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A Steiner triple system is a decomposition of $K_n$ into $K_3$, such as $S=\{013,026,045,124,156,235,346\}$. Steiner triple systems give rise to a Steiner Latin squares, such as $L$ below.

\[L=\left(\begin{matrix} 0 & 3 & 6 & 1 & \bf{5} & 4 & 2 \\\\ 3 & 1 & 4 & 0 & 2 & \bf{6} & 5 \\\\ 6 & 4 & 2 & 5 & 1 & 3 & \bf{0} \\\\ \bf{1} & 0 & 5 & 3 & 6 & 2 & 4 \\\\ 5 & \bf{2} & 1 & 6 & 4 & 0 & 3 \\\\ 4 & 6 & \bf{3} & 2 & 0 & 5 & 1 \\\\ 2 & 5 & 0 & \bf{4} & 3 & 1 & 6 \end{matrix}\right)\]

We define $L=(l_{ij})$ by $l_{ii}=i$ for all $i$ and $l_{ij}=k$ whenever $ijk$ is a triangle in $S$.

Note: Typically, Steiner Latin squares are viewed in an algebraic context and referred to as "Steiner quasigroups" -- Steiner quasigroups correspond to isomorphism classes of Steiner Latin squares, whereas for this question, I'm interested in the "labelled" case.

In some instances, such as $L$ above, the Latin square obtained is a diagonally-cyclic Latin square. That is, $L$ satisfies the identity $l_{(i+1)(j+1)}=l_{ij}+1 \pmod n$, for all $i,j \in \mathbb{Z}_n$, where the indices are taken modulo $n$ also. I've highlighted (in bold) an orbit of an entry of $L$ under this symmetry.

Which Steiner triple systems give rise to diagonally-cyclic Steiner Latin squares?

The above question was my original question, however, as Douglas Zare points out, these are precisely the Steiner triple systems that admit the automorphism $(0,1,\ldots,n-1)$.

Proof: If $L$ is a Steiner Latin square derived from $S$ then $(i,j,k)$ is an entry in $L$ if and only if it $ijk$ is an element of $S$. For $L$ to be diagonally-cyclic, if $(i,j,k)$ is an entry of $L$ then so is $(i+1,j+1,k+1) \pmod n$. Therefore $(i+1)(j+1)(k+1)$ is also an element of $S$. The converse is true by definition.

Since that was such an easy task, lets look at a (hopefully) more interesting question.

Let $L=(l_{ij})$ be a diagonally-cyclic Steiner Latin square. Let $\sigma$ be the permutation defined by $\sigma(j)=l_{0j}$ (that is, the first row of $L$). Then $\sigma$ is an orthomorphism of $\mathbb{Z}_n$. That is, $\sigma$ is a permutation of $\mathbb{Z}_n$ and the map defined by $i \mapsto \sigma(i)-i \pmod n$ is also a permutation of $\mathbb{Z}_n$.

In the above example $\sigma=(0)(13)(26)(45)$.

Which orthomorphisms of $\mathbb{Z}_n$ arise from diagonally-cyclic Steiner Latin squares?

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1 Answer 1

The triple 0ab is equal to 0ba. If 0ab occurs in triple system, then in the corresponding Steiner Latin square the (0,a) entry is b and the (0,b) entry is a. The permutation corresponding the 0 row must have the transposition $(a\ b)$. In fact, row $j$ written a permutation must be a product of disjoint transpositions (a.k.a an involution) that fixes only $j$.

If a Latin square admits an orthomorphism $\sigma$ in one row, then it admits an orthomorphism in every row. This follows from every row being a permutation of the first row and so the differences $\sigma(i)-i$ are all permuted, too.

Let $\rho$ be the permutation $(1\ 2\ \dots\ n)$. In a diagonally-cyclic Steiner Latin square, let the permutation $\pi_1$ correspond to row $1$ and then row j corresponds to the permutation be $\rho^j(\pi_1)$. If $\pi_1$ is an orthomorphism corresponding to the first row of a diagonally-cyclic Latin square, then $\rho^j(\pi_1)$ is required to be a orthomorphism, too.

If $\pi_1$ is an involutary orthomorphism corresponding to the first row of a diagonally-cyclic Steiner Latin square, then $\rho^j(\pi_1)$ must be an involutary orthomorphism with a single fixed point. So these are the necessary conditions for an orthomorphism to arise from a diagonally-cyclic Steiner Latin square. The question now becomes "Are they sufficient?". I suspect the answer is no. You need to find some starter cycles to produce a triple system. Maybe looking at the Latin subsquares will help.

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