Block Designs and Finite Projective Planes — Examples

Each displayed line has three points.

Let $T:=\{0,1,3\}$. Its nonzero differences are $\pm 1,\pm 2,\pm 3$, so every nonzero residue modulo $7$ occurs exactly once as $t_2-t_1$ with $t_1,t_2\in T$. Therefore any two distinct points of $\mathbb{Z}\mathrm{/}7$ lie in exactly one translate of $T$. By [L1], the seven triples form a Steiner triple system on seven points.

The same difference calculation shows that any two distinct displayed lines meet in exactly one point. The four points $0,1,2,5$ contain no displayed line, so no three of them are collinear. Thus the same seven triples, read as lines, satisfy [L2]; every line has three points, so they form a finite projective plane of order $2$. This projective plane is the Fano plane.

Each row of $N$ has three ones, because each point lies on three of the seven displayed lines. Hence every diagonal entry of $N{N}^{\mathsf{T}}$ is $3$.

Distinct rows of $N$ have dot product $1$, because [L1] says that any two distinct points lie on exactly one common line. Hence every off-diagonal entry of $N{N}^{\mathsf{T}}$ is $1$.

Therefore $N{N}^{\mathsf{T}}$ has diagonal entries $3$ and off-diagonal entries $1$, so $N{N}^{\mathsf{T}}=2I_7+J_7$. In particular this example realizes equality in Fisher's inequality, since the plane has $7$ points and $7$ lines.

The first three displayed blocks are the vertical blocks $V_x$, and the remaining nine are the blocks $B_{x,y,i}$ for the three unordered pairs $\{0,1\}$, $\{0,2\}$, and $\{1,2\}$ and the three layers $i\in \mathbb{Z}\mathrm{/}3$.

Theorem Bose's construction gives a Steiner triple system of order $6m+3$ for $m\ge 1$ therefore applies verbatim and shows that these twelve blocks form a Steiner triple system on nine points.

Applying [L2] with $n=3$ gives $3^2+3+1=13$ points and $13$ lines.

The line $⟨e_1,e_2⟩$ consists of the one-dimensional subspaces represented by $(1,0,0)$, $(0,1,0)$, $(1,1,0)$, and $(1,2,0)$, so it has $4$ projective points, exactly as order $3$ predicts.

Reading the two arrays entrywise produces the nine ordered pairs $(0,0),(1,1),(2,2),(1,2),(2,0),(0,1),(2,1),(0,2),(1,0)$, which are exactly the elements of ${\mathbb{F}}_3^2$ without repetition.

Therefore the two arrays are orthogonal. This is the $q=3$ instance of The linear Latin squares $L_a(i,j)=ai+j$ over ${\mathbb{F}}_q$ are pairwise orthogonal.

Let $\mathcal{C}$ be the family of complements of the seven Fano lines inside the seven-point set $\mathbb{Z}\mathrm{/}7$. Each member of $\mathcal{C}$ has size $4$, and there are seven of them.

Fix distinct points $a,b\in \mathbb{Z}\mathrm{/}7$. Among the seven displayed Fano lines, exactly one contains both $a$ and $b$, and inspection of the list shows that each of $a$ and $b$ lies on exactly three lines. So exactly five lines meet $\{a,b\}$, and therefore exactly two Fano lines avoid both points. Hence exactly two members of $\mathcal{C}$ contain $\{a,b\}$, so $\mathcal{C}$ is a $2$-$(7,4,2)$ design.

Since $\mathcal{C}$ has seven blocks on seven points, it is symmetric. If it were a projective plane, then [L1] would make it a symmetric $2$-$(7,4,1)$ design, contradicting step 1.2. Therefore this symmetric design is not a projective plane.

The addition table of ${\mathbb{F}}_3$ is a Latin square, so using it twice gives two Latin squares of the same order.

In the paired array, every cell has the form $(u,u)$ because the same Latin square is used twice. So the ordered pair $(0,1)$ never appears, and [L1] shows that the two squares are not orthogonal.