2.5. Gröbner basis

• Degree-\(2\) terms: From the non-commutative Buchberger theory [11], the set of generators of the defining ideal of \(\overline{FK}_{C_n}(n)\) in (2.2) establish the degree \(2\) elements of the Gröbner basis.

• Degree-\(3\) terms: To find the degree-\(3\) terms, we need to reduce the \(S\)-polynomials between any two pairs of degree-\(2\) terms with respect to lower degree terms.

  We observe that the \(S\)-polynomial between two monomials is always zero. Moreover, the only \(S\)-polynomials that reduce to non-zero elements for Gröbner basis are:

  1. \(S(\underline{x_{m,m+1}x_{1n}}-x_{1n}x_{m,m+1}, x_{1n}x_{12}, 3)\),
     

  2. \(S(\underline{x_{1,n}x_{12}}, x_{m,m+1}x_{l,l+1}-x_{l,l+1}x_{m,m+1}, 3)\).

  Item \(1\) reduces to: \[A=\{ x_{1n}x_{m,m+1}x_{12},~ 3\leq m\leq n-2\},\] The range of \(m\) is due to the fact that if \(m\) is less than or equal to \(2\), it would be divisible by the relation \(x_{m+1,m+2}x_{m,m+1}=0, (\text{for}~1\leq m \leq n-2\)) when \(m=1\).
  Item \(2\) reduces to: \[B=\{x_{1n}x_{l,l+1}x_{12},~ 3\leq l\leq n-2\}.\] The range of \(l\) ensures that \(x_{1n}x_{l,l+1}x_{12}\) is not divisible by the relation \(x_{1n}x_{n-1,n}=0\).

  Interestingly, sets \(A\) and \(B\) are the same. Since the set of degree-\(3\) elements in relations of \(\overline{FK}_{C_n}(n)\) is empty, the following set constitutes the entire degree-\(3\) Gröbner basis: \[\label{3terms} \text{degree-$3$ terms}: \{x_{1n}x_{m,m+1}x_{12}: ~ 3\leq m\leq n-2\}.----(2.9)\]

  Example 2.5. For \(n=4\), there are no degree-\(3\) terms.
  For \(n=5\), the only degree-\(3\) term in the Gröbner basis is \(x_{15}x_{34}x_{12}\).
  For \(n=6\), there are two degree-\(3\) terms: \(x_{16}x_{34}x_{12}\) and \(x_{16}x_{45}x_{12}\).

• Degree-\(4\) terms: To find degree-\(4\) terms, we only need to consider one \(S\)-polynomial (since the rest vanish): \[S(\underline{x_{m_2,m_2+1}x_{1n}}-x_{1n}x_{m_2,m_2+1},x_{1n}x_{m_1,m_1+1}x_{12},4).\] This \(S\)-polynomial, with respect to terms of degree less than \(4\), reduces to: \[x_{1n}x_{m_2,m_2+1}x_{m_1,m_1+1}x_{12},\] However, these terms reduce to zero unless the following inequalities hold: \[m_2-m_1\geq 2, ~~~m_1\geq 3,~~~ m_2\leq n-2.\] The terms \(x_{1n}x_{m_2,m_2+1}x_{m_1,m_1+1}x_{12}\) constitute the only degree-\(4\) terms, as the set of elements of degree \(4\) in the relations of \(\overline{FK}_{C_n}(n)\) is empty. Therefore, we have: \[\label{4terms}\text{\textbf{Degree-$4$ terms}}: \{x_{1n}x_{m_2,m_2+1}x_{m_1,m_1+1}x_{12}:~ m_2-m_1\geq 2,~m_1\geq 3 ~\text{and} ~m_2\leq n-2\}.----(2.10)\]

The terms of degrees \(3\) and \(4\) in Equations (2.9) and (2.10) provide insights into a general form for the elements of degree \(k\geq 4\) in the basis of \(\overline{FK}_{C_n}(n)\). Let’s explore this further with the following proposition:

Proposition 2.6. For \(k\geq 4\), any degree-\(k\) element of Gröbner basis for the ideal associated with \(\overline{FK}_{C_n}(n)\) is of the form: \[\label{kterms} \begin{split} &x_{1n}x_{m_{k-2}m_{k-2}+1}x_{m_{k-3}m_{k-3}+1} \cdots x_{m_2m_2+1}x_{m_1m_1+1}x_{12},~ \text{where},\\& m_{i}-m_{i-1}\geq 2,~i=2, 3, \cdots, k-2, ~\text{and}~ ~m_1\geq 3 ~\text{and} ~m_{k-2}\leq n-2. \end{split}----(2.11)\]

Proof. (by induction)

• Base of induction: Equation (2.10) for \(k=4\) serves as the base of induction.

• Inductive step: Assume that the statement is valid for degree-\(k\) elements of the Gröbner basis. We need to show that it is valid for degree-\((k+1)\) elements. Specifically, we want to prove that for degree-\((k+1)\) elements, we have: \[\label{kterms''} \begin{split} x_{1n}x_{m_{k-1}m_{k-1}+1}x_{m_{k-2}m_{k-2}+1}x_{m_{k-3}m_{k-3}+1} \cdots x_{m_2m_2+1}x_{m_1m_1+1}x_{12},~\text{where}\\m_i-m_{i-1}\geq 2,~i=2, 3, \cdots, k-1, ~\text{and}~ ~m_1\geq 3 ~\text{and} ~m_{k-1}\leq n-2. \end{split}----(2.12)\] However, the only \(S\)-polynomials available between a degree-\(k\) element in the statement and a lower degree term, which could reduce to a degree-\((k+1)\) element of the Gröbner basis is (as the rest vanish), are given by: \[\label{1}\begin{split}S(x_{m_{k-1}m_{k-1}+1}x_{1n}-x_{1n}x_{m, m+1},~ x_{1n}x_{m_{k-2}m_{k-2}+1}x_{m_{k-3}m_{k-3}+1} \cdots x_{m_1m_1+1}x_{12},~ k+1)\\=x_{1n}\underline{x_{m_{k-1}m_{k-1}+1}x_{m_{k-2}m_{k-2}+1}} \cdots x_{m_1m_1+1}x_{12}. \end{split}----(2.13)\] We now focus on the underlined product on the right hand side of (2.13): \[C=\underline{x_{m_{k-1},m_{k-1}+1}x_{m_{k-2},m_{k-2}+1}}.\] To ensure that \(C\) does not vanish or reduce to something else when reduced with any of the degree-\(2\) terms in (2.1), we need to avoid the following cases.

  1. If \(m_{k-1}=m_{k-2}\) then \(C\to 0\) due to the relation \(x_{m,m+1}^2=0\).

  2. If \(m_{k-1}=m_{k-2}-1\), then \(C\to 0\) due to the relation \(x_{m,m+1}x_{m+1,m+2}=0\).

  3. If \(m_{k-1}\leq m_{k-2}-2\), then \(C\to x_{m_{k-2}m_{k-2}+1}x_{m_{k-1}m_{k-1}+1}\) (by reduction with the leading term of the relation \(\underline{x_{m,m+1}x_{l,l+1}}-x_{l,l+1}x_{m,m+1}=0)\)

  4. If \(m_{k-1}\leq m_{k-2}+1\), then \(C\to 0\) by the relation \(x_{m+1,m+2}x_{m,m+1}=0\).

  To avoid these cases, we require the inequality \(n-2\geq m_{k-1}>m_{k-2}+1\), which is equivalent to \(m_{k-1}-m_{k-2}\geq 2\). Additionally, we need \(m_1\ge 3\) to ensure that \(x_{m_1,m_{m_1+1}x_{12}}\) does not become zero due to \(x_{m+1, m+2}x_{m,m+1}=0\) for \(1\le m\le n-2.\)
  Therefore the following three inequalities meet the requirement of (2.12) for validity at degree \(k+1\): \[m_{k-1}-m_{k-2}\geq 2,~ n-2\geq m_{k-1},~ \text{and}~m_1\geq 3.\]

This completes the proof. ◻

By combining the degree-\(k\) terms for \(k\geq 4\) from (2.11), the degree-\(3\) term from (2.9) and the degree-\(2\) terms from (2.1), we obtain the Gröbner basis \(GB_{\overline{FK}_{C_n}(n)}\) for the ideal associated with \(\overline{FK}_{C_n}(n)\): \[\label{gb'} \begin{split} GB_{\overline{FK}_{C_n}(n)}=&\{x_{m,m+1}^2:~1\leq m\leq n-1; ~x_{1,n}^2,\\& x_{m,m+1}x_{m+1, m+2}, ~x_{m+1,m+2}x_{m,m+1}, ~1\leq m \leq n-2\\&x_{n-1,n}x_{1n}, ~x_{1n}x_{n-1,n},~x_{12}x_{1n},~x_{1n}x_{12}\\&x_{m,m+1}x_{l,l+1}-x_{l,l+1}x_{m,m+1}, ~1\leq m\leq l-2,~ 3\leq l\leq n-1\\& x_{m,m+1}x_{1,n}-x_{1n}x_{m,m+1},~ 2\leq m\leq n-2,\\& x_{1n}x_{m,m+1}x_{12}, ~ 3\leq m\leq n-2,~\text{and}\\& x_{1n}x_{m_{k-2}m_{k-2}+1}x_{m_{k-3}m_{k-3}+1} \cdots x_{m_2m_2+1}x_{m_1m_1+1}x_{12},\\&\text{where}~ m_i-m_{i-1}\geq 2,~i=2, 3, \cdots, k-1, ~\text{and}~ ~m_1\geq 3 ~\text{and} ~m_{k-1}\leq n-2\}. \end{split}----(2.14)\]

Example 2.7. For \(n=3\) we have \[GB_{\overline{FK}_{C_3}(3)}=\{x_{12}^2,x_{23}^2,x_{13}^2, x_{12}x_{23}, x_{23}x_{13}, x_{12}x_{13}, x_{23}x_{12}, x_{13}x_{23}, x_{13}x_{12}\}.\]

2.6. Basis and dimension

Definition 2.8.

1. A matching in a graph is a subset of the set of edges in the graph where no two edges share a common vertex.

2. A matching on a set of line segments \(V_n=\{\overline{12},\overline{23},\cdots \overline{n-1,n}\}\) is a subset of \(V_n\) where no two line segments have a common vertex.

Example 2.9. Consider a square with successive sides/edges labelled as \(a, b, c,\text{and} ~d\). We have \(7\) sets of sides/edges where no two of them share a common vertex: \[\phi, \{a\}, \{b\}, \{c\}, \{d\}, \{a,c\}, \{b,d\}.\] Therefore, the number of matchings in a square is \(7\).

Theorem 2.10

1. There exists a one-to-one correspondence between the basis of \(\overline{FK}_{C_n}(n)\) and the set of matchings in an \(n\)-cycle \(C_n\).

2. The dimensions of \(\overline{FK}_{C_n}(n)\) equals the number of matchings in an \(n\)-cycle \(C_n\).

Proof. The basis of the algebra \(\overline{FK}_{C_n}(n)\) consists of all the words formed from generators
\(\{x_{12},x_{23},x_{34},\cdots, x_{n-1,n}; x_{1,n}\}\) that are not divisible by any leading term of the elements in the Gröbner basis from (2.14). Specifically:

• The element of degree \(0\) is \(1\),

• The elements of degree \(1\) are the generators \(x_{1n}\) and \(x_{m,m+1},\text{for} ~m=1, \cdots, n-1\).

However, for higher degree terms in \(\overline{FK}_{C_n}(n)\), the structure of the degree-\(2\) terms and higher degree terms in the Gröbner basis from (2.14) leads to the following observations:

1. Any monomial formed from our generators vanish due to the degree-\(2\) terms in the Gröbner basis if it has two adjacent \(1\)st indices differing by \(\pm{1}\) or \(0\).

2. Any monomial formed from our generators vanishes if two adjacent \(1\)st indices differ by more than or equal to \(2\) but are increasing in \(1\)st indices.

3. If any two adjacent \(2\)nd indices in a monomial share the common value \(n\), then the monomial vanishes by \(x_{n-1,n}x_{1n}\).

4. If any two adjacent \(2\)nd indexes share a common value less than \(n\), then since all the generators are of the form \(x_{m,m+1}\), they also share a common 1st index, leading to vanishing by \(x_{ij}x_{ij}\).

To avoid the above \(4\) cases, the basis elements of \(\overline{FK}_{C_n}(n)\) must consist of monomials in variables with decreasing \(1\)st indices (with the exception of \(x_{1n}\) when appears as the leading variable), such that any two successive \(1\)st indices differ by at least \(2\).

Considering the above points we come up with \(B(\overline{FK}_{C_n}(n))\), the basis of \(\overline{FK}_{C_n}(n)\): \[\label{b} \begin{split} B&(\overline{FK}_{C_n}(n))=\{1,~x_{1n},~x_{m,m+1},~\text{where}~ 1\leq m \leq n-1,\\& x_{1n}x_{m_1m_1+1},~~\text{where}~ 2\leq m_1 \leq n-2,\\&x_{m_1m_1+1}x_{m_2m_2+1}, ~\text{where}~m_1-m_2\geq 2,~m_2\geq1,~m_1\leq n-1, \\&x_{m_1m_1+1}x_{m_2m_2+1}\cdots x_{m_km_k+1},~\text{where}~ m_i-m_{i+1}\geq 2,~m_k\geq 1,~m_1\leq n-1, \\& x_{1n}x_{m_1m_1+1}x_{m_2m_2+1} \cdots x_{m_{k-1}m_{k-1}+1}, ~\text{where}~ m_i-m_{i+1}\geq 2,~ m_{k-1}\geq 2,\\&m_1\leq n-2 \}. \end{split}----(2.15)\] Consider the algebra \(\overline{FK}_{C_n}(n)\) and its basis \(B[\overline{FK}_{C_n}(n)]\), we want to show that this basis consists of all the matchings in the \(n\)-cycle graph \(C_n\).

1. Viewing Variables as Edges:

   In the basis of \(\overline{FK}_{C_n}(n)\) given in (2.15), let’s interpret the variables \(x_{m,m+1}\) for \(m=1, 2, \cdots, n-1\) and \(x_{1n}\), as representing the edges \(\overline{12}, \overline{23},\cdots \overline{n-1,n}\) and \(\overline{1n}\) of the \(n\)-cycle \(C_n\). Each element in the basis corresponds to a subset of the edges of \(C_n\).
   

2. Matching Interpretation: An element in the basis of \(\overline{FK}_{C_n}(n)\) can be viewed as a matching in the graph \(C_n\). The reason is that no monomial in the basis includes two variables with a common index. Therefore, we can view an element of the basis as a subset of the set of \(C_n\) edges with no common vertex. Hence, an element of the basis corresponds to a subset of the edges of \(C_n\) with no common vertex, which is precisely the definition of a matching.

3. Converse Interpretation: Conversely, any matching in \(C_n\) can be represented as an element in the basis of \(\overline{FK}_{C_n}(n)\). Furthermore, considering the domain of the indexes in (2.15), we see that all the monomials in the generators of \(\overline{FK}_{C_n}(n)\) that have no repeated indices are covered in the basis.
   

4. one-to-on correspondence, and the dimension of \(\overline{FK}_{C_n}(n)\): The above considerations establish a one-to-one correspondence between the set of all matchings in the \(n\)-cycle \(C_n\) and the basis of \(\overline{FK}_{C_n}(n)\). Therefore the dimension of algebra \(\overline{FK}_{C_n}(n)\) is equal to the number of matching in the \(n\)-cycle \(C_n\).

This completes the proof. ◻

Corollary 2.11. The dimension of \(\overline{FK}_{C_n}(n)\) is equal to the Lucas number \(L_n\).

Proof. From Theorem 2.10 we know that the dimensions of \(\overline{FK}_{C_n}(n)\) correspond to the number of matchings in an \(n\)-cycle \(C_n\), which precisely equals the Lucas number \(L_n\) [13]. This completes the proof. ◻

2.7. The highest degree elements in the basis

Consider the basis of \(\overline{FK}_{C_n}(n)\) as given in (2.15). We want to find the maximum degree of a monomial in the basis, which corresponds to the maximum size of a matching. This maximum degree occurs when any two adjacent 1st indices differ by exactly \(2\) (equivalently, when any two sides in the corresponding matching are separated by exactly one side in \(C_n\)).

We can calculate this maximum degree as follows:

\[\label{maxdeg}\begin{split}\text{For}~n\geq 4, ~ \text{the maximum degree/size}~= \lfloor\frac{n}{2}\rfloor,\\ \text{The multiplicity of the maximum degree/size}~=\begin{cases} 2, &\text{for even}~n\\ n, &\text{for odd}~n \end{cases} \end{split}----(2.16)\]

Example 2.12. For \(n=5\), the highest degree for elements in \(B[\overline{FK}(5)]\) is \(\lfloor\frac{n}{2}\rfloor=\lfloor\frac{5}{2}\rfloor=2\), and its multiplicity \(5\).

2.8. Matchings’ and Fibonacci number

The following Lemma establishes the connection of Fibonacci number to the number of matching in a set of line segments.

Lemma 2.13. Let \(M(n)\) denote the number of matchings in the set of line segments
\(V_n=\{\overline{12},\overline{23},\cdots \overline{n-1,n}\}\), \(n\geq 2\). Then \[\label{chi5'} M(n)=F_n,~ n\geq 2,----(2.17)\] where \(F_n\) is \(n\)-th Fibonacci number with initial values \(F_0=1,~F_1=1\).

Proof. We need to show that:

1. \(M(2)=F_2\), and

2. \(M(n+1)=M(n)+M(n-1)\), i.e., the recurrence relation for Fibonacci numbers , .

1. Base Case: For \(n=2\) the set \(V_2=\{\overline{12}\}\) contains two subsets: the empty set \(\phi\), and itself \(\{\overline{12}\}\). Thus \(M(2)=2\), which matches the Fibonacci number \(F(2)=2\) (with initial values \(F_0=1\) and \(F_1=1\)).

2. Inductive Step: Consider \(V_n\) as a union: \[V_n=\{\overline{n-1, n}\}\cup V_{n-1},~ \text{where}~ V_{n-1}=\{\overline{12},\overline{23},\cdots\overline{n-2, n-1}\}.\] Adding a line segment \(\overline{n,n+1}\) to \(V_n\) creates the set \(V_{n+1}\). This addition increases \(M(n)\), the number of matchings in \(V_n\), by introducing new matchings. To avoid common vertices, we add \(\overline{n,n+1}\) only to each matching in \(V_{n-1}\). Note that \(\overline{n,n+1}\) shares no common vertex with elements in \(V_{n-1}\) but does share a common vertex \(n\) with \(\overline{n-1,n}\). Therefore the number of new matchings added to \(M(n)\) is precisely the number of matchings in \(V_{n-1}\), which is \(M(n-1)\). Thus we have \(M(n+1)=M(n)+M(n-1)\). This completes the proof.

◻

Lemma 2.14. Let \(M(n,k)\) denote the number of matchings of size \(k\) in the set of line segments \(V_n=\{\overline{12},\overline{23},\cdots \overline{n-1,n}\}\). Then

1. \(M(n+1,k)=M(n,k)+M(n-1,k-1), ~\text{for}~ n\geq 2,~k\geq 0\),

2. \(M(n, k)={n-k\choose k}\).

Proof.

1. Consider \(V_n\) as \[V_n=V_{n-1}\cup \{\overline{n-1,n}\},\] where \(V_{n-1}=\{\overline{12},\overline{23},\cdots \overline{n-2,n-1}\}\). Addition of the line segment \(\overline{n,n+1}\) to \(V_n\) makes \[V_{n+1}=\{\overline{n,n+1}\}\cup V_{n-1}\cup \{\overline{n-1,n}\}.\] This addition increases \(M(n, k)\). To avoid common vertices, we add \(\overline{n,n+1}\), as before, only to each matching in \(V_{n-1}\). The reason is that \(\overline{n,n+1}\) does not share a common vertex with elements in \(V_{n-1}\) but have a common vertex \(n\) the elements in \(\{\overline{n,n+1}\}\) and \(\{\overline{n-1,n}\}\). So the number of the new matchings is exactly the number of elements in \(V_{n-1}\) of size \(k-1\), i.e., \(M(n-1, k-1)\). Hence we have: \[M(n+1,k)=M(n,k)+M(n-1,k-1).\]

2. To prove \(M(n, k)={n-k\choose k}\) we use double induction. We need to show that

   1. Base Case: For \(n=2\) and \(k=1\) the statement \(M(2,1)={2-1\choose 1}\) is true, as \(M(2,1)=1\) and \({2-1\choose 1}={1\choose 1}=1\).

   2. Inductive Step: We need to show that \[\begin{split} \text{If}~ &M(n,k)={n-k\choose k} \forall n\geq 2,~k>1,~\text{then}~ M(n+1,k)={n+1-k\choose k},\\ \text{and}\\ \text{If}~ &M(n,k)={n-k\choose k} \forall n\geq 2,~k>1,~\text{then}~ M(n,k+1)={n-(k+1)\choose k+1}. \end{split}\] By part (1) of the Lemma we have: \[\begin{split} M(n+1,k)&=M(n, k)+M(n-1, k-1)\\&={n-k\choose k}+{n-1-(k-1)\choose k-1}={n-k\choose k}+{n-k\choose k-1}\\ &=\frac{(n-k)!}{k!(n-2k)!}+\frac{(n-k)!}{(k-1)!(n-2k+1)!}=\cdots=\frac{(n-k+1)!}{k!(n-2k+1)!}\\&={n+1-k\choose k}.\\ \text{Also,}\\ % \text{if}~ &M(n,k)={n-k\choose k} \forall n\geq 2,~k>1,~\text{then}~ M(n,k+1)={n-k-1\choose k+1}.\\ M(n,k+1)&=M(n-1, k+1)+M(n-2, k)\\&={n-1-(k+1)\choose k+1}+{n-2-k\choose k}={n-k-2\choose k+1}+{n-k-2\choose k}\\ &=\frac{(n-k-2)!}{(k+1)!(n-2k-3)!}+\frac{(n-k-2)!}{k!(n-2k-2)!}=\cdots=\frac{(n-k-1)!}{(k+1)!(n-2k-21)!}\\&={n-k-1\choose k+1}={n-(k+1)\choose k+1}. \end{split}\]

This completes the proof. ◻

The following proposition establishes the connection of the number of matchings in a set of line segments with that of an \(n\)-cycle graph \(C_n\).

Proposition 2.15. . Consider an \(n\)-cycle graph denoted \(C_n\). Let \(M_{C_n}(n,k)\) represent the number of matchings of size \(k\) in \(C_n\). Additionally, let \(M(n, k)\) be the number of matchings of size \(k\) in the set of line segments \(V_n=\{\overline{12},\overline{23},\cdots \overline{n-1,n}\}\). Then we have: \[\label{123'} \begin{split} M_{C_n}(n, k)=&M(n, k)+M(n-2, k-1),~n\geq 4, ~k=2, 3, \cdots, \lfloor\frac{n}{2}\rfloor,~and\\ &M_{C_n}(n, k)=\frac{n}{n-k}{n-k\choose k}, \end{split}----(2.18)\]

Proof. \(M(n, k)\) could be viewed as the number of matchings of size \(k\) in \(C_n\setminus{\{\overline {1n}\}}\). We rewrite \(C_n\setminus{\{\overline {1n}\}}\) as \[C_n\setminus{\{\overline {1n}\}}=\{\overline{12} \}\cup V_{n-2}\cup \{\overline{n-1,n}\},\] where \(V_{n-2}=\{\overline{23},\overline{34}, \cdots, \overline{n-2,n-1}\}\). Now adding \(\overline{1n}\) to \(C_n\setminus{\{\overline {1n}\}}\) completes the \(n\)-cycle, as well increases \(M(n, k)\) to \(M_{C_n}(n, k)\) by making new matchings of size \(k\). As before, to avoid common vertices we need to add the edge \(\overline{1n}\) to the elements of size \(k-1\) only in \(V_{n-2}\).

Therefore, the number of the new matchings equals the number of matchings of size \(k-1\) in \(V_{n-2}\), that is, \(M(n-2, k-1)\). Hence, we have \(M_{C_n}(n, k)=M(n, k)+M(n-2, k-1)\). This completes the proof of the first part.
To prove the second part, we substitute \(M(n, k)={n-k\choose k}\) from Lemma 15 into the first part, as follows. \[\begin{aligned} \begin{split} M_{C_n}(n, k)&=M(n, k)+M(n-2, k-1)={n-k\choose k}+{n-1-k\choose k-1}\\&={n-k\choose k}+\frac{(n-1-k)!}{(k-1)!(n-2k)!}={n-k\choose k}+\frac{\frac{(n-k)!}{(n-k)}}{\frac{k!}{k}(n-2k)!}\\&={n-k\choose k}+\frac{k}{n-k}\frac{(n-k)!}{k!(n-2k)!}={n-k\choose k}+\frac{k}{n-k}{n-k\choose k}\\&=\frac{n}{n-k}{n-k\choose k}. \end{split} \end{aligned}----(2.19)\] This completes the proof. ◻

2.9. Hilbert series and \(q\)-Fibonacci/\(q\)-Lucas polynomials

[4, 5]. Putting together Proposition 2.15, Theorem 2.10, Corollary 2.11, and equation (2.16), we obtain: \[L_n=M_{C_n}(n)=\sum_{k=0}^{\lfloor\frac{n}{2}\rfloor}M_{C_n}(n,k)=\sum_{k=0}^{\lfloor\frac{n}{2}\rfloor}\frac{n}{n-k}{n-k\choose k}.\] The upper limit in the sum is due to the fact that maximum size of a matching in \(C_n\) or the maximum degree of an element in the basis of \(\overline{FK}_{C_n}(n)\) is \(\lfloor\frac{n}{2}\rfloor\).

Hence taking sum over all the matchings of different size \(k\) in \(C_n\) (elements of different degrees \(k\) in the basis, we have \(q\)-Lucas polynomial [4, 5]: \[\label{q-lucas} L_n(q)=\sum_{k=0}^{\lfloor\frac{n}{2}\rfloor}\frac{n}{n-k}{n-k\choose k}q^k,~L_0=2.----(2.20)\] From Lemmas 2.13 and 2.14 we have the total number of matchings in a set of line segments \(V_n=\{\overline{12},\overline{23},\cdots \overline{n-1,n}\}\) to be \(M_n=F_n\) and the number of matchings of size \(k\) in \(V_n\) to be \(M(n,k)={n-k\choose k}\). This results in \[F_n=M(n)=\sum_{k=0}^{\lfloor\frac{n}{2}\rfloor}M(n,k)=\sum_{k=0}^{\lfloor\frac{n}{2}\rfloor}{n-k\choose k}.\] Hence taking sum over all matchings of different sizes we have \(q\)-Fibonacci polynomial [4, 5]: \[\label{q-fibo} F_n(q)=\sum_{k=0}^{\lfloor\frac{n}{2}\rfloor}{n-k\choose k}q^k.----(2.21)\] Considering the information above, we can establish a connection between the \(q\)-Lucas polynomial and the Hilbert series of \(\overline{FK}_{C_n}(n)\) in the following proposition:

Proposition 2.16. The Hilbert series of \(\overline{FK}_{C_n}(n)\) is given by the \(q\)-Lucas number \(L_n(q)\): \[H_n(q)=L_n(q).----(2.22)\]

Proof. Recall the expression for the \(q\)-Lucas number from Equation (2.20): \[L_n(q)=\sum_{k=0}^{\lfloor\frac{n}{2}\rfloor}\frac{n}{n-k}{n-k\choose k}q^k,~L_0=2.\] In the above expression, the degree \(k\), representing the size of the matching, is summed over to yield the Lucas number. By Corollary 2.11, this Lucas number is equal to the dimension of \(\overline{FK}_{C_n}(n)\). Thus, the expression demonstrates that the Hilbert series is indeed given by Lucas \(q\)-number: \[H_n(q)=\sum_{k=0}^{\lfloor\frac{n}{2}\rfloor}\frac{n}{n-k}{n-k\choose k}q^k=L_n(q).\] This completes the proof. ◻

Example 2.17. \(H_5(q)=\sum_{k=0}^{\lfloor \frac{5}{2}\rfloor}\frac{5}{5-k}{5-k\choose k}q^k= %{5\choose 0}+\frac{5}{4}{4\choose 1}q+\frac{5}{3}{3\choose 2}q^2= 1+5q+5q^2\). Similarly \(H_6=1+6q+9q^2+2q^3\).

2.10. The action of \(D_n\) and decomposition

In this section we derive the character of \(\overline{FK}_{C_n}(n)\) over \(D_n\) for even and odd \(n\).

Remark 2.18. As previously mentioned, \(D_n\) defines an action on the algebra \(\overline{FK}_{C_n}(n)\). Under this action, the component of the character of \(\overline{FK}_{C_n}(n)\) associated with the conjugacy class indexed by \(\sigma\in D_n\) is the sum of the coefficients of \(z\) in \(\sigma(z)\) when \(z\) runs through the basis \(B[\overline{FK}_{C_n}(n)]\).

Lemma 2.19. Let \(r\in D_n\) realized in permutation group \(S_n\) as \(r=(1,2,\cdots,n)\). Then the trace of the representation of \(r^p\in D_n\), on \(\overline{FK}_{C_n}(n)\) is given by: \[\label{r^p} [\chi(r^p)](q)=\begin{cases} 1 ~$if$~ p=1\\ 1+\begin{cases} gcd(n,p)q^\frac{n}{gcd(n, p)},\text{if} ~~ 2\le p\le n-1,\text{and} ~ gcd(n, p)\ge\begin{cases} 2,~ n=\text{even}\\ 3,~n=\text{odd} \end{cases}\\ 0 ~~~~~~~~~~\text{otherwise} \end{cases}\\ H_n(q), \text{Hilbert series}, ~~~~~\text{if} ~~ p=n \end{cases}----(2.23)\]

Proof. In the following discussion, we consider \(M\) both as a basis element of \(\overline{FK}_{C_n}(n)\) and as a matching in \(C_n\). Similarly we treat \(r\) alternatively as a clockwise rotation of angle \(\frac{2\pi}{n}\) in \(C_n\) or as an operator acting on \(M\) as an eigenvector in \(r^pM=M\) with eigenvalue \(1\).

We consider the following points.

1. For any integer \(1\le p\le n\), there always exists an empty matching (equivalently, the identity element \(M=1\) in the basis), such that \(r^pM=M\).

2. \(rM=M\) does not hold for non-empty \(M\). The reason is that since \(r\) represents a clockwise rotation of \(\frac{2\pi}{n}\) in \(C_n\), it takes an edge in the matching \(M\) to the neighbouring edge in \(C_n\). However this neighbouring edge is not part of \(M\), as it could violate the definition of a matching (which forbids common vertices). Thus, \(rM=M\) cannot hold for a non-empty matching \(M\) (or any basis element \(M\neq 1\)). In other word, \(r^pM=M\) is ruled out for \(p=1\) and non-empty matchings.

3. The possibility of \(r^pM=-M\) is absurd. If we consider \(x_{m, m+1}\) as a general variable in a basis element \(M\), where \(1\leq m< n\), then for \(1< p\le n-1\) (since \(p=1\) is ruled out), we have: \[r^px_{m,m+1}=x_{(m+p)(mod~ n), (m+1+p)(mod~ n)}.\] We focus on the indices of \(x_{(m+p)(mod~ n), (m+1+p)(mod~ n)}\) and examine the following possibilities:

   • \((m+p)(mod~n)\le n-1\), then \((m+1+p) (mod~n)\le n.\) Then\[r^px_{m,m+1}=x_{(m+p)(mod~n),(m+1+p)(mod~n))}=x_{m', m'+1}~\] where \(m'=(m+p)(mod~n)\le n-1\).
     No negative sign develops in this case. Therefore we are left with the following case:

   • \((m+p)(mod~n)=n\), then \((m+1+p)(mod~n)=1.\) We have: \[r^px_{m,m+1}=x_{(m+p)(mod~n),(m+1+p)(mod~n))}=x_{n, 1}=-x_{1,n}.\] Here, a minus sign does develop. However, for \(r^pM=-M\) to hold, we would require that \(r^px_{1,n}=x_{m',m'+1}\) for some \(m'\) (to compensate for the \(x_{1n}\) generated by the operation of \(r^p\) on \(M\)). But then \[r^px_{1,n}=x_{1+p, (n+p)(mod~n)}=x_{1+p, p}=-x_{p,p+1}.\] Here, a second minus sign cancels out the first one. Therefore \(r^pM=-M\) is absurd. This is why \(gcd(n, p)\) was mentioned in the statement of the Lemma, as the number of eigenvectors \(M\).

4. The second item above rules out \(p=1\) in the equation \(r^pM=M\) for nonempty matching \(M\).
   For \(r^pM=M\) to hold, we require that the matching \(M\) be periodic with period \(p\). In fact, a matching can be periodic with period \(gcd(n, p)\) and still be fixed by \(r^p\). Specifically, a matching \(M\) in \(C_n\), after a clockwise rotation of \(gcd(n, p)\), coincides itself. Therefore the number of edges in \(M\) is \(\frac{n}{gcd(n, p)}\), and the number of such matchings \(M\) equals \(gcd(n,p)\). Alternatively, we can express this as the number of non-identity elements in the basis of degree \(\frac{n}{gcd(n, p)}\), that are fixed by \(r^p\), which equals \(gcd(n,p)\). Thus, we can write the character as \(gcd(n,p)q^{\frac{n}{gcd(n,p)}}\), where, we also include the identity element(empty matching) that is always fixed.

   However, [maxdeg], imposes a restriction on the highest degree: \[\frac{n}{gcd(n, p)}\le \lfloor\frac{n}{2}\rfloor\]or equivalently: \[\begin{cases} gcd(n,p)\ge 2,~ \text{for even}~ n\\ gcd(n,p)\ge \frac{2n}{n-1},~ \text{for odd}~n \end{cases}\] Since \(\frac{2n}{n-1}>2\) for finite \(n\), and \(gcd(n,p)\) is a natural number, we have \(\frac{2n}{n-1}\ge 3\). Therefore, the above restriction is modified as follows: \[\begin{cases} gcd(n,p)\ge 2,~ \text{for even}~ n\\ gcd(n,p)\ge 3,~ \text{for odd}~n \end{cases}\]

   putting everything together, we obtain the following expression for the character: \[\label{char11'''''} \chi(r^p)= 1+\begin{cases} gcd(n,p)q^\frac{n}{gcd(n, p)},&\text{if} ~~ 2\le p\le n-1,\text{and} ~ gcd(n, p)\ge\begin{cases} 2,~ n=\text{even}\\ 3,~n=\text{odd} \end{cases}\\ 0 ~~~~~~~~~~&\text{otherwise} \end{cases}----(2.24)\]

5. For \(p=n\), where \(r^p=r^n=\epsilon\), represents the identity element of \(D_n\), we have \(\chi(r^n=\epsilon)= dim[\overline{FK}_{C_n}(n)]=L_n\) by corollary 12. Alternatively, in the degree decomposition \(q\)-Lucas form, we express \(L_n(q)=H_n(q)\), which corresponds to the Hilbert series of \(\overline{FK}_{C_n}(n)\) according to Proposition 2.16. Therefore, the character of identity element of the group over the quotient algebra would be: \[\label{char11''''''} \chi(r^n=\epsilon)=L_n(q)=H_n(q)----(2.25)\]

The combination of the above items \(1-5\), along with Equations (2.24) and (2.25) provides the proof of the lemma. ◻

Example 2.20. Consider \(\overline{FK}_{C_6}(6)\). Then \[\begin{split}B[\overline{FK}_{C_n}(n)]=\{1, x_{12}, x_{23}, x_{34}, x_{45}, x_{56}, x_{16},\\ x_{34}x_{12}, x_{45}x_{12}, x_{45}x_{23}, x_{56}x_{12}, x_{56}x_{23}, x_{56}x_{34}, x_{16}x_{23}, x_{16}x_{34}, x_{16}x_{45},\\ x_{16}x_{45}x_{23}, x_{56}x_{34}x_{12}\}\end{split}\] Then, for \(2\le p\le 5\), such that \(\frac{6}{gcd(6,p)}\le \lfloor \frac{6}{2}\rfloor\) , i.e. \(gcd(6,p)\ge 2\), we will have the options \(p=2, 3, 4\). Therefore we will have:

• \(\chi(r)=1\)

• \(\chi(r^2)=3\) or \(\chi(r^2)(q)=1+2q^3\),
  

• \(\chi(r^3)=4\) or \(\chi(r^3)(q)=1+3q^2\),
  

• \(\chi(r^4)=3\) or \(\chi(r^4)(q)=1+2q^3\),
  

• \(\chi(r^5)=1\)
  

• \(\chi(r^6=\epsilon)=18\) or \(\chi(r^6)(q)=L_6(q)=H_6(q)=1+6q+9q^2+2q^3\).

Lemma 2.21. For even \(n\geq 4\), let \(s\in D_n\), be realized in permutation group as
\(s=(1n)(2, n-1)\cdots (\frac{n}{2}, \frac{n}{2}+1)\) in an \(n\)-cycle graph \(C_n\). Let \(\chi(s)\) be the trace of the representation of \(s\) over \(\overline{FK}_{C_n}(n)\). Then we have: \[\label{char1} [\chi(s)](q)=\sum_{k=0}^{\lfloor\frac{n}{4}\rfloor}{\frac{n}{2}-k\choose k}q^{2k}-2\sum_{k=0}^{\lfloor\frac{\frac{n}{2}-1}{2}\rfloor}{\frac{n}{2}-1-k\choose k}q^{2k+1}+\sum_{k=0}^{\lfloor\frac{\frac{n}{2}-2}{2}\rfloor}{\frac{n}{2}-2-k\choose k}q^{2k+2}.----(2.26)\] In terms of \(q\)-Fibonacci polynomials, according to the total degree decomposition we have: \[(q)=F_{\frac{n}{2}}(q^2)-2qF_{\frac{n}{2}-1}(q^2) +q^2F_{\frac{n}{2}-2}(q^2),----(2.27)\]where \(F\) represents the appropriate \(q\)-Fibonacci polynomial.

Proof.

• By Theorem 11 to any element of the basis of \(\overline{FK}_{C_n}(n)\) there corresponds a matching in \(C_n\). Let \(T_1=\{\overline{i_1,i_1+1}, \overline{i_2,i_2+1}\cdots \overline{i_m,i_m+1}\}\) be a matching of the set of line segments \(S_1=\{\overline{12}, \overline{23}, \cdots,\overline{\frac{n}{2}-1, \frac{n}{2}}\}\) in \(C_n\) as defined in Definition 9. It has a corresponding monomial: \[u=x_{i_1,i_1+1}x_{i_2,i_2+1}\cdots x_{i_m,i_m+1}\in B[\overline{FK}_{C_n}(n)].\] Now consider the reflection in \(s\) of the matching \(T_1\) ( and its corresponding monomial \(u\)), Since \(n\) is even, this reflection is itself a matching. let’s denote it as: \[T'_1=\{\overline{j_1,j_1+1}, \overline{j_2,j_2+1}\cdots \overline{j_m,j_m+1}\}.\]Also denote its corresponding monomial as: \[u'=x_{j_1j_1+1}x_{j_2j_2+1}\cdots x_{j_mj_m+1}.\]

  While the matching \(T_1\) is not symmetric about \(s\), the union of the matchings \(T_1\cup T'_1\) is symmetric, and its corresponding monomial \(uu' \in B[\overline{FK}_{C_n}(n)]\) is invariant under \(s\). This can be seen as follows: \[s(uu')=s(usu)=(su)s^2u=(su)u=u'u \xrightarrow{\mbox{\tiny{reduction w.r.t degree-2 terms}}} uu'.\] The last step in the above equation involves reduction with respect to degree-\(2\) terms in the Gröbner basis (2.14), since \(u\) and \(u'\) commute (there are no common indices among variables in \(uu'\)). The number of such eigenvectors in \(\overline{FK}_{C_n}(n)\) equals the number of matchings \(T_1\cup T_1'\) in \(C_n\). By construction, this is equivalent to the number of matchings \(T_1\) in the set \(\{\overline{12}, \overline{23}, \cdots,\overline{\frac{n}{2}-1,\frac{n}{2}}\}\). Using Equations (2.17) and (2.21) we find that this number is equal to: \[\label{chi1'} [M(\frac{n}{2})](q)=F_\frac{n}{2}(q)=\sum_{k=0}^{\lfloor\frac{n}{4}\rfloor}{\frac{n}{2}-k\choose k}q^{k}.----(2.28)\] However, each matching in \(T_1\cup T_1'\) contains twice as many edges as \(T_1\) does.

  Therefore, our first contribution to \(\chi(s)\), denoted as \(\chi_1(s)\) is given by: \[\label{chi1} [\chi_1^{}(s)](q)=\sum_{k=0}^{\lfloor\frac{n}{4}\rfloor}{\frac{n}{2}-k\choose k}q^{2k}=F_{\frac{n}{2}}(q^2).----(2.29)\] Next, to fine the number of eigenvectors with eigenvalue \(-1\) we consider the following cases:

  1. We consider the matching \(\{\overline{1n}\}\cup T_2\), where \(T_2\subset\{\overline{23},\cdots, \overline{\frac{n}{2}-1,\frac{n}{2}}\}\). We denote the reflection of \(T_2\) in \(s\) as \(T_2'\). we also consider \(v\) and \(v'\) as the corresponding monomials in \(B[\overline{FK}_{C_n}(n)]\), for \(T_2\) and \(T_2'\), respectively.

  2. We consider the matching \(\{\overline{\frac{n}{2}, \frac{n}{2}+1}\}\cup T_3\) where \(T_3\) is a matching in \(\{\overline{12},\cdots, \overline{\frac{n}{2}-2,\frac{n}{2}-1}\}\). We also consider its reflection \(T_3'\) in \(s\).

  In case \(1\), while the matching \(\{\overline{1n}\}\cup T_2\) is not symmetric about \(s\), the union \(\{\overline{1n}\}\cup T_2\cup T_2'\) forms a matching in \(C_n\) that is symmetric about \(s\). The corresponding monomial \(x_{1n}vv'\) forms an eigenvector of \(s\) with eigenvalue \(-1\), because: \[s(x_{1n}vv')=-x_{1n}s(vv')=-x_{1n}vv'.\]

  By construction, the number of matchings \(\{\overline{1n}\}\cup T_2\cup T_2'\) in \(C_n\) equals the number of matchings \(T_2\) in \(\{\overline{23},\cdots, \overline{\frac{n}{2}-1,\frac{n}{2}}\}\), which according to (2.17) and (2.21) is equal to: \[\label{chi2'} [M(\frac{n}{2}-1)](q)=F_{\frac{n}{2}-1}(q)=\sum_{k=0}^{\lfloor\frac{\frac{n}{2}-1}{2}\rfloor}{\frac{n}{2}-1-k\choose k}q^{k}.----(2.30)\]

  However, while the number of matchings \(T_2\) equals that of \(\{\overline{1n}\}\cup T_2\cup T_2'\), but the number of edges in each matching of \(\{\overline{1n}\}\cup T_2\cup T_2'\) is \(2k+1\) compared to \(k\) edges of \(T_2\). The odd number \(2k+1\), reflects the odd size of the matching \(\{\overline{1n}\}\cup T_2\cup T_2'\) in \(C_n\), as well the negative eigenvalue of \(s\). Therefore we have: \[-\sum_{k=0}^{\lfloor\frac{\frac{n}{2}-1}{2}\rfloor}{\frac{n}{2}-1-k\choose k}q^{2k+1}= -qF_{\frac{n}{2}-1}(q^2).\]where the rearrangement of \(q\) on the right hand side of the equation is for the sake of consistency with to definition of \(q\)-Fibonacci. Also the negative sign in the above is due the negative eigenvalue of \(s\). Having this for the case \((1)\), and considering the fact that the case \((2)\) also gives exactly the same result as case \((1)\), we apply a factor \(2\) for the above and hence we come up with a second contribution to \(\chi(s)\): \[\label{chi2} [\chi_2^{}(s)](q)=-2\sum_{k=0}^{\lfloor\frac{\frac{n}{2}-1}{2}\rfloor}{\frac{n}{2}-1-k\choose k}q^{2k+1}=-2qF_{\frac{n}{2}-1}(q^2).----(2.31)\]

• To cover the last contribution to \(\chi(s)\), we consider the matching \[\{\overline{1n}\}\cup\{\frac{n}{2}, \frac{n}{2}+1\}\cup T_4\cup T'_4.\] Here \(T_4\) is a matching in \(\{\overline{23},\overline{34},\cdots, \overline{\frac{n}{2}-2,\frac{n}{2}-1}\}\) (to avoid developing common vertex with \(\overline{1n}\) or \(\overline{\frac{n}{2}, \frac{n}{2}+1}~)\) and \(T'_4\) its reflection in \(s\). We also consider \(w\) and \(w'\), which are the corresponding monomials to \(T_4\) and \(T_4'\), respectively.
  By similar arguments as before, we have eigenvectors with eigenvalues \(+1\). The number of such eigenvectors equals the number of matchings \(T_4\subset\{\overline{23},\overline{34}, \cdots,\overline{\frac{n}{2}-2,\frac{n}{2}-1}\}\) which is given by Equations (2.17)and (2.21): \[\label{chi3'} [M(\frac{n}{2}-2)](q)=F_{\frac{n}{2}-2}(q)=\sum_{k=0}^{\lfloor\frac{\frac{n}{2}-2}{2}\rfloor}{\frac{n}{2}-2-k\choose k}q^{k}.----(2.32)\] However, while the number of matchings \(T_4\) equals that of \(\{\overline{1n}\}\cup\{\frac{n}{2}, \frac{n}{2}+1\}\cup T_4\cup T'_4\), the number of edges in each matching of the latter is \(2k+2\) compared to \(k\) edges in \(T_4\).
  Hence our last contribution to \(\chi(s)\) is \[\label{chi3} [\chi_3^{}(s)](q)=\sum_{k=0}^{\lfloor\frac{\frac{n}{2}-2}{2}\rfloor}{\frac{n}{2}-2-k\choose k}q^{2k+2}=q^2F_{\frac{n}{2}-2}(q^2).----(2.33)\]

Now, adding up the contributions from Equations (2.29), (2.31) and (2.33) we arrive at: \[(q)=\sum_{k=0}^{\lfloor\frac{n}{4}\rfloor}{\frac{n}{2}-k\choose k}q^{2k}-2\sum_{k=0}^{\lfloor\frac{\frac{n}{2}-1}{2}\rfloor}{\frac{n}{2}-1-k\choose k}q^{2k+1}+\sum_{k=0}^{\lfloor\frac{\frac{n}{2}-2}{2}\rfloor}{\frac{n}{2}-2-k\choose k}q^{2k+2}.\] This expresion is written in the following \(q\)-Fibonacci form using (2.21): \[(q)=F_{\frac{n}{2}}(q^2)-2qF_{\frac{n}{2}-1}(q^2) +q^2F_{\frac{n}{2}-2}(q^2).\] This completes the proof. ◻

Example 2.22.. For \(n=6\) we have \([\chi(s)](q)=F_3(q^2)-2qF_2(q^2) +q^2F_1(q^2)\).
Let \(q=1\), then \(\chi(s)\to F_3-2F_2+F_1=3-2(2)+1=0\).

Lemma 2.23.. For even \(n\), Let \(s, r\in D_n\), be realized in permutation group \(S_n\) as
\(s=(1,n)(2, n-1)\cdots (\frac{n}{2}, \frac{n}{2}+1)\) and \(r=(1,2,\cdots,n)\), in an \(n\)-cycle graph \(C_n\). Then we have: \[\label{char3} [\chi(sr)](q)=\sum_{k=0}^{\lfloor\frac{\frac{n}{2}-1}{2}\rfloor}{\frac{n}{2}-1-k\choose k}q^{2k}, ~n\geq 4,----(2.34)\] and in \(q\)-Fibonacci form according to total degree decomposition, we have: \[(q)=F_{\frac{n}{2}-1}(q^2),~n\geq 4,----(2.35)\] where \(\chi(sr)\) is the trace of the representation of \(sr\) over \(\overline{FK}_{C_n}(n)\).

Proof. Considering \(s\) and \(r\) as defined in the statement we will have: \[sr=(1,n-1)(2,n-2)\cdots (\frac{n}{2}-1,\frac{n}{2}+1).\] Therefore the only matchings in the \(n\)-cycle \(C_n\), symmetric about \(sr\), are \(T\cup T'\), where \(T\) is a matching in \(\{\overline{12},\overline{23},\cdots, \overline{\frac{n}{2}-2,\frac{n}{2}-1}\}\) and \(T'\) its reflection in \(sr\). The corresponding monomials to \(T\) and \(T'\) are \(w\) and \(w'\), respectively.

we have eigenvectors of \(sr\) with eigenvalue \(+1\), as \(sr(ww')=ww'\) as before. The number of such eigenvectors equals the number of matchings \(T\cup T'\) in \(C_n\) which equals the number of matchings \(T\) in \(\{\overline{12},\overline{23},\cdots, \overline{\frac{n}{2}-2,\frac{n}{2}-1}\}\) which is by (2.17) and (2.21) equal to: \[(q)=M(\frac{n}{2}-1)(q)=F_{\frac{n}{2}-1}(q)=\sum_{k=0}^{\lfloor\frac{\frac{n}{2}-1}{2}\rfloor}{\frac{n}{2}-1-k\choose k}q^k.\] However, while the number of matchings \(T\cup T'\) equals that of \(T\), but the number of edges in each matching \(T\cup T'\) is \(2k\) compared to the \(k\) edges of \(T\). Therefore we have: \[(q)=\sum_{k=0}^{\lfloor\frac{\frac{n}{2}-1}{2}\rfloor}{\frac{n}{2}-1-k\choose k}q^{2k}=F_{\frac{n}{2}-1}(q^2).\] This completes the proof. ◻

Example 2.24. For \(n=6\), we have \(\chi(sr)=\sum_{k=0}^1{2-k\choose k}q^{2k}={2\choose 0}+{1\choose 1}q=1+q^2\).

Lemma 2.25. For odd \(n\), Let \(s\in D_n\) realized in permutation group \(S_n\) as
\(s=(1,n)(2, n-1)\cdots (\frac{n-1}{2}, \frac{n+3}{2})(\frac{n+1}{2})\) in an \(n\)-cycle graph \(C_n\). Then \[\label{char2} [\chi(s)](q)=\sum_{k=0}^{\lfloor\frac{n-1}{4}\rfloor}{\frac{n-1}{2}-k\choose k}q^{2k}-\sum_{k=0}^{\lfloor\frac{n-3}{4}\rfloor}{\frac{n-3}{2}-k\choose k}q^{2k+1},----(2.36)\] with the \(q\)-Fibonacci version: \[(q)=F_\frac{n-1}{2}(q^2)-qF_\frac{n-3}{2}(q^2),\] where \(\chi(s)\) is the trace of the representation of \(s\) over \(\overline{FK}_{C_n}(n)\).

Proof.

• Consider a matching \(T_1\): \[T_1=\{\overline{i_1,i_1+1}, \overline{i_2,i_2+1}\cdots \overline{i_m,i_m+1}\}\] of the following set of line segments. \[S=\{\overline{12}, \overline{23}, \cdots,\overline{\frac{n-1}{2}-1, \frac{n-1}{2}}\}.\] where the corresponding monomial of \(S\) is: \[u=x_{i_1,i_1+1}x_{i_2,i_2+1}\cdots x_{i_m,i_m+1}.\] Let the reflection of \(T_1\) in \(s\), itself a matching, be \(T'_1\): \[T'_1=\{\overline{j_1,j_1+1}, \overline{j_2,j_2+1}\cdots \overline{j_m,j_m+1}\},\] with its corresponding monomial: \[u'=x_{j_1j_1+1}x_{j_2j_2+1}\cdots x_{j_mj_m+1}.\]

  Although the matching \(T_1\) is not symmetric about \(s\), the matching \(T_1\cup T'_1\) is symmetric. The monomial \(uu'\) associated with it forms an eigenvector of \(s\) with eigenvalue \(+1\), as shown previously: \(s(uu')=uu'\).

  The number of such eigenvectors equals the number of the matchings \(T_1\cup T_1'\) in \(C_n\) which equals the number of matchings \(T_1\) in \(\{\overline{12}, \overline{23}, \cdots,\overline{\frac{n-1}{2}-1,\frac{n-1}{2}}\}\) which is by (2.17) and (2.21) equal to: \[\label{char2'} [M(\frac{n-1}{2})](q)=F_\frac{n-1}{2}(q)=\sum_{k=0}^{\lfloor\frac{n-1}{4}\rfloor}{\frac{n-1}{2}-k\choose k}q^{k},----(2.37)\] While the number of matchings \(T\cup T'\) equals that of \(T\), but the number of edges in each matching \(T\cup T'\) is \(2k\) compared to \(k\) edges in \(T\). Hence our first contribution to \(\chi(s)\) is \[\label{char3_''} [\chi_1^{}(s)](q)=\sum_{k=0}^{\lfloor\frac{n-1}{4}\rfloor}{\frac{n-1}{2}-k\choose k}q^{2k}=F_{\frac{n-1}{2}}(q^2).----(2.38)\]

• We also have another set of matchings symmetric about \(s\). These matchings are formed by adding the edge \(\overline{1n}\) (corresponding variable \(x_{1n}\)), to avoid common vertex with \(\overline{1n}\), but only to each matching: \[T_2\subset\Bigg\{\overline{23},\overline{34}, \cdots,\overline{\frac{n-1}{2}-1,\frac{n-1}{2}}\Bigg\}.\] Let the reflection of \(T_2\) in \(s\) be \(T_2'\), then \(\{\overline{1n}\}\cup T_2\cup T_2'\) is symmetric about \(s\) and its correspondent monomial \(x_{12}ww'\) is an eigenvector of \(s\) with eigenvalue \(-1\), because, as before, we have: \[s(x_{1n}ww')=-x_{12}s(ww')=-x_{12}ww'.\]

  The number of such eigenvectors is equal to the number of matchings \(\{\overline{1n}\}\cup T_2\cup T_2'\) in \(C_n\) which is equal to the number of matchings \(T_2\subset\{\overline{23},\overline{34}, \cdots,\overline{\frac{n-1}{2}-1,\frac{n-1}{2}}\}\) which by (2.17) and (2.21) is equal to: \[M(\frac{n-1}{2}-1)=M(\frac{n-3}{2})=F_\frac{n-3}{2}=\sum_{k=0}^{\lfloor\frac{n-3}{4}\rfloor}{\frac{n-3}{2}-k\choose k}q^{k}.\]

  However, while the number of matchings \(\{\overline{1n}\}\cup T_2\cup T_2'\) equals that of \(T_2\), but the number of edges in each matching \(\{\overline{1n}\}\cup T_2\cup T_2'\) is \(2k+1\) compared to \(k\) edges in \(T_2\). Hence our 2nd contribution to \(\chi(s)\) is: \[\label{char2''} [\chi_2^{}(s)](q)=-\sum_{k=0}^{\lfloor\frac{n-3}{4}\rfloor}{\frac{n-3}{2}-k\choose k}q^{2k+1}=-qF_{\frac{n-3}{2}}(q^2).----(2.39)\]where the negative sign is due to the \(-1\) eigenvalue above.

Adding Equations (2.38) and (2.39) we arrive at: \[(q)=\sum_{k=0}^{\lfloor\frac{n-1}{4}\rfloor}{\frac{n-1}{2}-k\choose k}q^{2k}-\sum_{k=0}^{\lfloor\frac{n-3}{4}\rfloor}{\frac{n-3}{2}-k\choose k}q^{2k+1},\] with its \(q\)-Fibonacci form according to the total degree decomposition: \[(q)=F_\frac{n-1}{2}(q^2)-qF_\frac{n-3}{2}(q^2).\] This completes the proof. ◻

Example 2.26. For \(n=5\) we have \([\chi(s)](q)=\sum_{k=0}^1{2-k\choose k}q^{2k}-\sum_{k=0}^0{1-k\choose k}q^{2k+1}={2\choose 0}+{1\choose 1}q^2-{1\choose 0}q=1-q+q^2\).
Let \(q=1\), then \(\chi(s)=1\).

Remark 2.27. For odd \(n\), there is no symmetry in \(C_n\) about \(sr\), therefore we do not discuss the case of \(\chi(sr)\).

Character of \(\overline{FK}_{C_n}(n)\) over \(D_n\) and decomposition

Character of \(\overline{FK}_{C_n}(n)\) over \(D_n\) for even \(n\)

We combine (2.23), (2.26) and (2.34) to obtain the character of \(\overline{FK}_{C_n}(n=\text{even})\) over \(D_n\): \[\label{ceven1}\begin{split} char_{D_n}&[\overline{FK}_{C_n}(n=\text{even})]=\Bigg(\sum_{k=0}^{\frac{n}{2}}\frac{n}{n-k}{n-k\choose k}q^k, 1, 1+gcd(n,p)q^\frac{n}{gcd(n, p)}, 1+\frac{n}{2}q^2,\\&\sum_{k=0}^{\lfloor\frac{n}{4}\rfloor}{\frac{n}{2}-k\choose k}q^{2k}-2\sum_{k=0}^{\lfloor\frac{\frac{n}{2}-1}{2}\rfloor}{\frac{n}{2}-1-k\choose k}q^{2k+1}+\sum_{k=0}^{\lfloor\frac{\frac{n}{2}-2}{2}\rfloor}{\frac{n}{2}-2-k\choose k}q^{2k+2}, \\&\sum_{k=0}^{\lfloor\frac{n-2}{4}\rfloor}{\frac{n-2}{2}-k\choose k}q^{2k}\Bigg)----(3.1). \end{split}\] In terms of \(q\)-Fibonacci/\(q\)-Lucas polynomials, with initial values \(F_0=1\), \(F_1=1\), \(L_0=2\), \(L_1=1\) we obtain: \[\label{ceven1f} \begin{split} char_{D_n}[\overline{FK}_{C_n}(n=\text{even})]=\\\Bigg(L_n(q), ~1, ~\overbrace{1+gcd(n,p)q^{\frac{n}{gcd(n, p)}}}^{(\frac{n}{2}-2)~\text{columns}},~1+\frac{n}{2}q^2,~ F_\frac{n}{2}(q^2)-2qF_{\frac{n}{2}-1}(q^2)+q^2F_{\frac{n}{2}-2}(q^2),~F_\frac{n-2}{2}(q^2)\Bigg), \end{split}\] where \(F_n(q)=\sum_0^{\lfloor\frac{n}{2}\rfloor}{n-k\choose k}q^k\) is the \(q\)-Fibonacci polynomial, and \(1+gcd(n,p)q^{\frac{n}{gcd(n, p)}}\) is a sequence of \((\frac{n}{2}-2)\) terms forming the characters \(\chi(r^p)\) for \(p=2,\cdots, (\frac{n}{2}-1)\).

3.2. Decomposition in irreducible characters

In general a decomposition of a character \(\chi\) in irreducible characters \(\chi^{(i)}\) of a group \(G\), is done according to the following formula .
\(\chi=\sum_im_i\chi^{(i)},~ \text{with the coefficients}~ m_i=\langle\chi,\chi^{(i)}\rangle=\frac{1}{|G|}\sum_{K}|K|\chi_K^{}\overline{\chi}_K^{(i)},\) where the sum is over conjugacy classes.
We write our character in terms of irreducible characters of \(D_n\): \[\label{char5}\begin{split} char_{D_n}[\overline{FK}_{C_n}(n=\text{even})]=m_{1^n}\chi^{1^n}+m_{2^{\frac{n}{2}}}\chi^{2^{\frac{n}{2}}}&\\+\sum_{2\leq t\leq \frac{n}{2}-1,gcd(n,t)\ge 2}m_{(\frac{n}{gcd(n,t)})^{gcd(n,t)}}\chi^{(\frac{n}{gcd(n,t)})^{gcd(n,t)}}+m_{2^{(\frac{n}{2}-1)}11}\chi^{2^{(\frac{n}{2}-1)}11}+m_{2^{\frac{n}{2}'}}\chi^{2^{\frac{n}{2}'}} \end{split}----(3.3)\] The coefficients of the irreducible characters are calculated using the character table of \(D_n\), Table 3.2, with the character \(char_{D_n}[\overline{FK}_{C_n}(n=\text{even})]\) of Equation 3.2 appearing in the bottom row. We obtain:

Table: 1

\[\label{coef''}\begin{split} m_{1^n}=\frac{1}{2n}\Bigg\{L_n(q)+2+2\sum_{p=2}^{\frac{n}{2}-1}(1+gcd(n,p)q^\frac{n}{gcd(n, p)})+(1+\frac{n}{2}q^2)\\+\frac{n}{2}\Big[F_{\frac{n}{2}}(q^2)+(1-2q)F_{\frac{n}{2}-1}(q^2)+q^2F_{\frac{n}{2}-2}(q^2)\Big]\Bigg\}\\ m_{n}=\frac{1}{2n}\Bigg\{2L_n+4cos\frac{2\pi}{n}+4\sum_{p=2}^{\frac{n}{2}-1}(1+gcd(n,p)q^\frac{n}{gcd(n,p)})cos\frac{2p\pi}{n}-2(1+\frac{n}{2}q^2)\Bigg\}\\ m_{(\frac{n}{gcd(n,t)})^{gcd(n,t)}}=\frac{1}{2n}\Bigg\{2L_n+4cos\frac{2t\pi}{n}+4\sum_{p=2}^{\frac{n}{2}-1}(1+gcd(n,p)q^\frac{n}{gcd(n,p)})cos\frac{2pt\pi}{n}\\+2(-1)^t(1+\frac{n}{2}q^2)\Bigg\},~2\leq t\leq\frac{n}{2}-1\\ m_{2^{\frac{n}{2}}}=\frac{1}{2n}\Bigg\{L_n(q)+2+2\sum_{p=2}^{\frac{n}{2}-1}(1+gcd(n,p)q^\frac{n}{gcd(n, p)})+(1+\frac{n}{2}q^2)\\-\frac{n}{2}\Big[F_{\frac{n}{2}}(q^2)+(1-2q)F_{\frac{n}{2}-1}(q^2)+q^2F_{\frac{n}{2}-2}(q^2)\Big]\Bigg\}\\ m_{2^{\frac{n}{2}'}}= \frac{1}{2n}\Bigg\{L_n(q)-2+2\sum_{p=2}^{\frac{n}{2}-1}(-1)^p(1+gcd(n,p)q^\frac{n}{gcd(n, p)})+(-1)^{\frac{n}{2}}(1+\frac{n}{2}q^2)\\+\frac{n}{2}\Bigg[F_\frac{n}{2}(q^2)-(1+2q)F_{\frac{n}{2}-1}(q^2)+q^2F_{\frac{n}{2}-2}(q^2)\Bigg]\Bigg\}\\ m_{2^{(\frac{n}{2}-1)}1^2}= \frac{1}{2n}\Bigg\{L_n(q)-2+2\sum_{p=2}^{\frac{n}{2}-1}(-1)^p(1+gcd(n,p)q^\frac{n}{gcd(n, p)})+(-1)^{\frac{n}{2}}(1+\frac{n}{2}q^2)\\-\frac{n}{2}\Bigg[F_{\frac{n}{2}}(q^2)-(1+2q)F_{\frac{n}{2}-1}(q^2)+q^2F_{\frac{n}{2}-2}(q^2)\Bigg]\Bigg\}-----(3.4) \end{split}\]

Example 3.1. Let
\(char_{C_6}[\overline{FK}_{C_6}(6, q)]=m_{1^6}\chi^{1^6}+m_{2^3}\chi^{2^3}+m_{2^{3'}}\chi^{}_{2^{3'}}+m_{2^21^2}\chi^{2^21^2}+m_{6}\chi^{6}+m_{3^2}\chi^{3^2}\). Then from Equation 3.4 we have \[m_{1^6}= 1+2q^2, ~m_6=q+q^2,~m_{3^2}=q+2q^2,~m_{2^3}=q+q^3, m_{2^{3'}}=q^2,~m_{2^2 1^2}=q+q^3.\] Substitution of the above coefficients into the character Equation (3.3) yields: \[\label{char_'}\begin{split}char_{D_6}[\overline{FK}_{C_6}(6,~ q)]=&(1+2q^2)\chi^{1^6}+(q+q^3)\chi^{2^3}+(q^2)\chi^{2^{3'}}\\&+(q+q^3)\chi^{2^21^2}+(q+q^2)\chi^{6}+(q+2q^2)\chi^{3^2}. \end{split}----(3.5)\]

Table: 2 character table of \(D_n\) for odd \(n\).

Sorting in \(q\) we have the decomposition of \(char_{D_6}[\overline{FK}_{C_6}(6)]\) in usual degrees. \[\label{dec_'}\begin{split} char_{D_6}[\overline{FK}_{C_6}(6)]=&\chi^{}_{1^6}+(\chi^{}_{222}+\chi^{}_{2211}+\chi^{}_{6}+\chi^{}_{33})q\\&+(2\chi^{}_{1^6}+\chi^{}_{222'}+\chi^{}_{6}+2\chi^{}_{33})q^2+(\chi^{}_{222}+\chi^{}_{2211})q^3, \end{split}----(3.6)\] which upon putting \(q=1\) reduces to: \[char_{D_6}\overline{FK} _{C_6}(6) =3\chi_{1^6}^{}+2\chi_{222}^{}+\chi_{222'}^{}+2\chi_{2211}^{}+2\chi_{6}^{}+3\chi_{33}^{}.\]

Character of \(\overline{FK}_{C_n}(n)\) over \(D_n\) for odd \(n\) and decomposition

We combine (2.23) and (2.36), to obtain the character of \(\overline{FK}_{C_n}(n=odd)\) over \(D_n\): \[\label{codd}\begin{split} char_{D_n}[\overline{FK}_{C_n}(n=\text{odd})]=&\Bigg( \sum_{k=0}^\frac{n-1}{2}\frac{n}{n-k}{n-k\choose k}q^k,~ 1,~ 1+pq^\frac{n}{p}\delta_{\frac{n}{p},\text{integer}},\\&\sum_{k=0}^{\lfloor\frac{n-1}{4}\rfloor}{\frac{n-1}{2}-k\choose k}q^{2k}-\sum_{k=0}^{\lfloor\frac{n-3}{4}\rfloor}{\frac{n-3}{2}-k\choose k}q^{2k+1}\Bigg)----(3.7). \end{split}\] In terms of \(q\)-Fibonacci/\(q\)-Lucas polynomials, with initial values \(F_0=1\), \(F_1=1\), \(L_0=2\), \(L_1=1\) we have: \[\label{codd'}\begin{split} char_{D_n}[\overline{FK}_{C_n}(n=\text{odd})]=&\Bigg( L_n(q),~1,~\overbrace{1+gcd(n,p)q^{\frac{n}{gcd(n, p)}}}^{(\frac{n-3}{2})~\text{columns}},~F_\frac{n-1}{2}(q^2)-qF_\frac{n-3}{2}(q^2) \Bigg)----(3.8). \end{split}\]

3.4. Decomposition in irreducible characters

Consider the quotient of the Fomin-Kirillov algebra \(\overline{FK}_{C_n}(n=\text{odd})\) associated with the n-cycle subgraph of the complete graph on \(n\) vertices, we have: \[\label{char5'}\begin{split} char_{D_n}[\overline{FK}_{C_n}(n=\text{odd})]=m_{1^n}\chi^{1^n}+ m_{n}\chi^n\\+\sum_{2<t\leq \frac{n-3}{2}}m_{(\frac{n}{gcd(n,t)})^{gcd(n,t)}}\chi^{(\frac{n}{gcd(n,t)})^{gcd(n,t)}}+ m_{(\frac{n}{2})^2}\chi^{(\frac{n}{2})^2}+m_{2^{\frac{n-1}{2}}}\chi_{2^{\frac{n-1}{2}}}. \end{split}----(3.9)\] The coefficients \(m_i\) are calculated using the irreducible characters of \(D_n\) for odd \(n\), as shown in Table 2, where the character of \(\overline{FK}_{C_n}(n)\) appears in the bottom row, and by using formula: \(m_i=\langle\chi,\chi^{(i)}\rangle=\frac{1}{|G|}\sum_{K}|K|\chi_K^{}\overline{\chi}_K^{(i)}\). We obtain the following coefficients.

Example 3.2. Let \(n=5\), then \[char_{D_5}[\overline{FK}_{C_5}(5)]=m_{1^5}\chi_{1^5}^{}+ m_{2^21}\chi^{2^21}+m_{5}\chi^5+m_{5'}\chi^{5'},\] where the \(m\)-coefficients derived from Equation (3.10) are: \[m_{1^5}=1+q^2,~ m_{2^21}=q,~m_{5}=q+q^2,~ m_{5'}=q+q^2.\] Substituting these m-coefficients into the character, Equation (3.9), we obtain: \[char_{D_5}[\overline{FK}_{C_5}(5)](q)=(1+q^2)\chi^{1^5}+ q\chi^{2^21}+(q+q^2)\chi^{5}+(q+q^2)\chi^{5'}.\]Sorting the above in terms of \(q\) yields the decomposition in usual degree: \[char_{D_5}[\overline{FK}_{C_5}(5)](q)=\chi^{1^5}+ (\chi^{2^21}+\chi^{5}+\chi^{5'})q+(\chi^{1^5}+\chi^{5}+\chi^{5'})q^2,\] which upon putting \(q=1\) reduces to \[char_{D_5}\overline{FK}_{C_5}(5)=2\chi^{1^5}+\chi^{2^21}+2\chi^{5'}+2\chi^{5}.\]

3.4.1. Representation decomposition of \(D_n\) by conjugation class Since the basis of \(\overline{FK}_{C_n}(n)\) consists of monomials in variables with no common index (alternatively, matchings in an \(n\)-cycle), the \(S_n\)-degree of an element of the basis is product of disjoint \(2\)-cycles and \(1\)-cycles. So it belongs to the \(S_n\)-conjugacy class denoted by \(t_2^kt_1^{n-2k}\) where \(t_2\) and \(t_1\) stand for \(2\)-cycle and \(1\)-cycle respectively and where \(k\) is the degree of monomial. Since we have the same partition of the basis invariant under \(S_n\)-conjugacy class as under usual degree, we have essentially the same decomposition as by usual degree.

3.4.2. Representation decomposition of \(D_n\) by set partition typeSince no two variable appearing in a monomial \(M\in B[\overline{FK}_{C_n}(n)]\) share common indexes, each part of the set partition contains at most \(2\) indexes. Therefore, the set partition type for a degree \(k\) monomial \(M\) is of the form: \[\alpha=(\underbrace{2, 2, \cdots, 2}_k,\underbrace{1, 1, \cdots, 1}_{n-2k}), ~\text{denoted by}~ 2^k1^{n-2k},\] In this notation, each \('2'\) in \(\alpha\) represents an index of a variable if the variable appears in \(M\), while each \('1'\) represents an index that does not appear in any variable in \(M\). Since the partition of the basis remains invariant under set partition type as under usual degree, we essentially have the same decomposition as by usual degree.

Conclusion

We introduced a quotient of the Fomin-Kirillov algebra \(FK(n)\), denoted by \(\overline{FK}_{C_n}(n)\), associated with the \(n\)-cycle subgraph of the complete graph on \(n\) vertices. Through this quotient, We established an interesting connection between the Fomin-Kirillov algebra \(FK(n)\) and the theory of \(q\)-Fibonacci/\(q\)-Lucas polynomials. Specially:

• The Hilbert series of this quotient algebra corresponds to the \(q\)-Lucas polynomial.

• The dimension of the quotient algebra equals the Lucas number \(L_n\).

• We derived general formulas for the character of \(\overline{FK}_{C_n}(n)\) over the Dihedral group \(D_n\) for both even and odd \(n\).

• Notably, the representation decomposition of this quotient algebra for usual degree remains essentially the same as for both \(S_n\) degree and set partition degree.

Acknowledgement

I extend my sincere gratitude to Professor Bergeron for reviewing this paper and providing valuable insights.

Conflict of Interests:

The author declares that there is no conflicts of interest.

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