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I'm reading Distance Regular Graphs by Brouwer, Cohen, and Neumaier. In section 1.8, they explained Hadamard graphs.

Conversion from a Hadamard Matrix into a Hadamard Graph

An $n$-Hadamard graph $G$ is a graph on $4n$ vertices defined in terms of a Hadamard matrix of order $n$ $H_n = h_{ij}$ as follows:

  1. Define $4n$ symbols $r_i^+$, $r_i^-$, $c_i^+$, and $c_i^-$, where $r$ stands for row and $c$ stands for column and take these as the vertices of the graph.
  2. Then add two types of edges between row vertices and column vertices based on the sign of $h_{ij}$: \begin{equation*} \text{parallel edges $(r_i^+, c_j^+)$ and $(r_i^-, c_j^-)$ if $h_{ij} = +1$} \\ \text{crossing edges $(r_i^+, c_j^-)$ and $(r_i^-, c_j^+)$ if $h_{ij} = -1$} \end{equation*}

Then the graph $G$ will be a bipartite graph where the set of vertices is partitioned into row vertex set of $2n$ vertices and column vertex set of $2n$ vertices. And there will be $2n^2$ edges.

Equivalence between Graph and Matrix

In theorem 1.8.1 in their book, they showed that $G$ is a distance-regular graph with an intersection array $\{n,n-1,\frac{n}{2},1;1,\frac{n}{2},n-1,n\}$ if and only if the matrix $H$ is Hadamard matrix of order $n$.

My Question

Their proof of this theorem seems rather brief and I have hard time in understanding the equivalence. Especially, I don't understand what role two orthogonal rows or columns in the matrix $H_n$ play in the graph $G$ so that distance-regularity is achieved. The book cited three papers for the proof, but I cannot find any of them in the Internet.

Can anyone explain what's the idea or intuition behind the proof of equivalence?

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First, the book by Brouwer, Cohen and Neumaier is known for a degree of terseness; any one who uses it will have struggled with it at some point.

You describe a construction that, from an $n\times n$ Hadamard matrix, produces a bipartite graph on $4n$ vertices that is regular of degree $n$. This graph has diameter four; moreover for each vertex $u$ there is a unique vertex $u'$ at distance four from $u$. It follows that the the $4n$ vertices may be partitioned into $2n$ pairs.

So the way to proceed is to prove that these comments about pairs of vertices at distance four are correct and then, using this, prove that the parameters $b_i$ and $c_i$ are well-defined. This reduces to showing that if $v$ is at distance two from a vertex $u$, then $v$ has exactly $n/2$ neighbours in common with $u$. (This is where the orthogonality of the rows of $H$ enters.)

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