On the non-existence of $srg(76,21,2,7)$

TL;DR

This paper proves that a strongly regular graph with parameters (76,21,2,7) cannot exist by employing a novel approach based on the graph's unit vector representation.

Contribution

It introduces a new non-existence proof for the specific strongly regular graph using unit vector representation methods.

Findings

Proves non-existence of srg(76,21,2,7)

Introduces a novel proof technique based on vector representations

Enhances understanding of parameter constraints for strongly regular graphs

Abstract

We present a new non-existence proof for the strongly regular graph $G$ with parameters $(76,21,2,7)$, using the unit vector representation of the graph.

Equations

$$(\hat{v},\hat{w})=\left\{\begin{array}[]{r l}2,&\mbox{if $v=w$,}\\ -1,&\mbox{if $v$ and $w$ are adjacent,}\\ 0,&\mbox{if they are not adjacent.}\end{array}\right.$$

$$A_{t-1}=\left(\begin{array}[]{rrrrrr}2&-1&0&0&.&0\\ -1&2&-1&0&.&0\\ 0&-1&2&-1&.&0\\ .&.&.&.&.&.\\ 0&.&.&-1&2&-1\\ 0&.&.&.&-1&2\end{array}\right).$$

$$0=(0,w)=(\hat{v},w)=({\hat{v}}_1,w)+({\hat{v}}_2,v)+\cdots +({\hat{v}}_t,x).$$

$$0=-\frac{4}{9}s+\frac{2}{9}(t-s),$$

$$(\hat{v},\hat{w})=\left\{\begin{array}[]{r l}5,&\mbox{if $v=w$,}\\ -\frac{7}{4},&\mbox{if $v$ adjacent to $w$,}\\ \frac{1}{2},&\mbox{if $v$ is not adjacent to $w$.}\end{array}\right.$$

$$(\hat{v},\hat{w})=\left\{\begin{array}[]{r l}-1,&\mbox{if $v$ is adjacent to $w$,}\\ \frac{1}{2},&\mbox{if $v$ is not adjacent to $w$.}\end{array}\right.$$

$$(w^{\circ},(w^{\prime})^{\circ})=\left\{\begin{array}[]{l l}2,&\mbox{if $w=w^{\prime}$,}\\ -1,&\mbox{if $w$ and $w^{\prime}$ are adjacent,}\\ 3-\beta,&\mbox{if and $w$ and $w^{\prime}$ are non-adjacent.}\end{array}\right.$$

Full text

On the non-existence of $\mathbf{srg(76,21,2,7)}$

Monther R. Alfuraidan1, Ibrahim O. Sarumi2, Sergey Shpectorov3

*(1,2)*Department of Mathematics & Statistics, King Fahd University of Petroleum and Minerals

Dhahran 31261, Saudi Arabia

monther@kfupm.edu.sa, g201304910@kfupm.edu.sa,

*(3)*School of Mathematics, University of Birmingham, Birmingham, United Kingdom

s.shpectorov@bham.ac.uk

Abstract. We present a new non-existence proof for the strongly regular graph $G$ with parameters $(76,21,2,7)$, using the unit vector representation of the graph.

Keywords: Strongly regular graph, distance regular graph, unit vector representation.

AMS Subject Classification: Primary: 05E30, Secondary: 05C30

1. Introduction

A graph $G$ is said to be strongly regular with parameters $(v,k,\lambda,\mu)$ if the following condition holds: $G$ has $v$ vertices (i.e., $|V(G)|=v$) and, for $u,w\in V(G)$, the number of common neighbours of $u$ and $w$ in $G$ is $k$ if $u=w$ (so $G$ is regular of valency $k$), $\lambda$ if $u$ and $w$ are adjacent, and $\mu$ if $u$ and $w$ are non-adjacent. Strongly regular graphs are among the central objects in graph theory and its applications. We write $srg(v,k,\lambda,\mu)$ for any strongly regular graph with parameters $(v,k,\lambda,\mu)$.

Haemers [4] proved non-existence of $srg(76,21,2,7)$. His proof is very efficient (perhaps even a bit terse), and it relies on clever edge counting to establish that such $G$ must locally be a union of $3$-cliques. This means that $G$ is the collinearity graph of a point-line geometry $pg(3,6,1)$ (a generalized quadrangle of order $(3,6)$). At this point Haemers quotes the non-existence result for $pg(3,6,1)$ by Dixmier and Zara [2].

In this note, we give an alternative proof of Haemers’ theorem based on the well-known fact that every distance regular graph (and in particular, every strongly regular graph) admits a Euclidean realization as a set of unit vectors in an eigenspace of the adjacency matrix of $G$. In this realization, the value of the inner product of two vectors (the cosine of the angle between them) is fully determined by the mutual distance of the corresponding vertices. This is encoded in the so-called cosine sequence. Note that the eigenvalues of the adjacency matrix, dimension of each eigenspace, and the cosine sequence can be easily deduced from the parameters of $G$ via the readily available formulas (for example, see [3]).

There are many open cases of strongly regular graphs even for relatively small values of $v$ (see the table of feasible parameters up to $v=100$ in [BH]). Of course, the aim of our project is to contribute to one of the open cases. In this sense, the proof in this note is just a sample of things to come. However, we think that even this taster proof demonstrates efficiency of the method and it exhibits interesting features, such as the relation to roots systems, which arise in our proof not once, but twice.

Just like Haemers, we aim to show that $G$ is locally a union of cliques. However, once we achive this, we do not stop, but rather use our unit vector setup to achieve an outright contradiction. In this sense, we also provide an alternative proof of the result of Dixmier and Zara.

2. Starting point

Suppose $G$ is $srg(76,21,2,7)$. Then the adjacency matrix of $G$ has eigenvalues $21$, $2$, and $-7$ with multiplicities $1$, $56$, and $19$, respectively. We focus on the $19$-dimensional eigenspace corresponding to the eigenvalue $-7$. The cosine sequence for this eigenspace is $(1,-\frac{1}{3},\frac{1}{9})$. This means that our graph $G$ can be realized as a set of $76$ unit vectors $x_{v}$, $v\in V(G)$, in the Euclidean space $\mathbb{R}^{19}$ such that $(x_{u},x_{v})=-\frac{1}{3}$ if the distinct vertices $u$ and $v$ are adjacent, and $(x_{u},x_{v})=\frac{1}{9}$, if they are not.

From now on we identify vertices of $G$ with the corresponding unit vectors. Hence we simply write $u$ and $v$ in place of $x_{u}$ and $x_{v}$.

3. Neighbourhood

Fix an arbitrary $u\in V(G)$. The subgraph induced on the $21$ vertices in $G_{1}(u)$ is a union of cycles $C_{1},\ldots,C_{r}$, since its degree $\lambda$ is $2$. Let us slightly alter vectors $v\in G_{1}(u)$ to make them perpendicular to $u$. Namely, we set $\hat{v}:=\frac{3}{2}v+\frac{1}{2}u$ for each $v\in G_{1}(u)$. Clearly, $(u,\hat{v})=\frac{3}{2}(u,v)+\frac{1}{2}(u,u)=\frac{3}{2}(-\frac{1}{3})+\frac{1}{2}1=0$, as desired. Also, for $v,w\in G_{1}(u)$, we have $(\hat{v},\hat{w})=\frac{9}{4}(v,w)+\frac{3}{4}(u,w)+\frac{3}{4}(v,u)+\frac{1}{4}(u,u)=\frac{9}{4}(v,w)-\frac{1}{4}$. Therefore,

[TABLE]

Let $V_{i}:=\langle\hat{v}\mid v\in V(C_{i})\rangle$ be the subspace of $\mathbb{R}^{19}$ spanned by the vectors corresponding to the vertices of the $i$th cycle $C_{i}$ in $G_{1}(u)$. It follows from the above inner product values that $u\perp V_{i}$ for all $i$ and that $V_{i}\perp V_{j}$ for all $i\neq j$.

Lemma 3.1**.**

Let $V(C_{i})=\{v_{1},v_{2},\ldots,v_{t}\}$. Then we have:

1. (i)

$\hat{v}_{1}+\hat{v}_{2}+\cdots+\hat{v}_{t}=0$; and 2. (ii)

$\dim V_{i}=t-1$.

Proof.

(i) Let $\hat{v}=\hat{v}_{1}+\hat{v}_{2}+\cdots+\hat{v}_{t}$. Then, for each $j$, we have that $(\hat{v},\hat{v}_{j})=0$, since $\hat{v}_{j}$ itself contributes $2$ to the sum, and its two neighbours contribute $-1$ each, while all the other vertices of $C_{i}$ contribute naught. Therefore, $(\hat{v},\hat{v})=\sum_{j=1}^{t}(\hat{v},\hat{v}_{j})=0$, proving that $\hat{v}=0$.

(ii) Assuming that the vertices $v_{1},v_{2},\ldots,v_{t}$ appear in this order on the cycle $C_{i}$, let $A_{t-1}$ be the Gram matrix of the vectors $\hat{v}_{1},\hat{v}_{2},\ldots,\hat{v}_{t-1}$. Then

[TABLE]

Let $d_{t-1}$ be the determinant of $A_{t-1}$. Viewing $r=t-1$ as variable, we obtain the recursive relation $d_{r}=2d_{r-1}-d_{r-2}$ by expanding the determinant along the bottom row. Taking into account that $d_{1}=2$ and $d_{2}=3$, we easily deduce that $d_{r}=r+1\neq 0$, and so $\hat{v}_{1},\hat{v}_{2},\ldots,\hat{v}_{t-1}$ are linearly independent. ∎

We included this proof for completeness; however, we need to mention that these facts are well known. Indeed, the matrix above is the Gram matrix of a basis from the root system of type $A_{t-1}$, and if we add the missing vector $\hat{v}_{t}$ then this gives the basis of the affine root system $\tilde{A}_{t-1}$.

We now focus on a vertex $w$ from $G_{2}(u)$ and study the $\mu=7$ neighbours of $w$ in $G_{1}(u)$. Let $s_{i}$ be the number of such neighbours on the cycle $C_{i}$.

Lemma 3.2**.**

The length $t_{i}$ of $C_{i}$ is a multiple of $3$; namely, $t_{i}=3s_{i}$.

Proof.

Let again $V(C_{i})=\{v_{1},v_{2},\ldots,v_{t}\}$, where $t=t_{i}$, and $\hat{v}=\hat{v}_{1}+\hat{v}_{2}+\cdots+\hat{v}_{t}$.

Note that $(\hat{v}_{j},w)=(\frac{3}{2}v_{j}+\frac{1}{2}u,w)=\frac{3}{2}(v_{j},w)+\frac{1}{2}(u,w)$. If $w$ is adjacent to $v_{j}$, this results in $-\frac{1}{2}+\frac{1}{18}=-\frac{4}{9}$, and otherwise, the result is $\frac{1}{6}+\frac{1}{18}=\frac{2}{9}$. Now consider the equality

[TABLE]

Since $w$ is adjacent to $s=s_{i}$ vertices and non-adjacent to $t-s$ vertices, we obtain from here that

[TABLE]

which gives $t=3s$, as claimed. ∎

4. Second layer

We alter the vertices in $G_{2}(u)$ in a similar way to make them perpendicular to $u$. For $w\in G_{2}(u)$, we set $\hat{w}:=\frac{9}{4}w-\frac{1}{4}u$. Then $(\hat{w},u)=\frac{9}{4}\frac{1}{9}-\frac{1}{4}1=0$, as claimed. Similarly, we compute, for $v,w\in G_{2}(u)$,

[TABLE]

Finally, we also compute, and also in a very similar way, the inner products $(\hat{v},\hat{w})$ for $v\in G_{1}(u)$ and $w\in G_{2}(u)$. These are:

[TABLE]

Recall that every vertex $w\in G_{2}(u)$ has seven neighbours in $G_{1}(u)$. Let us first describe the subgraph $M=M_{w}$ induced on these seven vertices.

Lemma 4.1**.**

Each connected component of $M$ is of size $1$ or $2$.

Proof.

If $xyz$ is a $2$-path in $M$, with $x\neq z$, then $uxwz$ is a $4$-cycle in $G_{1}(y)$, a contradiction with Lemma 3.2 with $y$ in place of $u$. ∎

If $x$ is a size $1$ component of $M$ then the projection $p_{x}$ of $\hat{w}$ to the $1$-space spanned by $\hat{x}$ coincides with $\frac{(\hat{w},\hat{x})}{(\hat{x},\hat{x})}\hat{x}=-\frac{1}{2}\hat{x}$. Hence $(p_{x},p_{x})=\frac{1}{4}(\hat{x},\hat{x})=\frac{1}{2}$. If $xy$ is a size $2$ component of $M$ then by symmetry the projection $p_{xy}$ of $\hat{w}$ to the subspace spanned by $\hat{x}$ and $\hat{y}$ is a multiple of $d=\hat{x}+\hat{y}$. Note that $(d,d)=(\hat{x}+\hat{y},\hat{x}+\hat{y})=2-1-1+2=2$ and $(\hat{w},d)=(\hat{w},\hat{x}+\hat{y})=-1-1=-2$. Hence $p_{xy}=\frac{(\hat{w},d)}{(d,d)}d=-d$, and so $(p_{xy},p_{xy})=(-d,-d)=(d,d)=2$.

Projections corresponding to different components of $M$ are orthogonal. Hence, if we have $k$ components of size $2$ and, correspondingly, $7-2k$ components of size $1$ then the length of the projection of $\hat{w}$ to the subspace of $V$ spanned by all $\hat{x}$, $x\in M$, is $2k+\frac{1}{2}(7-2k)=2k-k+\frac{7}{2}=k+\frac{7}{2}$. Since this must be at most $(\hat{w},\hat{w})=5$, we conclude that $k=0$ or $1$.

Consider one of the cycles $C=C_{i}$ of length $t=t_{i}$ and $N=C\cap M$ consisting of $s=s_{i}$ vertices. We know that $t=3s$. Let $U=V_{i}$, the subspace spanned by the vectors $\hat{v}$, $v\in C$, and let $p$ be the projection of $\hat{w}$ onto $U$.

Lemma 4.2**.**

We have $(p,p)\geq\frac{s}{2}$. Furthermore, $(p,p)=\frac{s}{2}$ if and only the vertices of $N$ are evenly spaced in $C$, containing every third vertex along $C$ and $p=-\frac{1}{2}\sum_{v\in N}\hat{v}$.

Proof.

Let $p^{\prime}$ be the projection of $\hat{w}$ to the subspace spanned by $\hat{v}$ for $v\in N$. Then certainly $(p,p)\geq(p^{\prime},p^{\prime})$ with equality holding only if $p=p^{\prime}$. The graph $N$ consists of several components of $M$. The computation before the lemma shows that a component of size $1$ contributes $\frac{1}{2}$ to $(p^{\prime},p^{\prime})$ and a component of size $2$ contributes $2>2\cdot\frac{1}{2}$. Hence $(p,p)\geq(p^{\prime},p^{\prime})\geq s\cdot\frac{1}{2}=\frac{s}{2}$, as claimed. Furthermore, the equality only holds when every component is of size $1$ and also $p=p^{\prime}=\sum_{v\in N}p_{v}=-\frac{1}{2}\sum_{v\in N}\hat{v}$. In particular, $N$ is an independent subset of $C$. Taking now a vertex $x\in C\setminus N$, we see that $\frac{1}{2}=(\hat{w},\hat{x})=(p,\hat{x})=(p^{\prime},\hat{x})=-\frac{1}{2}(\sum_{v\in N}\hat{v},\hat{x})=-\frac{1}{2}(-m)$, where $m$ is the numbers of vertices of $N$ adjacent to $x$. This shows that $m=1$ for every $x\in C\setminus N$, and hence $N$ is evenly spaced in $C$.

Conversely, if $N$ is evenly spaced then $(p^{\prime},\hat{x})=\frac{1}{2}=(\hat{w},\hat{x})$ for every $x\in C\setminus N$ and $(p^{\prime},\hat{x})=-\frac{1}{2}(\hat{x},\hat{x})=-\frac{1}{2}\cdot 2=-1=(\hat{w},\hat{x})$ for every $x\in N$. Hence $p=p^{\prime}$. ∎

We now assume that $k=1$ and focus on the cycle $C=C_{i}$ containing the only component of $M$ of size $2$. Without loss of generality, we may assume that $C=v_{1}v_{2}\cdots v_{t}$ and $v_{1}v_{2}$ is the size $2$ component of $N=C\cap M$.

Clearly, $t\geq 6$ and it follows from Lemma 3.1 that $t\leq 15$, since $\dim V\leq 18$. Also, it follows from Lemma 4.2 that the length $(p,p)$ of the projection of $\hat{w}$ onto the subspace $U$ corresponding to $C$ is at most $5-\frac{1}{2}(7-s)=\frac{s+3}{2}$.

Lemma 4.3**.**

If $w$ is adjacent to $v_{1}$ and $v_{2}$ then $\hat{w}\in V$, $p=-(\hat{v}_{1}+\hat{v}_{2})+\sum_{m=1}^{s-1}\frac{1}{2}(\hat{v}_{3m+1}+\hat{v}_{3m+2})$, and $(p,p)=\frac{s+3}{2}$. Furthermore, the subgraph $N_{j}=C_{j}\cap M$ is evenly spaced in $C_{j}$ for each $j\neq i$.

Proof.

The set of $t-4=3s-4$ vertices $\{v_{4},v_{5},\ldots,v_{t-1}\}$ consists of $s-2$ vertices from $N$ (neighbours of $w$) and $(3s-4)-(s-2)=2(s-1)$ other vertices (non-neighbours). Let us view the $s-2$ neighbours as dividers splitting the non-neighbours into $s-1$ continuous parts $R_{j}$ of lengths $r_{1},r_{2},\ldots,r_{s-1}$. Let $d=\hat{v}_{1}+\hat{v}_{2}$ and, for $j=1,\ldots,s-1$, we set $d_{j}=\sum_{v\in R_{j}}\hat{v}$. We let $U^{\prime}$ be the subspace of $U$ spanned by $d,d_{1},\ldots,d_{s-1}$ and let $p^{\prime}$ be the projection of $\hat{w}$ onto $U^{\prime}$. Then, clearly, $(p,p)\geq(p^{\prime},p^{\prime})$ and, since the vectors $d,d_{1},\ldots,d_{s-1}$ are pairwise orthogonal and of length $2$, we have that $p^{\prime}=-d+\frac{1}{4}\sum_{j=1}^{s-1}r_{j}d_{j}$, which means that $(p^{\prime},p^{\prime})=2+\frac{1}{8}\sum_{j=1}^{s-1}r_{j}^{2}$. Hence, to find the minimum of the latter, we need to minimize $\sum_{j=1}^{s-1}r_{j}^{2}$ under the restriction that $\sum_{j=1}^{s-1}r_{j}=2(s-1)$. Clearly, the minimum is achieved when all $r_{j}$ are equal, that is when all $r_{j}$ are equal to $\frac{2(s-1)}{s-1}=2$. The minimum value $(p^{\prime},p^{\prime})$ is, therefore, $2+\frac{1}{8}(s-1)2^{2}=2+\frac{s-1}{2}=\frac{s+3}{2}$.

Hence $\frac{s+3}{2}\geq(p,p)=(p^{\prime},p^{\prime})\geq\frac{s+3}{2}$. Clearly, this means that $p=p^{\prime}$ is of length $\frac{s+3}{2}$, and so every part $R_{j}$ is of size $2$, which leads to the vectors in the statement of the lemma. Also for every cycle $C_{m}$ other than $C$ we must have the minimum length value $\frac{s_{m}}{2}$ and so the vertices $C_{m}\cap M$ must be evenly spaced in $C_{m}$. ∎

Let us adopt the following terminology: the vectors $d=\hat{v}_{i}+\hat{v}_{i+1}$ will be called pairs, while the edge $v_{i}v_{i+1}$ will be called the base of the pair $d$. Using these terms, $p$ in the lemma above is the sum of the unique minus-pair $-(\hat{v}_{1}+\hat{v}_{2})$ and $s-1$ half-pairs $\frac{1}{2}(\hat{v}_{3m+1}+\hat{v}_{3m+2})$.

Lemma 4.4**.**

Every cycle $C_{i}$ in $G_{1}(u)$ has length $3$.

Proof.

Suppose, by contradiction, that $C=C_{i}=v_{1}v_{2}\ldots v_{t}$ has length $t\geq 6$. In $G_{1}(v_{1})$, $u$ is adjacent to $v_{t}$ and $v_{2}$, which are not adjacent to each other. Hence $v_{t}uv_{2}$ is part of a cycle $D$ in $G_{1}(v_{1})$ of length at least $6$. Let $w\neq u$ be the second neighbour of $v_{2}$ in $D$, and let $w^{\prime}\neq v_{2}$ be the second neighbour of $w$ in $D$. Note that $w$ is adjacent to $v_{1}$ and $v_{2}$, and hence $\hat{w}$ is as in Lemma 4.3. In particular, $p=-(\hat{v}_{1}+\hat{v}_{2})+\sum_{m=1}^{s-1}\frac{1}{2}(\hat{v}_{3m+1}+\hat{v}_{3m+2})$ is the projection of $\hat{w}$ to the subspace $U=V_{i}$.

We obtain a contradiction by computing $(\hat{w},\hat{w}^{\prime})$. Since $w$ and $w^{\prime}$ are adjacent vertices in $G_{2}(u)$ (note that $w$ and $w^{\prime}$ are not adjacent to $u$, since $D$ has length at least $6$), the value of the inner product must be $-\frac{7}{4}$. On the other hand, we can estimate the value as follows. Recall that $\hat{w}\in V$ by Lemma 4.3 and so $\hat{w}=\sum_{j=1}^{r}p_{j}$, where $p_{j}$ is the projection of $\hat{w}$ to the subspace $V_{j}$ corresponding to the cycle $C_{j}$ in $G_{1}(u)$. We already know $p=p_{i}$ and, by Lemmas 4.3 and 4.2, if $j\neq i$ then $p_{j}=-\frac{1}{2}\sum_{v\in N_{j}}\hat{v}$, where $N_{j}=M\cap C_{j}$ is evenly spaced in $C_{j}$. It follows that $(\hat{w},\hat{w}^{\prime})=\sum_{j=1}^{r}(p_{j},\hat{w})$. Clearly, $w^{\prime}$ is adjacent to $v_{1}$ but not to $v_{2}$. Hence $(-(\hat{v}_{1}+\hat{v}_{2}),\hat{w}^{\prime})=-(-1+\frac{1}{2})=\frac{1}{2}$. Consider now a half-pair $\frac{1}{2}(\hat{v}_{3m+1}+\hat{v}_{3m+2})$. If $w^{\prime}$ is adjacent to both $v_{3m+1}$ and $v_{3m+2}$ then $\hat{w}^{\prime}$ is described as in Lemma 4.3 with the minus-pair base $-$ $v_{3m+1}v_{3m+2}$. This means, however, that $v_{1}v_{2}$ is the base of a half-pair for $w^{\prime}$. Hence $w^{\prime}$ cannot be adjacent to $v_{1}$, a contradiction. Therefore, $w^{\prime}$ is adjacent to at most one of $v_{3m+1}$ and $v_{3m+2}$. If $w^{\prime}$ is adjacent to one of these then $(\frac{1}{2}(\hat{v}_{3m+1}+\hat{v}_{3m+2}),\hat{w}^{\prime})=\frac{1}{2}(-1+\frac{1}{2})=-\frac{1}{4}$. If $w^{\prime}$ is adjacent to neither of them then $(\frac{1}{2}(\hat{v}_{3m+1}+\hat{v}_{3m+2}),\hat{w}^{\prime})=\frac{1}{2}(\frac{1}{2}+\frac{1}{2})=\frac{1}{2}$. Hence the smallest possible value of $(p,\hat{w}^{\prime})$ is $\frac{1}{2}+(s-1)(-\frac{1}{4})=\frac{3}{4}-\frac{s}{4}$. In all $C_{j}\neq C$, $w$ is adjacent to $7-s$ vertices $v$, and for each such vertex, $\hat{v}$ appears in $\hat{w}$ with coefficient $-\frac{1}{2}$. If $w^{\prime}$ is adjacent to $v$ then this gives contribution $-\frac{1}{2}(-1)=\frac{1}{2}$ to the value of $(\hat{w},\hat{w}^{\prime})$. If $w^{\prime}$ and $v$ are not adjacent then the contribution is $-\frac{1}{2}\frac{1}{2}=-\frac{1}{4}$. Hence the smallest possible contribution from all vectors $\hat{v}$ appearing in $\hat{w}$, where $v\not\in C$, is $(7-s)(-\frac{1}{4})=-\frac{7}{4}+\frac{s}{4}$. Putting all of the above together, we conclude that $(\hat{w},\hat{w}^{\prime})\geq(\frac{3}{4}-\frac{s}{4})+(-\frac{7}{4}+\frac{s}{4})=-1$. This clearly is a contradiction since $(\hat{w},\hat{w}^{\prime})=-\frac{7}{4}$. ∎

5. Contradiction

Vertices and $4$-cliques of $G$ form a point-line geometry. It follows from Lemma 4.4 that every point lies in seven lines and then, using the parameters of $G$, it is easy to deduce that this geometry is a generalized quadrangle of order $(3,6)$, which cannot exist due to a theorem of Dixmier and Zara [2]. However, with the wealth of information that we have collected, we can achieve a quick contradiction not using [2].

Let $T=S^{\perp}$, where $S$ is the span of the vectors in $\{u\}\cup G_{1}(u)$. That is, $S=\langle u\rangle\oplus V$. Since all cycles in $G_{1}(u)$ are of length $3$, Lemma 3.1 shows that $\dim S=1+7\cdot 2=15$, and so $\dim T=4$.

If $w\in G_{2}(u)$ then the projection of $\hat{w}$ onto $V$ coincides with $-\frac{1}{2}\sum_{v\in M}\hat{v}$ and it has length $\frac{7}{2}$. It follows that the projection of $\hat{w}$ onto $T$ has length $5-\frac{7}{2}=\frac{3}{2}$. Let $w^{\circ}$ denote $\frac{2}{\sqrt{3}}$ times the projection of $\hat{w}$ onto $T$. Then $(w^{\circ},w^{\circ})=2$. We now compute $(w^{\circ},(w^{\prime})^{\circ})$ for distinct $w,w^{\prime}\in G_{2}(u)$.

If $w$ and $w^{\prime}$ are adjacent then $(\hat{w},\hat{w}^{\prime})=-\frac{7}{4}$. Note that the edge $ww^{\prime}$ lies in a unique $4$-clique and so $w$ and $w^{\prime}$ have a unique common neighbour in $G_{1}(u)$. It follows that if $p$ and $p^{\prime}$ are the projections of $\hat{w}$ and $\hat{w}^{\prime}$ onto $V$ then $(p,p^{\prime})=\frac{1}{2}+6(-\frac{1}{4})=-1$. Hence $(w^{\circ},(w^{\prime})^{\circ})=\frac{4}{3}(-\frac{7}{4}+1)=-1$.

If $w$ and $w^{\prime}$ are non adjacent then $(\hat{w},\hat{w}^{\prime})=\frac{1}{2}$. Let $\beta$ the the number of common neighbours of $w$ and $w^{\prime}$ in $G_{1}(u)$. Then $(p,p^{\prime})=\beta\frac{1}{2}+(7-\beta)(-\frac{1}{4})=-\frac{7}{4}+\frac{3\beta}{4}$ and $(w^{\circ},(w^{\prime})^{\circ})=\frac{4}{3}(\frac{1}{2}-(-\frac{7}{4}+\frac{3\beta}{4}))=3-\beta$.

To summarize, if $w,w^{\prime}\in G_{2}(u)$ and $\beta=|G_{1}(u)\cap G_{1}(w)\cap G_{1}(w^{\prime})|$ then

[TABLE]

Clearly, it follows that $1\leq\beta\leq 5$.

Notice that the above values of inner products mean that all vectors $w^{\circ}$, $w\in G_{2}(u)$ are contained in a root system in $T$. Indeed, since all values are integers, the vectors $w^{\circ}$ span an integral lattice and all vectors of length $2$ from that lattice form a root system.

The largest root system in dimension $4$ is $D_{4}$ having $24$ vectors splitting into $12$ pairs of opposite roots. Since $|G_{2}(u)|=54>4\cdot 12$, we must have five vertices $\{w_{1},\ldots,w_{5}\}$ such that all vectors $(w_{i})^{\circ}$ belong to the same pair of opposite roots.

Lemma 5.1**.**

There is no strongly regular graph with parameters $(76,21,2,7)$.

Proof.

Consider the five vertices $\{w_{1},\ldots,w_{5}\}$ such that all vectors $(w_{i})^{\circ}$ are in the same pair of opposite roots $\{r,-r\}$. Without loss of generality, let the first $s\geq 3$ of the vectors $(w_{i})^{\circ}$ be $r$ and the remaining $5-s$ be $-r$.

From the calculations above, the vertices $w_{i}$ are pairwise non-adjacent. Furthermore, if $(w_{i})^{\circ}=(w_{j})^{\circ}$ then $w$ and $w^{\prime}$ have exactly one common neighbour in $G_{1}(u)$, and if $(w_{i})^{\circ}=-(w_{j})^{\circ}$ then $w_{i}$ and $w_{j}$ have exactly five common neighbours in $G_{1}(u)$.

If $s\neq 5$ then $(w_{5})^{\circ}=-r$ and so both $w_{1}$ and $w_{2}$ must have five neighbours among the seven vertices from $M=G_{1}(u)\cap G_{1}(w_{5})$. However, this means that $w_{1}$ and $w_{2}$ have at least two common neighbours in $M$; a contradiction. Therefore, $s=5$ and any two vectors $w_{i}$ and $w_{j}$ share a unique common neighbour in $G_{1}(u)$.

For the final contradiction, note that there are at most three $3$-cycles in $G_{1}(u)$ where $w_{1}$, $w_{2}$, and $w_{3}$ may have common neighbours. It follows that there are at least four $3$-cycles $C$ where $w_{1}$, $w_{2}$ and $w_{3}$ are adjacent to the three distinct vertices of $C$. This means that, in each of these $3$-cycles $C$, the vertex $w_{4}$ would have the same neighbour as one of the vertices $w_{1}$, $w_{2}$, and $w_{3}$. Clearly, this means that $w_{4}$ must share at least two common neighbours with one of the vectors $w_{1}$, $w_{2}$, or $w_{3}$; a contradiction. ∎

Acknowledgement

The authors are grateful to King Fahd University of Petroleum and Minerals for supporting this research.