{m_{2}}_{a_{2}=1}k^{\prime (2)}_{a_{2}}\llbracket \omega ^{(2)}_{a_{2}}\rrbracket +\sum ^{d_{2}}_{b_{2}=1} k^{\prime \prime (2)}_{b_{2}}\llbracket u_{b_{2}}\rrbracket$.

Continuing the above procedure for $\llbracket \omega _{s_{1}+1}\rrbracket$ and the following forms we obtain the required presentation of $\llbracket \omega \rrbracket$. $\square$

The above proposition indicates an importance of R-relative differential 2-forms that are exact.

Example 8.4

Consider the algebra $B_{1,3}$. Then, we deduce from Corollary 4.6 that $\text{ dim}_{K}\text{ Der}_{R}(B_{1,3})=2$. Moreover, $DR^{1}_{R}(B_{1,3})$ has nonzero elements $\llbracket e_{0}\otimes \overline{c_{0}}\rrbracket$, $\llbracket \alpha _{1}\otimes \overline{\alpha _{0}}\rrbracket$ in view of Proposition 6.5. Using Lemma 6.7 and Lemma 6.1, it is easy to see that the above two elements form a basis od $DR^{1}_{R}(B_{1,3})$. Hence, $\text{ dim}_{K}DR^{1}_{R}(B_{1,3})=2$. Therefore, it can happen that there is a symplectic manifold $(B_{1,3},\llbracket \omega \rrbracket )$. But $B_{1,3}$ is a preprojective algebra for the hereditary algebra $K(0{\mathop {\longrightarrow }\limits ^{\alpha _{0}}} 1)$. Hence, it is known that such manifolds exist in the differential calculus $(DR^{\bullet }_{R}(B_{1,3}),d)$.

Example 8.5

Consider the algebra $B_{2,3}$. Then, one can check that $\text{ dim}_{K}\text{ Der}_{R}(B_{2,3})=3$. Moreover, we can check straightforward that the elements $\llbracket \alpha _{2}\alpha _{1}\otimes \overline{\alpha _{0}}\rrbracket$, $\llbracket \alpha _{0}\alpha _{2}\otimes \overline{\alpha _{1}}\rrbracket$, $\llbracket \alpha _{1}\alpha _{0}\otimes \overline{\alpha _{2}}\rrbracket$ span the K-linear space $DR^{1}_{R}(B_{2,3})$. But these elements are linearly dependent. Thus, $\text{ dim}_{K}DR^{1}_{R}(B_{2,3})<3$. Consequently, there is no symplectic manifold $(B_{2,3},\llbracket \omega \rrbracket )$ in the differential calculus $(DR^{\bullet }_{R}(B_{2,3}),d)$.

Lemma 8.6

For any algebra $B_{r,r+2}$, $r\ge 0$, and for any generalized cycle $e_{i}\otimes \overline{u_{1}}\otimes \overline{u_{2}}$ in $\Omega ^{2}_{R}(B_{r,r+2})$, we have the following equality $\llbracket e_{i}\otimes \overline{u_{1}}\otimes \overline{u_{2}}\rrbracket = -\llbracket e_{j}\otimes \overline{u_{2}}\otimes \overline{u_{1}}\rrbracket$ in $DR^{2}_{R}(B_{r,r+2})$, where $i=t(u_{1})$, $j=t(u_{2})$.

Proof

Consider the commutator $[e_{i}\otimes \overline{u_{1}},e_{j}\otimes \overline{u_{2}}] = (e_{i}\otimes \overline{u_{1}})(e_{j}\otimes \overline{u_{2}}) -$

$(-1)^{1}(e_{j}\otimes \overline{u_{2}})(e_{i}\otimes \overline{u_{1}})= u_{1}\otimes \overline{e_{j}}\otimes \overline{u_{2}} - e_{i}\otimes \overline{u_{1}e_{j}}\otimes \overline{u_{2}}+ u_{2}\otimes \overline{e_{i}}\otimes \overline{u_{1}} - e_{j}\otimes \overline{u_{2}e_{i}}\otimes \overline{u_{1}}=$ $-e_{i}\otimes \overline{u_{1}}\otimes \overline{u_{2}} - e_{j}\otimes \overline{u_{2}}\otimes \overline{u_{1}}$. Then, the required condition follows. $\Box$

9 Exact R-relative differential 2-forms

Lemma 9.1

For an algebra $B_{r,r+2}$, $r\ge 0$, let $\llbracket \omega \rrbracket =\sum ^{r}_{i=0}k_{i}\llbracket e_{i}\otimes \overline{h_{i}}\otimes \overline{\alpha _{i}}\rrbracket$, $0\not =k_{i}\in K$, $i=0,1,\ldots ,r$, and $0\not =\theta = \sum ^{r}_{j=0}k^{\prime }_{j}\theta ^{0}_{(\alpha _{j})}\in \text{ Der}_{R}(B_{r,r+2})$, $k^{\prime }_{j}\in K$, $j=0,1,\ldots ,r$. If $i_{\llbracket \omega \rrbracket }(\theta )=0$ in $DR^{1}_{R}(B_{r,r+2})$ then $\theta =0$.

Proof

First notice that we have $i_{\llbracket \omega \rrbracket }(\theta )= \sum ^{r}_{i=0}\sum ^{r}_{j=0}k_{i}k^{\prime }_{j}(\llbracket \theta ^{0}_{(\alpha _{j})}(h_{i})\otimes \overline{\alpha _{i}}\rrbracket -$ $\llbracket e_{i}\otimes \overline{h_{i}\theta ^{0}_{(\alpha _{j})}(\alpha _{i})}\rrbracket )= \sum _{i\not = j}k_{i}k^{\prime }_{j}\llbracket h_{i}\otimes \overline{\alpha _{i}}\rrbracket - \sum _{i=j}k_{i}k^{\prime }_{j}\llbracket e_{i}\otimes \overline{c_{i}}\rrbracket =$ $\sum _{i\not = j}k_{i}k^{\prime }_{j}\llbracket h_{i}\otimes \overline{\alpha _{i}}\rrbracket -\sum ^{r}_{i=0}k_{i}k^{\prime }_{i}(\llbracket h_{0}\otimes \overline{\alpha _{0}}\rrbracket + \llbracket h_{1}\otimes \overline{\alpha _{1}}\rrbracket +\cdots +\llbracket h_{r}\otimes \overline{\alpha _{r}}\rrbracket )$, because we infer by Lemma 6.1 and Lemma 6.3 that $\llbracket e_{i}\otimes \overline{c_{i}}\rrbracket = \llbracket h_{0}\otimes \overline{\alpha _{0}}\rrbracket +$ $\llbracket h_{1}\otimes \overline{\alpha _{1}}\rrbracket +\cdots +\llbracket h_{r}\otimes \overline{\alpha _{r}}\rrbracket$.

If $i_{\llbracket \omega \rrbracket }(\theta )=0$, then for each $i=0,1,\ldots ,r$ we have the equality $0=k_{i}\sum _{j\not = i}(k^{\prime }_{j}-k^{\prime }_{i})-k_{i}k^{\prime }_{i}$. Since all $k_{i}\not =0$, $i=0,1,\ldots ,r$, so we obtain the following system of equations:

$$\begin{aligned} \begin{array}{l} 0=\sum _{j\not =0}(k^{\prime }_{j}-k^{\prime }_{0})-k^{\prime }_{0} =\sum _{j\not =0}k^{\prime }_{j} -(r+1)k^{\prime }_{0} \\ 0= \sum _{j\not = 1}(k^{\prime }_{j}-k^{\prime }_{1})-k^{\prime }_{1}= \sum _{j\not =1}k^{\prime }_{j}-(r+1)k^{\prime }_{1} \\ ~~~~ \vdots \\ 0=\sum _{j\not =r}(k^{\prime }_{j}-k^{\prime }_{r})-k^{\prime }_{r}=\sum _{j\not =r}k^{\prime }_{j}-(r+1)k^{\prime }_{r}.\end{array} \end{aligned}$$

The matrix of this system of equations has the following form

$$\begin{aligned} M=\left[ \begin{array}{ccccc} -(r+1) &{} 1 &{} 1 &{} \cdots &{} 1 \\ 1&{} -(r+1) &{} 1 &{} \cdots &{} 1 \\ 1 &{} 1 &{} -(r+1) &{} \cdots &{} 1 \\ \vdots &{} \vdots &{} \vdots &{} &{}\vdots \\ 1 &{} \cdots &{} 1 &{} 1 &{} -(r+1) \end{array}\right] . \end{aligned}$$

Since $\text{ char }(K)=0$, $\text{ det }(M)\not =0$ and $k^{\prime }_{0}=k^{\prime }_{1}=\cdots =k^{\prime }_{r}=0$. Thus, $\theta =0$. $\square$

Lemma 9.2

For an algebra $B_{r,r+2}$, $r\ge 0$, let $\llbracket \omega \rrbracket =\sum ^{r}_{i=0}k_{i}\llbracket e_{i}\otimes \overline{h_{i}}\otimes \overline{\alpha _{i}}\rrbracket$, $k_{i}\in K$, $i=0,1,\ldots ,r$. If there are $i_{0}\not = i_{1}$ in $\{0,1,\ldots ,r\}$ such that $k_{i_{0}}=0=k_{i_{1}}$, then $i_{\llbracket \omega \rrbracket }$ is not a monomorphism.

Proof

Let $\llbracket \omega \rrbracket =\sum ^{r}_{i=0} k_{i}\llbracket e_{i}\otimes \overline{h_{i}}\otimes \overline{\alpha _{i}}\rrbracket$ and suppose that $k_{i_{0}}=0=k_{i_{1}}$ for some $i_{0}\not =i_{1}$. Then, $i_{\llbracket \omega \rrbracket }(\theta ^{0}_{(\alpha _{i_{0}})})=\sum _{i\not =i_{0}}k_{i}\llbracket h_{i}\otimes \overline{\alpha _{i}}\rrbracket -k_{i_{0}}\llbracket e_{i_{0}}\otimes \overline{c_{i_{0}}}\rrbracket =$ $\sum _{i\not = i_{0},i_{1}}k_{i}\llbracket h_{i}\otimes \overline{\alpha _{i}}\rrbracket +$ $k_{i_{1}}\llbracket h_{i_{1}}\otimes \overline{\alpha _{i_{1}}}\rrbracket -k_{i_{0}}\llbracket e_{i_{0}}\otimes \overline{c_{i_{0}}}\rrbracket = \sum _{i\not =i_{0},i_{1}}k_{i}\llbracket h_{i}\otimes \overline{\alpha _{i}}\rrbracket$. Further, $i_{\llbracket \omega \rrbracket }(\theta ^{0}_{(\alpha _{i_{1}})})= \sum _{i\not =i_{1}}k_{i}\llbracket h_{i}\otimes \overline{\alpha _{i}}\rrbracket - k_{i_{1}}\llbracket e_{i_{1}}\otimes \overline{c_{i_{1}}}\rrbracket =\sum _{i\not =i_{0},i_{1}}k_{i}\llbracket h_{i}\otimes \overline{\alpha _{i}}\rrbracket + k_{i_{0}}\llbracket h_{i_{0}}\otimes \overline{\alpha _{i_{0}}}\rrbracket -k_{i_{1}}\llbracket e_{i_{1}}\otimes \overline{c_{i_{1}}}\rrbracket =\sum _{i\not =i_{0},i_{1}}k_{i}\llbracket h_{i}\otimes \overline{\alpha _{i}}\rrbracket$. Thus, $i_{\llbracket \omega \rrbracket }(\theta ^{0}_{(\alpha _{i_{0}})})=i_{\llbracket \omega \rrbracket }(\theta ^{0}_{(\alpha _{i_{1}})})$. Hence, $i_{\llbracket \omega \rrbracket }$ is not a monomorphism. $\square$

Lemma 9.3

For an algebra $B_{r,r+2}$, $r\ge 0$, the following condition holds: if $\llbracket \omega \rrbracket =\sum ^{r}_{i=0}k_{i}\llbracket e_{i}\otimes \overline{h_{i}}\otimes \overline{\alpha _{i}}\rrbracket$, $k_{i}\in K$, $i=0,1,\ldots ,r$, and there is $i_{0}\in \{0,1,\ldots ,r\}$ such that $k_{i_{0}}=0$ and every $k_{i}\not =0$ for $i\not =i_{0}$ then $i_{\llbracket \omega \rrbracket }$ is an epimorphism.

Proof

Without loss of generality, we may assume that $i_{0}=0$. Then, $k_{0}=0$ and $k_{i}\not =0$ for $i\not =0$. Let $\theta =\sum ^{r}_{j=0}k^{\prime }_{j}\theta ^{0}_{(\alpha _{j})}$, $k^{\prime }_{j}\in K$, $j=0,1,\ldots ,r$. Fix $p\in \{0,1,\ldots ,r\}$. Let $\llbracket h_{p}\otimes \overline{\alpha _{p}}\rrbracket =i_{\llbracket \omega \rrbracket }(\theta )= \sum _{i\not =j,i\not =0} k_{i}k^{\prime }_{j}\llbracket h_{i}\otimes \overline{\alpha _{i}}\rrbracket -\sum ^{r}_{i=1}k_{i}k^{\prime }_{i}\llbracket e_{i}\otimes \overline{c_{i}}\rrbracket =$

$\sum _{i\not =j,i\not =0} k_{i}k^{\prime }_{j} \llbracket h_{i}\otimes \overline{\alpha _{i}}\rrbracket - \sum ^{r}_{i=1}k_{i}k^{\prime }_{i}(\llbracket h_{0}\otimes \overline{\alpha _{0}}\rrbracket +\llbracket h_{1}\otimes \overline{\alpha _{1}}\rrbracket + \cdots +\llbracket h_{r}\otimes \overline{\alpha _{r}}\rrbracket )$.

We deduce from the above equality that $\llbracket h_{p}\otimes \overline{\alpha _{p}}\rrbracket = k_{1}\sum _{j\not =1}k^{\prime }_{j}\llbracket h_{1}\otimes \overline{\alpha _{1}}\rrbracket -k_{1}k^{\prime }_{1}\sum ^{r}_{a=0}\llbracket h_{a}\otimes \overline{\alpha _{a}}\rrbracket + k_{2}\sum _{j\not =2}k^{\prime }_{j}\llbracket h_{2}\otimes \overline{\alpha _{2}}\rrbracket -k_{2}k^{\prime }_{2}\sum ^{r}_{a=0}\llbracket h_{a}\otimes \overline{\alpha _{a}}\rrbracket + \cdots +$ $k_{r}\sum _{j\not =r}k^{\prime }_{j}\llbracket h_{r}\otimes \overline{\alpha _{r}}\rrbracket - k_{r} k^{\prime }_{r}\sum ^{r}_{a=0}\llbracket h_{a}\otimes \overline{\alpha _{a}}\rrbracket$. Further, we have $\llbracket h_{p}\otimes \overline{\alpha _{p}}\rrbracket =$ $(k_{1}\sum _{j\not =1}k^{\prime }_{j} -\sum ^{r}_{i=1}k_{i}k^{\prime }_{i})\llbracket h_{1}\otimes \overline{\alpha _{1}}\rrbracket + (k_{2}\sum _{j\not =2}k^{\prime }_{j}-\sum ^{r}_{i=1}k_{i}k^{\prime }_{i})\llbracket h_{2}\otimes \overline{\alpha _{2}}\rrbracket +\cdots +$

$(k_{r}\sum _{j\not =r}k^{\prime }_{j}-\sum ^{r}_{i=1}k_{i}k^{\prime }_{i})\llbracket h_{r}\otimes \overline{\alpha _{r}}\rrbracket -\sum ^{r}_{i=1}k_{i}k^{\prime }_{i}\llbracket h_{0}\otimes \overline{\alpha _{0}}\rrbracket$.

Since the system $\{\llbracket h_{i}\otimes \overline{\alpha _{i}}\rrbracket \}_{i=0,1,\ldots ,r}$ is a basis of the K-linear space $DR^{1}_{R}(B_{r,r+2})$ by Proposition 6.8, for any $p=0,1,\ldots ,r$ we obtain the following system of equations with variables $k^{\prime }_{0},k^{\prime }_{1},\ldots ,k^{\prime }_{r}$:

$$\begin{aligned} \begin{array}{l} 0=-\sum ^{r}_{i=1} k_{i}k^{\prime }_{i} \\ 0= k_{1}\sum _{j\not =1}k^{\prime }_{j} -\sum ^{r}_{i=1}k_{i}k^{\prime }_{i} \\ ~~~~ \vdots \\ 0=k_{p-1}\sum _{j\not =p-1}k^{\prime }_{j}-\sum ^{r}_{i=1}k_{i}k^{\prime }_{i} \\ 1=k_{p}\sum _{j\not =p}k^{\prime }_{j}-\sum ^{r}_{i=1}k_{i}k^{\prime }_{i} \\ 0=k_{p+1}\sum _{j\not =p+1}k^{\prime }_{j}-\sum ^{r}_{i=1}k_{i}k^{\prime }_{i} \\ ~~~~ \vdots \\ 0=k_{r}\sum _{j\not =r}k^{\prime }_{j} -\sum ^{r}_{i=1}k_{i}k^{\prime }_{i} \end{array} \end{aligned}$$

that gives us the following system:

$$\begin{aligned} \begin{array}{l} 0=\sum ^{r}_{i=1}k_{i}k^{\prime }_{i} \\ 0=k_{1}\sum _{j\not =1}k^{\prime }_{j} \\ ~~~~\vdots \\ 0=k_{p-1}\sum _{j\not =p-1}k^{\prime }_{j}\\ 1=k_{p}\sum _{j\not =p}k^{\prime }_{j} \\ 0=k_{p+1}\sum _{j\not =p+1}k^{\prime }_{j} \\ ~~~~\vdots \\ 0=k_{r}\sum _{j\not =r}k^{\prime }_{j} \end{array} \end{aligned}$$

for $p=1,2,\ldots ,r$ and the above system for $p=0$ has the form:

$$\begin{aligned} \begin{array}{l} -1=\sum ^{r}_{i=1}k_{i}k^{\prime }_{i} \\ -1=k_{1}\sum _{j\not =1}k^{\prime }_{j} \\ -1=k_{2}\sum _{j\not =2}k^{\prime }_{j} \\ ~~~~ \vdots \\ -1=k_{r}\sum _{j\not =r}k^{\prime }_{j} .\end{array} \end{aligned}$$

The matrix M of these two systems has the form

$$\begin{aligned} M=\left[ \begin{array}{cccccccc} 0 &{} k_{1} &{} k_{2} &{} k_{3} &{} \cdots &{} k_{r-2} &{} k_{r-1} &{} k_{r} \\ k_{1} &{} 0 &{} k_{1} &{} k_{1} &{} \cdots &{} k_{1} &{} k_{1} &{} k_{1} \\ k_{2} &{} k_{2} &{} 0 &{} k_{2} &{} \cdots &{} k_{2} &{} k_{2} &{} k_{2} \\ \vdots &{} \vdots &{} &{} &{} &{} \vdots &{} &{} \vdots \\ k_{p} &{} k_{p} &{} &{} \cdots &{} k_{p} &{} 0 &{} \cdots &{} k_{p} \\ \vdots &{} \vdots &{} &{} &{} &{} &{} &{} \vdots \\ k_{r} &{} k_{r} &{} &{} \cdots &{} &{} &{} k_{r} &{} 0 \end{array}\right] . \end{aligned}$$

Since all $k_{1},k_{2},\ldots ,k_{r}\not =0$, $\text{ det }(M)\not =0$ and each of these systems has exactly one solution.

Consequently, each basis member $\llbracket h_{p}\otimes \overline{\alpha _{p}}\rrbracket \in \text{ im }(i_{\llbracket \omega \rrbracket })$. Therefore, $i_{\llbracket \omega \rrbracket }$ is an epimorphism. $\square$

10 Proofs of the main results

Proof of Theorem 1.1

In order to prove condition (1) notice that we infer by Lemma 9.1 and Lemma 9.3 that there are exact R-relative differential 2-forms $\llbracket \omega \rrbracket \in DR^{2}_{R}(B_{r,r+2})$ such that $i_{\llbracket \omega \rrbracket }:\text{ Der}_{R}(B_{r,r+2}) \rightarrow DR^{1}_{R}(B_{r,r+2})$ is a monomorphism or an epimorphism. Hence, they are isomorphisms, because $\text{ dim}_{K}\text{ Der}_{R}(B_{r,r+2})=\text{ dim}_{R}DR^{1}_{R}(B_{r,r+2})$ by Corollary 4.6 and Corollary 6.9. Consequently, the forms $\llbracket \omega \rrbracket$ are non-degenerate. Their forms show that they are exact, and so they are closed as well. Therefore, there are exact symplectic manifolds $(B_{r,r+2},\llbracket \omega \rrbracket )$.

For a proof of condition (2) notice that a 2-form $\llbracket \omega \rrbracket$ is exact provided that it is of the form $\sum ^{r}_{i=0}k_{i}\llbracket e_{i}\otimes \overline{u_{i}}\otimes \overline{v_{i}}\rrbracket$, because we infer by Proposition 6.8 that $\text{ im }(d_{1})$ is spanned by $\{\llbracket e_{i}\otimes \overline{h_{i}}\otimes \overline{\alpha _{i}}\rrbracket \}_{i=0,1,\ldots ,r}$. Furthermore, we deduce from Lemmas 9.1, 9.2, 9.3 that if $(B_{r,r+2},\llbracket \omega \rrbracket )$ is an exact symplectic manifold then there exists at most one index $i_{0}\in \{0,1,\ldots ,r\}$ such that $k_{i_{0}}=0$.

The opposite implication is a consequence of Lemmas 9.1, 9.2, 9.3. $\square$

We say that two symplectic manifolds $(B_{r_{1},l_{1}},\llbracket \omega _{1}\rrbracket )$, $(B_{r_{2},l_{2}},\llbracket \omega _{2}\rrbracket )$ are symplectomorphic if there is a K-algebra isomorphism $f:B_{r_{1},l_{1}} \rightarrow B_{r_{2},l_{2}}$ such that $f^{*}(\llbracket \omega _{1}\rrbracket )=\llbracket \omega _{2}\rrbracket$. If this is the case, then we call the map f a symplectomorphism.

Proof of Theorem 1.2

Let $(B_{r,r+2},\llbracket \omega \rrbracket )$ be an exact symplectic manifold. Then, we infer by Theorem 1.1 that $\llbracket \omega \rrbracket =\sum ^{r}_{i=0}k_{i}\llbracket e_{i}\otimes \overline{h_{i}}\otimes \overline{\alpha _{i}}\rrbracket \not =0$ and there exists at most one index $i_{0}\in \{0,1,\ldots ,r\}$ such that $k_{i_{0}}=0$ Now, consider a K-algebra automorphism $f:B_{r,r+2}\rightarrow B_{r,r+2}$ such that $f(e_{i})=e_{i+s}$, $f(\alpha _{i})=\alpha _{i+s}$ for $s=r+1-i_{0}$. Then, $f(\alpha _{i_{0}})=\alpha _{0}$ and $f(e_{i_{0}})=e_{0}$. Consequently, $f(h_{i})=h_{i+s}$, $i\not =i_{0}$, and $f(h_{i_{0}})=h_{0}$. Therefore, the induced map $f^{*}:DR^{2}_{R}(B_{r,r+2}) \rightarrow DR^{2}_{R}(B_{r,r+2})$ is the K-linear transformation such that $f^{*}(\llbracket e_{i}\otimes \overline{h_{i}}\otimes \overline{\alpha _{i}}\rrbracket )=\llbracket f(e_{i})\otimes \overline{f(h_{i})}\otimes \overline{f(\alpha _{i})}\rrbracket = \llbracket e_{i+s} \otimes \overline{h_{i+s}}\otimes \overline{\alpha _{i+s}}\rrbracket$. Hence, $f^{*}(\llbracket \omega \rrbracket ) =f^{*}(\sum ^{r}_{i=0}k_{i}\llbracket e_{i}\otimes \overline{h_{i}}\otimes \overline{\alpha _{i}}\rrbracket ) =\sum ^{r}_{i=0}k_{i}f^{*}(\llbracket e_{i}\otimes \overline{h_{i}}\otimes \overline{\alpha _{i}}\rrbracket =$ $\sum ^{r}_{i=0}k_{i}\llbracket e_{i+s}\otimes \overline{h_{i+s}}\otimes \overline{\alpha _{i+s}}\rrbracket$. Notice that in this sum coefficient $k_{i_{0}}=0$ appears at $\llbracket e_{0}\otimes \overline{h_{0}}\otimes \overline{\alpha _{0}}\rrbracket$. Then, the required condition is satisfied.

If all coefficients $k_{i}\not =0$, $i=0,1,\ldots ,r$, then consider a K-algebra automorphism $f:B_{r,r+2} \rightarrow B_{r,r+2}$ such that $f(e_{i})=e_{i}$, $i=0,1,\ldots ,r$, and $f(\alpha _{i})=\alpha _{i}$ for $i=1,2,\ldots ,r$. Moreover, we put $f(\alpha _{0})=\frac{1}{k_{0}}\alpha _{0}$. It is easy to see that this is a well-defined K-algebra automorphism of $B_{r,r+2}$. Further, we have $f^{*}(\llbracket e_{i}\otimes \overline{h_{i}}\otimes \overline{\alpha _{i}}\rrbracket ) =\llbracket e_{i}\otimes \overline{f(h_{i})}\otimes \overline{f(\alpha _{i})}\rrbracket =\frac{1}{k_{0}}\llbracket e_{i}\otimes \overline{h_{i}}\otimes \overline{\alpha _{i}}\rrbracket$. Therefore, $f^{*}(\llbracket \omega \rrbracket ) =\sum ^{r}_{i=0}k_{i}f^{*}(\llbracket e_{i}\otimes \overline{h_{i}}\otimes \overline{\alpha _{i}}\rrbracket )=$ $\sum ^{r}_{i=0}\frac{k_{i}}{k_{0}}\llbracket e_{i}\otimes \overline{h_{i}}\otimes \overline{\alpha _{i}}\rrbracket$ and $\frac{k_{0}}{k_{0}}=1$ is the zero coefficient of the form. Consequently, the required condition holds. $\square$

11 Final results and comments

Lemma 11.1

For any algebra $B_{r,r+1}$, $r\ge 1$, the following conditions are satisfied:

• (1) The system $\{\llbracket h_{i}\otimes \overline{\alpha _{i}}\rrbracket \}_{i=0,1,\ldots ,r}$ spans the K-linear space $DR^{1}_{R}(B_{r,r+1})$.

• (2) The system $\{\llbracket h_{i}\otimes \overline{\alpha _{i}}\rrbracket \}_{i=0,1,\ldots ,r}$ is linearly dependent.

• (3) $\text{ dim}_{K}DR^{1}_{R}(B_{r,r+1})\le r$.

Proof

Condition (1) is a straightforward consequence of Lemma 6.3 and Lemma 6.1.

For a proof of condition (2) notice that $\llbracket h_{i}\otimes \overline{\alpha _{i}}\rrbracket + \llbracket \alpha _{i}\otimes \overline{h_{i}}\rrbracket = \llbracket e_{i}\otimes \overline{c_{i}}\rrbracket$, because $[\alpha _{i},e_{i}\otimes \overline{h_{i}}]=\alpha _{i}\otimes \overline{h_{i}} + h_{i}\otimes \overline{\alpha _{i}} - e_{i}\otimes \overline{c_{i}}$. But $l=r+1$, so $c_{i}=0$ in $B_{r,r+1}$, and so $\llbracket h_{i}\otimes \overline{\alpha _{i}}\rrbracket + \llbracket \alpha _{i}\otimes \overline{h_{i}}\rrbracket =0$. Further, applying Lemma 6.3 we obtain that $\llbracket \alpha _{i}\otimes \overline{h_{i}}\rrbracket =\llbracket h_{i+1}\otimes \overline{\alpha _{i+1}}\rrbracket + \llbracket h_{i+2}\otimes \overline{\alpha _{i+2}}\rrbracket + \cdots + \llbracket h_{r}\otimes \overline{\alpha _{r}}\rrbracket + \llbracket h_{0}\otimes \overline{\alpha _{0}}\rrbracket + \cdots +$ $\llbracket h_{i-1}\otimes \overline{\alpha _{i-1}}\rrbracket$. Therefore, $\llbracket h_{i}\otimes \overline{\alpha _{i}}\rrbracket + \llbracket \alpha _{i}\otimes \overline{h_{i}}\rrbracket = \llbracket h_{i}\otimes \overline{\alpha _{i}}\rrbracket + \llbracket h_{i+1}\otimes \overline{\alpha _{i+1}}\rrbracket +\cdots +$ $\llbracket h_{r}\otimes \overline{\alpha _{r}}\rrbracket + \llbracket h_{0}\otimes \overline{\alpha _{0}}\rrbracket +\cdots +\llbracket h_{i-1}\otimes \overline{\alpha _{i-1}}\rrbracket =0$. Consequently, the system $\{\llbracket h_{i}\otimes \overline{\alpha _{i}}\rrbracket \}_{i=0,1,\ldots ,r}$ is linearly dependent.

In order to prove (3) notice that this system spans $DR^{1}_{R}(B_{r,r+1})$ by (1) and is linearly dependent by (2). Then, $\text{ dim}_{K}DR^{1}_{R}(B_{r,r+1})\le r$. $\square$

Theorem 11.2

For any algebra $B_{r,r+1}$, $r\ge 0$, and the Karoubi-de Rham differential calculus $(DR^{\bullet }_{R}(B_{r,r+1}),d)$ there is no symplectic manifold $(B_{r,r+1},\llbracket \omega \rrbracket )$.

Proof

Suppose to the contrary that for some algebra $B_{r,r+1}$ there is a symplectic manifold $(B_{r,r+1},\llbracket \omega \rrbracket )$. Then, $\llbracket \omega \rrbracket \in DR^{2}_{R}(B_{r,r+1})$ is an R-relative differential 2-form that is closed and non-degenerate. Hence, $i_{\llbracket \omega \rrbracket }:\text{ Der}_{R}(B_{r,r+1}) \rightarrow DR^{1}_{R}(B_{r,r+1})$ is an isomorphism. But we know from Lemma 11.1 that $\text{ dim}_{K}DR^{1}_{R}(B_{r,r+1})\le r$ and $\text{ dim}_{K}\text{ Der}_{R}(B_{r,r+1})=r+1$ by Corollary 4.6. Consequently, $i_{\llbracket \omega \rrbracket }$ cannot be an isomorphism, which shows that there is no symplectic manifold $(B_{r,r+1},\llbracket \omega \rrbracket )$. $\square$

Remark 11.3

Theorem 11.2 is proved without computing the Karoubi-de Rham complex. It is not easy to check whether this complex is non-trivial for algebras $B_{r,r+1}$. Vanishing of the elements $\llbracket e_{i}\otimes \overline{c_{i}}\rrbracket$ in $DR^{1}_{R}(B_{r,r+1})$ is a reason that we cannot repeat the arguments in proving the non-triviality of the complex $(DR^{\bullet }_{R}(B_{r,r+1}),d)$.

Remark 11.4

In many examples of algebras, non-existence of symplectic manifolds is a consequence of trivialisation of their Karoubi-de Rham complex. This is the case for hereditary algebras or canonical algebras (see [11]).