\section{Introduction}\label{sone}
Let $A$ be a finite dimensional algebra over a field $k$ with finite global dimension.
When $A$ has indecomposable projective modules $P_i$ for $i=1,...,n$, then the \emph{Cartan matrix} $M$ of $A$ is defined as the $n \times n$-matrix with entries $m_{i,j}:=\operatorname{Hom}_A(P_i, P_j)$. The \emph{Coxeter matrix} (or also Coxeter transformation) $C$ of $A$ is then defined by $C:=-M^{-1} M^T$.
The Coxeter matrix of a finite dimensional algebra with finite global dimension is the main object of study in the spectral theory of finite dimensional algebras, we refer for example to the survey article \cite{LP}. Of special interest in homological algebra are algebras with periodic Coxeter matrix, that is $C^l=id$ for some $l \geq 1$. Algebras with periodic Coxeter matrix include for example fractionally Calabi-Yau algebras that arise in geometric and combinatorial contexts, we refer for example to \cite{CDIM} and \cite{Y}.

When $P$ is a finite poset then the set of order ideals of $P$ is a distributive lattice $L$ and every finite distributive lattice arises this way and is uniquely determined by $P$.
The Coxeter matrix of the incidence algebra $kL$ can be directly read off from elementary combinatorial data involving the M\"obius function of $L$.
When $S$ is an antichain of $P$ then set $I(S)$ to be the order ideal whose maximal elements are given by $S$ and set $M(S)$ to be the order ideal whose minimal non-elements are given by $S$. The \emph{rowmotion bijection} is given on the elements of $L$ by sending $I(S)$ to $M(S)$, which defines a bijection. The rowmotion bijection is one of the main attractions in dynamical algebraic combinatorics, we refer for example to the articles \cite{H}, \cite{Str} and \cite{TW}. When $L$ has $n$ elements, we can associate to the rowmotion bijection an $n \times n$-permutation matrix in a natural way.
While the Coxeter matrix and the rowmotion bijection are well studied objects in representation theory and combinatorics respectively, it seems that no relation between them has been shown before.

The following theorem, which is our first main result, was first conjectured by Sam Hopkins, who noted the identity in several examples:
\begin{theorem}
Let $L$ be a finite distributive lattice with $C$ the Coxeter matrix of the incidence algebra of $kL$ and $\rho$ the rowmotion bijection for $L$ viewed as a permutation matrix.
Then $\rho^{-1} C$ has order two, that is $(\rho^{-1} C)^2=id$.

\end{theorem}

In the article \cite{IM} it was noted that the rowmotion bijection for a distributive lattice $L$ is a special case of a much more general bijection that exists for any Auslander regular algebra.
Recall here that a finite dimensional algebra $A$ is called \emph{Auslander regular} if $A$ has finite global dimension and in the minimal injective coresolution 
$$0 \rightarrow A \rightarrow I_0 \rightarrow I_1 \rightarrow \cdots I_n \rightarrow 0$$
we have that the projective dimension of $I_i$ is bounded by $i$ for all $i \geq 0$.
In \cite{I}, Iyama defined the so-called \emph{grade bijection} of an Auslander
regular algebra $A$. It is a permutation of the simple modules of $A$, whose precise definition we will recall in the next section. 
Again, we can associate a permutation matrix to the grade bijection that we will usually denote by $R_A$, or $R$ for short, for an Auslander regular algebra $A$.
One of the main results in \cite{IM} states that a finite lattice $L$ is distributive if and only if the incidence algebra $kL$ is Auslander regular. 
Moreover, when $L$ is distributive the grade bijection gives a homological realisation of the rowmotion bijection, when one identifies the elements of the lattice $L$ with the simple $kL$-modules in a natural way.
This leads to the natural question, whether for other Auslander regular algebras
there is a simple relation between the Coxeter matrix and the grade bijection.


We give a positive answer for an important class of algebras that also appears in combinatorics. Namely, an algebra $A$ is called $n$-representation-finite if $A$ has global dimension at most $n$ and there is an $n$-cluster tilting object $M$ in $\mod A$, which is then uniquely given when we assume that $M$ is basic. 
The notion of $n$-representation finite algebras was introduced in \cite{Iya2} and includes many important classes of algebras such as all path algebras of Dynkin type, higher Auslander algebras of Dynkin type $\mathcal{A}$ that have strong relations to the combinatorics of cyclic  polytopes \cite{OT} and the 2-representation finite algebras that have strong relations to quivers with potential and Jacobian algebras \cite{HI}.  
For an $n$-representation finite $A$ with $n$-cluster tilting module $M$, the endomorphism algebra $B:=\operatorname{End}_A(M)$ will be an higher Auslander algebra that is thus Auslander regular. Our second main result is as follows for such algebras $B$.
\begin{theorem}
Let $A$ be an $n$-representation finite algebra with $n$-cluster tilting module $M$.
Let $B=\operatorname{End}_A(M)$ with grade bijection $R$ and Coxeter matrix $C$.
If $n$ is even then $(R^{-1}C)^2=id$ and if $n$ is odd then $(R^{-1}C+id)^2=0$.
\end{theorem}




\section{Preliminaries}
We assume that algebras are finite dimensional over a field $k$ and that they are non-semisimple and connected unless stated otherwise. We assume that modules are right modules unless otherwise stated. $D=\operatorname{Hom}_k(-,k)$ denotes the duality and $J$ denotes the Jacobson radical of an algebra. We assume that the reader is familiar with the basics of representation theory and homological algebra of finite dimensional algebras and we refer for example to the textbooks \cite{ARS} and \cite{SkoYam}.
Let $\nu=\nu_A:=D \operatorname{Hom}_A(-,A)$ denote the Nakayama functor of an algebra $A$ and $\nu^{-1}=\operatorname{Hom}_A(D(A),-)$ its inverse. It is well known that $\nu$ induces an equivalence between the category of projective $A$-modules and the category of injective $A$-modules with inverse $\nu^{-1}$.
The \emph{global dimension} of an algebra $A$ is defined as the supremum of all projective dimensions of $A$-modules. The \emph{dominant dimension} of an algebra $A$ with minimal injective coresolution 
$$0 \rightarrow A \rightarrow I_0 \rightarrow I_1 \rightarrow \cdots $$
is defined as the smallest $n \geq 0$ such that $I_n$ is not projective or as infinite if there is no such $I_n$. For an $A$-module $M$, $\operatorname{\mathrm{add}}(M)$ will denote the full subcategory of $\mod-A$ consisting of all direct summands of $M^n$ for some natural number $n$.
Let $K_0(A)$ denote the Grothendieck group of an algebra $A$ with finite global dimension with basis given by the indecomposable projective modules $[P]$. For simplicity, we will often omit the usual brackets $[P]$ for an element in the Grothendieck group $K_0(A)$ when there is no danger of confusion. When $A$ has finite global dimension, the \emph{Coxeter transformation} of $A$ is defined as $C_A([P]):=-[\nu(P)]$. When this linear transformation is expressed as a matrix with respect to the basis of $K_0(A)$ given by the classes of the indecomposable projective modules, we recover the matrix $C$ from Section \ref{sone}.
An $A$-module $M$ is called \emph{$n$-cluster-tilting} when $M$ is a generator-cogenerator and $M^{ \perp n}=\operatorname{\mathrm{add}}(M)=^{\perp n}M$, where $M^{ \perp n}= \{ N \in \mod A | \operatorname{Ext}_A^i(M,N)=0 $ for all $1 \leq i < n \}$ and $^{\perp n}M= \{ N \in \mod A | \operatorname{Ext}_A^i(N,M)=0 $ for all $1 \leq i < n \}$. A 1-cluster tilting module $M$ exists if and only if $A$ is representation-finite; in this case $\operatorname{\mathrm{add}}(M)=\mod A$.
By the higher Auslander correspondence, see for example \cite{Iya4}, $M$ is $n$-cluster tilting if and only if the algebra $B:=\operatorname{End}_A(M)$ is a \emph{higher Auslander algebra} of global dimension $n+1$, that is it has global dimension and dominant dimension equal to $n+1$.

An algebra $A$ is called \emph{Auslander regular} when $A$ has finite global dimension and there exists an injective coresolution
$$0 \rightarrow A \rightarrow I^0 \rightarrow I^1 \rightarrow \cdots \rightarrow I^n \rightarrow 0$$
of the regular module such that $\operatorname{pdim} I^i \leq i$ for all $i \geq 0$.
Important classes of Auslander regular algebras are incidence algebras of distributive lattices (see \cite{IM}), higher Auslander algebras (see \cite{Iya4}) and blocks of category $\mathcal{O}$ (see \cite{KMM}).
The \emph{grade} of an $A$-module $M$ is defined by $\operatorname{grade} M:= \inf \{ i \geq 0 \mid \operatorname{Ext}_A^i(M,A) \neq 0 \}$. Dually, the \emph{cograde} of a module $M$ is defined as $\inf \{i \geq 0 \mid \operatorname{Ext}_A^i(D(A),M) \neq 0 \}$. For every Auslander regular algebra, there exists a bijection on the simple $A$-modules as follows:
\begin{theorem} (\cite[Theorem 2.10]{I}) \label{graderowmotiontheorem}
Let $A$ be an Auslander regular algebra.
Then there is a bijection $R_A$ sending a simple $A$-module $S$ to the simple $A$-module $R_A(S)=\top(D \operatorname{Ext}_A^{g_S}(S,A))$, where $g_S:= \operatorname{grade} S$.
The grade of $S$ equals the cograde of $R_A(S)$.
\end{theorem}
We call the bijection $R_A$ as in the previous theorem for Auslander regular algebras the \emph{grade bijection}.
We refer to \cite[Theorem 2.10]{I} for a more general statement and proofs. 
In \cite{IM} it was shown that the grade bijection gives a categorification of the rowmotion map for incidence algebras of distributive lattices.
We define the grade bijection on the Grothendieck group of an Auslander regular algebra $A$ by $R_A([P_S])=[P_{R_A(S)}]$, where $P_S$ denotes the projective cover of a simple module $S$. 
We define the \emph{rowmotion Coxeter transformation} of an Auslander regular algebra $A$ as $R_A^{-1} C_A$.  Note that since $R_A$ is a permutation matrix, we have $R_A^{-1}=R_A^T$. We will be mainly interested in the minimal polynomial of $R_A^{-1} C_A$ and the relation $(R_A^{-1} C_A)^2=id$. Note that since $  C_A R_A^{-1}= R_A (R_A^{-1} C_A) R_A^{-1}$, the operators $C_A R_A^{-1}$ and $R_A^{-1} C_A$ are similar and thus they have the same minimal polynomials and we have $(C_A R_A^{-1})^2=id$ if and only if $(R_A^{-1} C_A)^2=id$.
For this reason we will also look at the operator $C_A R_A^{-1}$ instead of $R_A^{-1} C_A$ in certain situations.

We denote by $\tau_n:=\tau \Omega^{n-1}$ for $n \geq 1$ the \emph{$n$-Auslander-Reiten translate} and by $\tau_n^{-1}:= \Omega^{-(n-1)}\tau^{-1}$ the \emph{inverse $n$-Auslander-Reiten translate}.




\section{The Coxeter transformation and the rowmotion bijection for distributive lattices}
Let $P$ denote a finite poset and $L$ the distributive lattice of order ideals of $P$.
Let $S$ be an antichain of $P$ and $I(S)$ the order ideal whose maximal elements are given by $S$ and $M(S)$ the order ideal whose minimal non-elements are given by $S$.
The \emph{Hasse quiver} of a poset $P$ is the finite quiver with points $x \in P$ and an arrow $x \rightarrow y$ if $x$ covers $y$ (note that we use here the opposite convention compared to \cite{IM}). The \emph{incidence algebra} $kP$ of a poset $P$ over a field $k$ is defined as the quiver algebra $kQ/\mathcal I$ with $Q$ the Hasse quiver of $P$ and the relations $\mathcal I$ such that any two paths that start and end at the same points get identified. We multiply arrows in $Q$ as paths rather than as functions, i.e., if we write $p_x^y$ for the element of $kP$ corresponding to elements $x\leq y$, then for $x\leq y\leq z$, we have
$p_x^yp_y^z=p_x^z$. Thus, right modules for $kP$ define representations of the
Hasse quiver of $P$.

The \emph{rowmotion bijection} $\rho$ for a distributive lattice $L$ given as the set of order ideals of a poset $P$ is defined as the permutation sending $I(S)$ to $M(S)$.
As explained in the preliminaries, the rowmotion bijection is a special case of the more general grade bijection when viewing the incidence algebra of a distributive lattice as an Auslander regular algebra. We will denote by $P_{O}$ the indecomposable projective module in the incidence algebra $kL$ of a distributive lattice $L$ corresponding to the order ideal $O$ and $J_O:=\nu(P_O)$ (we use $J$ instead of $I$ for the indecomposable injectives in this section to avoid confusion with $I(S)$). When $A=kL$ is the incidence algebra of a distributive lattice, we set $R_A([P_{I(S)}])=[P_{M(S)}]$ motivated by the fact that the grade bijection gives a homological realisation of the rowmotion bijection as explained after Theorem \ref{graderowmotiontheorem}.
We will use the following result, whose proof can be found in \cite[Theorem 3.2]{IM} and where one can also find a description of the differentials. See also \cite[Definition 2.3]{Y}, where this construction was anticipated, though restricted to a particular case.
\begin{theorem}\label{resn}
Let $L$ be a distributive lattice given as the set of order ideals of a poset $P$.
Then a minimal projective resolution of the indecomposable injective module $J_{I(S)}$ for an antichain $S$ is given as follows in $kL$:
$$0 \rightarrow P_{M(S)} \rightarrow \bigoplus_{T \subseteq S ,|T|=r-1}^{}{P_{M(T)}} \rightarrow \cdots \rightarrow \bigoplus_{T \subseteq S ,|T|=i}^{}{P_{M(T)}} \rightarrow \cdots \rightarrow \bigoplus_{T \subseteq S ,|T|=1}^{}{P_{M(T)}} \rightarrow P_{M(\emptyset)} \rightarrow J_{I(S)} \rightarrow 0.$$

\end{theorem}

\begin{example} For an example of the previous theorem, consider the case where $P$ is a poset consisting of three incomparable elements $\{x_1,x_2,x_3\}$. The lattice $L$ is
  then a Boolean lattice consisting of the eight subsets of $\{x_1,x_2,x_3\}$,
  all of which
  are order ideals of $P$. The Hasse diagram of $L$ is given by:

  $$\begin{tikzpicture}[stealth-,xscale=1.3] \node(a) at (0,0) {$\emptyset$};
    \node (b) at (-1,1) {$\{x_1\}$};
    \node (c) at (0,1) {$\{x_2\}$};
    \node (d) at (1,1) {$\{x_3\}$};
    \node (e) at (-1,2) {$\{x_1,x_2\}$};
    \node (f) at (0,2) {$\{x_1,x_3\}$};
    \node (g) at (1,2) {$\{x_2,x_3\}$};
    \node (h) at (0,3) {$\{x_1,x_2,x_3\}$};
    \draw (a) -- (b);\draw (a)--(c);\draw (a)--(d);\draw (b)--(e);\draw (b)--(f);\draw (c)--(e);\draw (c)--(g);\draw
    (d)--(f);\draw (d)--(g);\draw (e)--(h);\draw (f)--(h);\draw (g) --(h);
    \end{tikzpicture}$$


  Let $S=\{x_1,x_2\}$. The projective resolution of $J_{I(S)}$ given by Theorem
  \ref{resn} is as follows, where we represent $kL$ modules by the corresponding representations of the Hasse quiver.

  $$ 0 \rightarrow \vcenter{\hbox{\begin{tikzpicture}[stealth-,scale=.7] \node(a) at (0,0) {$k$};
    \node (b) at (-1,1) {$0$};
    \node (c) at (0,1) {$0$};
    \node (d) at (1,1) {$k$};
    \node (e) at (-1,2) {$0$};
    \node (f) at (0,2) {$0$};
    \node (g) at (1,2) {$0$};
    \node (h) at (0,3) {$0$};
    \draw (a) -- (b);\draw (a)--(c);\draw (a)--(d);\draw (b)--(e);\draw (b)--(f);\draw (c)--(e);\draw (c)--(g);\draw
    (d)--(f);\draw (d)--(g);\draw (e)--(h);\draw (f)--(h);\draw (g) --(h);
    \end{tikzpicture}}}
 \rightarrow 
\vcenter{\hbox{  \begin{tikzpicture}[stealth-,scale=.7] \node(a) at (0,0) {$k$};
    \node (b) at (-1,1) {$k$};
    \node (c) at (0,1) {$0$};
    \node (d) at (1,1) {$k$};
    \node (e) at (-1,2) {$0$};
    \node (f) at (0,2) {$k$};
    \node (g) at (1,2) {$0$};
    \node (h) at (0,3) {$0$};
    \draw (a) -- (b);\draw (a)--(c);\draw (a)--(d);\draw (b)--(e);\draw (b)--(f);\draw (c)--(e);\draw (c)--(g);\draw
    (d)--(f);\draw (d)--(g);\draw (e)--(h);\draw (f)--(h);\draw (g) --(h);
  \end{tikzpicture}}}
  \oplus
\vcenter{\hbox{  \begin{tikzpicture}[stealth-,scale=.7] \node(a) at (0,0) {$k$};
    \node (b) at (-1,1) {$0$};
    \node (c) at (0,1) {$k$};
    \node (d) at (1,1) {$k$};
    \node (e) at (-1,2) {$0$};
    \node (f) at (0,2) {$0$};
    \node (g) at (1,2) {$k$};
    \node (h) at (0,3) {$0$};
    \draw (a) -- (b);\draw (a)--(c);\draw (a)--(d);\draw (b)--(e);\draw (b)--(f);\draw (c)--(e);\draw (c)--(g);\draw
    (d)--(f);\draw (d)--(g);\draw (e)--(h);\draw (f)--(h);\draw (g) --(h);
  \end{tikzpicture}}}
  \rightarrow
\vcenter{\hbox{   \begin{tikzpicture}[stealth-,scale=.7] \node(a) at (0,0) {$k$};
    \node (b) at (-1,1) {$k$};
    \node (c) at (0,1) {$k$};
    \node (d) at (1,1) {$k$};
    \node (e) at (-1,2) {$k$};
    \node (f) at (0,2) {$k$};
    \node (g) at (1,2) {$k$};
    \node (h) at (0,3) {$k$};
    \draw (a) -- (b);\draw (a)--(c);\draw (a)--(d);\draw (b)--(e);\draw (b)--(f);\draw (c)--(e);\draw (c)--(g);\draw
    (d)--(f);\draw (d)--(g);\draw (e)--(h);\draw (f)--(h);\draw (g) --(h);
   \end{tikzpicture}}}
   \rightarrow
 \vcenter{\hbox{   \begin{tikzpicture}[latex-,scale=.7] \node(a) at (0,0) {$0$};
    \node (b) at (-1,1) {$0$};
    \node (c) at (0,1) {$0$};
    \node (d) at (1,1) {$0$};
    \node (e) at (-1,2) {$k$};
    \node (f) at (0,2) {$0$};
    \node (g) at (1,2) {$0$};
    \node (h) at (0,3) {$k$};
    \draw (a) -- (b);\draw (a)--(c);\draw (a)--(d);\draw (b)--(e);\draw (b)--(f);\draw (c)--(e);\draw (c)--(g);\draw
    (d)--(f);\draw (d)--(g);\draw (e)--(h);\draw (f)--(h);\draw (g) --(h);
    \end{tikzpicture}}}
    \rightarrow 0$$
\end{example}
  
\begin{corollary}
Let $L$ be a distributive lattice given as the set of order ideals of a poset $P$.
Let $S$ denote an antichain of $P$.
Then the Coxeter transformation $C_A$ for $A=kL$ is given by 
$$C_A( [P_{I(S)}])=-\sum\limits_{T \subseteq S}^{}{ (-1)^{|T|} [ P_{M(T)}}].$$
\end{corollary}


\begin{theorem}
Let $A=kL$ be the incidence algebra of a distributive lattice $L$ given as the set of order ideals of a poset $P$.
Then $(R_A^{-1} C_A)^2=id$.

\end{theorem}
\begin{proof}
We first look at $C_A R_A^{-1} C_A$ and determine the values on a basis:
$$ C_A R_A^{-1} C_A ([P_{I(S)}])= C_A R_A^{-1} (-\sum\limits_{T \subseteq S}^{}{ (-1)^{|T|} [ P_{M(T)}}])$$
$$= C_A ( -\sum\limits_{T \subseteq S}^{}{ (-1)^{|T|} [ P_{I(T)}}])$$
$$= \sum\limits_{T \subseteq S}^{}{ \sum\limits_{R \subseteq T}^{}{ (-1)^{|T|} (-1)^{|R|} [P_{M(T)}]}}.$$
Now we consider all ways to obtain a particular $R$ as in the sum.
If $R$ is properly contained in $S$, there are $2^{|S|-|R|}$ choices for $T$, half of even cardinality and half of odd cardinality.
These terms therefore cancel. The only surviving term is $R=S$, in which case we also have $T=S$.
The result is that $C_A R_A^{-1} C_A([P_{I(S)}])=[P_{M(S)}]$ and if we now apply $R_A^{-1}$ to both sides, we get that $R_A^{-1} C_A R_A^{-1} C_A ([P_{I(S)}])=[P_{I(S)}]$, as desired.




\end{proof}



The next example shows that the identity $(R^{-1}C)^2=id$, which is true for distributive lattices, does not hold for Auslander regular incidence algebras of posets that are not lattices:
\begin{example} \label{examplenonlattice}
The following poset with incidence algebra $A$ is Auslander regular:
\begin{center}
\begin{tikzpicture}[>=latex,line join=bevel,]
\node (node_0) at (21.0bp,6.5bp) [draw,draw=none] {$6$};
  \node (node_1) at (6.0bp,55.5bp) [draw,draw=none] {$4$};
  \node (node_2) at (36.0bp,55.5bp) [draw,draw=none] {$5$};
  \node (node_3) at (6.0bp,104.5bp) [draw,draw=none] {$2$};
  \node (node_4) at (36.0bp,104.5bp) [draw,draw=none] {$3$};
  \node (node_5) at (21.0bp,153.5bp) [draw,draw=none] {$1$};
  \draw [black,<-] (node_0) ..controls (17.092bp,19.746bp) and (13.649bp,30.534bp)  .. (node_1);
  \draw [black,<-] (node_0) ..controls (24.908bp,19.746bp) and (28.351bp,30.534bp)  .. (node_2);
  \draw [black,<-] (node_1) ..controls (6.0bp,68.603bp) and (6.0bp,79.062bp)  .. (node_3);
  \draw [black,<-] (node_1) ..controls (13.953bp,68.96bp) and (21.16bp,80.25bp)  .. (node_4);
  \draw [black,<-] (node_2) ..controls (28.047bp,68.96bp) and (20.84bp,80.25bp)  .. (node_3);
  \draw [black,<-] (node_2) ..controls (36.0bp,68.603bp) and (36.0bp,79.062bp)  .. (node_4);
  \draw [black,<-] (node_3) ..controls (9.9083bp,117.75bp) and (13.351bp,128.53bp)  .. (node_5);
  \draw [black,<-] (node_4) ..controls (32.092bp,117.75bp) and (28.649bp,128.53bp)  .. (node_5);
\end{tikzpicture}
\end{center}
Let $N_i$ denote the indecomposable projective $A$-modules.
For $R_A$ and $C_A$ we obtain the values as in the following tables.
We emphasize that $R_A$ here is given by the grade bijection. Currently, no purely combinatorial description of this bijection is available for Auslander regular incidence algebras of posets except in the case that the poset is a distributive lattice. Such a combinatorial description would be a generalization of rowmotion on distributive lattices, and would be very interesting to have.
\begin{center}
\begin{table}[h!]
  \begin{center}
    \caption{}
    \label{tab:table1}
    \begin{tabular}{l|c|r|c|c|c|c} 
      $R_A$ & $N_1$ & $N_2$ & $N_3$ & $N_4$ & $N_5$ & $N_6$ \\
      \hline
      $N_1$ & 0 & 0& 0&0&0&1\\
      $N_2$ & 0 & 0& 1&0&0&0\\
      $N_3$ & 0 & 1& 0&0&0&0\\
      $N_4$ & 0 & 0& 0&0&1&0\\
      $N_5$ & 0 & 0& 0&1&0&0\\
      $N_6$ & 1 & 0& 0&0&0&0\\
    \end{tabular}
  \end{center}
\end{table} 
\begin{table}[h!]
  \begin{center}
    \caption{}
    \label{tab:table2}
    \begin{tabular}{l|c|r|c|c|c|c}
      $C_A$ & $N_1$ & $N_2$ & $N_3$ & $N_4$ & $N_5$ & $N_6$ \\
      \hline
      $N_1$ & -1 & -1& -1&-1&-1&-1\\
      $N_2$ & 1 & 0& 1&0&0&0\\
      $N_3$ & 1 & 1& 0&0&0&0\\
      $N_4$ & -1 & 0& 0&0&1&0\\
      $N_5$ & -1 &0& 0&1&0&0\\
      $N_6$ & 1 & 0& 0&0&0&0\\
     
     
     
     
     
     
    \end{tabular}
  \end{center}
\end{table} 
\end{center}
The matrix $R_A^{-1} C_A $ has minimal polynomial $x^3-x^2-x+1$ and thus the identity $(R_A^{-1} C_A)^2=id$ is not true for this poset.

\end{example}


\section{$n$-representation finite algebras}

Let $A$ be an algebra and let $M$ be a generator-cogenerator of $\mod A$ and $B:=\operatorname{End}_A(M)$. We will summarize several properties of $B$ in this situation in the following and refer for example to \cite[Chapter VI.5]{ARS} for more information. There is an equivalence of categories $\operatorname{\mathrm{add}}(M) \cong \operatorname{proj} B$, given by $\operatorname{Hom}_A(M,-)$. We denote the indecomposable projective $B$-modules associated to an indecomposable $A$-module $N \in \operatorname{\mathrm{add}}(M)$ by $L_N:=\operatorname{Hom}_A(M,N)$. Note that $\nu(L_N)=D \operatorname{Hom}_B(L_N,B)=D \operatorname{Hom}_B(\operatorname{Hom}_A(M,N), \operatorname{Hom}_A(M,M)) \cong D \operatorname{Hom}_A(N,M)$ and thus the indecomposable injective $B$-modules are given by $T_N := D \operatorname{Hom}_A(N,M)$ for an indecomposable module $N \in \operatorname{\mathrm{add}}(M)$.
We denote by $S_N$ the simple $B$-module with projective cover $\operatorname{Hom}_A(M,N)$ and dually $U_N=D(S_N)$ the left simple $B$-module. 

An algebra $A$ is called \emph{$n$-representation-finite} for some $n \geq 1$ if $\operatorname{gldim} A \leq n$ and there is an $n$-cluster tilting module $M$ in $\mod A$, see for example \cite{Iya2} where such algebras were studied systematically for the first time.
Note that an $n$-cluster tilting module $M$ in a $n$-representation-finite algebra is unique when we assume that $M$ is basic. Note also that $M$
necessarily contains the indecomposable projective and injective modules as
direct summands. 
$1$-representation-finite algebras are exactly the representation finite hereditary algebras that are classified by Dynkin diagrams.

We collect several results that we will need. For a survey on $n$-cluster tilting categories and higher Auslander algebras, we refer to section 2 of \cite{Iya5}. For the definition
of $n$-almost split sequences and their basic properties we refer to section 2.3 in \cite{Iya5}, which also contains the next lemma.
\begin{lemma}
  Let $A$ be an algebra with $n$-cluster-tilting module $M$ and $B:=\operatorname{End}_A(M)$.
  \begin{enumerate} \item
Let $N$ be an indecomposable non-projective
summand of $M$. There is therefore an $n$-almost split sequence
$$ 0 \rightarrow \tau_n(N) \rightarrow \dots \rightarrow N \rightarrow 0.$$
Applying $\operatorname{Hom}_A(M,-)$ to it yields a minimal projective resolution of $S_N$:
\begin{align}\label{SPres}
  0\rightarrow \operatorname{Hom}(M,\tau_n(N))\rightarrow \cdots \rightarrow \operatorname{Hom}(M,N)
  \rightarrow S_N\rightarrow 0.\end{align}
\item Let $N$ be an indecomposable non-injective summand of $M$. There
  is an $n$-almost split sequence
  $$ 0 \rightarrow N \rightarrow \dots \rightarrow \tau_n^{-1}(N)
  \rightarrow 0.$$
  Applying $D\operatorname{Hom}_A(-,M)$, we obtain a minimal injective coresolution of
  $S_N$:
  \begin{align}\label{SIres}
  0\rightarrow S_N \rightarrow D\operatorname{Hom}(N,M)
  \rightarrow \cdots \rightarrow D\operatorname{Hom}(\tau_n^{-1}(N),M)\rightarrow 0.\end{align}
\end{enumerate}
  \end{lemma}

  As an immediate consequence of this lemma, we deduce that if $N$ is
  an indecomposable summand of $M$ which is not projective, then the
  projective dimension of $S_N$ is $n+1$; similarly, if $N$ is not
  injective, then the injective dimension of $S_N$ is $n+1$. 



\begin{lemma} \label{higherausreiformulas}
Let $A$ be an $n$-representation-finite algebra with $n$-cluster tilting module $M$. Let $X,Y \in \operatorname{\mathrm{add}}(M)$.
\begin{enumerate}
\item $\operatorname{Hom}_A(\tau_n^{-1}(Y),X)=D \operatorname{Ext}_A^n(X,Y)$.
\item $\operatorname{Hom}_A(Y,\tau_n(X))=D \operatorname{Ext}_A^n(X,Y)$.
\end{enumerate}
\end{lemma}
\begin{proof}
We show (1), the proof of (2) is dual. 
We have in general that $\underline{\operatorname{Hom}}_A(\tau_n^{-1}(Y),X)=D \operatorname{Ext}_A^n(X,Y)$ for any $X,Y \in \operatorname{\mathrm{add}}(M)$ for a cluster tilting module $M$ (without the assumption that $A$ has global dimension at most $n$), by \cite[Theorem 2.3.1]{Iya1}. 
Now the assumption that $A$ has additionally global dimension at most $n$ gives us that $\underline{\operatorname{Hom}}_A(\tau_n^{-1}(Y),X)=\operatorname{Hom}_A(\tau_n^{-1}(Y),X)$ by \cite[Lemma 2.4(d)]{Iya2}.
\end{proof}




\begin{lemma}
Let $N$ be an $A$-module of finite projective dimension.
Then $\operatorname{pdim} N= \sup \{ i \geq 0 \mid \operatorname{Ext}_A^i(N,A) \neq 0 \}$.
\end{lemma}
\begin{proof}
See for example \cite[Lemma VI.5.5]{ARS}.
\end{proof}

Given this lemma, it follows from the definition of $n$-representation finite algebra
that the
summands of $M$ are either projective or of projective
dimension $n$.

\begin{lemma} \label{big} Let $A$ be an algebra with $n$-cluster-tilting module $M$ and $B:=\operatorname{End}_A(M)$.
  \begin{enumerate}
    \item \label{point-one} Let $N$ be an indecomposable projective summand of $M$. Then $T_N$ is projective as a $B$-module. 
  \item
Let $N$ be an indecomposable non-projective
summand of $M$. Suppose that its minimal projective resolution is
$$0\rightarrow P_n \rightarrow \cdots \rightarrow P_1 \rightarrow P_0 \rightarrow N \rightarrow 0.$$
Then the projective resolution as a $B$-module of $T_N$ is given by:
\begin{align}\label{Bres} 0 \rightarrow \operatorname{Hom}_A(M,\tau_n(N)) \rightarrow D \operatorname{Hom}_A(P_n,M) \rightarrow \cdots \rightarrow D\operatorname{Hom}_A(P_0,M) \rightarrow T_N \rightarrow 0.\end{align}
\item Let $N'$ be an injective indecomposable direct summand of $M$. Then
  $L_{N'}$ is itself injective, so it is its own injective resolution.
\item Let $N'$ be a non-injective indecomposable direct summand of $M$.
  Let $N=\tau_n^{-1}(N')$. Then 
  the leftmost term of (\ref{Bres}) is $L_{N'}$, while the remaining terms are
  injective, so (\ref{Bres}) is the injective coresolution of $L_{N'}$.
\item \label{bij} There is a bijection between the indecomposable projective $B$-modules of injective dimension $n+1$ and the indecomposable injective $B$-modules of projective dimension $n+1$ given by $\Omega^{-(n+1)}$ with inverse $\Omega^{n+1}$.
  \end{enumerate}
\end{lemma}

  \begin{proof}\begin{enumerate}
      \item Since $N$ is projective, $T_N=D \operatorname{Hom}_B(N,M) \cong \operatorname{Hom}(M, \nu(N)) = L_{\nu(N)}$ by \cite[Chapter III, Corollary 6.2]{SkoYam}, so $T_N$ is itself projective.
      
      \item Since $N$ is non-projective, we have already established that
        $\operatorname{pdim} N=n$. 
In order to calculate $\operatorname{Ext}^i(N,M)$, we would apply $\operatorname{Hom}(-,M)$ to the projective resolution of $N$, and since
$\operatorname{Ext}^i(N,M)=0$ except for $i=0$ and $i=n$, we obtain an exact sequence:
\begin{align} \label{exactsequence}
0 \rightarrow \operatorname{Hom}_A(N,M) \rightarrow \operatorname{Hom}_A(P_0,M) \rightarrow \operatorname{Hom}_A(P_1,M) \rightarrow \cdots \rightarrow \operatorname{Hom}_A(P_n,M) \rightarrow \operatorname{Ext}_A^n(N,M) \rightarrow 0.
\end{align}
Now by the higher Auslander-Reiten formulas in Lemma \ref{higherausreiformulas}, we have
$\operatorname{Ext}_A^n(N,M) \cong D \operatorname{Hom}_A(M, \tau_n(N))$. Note also that
$D\operatorname{Hom}_A(N,M) =: T_N$. Thus, applying the duality $D$ to (\ref{exactsequence}) we obtain the exactness of (\ref{Bres}). It is a projective resolution
since the terms $D\operatorname{Hom}_A(P_i,M)$ are projective-injective.

\item This is dual to (\ref{point-one}).

\item This is immediate.

\item The indecomposable projective $B$-modules which are not injective are the
  modules $L_N$ for $N$ an indecomposable direct summand of $M$ which is not
  injective. By (\ref{Bres}), we see that $\Omega^{-(n+1)} L_N=T_{\tau_n(N)}$.
  The modules of the form $T_{\tau_n(N)}$ are exactly the indecomposable
  injective $B$-modules which are not projective, and (\ref{Bres}) shows that
  $\Omega^{n+1} T_{\tau_n(N)}=L_N$.
  \end{enumerate}
  \end{proof}



\begin{proposition} \label{propograde}
Let $A$ be an algebra with $n$-cluster-tilting module $M$ and $B:=\operatorname{End}_A(M)$.
Let $N \in \operatorname{\mathrm{add}}(M)$ be an indecomposable module with corresponding simple $B$-module $S_N$. Then we have $\operatorname{grade} S_N=0$ if $N$ is projective and 
$\operatorname{grade} S_N=n+1$ if $N$ is non-projective.
\end{proposition}
\begin{proof}
 
Suppose $N$ is projective, so $D\operatorname{Hom}_A(N,M)$ is projective.
  $S_N$ injects
into $\nu(L_N)=D\operatorname{Hom}_A(N,M)$. Thus $\operatorname{Hom}_B(S_N,B)\ne 0$, so $\operatorname{grade} S_N=0$.

Suppose now that $N$ is not projective.
To calculate $\operatorname{Ext}_B^i(S_N,B)$, we
apply $\operatorname{Hom}_B(-,B)$ to the resolution in (\ref{SPres}). 
Because $\operatorname{Hom}_B(\operatorname{Hom}_A(X,M),B)=\operatorname{Hom}_B(\operatorname{Hom}_A(X,M),\operatorname{Hom}_A(M,M))=\operatorname{Hom}(M,X)$, 
we obtain that $\operatorname{Ext}^i_B(S_N,B)$ is isomorphic to the $i$-th homology of
$$ 0\rightarrow \operatorname{Hom}(N,M) \rightarrow \dots \rightarrow \operatorname{Hom}(\tau_n(N),M)\rightarrow 0$$
This is dual to the injective coresolution of $S_{\tau_n(N)}$ obtained from
(\ref{SIres}),
so it is exact except in the position corresponding to
$\operatorname{Ext}^{n+1}_B(S_N,B)$, and we see that $\operatorname{grade} S_N = n+1$.
\end{proof}









In the next theorem we calculate the Coxeter transformation of higher Auslander algebras coming from $n$-cluster tilting modules in $n$-representation-finite algebras.
\begin{theorem} \label{coxetercalc}
Let $A$ be a $n$-representation-finite algebra with $n$-cluster tilting module $M$ and indecomposable $A$-module $N \in \operatorname{\mathrm{add}}(M)$.
\begin{enumerate}
\item If $N$ is projective, $C([L_N])= - [L_{\nu(N)}]$.
\item If $N$ is non-projective, let $0 \rightarrow P_n\rightarrow \cdots \rightarrow P_1 \rightarrow P_0 \rightarrow N \rightarrow 0$ be a minimal projective resolution of $N$.
Then $C([L_N])= \sum\limits_{i=0}^{n}{(-1)^{i+1} [L_{\nu(P_i)}]}+(-1)^n[L_{\tau_n(N)}]$.

\end{enumerate}

\end{theorem}
\begin{proof}
  By definition $C_B([L_N])= -[T_N]$. To express $[T_N]$ with respect to the basis of
  indecomposable projective $B$-modules, it suffices to calculate a projective
  resolution of $T_N$ and take the alternating sum. We have
  found these projective resolutions in Lemma \ref{big}, and the claim follows.
  \end{proof}


We now calculate the grade bijection for higher Auslander algebras that are endomorphism rings of an $n$-cluster tilting module.

\begin{theorem}
Let $A$ be an algebra with $n$-cluster tilting module $M$ and $B:=\operatorname{End}_A(M)$. Let $N \in \operatorname{\mathrm{add}}(M)$.
Then the grade bijection $R$ for $B$ is given by $R(L_N)=L_{\nu(N)}$ when $N$ is projective indecomposable and $R(L_N)=L_{\tau_n(N)}$ else.

\end{theorem} 
\begin{proof}
First assume that $N$ is projective indecomposable. By Proposition \ref{propograde}, $\operatorname{grade} S_N=0$ and thus we have to calculate the simple $B$-module $\top( D\operatorname{Hom}_B(S_N,B))$.
By definition we have the exact sequence
$$\operatorname{Hom}_A(M,N) \rightarrow S_N \rightarrow 0.$$
Applying the functor $\operatorname{Hom}_B(-,B)$ to it gives:
$$0 \rightarrow \operatorname{Hom}_B(S_N,B) \rightarrow \operatorname{Hom}_B(\operatorname{Hom}_A(M,N) , \operatorname{Hom}_A(M,M)).$$
Now  $\operatorname{Hom}_B(\operatorname{Hom}_A(M,N) , \operatorname{Hom}_A(M,M)) \cong \operatorname{Hom}_A(N,M)$ and applying $D$ to the previous exact sequence gives:
$$D \operatorname{Hom}_A(N,M) \rightarrow D \operatorname{Hom}_A(S_N,B) \rightarrow 0.$$
Now the indecomposable injective $B$-module $D \operatorname{Hom}_A(N,M)$ is also projective since $N$ is projective by assumption.
We have then $T_N=D \operatorname{Hom}_A(N,M) \cong \operatorname{Hom}_A(M, \nu(N)) = L_{\nu(N)}$.
Thus $\top ( D\operatorname{Hom}_B(S_N,B))=S_{\nu(N)}$.

Now assume that $N$ is non-projective indecomposable.
Then $\operatorname{grade} S_N=n+1$ by Proposition \ref{propograde} we have to calculate $\top( D \operatorname{Ext}_B^{n+1}(S_N,B))$
Let $I^i(B)$ denote the terms of a minimal injective coresolution of the regular module $B$ for $i=0,1,...,n+1.$
Note that by Lemma \ref{big} (\ref{bij}) we have $\dim(\operatorname{Ext}_B^{n+1}(S_N,B))=1$ since every indecomposable injective module that is a summand of $I^{n+1}(B)$ appears exactly once. 
Thus $\operatorname{Ext}_B^{n+1}(S_N,B)=\top(\operatorname{Ext}_B^{n+1}(S_N,B))=\operatorname{\mathrm{soc}}( \operatorname{Ext}_B^{n+1}(S_N,B))$.
From (\ref{SPres}), we know a minimal projective resolution of $S_N$ is given by 
$$0 \rightarrow \operatorname{Hom}_A(M,\tau_n(N)) \rightarrow \cdots \rightarrow \operatorname{Hom}_A(M,N)\rightarrow S_N \rightarrow 0.$$
Applying the functor $D\operatorname{Hom}_B(-,B)$ and remembering that $\operatorname{grade} S_N =n+1$ we obtain the exact sequence:
$$0\rightarrow D\operatorname{Ext}_B^{n+1}(S_N,B) \rightarrow D\operatorname{Hom}(\tau_n(N),M) \rightarrow
\cdots\rightarrow D\operatorname{Hom}(N,M) \rightarrow 0.$$
This is an injective resolution of $D\operatorname{Ext}_B^{n+1}(S_N,B)$. The socle of
$D\operatorname{Hom}(\tau_n(N),M)$ is $S_{\tau_n(N)}$, so this must also be the socle of
$D\operatorname{Ext}^{n+1}_B(S_N,B)$. What we wanted to calculate was the top of
$D\operatorname{Ext}^{n+1}_B(S_N,B)$. However, since $\operatorname{Ext}^{n+1}_B(S_N,B)$ is
one-dimensional, the two coincide, and the top of $D\operatorname{Ext}^{n+1}_B(S_N,B)$ is
$S_{\tau_n(N)}$, as desired.
\end{proof}


\begin{theorem} \label{maintheoremstrongnrepfin}
Let $A$ be an $n$-representation-finite algebra with $n$-cluster-tilting module $M$ and $B:=\operatorname{End}_A(M)$.
Then we have $(C_B R_B^{-1})^2=id $ if $n$ is even and $(C_B R_B^{-1}+id)^2=0$ when $n$ is odd.

\end{theorem} 
\begin{proof}
If $N$ is injective indecomposable then we have $C_B R_B^{-1}([L_N])=C_B([L_{\nu^{-1}(N)}])=-[L_N]$, since $\nu^{-1}(N)$ is projective.
If $N$ is indecomposable and not injective we have:
$C_B R_B^{-1}([L_N])=C_B([L_{\tau_n^{-1}(N)}])= \sum\limits_{i=0}^{n}{(-1)^{i+1} [L_{\nu(P_i)}]}+(-1)^n[L_N]$ when $P_i$ denotes a minimal projective resolution of $\tau_n^{-1}(N)$.

Now assume that $n$ is even.
Then $(C_B R_B^{-1})^2([L_N])=C_B R_B^{-1}(-[L_N])=[L_N]$ when $N$ is indecomposable injective and when $N$ is non-injective we obtain: 

\begin{eqnarray*}
  (C_B R_B^{-1})^2([L_N])&=&C_B R_B^{-1}\left( \sum\limits_{i=0}^{n}{(-1)^{i+1} [L_{\nu(P_i)}]}+(-1)^n[L_{N}]\right)\\
  &=& C_B \left( \sum\limits_{i=0}^{n}{(-1)^{i+1} [L_{P_i}]} + (-1)^n [L_{\tau_n^{-1}(N)}]\right)\\

  &=&-\sum\limits_{i=0}^{n}{(-1)^{i+1} [L_{\nu(P_i)}]} + (-1)^n\left(\sum\limits_{i=0}^{n}{(-1)^{i+1} [L_{\nu(P_i)}]}+(-1)^n[L_{N}]\right)\\
  &=&[L_N].\end{eqnarray*}
Thus $(C_B R_B^{-1})^2=id$.

Now assume that $n$ is odd.
When $N$ is injective we obtain $(C_B R_B^{-1}+id)([L_N])=-[L_N]+[L_N]=0$.
When $N$ is non-injective we have 
$$(C_B R_B^{-1}+id)([L_N])=  \sum\limits_{i=0}^{n}{(-1)^{i+1} [L_{\nu(P_i)}]}+(-1)^n[L_{N}] + [L_N]= \sum\limits_{i=0}^{n}{(-1)^{i+1} [L_{\nu(P_i)}]}.$$
Thus 
$$(C_B R_B^{-1}+id)^2([L_N])=(C_B R_B^{-1}+id) \left(\sum\limits_{i=0}^{n}{(-1)^{i+1} [L_{\nu(P_i)}}]\right)=0,$$
 since we saw that $(C_B R_B^{-1}+id)([L_N])=0$ when $N$ is injective and $\nu(P_i)$ is always injective. Thus $(C_B R_B^{-1}+id)^2=0$.
\end{proof}

Specialising to path algebras of Dynkin type gives the following special case:

\begin{example}
Let $A=KQ$ be a path algebra of Dynkin type and $M$ the direct sum of all indecomposable $A$-modules, which is a 1-cluster tilting module. Let $B:=\operatorname{End}_A(M)$, which is the Auslander algebra of $A$.
By Theorem \ref{maintheoremstrongnrepfin}, we have $(C_B R_B^{-1}+id)^2=0$.

\end{example}

\subsection*{Acknowledgements}
The authors would like to express their gratitude
to Sam Hopkins for communicating his conjecture
to us. This project began at a workshop on Dynamical Algebraic Combinatorics
hosted (online) by the Banff International
Research Station in 2020.
The authors are grateful to BIRS and to the organizers of
the workshop, who provided the opportunity for the project to begin.

Ren\'e Marczinzik is funded by the DFG with the project number 428999796.
Hugh Thomas is partially supported by an NSERC Discover Grant RGPIN-2016-04872
and the Canada Research Chairs program, grant number 230635.
Part of the work on this paper was undertaken during Emine Y\i{}ld\i{}r\i{}m's tenure as INI-Simons Post Doctoral Research Fellow hosted by the Isaac Newton Institute for Mathematical Sciences (INI) participating in programme `Cluster algebras and representation theory', and by the Department of Pure Mathematics and Mathematical Statistics (DPMMS) at the University of Cambridge.   This author would like to thank INI and DPMMS for support and hospitality during this fellowship, which was supported by Simons Foundation (Award ID 316017) and by EPSRC (grant number EP/R014604/1). 


