text:
\section{Introduction}
\subsection{Motivation}
The study of the spectrum of graphs is an important theme in the theory of expanders. It was remarked by Breuillard--Green--Guralnick--Tao that the eigenvalues of the normalised Laplacian operator of non-bipartite, finite Cayley graphs are bounded away from $2$ (see \cite[Appendix E]{BGGTExpansionSimpleLie}). Recently, the first author established an explicit upper bound.

Given a subset $S$ of a finite group $G$ with $|S| = d\geq 1$, the associated Cayley graph $C(G, S)$ has $G$ as its set of vertices and for $x, y\in G$, there is an edge from $x$ to $y$ if $y = x s$ for some $s\in S$. This graph is undirected if and only if $S$ is symmetric. In \cite[Theorem 1.4]{BiswasCheegerCayley} (cf. \cite[Theorem 2.11]{CheegerCayleySum}), it is established that if the Cayley graph $C(G, S)$ is undirected and connected, then the nontrivial spectrum of its adjacency operator lies in the interval
$$\left( -1 + \frac{h^4}{2^9d^8}
,
1 - \frac{h^2}{2d^2}
\right]$$
where $h$ denotes the vertex Cheeger constant of $C(G, S)$.

It turns out that a similar result holds for the Cayley sum graphs, which are classical combinatorial objects, e.g., see \cite{HandBookCombi}. The Cayley sum graph $C_\Sigma(G, S)$ has $G$ as its set of vertices, and for $x, y\in S$, there is an edge from $x$ to $y$ if $y = x^{-1} s$ for some $s\in S$. This graph is undirected if and only if $S$ is closed under conjugation (see \cite[Lemma 2.6]{CheegerCayleySum}). In \cite[Theorem 1.3]{CheegerCayleySum}, it is established that if the Cayley sum graph $C_\Sigma(G, S)$ is undirected and connected, then the nontrivial spectrum of its adjacency operator lies in the interval
$$\left( -1 + \frac{h_\Sigma^4}{2^9d^8}
,
1 - \frac{h_\Sigma^2}{2d^2}
\right]$$
where $h_\Sigma$ denotes the vertex Cheeger constant of $C_\Sigma(G, S)$.

\subsection{Results obtained}
Given a group automorphism $\sigma$ of $G$, one can consider variants of the Cayley graph and the Cayley sum graph, viz., the twisted Cayley graph $C(G, S)^\sigma$ and the twisted Cayley sum graph $C_\Sigma (G, S)^\sigma$. The twisted Cayley graph $C(G, S)^\sigma$ has $G$ as its set of vertices, and for $x, y\in G$, there is an edge from $x$ to $y$ if $y = \sigma(xs)$ for some $s\in S$. The twisted Cayley sum graph $C_\Sigma (G, S)^\sigma$ has $G$ as its set of vertices, and for $x, y\in G$, there is an edge from $x$ to $y$ if $y = \sigma(x^{-1} s)$ for some $s\in S$.

Note that the twisted Cayley graphs, and the twisted Cayley sum graphs provide examples of graphs which are neither Cayley graphs, nor Cayley sum graphs (if we focus on the twists by automorphisms of order two only). In fact, there are twisted Cayley graphs which are isomorphic to no Cayley graphs, no Cayley sum graphs, no twisted Cayley sum graphs, and there are twisted Cayley sum graphs which are isomorphic to no Cayley graphs, no Cayley sum graphs, no twisted Cayley graphs, as the following examples illustrate.

Let $p$ be an odd prime. Let $D_{2p}$ denote the dihedral group of order $2p$. Let $s$ denote an element of $D_{2p}$ of order two. Let $\sigma$ be the involution of $D_{2p}$ which fixes $s$ and sends any element of order $p$ to its inverse. Then the twisted Cayley graph $C(D_{2p}, \{s\})^\sigma$ is isomorphic to no Cayley graph, no Cayley sum graph, no twisted Cayley sum graph on any group of order $2p$.

Let $\tau$ denote the involution of $\ensuremath{\mathbb{Z}}/2p\ensuremath{\mathbb{Z}}$ of order two. Then, for any symmetric subset $S$ of $\ensuremath{\mathbb{Z}}/2p\ensuremath{\mathbb{Z}}$ containing the identity element and having size $\leq p-1$, the twisted Cayley sum graph $C_\Sigma(\ensuremath{\mathbb{Z}}/2p\ensuremath{\mathbb{Z}}, S)^\tau$ is isomorphic to no Cayley graph, no Cayley sum graph, no twisted Cayley graph on any group of order $2p$.

In this article, one of our aim is to show that the spectrum of the twisted Cayley graph $C(G, S)^\sigma$ and twisted Cayley sum graph $C_\Sigma(G,S)^\sigma$ are bounded away from $-1$.

\begin{theorem}\label{Thm:Bdd}
Let $S$ be a subset of a finite group $G$ with $|S|= d$. Suppose $\sigma$ is an automorphism of $G$.
\begin{enumerate}
\item
Suppose $\sigma^{2}$ is the trivial automorphism of $G$.
If the twisted Cayley graph $C(G, S)^\sigma$ is connected and undirected and $|G| \geq 4$, then the nontrivial spectrum of its normalised adjacency operator lies in the interval
$$\left( -1 + \frac{h_\sigma^4}{2^{12}d^8}
,
1 - \frac{h_\sigma^2}{2d^2}
\right]$$
where $h_\sigma$ denotes the vertex Cheeger constant of $C(G, S)^\sigma$.

\item
Suppose $\sigma^{2k}$ is the trivial automorphism of $G$, where $k\geq 1$ is an odd integer.
If the twisted Cayley graph $C(G, S)^\sigma$ is connected and undirected and $|G| \geq 4$, then the nontrivial spectrum of its normalised adjacency operator lies in the interval
$$\left(
\left(
-1 + \frac {1}{2^{12} d^{8k}}
\left(
\frac 12
\left(
1 -
\left(
1 - \frac{h_\sigma^2}{2d^2}
\right)^k
\right)
\right)
^4
\right)^{1/k}
,
1 - \frac{h_\sigma^2}{2d^2}
\right]$$
where $h_\sigma$ denotes the vertex Cheeger constant of $C(G, S)^\sigma$.

\item If the twisted Cayley sum graph $C_\Sigma (G, S)^\sigma$ is connected and undirected and $|G| \geq 4$, then the nontrivial spectrum of its normalised adjacency operator lies in the interval
$$\left( -1 + \frac{h_{\Sigma, \sigma}^4}{2^{12}d^8}
,
1 - \frac{h_{\Sigma, \sigma}^2}{2d^2}
\right]$$
where $h_{\Sigma, \sigma}$ denotes the vertex Cheeger constant of $C_\Sigma (G, S)^\sigma$.
\end{enumerate}
\end{theorem}

More generally, we study the spectrum of undirected, connected, non-bipartite graphs carrying a suitable action of a group and establish Theorem \ref{thmPrincipal}. Using Theorem \ref{thmPrincipal}, we prove Theorem \ref{Thm:Bdd}. The proof of Theorem \ref{thmPrincipal} is motivated by the strategy used in \cite[Appendix E]{BGGTExpansionSimpleLie}.

In Section \ref{Sec:TwistsAnti}, we consider twisted Cayley graphs and twisted Cayley sum graphs with respect to an anti-automorphism of the underlying group. We establish that if these graphs are undirected and connected, then the nontrivial spectrum of their normalised adjacency operators are bounded away from $-1$ (see Theorem \ref{Thm:BddAnti}).

We consider Schreier graphs in Section \ref{Sec:Schreier}. We prove under certain hypothesis that the nontrivial eigenvalues of the normalised adjacency operator of a given Schreier graph is bounded away form $-1$ (see Theorem \ref{Thm:Schreier}).

\subsection{Acknowledgements}
We wish to thank Emmanuel Breuillard for a number of helpful discussions during the opening colloquium of the M\"unster Mathematics Cluster,
and the MFO for their hospitality, where a part of this work was initiated. The first author is supported by the ISF Grant no. 662/15 at the Technion. The second author would like to acknowledge the Initiation Grant from the Indian Institute of Science Education and Research Bhopal and the INSPIRE Faculty Award from the Department of Science and Technology, Government of India.

\section{Preliminaries}
In the following, all the graphs considered are undirected. However, these graphs may contain multiple edges and even multiple loops at certain vertices. Given a finite $d$-regular multi-graph $\mathbb{G} = (V,E)$ having $V$ as its set of vertices and $E$ as its multiset of edges, we have the normalised adjacency operator $T$ of size $|V|\times |V|$. The normalised Laplacian operator of $\mathbb{G}$ is defined by
$$L:= I_{|V|} - T,$$
where $I_{|V|}$ denotes the identity matrix of size $|V|\times |V|$. Let $n$ denote the number of elements of $V$. Denote the eigenvalues of $T$ and the eigenvalues of $L$ by $\lbrace t_{i} : i= 1, \cdots, n\rbrace $ and $\lbrace \lambda_{i} : i= 1, \cdots, n\rbrace $ respectively such that $\lambda_{i} = 1 - t_{i}$ and
$$
 0 = \lambda_{1} \leqslant \lambda_{2} \leqslant \cdots \leqslant \lambda_{n-1}\leqslant \lambda_{n}  \leqslant 2.
 $$

Let $\mathbb{G} = (V,E)$ be a multi-graph. For a subset $V_{1}\subseteq V$, its neighbourhood in $\mathbb G$ is denoted by $N(V_{1})$ and is defined as
$$N(V_{1}) := \lbrace v\in V : (v, v_{1})\in E \text{ for some } v_{1}\in V_{1}\rbrace.$$
The boundary of $V_{1}$ is defined as $ \partial(V_{1}) := N(V_{1})\backslash V_{1}$.

\begin{definition}[Vertex Cheeger constant]
The vertex Cheeger constant $h(\mathbb{G})$ of a multi-graph $\mathbb{G} = (V,E)$ is defined as
$$h(\mathbb{G}) := \inf \left\lbrace \frac{|\partial(V_{1})|}{|V_{1}|} : \emptyset \neq V_{1}\subseteq V, |V_{1}|\leqslant \frac{|V|}{2} \right\rbrace.$$
\end{definition}

We recall the notion of expander graphs as stated in \cite{AlonEigenvalueExpand}.

\begin{definition}[$(n,d,\varepsilon)$-expander]
\label{vexp}
Let $\varepsilon>0$. An $(n,d,\varepsilon)$-expander is a graph $(V,E)$ on $n$ vertices, having maximal degree $d$, such that for every set $\emptyset \neq V_{1}\subseteq V$ satisfying $|V_{1}|\leqslant \frac{|V|}{2}$, the inequality $|\partial(V_{1})|\geqslant \varepsilon|V_{1}|$ holds (equivalently, $h((V, E))\geqslant \varepsilon).$
\end{definition}

The degree of a vertex of a multi-graph is the number of half-edges adjacent to it (in the absence of loops). The presence of a loop at a vertex increases its degree by one. A multi-graph is called $r$-regular if each vertex has degree $r$. Apart from the notion of vertex expansion as in Definition \ref{vexp}, there is the notion of edge expansion.

\begin{definition}[Edge expansion]
Let $\mathbb{G} = (V,E)$ be a $d$-regular multi-graph with vertex set $V$ and edge multiset $E$. For any nonempty subset $V_1$ of $V$, the edge boundary $E(V_{1},V\backslash V_{1})$ of $V_{1}$ is defined as the multiset
$$E(V_{1},V\backslash V_{1}) := \lbrace (v_{1},v_2)\in E: v_{1}\in V, v_2\in V\backslash V_{1} \rbrace ,$$
and the edge expansion ratio $\phi(V_{1})$ of $V_1$ is defined as
$$\phi(V_{1}) := \frac{|E(V_{1},V\backslash V_{1})|}{d|V_{1}|}.$$
\end{definition}

\begin{definition}[Edge Cheeger constant]
The edge Cheeger constant $\mathfrak{h}(\mathbb{G})$ of a multi-graph $\mathbb G = (V, E)$ is defined by
$$\mathfrak{h}(\mathbb{G}):= \inf_{\emptyset \neq V_{1}\subseteq V, |V_{1}|\leqslant |V|/2} \phi(V_{1}).$$
\end{definition}

The two Cheeger constants are related by the following lemma.

\begin{lemma}
\label{Lemma:VertexEdgeCons}
For $d$-regular multi-graph $\mathbb{G} = (V,E)$, the inequalities
$$ \frac{h(\mathbb{G})}{d} \leqslant \mathfrak{h}(\mathbb{G}) \leqslant h(\mathbb{G})$$
hold.
\end{lemma}

\begin{proof}
For $\emptyset \neq V_{1}\subseteq V$, consider the map $$\psi:E(V_{1},V\backslash V_{1}) \rightarrow \partial(V_{1}) \text{ given by }(v_{1},v_2)\mapsto v_{2}.$$
This map is surjective, and hence the first inequality follows. For the second inequality, note that the fibre of any point of $\partial(V_1)$ under the map $\psi$ contains at most $d$ elements.
\end{proof}

The following Proposition relates the edge Cheeger constant of a multi-graph with the second smallest eigenvalue of its Laplacian operator. It is the version for graphs of the corresponding inequalities for the Laplace--Beltrami operator on compact Riemannian manifolds. It was first established by Cheeger \cite{CheegerLowerBddSmallest} (the lower bound) and by Buser \cite{BuserNoteIsoperimetric} (the upper bound). The discrete version was proved by Alon and Millman \cite{AlonMilmanIsoperiIneqSupConcen} (Proposition \ref{Prop:chin}).

\begin{proposition}[Discrete Cheeger--Buser inequality]
\label{Prop:chin}
Let $\mathbb{G} = (V,E)$ be a finite $d$-regular multi-graph.
Let $\mathfrak{h}(\mathbb{G})$ denote its edge Cheeger constant and $\lambda_{2}$ denote the second smallest eigenvalue of its normalised Laplacian operator. Then
$$ \frac{\mathfrak{h}(\mathbb{G})^{2}}{2} \leqslant \lambda_{2} \leqslant 2\mathfrak{h}(\mathbb{G}).$$
\end{proposition}

\begin{proof}
See \cite[Propositions 4.2.4, 4.2.5]{LubotzkyDiscreteGroups} or
\cite{ChungLaplacianOfGraph}.
\end{proof}

\section{Generalities}

Let $(V, E)$ be a finite graph (which may contain multiple edges, and even multiple loops at certain vertices) of degree $d$.
The neighbourhood of a subset $V'$ of $V$ in $(V, E)$ is denoted by $\ensuremath{\mathcal{N}}(V')$. Assume that there exist an integer $d\geq 1$ and permutations $\theta_1, \cdots, \theta_d: V\to V$ such that the vertices $v, \theta_i(v)$ are adjacent in $(V, E)$ for any $v\in V$ and $1\leq i \leq d$, and $\ensuremath{\mathcal{N}}(v)$ is equal to $\cup_{i=1}^d \{\theta_i(v)\}$ for any $v\in V$. For a subset $V'$ of $V$, denote the subset $\theta_i(V')$ of $\ensuremath{\mathcal{N}}(V')$ by $\ensuremath{\mathcal{N}}^i(V')$.

\begin{proposition}
\label{Prop:AExists}
Assume that the graph $(V, E)$ is undirected. Suppose $|V|\geq 4$, the graph $(V, E)$ is an $\varepsilon$-vertex expander for some $\varepsilon>0$ and its normalised adjacency operator has an eigenvalue in the interval $(-1, -1+\zeta]$ for some $\zeta$ satisfying $0<\zeta \leq \frac{\varepsilon^2}{4d^4}$. Then for some subset $A$ of $V$, the inequalities
$$\left(\frac 1{2 + \beta + \frac{d\beta}{\varepsilon}}\right) |V| \leq |A| \leq \frac 12 |V|$$
hold with $\beta = d^2 \sqrt{2\zeta (2-\zeta)}$. Let $\tau$ be an automorphism of the set $V$ such that
$$
\ensuremath{\mathcal{N}} (\ensuremath{\mathcal{N}}(\tau(A))) \subseteq \tau(\ensuremath{\mathcal{N}}(\ensuremath{\mathcal{N}}(A))).
$$
If
$$\beta < \frac{\varepsilon^2}{2d(2d+1)},$$
then exactly one of the inequalities
$$|A \cap \tau(A)| \leq \frac{d \beta}{\varepsilon^2} (\varepsilon + d + 2)|A|,
\quad
|A \cap \tau(A)| \geq \left( 1 - \frac {d \beta}{ \varepsilon^2}  ( \varepsilon + d + 2) \right) |A|$$
holds.
\end{proposition}

\begin{proof}
Since the graph $(V, E)$ is an $\varepsilon$-vertex expander with $\varepsilon>0$, it follows that for some vertex $v\in V$ and for some $1\leq i \leq d$, $\theta_i(v) \neq v$. Using $|V| \geq 4$, one obtains
\begin{align*}
\varepsilon |\{v, \theta_i(v)\}|
& \leq |(\ensuremath{\mathcal{N}}(v) \cup \ensuremath{\mathcal{N}}(\theta_i(v)) ) \setminus \{v, \theta_i(v)\}| \\
& \leq |\ensuremath{\mathcal{N}}(v) \setminus \{\theta_i(v)\}| + |\ensuremath{\mathcal{N}}(\theta_i(v))\setminus \{v\}|\\
& \leq 2(d-1),
\end{align*}
which implies
\begin{equation}
\label{Eqn:EpsilonBoundSch}
\varepsilon \leq d-1.
\end{equation}
and hence $\zeta < 1$. Let $T$ denote the normalised adjacency operator of the graph $(V, E)$. Since $T$ has an eigenvalue in $(-1, -1+\zeta]$ and $\zeta <1$, it follows that $T^2$ has an eigenvalue $\nu$ in $[(1-\zeta)^2, 1)$.

Consider the undirected multi-graph $\ensuremath{\mathcal{M}}$ (which may contain multiple edges, and multiple loops) with $V$ as its set of vertices and its edges are obtained by drawing an edge from $v$ to each element of $\ensuremath{\mathcal{N}}(\ensuremath{\mathcal{N}}(\{v\}))$ (considered as a multiset). In other words, $\ensuremath{\mathcal{M}}$ is the undirected multi-graph having $V$ as its set of vertices and $T^2$ as its normalised adjacency operator. Thus the second largest eigenvalue of the normalised adjacency operator of $\ensuremath{\mathcal{M}}$ is $\geq \nu\geq (1-\zeta)^2 = 1 - \zeta(2-\zeta)$. Hence the second smallest eigenvalue of the normalised Laplacian operator of $\ensuremath{\mathcal{M}}$ is $\leq \zeta(2-\zeta)$. By the discrete Cheeger--Buser inequality (Proposition \ref{Prop:chin}), it follows that the edge Cheeger constant of $\ensuremath{\mathcal{M}}$ satisfies
$$\frac 12 \ensuremath{\mathfrak{h}}(\ensuremath{\mathcal{M}})^2 \leq \zeta(2-\zeta),$$
which yields
$$\ensuremath{\mathfrak{h}}(\ensuremath{\mathcal{M}})\leq  \sqrt{2\zeta(2-\zeta)}.$$
Since $(V, E)$ has degree $d$, from Lemma \ref{Lemma:VertexEdgeCons}, it follows that the vertex Cheeger constant of $\ensuremath{\mathcal{M}}$ satisfies
$$h(\ensuremath{\mathcal{M}}) \leq
d^2 \ensuremath{\mathfrak{h}}(\ensuremath{\mathcal{M}}) \leq d^2\sqrt{2\zeta(2-\zeta)}.$$
This implies that for some nonempty subset $A$ of $V$ with $|A| \leq \frac 12 |V|$,
$$
|\ensuremath{\mathcal{N}}(\ensuremath{\mathcal{N}}(A))\setminus A|
\leq |A| d^2 \sqrt{2\zeta(2-\zeta)}
= |A| \beta
$$
holds. Note that $|A \cup \ensuremath{\mathcal{N}}(A)| \geq \frac{|V|}{2}$. Otherwise, we obtain
\begin{align*}
\varepsilon|A|
& \leq \varepsilon |A \cup \ensuremath{\mathcal{N}}(A)| \\
& \leq |\ensuremath{\mathcal{N}}(A \cup \ensuremath{\mathcal{N}}(A)) \setminus (A \cup \ensuremath{\mathcal{N}}(A))|\\
& \leq |\ensuremath{\mathcal{N}}(\ensuremath{\mathcal{N}}(A))\setminus A| \\
& \leq |A| d^2 \sqrt{2\zeta(2-\zeta)}\\
& = |A| \beta.
\end{align*}
This implies
$\varepsilon\leq d^2 \sqrt{2\zeta(2-\zeta)} < d^2 \sqrt{4\zeta}$, which contradicts the assumption that $\zeta \leq \frac{\varepsilon^2}{4d^4}$. It follows that
\begin{align*}
\varepsilon |V \setminus (A \cup \ensuremath{\mathcal{N}}(A))|
& \leq  \varepsilon|(A \cup \ensuremath{\mathcal{N}}(A))^c|\\
& \leq |\ensuremath{\mathcal{N}}((A \cup \ensuremath{\mathcal{N}}(A))^c) \setminus (A \cup \ensuremath{\mathcal{N}}(A))^c|  \\
& \leq |(A \cup \ensuremath{\mathcal{N}}(A)) \cap \ensuremath{\mathcal{N}}((A \cup \ensuremath{\mathcal{N}}(A))^c)| \\
& \leq \sum_{i = 1}^d
|(A \cup \ensuremath{\mathcal{N}}(A)) \cap \theta_i((A \cup \ensuremath{\mathcal{N}}(A))^c)| \\
& \leq \sum_{i = 1}^d |\ensuremath{\mathcal{N}}(A \cup \ensuremath{\mathcal{N}}(A)) \cap (A \cup \ensuremath{\mathcal{N}}(A))^c|\\
& \leq d |\ensuremath{\mathcal{N}}(A \cup \ensuremath{\mathcal{N}}(A)) \cap (A \cup \ensuremath{\mathcal{N}}(A))^c|\\
& \leq d |\ensuremath{\mathcal{N}}(A \cup \ensuremath{\mathcal{N}}(A)) \setminus (A \cup \ensuremath{\mathcal{N}}(A))|\\
& \leq d|\ensuremath{\mathcal{N}}(\ensuremath{\mathcal{N}}(A))\setminus A| \\
& \leq d|A| d^2 \sqrt{2\zeta(2-\zeta)}\\
& = d |A| \beta,
\end{align*}
which implies
\begin{align*}
|V|
& \leq \frac{d\beta}{\varepsilon}|A| + |A \cup \ensuremath{\mathcal{N}}(A)|\\
& \leq \frac{d\beta}{\varepsilon}|A| + |A| + |\ensuremath{\mathcal{N}}(A)| \\
& \leq \frac{d\beta}{\varepsilon}|A| + |A| + |\ensuremath{\mathcal{N}}(\ensuremath{\mathcal{N}}(A))| \\
& \leq \frac{d\beta}{\varepsilon}|A| + 2|A| + |\ensuremath{\mathcal{N}}(\ensuremath{\mathcal{N}}(A))\setminus A|\\
& \leq \frac{d\beta}{\varepsilon}|A| + 2|A| + \beta|A|,
\end{align*}
and hence
$$\frac{|V|}{2 + \beta + \frac{d\beta}{\varepsilon}} \leq |A| \leq \frac{|V|}{2}.$$

Note that the inequalities
$$
\frac {2d\beta} { \varepsilon^2} ( \varepsilon + d + 2)
\leq \frac {2d\beta} { \varepsilon^2}( d -1 + d + 2)
= \frac {2d\beta} { \varepsilon^2} ( 2d + 1)
<1
$$
imply that
$$\frac{d \beta}{\varepsilon^2} (\varepsilon + d + 2)
<
1 - \frac {d \beta}{ \varepsilon^2}  ( \varepsilon + d + 2).$$
Hence it suffices to show that one of the inequalities
$$|A \cap (\tau(A))| \leq \frac{d \beta}{\varepsilon^2} (\varepsilon + d + 2)|A|,
\quad
|A \cap (\tau(A))| \geq \left( 1 - \frac {d \beta}{ \varepsilon^2}  ( \varepsilon + d + 2) \right) |A|$$
holds.

Let $B = A\Delta(\tau(A))^c$. Let $\ensuremath{\mathrm{id}}$ denote the identity map from $V$ to $V$. Note that
\begin{align*}
|\ensuremath{\mathcal{N}}^i(B) \Delta B|
& \leq |\ensuremath{\mathcal{N}}^i(A) \Delta \ensuremath{\mathcal{N}}^i((\tau(A))^c) \Delta A \Delta(\tau(A))^c| \\
& \leq |\ensuremath{\mathcal{N}}^i(A) \Delta \ensuremath{\mathcal{N}}^i((\tau(A))^c) \Delta A^c \Delta \tau(A)| \\
& \leq |\ensuremath{\mathcal{N}}^i(A) \Delta A^c \Delta \ensuremath{\mathcal{N}}^i((\tau(A))^c) \Delta \tau(A)| \\
& \leq |\ensuremath{\mathcal{N}}^i(A) \Delta A^c| + |\ensuremath{\mathcal{N}}^i((\tau(A))^c) \Delta \tau(A)|\\
& = |\ensuremath{\mathcal{N}}^i(A) \Delta A^c| + |\ensuremath{\mathcal{N}}^i(\tau(A)) \Delta \tau(A^c)|\\
& = \sum_{\phi\in \{\ensuremath{\mathrm{id}}, \tau\}} |\ensuremath{\mathcal{N}}^i(\phi(A)) \Delta \phi(A^c)|\\
& = \sum_{\phi\in \{\ensuremath{\mathrm{id}}, \tau\}} \left( |\ensuremath{\mathcal{N}}^i(\phi(A))| + |\phi(A^c)| - 2|\ensuremath{\mathcal{N}}^i(\phi(A)) \cap \phi(A^c)|\right)\\
& = \sum_{\phi\in \{\ensuremath{\mathrm{id}}, \tau\}} \left( |V| - 2|\ensuremath{\mathcal{N}}^i(\phi(A)) \cap \phi(A^c)|\right)\\
& = \sum_{\phi\in \{\ensuremath{\mathrm{id}}, \tau\}} \left( |V| - 2|\ensuremath{\mathcal{N}}^i(\phi(A))| + 2|\ensuremath{\mathcal{N}}^i(\phi(A))|  - 2|\ensuremath{\mathcal{N}}^i(\phi(A)) \cap \phi(A^c)|\right)\\
& = \sum_{\phi\in \{\ensuremath{\mathrm{id}}, \tau\}} \left( |V| - 2|A| + 2|\ensuremath{\mathcal{N}}^i(\phi(A)) \cap \phi(A)|\right)\\
& = 2(|V| - 2|A|) + 2\sum_{\phi\in \{\ensuremath{\mathrm{id}}, \tau\}} |\ensuremath{\mathcal{N}}^i(\phi(A)) \cap \phi(A)| \\
& \leq 2(|V| - 2|A|) +
\frac 2\varepsilon\sum_{\phi\in \{\ensuremath{\mathrm{id}}, \tau\}} \varepsilon|\ensuremath{\mathcal{N}}(\phi(A)) \cap \phi(A)| \\
& \leq 2(|V| - 2|A|) +
\frac 2\varepsilon\sum_{\phi\in \{\ensuremath{\mathrm{id}}, \tau\}} |\ensuremath{\mathcal{N}}(\ensuremath{\mathcal{N}}(\phi(A)) \cap \phi(A)) \setminus (\ensuremath{\mathcal{N}}(\phi(A)) \cap \phi(A))| \\
& \leq 2(|V| - 2|A|) +
\frac 2\varepsilon\sum_{\phi\in \{\ensuremath{\mathrm{id}}, \tau\}} |\ensuremath{\mathcal{N}}(\ensuremath{\mathcal{N}}(\phi(A)) \setminus \phi(A)| \\
& \leq 2(|V| - 2|A|) +
\frac 2\varepsilon\sum_{\phi\in \{\ensuremath{\mathrm{id}}, \tau\}} |\phi(\ensuremath{\mathcal{N}}(\ensuremath{\mathcal{N}}(A))) \setminus \phi(A)| \\
& = 2(|V| - 2|A|) +
\frac 2\varepsilon\sum_{\phi\in \{\ensuremath{\mathrm{id}}, \tau\}} |\ensuremath{\mathcal{N}}(\ensuremath{\mathcal{N}}(A)) \setminus A| \\
& \leq 2\left(\beta + \frac{d\beta}\varepsilon \right)|A| +\frac 2\varepsilon 2|A|\beta\\
& =  \frac{2\beta}\varepsilon (\varepsilon + d + 2)|A| .
\end{align*}
It follows that
$$
|\ensuremath{\mathcal{N}}(B) \Delta B|
\leq
\frac{2d\beta}\varepsilon (\varepsilon + d + 2)|A| .
$$

Note that
\begin{align*}
|\ensuremath{\mathcal{N}}^i(B^c) \Delta B^c|
& \leq |\ensuremath{\mathcal{N}}^i(A) \Delta \ensuremath{\mathcal{N}}^i(\tau(A)) \Delta A \Delta\tau(A)| \\
& \leq |\ensuremath{\mathcal{N}}^i(A) \Delta \ensuremath{\mathcal{N}}^i(\tau(A)) \Delta A^c \Delta (\tau(A))^c| \\
& \leq |\ensuremath{\mathcal{N}}^i(A) \Delta A^c \Delta \ensuremath{\mathcal{N}}^i(\tau(A)) \Delta (\tau(A))^c| \\
& \leq |\ensuremath{\mathcal{N}}^i(A) \Delta A^c| + |\ensuremath{\mathcal{N}}^i(\tau(A)) \Delta (\tau(A))^c| )\\
& \leq |\ensuremath{\mathcal{N}}^i(A) \Delta A^c| + |\ensuremath{\mathcal{N}}^i((\tau(A)^c)) \Delta \tau(A)| )\\
& \leq \frac{2\beta}\varepsilon (\varepsilon + d + 2)|A| .
\end{align*}
It follows that
$$
|\ensuremath{\mathcal{N}}(B^c) \Delta B^c|
\leq
\frac{2d\beta}\varepsilon (\varepsilon + d + 2)|A| .
$$

We consider the following cases, viz., $|B|\leq \frac{|V|}2, |B| > \frac{|V|}2$. When $|B|\leq \frac{|V|}2$ holds, we obtain
$$
\varepsilon |B| \leq |\ensuremath{\mathcal{N}}(B) \setminus B| \leq |\ensuremath{\mathcal{N}}(B) \Delta B|  \leq \frac{2d \beta}\varepsilon (\varepsilon + d + 2)|A|,
$$
which yields
$$|B| \leq \frac {2d\beta}{ \varepsilon^2} ( \varepsilon + d + 2) |A|.
$$
Since
\begin{align*}
|V| - |B|
& = |B^c| \\
& = |A \Delta (\tau (A))| \\
& = |A| - |A\cap (\tau(A))| + |\tau(A)| - |A\cap (\tau(A))| \\
& = 2|A| - 2|A\cap (\tau(A))|
\end{align*}
holds, we obtain
$$
2|A \cap (\tau(A)) |
\leq |V| - 2|A| + 2|A \cap (\tau(A))|
= |B|
\leq \frac{2d \beta}{\varepsilon^2} (\varepsilon + d + 2)|A|.
$$
While $|B| > \frac{|V|}2$ holds, we obtain
$$
\varepsilon |B^c| \leq |\ensuremath{\mathcal{N}}(B^c)\setminus B^c| \leq |\ensuremath{\mathcal{N}}(B^c)\Delta B^c|
\leq \frac{2d\beta}{\varepsilon}(\varepsilon + d + 2) |A|,
$$
which yields
$$|B^c| \leq \frac{2d\beta}{\varepsilon^2}(\varepsilon + d + 2) |A|.
$$
Since
$$
|B^c|
= |A \Delta (\tau(A))|
= |A| - |A\cap (\tau(A))|  + |(\tau(A)) | - |A\cap (\tau(A))|
= 2|A| - 2|A\cap (\tau(A))| $$
holds, we obtain
$$
|A \cap (\tau(A)) |
\geq |A| - \frac {d\beta}{ \varepsilon^2} ( \varepsilon + d + 2)|A|
= \left( 1- \frac {d\beta}{ \varepsilon^2} ( \varepsilon + d + 2)\right)|A|.
$$
This completes the proof.
\end{proof}

\begin{theorem}\label{thmPrincipalk1}
Suppose $V$ carries a left action of a group $\ensuremath{\mathcal{G}}$ such that the following conditions hold.
\begin{enumerate}
\item No index two subgroup of $\ensuremath{\mathcal{G}}$ acts transitively on $V$.

\item The action of $\ensuremath{\mathcal{G}}$ on the set $V$ is ``transitive of order $t$'' in the sense that for each $(u, v)\in V\times V$, the equation $gu = v$ has exactly $t$ distinct solutions for $g\in \ensuremath{\mathcal{G}}$. In other words, the action of $\ensuremath{\mathcal{G}}$ on $V$ is transitive and the stabilizer of each element of $V$ have the same size (equal to $t$).

\item For each $\theta_i, 1\leq i\leq d$ and $v\in V$, there is an automorphism or an anti-automorphism $\psi_{i,v}$ of the group $\ensuremath{\mathcal{G}}$ such that one of
$$\theta_i(g\cdot v) =  \psi_{i,v}(g) \cdot \theta_i(v)$$
and
$$\theta_i(g\cdot v) =  \psi_{i, v}(g^{-1}) \cdot \theta_i(v)$$
holds for any $g\in \ensuremath{\mathcal{G}}$.

\item For any $\tau\in \ensuremath{\mathcal{G}}$,
$
\ensuremath{\mathcal{N}} (\ensuremath{\mathcal{N}}(\tau(A))) \subseteq \tau(\ensuremath{\mathcal{N}}(\ensuremath{\mathcal{N}}(A)))
$
holds.
\end{enumerate}
Assume that the graph $(V, E)$ is undirected and non-bipartite. Assume further that $|V|\geq 4$, the graph $(V, E)$ is an $\varepsilon$-vertex expander for some $\varepsilon>0$. Then the nontrivial eigenvalues of the normalised adjacency operator of this graph are greater than $-1 + \ell_{\varepsilon, d}$ with
$$\ell_{\varepsilon, d}
=\frac{\varepsilon^4}{2^{12} d^8}.$$
\end{theorem}

\begin{proof}
On the contrary, let us assume that a nontrivial eigenvalue of the normalised adjacency operator of the graph $(V, E)$ lies in the interval $\left[-1, -1 +  \ell_{\varepsilon, d}\right]$. Since $(V, E)$ is non-bipartite, it follows that $-1$ is not an eigenvalue of its normalised adjacency operator. Hence an eigenvalue of the normalised adjacency operator of the graph $(V, E)$ lies in the interval $(-1, -1+ \ell_{\varepsilon, d}]$. Set
\begin{align*}
\gamma & = d^2 \sqrt{2 \ell_{\varepsilon, d}(2- \ell_{\varepsilon, d})},\\
r & =  1- \frac {d \gamma}{ \varepsilon^2}  ( \varepsilon + d + 2).
\end{align*}
Since $\ell_{\varepsilon, d} = \frac{\varepsilon^4}{2^{12} d^8}$, we have
$$\gamma =  d^2 \sqrt{2 \ell_{\varepsilon, d}(2- \ell_{\varepsilon, d})} <  d^2 \sqrt{4 \ell_{\varepsilon, d}} \leqslant \frac{\varepsilon^{2}}{2^5d^2}.$$
$$
1 - r = \frac {d \gamma}{ \varepsilon^2}  ( \varepsilon + d + 2)
\leq \frac {d \gamma}{ \varepsilon^2}  (2d+1)
< \frac 3{2^3\sqrt 2} < \frac 13.$$
Consequently,
\begin{equation}
\label{Eqn:BoundssigmaCayleySum}
\ell_{\varepsilon, d}\leq \frac{\varepsilon^2}{4d^4},
\gamma < \frac{\varepsilon^2}{2d(2d+1)} \text{ and }  r> \frac 23.
\end{equation}

Define the subset $H$ of $\ensuremath{\mathcal{G}}$ by
$$H  :=
\{
\tau\in \ensuremath{\mathcal{G}} \,:\,\, |A \cap  (\tau(A))| \geq r|A|
\}.$$
Note that $H$ contains the identity element of $\ensuremath{\mathcal{G}}$. By the triangle inequality,
\begin{align*}
|A \setminus (\tau_1(\tau_2(A)))|
& \leq |A \setminus (\tau_1(A)) | + |(\tau_1(A)) \setminus (\tau_1(\tau_2(A)))| \\
& =  |A \setminus  (\tau_1(A)) | + |A  \setminus (\tau_2(A)) |\\
& = |A | -  |A \cap (\tau_1(A))  | + |A | - |A  \cap (\tau_2(A)) |\\
& \leq 2|A| - 2r |A| .
\end{align*}
Consequently,
\begin{align*}
|A \cap (\tau_1(\tau_2(A)))|
& = |A | - |A \setminus (\tau_1(\tau_2(A)))| \\
& \geq |A | - 2|A| + 2r |A| \\
& = (2r - 1) |A|.
\end{align*}
If $|A \cap (\tau_1(\tau_2(A)))| \leq (1-r) |A|$, then we obtain
$$(1-r) |A|
\geq |A \cap (\tau_1(\tau_2(A)))|
\geq  (2r - 1) |A|,$$
which implies $r\leq \frac 23$. Since $r>\frac 23$, it follows that the inequality
$|A \cap (\tau_1(\tau_2(A)))| \leq (1-r) |A|$
does not hold. Hence, by Proposition \ref{Prop:AExists}(1), $H$ contains $\tau_1\tau_2$. So $H$ is a subgroup of $\ensuremath{\mathcal{G}}$.

For any $g\in \ensuremath{\mathcal{G}}$, the map
$$A\cap g^{-1} A \to A\times A, \quad
x\mapsto (x, gx)$$
is well-defined, and induces a map
$$\varphi: \coprod _{g\in \ensuremath{\mathcal{G}}} A\cap g^{-1} A \to A\times A.$$
Each of its fibre contains exactly $t$ elements and hence
$$
|A|^2
= im(\varphi)
= \sum_{x\in im(\varphi) } \frac{|\varphi^{-1} (x)|} {|\varphi^{-1} (x)|}
= \frac 1t \sum_{x\in im(\varphi)} |\varphi^{-1} (x)|
= \frac 1t \sum_{g\in \ensuremath{\mathcal{G}}} |A\cap g^{-1} A|.$$

If $\ensuremath{\mathcal{G}} = H$, we obtain
$$|A| \cdot \frac{|V|}{2}
\geq |A|^2 = \frac 1t \sum_{g\in \ensuremath{\mathcal{G}}} |A\cap g^{-1} A|
\geq \frac 1t |\ensuremath{\mathcal{G}}| \cdot r|A|,$$
which gives
$$|\ensuremath{\mathcal{G}}| \leq \frac t{2r}{|V|}.$$
This contradicts that $|\ensuremath{\mathcal{G}}| \geq t|V|$. Hence $H$ is a proper subgroup of $\ensuremath{\mathcal{G}}$.

The following estimate
$$
t |A|^2
= \sum_{g\in \ensuremath{\mathcal{G}}} |A\cap g^{-1} A|
\leq |H| |A| + \frac{d\gamma}{\varepsilon^2}(\varepsilon + d+ 2)|A||\ensuremath{\mathcal{G}}\setminus H| $$
implies
$$t |A|
\leq |H| + \frac{d\gamma}{\varepsilon^2}(\varepsilon + d+ 2)(|\ensuremath{\mathcal{G}}| - |H|).$$
Using Proposition \ref{Prop:AExists}(1), we obtain
$$
\left(\frac{1}{2+ \gamma + \frac{d\gamma}{\varepsilon}}\right) |\ensuremath{\mathcal{G}}|
= \left(\frac{t}{2+ \gamma + \frac{d\gamma}{\varepsilon}}\right) |V|
\leq \frac{d\gamma}{\varepsilon^2}(\varepsilon + d+ 2)|\ensuremath{\mathcal{G}}|
+  \left(1 - \frac{d\gamma}{\varepsilon^2}(\varepsilon + d+ 2)\right) |H|.$$
We claim that $H$ is a subgroup of $\ensuremath{\mathcal{G}}$ of index two. To prove this claim, it suffices to show that
\begin{equation}
\label{Eqn:13rdInePri}
\frac 13 \left(1 - \frac{d\gamma}{\varepsilon^2}(\varepsilon + d+ 2)\right)
<
\left(\frac{1}{2+ \gamma + \frac{d\gamma}{\varepsilon}}\right) - \frac{d\gamma}{\varepsilon^2}(\varepsilon + d+ 2),
\end{equation}
i.e.,
$$\left(2 + \gamma + \frac{d\gamma}{\varepsilon}\right) \left( 1 + \frac{2d\gamma}{\varepsilon^2}(\varepsilon + d+ 2) \right) < 3,$$
which is equivalent to
\begin{equation}
\label{Eqn:13rdInequality}
\left(\gamma + \frac{d\gamma}{\varepsilon}\right) + \frac{2d\gamma}{\varepsilon^{2}}(\varepsilon + d+ 2)\left(2 + \gamma + \frac{d\gamma}{\varepsilon}\right) < 1.
\end{equation}

Assume that $\zeta \leq \frac{\varepsilon^4}{2^9 d^8}$. Since
$$\gamma \leq
\frac{\varepsilon^{2}}{2^3\sqrt{2}d^2}
$$
and $d \geq 2$ (by Equation \eqref{Eqn:EpsilonBoundSch}), we obtain
\begin{align*}
& \gamma + \frac{d\gamma}{\varepsilon} + \frac{2d\gamma}{\varepsilon^{2}}(\varepsilon + d+ 2)\left(2 + \gamma + \frac{d\gamma}{\varepsilon}\right)\\
& = \frac{\gamma}{\varepsilon}(\varepsilon + d) + \frac{2d\gamma}{\varepsilon^{2}}(\varepsilon + d+ 2)\left(2 + \frac{\gamma}{\varepsilon}(\varepsilon+ d)\right) \\
& = \frac{\gamma}{\varepsilon}\left(\varepsilon +d+ \frac{4d(\varepsilon +d+ 2)}{\varepsilon} \right) + \frac{2d\gamma^2}{\varepsilon^3}(\varepsilon+d)(\varepsilon +  d + 2)\\
& = \frac{\gamma}{\varepsilon^2}\left(\varepsilon(\varepsilon +d)+ 4d(\varepsilon +d+ 2)\right) + \frac{2d\varepsilon \gamma^2}{\varepsilon^4}(\varepsilon+d)(\varepsilon +  d + 2)\\
& \leq \frac{\gamma}{\varepsilon^2}\left((d-1) (2d-1) + 4d(2d+1) \right) + \frac{2d\gamma^2}{\varepsilon^4} (d-1)(2d-1)(2d+1)\\
& = \frac{1}{8\sqrt 2 d^2}(10d^2 + d +1) + \frac{1}{64d^3}(d-1)(2d-1)(2d+1)\\
& = \frac{(d-1)(4d^2-1) + 4\sqrt 2 d (10d^2 + d + 1)}{64d^3}\\
& \leq \frac{(d-1)(4d^2-1) + 41\sqrt 2 d^3 + 8\sqrt 2 d}{64d^3}\\
& \leq \frac{(41\sqrt 2 + 4) d^3 + 8\sqrt 2 d - 4d^2 - d + 1}{64d^3}\\
& \leq \frac{(41\sqrt 2 + 4) d^3 + 4d(2 \sqrt 2  - d) + (1-d)}{64d^3}\\
& < \frac{41\sqrt 2 + 4}{64} + \frac{\sqrt 2 - 1} {32} \\
& < 1.
\end{align*}
So the claim that $H$ is a subgroup of $\ensuremath{\mathcal{G}}$ of index two follows. Since no index two subgroup of $\ensuremath{\mathcal{G}}$ acts transitively on $V$, the action of $H$ on $V$ has at least two distinct orbits. Since the action of $\ensuremath{\mathcal{G}}$ on $V$ is transitive, the action of $H$ on $V$ has exactly two orbits. Let $\ensuremath{\mathcal{O}}_1, \ensuremath{\mathcal{O}}_2$ denote the orbits of the action of $H$ on $V$.

Note that for any $g\in H$ and for any subset $B$ of $V$ contained in $\ensuremath{\mathcal{O}}_1$ or $\ensuremath{\mathcal{O}}_2$, the map
$$B\cap g^{-1} B \to B\times B, \quad
x\mapsto (x, gx)$$
is well-defined, and induces a map
$$\varphi: \coprod _{g\in H} B\cap g^{-1} B \to B\times B.$$
Each of its fibre contains exactly $t$ elements and hence
$$
|B|^2
= im(\varphi)
= \sum_{x\in im(\varphi) } \frac{|\varphi^{-1} (x)|} {|\varphi^{-1} (x)|}
= \frac 1t \sum_{x\in im(\varphi)} |\varphi^{-1} (x)|
= \frac 1t \sum_{g\in H} |B\cap g^{-1} B|.$$
For any $g\in H$, we have
\begin{align*}
A\cap g^{-1} A
& =
((A\cap \ensuremath{\mathcal{O}}_1) \cup (A\cap \ensuremath{\mathcal{O}}_2))
\cap
(g^{-1} ((A\cap \ensuremath{\mathcal{O}}_1) \cup (A\cap \ensuremath{\mathcal{O}}_2)))\\
& =
((A\cap \ensuremath{\mathcal{O}}_1) \cap
(g^{-1} (A\cap \ensuremath{\mathcal{O}}_1))
\cup
((A\cap \ensuremath{\mathcal{O}}_2) \cap
(g^{-1} (A\cap \ensuremath{\mathcal{O}}_2)),
\end{align*}
which yields
$$
|A \cap g^{-1} A|  = \sum_{B = A\cap \ensuremath{\mathcal{O}}_1, A\cap \ensuremath{\mathcal{O}}_2}|B\cap g^{-1} B| .$$
So
\begin{align*}
|A\cap \ensuremath{\mathcal{O}}_1|^2 + |A\cap \ensuremath{\mathcal{O}}_2|^2
& = \sum_{B = A\cap \ensuremath{\mathcal{O}}_1, A\cap \ensuremath{\mathcal{O}}_2} |B|^2\\
& = \frac 1t \sum_{B = A\cap \ensuremath{\mathcal{O}}_1, A\cap \ensuremath{\mathcal{O}}_2}\sum_{g\in H} |B\cap g^{-1} B| \\
& = \frac 1t \sum_{g\in H} \sum_{B = A\cap \ensuremath{\mathcal{O}}_1, A\cap \ensuremath{\mathcal{O}}_2}|B\cap g^{-1} B| \\
& = \frac 1t \sum_{g\in H} |A \cap g^{-1} A| .
\end{align*}
It follows that
\begin{align*}
|A|^2 - 2|A\cap \ensuremath{\mathcal{O}}_1||A\cap \ensuremath{\mathcal{O}}_2|
& = (|A\cap \ensuremath{\mathcal{O}}_1| + |A\cap \ensuremath{\mathcal{O}}_2|)^2 - 2|A\cap \ensuremath{\mathcal{O}}_1||A\cap \ensuremath{\mathcal{O}}_2| \\
& = |A\cap \ensuremath{\mathcal{O}}_1|^2 + |A\cap \ensuremath{\mathcal{O}}_2|^2  \\
& = \frac 1t \sum_{g\in H} |A\cap gA| \\
& = \frac 1t \sum_{g\in \ensuremath{\mathcal{G}}} |A\cap gA| - \frac 1t \sum_{g\in H^c} |A\cap gA| \\
& = |A|^2 - \frac 1t \sum_{g\in H^c} |A\cap gA| \\
& \geq |A|^2 - \sum_{g\in H^c} \frac{d \gamma}{t\varepsilon^2} (\varepsilon + d + 2)|A| \\
& = |A|^2 - \frac{d \gamma}{t\varepsilon^2} (\varepsilon + d + 2)|A||H|.
\end{align*}
This implies that
$$
|A\cap \ensuremath{\mathcal{O}}_1||A\cap \ensuremath{\mathcal{O}}_2|
\leq \frac{d \gamma}{4t\varepsilon^2} (\varepsilon + d + 2)|A||\ensuremath{\mathcal{G}}|
\leq \frac{d \gamma}{4t\varepsilon^2} (\varepsilon + d + 2) \frac{t|V|^2}{2}\leq \frac{d \gamma}{8\varepsilon^2} (\varepsilon + d + 2)|V|^2.
$$
Hence, for the orbit $\ensuremath{\mathcal{O}}$ of some element of $V$ under the action of $H$, we have
$$
|A\cap \ensuremath{\mathcal{O}}^c| \leq
\sqrt{\frac{d\gamma}{8\varepsilon^2} (\varepsilon + d + 2) }|V|.$$
For any $1\leq i \leq d$,
\begin{align*}
|\theta_i(\ensuremath{\mathcal{O}})\cap \ensuremath{\mathcal{O}} |
& = |\theta_i(\ensuremath{\mathcal{O}}\cap A) \cap \ensuremath{\mathcal{O}}| + |\theta_i(\ensuremath{\mathcal{O}}\setminus A) \cap \ensuremath{\mathcal{O}}|\\
& \leq |\theta_i(A) \cap \ensuremath{\mathcal{O}}| + |\ensuremath{\mathcal{O}}\setminus A|\\
& = |\theta_i(A) \cap (\ensuremath{\mathcal{O}}\cap A) | + |\theta_i(\ensuremath{\mathcal{O}}\cap A) \cap (\ensuremath{\mathcal{O}}\setminus A) | + |\ensuremath{\mathcal{O}}\setminus A|\\
& \leq |\theta_i(A) \cap A| + 2|\ensuremath{\mathcal{O}}\setminus A|\\
& = |\theta_i(A) \cap A| + 2|\ensuremath{\mathcal{O}}| - 2|\ensuremath{\mathcal{O}}\cap A|\\
& = |\theta_i(A) \cap A| + |V| - 2|A| + 2|A\cap \ensuremath{\mathcal{O}}^c|\\
& \leq \frac \gamma\varepsilon |A| + \left(2 + \gamma + \frac{d\gamma}{\varepsilon}\right)|A| - 2|A| + 2\sqrt{\frac{d\gamma}{8\varepsilon^2} (\varepsilon + d + 2) }|V| \\
& = \frac \gamma\varepsilon |A| + \left(\gamma + \frac{d\gamma}{\varepsilon}\right)|A|  + 2\sqrt{\frac{d\gamma}{8\varepsilon^2} (\varepsilon + d + 2) }|V| \\
& \leq \left(\frac \gamma\varepsilon (d+1 + \varepsilon)  + \sqrt{\frac{2d\gamma}{\varepsilon^2} (\varepsilon + d + 2) }\right) \frac{|V|}{2} \\
& \leq \left(\frac {2d\gamma}\varepsilon |H| + \sqrt{\frac{2d\gamma}{\varepsilon^2} (2d + 1) }\right) \frac{|V|}{2} \\
& \leq \left(\frac {2d\gamma}\varepsilon  + \sqrt{\frac{6d^2\gamma}{\varepsilon^2}} \right) \frac{|V|}{2}\\
& < \left(\frac {1}{2^4}  + \sqrt{\frac{6}{2^5}} \right) \frac{|V|}{2}\\
& < \frac{|V|}{4} \\
& = \frac{|\ensuremath{\mathcal{G}}|}{4t} \\
& = \frac{|H|}{2t}.
\end{align*}

Note that for any two subgroup $H_1, H_2$ of $\ensuremath{\mathcal{G}}$ of index two, the inequalities
$$|H_1\cap H_2| \geq \frac{|H_1|}{2}, |H_1^c\cap H_2^c| \geq \frac{|H_1|}{2}$$
hold. Indeed, for any $x\in H_1 \cap H_2^c$, the `left multiplication by $x$' map induces an injection from $H_1\cap H_2^c\to H_1 \cap H_2$. This implies that
$$ |H_1\cap H_2|
\geq |H_1 \cap H_2^c|,$$
which yields
$$ |H_1\cap H_2|
\geq \frac 12 ( |H_1\cap H_2^c|  + |H_1\cap H_2| ) = \frac{|H_1|}{2}.$$
It follows that $|H_1\cap H_2^c| \leq \frac{|H_1|}2$, which implies $|H_1^c\cap H_2^c| \geq \frac{|H_1|}2$.

Note that no two elements of $\ensuremath{\mathcal{O}}$ are adjacent in $(V, E)$. Otherwise, for $u, v\in \ensuremath{\mathcal{O}}$, $u = \theta_i(v)$ for some $i$ and hence
\begin{align*}
|\theta_i(\ensuremath{\mathcal{O}})\cap \ensuremath{\mathcal{O}}|
& = |\theta_i(Hv) \cap Hu | \\
& = | \psi_{i,v} (H) \theta_i(v) \cap H u| \\
& = |\psi_{i,v} (H) u \cap H u|\\
& \geq |(\psi_{i,v} (H) \cap H) u|\\
& \geq \frac{|\psi_{i,v} (H) \cap H|}{t} \\
& \geq \frac{|H|}{2t}.
\end{align*}
Moreover, no two elements of $\ensuremath{\mathcal{O}}^c$ are adjacent in $(V, E)$. Otherwise, for $u, v\in \ensuremath{\mathcal{O}}^c$, $u = \theta_i(v)$ for some $i$ and hence
\begin{align*}
|\theta_i(\ensuremath{\mathcal{O}})\cap \ensuremath{\mathcal{O}}|
& = |\theta_i(H^c v) \cap H^c u | \\
& = |\psi_{i,v} (H^c) \theta_i(v) \cap H^c u|\\
& = |\psi_{i,v} (H^c) u \cap H^c u| \\
& \geq |(\psi_{i,v} (H^c) \cap H^c) u| \\
& \geq \frac{|\psi_{i,v} (H^c) \cap H^c|}{t} \\
& \geq \frac{|H|}{2t}.
\end{align*}
So $(V, E)$ is bipartite, contradicting the hypothesis. Hence, the nontrivial eigenvalues of the normalised adjacency operator of this graph are greater than $-1 + \ell_{\varepsilon, d}$ with
$$\ell_{\varepsilon, d}
=\frac{\varepsilon^4}{2^{12} d^8}.$$
\end{proof}

In the following, $(V, E)^k$ denotes the graph having $V$ as its set of vertices and the $k$-th power of the adjacency operator of $(V,E)$ as its adjacency operator.

\begin{theorem}
\label{thmPrincipal}
Suppose $V$ carries a left action of a group $\ensuremath{\mathcal{G}}$, $(V, E)$ is undirected, $|V|\geq 4$ and $k\geq 1$ be an odd integer such that the following conditions hold.
\begin{enumerate}
\item No index two subgroup of $\ensuremath{\mathcal{G}}$ acts transitively on $V$.

\item The action of $\ensuremath{\mathcal{G}}$ on the set $V$ is ``transitive of order $t$'' in the sense that for each $(u, v)\in V\times V$, the equation $gu = v$ has exactly $t$ distinct solutions for $g\in \ensuremath{\mathcal{G}}$. In other words, the action of $\ensuremath{\mathcal{G}}$ on $V$ is transitive and the stabilizer of each element of $V$ have the same size (equal to $t$).

\item
$(V, E)^k$ is an $\varepsilon_k$-vertex expander with $\varepsilon_k >0$.

\item
For $1\leq i_1, \cdots, i_k \leq d$ and $v\in V$, there is an automorphism or an anti-automorphism $\psi_{(i_1, \cdots, i_k), v}$ of the group $\ensuremath{\mathcal{G}}$ such that one of
$$
(\theta_{i_1} \circ \cdots \circ \theta_{i_k} )(g\cdot v)
=
\psi_{(i_1, \cdots, i_k),  v}(g) \cdot
(\theta_{i_1} \circ \cdots  \circ \theta_{i_k} )(v)
$$
and
$$
(\theta_{i_1} \circ \cdots  \circ \theta_{i_k} )(g\cdot v)
=
\psi_{(i_1, \cdots, i_k),  v}(g^{-1}) \cdot
(\theta_{i_1} \circ \cdots \circ \theta_{i_k} )(v)
$$
holds for any $g\in \ensuremath{\mathcal{G}}$.

\item For any $\tau\in \ensuremath{\mathcal{G}}$ and for any subset $A$ of $V$,
$
\ensuremath{\mathcal{N}}^k (\ensuremath{\mathcal{N}}^k(\tau(A))) \subseteq \tau(\ensuremath{\mathcal{N}}^k(\ensuremath{\mathcal{N}}^k(A)))
$
holds.
\end{enumerate}
Then the nontrivial eigenvalues of the normalised adjacency operator of $(V, E)$ are greater than
$$
\left(
-1 + \frac {1}{2^{12} d^{8k}}
\varepsilon_k^4
\right)^{1/k}.
$$
If $(V, E)$ is an $\varepsilon$-vertex expander with $\varepsilon >0$, then the vertex Cheeger constant of $(V, E)^k$ is
$$
\geq
\frac 12
\left(
1 -
\left(
1 - \frac{\varepsilon^2}{2d^2}
\right)^k
\right)
.$$
\end{theorem}

\begin{proof}
By Theorem \ref{thmPrincipalk1}, the nontrivial spectrum of the adjacency operator of $(V, E)^k$ is bounded
away from $-1 + \frac {\varepsilon_k^4}{2^{12} d^{8k}}$. Since $k$ is odd, the nontrivial spectrum of $(V, E)$ is bounded away from
$$
\left(
-1 + \frac {1}{2^{12} d^{8k}}
\varepsilon_k^4
\right)^{1/k}
.$$

Since $(V, E)$ is an $\varepsilon$-vertex expander, the largest nontrivial eigenvalue of its adjacency operator is bounded by $1 - \frac{\varepsilon^2}{2d^2}$. Since $k$ is odd, the largest nontrivial eigenvalue of the adjacency operator of $(V, E)^k$ is bounded by
$$
\left(
1 - \frac{\varepsilon^2}{2d^2}
\right)^k
.$$
By Lemma \ref{Lemma:VertexEdgeCons} and the discrete Cheeger--Buser inequality (Proposition \ref{Prop:chin}), the vertex Cheeger constant of $(V, E)^k$ is
$$
\geq
\frac 12
\left(
1 -
\left(
1 - \frac{\varepsilon^2}{2d^2}
\right)^k
\right)
.$$
\end{proof}

\section{Spectral expansion of the Cayley and Cayley sum graphs twisted by automorphisms}
\label{Sec:Twists}

Let $G$ be a finite group, $S$ be a subset of $G$ and $\sigma$ be a group automorphism of $G$.

Consider the twist $C(G, S)^\sigma$ of the Cayley graph $C(G, S)$ by the automorphism $\sigma$. The graph $C(G, S)^\sigma$ has $G$ as its set of vertices, and there is an edge from $x$ to $y$ whenever $y = \sigma(xs)$ for some $s\in S$. Roughly speaking, the twisted Cayley graph $C(G, S)^\sigma$ has the same set of vertices as that of the Cayley graph $C(G, S)$, and given a vertex $x$ in $C(G, S)^\sigma$, its adjacent vertices are precisely the translates of the adjacent vertices of $x$ in $C(G, S)$ under $\sigma$.

Consider the twist $C_\Sigma(G, S)^\sigma$ of the Cayley sum graph $C_\Sigma(G, S)$ by the automorphism $\sigma$. The graph $C_\Sigma(G, S)^\sigma$ has $G$ as its set of vertices, and there is an edge from $x$ to $y$ whenever $y = \sigma(x^{-1} s)$ for some $s\in S$. Roughly speaking, the twisted Cayley sum graph $C_\Sigma(G, S)^\sigma$ has the same set of vertices as that of the Cayley sum graph $C_\Sigma(G, S)$, and given a vertex $x$ in $C_\Sigma(G, S)^\sigma$, its adjacent vertices are precisely the translates of the adjacent vertices of $x$ in $C_\Sigma(G, S)$ under $\sigma$.

\subsection{The twisted Cayley graph}

For a subset $A$ of $G$, its neighbourhood in $C(G, S)^\sigma$ is denote by $\ensuremath{\mathscr{N}}(A)$. Given $g_1, \cdots, g_r \in G$,
$\prod_{i = 1}^r g_i$
denotes the product
$g_1 \cdots g_r$.

\begin{proof}[Proof of Theorem \ref{Thm:Bdd}(1), (2)]
Let $\ensuremath{\mathcal{G}}$ denote the group $G$ and $m$ be a positive integer. Consider the action of $\ensuremath{\mathcal{G}}$ on the set of vertices of $(C(G, S)^\sigma)^m$ by left multiplication, i.e.,
$$g\cdot v = gv$$
for any $g\in \ensuremath{\mathcal{G}}, v\in G$.

For an element $(s_1, \cdots, s_m)\in S^m$, let $\theta:G\to G$ denote the bijection defined by
$$\theta(v)
=
\sigma^m(v)
\prod_{i=1}^m
\sigma^{m+1-i} (s_i)$$
for $v\in G$.
For each $(s_1, \cdots, s_m)\in S^m$ and $v\in G$, note that
\begin{align*}
\theta(g\cdot v)
& =
\sigma^m(g) \cdot \theta(v) \\
& =
\psi_{(s_1, \cdots, s_m), v}(g) \cdot \theta(v)
\end{align*}
for any $g\in \ensuremath{\mathcal{G}}$, where $\psi_{(s_1, \cdots, s_m), v}$ denotes the automorphism
$$g\mapsto
\sigma^m(g)
$$
of the group $\ensuremath{\mathcal{G}}$. Moreover, for any subset $A$ of $G$,
$\ensuremath{\mathscr{N}}^m(g\cdot A) = \sigma^m(g) \cdot \ensuremath{\mathscr{N}}^m(A)$.

Since $C(G, S)^\sigma$ is connected, its vertex Cheeger constant $h_\sigma$ is positive. Thus $C(G, S)^\sigma$ is an $h_\sigma$-expander with $h_\sigma>0$.
If $\sigma^2$ is the trivial automorphism of $G$, then from Theorem \ref{thmPrincipal}, the nontrivial eigenvalues of the normalised adjacency operator of $C(G, S)^\sigma$ are greater than
$$
-1 + \frac {h_\sigma^4}{2^{12} d^{8}}
.
$$
If $\sigma^{2k}$ is the trivial automorphism of $G$ for some odd integer $k\geq 1$, then from Theorem \ref{thmPrincipal}, the nontrivial eigenvalues of the normalised adjacency operator of $C(G, S)^\sigma$ are greater than
$$
\left(
-1 + \frac {1}{2^{12} d^{8k}}
\left(
\frac 12
\left(
1 -
\left(
1 - \frac{h_\sigma^2}{2d^2}
\right)^k
\right)
\right)
^4
\right)^{1/k}.
$$
By the discrete Cheeger--Buser inequality (Proposition \ref{Prop:chin}), the result follows.
\end{proof}

\subsection{The twisted Cayley sum graph}

For a subset $A$ of $G$, its neighbourhood in $C_\Sigma(G, S)^\sigma$ is denote by $\ensuremath{\mathscr{N}}_\Sigma(A)$. Given $g_1, \cdots, g_r \in G$,  $\prod_{i = 1}^r g_i$ denotes the product $g_1 \cdots g_r$.

\begin{lemma}
\label{Lemma:UndirectedsigmaCayleySum}
The twisted Cayley sum graph $C_\Sigma (G, S)^\sigma$ is undirected if and only if $S$ contains $\sigma^2(g)\sigma(s)g^{-1}$ for any $s\in S, g\in G$.
\end{lemma}

\begin{proof}
If $h$ is adjacent to $g$, then $h = \sigma(g^{-1} s)$ for some $s\in S$. Note that
\begin{align*}
g
& = s \sigma^{-1}(h^{-1}) \\
& = \sigma(h^{-1}) \sigma(h) s\sigma^{-1}(h^{-1}),
\end{align*}
which implies that $g$ is adjacent to $h$ if and only if $\sigma(h) s\sigma^{-1}(h^{-1}) \in \sigma^{-1}(S)$, i.e.,
$\sigma^2(h) \sigma(s) h^{-1} \in S$. Hence $g$ is adjacent to each of its adjacent vertices if and only if $S$ contains
$$\sigma^2(\sigma(g^{-1} s)) \sigma(s) (\sigma(g^{-1} s))^{-1}
$$
for any $s\in S$. So $C_\Sigma (G, S)^\sigma$ is undirected if and only if $S$ contains $
\sigma^2(\sigma(g^{-1} s)) \sigma(s) (\sigma(g^{-1} s))^{-1}
$ for any $s\in S, g\in G$. Note that for $s\in S$ and $x\in G$ and $g = (\sigma^{-1}(x)s^{-1})^{-1}$,
$$
\sigma^2(\sigma(g^{-1} s)) \sigma(s) (\sigma(g^{-1} s))^{-1}
=
\sigma^2(x) \sigma(s) x^{-1}.
$$
So $C_\Sigma (G, S)^\sigma$ is undirected if and only if $S$ contains $\sigma^2(g)\sigma(s)g^{-1}$ for any $s\in S, g\in G$. Hence the Lemma follows.
\end{proof}

\begin{proof}[Proof of Theorem \ref{Thm:Bdd}(3)]
Let $\ensuremath{\mathcal{G}}$ denote the group $G$ and $m$ be a positive integer. Consider the action of $\ensuremath{\mathcal{G}}$ on the set of vertices of $(C_\Sigma(G, S)^\sigma)^m$ by right multiplication via the inverse, i.e.,
$$g\cdot v = vg^{-1}$$
for any $g\in \ensuremath{\mathcal{G}}, v\in G$.

For an element $(s_1, \cdots, s_m)\in S^m$, let $\theta:G\to G$ denote the bijection defined by
$$\theta(v)
=
\left(
\prod_{i = 1}^{\lfloor m/2\rfloor} \sigma^{2i} (s_{m+1 -2i}^{-1})
\right)
\sigma(v^{(-1)^m})
\left(
\prod_{i = 1}^{\lceil m/2\rceil} \sigma^{2\lceil m/2\rceil +1 - 2i} (s_{m+1 -(2\lceil m/2\rceil +1 - 2i)})
\right)
$$
for $v\in G$.
For each $(s_1, \cdots, s_m)\in S^m$ and $v\in G$, set
$$
\alpha =
\left(
\prod_{i = 1}^{\lfloor m/2\rfloor} \sigma^{2i} (s_{m+1 -2i}^{-1})
\right),
\beta =
\left(
\prod_{i = 1}^{\lceil m/2\rceil} \sigma^{2\lceil m/2\rceil +1 - 2i} (s_{m+1 -(2\lceil m/2\rceil +1 - 2i)})
\right),
$$
and note that
\begin{align*}
\theta(g\cdot v)
& =
\alpha \sigma(gv^{-1}) \beta \\
& =
\alpha \sigma(v^{-1}) \beta (\sigma(v^{-1}) \beta)^{-1} \sigma(gv^{-1}) \beta \\
& =
\alpha \sigma(v^{-1}) \beta
\left(
(\sigma(v^{-1}) \beta)^{-1} \sigma(g^{-1}) \sigma(v^{-1}) \beta
\right)^{-1} \\
& =
\theta(v)
\left(
(\sigma(v^{-1}) \beta)^{-1} \sigma(g^{-1}) \sigma(v^{-1}) \beta
\right)^{-1} \\
& =
\left(
(\sigma(v^{-1}) \beta)^{-1} \sigma(g^{-1}) \sigma(v^{-1}) \beta
\right)\cdot \theta(v) \\
& =
\psi_{(s_1, \cdots, s_m), v}(g^{-1}) \cdot \theta (v)
\end{align*}
for any $g\in \ensuremath{\mathcal{G}}$, where $\psi_{(s_1, \cdots, s_m), v}$ denotes the automorphism
$$g\mapsto
(\sigma(v^{-1}) \beta)^{-1} \sigma(g) \sigma(v^{-1}) \beta
$$
of the group $\ensuremath{\mathcal{G}}$.

For any subset $A$ of $G$, note that
$$\ensuremath{\mathscr{N}}_\Sigma^{m}
(A)
=
\sigma^2(S^{-1}) \sigma^4(S^{-1}) \cdots \sigma^{2\lfloor m/2\rfloor}(S^{-1})
\sigma^{m}(A^{(-1)^m})
\sigma^{2\lceil m/2\rceil -1}(S) \cdots \sigma^3(S) \sigma(S).
$$
For any $g\in \ensuremath{\mathcal{G}}$, one obtains
\begin{align*}
& \sigma^{2m}(g)
\sigma^{2m-1}(S) \cdots \sigma^3(S) \sigma(S)
g^{-1}
\\
& =
\left(
\sigma^{2m}(g)
\sigma^{2m-1}(S)
\sigma^{2m-2}(g^{-1})
\right)
\left(
\sigma^{2m-2}(g)
\sigma^{2m-3}(S)
\sigma^{2m-4}(g^{-1})
\right)
\cdots \\
& \qquad \qquad
\left(
\sigma^6(g)
\sigma^5(S)
\sigma^4(g^{-1})
\right)
\left(
\sigma^4(g)
\sigma^3(S)
\sigma^2(g^{-1})
\right)
\left(
\sigma^2(g)
\sigma(S)
g^{-1}
\right)
\\
& =
\sigma^{2m-2}(S)
\cdots
\sigma^4(S)
\sigma^2(S)
S \\
& =
\sigma^{2m-1}(S)
\cdots
\sigma^5(S)
\sigma^3(S)
\sigma(S).
\end{align*}
So, for any subset $A$ of $G$,
\begin{align*}
\ensuremath{\mathscr{N}}_\Sigma^m(\ensuremath{\mathscr{N}}_\Sigma^m(g\cdot A))
& =
\sigma^2(S^{-1}) \sigma^4(S^{-1}) \cdots \sigma^{2m}(S^{-1})
\sigma^{2m}(g\cdot A)
\sigma^{2m-1}(S) \cdots \sigma^3(S) \sigma(S) \\
& =
\sigma^2(S^{-1}) \sigma^4(S^{-1}) \cdots \sigma^{2m}(S^{-1})
\sigma^{2m}(A) \sigma^{2m}(g^{-1})
\sigma^{2m-1}(S) \cdots \sigma^3(S) \sigma(S) \\
& =
\sigma^2(S^{-1}) \sigma^4(S^{-1}) \cdots \sigma^{2m}(S^{-1})
\sigma^{2m}(A) \sigma^{2m}(g^{-1})
\sigma^{2m-1}(S) \cdots \sigma^3(S) \sigma(S) gg^{-1}\\
& =
\sigma^2(S^{-1}) \sigma^4(S^{-1}) \cdots \sigma^{2m}(S^{-1})
\sigma^{2m}(A)
\sigma^{2m-1}(S) \cdots \sigma^3(S) \sigma(S) g^{-1}\\
& =
\ensuremath{\mathscr{N}}_\Sigma^m(\ensuremath{\mathscr{N}}_\Sigma^m(A)) g^{-1} \\
& =
g \cdot \ensuremath{\mathscr{N}}_\Sigma^m(\ensuremath{\mathscr{N}}_\Sigma^m(A))
\end{align*}
holds for any $g\in \ensuremath{\mathcal{G}}$.

Since $C_\Sigma (G, S)^\sigma$ is connected, its vertex Cheeger constant $h_{\Sigma, \sigma}$ is positive. Thus $C_\Sigma (G, S)^\sigma$ is an $h_{\Sigma, \sigma}$-expander with $h_{\Sigma, \sigma} >0$.
By Theorem \ref{thmPrincipal}, the nontrivial eigenvalues of the normalised adjacency operator of $C_\Sigma(G, S)^\sigma$ are greater than
$$
-1 + \frac {h_{\Sigma, \sigma}^4}{2^{12} d^{8}}
.
$$
By the discrete Cheeger--Buser inequality (Proposition \ref{Prop:chin}), the result follows.
\end{proof}

\section{Spectral expansion of the Cayley and Cayley sum graphs twisted by anti-automorphisms}
\label{Sec:TwistsAnti}

Let $G$ be a finite group, $S$ be a subset of $G$ and $\sigma$ be an group anti-automorphism of $G$.

Consider the twist $C(G, S)^\sigma$ of the Cayley graph $C(G, S)$ by the anti-automorphism $\sigma$. The graph $C(G, S)^\sigma$ has $G$ as its set of vertices, and there is an edge from $x$ to $y$ whenever $y = \sigma(xs)$ for some $s\in S$. Roughly speaking, the twisted Cayley graph $C(G, S)^\sigma$ has the same set of vertices as that of the Cayley graph $C(G, S)$, and given a vertex $x$ in $C(G, S)^\sigma$, its adjacent vertices are precisely the translates of the adjacent vertices of $x$ in $C(G, S)$ under $\sigma$.

Consider the twist $C_\Sigma(G, S)^\sigma$ of the Cayley sum graph $C_\Sigma(G, S)$ by the anti-automorphism $\sigma$. The graph $C_\Sigma(G, S)^\sigma$ has $G$ as its set of vertices, and there is an edge from $x$ to $y$ whenever $y = \sigma(x^{-1} s)$ for some $s\in S$. Roughly speaking, the twisted Cayley sum graph $C_\Sigma(G, S)^\sigma$ has the same set of vertices as that of the Cayley sum graph $C_\Sigma(G, S)$, and given a vertex $x$ in $C_\Sigma(G, S)^\sigma$, its adjacent vertices are precisely the translates of the adjacent vertices of $x$ in $C_\Sigma(G, S)$ under $\sigma$.

\begin{theorem}\label{Thm:BddAnti}
Let $S$ be a subset of a finite group $G$ with $|S|= d$. Suppose $\sigma$ is an anti-automorphism of $G$.
\begin{enumerate}
\item If the twisted Cayley graph $C (G, S)^\sigma$ is connected and undirected and $|G| \geq 4$, then the nontrivial spectrum of its normalised adjacency operator lies in the interval
$$\left( -1 + \frac{h_\sigma^4}{2^{12}d^8}
,
1 - \frac{h_\sigma^2}{2d^2}
\right]$$
where $h_\sigma$ denotes the vertex Cheeger constant of $C (G, S)^\sigma$.

\item
Suppose $\sigma^{2}$ is the trivial automorphism of $G$.
If the twisted Cayley sum graph $C_\Sigma(G, S)^\sigma$ is connected and undirected and $|G| \geq 4$, then the nontrivial spectrum of its normalised adjacency operator lies in the interval
$$\left( -1 + \frac{h_{\Sigma, \sigma}^4}{2^{12}d^8}
,
1 - \frac{h_{\Sigma, \sigma}^2}{2d^2}
\right]$$
where $h_{\Sigma, \sigma}$ denotes the vertex Cheeger constant of $C_\Sigma(G, S)^\sigma$.

\item
Suppose $\sigma^{2k}$ is the trivial automorphism of $G$, where $k\geq 1$ is an odd integer.
If the twisted Cayley graph sum $C_\Sigma(G, S)^\sigma$ is connected and undirected and $|G| \geq 4$, then the nontrivial spectrum of its normalised adjacency operator lies in the interval
$$\left(
\left(
-1 + \frac {1}{2^{12} d^{8k}}
\left(
\frac 12
\left(
1 -
\left(
1 - \frac{h_{\Sigma, \sigma}^2}{2d^2}
\right)^k
\right)
\right)
^4
\right)^{1/k}
,
1 - \frac{h_{\Sigma, \sigma}^2}{2d^2}
\right]$$
where $h_{\Sigma, \sigma}$ denotes the vertex Cheeger constant of $C_\Sigma(G, S)^\sigma$.
\end{enumerate}
\end{theorem}

\subsection{Cayley graph twisted by anti-automorphisms}

For a subset $A$ of $G$, its neighbourhood in $C(G, S)^\sigma$ is denote by $\ensuremath{\mathscr{N}}(A)$. Given $g_1, \cdots, g_r \in G$,
$\prod_{i = 1}^r g_i$
denotes the product
$g_1 \cdots g_r$.

\begin{lemma}
\label{Lemma:UndirectedsigmaCayleySum}
The twisted Cayley graph $C (G, S)^\sigma$ is undirected if and only if $S$ contains $\sigma^2(g) \sigma(s^{-1}) g^{-1}$ for any $s\in S, g\in G$.
\end{lemma}

\begin{proof}
If $h$ is adjacent to $g$, then $h = \sigma(g s)$ for some $s\in S$. Note that
\begin{align*}
g
& = \sigma^{-1}(h)s^{-1} \\
& = \sigma^{-1}(h)s^{-1} \sigma(h^{-1}) \sigma(h) \\
& = \sigma(h^{-1} \sigma^{-1}(s^{-1}) \sigma^{-2}(h))\sigma(h) \\
& = \sigma(h (h^{-1} \sigma^{-1}(s^{-1}) \sigma^{-2}(h))),
\end{align*}
which implies that $g$ is adjacent to $h$ if and only if $h^{-1} \sigma^{-1}(s^{-1}) \sigma^{-2}(h) \in S$. Hence $g$ is adjacent to each of its adjacent vertices if and only if $S$ contains
$$
(\sigma(s) \sigma(g))^{-1} \sigma^{-1}(s^{-1}) \sigma^{-2}(\sigma(s) \sigma(g))
$$
for any $s\in S$. So $C (G, S)^\sigma$ is undirected if and only if $S$ contains
$$(\sigma(s) \sigma(g))^{-1} \sigma^{-1}(s^{-1}) \sigma^{-2}(\sigma(s) \sigma(g))
=
\sigma(g)^{-1} \sigma(s^{-1}) \sigma^{-1}(s^{-1}) \sigma^{-1}(s) \sigma^{-1}(g)
=
\sigma(g)^{-1} \sigma(s^{-1}) \sigma^{-1}(g)
$$
for any $s\in S, g\in G$.
So $C (G, S)^\sigma$ is undirected if and only if $S$ contains $\sigma^2(g)\sigma(s^{-1})g^{-1}$ for any $s\in S, g\in G$. Hence the Lemma follows.
\end{proof}

\begin{proof}[Proof of Theorem \ref{Thm:BddAnti}(1)]
Let $\ensuremath{\mathcal{G}}$ denote the group $G$. Consider the action of $\ensuremath{\mathcal{G}}$ on the set of vertices of $C(G, S)^\sigma$ by left multiplication, i.e.,
$$g\cdot v = gv$$
for any $g\in \ensuremath{\mathcal{G}}, v\in G$.

For an element $s\in S$, let $\theta:G\to G$ denote the bijection defined by
$$\theta(v)
=
\sigma(vs)$$
for $v\in G$.
For each $s\in S$ and $v\in G$, note that
\begin{align*}
\theta(g\cdot v)
& =
\sigma(gvs) \\
& =
\sigma(gvs) (\sigma(vs))^{-1} \sigma(vs) \\
& =
(\sigma(vs) \sigma(g) (\sigma(vs))^{-1}) \sigma(vs) \\
& =
\psi_{s, v}(g) \cdot \theta(v)
\end{align*}
for any $g\in \ensuremath{\mathcal{G}}$, where $\psi_{s, v}$ denotes the anti-automorphism
$$g\mapsto
\sigma(vs) \sigma(g) (\sigma(vs))^{-1}
$$
of the group $\ensuremath{\mathcal{G}}$.

For any subset $A$ of $G$, note that
\begin{align*}
\ensuremath{\mathscr{N}}^{2}(A)
& = \sigma(S) \sigma^2(A) \sigma^2(S),
\end{align*}
which yields
\begin{align*}
\ensuremath{\mathscr{N}}^{2}(g\cdot A)
& =
\sigma(S) \sigma^2(g \cdot A) \sigma^2(S)\\
& =
\sigma(S) \sigma^2(g)\sigma^2(A) \sigma^2(S)\\
& =
g g^{-1} \sigma(S) \sigma^2(g)\sigma^2(A) \sigma^2(S)\\
& =
g (\sigma^2(g^{-1})\sigma(S^{-1})g)^{-1} \sigma^2(A) \sigma^2(S)\\
& =
g S^{-1} \sigma^2(A) \sigma^2(S)\\
& =
g \sigma(S) \sigma^2(A) \sigma^2(S)\\
& =
g \cdot \ensuremath{\mathscr{N}}^2(A)  .
\end{align*}

Since $C(G, S)^\sigma$ is connected, its vertex Cheeger constant $h_\sigma$ is positive. Thus $C(G, S)^\sigma$ is an $h_\sigma$-expander with $h_\sigma>0$.
By Theorem \ref{thmPrincipal}, the nontrivial eigenvalues of the normalised adjacency operator of $C(G, S)^\sigma$ are greater than
$$
-1 + \frac {h_\sigma^4}{2^{12} d^{8}}
.
$$
By the discrete Cheeger--Buser inequality (Proposition \ref{Prop:chin}), the result follows.
\end{proof}

\subsection{Cayley sum graph twisted by anti-automorphisms}

For a subset $A$ of $G$, its neighbourhood in $C_\Sigma(G, S)^\sigma$ is denote by $\ensuremath{\mathscr{N}}_\Sigma(A)$. Given $g_1, \cdots, g_r \in G$,
$\prod_{i = 1}^r g_i$
denotes the product
$g_1 \cdots g_r$.

\begin{proof}[Proof of Theorem \ref{Thm:BddAnti}(2), (3)]
Let $\ensuremath{\mathcal{G}}$ denote the group $G$. Consider the action of $\ensuremath{\mathcal{G}}$ on the set of vertices of $C_\Sigma(G, S)^\sigma$ by right multiplication via the inverse, i.e.,
$$g\cdot v = vg^{-1}$$
for any $g\in \ensuremath{\mathcal{G}}, v\in G$.

For an element $(s_1, \cdots, s_m)\in S^m$, let $\theta:G\to G$ denote the bijection defined by
$$\theta(v)
=
\prod_{i=1}^m
\sigma^i (s^{(-1)^{i-1}}_{m+1-i})
\sigma^m(v^{(-1)^m}) $$
for $v\in G$.
For each $(s_1, \cdots, s_m)\in S^m$ and $v\in G$, note that
\begin{align*}
\theta(g\cdot v)
& =
\prod_{i=1}^m
\sigma^i (s^{(-1)^{i-1}}_{m+1-i})
\sigma^m((g\cdot v) ^{(-1)^m})\\
& =
\prod_{i=1}^m
\sigma^i (s^{(-1)^{i-1}}_{m+1-i})
\sigma^m(v^{(-1)^m})
(\sigma^m(v^{(-1)^m}))^{-1}
\sigma^m((g\cdot v) ^{(-1)^m})\\
& =
\theta(v)
(\sigma^m(v^{(-1)^m}))^{-1}
\sigma^m((g\cdot v) ^{(-1)^m})\\
& =
\theta(v)
\sigma^m(g^{(-1)^{m-1}}) \\
& =
\sigma^m(g^{(-1)^m})
\cdot \theta(v) \\
& =
\psi_{(s_1, \cdots, s_m)}(g) \cdot \theta(v)
\end{align*}
for any $g\in \ensuremath{\mathcal{G}}$, where $\psi_{(s_1, \cdots, s_m), v}$ denotes the automorphism
$$g\mapsto
\sigma^m(g^{(-1)^m})
$$
of the group $\ensuremath{\mathcal{G}}$.

For any integer $m\geq 1$ and any subset $A$ of $G$, note that
\begin{align*}
\ensuremath{\mathscr{N}}_{\Sigma}^{2m}(A)
& = \sigma(S) \sigma^2(S^{-1}) \sigma^3(S) \sigma^4(S^{-1}) \cdots \sigma^{2m-1} (S) \sigma^{2m} (S^{-1})\sigma^{2m}(A) ,
\end{align*}
which yields
\begin{align*}
\ensuremath{\mathscr{N}}_{\Sigma}^{2m}(g\cdot A)
& = \sigma(S) \sigma^2(S^{-1}) \sigma^3(S) \sigma^4(S^{-1}) \cdots \sigma^{2m-1} (S) \sigma^{2m} (S^{-1})\sigma^{2m}(g\cdot A) \\
& = \sigma(S) \sigma^2(S^{-1}) \sigma^3(S) \sigma^4(S^{-1}) \cdots \sigma^{2m-1} (S) \sigma^{2m} (S^{-1})\sigma^{2m}(Ag^{-1}) \\
& = \sigma^{2m}(g) \cdot (\sigma(S) \sigma^2(S^{-1}) \sigma^3(S) \sigma^4(S^{-1}) \cdots \sigma^{2m-1} (S) \sigma^{2m} (S^{-1})\sigma^{2m}(A)) \\
& = \sigma^{2m}(g) \cdot \ensuremath{\mathscr{N}}_{\Sigma}^{2m}(A) .
\end{align*}

Since $C_\Sigma(G, S)^\sigma$ is connected, its vertex Cheeger constant $h_{\Sigma, \sigma}$ is positive. Thus $C_\Sigma(G, S)^\sigma$ is an $h_{\Sigma, \sigma}$-expander with $h_{\Sigma, \sigma}>0$.
If $\sigma^2$ is the trivial automorphism of $G$, then from Theorem \ref{thmPrincipal}, the nontrivial eigenvalues of the normalised adjacency operator of $C_\Sigma(G, S)^\sigma$ are greater than
$$
-1 + \frac {h_{\Sigma, \sigma}^4}{2^{12} d^{8}}
.
$$
If $\sigma^{2k}$ is the trivial automorphism of $G$ for some odd integer $k\geq 1$, then from Theorem \ref{thmPrincipal}, the nontrivial eigenvalues of the normalised adjacency operator of $C_\Sigma(G, S)^\sigma$ are greater than
$$
\left(
-1 + \frac {1}{2^{12} d^{8k}}
\left(
\frac 12
\left(
1 -
\left(
1 - \frac{h_{\Sigma, \sigma}^2}{2d^2}
\right)^k
\right)
\right)
^4
\right)^{1/k}.
$$
By the discrete Cheeger--Buser inequality (Proposition \ref{Prop:chin}), the result follows.
\end{proof}

\section{Spectral expansion of Schreier graphs}
\label{Sec:Schreier}

Given a subgroup $H$ of a finite group $G$, and a symmetric subset $S$ of $G$ with $|S| = d$, the Schreier graph $\mathrm{Sch} (G, H, S)$ has the set $H\backslash G$ of right cosets of $H$ in $G$ as its set of vertices and there is an edge from $Hg$ to $Hg'$ for $g, g'\in G$, if $Hg' = Hgs$ for some $s\in S$.

\begin{theorem}
\label{Thm:Schreier}
Suppose no index two subgroup of $G$ acts transitively on the vertex set of $\mathrm{Sch} (G, H, S)$. Assume that the index of $H$ in $G$ is at least $4$, and $S\cdot S$ contains its conjugates by the elements of $G$. If the Schreier graph $\mathrm{Sch} (G, H, S)$ is connected, then the nontrivial spectrum of its normalised adjacency operator lies in the interval
$$\left( -1 + \frac{h_{\mathrm{Sch}}^4}{2^{12}d^8}
,
1 - \frac{h_{\mathrm{Sch}}^2}{2d^2}
\right]$$
where $h_{\mathrm{Sch}}$ denotes the vertex Cheeger constant of $\mathrm{Sch} (G, H, S)$.
\end{theorem}

\begin{proof}
Let $\ensuremath{\mathcal{G}}$ denote the group $G$. Consider the action of $\ensuremath{\mathcal{G}}$ on the set of vertices $H\backslash G$ of $\mathrm{Sch} (G, H, S)$ by right multiplication via the inverse, i.e.,
$$\tau \cdot Hg
= Hg \tau^{-1}
$$
for any $\tau \in \ensuremath{\mathcal{G}}$ and any right coset $Hg$ of $H$ in $G$.
Note that this action is transitive of order $|H|$.

For an element $s\in S$, let $\theta:
H\backslash G
\to
H\backslash G$
denote the bijection defined by
$$\theta(Hg)
= Hgs.$$
For any $Hg\in H\backslash G$, note that
\begin{align*}
\theta(\tau \cdot Hg)
& = Hg\tau^{-1} s\\
& = Hgs (s^{-1} \tau s)^{-1} \\
& = (s^{-1} \tau s) \cdot Hgs\\
& = \psi_s(\tau) \cdot Hgs\\
\end{align*}
for any $\tau \in \ensuremath{\mathcal{G}}$, where $\psi_s$ denote the automorphism
$$\tau \mapsto s^{-1} \tau s$$
of the group $\ensuremath{\mathcal{G}}$.

For any element $Hg\in H\backslash G$, note that
\begin{align*}
\ensuremath{\mathscr{N}}_S(\ensuremath{\mathscr{N}}_S(\tau\cdot Hg))
& = \{Hg\tau^{-1} x\,|\, x\in S\cdot S\} \\
& = \{Hg\tau^{-1} x\,|\, x\in \tau S\cdot S \tau^{-1} \} \\
& = \{Hgx\tau^{-1} \,|\, x\in S\cdot S\} \\
& = \{\tau \cdot Hgx \,|\, x\in S\cdot S\} \\
& = \tau \cdot \ensuremath{\mathscr{N}}_S(\ensuremath{\mathscr{N}}_S(Hg))
\end{align*}
holds for any $\tau \in \ensuremath{\mathcal{G}}$.

If the Schreier graph $\mathrm{Sch} (G, H, S)$ is an $\varepsilon$-vertex expander for some $\varepsilon>0$, then by Theorem \ref{thmPrincipalk1}, the nontrivial eigenvalues of its normalised adjacency operator are greater than
$$ - 1 + \frac{\varepsilon^4} {2^{12} d^8}.$$
By the discrete Cheeger--Buser inequality (Proposition \ref{Prop:chin}), the result follows.
\end{proof}

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  \setbox7\hbox{\accent"5E#1}\penalty 10000\relax\fi\raise 1\ht7
  \hbox{\lower1.15ex\hbox to 1\wd7{\hss\accent"7E\hss}}\penalty 10000
  \hskip-1\wd7\penalty 10000\box7}
  \def\polhk#1{\setbox0=\hbox{#1}{\ooalign{\hidewidth
  \lower1.5ex\hbox{`}\hidewidth\crcr\unhbox0}}}
\providecommand{\bysame}{\leavevmode\hbox to3em{\hrulefill}\thinspace}
\providecommand{\MR}{\relax\ifhmode\unskip\space\fi MR }
\providecommand{\MRhref}[2]{%
  \href{http://www.ams.org/mathscinet-getitem?mr=#1}{#2}
}
\providecommand{\href}[2]{#2}

redpajama_set_name: RedPajamaArXiv


Six-time European Youth Champion William Svensson averaged 266 for three games to set the tone in the second round of the finals in the Norwegian Open 2018 by Brunswick.
Svensson (right, and featured photo), who won two EYC titles each in 2016, 2017 and 2018, fired games of 242, 277 and 279 for 798 to beat out his compatriot Björn Lindqvist for the top spot by 25 pins. Lindqvist (left), who survived the first round cut in eighth place, qualified for the third round in second place with 773, including 277 and back-to-back 248 games.
Norwegian champion Jonas Dammen was third with 728 and was followed by World and European Youth champion Robert Lindberg in fourth place with 715 and 2013 Norwegian Open winner James Blomgren in fifth place with 711.
Norwegians Audun Boye Hansen and Tor Inge Jansen (left) made the fourth cut in sixth and seventh place with 695 and 691, respectively.
Former EYC champion Alida Molander (right) took the eighth and last spot to advance with 662, including eight bonus pins per game. Molander overtook defending champion Raymond Jansson (belwo left) of Sweden in the last game by nine pins, who finished in ninth place with 653 and was eliminated.
The top 8 will meet qualifiers 9 to 16 in the third round.
Top 8 advance to Final Round 3; places 9-16 eliminated, finish in 25th to 32nd place.


text:
About Medicinoxy
Associate / Full Professor in Rheumatology
University of Washington Division of Rheumatology
Associate or Full Professor WOT- Rheumatology (Clinical Director)
Organization: Medicine, Rheumatology
Title: Associate or Full Professor WOT- Rheumatology (Clinical Director)
The Division of Rheumatology, Department of Medicine at the University of Washington, School of Medicine is recruiting for a full-time faculty member at the Associate or full Professor level (without tenure due to funding, annual service period of 12 months [July 1-June 30], commensurate with qualifications) to serve as Clinical Director of the Division, overseeing clinical practice across all sites: Harborview Medical Center, University of Washington Medical Center, VA of Puget Sound, Northwest Hospital, and Eastside Specialty Center. Duties of the position will include experienced and effective leadership with a strong commitment to clinical operation efficiency, focusing on faculty productivity, access to care, and patient and physician satisfaction. This position will work with section heads, key clinical faculty, division revenue cycle manager, and clinic managers to achieve these goals. Additionally, this position will have clinical responsibilities across sites. University of Washington faculty engage in teaching, research, and service. This position has an anticipated start date of February 1, 2021.
Successful candidates must have an M.D. degree (or foreign equivalent) with board-certification in rheumatology (or foreign equivalent). In order to be eligible for University sponsorship for an H-1B visa, graduates of foreign (non-U.S.) medical schools must show successful completion of all three steps of the U.S. Medical Licensing Exam (USMLE), or equivalent as determined by the Secretary of Health and Human Services.
Positions will remain open until filled. Applicants should submit their curriculum vitae, a career statement and three letters of reference. For questions, please contact LyKheng Huot at lykheng@uw.edu.
University of Washington is an affirmative action and equal opportunity employer. All
qualified applicants will receive consideration for employment without regard to race, color, creed,
religion, national origin, sex, sexual orientation, marital status, pregnancy, genetic information,
gender identity or expression, age, disability, or protected veteran status.
The University of Washington is using Interfolio's Faculty Search to conduct this search. Applicants to this position receive a free Dossier account and can send all application materials, including confidential letters of recommendation, free of charge.
For help signing up, accessing your account, or submitting your application, please check out Interfolio's help and support section or get in touch via email at help@interfolio.com or phone at (877)997-8807.
If you apply for this position please say you saw it on Medicinoxy

redpajama_set_name: RedPajamaCommonCrawl


text:
package xszymo.controllers;

import org.springframework.stereotype.Controller;
import org.springframework.ui.Model;
import org.springframework.web.bind.annotation.GetMapping;
import org.springframework.web.bind.annotation.PostMapping;
import org.springframework.web.bind.annotation.RequestParam;
import xszymo.controllers.providers.Algorithm;
import xszymo.controllers.providers.CountersProvider;
import xszymo.rest.objects.ChampionInfo;
import xszymo.services.RestApiStatic;

import javax.servlet.http.HttpServletRequest;
import java.io.IOException;
import java.util.Collection;
import java.util.LinkedList;

@Controller
public class Counters {
	@GetMapping("counters")
	public static String counter(@RequestParam("allyChampion1") String allyChampion1,
	                             @RequestParam("allyChampion2") String allyChampion2, @RequestParam("allyChampion3") String allyChampion3,
	                             @RequestParam("allyChampion4") String allyChampion4, @RequestParam("allyChampion5") String allyChampion5,
	                             @RequestParam("enemyChampion1") String enemyChampion1,
	                             @RequestParam("enemyChampion2") String enemyChampion2,
	                             @RequestParam("enemyChampion3") String enemyChampion3,
	                             @RequestParam("enemyChampion4") String enemyChampion4,
	                             @RequestParam("enemyChampion5") String enemyChampion5, HttpServletRequest request, Model model) {

		Collection<ChampionInfo> ally = new LinkedList<>();
		Collection<ChampionInfo> enemy = new LinkedList<>();

		ChampionInfo pickedChampion = RestApiStatic.getOneChampion(request.getParameter("championRadio"));

		ally = RestApiStatic.getAllTeam(allyChampion1, allyChampion2, allyChampion3, allyChampion4, allyChampion5);
		enemy = RestApiStatic.getAllTeam(enemyChampion1, enemyChampion2, enemyChampion3, enemyChampion4, enemyChampion5);

		Collection<ChampionInfo> champions5 = Algorithm.algorithm(ally, enemy, pickedChampion, request.getParameter("line"));
		LinkedList<ChampionInfo> champions = new LinkedList<ChampionInfo>();
		champions.addAll(champions5);

		Collection<ChampionInfo> champions1 = new LinkedList<ChampionInfo>();
		Collection<ChampionInfo> champions2 = new LinkedList<ChampionInfo>();
		Collection<ChampionInfo> champions3 = new LinkedList<ChampionInfo>();
		Collection<ChampionInfo> champions4 = new LinkedList<ChampionInfo>();
		int a = champions.size() / 3;

		for (int i = 0; i < champions.size(); i++)
			if (i < a)
				champions1.add(champions.get(i));
			else if (i < a + a)
				champions2.add(champions.get(i));
			else if (i < a + a + a)
				champions3.add(champions.get(i));
			else
				champions4.add(champions.get(i));

		model.addAttribute("champions1", champions1);
		model.addAttribute("champions2", champions2);
		model.addAttribute("champions3", champions3);
		model.addAttribute("champions4", champions4);
		return "teamCounters";
	}

	@PostMapping("countersForOneChampion")
	public static String counter(@RequestParam("enemyName") String enemyName, HttpServletRequest request, Model model)
			throws IOException {
		model.addAttribute("champions", CountersProvider.getCounters(enemyName));
		model.addAttribute("champion", enemyName);
		model.addAttribute("tips", RestApiStatic.getEnemyTips(enemyName));
		return "oneChampionCounters";
	}
}

redpajama_set_name: RedPajamaGithub


text:
Amphixystis commatias är en fjärilsart som beskrevs av Edward Meyrick 1915. Amphixystis commatias ingår i släktet Amphixystis och familjen äkta malar.

Artens utbredningsområde är Sri Lanka. Inga underarter finns listade i Catalogue of Life.

Källor

Äkta malar
commatias

redpajama_set_name: RedPajamaWikipedia


Like it or loath it, the festive season is fast approaching, and before you know it Noddy Holder will start screaming "It's Chrrrrriiiiisstmmmaassss!" into your ear. So this is your starter for ten to ensure you pop some fizz into your party or celebration.
Rest assured it's the ultimate sound effect and a great way to indicate the good times are rolling. With more choices available than seats at a theatre, whatever you go for will certainly add a touch of sparkle. A quick twist of the bottle whilst holding the cork is all that's needed, who said it was labour intensive? However, there are those out there still trying to open sparkling wine with a corkscrew, an interesting struggle with a bang at the end.
Many of us answer this SOS (Sort Out Sparkles) by grabbing a bottle of reliable Prosecco, We adore its fresh and fruity charm thanks to what is known as the 'tank method', nothing to do with bottles being fired in battle I must add! If you do opt for this choice, make a big impression on your guests and splash out on a magnum for a great table centrepiece.
Whilst regular blasts of Champagne can sometimes have an effect on your wallet, cracking Cava is a great alternative to both this popular French tipple and Prosecco, if you want a fizz to dance the flamenco beat in your mouth. A wonderful blend of indigenous Spanish grape varieties it's brilliant bubbly for your buck.
After something with a catchy name? A glass of Bird in Hand Sparkling Pinot Noir with its delightfully delicate flavours of strawberries and cream is certainly one to savour, whilst keep your eyes peeled for Cremant, which literally means 'creamy' It's a sublime alternative to Champagne at a snip of the price but produced outside the Champagne region in France. It's also produced using the same 'traditional method' production technique with slightly less finesse.
Want to raise a glass to Rule, Britannia? It's no secret that I love English fizz. Pour a glass of this electrifying homemade delight and treat your loved one to a tour at your nearest vineyard for that perfect festive gift. There are over 500 vineyards in the UK and you can even pop down to your local B&Q store and buy English Sparkling Wine as a paint colour!! Approx 3.86 million bottles were produced in the UK last year, and the recent heatwave has certainly helped to boost this year's harvest.
Whether you're after a Champagne Supernova or help with the pick of the pops, jump on the bubbly bandwagon with a joyful bang and don't spare the horses!
Adored by hardnosed wine judges and customers alike, this Prosecco is the genuine article bursting with citrus-rich character and bundles of charm.
This is their flagship wine, a fresh and fruity fizz, perfect for all celebrations, with English hedgerow scents and a touch of honey on the palate.
La Folie translates as 'madness' but it doesn't mean you have to Suggs blaring away in the background or wear baggy trousers whilst enjoying this sublime tipple. Hailing from Luberon in central Provence, Mirabeau La Folie Rosé is the perfect aperitif and pairs beautifully with salads, seafood tapas and summer fruit desserts.
Fine, elegant and complex combining the freshness of Chenin Blanc, Chardonnay floral tones and a little red fruit from Cabernet Franc. Drink with olives & nuts as a classy aperitif.
Quite literally Pinot Noir on steroids! This light and refreshing fizz is salmon pink in colour with a lively palate that evokes flavours of fresh strawberry. A delicate bead leads to a crisp, clean finish. The perfect pre-dinner apéritif or celebratory drink.


text:
Donkey Kong Oddities
A 30th anniversary celebration of the arcade game the world went ape over.
Posted by Benj Edwards | Sunday, July 10, 2011
The Popeye Connection
Few know this, but Mario, Donkey Kong, and Pauline (the girl you rescue) were originally Popeye, Bluto, and Olive Oyl. In the early 1980s, Nintendo manufactured licensed Popeye products like playing cards and a Game & Watch unit, so Donkey Kong designer Shigeru Miyamoto asked to work on a Popeye video game.
At some point during development, the Popeye license fell through and Miyamoto had to come up with unique characters that would fit the same theme: villain abducts damsel, hero rescues damsel while dodging stuff hurled at him. Sounds like every Popeye cartoon, doesn't it? Donkey Kong was born. (Nintendo did end up releasing a completely different Popeye game in 1982.)
Slides: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
35 Comments For This Post
Arnum Says:
Oh Man, this is my favorite game, at least it was in 1982. At the age of twelve I would go to our local arcade (a gas station) and play Crazy Kong on a table top for 20 cents a game. I lived in trhe country, so we were behind the times, as The first games (also tabletop) were asteroid and space invaders. Next to the Crazy Kong table we had Frogger and Jungle Boy. Crazy Kong is the game I sunk most of mu cash into. I held the town record for a while too. How did I not get bored with that game?
ezhanebrand Says:
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http://www.classiccardybootsshop.com Hey, the young man anxious and said: "We made an appointment to run away!"
The artwork on the "Donkey Kong on Acid" actually looks like "Mr. Do's Castle" to me, the juxtaposition with the DK name makes a certain sense if you played them both!
Floor Tiles Sydney Says:
I always loved Donkey Kong. FAv game
Forrest Says:
The Screenshot of the IBM PC port on slide 13 looks very similar to the version that I grew up playing, I don't recall what type of computer it was, just that we were running Window's 3.0 and that I had to work very hard and be on my best behavior to earn time playing the game. Ironically, now that I am married I have to do the same exact thing to play my current video games.
Guppy06 Says:
Did the Intellivision box artist go on to produce the art for Mega Man?
This is why I will never give up my old stand-up Donky Kong that's in the basement now. My dad got it at an auction back in '87, then after floating around the family for years, I snatched it up at the first opportunity!
Wife hates it! But I'm taking it to the grave!
Dan.
May I recommend this excellent documentary on competitive Donkey Kong players:
http://en.wikipedia.org/wiki/The_King_of_Kong:_A_…
You couldn't make this stuff up.
The "troll" look of Donkey Kong on the PC is mostly due to the weird palette limitations of CGA graphics which were the only (or at least most common) color video adapters at the time.
I was technican with a company which distributed DK machines. Had a service call on a Sunday afternoon to a site where "couldn't insert coins". Arrived to find there was so many coins in the bin that thay had overflowed and then stacked inside the machine right up to the bottom of the coin acceptor. Those machines MINTED cash.
On a tech POV – the DK Motherboard was the first one I ever saw equipped with ECL (emitter coupled logic) RAM. This stuff, which was used in Cray supercomputers, drew so much current and hence ran really hot that it was mounted on a stand-off daughter board. I guess that the speed the barrel sprites moved required it?
And The Says:
No mention of the "Pac-Kong" game for the Atari 2600?!
@Apogalacticon Says:
October 2nd, 2011 at 7:03 pm
Oh man, who doesn't remember the old donkey kong and mario bros games? They will never again make games as simple and adictive as back then.
Starcraft 2 Tips
@Pseudofossils Says:
I was just explaining to my young kids how I played this growing up. The kept asking me why I only got to play as Mario when the game was called Donkey Kong. It reminded me of Patrick Stewart explaining that he thought "Star Trek" was the name of the ship when he auditioned for ST:TNG.
kvalitets flyverdragt
What a great read, I'm really glad that I stumbled upon this.
Minneapolis chiropractor
AlenCoder Says:
October 31st, 2011 at 4:13 am
Thanks for sharing this powerful article with us. I've been doing some research for my own link building service business, and I found your article by accident.
Oh wow, I remember this game. I was so hooked! 😀
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jones123peter Says:
I am very impressed. I do have a couple questions for you personally however. Do you think you're thinking about doing a follow-up posting about this? Will you be going to keep bringing up-to-date
Carnosine
this game is a 10/10 , if it had been called "mario country returns" it would have recieved a 10. This is possibly the best game in the wii and yes i have put that above galaxy and bros. i pulled out my old snes and played dkc and 2 and 3 next to this , this new dkc is the most faithful remake with added modernism i have ever seen! astroloji
this is my favorite game that i want to play.I'll be play this game soon .thanks for giving us these kind of information.thanks for it.
Deendid Says:
I don't recall what type of computer it was, just that we were running Window's 3.0 and that I had to work very hard and be on my best behavior to earn time playing the game. Ironically, now that I am married I have to do the same exact thing to play my current video games.
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Thirty years ago this month, Nintendo released Donkey Kong to arcades across the United States. The games American version went on to sell tens of thousands of units, saving the then-struggling US branch of the company and paving the way for Nintendos future success on Western shores.
Prix des Timbres Says:
Prix des Timbres
Cheers on a great post, I've bookmarked and I look forward to future postings.
TomPeris Says:
Oh man, who doesn't remember the old donkey kong and mario bros games? They will never again make games as simple and adictive as back then. cheap textbooks
Wedding Photography Says:
I can't believe this is 30 years old!!! I played this as a kid in San Diego California. I was at a wedding last year and the bride and groom had this as an arcade game at there reception. Good things are always good I guess.
Mike R Says:
November 21st, 2011 at 11:31 am
What cool little feature. I also have in my possession some Donkey Kong oddities. I have a series of animation cells of Donkey Kong and Mario from a little short that The Tennessee Valley Authority produced. I'd be happy to try to dig those up and provide scans if you are interested.
November 21st, 2011 at 11:58 pm
The game is definitely difficult, that's not in question. There is cheapness and some unpredictability that's maddening, but the worst and most *unacceptable* problem of the game is that when you die, you lose TWO lives instead of one. So not fair.gold coins
juhi123 Says:
November 23rd, 2011 at 11:26 pm
What a write!! Very informative also easy to understand. Looking for more such posts!! Do you have a my space? I recommended it on digg. The only thing that it's missing is a bit of new design. Anyway thank you for this blog. Cariszone
Roger Randor Says:
A classic that we never forget! Badly that it starts again after you have rescued the princess … But it doesn't bother me so much … billig seo
Brandon Na Says:
Another interesting titbit of trivia is that Mario made one of his first appearances in a game as a villain in Donkey Kong. However, he went on to become the world famous mascot, more famous than the game's protagonist itself. Nevertheless, the cultural impact of Donkey kong is undeniable.
jones Says:
I'm kinda curious that in write-ups like this, no one ever brings up the unusual idea of a "donkey" versus a "monkey". (Maybe it's on Wikipedia, I haven't looked) but as I was told, the title of the game got lost in translation between the Japanese developers and the Americans. Change one letter in the title and it goes from Donkey Kong to Monkey Kong. Seeing as "King Kong" was a notable monster at that time a la Godzilla, Mothra, and all the others, the idea made sense since King Kong scaled buildings and took the woman (Fay Wray)at the top of the building. So the game was initially meant to be titled "Monkey Kong".
At least as I was told. And to me, that makes a little more sense than the Popeye reference. Although I see how the two ideas could've been combined.
Rezart Taci
Memories! Donkey Kong, Mario. WOW, this bring me back in childhood. Thanks!
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Cat insurance Says:
It took me just under 41 hours to get 101% of everything in the game and beat the final boss. I encountered a strange glitch before I actually got everything. I went to the large Banana Fairy after photographing all the smaller ones in the game. She gave me the 201st banana (the one with the Rareware sticker), and my percentage total showed "101%"…but the thing is, I was still missing a banana from the Crystal Caves level at the time! My rating should have only been 100%. When I finally got that missing banana, my total stayed at 101%.
Office design Says:
Nintendo of America was struggling in 1980 and 1981. As the video game craze swept the country, none of Nintendo's game titles managed to make an impact in the booming industry. Space Firebird did modest business, but was licensed stateside to Sega/Gremlin, leaving Nintendo-labelled titles such as Sheriff and Heli-Fire to flounder.
interpoker review Says:
What a great post, I am now 29 years-old but can recall this nice video game which I played quite a few years back now! Nintendo even released it on Super Nintendo at some point, it was a superb game with better graphics than the very first one obviously! It brings back very good memories to me, thanks! interpoker review
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text:
← Bernie Ecclestone: "You know I don't like democracies…"
F1 bosses can learn from Formula E, says Prost →
May 14, 2015 · 7:02 am
Strategy Group set to make call on fifth engine
The ongoing debate over teams being allowed to use a fifth engine in 2015 without a grid penalty will be one of the highest profile elements of Thursday's F1 Strategy group meeting.
Last year the rules allowed teams to use five engines without penalty, but the figure went down to four this year as it was deemed that the manufacturers would be on top of reliability. However Renault has had a worse time than last season, while Honda is of course now in its own debut year.
The fifth engine originally emerged in the context of ensuring that teams do not start to sit in the garage on Fridays as the season goes on in order to save mileage, which is why Bernie Ecclestone gave his support.
However Renault and Honda's ongoing problems have meant that extra engines and grid penalties are now inevitable for RBR, STR and McLaren, and there's been a push for the fifth engine to simply be added to the pool, rather than reserved for Friday use.
The consensus is that its supporters will struggle to get the idea voted through. Mercedes will only support the fifth engine if it's strictly related to Fridays. Meanwhile engine customers – such as Force India – are adamant that they don't want to pay for the extra engine.
"We're not against it," Force India's Bob Fearnley told this writer. "But we are absolutely against paying extra money for it, so on that principle the answer would be we wouldn't vote for it.
"We've lost an engine with Nico [Hulkenberg], and we lost it very early in its life, so effectively we're running a three-engine programme for one of our drivers. So the question is can we somehow or other eke it out, assuming we don't lose another engine? If we're going to do that obviously we're doing it to try save money. We'd vote against the principle of it costing us more, that's for sure.
"The thing is we don't know what we're voting on, because the original concept for this fifth engine was based around Fridays, so until we know how it's addressing the Friday issue it's very difficult to make a decision. For me it's what are they trying to achieve with it? The original conversations were about making sure Fridays are still fully supported."
McLaren and Honda are obviously keen for the change to go ahead, although team principal Eric Boullier admitted that he's not hopeful.
"As far as McLaren is concerned obviously we would be happy to have a fifth engine," said the Frenchman. "I think having [us as] a new engine manufacturer in F1 it would be fair to have the same conditions that the other ones had when they were running last year. So yes, we are in favour of a fifth engine. But I'm not sure it will be that easy."
Meanwhile Ecclestone has admitted to this writer that the cost issue was the major problem: "We agreed, or all the teams agreed, to have five engines during the year. We've got four in the regulations, and now people are saying maybe it should stay four. People who supply the engines don't want to supply more unless they get more money, and the teams can't afford it."
One response to "Strategy Group set to make call on fifth engine"
Rafael CE
Bob F has a point, funny that you had to pay extra for a fifth engine because first or second did not last for at least five race weekends as required by regulations.


beautiful gift - your child - the visible sign of your love for each other. You have played an important part in God's plan of creation, and with Him have given life to a new member of His family. What more beautiful way can we welcome your child into God's family than through the sacrament of baptism.
Please call the Rectory for an appointment one-month prior baptism. This sacrament is administered usually on Sunday after 10:00 Mass (in English) and on Sunday after 11:30AM Mass (in Polish).
Every person being baptized must have a godparent. You must have at least one sponsor, but usually infants get one of each gender. Being a godparent really means actively giving good Christian witness and example and being a role model and support by regularly and faithfully practicing their Catholic faith.
Godparents must be practicing Catholics, 16 years of age and Confirmed. People who are divorced may be sponsors. People who are divorced and are remarried cannot be sponsors unless they received an annulment from the Church from their first marriage and were married in a Catholic Church.


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West Virginia is known for its breathtaking mountains and bountiful hills, earning the state the nickname "The Mountain State". Among these mountains includes the Allegheny Mountains, comprising over 10% of the state's land. Due to the scenery and historic monuments within the state, the tourism industry in West Virginia deeply impacts its economy.
Individuals can find opportunity for employment in the tourism and hospitality industry in this beautiful state. Search for training in West Virginia ranging from customer service and event support to financial and administration courses. Gain qualifications to enter and grow in the hospitality industry.
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text:
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Megan Sweeney
Associate and Affiliated Faculty
Arthur F Thurnau Professor; Associate Professor of English Language and Literature; Director Undergraduate Studies, English Language and Literature; Associate Professor of Afroamerican and African Studies; Associate Professor of Women's and Gender Studies; Associate Chair of the Department of Women's and Gender Studies
meganls@umich.edu
Office Information:
hours: Zoom office hours: Tuesdays, 12:00-1:30; by appointment
Afroamerican and African Studies
B.A., Northwestern University, 1989; Ph.D., Duke University, 2002
Megan Sweeney is Arthur F. Thurnau Associate Professor of English, Afroamerican and African Studies, and Women's and Gender Studies at the University of Michigan, Ann Arbor. Her publications include an award-winning monograph, Reading Is My Window: Books and the Art of Reading in Women's Prisons (2010); an edited collection, The Story Within Us: Women Prisoners Reflect on Reading (2012); a collection of lyric essays called Mendings, under contract with Duke University Press; numerous articles about African American literature, reading, and incarceration; and lyric essays published in Brevity, Entropy Magazine, The Normal School, and Bennington Review. Sweeney's writing has been supported by fellowships from the Radcliffe Institute for Advanced Study at Harvard University, the Institute for the Humanities at the University of Michigan, the Ford Foundation, and the Woodrow Wilson Foundation. Deeply committed to teaching, Sweeney has received the John D'Arms Award for Distinguished Graduate Mentoring in the Humanities (2021), a Class of 1923 Memorial Teaching Award (2010), and an Arthur F. Thurnau Professorship (2014), the university's highest award for undergraduate teaching. She currently serves as Director of Undergraduate Studies in the Department of English Language & Literature, as Associate Chair and Director of Graduate Studies in Women's and Gender Studies, and as Director of the Hopwood Program.
Ph.D., Duke University, Literature, 2002
Department of Afroamerican and African Studies
Department of Women's and Gender Studies
Department of American Culture
Department of Afroamerican and African Studies (DAAS)
Haven Hall, Room 4700
daas-info@umich.edu