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\title{\bf Mutually unbiased Bush-type Hadamard matrices and association schemes}

% input author, affilliation, address and support information as follows;
% the address should include the country, and does not have to include
% the street address 
\author{Hadi Kharaghani\thanks{Supported by an NSERC Discovery Grant}\qquad Sara Sasani\\
\small Department of Mathematics and Computer Science\\[-0.8ex]
\small University of Lethbridge\\[-0.8ex] 
\small Lethbridge, Alberta, T1K 3M4, Canada\\
\small\tt \{kharaghani,sasani\}@uleth.ca\\
%\and
%Sara Sasani\\
%\small Department of Mathematics and Computer Science\\[-0.8ex]
%\small University of Lethbridge\\[-0.8ex] 
%\small Lethbridge, Alberta, T1K 3M4, Canada\\
%\small\tt sasani@uleth.ca\\
\and
Sho Suda\thanks{Supported by JSPS KAKENHI Grant Number 15K21075}\\
\small Department of Mathematics Education\\[-0.8ex]
\small Aichi University of Education\\[-0.8ex] 
\small Kariya 448-8542, Japan\\
\small\tt suda@auecc.aichi-edu.ac.jp\\
}

% \date{\dateline{submission date}{acceptance date}\\
% \small Mathematics Subject Classifications: comma separated list of
% MSC codes available from http://www.ams.org/mathscinet/freeTools.html}

\date{\dateline{Dec **, 2014}{}\\
\small Mathematics Subject Classifications: 05E20, 05E30, 05B30}

\begin{document}

\maketitle

\begin{abstract}


It was shown by LeCompte, Martin, and Owens in 2010 that the existence of mutually unbiased Hadamard matrices and the identity matrix, which coincide with mutually unbiased bases, is equivalent to that of a $Q$-polynomial association scheme of class four which is both $Q$-antipodal and $Q$-bipartite. 
We prove that the existence of a set of mutually unbiased Bush-type Hadamard matrices is equivalent to that of an association scheme of class five.
As an application of this equivalence, we obtain an upper bound of the number of mutually unbiased Bush-type Hadamard matrices of order $4n^2$ to be $2n-1$. This is in contrast to the fact that the best general upper bound for the mutually unbiased Hadamard matrices of order $4n^2$ is $2n^2$. 
We also discuss a relation of our scheme to some fusion schemes which are $Q$-antipodal and $Q$-bipartite $Q$-polynomial of class $4$.  


\end{abstract}

\section{Introduction}
A {\it  Hadamard} matrix is a matrix $H$ of order $n$ with
entries in $\{-1,1\}$ and orthogonal rows in the usual
inner product on ${\mathbb{R}}^n$.
Two Hadamard matrices $H$ and $K$ of order $n$ are called {\it unbiased}
if $HK^{t}={\sqrt n} L$ for some Hadamard matrix $L$,
where $K^t$ denotes the transpose of $K$.
In this case, it follows that $n$ must be a perfect square.
A Hadamard matrix of order $n$ for which the row sums and column sums
are all the same, necessarily $\sqrt n$, is called {\it regular},
see \cite{ys}.
\begin{definition}
A \emph{Bush-type} Hadamard matrix
is a  block Hadamard matrix $H=[H_{ij}]$ of
order $4n^2$ with block size $2n$,
$H_{ii}=J_{2n}$
and $H_{ij}J_{2n}
= J_{2n}H_{ij}=0$, $i\ne j$, $1\le i \le 2n$, $1\le j\le 2n$
where $J_{\ell}$ is the $\ell \times \ell$ matrix of ones.
\end{definition}
It is known that for odd values of $n$ there is no pair of  unbiased Bush-type Hadamard matrices of order $4n^2$ \cite{bk}.
In contrast, 
for even values of $n$, there are unbiased Bush-type Hadamard matrices of order $4n^2$ \cite{hko}. 
(Note that a missing necessary assumption is needed in the proof of  \cite[Theorem 13]{hko}. The modified version will be presented in Section~\ref{5classas}.) 
One very important property
of unbiased Bush-type Hadamard matrices,  as is shown in section 3, is the fact that for any two unbiased pair of Bush-type Hadamard matrices $H$ and $K$ of order $4n^2$, $\frac{1}{2n}HK^{t}$ is also a Bush-type Hadamard matrix. 

It was shown by LeCompte, Martin, and Owens in \cite{lmo} that the existence of mutually unbiased Hadamard matrices and the identity matrix, which coincide with mutually unbiased bases, is equivalent to that of a $Q$-polynomial association scheme of class four which is both $Q$-antipodal and $Q$-bipartite. 

Our aim in this
paper is to show that the existence of unbiased Bush-type Hadamard matrices is equivalent to the existence
of a certain association scheme of class five. 
As an application of this equivalence, we obtain an upper bound for the number of mutually unbiased Bush-type Hadamard matrices of order $4n^2$ to be $2n-1$, whereas the best general upper bound for the mutually unbiased Hadamard matrices of order $4n^2$ is $2n^2$ \cite[Theorem 2]{hko}. 
Also we discuss a relation of our scheme to some association schemes of class four. 

\section{Association schemes}
A \emph{symmetric $d$-class association scheme}, see \cite{bi}, 
with vertex set $X$ of size $n$ and $d$ classes
is a set of nonzero symmetric $(0,1)$-matrices $A_0, \ldots, A_d$ with
rows and columns indexed by $X$, such that:
\begin{enumerate}
\item $A_0=I_n$, the identity matrix of order $n$.
\item $\sum_{i=0}^d A_i = J_n$.
\item For all $i$, $j$, $A_iA_j=\sum_{k=0}^d p_{ij}^k A_k$
for some $p_{ij}^k$.
\end{enumerate}
It follows from property (3) that the matrices $A_i$ necessarily commute.
The vector space spanned by the matrices $A_i$ forms a commutative algebra, denoted by $\mathcal{A}$ and called the \emph{Bose-Mesner algebra} or \emph{adjacency algebra}.
There exists a basis of $\mathcal{A}$ consisting of primitive idempotents, say $E_0=(1/n)J_n,E_1,\ldots,E_d$. 
Since  $\{A_0,A_1,\ldots,A_d\}$ and $\{E_0,E_1,\ldots,E_d\}$ are two bases in $\mathcal{A}$, there exist  change-of-bases matrices $P=(P_{ij})_{i,j=0}^d$, $Q=(Q_{ij})_{i,j=0}^d$ so that
\begin{align*}
A_j=\sum_{i=0}^d P_{ij}E_j,\quad E_i=\frac{1}{n}\sum_{i=0}^d Q_{ij}A_i.
\end{align*}   
Since disjoint $(0,1)$-matrices $A_i$'s form a basis of $\mathcal{A}$, the algebra $\mathcal{A}$ is closed under the entrywise multiplication denoted by $\circ$. 
The \emph{Krein parameters} $q_{ij}^k$ are defined by $E_i\circ E_j=\frac{1}{n}\sum_{k=0}^dq_{ij}^k E_k$. 
The \emph{Krein matrix} $B_i^*$ is defined as $B_i^*=(q_{ij}^k)_{j,k=0}^d$.

Each of the matrices $A_i$ can be considered as the adjacency matrix of some graph without multiedges. The scheme is \emph{imprimitive} if,
on viewing the $A_i$'s as the adjacency matrices
of graphs $G_i$ on vertex set $X$, at least one of the $G_i$, $i \ne 0$, is disconnected. 
Then there exists a set $\mathcal{I}$ of indices, under a suitable ordering, such that $0$ and such $i$ are elements of $\mathcal{I}$ and $\sum_{j\in\mathcal{I}}A_j=I_p\otimes J_q$ for some $p,q$ with $1<p<n$. 
Thus the set of $n$ vertices $X$ are partitioned into $p$ subsets called \emph{fibers}, each of which has size $q$. 
The set $\mathcal{I}$ defines an equivalence relation on $\{0,1,\ldots,d\}$ by $j\sim k$ if and only if $p_{ij}^k\neq 0$ for some $i\in \mathcal{I}$.  
Let $\mathcal{I}_0=\mathcal{I},\mathcal{I}_1,\ldots,\mathcal{I}_t$ be the equivalence classes on $\{0,1,\ldots,d\}$ by $\sim$. 
Then by \cite[Theorem~9.4]{bi} there exist $(0,1)$-matrices $\overline{A}_j$ ($0\leq j\leq t$) such that 
\begin{align*}
\sum_{i\in \mathcal{I}_j}A_i=\overline{A}_j\otimes J_q,
\end{align*}
and the matrices $\overline{A}_j$ ($0\leq j\leq t$) define an association scheme on the set of fibers. 
This is called the \emph{quotient association scheme} with respect to $\mathcal{I}$

For fibers $U$ and $V$, let $\mathcal{I}(U,V)$ denote the set of indices of adjacency matrices $A_i$ with $(A_i)_{u,v}$ for some $u\in U$, $v\in V$.
 
We define a $(0,1)$-matrix $A_i^{UV}$ by 
\begin{align*}
(A_i^{UV})_{xy}=\begin{cases}
1 &\text{ if } (A_i)_{xy}=1, x\in U, y\in V,\\
0 &\text{ otherwise}.
\end{cases}
\end{align*}
We define uniformity for imprimitive association schemes following \cite{dmm,h2}.
\begin{definition}
An imprimitive association scheme is called uniform if its quotient association scheme is class 1 and there exist integers $a_{ij}^k$ such that for  all fibers $U,V,W$ and $i\in\mathcal{I}(U,V),j\in\mathcal{I}(V,W)$, we have 
\begin{align*}
A_i^{UV}A_j^{VW}=\sum_{k}a_{ij}^k A_k^{UW}.
\end{align*}
\end{definition}

\section{Class 5 Association Scheme}\label{5classas}
Let $\{ H_1, H_2, \ldots, H_m\}$ be a set
of Mutually Unbiased Regular Hadamard (MURH) matrices of order $4n^2$ with $m\ge 2$.
Write
\begin{align}M =
\left[
\begin{array}{c}
I\\H_1/2n\\H_2/2n\\\ldots\\H_m/2n
\end{array}
\right]
\left[
\begin{array}{ccccc}
I&H_1^t/2n&H_2^t/2n&\ldots&H_m^t/2n
\end{array}
\right].\label{eq:M}
\end{align}
be the {\em Gramian} of the set of matrices $\{I, \frac{1}{2n}H_1, \frac{1}{2n}H_2, \ldots, \frac{1}{2n}H_m\}$.
Let $B=2n(M-I)$.
Then $B$ is
a symmetric $(0,-1,1)$-matrix. 
Let  $$B=B_1-B_2,$$ where
$B_1$ and $B_2$ are  disjoint 
$(0,1)$-matrices.  By reworking a result of Mathon, see \cite{d},
 we have the following:
\begin{lemma}\label{3-assoc}
Let $I=I_{4n^2(m+1)}$, $B_1, B_2$ and $B_3=I_{m+1}\otimes J_{4n^2} - I_{4n^2(m+1)}$. Then, $I,B_1,B_2,B_3$
form a 3-class association scheme.
\end{lemma}
\begin{proof}
The intersection numbers can be read off from the following easily verified equations:
\begin{align}
B_1^2&=
(2n^2+ n)mI+(n^2+\frac{3}{2}n)(m-1)B_1\notag\break\displaybreak[0]\\
&+(n^2+n/2)(m-1)B_2+(n^2+n)B_3, \notag\break\displaybreak[0]\\
B_2^2&=
(2n^2- n)mI+(n^2-\frac{1}{2}n)(m-1)B_1\notag\\ &+(n^2-\frac{3}{2}n)(m-1)B_2+(n^2-n)B_3,\notag\break\displaybreak[0]\\
B_1B_2 &=(n^2-n/2)(m-1)B_1\notag\break\displaybreak[0]\\
&+(n^2+n/2)(m-1)B_2+n^2mB_3,\notag\break\displaybreak[0]\\
B_1B_3 &=(2n^2+n-1)B_1+(2n^2+n)B_2,\notag\break\displaybreak[0]\\
B_2B_3 &=
(2n^2-n)B_1+(2n^2-n-1)B_2.\notag \qedhere
\end{align}
\end{proof}

We now impose a further structure on the regular Hadamard matrices $H_i$ and assume that 
they are all of Bush type. First we need the following.

\begin{lemma}\label{lem:bh}
Let $H$ and $K$ be two unbiased Bush-type Hadamard matrices of order $4n^2$. 
Let $L$ be a $(1,-1)$-matrix so that $HK^t=2nL$. 
Then $L$ is a Bush-type Hadamard matrix. 
\end{lemma}
\begin{proof}
Let $X=I_{2n}\otimes J_{2n}$, then $L$ is of Bush type if and only if  $LX=XL=2nX$.
We calculate $LX$. 
$$LX=\frac{1}{2n}HK^t X=2nX.$$
Similarly, we have $XL=2nX$. 
Thus $L$ is a Bush-type Hadamard matrix. 
\end{proof}

This enables us to add two more classes and we have
the following.

\begin{theorem}\label{mn5}
Let  $B_1, B_2$ denote the matrices defined above.
Let 
\begin{itemize}
  \item $A_0 = I_{4n^2(m+1)}$
  \item $A_1 = I_{m+1} \otimes I_{2n} \otimes (J_{2n} - I_{2n})$
  \item $A_2 = I_{m+1} \otimes (J_{2n} - I_{2n}) \otimes J_{2n}$
\item $A_3 = (J_{m+1} - I_{m+1} ) \otimes I_{2n} \otimes J_{2n}$
\item $A_4 = B_1 - A_3$ 
\item $A_5 = B_2$
\end{itemize}
Then, $A_0,A_1,A_2,A_3,A_4,A_5$ form a 5-class association scheme.
\end{theorem}

\begin{proof}
We work out the intersection numbers, using some of the relations in Lemma \ref{3-assoc}. Note that $A_0+A_1$, $A_3$, ($A_0+A_1+A_2$) are block matrices 
of block size $2n$, ($4n^2$, respectively), where each block
is either the zero or the matrix of ones. On the other hand $A_4$ and $A_5$ are block matrices of block size $2n$, where 
the blocks are either the zero matrix or of row and column sum $n$. So, it is straightforward computation to see the following:
\begin{align}
A_1A_1 &= (2n-1)A_0+(2n-2)A_1.\notag\break\displaybreak[0]\\
A_1A_2 &= (2n-1)A_2.\notag\break\displaybreak[0]\\
A_1A_3 &= (2n-1) A_3.\notag\break\displaybreak[0]\\
A_1A_4 &= (n-1)A_4 + nA_5.\notag\break\displaybreak[0]\\
A_1A_5 &= nA_4 +(n-1)A_5.\notag\break\displaybreak[0]\\
A_2A_2 &=  2n(2n-1)A_0+ 2n(2n-1)A_1 + 2n(2n-2) A_2. \notag\break\displaybreak[0]\\
A_2A_3 &= 2n(A_4+A_5).\notag\break\displaybreak[0]\\
A_2A_4 &=(2n-1)nA_3 + (2n-2)n(A_4+A_5). \notag\break\displaybreak[0]\\
A_2A_5 &=(2n-1)nA_3 + (2n-2)n(A_4+A_5).\notag\break\displaybreak[0]\\
A_3A_3 &= 2mn(A_0+A_1) + 2n(m-1)A_3.\notag\break\displaybreak[0]\\
A_3A_4 &= mnA_2 + (m-1)n(A_4+A_5).\notag\break\displaybreak[0]\\
A_3A_5 &=  mnA_2 + (m-1)n(A_4+A_5).\notag
\end{align}
Using these, the facts that $A_3+A_4=B_1$, $A_5(A_3+A_4)=B_2B_1$, and the intersection numbers in Lemma \ref{3-assoc} we have:
\begin{align}
A_4A_5 &= n^2mA_1 + m(n^2-n) A_2 +(n^2 - \frac{n}{2})(m-1)A_3 \notag\\
&+ (n^2 - \frac{3n}{2})(m-1)A_4 + (n^2 -\frac{n}{2} )(m-1)A_5.\notag
\end{align}
Finally, noting that $A_4-A_5$ is a block matrix of block size $2n$, where 
the blocks are either the zero matrix or of row and column sum zero, it follows that
 $$(A_4+A_5)(A_4-A_5)=0,$$ so we have
 \begin{align}
A_4A_4 &= A_5A_5 =(2n^2 - n)m I  + (n^2-n)m(A_1+A_2) +\notag\\
&(n^2 - \frac{n}{2})(m-1) ( A_3 + A_4) + (n^2 - \frac{3n}{2})(m-1) A_5.\notag
\end{align}
\end{proof}

\begin{definition}
Two Latin squares $L_1$ and $L_2$ of size $n$ on symbol set $\{0,1,2,\ldots,n-1\}$ are called
{\it suitable} if every superimposition of each row of $L_1$ on each row of $L_2$ results in
only one element of the form $(a,a)$.
A set of Latin squares in which every distinct pair of Latin squares
is mutually suitable is called
\emph{Mutually Suitable Latin Squares}, denoted MSLS.
\end{definition}

The existence and a construction method for MUBH  matrices to use mutually suitable Latin squares were given in \cite[Theorem 13]{hko}.
However, in order to obtain Bush-type Hadamard matrices as defined here, an additional assumption on the MSLS is needed as follows.
\begin{proposition}
If there are $m$ mutually suitable Latin squares of size $2n$ with all one entries on diagonal and a Hadamard matrix of order $2n$, then there are $m$ mutually unbiased Bush-type Hadamard matrices of order $4n^2$.
\end{proposition}
The construction is exactly same as \cite[Theorem 13]{hko}. 
The resulting mutually unbiased Hadamard matrices are all of Bush-type.
Indeed, each Latin square has the entries $1$ on diagonal, thus the resulting Hadamard matrix has the all ones matrices on diagonal blocks. 

The equivalence of MOLS and MSLS was given in \cite[Lemma 9]{hko}. 
The assumption on MOLS corresponding to MSLS with all ones entries on diagonal is that each Latin square has $(1,2,\cdots,n)$ as the first row. 
The MOLS having this property is constructed by the use of finite fields as follows. 
For each $\alpha\in \mathbb{F}_q\setminus\{0\}$, define $L_{\alpha}$ as $(i,j)$-entry equal to $\alpha i+j$, where $i,j\in\mathbb{F}_q$. 
By switching rows so that the first row corresponds to $0\in \mathbb{F}_q$ and mapping $\mathbb{F}_q$ to $\{1,2,\ldots,n\}$ such that each first row becomes $(1,2,\ldots,n)$, we obtain the desired MOLS. 
Thus we have the same conclusion as \cite[Corollary~15]{hko}.
 
\begin{remark}\label{rem:1}

\begin{enumerate}
 \item[(a)] Rewriting $A_4A_4=A_5A_5$ as:
\begin{align}
A_4A_4 &= A_5A_5= n^2m I +(n^2-n)mJ + n(\frac{m+1}{2}-n)(A_3+A_4)\notag\\
& + n(\frac{3}{2}-\frac{m}{2}-n) A_5.\notag
\end{align}
It is seen that, for $m=2n-1$, $A_5$ is the adjacency matrix of a strongly regular graph and $A_4$
is the adjacency matrix of a Deza Graph, see \cite{dd,efhhh}. 
There exists an example satisfying $m=2n-1$ for $n=2^{k-1}$ and $k\ge 1$ \cite[Corollary~15]{hko}.
\item[(b)] The association scheme of class $5$ is uniform.  
Any two fibers define a coherent configuration, which is a strongly regular design of the second kind, see \cite{h}.  
The first and second eigenmatrices and $B_5^*$ are as follows:
\begin{align*}
P&=\begin{pmatrix}
1&2n-1&2n(2n-1)&2nm&n(2n-1)m&n(2n-1)m\\
1&-1&0&0&nm&-nm\\
1&2n-1&-2n&2nm&-nm&-nm\\
1&2n-1&-2n&-2n&n&n\\
1&-1&0&0&-n&n\\
1&2n-1&2n(2n-1)&-2n&-n(2n-1)&-n(2n-1)
 \end{pmatrix},\displaybreak[0]\\
Q&=\begin{pmatrix}
1&2n(2n-1)&2n-1&(2n-1)m&2n(2n-1)m&m\\
1&-2n&2n-1&(2n-1)m&-2nm&m\\
1&0&-1&-m&0&m\\
1&0&2n-1&-2n+1&0&-1\\
1&2n&-1&1&-2n&-1\\
1&-2n&-1&1&2n&-1
 \end{pmatrix},\displaybreak[0]\\
B_5^*&=\begin{pmatrix}
0&0&0&0&0&1\\
0&0&0&0&1&0\\
0&0&0&1&0&0\\
0&0&m&m-1&0&0\\
0&m&0&0&m-1&0\\
m&0&0&0&0&m-1
\end{pmatrix}.
\end{align*}
Thus the association scheme certainly satisfies \cite[Proposition 4.7]{dmm}. 
Note that the scheme is not $Q$-polynomial since no column has all distinct entries.

Since the Krein number $q_{1,2}^1=\frac{2n-m-1}{m+1}$ must be positive, 
we obtain $m\leq 2n-1$ holds. 
This means that the number of MUBH matrices of order $4n^2$ is at most $2n-1$. 
The example attaining the upper bound is given in \cite[Corollary 15]{hko}.
\item[(c)]
The first, second eigenmatrices and the Krein matrix $B_1^*$ of the class $3$ association scheme are as follows:
\begin{align*}
P&=\begin{pmatrix}
1&n(2n+1)m&n(2n-1)m&4n^2-1\\
1&nm&-nm&-1\\
1&-n&n&-1\\
1&-n(2n+1)&-n(2n-1)&4n^2-1
 \end{pmatrix},\displaybreak[0]\\
Q&=\begin{pmatrix}
1&4n^2-1&(4n^2-1)m&m\\
1&2n-1 &-2n+1&1\\
1&-2n-1&2n+1&1\\
1&-1&-m&m
 \end{pmatrix},\displaybreak[0]\\
B_1^*&=\begin{pmatrix}
0&1&0&0\\
4n^2-1&\frac{2(2n^2-m-1)}{m+1}&\frac{4n^2}{m+1}&0\\
0&\frac{4n^2 m}{m+1}&\frac{(4n^2-2)m-2}{m+1}&4n^2-1\\
0&0&1&0
\end{pmatrix}.
\end{align*}
This association scheme is a $Q$-antipodal $Q$-polynomial scheme of class $3$.  
By \cite[Theorem 5.8]{d}, this scheme comes from a linked systems of symmetric designs.
\end{enumerate}
\end{remark}

Next we show the converse implication as follows.
\begin{theorem}
Assume that there exists an association scheme with the same eigenmatrices as in Remark~\ref{rem:1}(b). 
Then there exists a set of MUBH $\{H_1,\ldots.H_m\}$ of order $4n^2$.  
\end{theorem}
\begin{proof}
Let $A_0,\ldots,A_5$ be the adjacency matrices of an association scheme with the same eigenmatrices as in Remark~\ref{rem:1}. 
Let $B_0=A_0$, $B_1=A_3+A_4$, $B_2=A_5$ and $B_3=A_1+A_2$.
By Remark~\ref{rem:1} $B_i$'s form a linked system of symmetric designs. 
Thus we rearrange the vertices so that $B_3=I_{m+1}\otimes J_{4n^2}-I_{4n^2(m+1)}$. 
 
We first determine the form of $A_3$. 
Since $A_1$ is the adjacency matrix of an imprimitive strongly regular graph with eigenvalues $2n-1,-1$ with multiplicities $2n(m+1),2n(2n-1)(m+1)$,  
$A_1$ is $I_{2n(m+1)}\otimes (J_{2n}-I_{2n})$ after rearranging the vertices. 
By $B_3=I_{m+1}\otimes J_{4n^2}-I_{4n^2(m+1)}=A_1+A_2$, $A_2$ has the desired form. 
Since $B_3$ and $A_3$ are disjoint and $A_2A_3=2n(A_4+A_5)$, we obtain $A_3=(J_{m+1} - I_{m+1}) \otimes I_{2n} \otimes J_{2n}$. 

Letting $G=(m+1)(E_0+E_1+E_2)$, we have 
\begin{align*}
G&=(m+1)(E_0+E_1+E_2)\\
&=\frac{1}{4n^2}\sum_{i=0}^5(Q_{0,i}+Q_{1,i}+Q_{2,i})A_i\\
&=A_0+\frac{1}{2n}A_3+\frac{1}{2n}A_4-\frac{1}{2n}A_5.
\end{align*}
Since $A_3+A_4+A_5=(J_{m+1}-I_{m+1})\otimes J_{2n} \otimes J_{2n}$, $G$ is the following form
\begin{align*}
G=\begin{pmatrix}
I_{2n}&\frac{1}{2n}H_{1,2}&\ldots&\frac{1}{2n}H_{1,m+1}\\
\frac{1}{2n}H_{2,1}&I_{2n}&\ldots&\frac{1}{2n}H_{2,m+1}\\
\vdots&\vdots&\ddots&\vdots\\
\frac{1}{2n}H_{m+1,1}&\frac{1}{2n}H_{m+1,2}&\ldots&I_{2n}
\end{pmatrix}
\end{align*}
where $H_{i,j}$ ($i\neq j$) is a $(1,-1)$-matrix.

We claim that  $H_k:=H_{k+1,1}$ ($1\leq k\leq m$) are mutually unbiased Bush-type Hadamard matrices. 
Let $\bar{A}$ denote the submatrix of entries that lie in the rows and columns on the first and $(k+1)$th blocks.
We consider the principal submatrix $\bar{G}$.  
Since the association scheme is uniform, 
the restricting to indices on the first and second blocks yields an association scheme with the eigenmatrix $\bar{P}=(\bar{P}_{ij})_{i,j=0}^5$ obtained by putting $m=1$. 

Since $\bar{G}=(m+1)(\bar{E}_0+\bar{E}_1+\bar{E}_2)$ holds and $\frac{m+1}{2}\bar{E}_i$ ($i=0,1,2$) are primitive idempotents of  the subscheme, we have $\bar{G}^2=2\bar{G}$. 
Expanding  the left hand-side to use the form $\bar{G}=\left( \begin{smallmatrix}I_{2n}&\frac{1}{2n}H_{k}^t\\ \frac{1}{2n}H_{k}&I_{2n} \end{smallmatrix}\right)$, we obtain 
\begin{align*}
\begin{pmatrix}
I_{2n}+\frac{1}{4n^2}H_{k}^tH_{k}&\frac{1}{n}H_{k}^t\\
\frac{1}{n}H_{k}& I_{2n}+\frac{1}{4n^2}H_{k}H_{k}^t
\end{pmatrix}=
\begin{pmatrix}
2I_{2n}&\frac{1}{n}H_{k}^t\\ \frac{1}{n}H_{k}&2I_{2n}
\end{pmatrix}. 
\end{align*}
This implies that $H_{k}$ is a Hadamard matrix of order $4n^2$.

Next we show $H_k$ is of Bush-type.
Now we calculate $\bar{A}_3\bar{G}$ in two ways. First we have
\begin{align*}
\bar{A}_3\bar{G}&=(m+1)\bar{A}_3(\bar{E}_0+\bar{E}_1+\bar{E}_2)\\ 
&=(m+1)(\sum_{i=0}^5\bar{P}_{i3}\bar{E}_i)(\bar{E}_0+\bar{E}_1+\bar{E}_2)\\ 
&=(m+1)(\sum_{i=0}^2\bar{P}_{i3}\bar{E}_i)\\
&=2n(m+1)(\bar{E}_0+\bar{E}_2)\\ 
&=2n(m+1)(\frac{1}{4n^2(m+1)}\sum_{i=0}^5(\bar{Q}_{i,0}+\bar{Q}_{i,2})\bar{A}_i)\\ 
&=(A_0+A_1+A_3)\\
&=\begin{pmatrix}
I_{2n} \otimes J_{2n}&I_{2n} \otimes J_{2n}\\
I_{2n} \otimes J_{2n}&I_{2n} \otimes J_{2n}
\end{pmatrix}.
\end{align*}
Second we have 
\begin{align*}
\bar{A}_3\bar{G}&=\begin{pmatrix}
0&I_{2n} \otimes J_{2n}\\
I_{2n} \otimes J_{2n}&0
\end{pmatrix}\begin{pmatrix}
I_{2n}&\frac{1}{2n}H_{k}^t\\
\frac{1}{2n}H_{k}&I_{2n}
\end{pmatrix}\\
&=\begin{pmatrix}
\frac{1}{2n}(I_{2n} \otimes J_{2n})H_{k}&I_{2n} \otimes J_{2n}\\
I_{2n} \otimes J_{2n}& \frac{1}{2n}(I_{2n} \otimes J_{2n})H_{k}^t
\end{pmatrix}.
\end{align*}
Comparing these two equations yields 
\begin{align*}
(I_{2n} \otimes J_{2n})H_{k}=(I_{2n} \otimes J_{2n})H_{k}^t=2nI_{2n} \otimes J_{2n}.
\end{align*}
This implies that $H_{k}$ is of Bush-type by Lemma~\ref{lem:bh}.
 
Finally we show $H_1,\ldots,H_m$ are unbiased. 
Let $k,k'$ be integers such that $1\leq k<k'\leq m$.
Let $\tilde{A}$ denote the submatrix of vertices that lie in the rows and columns on the first, $(k+1)$th and $(k'+1)$th blocks.
We then have $\tilde{G}^2=3\tilde{G}$. 
Comparing the $(2,3)$-block, we obtain  
\begin{align*}
\frac{1}{4n^2}H_kH^t_{k'} +\frac{1}{2n} I_{2n}H_{k+1,k'+1} + \frac{1}{2n}H_{k+1,k'+1}I_{2n} = \frac{3}{2n}H_{k+1,k'+1},
\end{align*}  
namely $\frac{1}{4n^2}H_k H_{k'}^t=\frac{1}{2n}H_{k+1,k'+1,}$. 
Since $H_{k+1,k'+1}$ is a $(-1,1)$-matrix,  $H_k$ and $H_{k'}$ are unbiased. 
\end{proof}

\section{$8$ class association schemes}
Linked systems of symmetric designs with specific parameters have the extended $Q$-bipartite double which yields an association scheme of mutually unbiased bases \cite[Theorem 3.6]{mmw}. 
Next we show an association scheme from our association schemes of class $5$ has a double cover and show a relation to an association scheme of class $4$ as a fusion scheme. 
\begin{theorem}
Let $A_0,A_1,\ldots,A_5$ be the adjacency matrices of the association scheme in Theorem~\ref{mn5}. 
Define \begin{align*}
\tilde{A}_0&=\begin{pmatrix} A_0&0\\0&A_0\end{pmatrix},
\tilde{A}_1=\begin{pmatrix} A_1&0\\0&A_1\end{pmatrix},
\tilde{A}_2=\begin{pmatrix} 0&A_1\\A_1&0\end{pmatrix},
\tilde{A}_3=\begin{pmatrix} A_2&A_2\\A_2&A_2\end{pmatrix},\\
\tilde{A}_4&=\begin{pmatrix} A_3&0\\0&A_3\end{pmatrix},
\tilde{A}_5=\begin{pmatrix} 0&A_3\\A_3&0\end{pmatrix},
\tilde{A}_6=\begin{pmatrix} A_4&A_5\\A_5&A_4\end{pmatrix},
\tilde{A}_7=\begin{pmatrix} A_5&A_4\\A_4&A_5\end{pmatrix},\\
\tilde{A}_8&=\begin{pmatrix} 0&A_0\\A_0&0\end{pmatrix}.
\end{align*}  
Then $\tilde{A}_0,\ldots,\tilde{A}_8$ form an association scheme.
\end{theorem}
\begin{proof}
This follows from the calculation in Theorem~\ref{mn5}.  
\end{proof}

\begin{remark}
\begin{enumerate}
\item[(a)] The association scheme of class $8$ is also uniform.  
The second eigenmatrix and $B_8^*$ are as follows:
\begin{align*}
Q&=\left(\begin{smallmatrix}
1&2n(2n-1)&2n&2n-1&2n(2n-1)(m+1)&(2n-1)m&2nm&2n(2n-1)m&m\\
1&-2n&2n&2n-1&-2n(m+1)&(2n-1)m&2nm&-2nm&m\\
1&2n&-2n&2n-1&-2n(m+1)&(2n-1)m&-2nm&2nm&m\\
1&0&0&-1&0&-m&0&0&m\\
1&0&2n&2n-1&0&-2n+1&-2n&0&-1\\
1&0&-2n&2n-1&0&-2n+1&2n&0&-1\\
1&2n&0&-1&0&1&0&-2n&-1\\
1&-2n&0&-1&0&1&0&2n&-1\\
1&-2n(2n-1)&-2n&2n-1&2n(2n-1)(m+1)&(2n-1)m&-2nm&-2n(2n-1)m&m
\end{smallmatrix}\right),\displaybreak[0]\\
B_8^*&=\begin{pmatrix}
0&0&0&0&0&0&0&0&1\\
0&0&0&0&0&0&0&1&0\\
0&0&0&0&0&0&1&0&0\\
0&0&0&0&0&1&0&0&0\\
0&0&0&0&m&0&0&0&0\\
0&0&0&m&0&m-1&0&0&0\\
0&0&m&0&0&0&m-1&0&0\\
0&m&0&0&0&0&0&m-1&0\\
m&0&0&0&0&0&0&0&m-1
\end{pmatrix}.
\end{align*}
Thus the association scheme certainly satisfies \cite[Proposition 4.7]{dmm}. 
Note that the scheme is not $Q$-polynomial since no column has all distinct entries.  
\item[(b)] 
Letting $\tilde{B}_1=\tilde{A}_1+\tilde{A}_2+\tilde{A}_3$, $\tilde{B}_2=\tilde{A}_4+\tilde{A}_6$ and  $\tilde{B}_3=\tilde{A}_5+\tilde{A}_7$, $\tilde{B}_4=\tilde{A}_8$, 
we obtain a fusion association scheme of class $4$.  
The second eigenmatrix of the class $4$ fusion association scheme is as follows:
\begin{align*}
Q&=\begin{pmatrix}
1&4n^2&(4n^2-1)(m+1)&4n^2m&m\\
1&0&-m-1&0&m\\
1&2n&-0&-2n&-1\\
1&-2n&0&2n&-1\\
1&-4n^2&(4n^2-1)(m+1)&-4n^2m&m
 \end{pmatrix}.
\end{align*}
This association scheme is a $Q$-antipodal and $Q$-bipartite $Q$-polynomial scheme of class $4$, see \cite{lmo}.  
\end{enumerate}
\end{remark}
\subsection*{Acknowledgements}
Thanks to the anonymous referee for the time spent on reading this paper and for many nice suggestions.

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\end{document}
