Algebraic degree of Cayley graphs over abelian groups and dihedral groups

Abstract

For a graph $\Gamma$, let K be the smallest field containing all eigenvalues of the adjacency matrix of $\Gamma$. The algebraic degree $\deg (\Gamma )$ is the extension degree $[K:\mathbb {Q}]$. In this paper, we completely determine the algebraic degrees of Cayley graphs over abelian groups and dihedral groups.

1 Introduction

The algebraic degree of a graph was defined in [10] in order to generalize the concept of integral graphs. The spectrum of a graph $\Gamma$ is defined as the multiset of eigenvalues of the adjacency matrix of $\Gamma$. In particular, those eigenvalues are the roots of the monic characteristic polynomial of the adjacency matrix associated with $\Gamma$. Therefore, every eigenvalue of $\Gamma$ is an algebraic integer in some algebraic extension K of the rationals, where K is called the splitting field of $\Gamma$. The algebraic degree $\deg (\Gamma )$ is defined as the degree $[K : \mathbb Q]$. In particular, $\Gamma$ is called integral if $\deg (\Gamma ) = 1$.

The Cayley graph $\textrm{Cay}(G,S)$ is defined as the graph with vertex set G, where G denotes a finite group and $S \subseteq G$, and edges from $g \in G$ to $h \in G$ whenever $gh^{-1} \in S$. Note that $\textrm{Cay}(G,S)$ is an undirected graph if and only if $S = S^{-1}$, and has loops if and only if $e \in S$. If $S \ne S^{-1}$, then $\textrm{Cay}(G,S)$ is also called Cayley digraph.

In [8] and [9], Mönius precisely determined the algebraic degree of circulant digraphs, i.e. Cayley digraphs over cyclic groups. Moreover, integral Cayley graphs were studied intensively by several authors, e.g. Lu [7], Klotz and Sander [5, 6] and Ahmady et al. [1].

In this paper, we completely determine the splitting fields of Cayley graphs over abelian and dihedral groups. In particular, we precisely compute the algebraic degree of Cayley graphs and digraphs over abelian groups. We also give an upper bound for the algebraic degree of Cayley graphs and digraphs over dihedral groups, as well as a lower bound for the algebraic degree of Cayley graphs over dihedral groups.

2 Cayley graphs and digraphs over abelian groups

Let G be an abelian group of order n and let $S \subseteq G$ be a subset of G. Denote by $\Gamma =\textrm{Cay}(G,S)$ the respective Cayley (di)graph, and let K be the splitting field of $\Gamma$, i.e. the minimum field containing all eigenvalues of $\Gamma$. Without loss of generality, assume that $G=\mathbb {Z}_{n_1}\times \mathbb {Z}_{n_2}\times \cdots \times \mathbb {Z}_{n_r}$, where $n=n_1n_2\cdots n_r$. Therefore, each element $g\in G$ can be expressed as $g=(g_1,g_2,\ldots ,g_r)$. For a positive integer m, denote by $\zeta _{m}=e^{2\pi \textbf{i}/m}$ the primitive m-th root of unity, where $\textbf{i}=\sqrt{-1}$. The eigenvalues of $\Gamma$ were obtained by Babai [3]:

Lemma 1

([3]) The eigenvalues $\lambda _{g}$ of $\Gamma$ are given by $\lambda _g=\sum _{s\in S}\prod _{i=1}^r\zeta _{n_i}^{g_is_i}$, for $g \in G$.

It is clear that $\textrm{Gal}(\mathbb {Q}(\zeta _n)/\mathbb {Q})\cong \mathbb {Z}_n^*$. Let $\eta$: $\textrm{Gal}(\mathbb {Q}(\zeta _n)/\mathbb {Q})\rightarrow \mathbb {Z}_n^*$ be the isomorphism defined by $\eta (\sigma )=k$ for any $\sigma \in \textrm{Gal}(\mathbb {Q}(\zeta _n)/\mathbb {Q})$, where $k\in \mathbb {Z}_n^*$ is the integer such that $\sigma (\zeta _n)=\zeta _n^{k}$. Let $\mathbb {Z}_n^*$ act on G by $ag=a(g_1,g_2,\ldots ,g_r)=(ag_1,ag_2,\ldots ,ag_r)$ for any $a\in \mathbb {Z}_n^*$ and $g\in G$. This leads to $\sigma (\zeta _{n_i}^k)=\sigma (\zeta _{n}^{kn/n_i})=\zeta _n^{\eta (\sigma )kn/n_i}=\zeta _{n_i}^{\eta (\sigma )k}$. Therefore, for any $\sigma \in \textrm{Gal}(\mathbb {Q}(\zeta _n)/\mathbb {Q})$ and $g\in G$, we have

$$\begin{aligned} \sigma (\lambda _g)=\sigma \left( \sum _{s\in S}\prod _{i=1}^r\zeta _{n_i}^{g_is_i}\right) =\sum _{s\in S}\prod _{i=1}^r\sigma (\zeta _{n_i}^{g_is_i})=\sum _{s\in S}\prod _{i=1}^r\zeta _{n_i}^{\eta (\sigma )g_is_i}. \end{aligned}$$

(1)

Let $S = \{ (s_1, \ldots , s_r) \mid s_i \in \mathbb Z_{n_i} \}$. We say that a subgroup $H \subseteq \mathbb Z_n^*$ is fixing S if and only if $hS = \{(hs_1 \bmod n_1, \ldots , hs_r \bmod n_r) \mid s_i \in \mathbb Z_{n_i} \} = S$ for all $h \in H$.

Subsequently, let $H=\eta (\textrm{Gal}(\mathbb {Q}(\zeta _n)/K))$. According to (1), Li [4] showed the following result:

Lemma 2

([4]) For all $g \in G$, the eigenvalue $\lambda _g$ is contained in K if and only if S is a union of some orbits Hx for $x\in G$.

Note that $\varphi (n)=[\mathbb {Q}(\zeta _n):\mathbb {Q}]=[\mathbb {Q}(\zeta _n):K][K:\mathbb {Q}]$. From Lemma 2, we immediately get the following result:

Theorem 1

Let $\mathcal H=\{h\in \mathbb {Z}_n^*\mid hS=S\}$ be the largest subgroup of $\mathbb Z_n^*$ fixing S. Then, the splitting field of $\Gamma$ is given by

$$\begin{aligned} K=\mathbb {Q}(\zeta _n)^{\eta ^{-1}(\mathcal H)}=\{x\in \mathbb {Q}(\zeta _n)\mid \sigma x=x,\forall \sigma \in \eta ^{-1}(\mathcal H)\}. \end{aligned}$$

Therefore, $\mathcal H = H$ and the algebraic degree of $\Gamma$ is

$$\begin{aligned} \textrm{deg}(\Gamma )=\frac{\varphi (n)}{|H|}. \end{aligned}$$

Proof

Since $\mathcal H$ is a subgroup fixing S, we see that S is a union of some orbits and, therefore, by Lemma 2, all eigenvalues of $\Gamma$ belong to $\mathbb {Q}(\zeta _n)^{\eta ^{-1}(\mathcal H)}$. Now, let L be a field containing all eigenvalues of $\Gamma$, then, again by Lemma 2, S is a union of some orbits $\eta (\textrm{Gal}(\mathbb {Q}(\zeta _n)/L)x$ for $x\in G$. This means that $\eta (\textrm{Gal}(\mathbb {Q}(\zeta _n)/L)$ fixes S. Since $\mathcal H$ is the largest subgroup of $\mathbb Z_n^*$ fixing S, we have that $\eta (\textrm{Gal}(\mathbb {Q}(\zeta _n)/L)\le \mathcal H$ and, thus, $\mathbb {Q}(\zeta _n)^{\eta ^{-1}(\mathcal H)}\subseteq L$. Therefore, $\mathbb {Q}(\zeta _n)^{\eta ^{-1}(\mathcal H)}$ must be the smallest field containing all eigenvalues of $\Gamma$, i.e. $K = \mathbb {Q}(\zeta _n)^{\eta ^{-1}(\mathcal H)}$ and $\mathcal H = H$. $\square$

Example 1

(Integral Cayley graph over abelian group) Let $G=\mathbb {Z}_2\times \mathbb {Z}_4$ and $S=\{(0,1),(1,0),(0,-1)\}$. Note that $\mathbb {Z}_8^*=\{1,3,-3,-1\}$, and

$$\begin{aligned} 3S=-3S=\{(0,-1),(1,0),(0,1)\}=S=-S. \end{aligned}$$

Therefore, $H=\mathbb {Z}_8^*$ and $\textrm{deg}(\Gamma )=1$, i.e., $\Gamma$ is integral. In fact, the spectrum of $\Gamma$ is $\left\{ \pm 3,[\pm 1]^3\right\}$.

Example 2

(Cayley graph over abelian group of algebraic degree 2) Let $G=\mathbb {Z}_4\times \mathbb {Z}_6$ and $S=\{(1,1),(-1,-1),(0,1),(0,-1)\}$. Note that $\mathbb {Z}_{24}^*=\{1,5,7,11,-11,-7,-5,-1\}$, and

$$\begin{aligned} \left\{ \begin{array}{l} 5S=-5S=7S=-7S=\{(1,-1),(-1,1),(0,-1),(0,1)\}\ne S,\\ 11S=-11S=\{(-1,-1),(1,1),(0,1),(0,-1)\}=S=-S. \end{array}\right. \end{aligned}$$

Therefore, $H=\{1,11,-11,-1\}$ and $\textrm{deg}(\Gamma )=2$. In fact, the spectrum of $\Gamma$ is

$$\begin{aligned} \left\{ \pm 4, [\pm 2]^4, [\pm 1\pm \sqrt{3}]^2,[0]^6\right\} . \end{aligned}$$

Example 3

(Cayley digraph over abelian group of algebraic degree 4) Let $G = \mathbb Z_4 \times \mathbb Z_6$ and $S= \{(1,1),(0,1),(0,-1)\}$. We observe that

$$\begin{aligned} \left\{ \begin{array}{l} -7S=5S= \{(1,-1),(0,-1),(0,1)\}\ne S,\\ 7S =-5S = \{(-1,1), (0,1), (0,-1)\} \ne S, \\ 11S = \{(-1,-1),(0,-1),(0,1)\} \ne S,\\ -11S=\{(1,1),(0,1),(0,-1)\}=S. \end{array}\right. \end{aligned}$$

Thus, $H= \{1,-1\}$ and $\deg (\Gamma ) = 4$.

In [9], Mönius solved the Inverse Galois problem for circulant graphs showing that every finite abelian extension of the rationals is the splitting field of some circulant graph. A similar result can be obtained for (non-circulant) Cayley graphs over abelian groups: Let $G=\mathbb {Z}_{n_1}\times \mathbb {Z}_{n_2}\times \cdots \times \mathbb {Z}_{n_r}$ be a non-cyclic abelian group, i.e. $n = n_1n_2 \cdots n_r$ where each $n_i$ is a prime power. For any subgroup H of $\mathbb Z_n^*$, let

$$S = (H \bmod n_1) \times (H \bmod n_2) \times \cdots \times (H \bmod n_r)$$

for $(H \bmod n_i) = \{h \bmod n_i \mid h \in H \}$, $i = 1, \ldots , r$. Then, H is the largest subgroup of $\mathbb Z_n^*$ fixing S and, therefore, the splitting field of $\Gamma = \textrm{Cay}(G,S)$ equals $K = \mathbb Q(\zeta _n)^{\eta ^{-1}(H)}$. Together with the well-known Kronecker–Weber theorem, we get the following result.

Corollary 1

(Inverse Galois problem for Cayley graphs over abelian groups) Every finite abelian extension K of the rationals (of order n) is the splitting field of some Cayley graph over an abelian group. In particular, if n has at least one prime divisor of order $\ge 2$, then there is a non-circulant Cayley graph over an abelian group with splitting field K.

3 Cayley graphs over dihedral groups

In this section, we restrict our considerations to Cayley graphs over dihedral groups, i.e. we always assume that $G=D_n=C_n\rtimes C_2=\langle a,b\mid a^n=b^2=1,bab=a^{-1}\rangle$ and $S\subset G$ is a subset with $e\not \in S$ and $S=S^{-1}$. Let $S=S_1\cup S_2$, where $S_1\subseteq \langle a\rangle$ and $S_2\subseteq b\langle a\rangle$, and $I_1=\{i\in \mathbb Z_n \mid a^i\in S_1\}$. It is clear that $I_1=-I_1$ since $S=S^{-1}$. Moreover, let $\Gamma =\textrm{Cay}(G,S)$ denote the respective Cayley graph and let K be the minimum field containing all eigenvalues of $\Gamma$. Let $\chi _l$ be the irreducible characters of $D_n$ of degree 2 for $1\le l\le \lfloor \frac{n-1}{2}\rfloor$, where $\chi _l(a^k)=2\cos \frac{2\pi lk}{n}$ and $\chi _l(ba^k)=0$. For a subset $A\subseteq G$, let $\chi _l(A)=\sum _{x\in A}\chi _l(x)$ and $\chi _l(A^2)=\sum _{x,y\in A}\chi _l(xy)$. The eigenvalues of $\Gamma$ were obtained by Babai [3] and were restated by Lu [7].

Lemma 3

([3, 7]) The eigenvalues of $\Gamma$ consist of some integers and the roots of

$$\begin{aligned} f_l(x)=x^2-\chi _l(S_1)x+\frac{1}{2}\left( \chi _l(S_1)^2-\left( \chi _l(S_1^2)+\chi _l(S_2^2)\right) \right) , \end{aligned}$$

for $1\le l\le \lfloor \frac{n-1}{2}\rfloor$. In particular, all possibly non-integral eigenvalues are contained in the set

$$\begin{aligned} \left\{ \frac{b_l\pm \sqrt{c_l}}{2}\mid 1\le l\le \lfloor (n-1)/2\rfloor \right\} , \end{aligned}$$

where $b_l=\chi _l(S_1)$ and $c_l=2(\chi _l(S_1^2)+\chi _l(S_2^2))-(\chi _l(S_1))^2.$

Since $I_1=-I_1$, it is clear that

$$b_l=\chi _l(S_1)=\sum _{a^s\in S_1}2\cos \frac{2\pi ls}{n}=2\sum _{i\in I_1}\zeta _n^{li}$$

and $b_l,c_l\in \mathbb {Q}(\zeta _n)$. Let $K_0$ be a field such that $\mathbb {Q}\subseteq K_0\subseteq \mathbb {Q}(\zeta _n)$. Therefore, $\textrm{Gal}(\mathbb {Q}(\zeta _n)/K_0))\le \textrm{Gal}(\mathbb {Q}(\zeta _n)/\mathbb {Q})\cong \mathbb {Z}_n^*$. Recall that $\eta$ is the isomorphism from $\textrm{Gal}(\mathbb {Q}(\zeta _n)/\mathbb {Q})$ to $\mathbb {Z}_n^*$ such that $\sigma (\zeta _n)=\zeta _n^{\eta (\sigma )}$. In what follows, we always assume that $H=\eta (\textrm{Gal}(\mathbb {Q}(\zeta _n)/K_0))$. We first get the following result:

Lemma 4

If $b_1,c_1\in K_0$, then $b_l,c_l\in K_0$ for $1\le l\le \lfloor \frac{n-1}{2}\rfloor$.

Proof

For $1\le l\le \lfloor \frac{n-1}{2}\rfloor$, let $\sigma _l$: $\mathbb {Q}(\zeta _n)\rightarrow \mathbb {Q}(\zeta _n)$ be defined by $\sigma _l(\zeta _n)=\zeta _n^l$. It is clear that $\sigma _l$ is a homomorphism and $b_l=\sigma _l(b_1)$. Thus, for any $\sigma \in \textrm{Gal}(\mathbb {Q}(\zeta _n)/K_0)$, we have

$$\begin{aligned} \sigma (b_l)=\sigma (\sigma _l(b_1))=\sigma \left( \sigma _l\left( 2\sum _{i\in I_1}\zeta _n^{i}\right) \right) =2\sum _{i\in I_1}\zeta _n^{\eta (\sigma )li}=\sigma _l(\sigma (b_1))=\sigma _l(b_1)=b_l. \end{aligned}$$

This leads to $b_l\in K_0$. Analogously, we also get $c_l\in K_0$. $\square$

For a subset $A\subseteq \{1, \ldots , n\}$, denote by $\delta _A$ the characteristic vector of A, that is $\delta _A\in \mathbb {Q}^n$ with $\delta _A(i)=1$ if $i\in A$ and 0 otherwise.

Lemma 5

The number $b_1$ is an element of $K_0$ if and only if $I_1$ is a union of some orbits Hk for $k\in \mathbb {Z}_n$.

Proof

To show the sufficiency, we only need to consider the case where $I_1$ is exactly one orbit. Suppose that $I_1=Hk$. For any $\sigma \in \textrm{Gal}(\mathbb {Q}(\zeta _n)/K_0)$, we have

$$\begin{aligned} \begin{array}{lll} \sigma (b_1)&{}=&{}\sigma \left( 2\sum _{i\in I_1}\zeta _n^i\right) =\sigma \left( 2\sum _{hk\in Hk}\zeta _n^{hk}\right) \\ &{}=&{}2\sum _{hk\in Hk}\sigma (\zeta _n^{hk})=2\sum _{hk\in Hk}\zeta _n^{\eta (\sigma )hk}\\ &{}=&{}2\sum _{h'k\in \eta (\sigma )Hk}\zeta _n^{h'k}=2\sum _{h'k\in Hk}\zeta _n^{h'k}\\ &{}=&{}2\sum _{i\in I_1}\zeta _n^i=b_1.\end{array} \end{aligned}$$

This leads to $b_1\in K_0$.

Conversely, assume that $A_1,A_2,\ldots ,A_r$ have the form $A_i=H{k_i}$ for some $k_i \in \mathbb Z_n$. Let M be the $n\times n$ square matrix indexed by $\mathbb {Z}_n$ with (i, j)-entry being $\zeta _n^{ij}$. It is clear that M is non-singular. Let V, W be vector spaces over $K_0$ defined by $V=\{v\in K_0^n\mid Mv\in K_0^n\}$ and $W=\langle \delta _{A_1},\ldots ,\delta _{A_r}\rangle$, where $\langle \delta _{A_1},\ldots ,\delta _{A_r}\rangle$ denotes the span of the characteristic vectors $\delta _{A_1},\ldots ,\delta _{A_r}$ with $\delta _{A_i} \in K_0^n$. On the one hand, for any $v\in W$, we get $Mv\in K_0^n$ by the same arguments as above, which leads to $W\subseteq V$. On the other hand, if $s,t\in A_i=Hk_i$, then there exists $h\in H$ such that $t=hs$. Let $\sigma =\eta ^{-1}(h)$, i.e. $\sigma (\zeta _n)=\zeta _n^{h}$, and $v \in V$. Since $\sigma \in K_0^{\eta ^{-1}(H)}$, we have that $\sigma ((Mv)_s) = (Mv)_s$ where $(Mv)_s$ denotes the s-th entry of the vector Mv. Moreover, we get

$$\begin{aligned} (Mv)_s = \sigma ((Mv)_s)=\sigma \left( \sum _{x=0}^{n-1}\zeta _n^{sx}v(x)\right) =\sum _{x=0}^{n-1}\zeta _n^{hsx}v(x)=\sum _{x=0}^{n-1}\zeta _n^{tx}v(x)=(Mv)_t. \end{aligned}$$

Thus, for all $v \in V$ we have $(Mv)_s = (Mv)_t$ whenever $s,t \in A_i$. Therefore, Mv is a linear combination of $\delta _{A_1},\ldots ,\delta _{A_r}$, i.e. $Mv\in W$. Hence, $MV\subseteq W$ and, thus, $\dim V\le \dim W$. Since $W \subseteq V$, we get $V=W$. Moreover, since $b_1=2(M\delta _{I_1})_1\in K_0$, by Lemma 4, we have $b_l=2(M\delta _{I_1})_l\in K_0$. Therefore, $M\delta _{I_1}\in K_0^n$ and, thus, $\delta _{I_1}\in V$ by definition of V. Since $V=W$, it follows that $I_1$ is the union of some orbits. $\square$

Lemma 6

For $1\le l\le \lfloor \frac{n-1}{2}\rfloor$, the number $c_l$ is equal to $2\chi _l(S_2^2)$.

Proof

Since $I_1=-I_1$, we have $b_l=\chi _l(S_1)=2\sum _{i\in I_1}\zeta _n^{li}$. By simple calculations, we get

$$\begin{aligned}\begin{array}{lll} 2\chi _l(S_1^2)&{}=&{}2\sum _{a^i,a^j\in S_1}\chi _l(a^{i}a^j)=2\sum _{i,j\in I_1}2\cos \left( \frac{2\pi l(i+j)}{n}\right) \\ &{}=&{}2\sum _{i,j\in I_1}(\zeta _n^{l(i+j)}+\zeta _n^{-l(i+j)})=2\sum _{i\in I_1}\zeta _n^{li}\sum _{j\in I_1}\zeta _n^{lj}\\ {} &{}&{}+2\sum _{i\in I_1}\zeta _n^{-li}\sum _{j\in I_1}\zeta _n^{-lj}\\ &{}=&{}\sum _{i\in I_1}\zeta _n^{li} b_l+\sum _{i\in I_1}\zeta _n^{-li}b_l=b_l(2\sum _{i\in I_1}\zeta _n^i)=b_l^2. \end{array} \end{aligned}$$

Therefore,

$$\begin{aligned} c_l=2(\chi _l(S_1^2)+\chi _l(S_2^2))-(\chi _l(S_1))^2=2\chi _l(S_2^2). \end{aligned}$$

$\square$

A multiset X is a collection of elements where an element may appear more than once. For $x\in X$, denote by $m_X(x)$ the multiplicity of x in X. To avoid confusion, we use $[\cdot ]$ to denote a multiset. For example, $X=[1,1,2,3,3]$ is a multiset and $m_X(1)=2$. Given two multisets X, Y, their multiple XY is a multiset, that is, $XY=[xy\mid x\in X,y\in Y]$, where xy may occur more than once. The multi-union $X\sqcup Y$ is the multiset with $m_{X\sqcup Y}(z)=m_X(z)+m_Y(z)$ for any element z. For example, consider the two multisets of integers $X=[1,1,-1],Y=[1,2]$, then $XY=[1,2,1,2,-1,-2]$ and $X\sqcup Y=[1,1,1,-1,2]$. Denote by $I_2=[k\mid a^k\in S_2^2]$ the multiset of all indices k such that $a^k\in S_2^2$. By Lemma 6, we get the following result. Since the proof is very similar to the one of Lemma 5, we omit it.

Lemma 7

The number $c_1$ is contained in $K_0$ if and only if $I_2$ is a multi-union of some orbits Hk for $k\in \mathbb {Z}_n$.

Combining Lemma 5 and Lemma 7, we get the following result. The proof is very similar to the one of Theorem 1 and, therefore, we omit it, too.

Theorem 2

Let $H=\{h\in \mathbb {Z}_n^*\mid hI_1=I_1,hI_2=I_2\}$ be the subgroup fixing both, $I_1$ and $I_2$. Then, $K=K_0(\sqrt{c_1},\ldots ,\sqrt{c_l})$, where $K_0=\mathbb {Q}(\zeta _n)^{\eta ^{-1}(H)}=\{x\in \mathbb {Q}(\zeta _n)\mid \sigma x=x,\forall \sigma \in \eta ^{-1}(H)\}$.

Assume that $\{k_1,\ldots ,k_r\}$ is a maximum subset of $\mathbb Z_n$ such that all the orbits $Hk_1,Hk_2,\ldots ,Hk_r$ are distinct. The set $R(H)=\{k_1,\ldots ,k_r\}$ is called a representative of H. Suppose that $I_2=m_1\circ Hk_1\sqcup m_2\circ Hk_2\sqcup \cdots \sqcup m_r\circ Hk_r$, where $m_i\circ Hk_i$ indicates that the orbit $Hk_i$ appears $m_i$ times. By simple calculations, we immediately get

$$\begin{aligned} \chi _l(S_2^2)=2\sum _{i=1}^rm_i\sum _{hk_i\in Hk_i}\zeta _n^{lhk_i}. \end{aligned}$$

Note that, if $s,t\in Hk$, then there exists $h_0\in H$ such that $h_0s=t$. Let $\sigma =\eta ^{-1}(h_0)$. We have

$$\begin{aligned} \sigma (\chi _s(S_2^2))= & {} \sigma \left( 2\sum _{i=1}^rm_i\sum _{hk_i\in Hk_i}\zeta _n^{shk_i}\right) =2\sum _{i=1}^rm_i\sum _{hk_i\in Hk_i}\zeta _n^{h_0shk_i}\\= & {} 2\sum _{i=1}^rm_i\sum _{hk_i\in Hk_i}\zeta _n^{thk_i}=\chi _t(S_2^2). \end{aligned}$$

Since $\chi _s(S_2^2)\in K_0$, we have $\chi _s(S_2^2)=\chi _t(S_2^2)$. Let $\mathcal {N}=\{k_i\mid \{1,2,\ldots ,\lfloor (n-1)/2\rfloor \}\cap Hk_i\ne \emptyset \}$. Therefore, all possible values of $c_l$ are

$$\begin{aligned} c_{k_i}=4\sum _{j=1}^rm_j\sum _{hk_j\in Hk_j}\zeta _n^{k_ihk_j}, \end{aligned}$$

for $k_i\in \mathcal {N}$. The following result is obtained:

Corollary 2

The algebraic degree of $\Gamma$ is bounded by

$$\begin{aligned} \frac{\varphi (n)}{|H|}\le \textrm{deg}(\Gamma )\le \frac{\varphi (n)}{|H|}2^{|\mathcal {N}|}. \end{aligned}$$

Example 4

(Cayley graph over dihedral group of algebraic degree 2) Let $G=D_8$ and $S=\{a,a^7,b\}$. Then, $I_1=\{1,-1\}$ and $I_2=[0]$. Therefore, $H=\{1,-1\}\le \mathbb {Z}_8^*$, the representative is $R(H)=\{0,1,2,3,4\}$ and $\mathcal {N}=\{1,2,3\}$. By simple calculations, we have $c_1=c_2=c_3=4$. Thus, $K=\mathbb {Q}(\zeta _8)^{\eta ^{-1}(H)}=\mathbb {Q}(\sqrt{2})$ and $\textrm{deg}(\Gamma )=2=\frac{\varphi (8)}{|H|}$.

Example 5

(Cayley graph over dihedral group of algebraic degree 4) Let $G=D_{12}$ and $S=\{a,a^{-1},a^5,a^{-5},b,ba,ba^5\}$. Then, $I_1=\{1,-1,5,-5\}$ and $I_2=[0,0,0,1,4,5,-1,-4,-5\}$. Therefore, $H=\mathbb {Z}_{12}^*$, $R(H)=\{0,1,2,3,4,6\}$ and $\mathcal {N}=\{1,2,3,4\}$. By simple calculations, we get $c_1=8$, $c_2=16$, $c_3=20$ and $c_4=0$. Thus, $K=\mathbb {Q}(\sqrt{2},\sqrt{5})$ and $\textrm{deg}(\Gamma )=4$.

Though the field $K_0=\mathbb {Q}^{\eta ^{-1}(H)}$ is very clear, it is not easy to determine K. From the examples above, K completely relies on the values $c_{k}$ for $k\in \mathcal {N}$. However, it seems that such values could not be described clearly since H is just a subgroup of $\mathbb {Z}_n^*$. In what follows, we therefore consider a special case of H.

If $H=\mathbb {Z}_n^*$, then $K_0=\mathbb {Q}$. Moreover, $R(H)\backslash \{0\}$ consists of all divisors of n and, hence, $\mathcal {N}=\{1\le k\le \lfloor (n-1)/2\rfloor \mid k\mid n\}$. Furthermore, for each $d\mid n$, we have $Hd=\mathbb {Z}_{n/d}^*d$. Therefore, for any $k\in \mathcal {N}$, we have

$$\begin{aligned} c_k=4\sum _{d\mid n}m_d\sum _{t\in \mathbb {Z}_{n/d}}\zeta _n^{ktd}=4\sum _{d\mid n}m_d\sum _{t\in \mathbb {Z}_{n/d}}\zeta _{n/d}^{kt}=4\sum _{d\mid n}m_dr_{n/d}(k), \end{aligned}$$

where $r_q(m)=\sum _{1\le j\le q,gcd(j,q)=1}\zeta _q^{mj}$ is the famous Ramanujan sum. Note that

$$\begin{aligned} r_q(m)=\frac{\varphi (q)}{\varphi \left( \frac{q}{gcd(m,q)}\right) }\mu \left( \frac{q}{gcd(m,q)}\right) , \end{aligned}$$

where $\mu$ is the Möbius function. Thus, we have

$$\begin{aligned} c_k=4\sum _{d\mid n}m_d\frac{\varphi (n/d)}{\varphi \left( \frac{n/d}{gcd(k,n/d)}\right) }\mu \left( \frac{n/d}{gcd(k,n/d)}\right) \end{aligned}$$

and, in particular, $c_1=4\sum _{d\mid n}m_d\mu (n/d)$.

Example 6

(Integral Cayley graph over dihedral group) Let $G=D_8$ and $S=\{a,a^3,a^5,a^7,b,ba^4\}$. Then, $I_1=\{1,3,5,7\}$ and $I_2=[0,0,4,-4]$. This leads to $H=\mathbb {Z}_8^*$ and $R(H)=\{0,1,2,4\}$. Therefore, $\mathcal {N}=\{1,2\}$. Since $I_2=2\circ H8\sqcup 2\circ H4$, we get

$$\begin{aligned} c_1=4(2\mu (1)+2\mu (2))=0, c_2=4(2\mu (1)+2\frac{\varphi (2)}{\varphi (1)}\mu (1))=16. \end{aligned}$$

Thus, $K=\mathbb {Q}$ and $\textrm{deg}(\Gamma )=1$.

4 An upper bound for the algebraic degree of Cayley digraphs over dihedral groups

So far, we restricted our considerations to undirected Cayley graphs. If we omit the restrictions on S, then $I_1 = -I_1$ does not hold anymore in general. This makes the computation of the $c_l$’s and the field $K_0$ much more difficult. At least we could find an upper bound for the algebraic degree of Cayley digraphs over dihedral groups:

Theorem 3

Let $\Gamma$ denote a Cayley digraph over the dihedral group $D_n$, then

$$\deg (\Gamma ) \le \frac{\varphi (n)}{|H|}2^{|\mathcal {N}|}.$$

Proof

Note that Lemma 3 still holds for digraphs. For $1\le l\le \lfloor \frac{n-1}{2}\rfloor$, we now get $b_l = \chi _l(S_1) = \sum _{a^s\in S_1} 2 \cos \frac{2 \pi l s}{n} = \sum _{i \in I_1} (\zeta _n^{li} + \zeta _n^{-li}) = \sum _{i \in I_1 \sqcup -I_1} \zeta _n^{li}.$ Similar as in the proofs of Lemma 4 and Lemma 5, we can show that if $b_1, c_1 \in K_0$, then $b_l,c_l \in K_0$, and that $b_1 \in K_0$ if and only if $I_1 \sqcup -I_1$ is a multi-union of some orbits Hk for $k \in \mathbb Z_n$.

Again, let $I_2 = \{k \mid a^k \in S_2^2\}$. Note that $I_2 = -I_2$. With similar, but a bit more cumbersome computations as above, we now get

$$c_l = 2\chi _l(S_2^2) + b_l^2 - 4 \sum _{i \in I_1} \zeta _n^{li} \sum _{j \in I_1} \zeta _n^{-lj}.$$

It is clear that with $I_1$ being a union of orbits Hk, so are $-I_1$ and $I_1 \sqcup -I_1$. Therefore, if $I_1$ is a union of orbits and $I_2$ is a multi-union of orbits, then $b_l, c_l \in K_0$ for all l. Thus, if H denotes the subgroup fixing both, $I_1$ and $I_2$, then the splitting field K of $\Gamma$ must be contained in the field $K_0(\sqrt{c_1}, \ldots , \sqrt{c_l})$ where $K_0 = \mathbb Q(\zeta _n)^{\eta ^{-1}(H)}$. Hence, the statement follows. $\square$