Sufficient conditions for STS$(3^k)$ of 3-rank $\leq 3^k-r$ to be resolvable

TL;DR

This paper establishes sufficient conditions based on the structure and orthogonal Latin squares for the resolvability of certain Steiner triple systems with specific 3-rank constraints, enabling their partition into parallel classes.

Contribution

It provides new sufficient conditions for the resolvability of $STS(3^k)$ with limited 3-rank, linking their structure to orthogonal Latin squares and partitioning into parallel classes.

Findings

Conditions for resolvability of $STS(3^k)$ are established.

Block sets can be partitioned into parallel classes under these conditions.

Resolvability is proved when conditions are satisfied.

Abstract

Based on the structure of non-full-$3$-rank $STS(3^k)$ and the orthogonal Latin squares, we mainly give sufficient conditions for $STS(3^k)$ of $3$-rank $\leq 3^k-r$ to be resolvable in the present paper. Under the conditions, the block set of $STS(3^k)$ can be partitioned into $\frac{3^k-1}{2}$ parallel classes, i.e., $\frac{3^k-1}{2}$ $1$-$(v,3,1)$ designs. Finally, we prove that $STS(3^k)$ of 3-rank $\leq 3^k-r$ is resolvable under the sufficient conditions.

Tables

1	2	3
2	3	1
3	1	2

$1\alpha$	$2\beta$	$3\gamma$	$4\delta$	$5ϵ$	$6\epsilon$	$7\zeta$	$8\theta$	$9\eta$
$2\gamma$	$3\alpha$	$1\beta$	$5\epsilon$	$6\delta$	$4ϵ$	$8\eta$	$9\zeta$	$7\theta$
$3\beta$	$1\gamma$	$2\alpha$	$6ϵ$	$4\epsilon$	$5\delta$	$9\theta$	$7\eta$	$8\zeta$
$4\zeta$	$5\eta$	$6\theta$	$7\beta$	$8\alpha$	$9\gamma$	$1\delta$	$2ϵ$	$3\epsilon$
$5\theta$	$6\zeta$	$4\eta$	$8\gamma$	$9\beta$	$7\alpha$	$2\epsilon$	$3\delta$	$1ϵ$
$6\eta$	$4\theta$	$5\zeta$	$9\alpha$	$7\gamma$	$8\beta$	$3ϵ$	$1\epsilon$	$2\delta$
$7\delta$	$8ϵ$	$9\epsilon$	$1\zeta$	$2\theta$	$3\eta$	$4\alpha$	$5\beta$	$6\gamma$
$8\epsilon$	$9\delta$	$7ϵ$	$2\eta$	$3\zeta$	$1\theta$	$5\gamma$	$6\alpha$	$4\beta$
$9ϵ$	$7\epsilon$	$8\delta$	$3\theta$	$1\eta$	$2\zeta$	$6\beta$	$4\gamma$	$5\alpha$

Equations

$$\bigcup _{i,j,k}LS(H_i,H_j,H_k)=\bigcup _{l=1}^{\frac{g}{3^{r-2}}}\bigcup _{(i,j,k)\in J_l}LS(H_i,H_j,H_k),$$

$$(k,t)\ge \frac{{\widetilde{N_1}(T)}^M\cdot {\widetilde{N_3}(T)}^{M(M-1)/6}}{(T!{)}^M\cdot |AGL(k-t,3)|}-\frac{{N_1(T^{\prime })}^{M^{\prime }}\cdot {N_3(T^{\prime })}^{M^{\prime }(M^{\prime }-1)/6}}{((T^{\prime })!{)}^{M^{\prime }-k+t-2}},$$

Full text

Sufficient conditions for STS$(3^{k})$ of 3-rank $\leq 3^{k}-r$ to be resolvable

Yaqi Lu, Minjia Shi Yaqi Lu and Minjia Shi, School of Mathematical Sciences, Anhui University, Hefei, Anhui, 230601, China, smjwcl.good@163.com, LyqSunshine8@163.com.

Abstract: Based on the structure of non-full-$3$-rank $STS(3^{k})$ and the orthogonal Latin squares, we mainly give sufficient conditions for $STS(3^{k})$ of $3$-rank $\leq 3^{k}-r$ to be resolvable in the present paper. Under the conditions, the block set of $STS(3^{k})$ can be partitioned into $\frac{3^{k}-1}{2}$ parallel classes, i.e., $\frac{3^{k}-1}{2}$ $1$-$(v,3,1)$ designs. Finally, we prove that $STS(3^{k})$ of 3-rank $\leq 3^{k}-r$ is resolvable under the sufficient conditions.

Keywords: Steiner triple system; Latin squares; Parallel classes; Resolvable.

MSC (2010) : 05B15.

1 Introduction

A Steiner triple system ($STS$ in short) is a pair $(\mathcal{P},\mathcal{B})$ [5], where $\mathcal{P}$ is a set of $v$ elements, called points, and $\mathcal{B}$ is a collection of distinct subsets of $\mathcal{P}$ of size $3$, called blocks, such that every subset of points of size $2$ is contained in precisely $1$ block. In fact, a $STS$ is a $2$-$(v,3,1)$ design.

Whether before or now, many scholars are interested in $STS(v)$’s structure and solvability, and they obtained some results when $v$ takes a smaller value, such as $STS(7)$, $STS(9)$, $STS(31)$ and so on. In [1] and [2], Bryant and Horsley already gave Steiner triple systems without parallel classes, an infinite family of Steiner triple systems without parallel classes and the solvability of $STS(7)$, $STS(9)$, $STS(13)$, $STS(15)$, $STS(19)$ (see also in [9]) and $STS(21)$. In 1999, Lam et al., studied cyclically resolvable cyclic Steiner triple systems of order $21$ and $39$ [10]. Jungnickel et al. in [6] gave the classification of Steiner triple systems on $27$ points with $3$-rank $24$. On the other hand, the existence of resolvable Steiner Quadruple Systems was also studied in [4]. Recently, Jungnickel et al. [7] gave the structure of non-full-$3$-rank STS. Moreover, Kirkman [8] proved that there exists a Steiner triple system of order $v$ if and only if $v\equiv 1$ or $3$ mod $6$ (see [3]) and $3^{k}\equiv 3$ mod $6$, that is, there exist $STS(3^{k})$. Therefore, the existence of STS$(3^{k})$ is guaranteed.

Motivated by the work listed above, a natural question is that: for any positive integer $k$, when $STS(3^{k})$ is resolvable ? This paper is devoted to giving sufficient conditions for the solvability of $STS(3^{k})$.

The material is organised as follows. The next section contains the preliminaries of $STS$, which are necessary for the rest of the paper. In Section 3, based on the theorem of the structure of non-full-3-rank $STS$, we prove that $STS$ $(3^{k})$ is resolvable under some conditions and find the sufficient conditions for its solvability.

2 Preliminaries

Definition 2.1**.**

A $t$-$(v,k,\lambda)$ design, or briefly a $t$-design, is a pair $(\mathcal{P},\mathcal{B})$, where $\mathcal{P}$ is a set of $v$ elements, called points, and $\mathcal{B}$ is a collection of distinct subsets of $\mathcal{P}$ of size $k$, called blocks, such that every subset of points of size $t$ is contained in precisely $\lambda$ blocks.

A $STS$ is a pair $(\mathcal{P},\mathcal{B})$ where $\mathcal{P}$ is a set of $v$ elements, called points, and $\mathcal{B}$ is a collection of distinct subsets of $\mathcal{P}$ of size $3$, called blocks, such that every subset of points of size $2$ is contained in precisely $1$ block.

Lemma 2.2**.**

(Theorem 8.1.3 in [5]) Let $(\mathcal{P},\mathcal{B})$ be a $t$-$(v,k,\lambda)$ design. Let $0\leq i\leq t$. Then $(\mathcal{P},\mathcal{B})$ is an $i$-$(v,k,\lambda_{i})$ design, where \lambda_{i}=\lambda\frac{\left(\begin{matrix}v-i$\\$t-i\end{matrix}\right)}{\left(\begin{matrix}k-i$\\$t-i\end{matrix}\right)}=\lambda\frac{(v-i)(v-i-1)\cdots(v-t+1)}{(k-i)(k-i-1)\cdots(k-t+1)}.

By Lemma 2.2, we know that $STS(v)$ is a $1$-$(v,3,\frac{v-1}{2})$ design.

Definition 2.3**.**

$P$* is called a parallel class if $P\subset\mathcal{B}$ and $(S,P)$ is a $1$-$(v,3,1)$ design.*

Definition 2.4**.**

$STS(v)$* is called resolvable if $\mathcal{B}$ can be partitioned into parallel classes.*

From Definitions 2.3 and 2.4, we know that $STS(v)$ is resolvable if and only if $STS(v)$ can be partitioned into $\frac{v-1}{2}$ $1$-$(v,3,1)$ designs.

Definition 2.5**.**

A Latin square is an $n\times n$ array filled with $n$ different symbols, each occurring exactly once in each row and exactly once in each column.

Let $A$, $B$, $C$ be three sets. From the definition of Latin square, we know that a Latin square which is denoted by $LS(A,B,C)$ is a set of triples $\{(x,y,z)|(x,y,z)\in A\times B\times C\}$ such that $\forall x\in A$, $y\in B$, there eixsts unique $z\in C$, $(x,y,z)\in LS(A,B,C)$. Similarly, $\forall x\in A$, $z\in C$, there eixsts unique $y\in B$, $(x,y,z)\in LS(A,B,C)$, $\forall y\in B$, $z\in C$, there eixsts unique $x\in A$, $(x,y,z)\in LS(A,B,C)$.

3 $STS(3^{k})$ of $3$-rank $\leq 3^{k}-r$

In this section, $v=3^{k}$, that is, there exist $STS(v)$ of $3$-rank $\leq 3^{k}-r$ for arbitrary $r$. By $V^{v}$, we denote the vector space of all $v$-tuples over $\mathbb{F}_{3}$. Denote by $\mathcal{D}$ the set of subspaces of $V^{v}$, each including the all-one vector and being orthogonal to at least one $STS(v)$; denote $\mathcal{D}_{j}=\{D\in\mathcal{D}:dim(D)=j+1\}$.

Lemma 3.1**.**

*(Lemma $3.8$, [7]) Let $S$ be an arbitary Steiner Triple system $STS(N)$ contained in the triple system $\mathcal{D}$. Then the block set $\mathcal{B}$ of $\mathcal{S}$ splits as follows:

(1) for all $i=1,\ldots,M$, a set $\mathcal{B}_{i}$ of $T(T-1)/6$ blocks such that $(\mathcal{G}_{i},\mathcal{B}_{i})$ is a Steiner triple systems $\mathcal{S}_{i}$ on the $T$ points in $\mathcal{G}_{i}$;

(2) for each line $l$ of the affine geometry $\sum=AG(k-t,3)$, a set $\mathcal{B}_{l}$ of $T^{2}=3^{2t}$ blocks forming a transversal design $TD[3;T]$ on the three groups determined by the points of $l$.*

Proposition 3.2**.**

STS$(9)$ of $3$-rank $\leq 7$ is resolvable.

Proof.

By Lemma 3.1, given a subspace $D$ from $\mathcal{D}_{1}$, the set of $STS(9)$ orthogonal to D is in one-to-one correspondence with the collections of $3$ Steiner triple systems of order $3$ and $1$ Latin square of order $3$. A generator matrix for a subspace $D$ from $\mathcal{D}_{1}$ is

\left[\begin{array}[]{ccccccccc}1&1&1&1&1&1&1&1&1\\ 0&0&0&1&1&1&2&2&2\\ \end{array}\right].

Let the Latin square of order $3$ be

Then the indence matrix of $STS(9)$ is shown as follow:

I_{9}=\left[\begin{array}[]{ccccccccc}1&1&1&0&0&0&0&0&0\\ 0&0&0&1&1&1&0&0&0\\ 0&0&0&0&0&0&1&1&1\\ 1&0&0&1&0&0&1&0&0\\ 1&0&0&0&1&0&0&1&0\\ 1&0&0&0&0&1&0&0&1\\ 0&1&0&1&0&0&0&1&0\\ 0&1&0&0&1&0&0&0&1\\ 0&1&0&0&0&1&1&0&0\\ 0&0&1&1&0&0&0&0&1\\ 0&0&1&0&1&0&1&0&0\\ 0&0&1&0&0&1&0&1&0\\ \end{array}\right].

Therefore, the nonzero positions of each row of $I_{9}$ are \mathcal{B}=\big{\{}\{1,2,3\},\{4,5,6\},\{7,8,9\},\{1,4,7\},\\ \{1,5,8\},\{1,6,9\},\{2,4,8\},\{2,5,9\},\{2,6,7\},\{3,4,9\},\{3,5,7\},\{3,6,8\}\big{\}}. Obviously, $STS(9)$’s block set $\mathcal{B}$ is partitioned into four parallel classes, i.e., {{1,2,3},{4,5,6},{7,8,9}},{{1,4,7},{2,5,

9},{3,6,8}}, {{1,5,8},{2,6,7},{3,4,9}},{{1,6,9},{2,4,8},{3,5,7}}. So $STS(9)$ of $3$-rank $\leq 7$ is resolvable. ∎

Proposition 3.3**.**

STS$(27)$ of $3$-rank $\leq 24$ is resolvable.

Proof.

The proof is similar to that of Proposition 3.2. ∎

In order to give sufficient conditions for solvability of $STS(3^{k})$, we first introduce the concept of orthogonal Latin square.

Definition 3.4**.**

A orthogonal Latin square of order $n$ over two sets $Q$ and $E$, each consisting of $n$ symbols, is an $n\times n$ arrangement of cells, each cell containing an ordered pair $(q,e)$, where $q$ is in $Q$ and $e$ is in $E$, such that every row and every column contains each element of $Q$ and each element of $E$ exactly once, and that no two cells contain the same ordered pair.

Remark 3.5**.**

Every of two orthogonal Latin squares of order $n$ includes $n$ parallel classes.

Proposition 3.6**.**

STS$(9)$ of $3$-rank $\leq 8$ is resolvable. Moreover, if there always exsits an orthogonal Latin square for all the Latin squares of order $9$ which appear in the decomposition of $STS(27)$, then STS$(27)$ of $3$-rank $\leq 25$ is resolvable.

Proof.

By Lemma 3.1, given a subspace $D$ from $\mathcal{D}_{1}$, the set of $STS(27)$ orthogonal to D is in one-to-one correspondence with the collections of $3$ $STS$ of order $9$ and $1$ Latin square of order $9$. Let the (orthogonal) Latin square of order $9$ be

The first part of the decomposition produces $4$ parallel classes, and the other part of the decomposition produces $9$ parallel classes since there exists an orthogonal Latin square for the Latin square of order $9$. It means that $STS(27)$’s block set $\mathcal{B}$ is partitioned into $13$ parallel classes. Therefore, $STS(27)$ of $3$-rank $\leq 25$ is resolvable. ∎

All in all, if the $STS(3^{k})$ is resolvable, then $\frac{3^{k}}{3\times 3^{k-r+1}}=3^{r-2}$ Latin squares can be considered as a group to produce $3^{k-r+1}$ parallel classes in $STS(3^{k})$ of $3$-rank $\leq 3^{k}-r$.

The following theorem is the main result in this paper.

Theorem 3.7**.**

Let $S$ be an $STS(3^{k})$ of $3$-rank at most $3^{k}-r$. If there always exsits an orthogonal Latin square for each Latin square of order $3^{k-r+1}$ that appears in the decomposition of $S$ and all STS$(3^{k-r+1})$ which appear in the decomposition of $S$ are resolvable, then $S$ is resolvable.

Proof.

By Lemma 3.1, given a subspace $D$ from $\mathcal{D}_{r-1}$, the set of $S$ orthogonal to D is in one-to-one correspondence with the collections of $3^{r-1}$ $S$ of order $3^{k-r+1}$ and $\frac{3^{r-1}(3^{r-1}-1)}{6}$ Latin squares of order $3^{k-r+1}$. Let $G_{1}=\{1,2,3,\cdots,3^{k-r+1}-1,3^{k-r+1}\}$, $G_{2}=\{3^{k-r+1}+1,3^{k-r+1}+2,3^{k-r+1}+3,\cdots,2\times 3^{k-r+1}-1,2\times 3^{k-r+1}\}$, $\ldots$, $G_{3^{r-1}}=\{3^{k}-3^{k-r+1}+1,3^{k}-3^{k-r+1}+2,3^{k}-3^{k-r+1}+3,\cdots,3^{k}\}$, the first part $3^{r-1}$ STS$(3^{k-r+1})$ can be denoted by $STS(G_{1})\cup STS(G_{2})\cup\cdots\cup STS(G_{3^{r-1}})$, this $1$-$(3^{k},3,\frac{3^{k-r+1}-1}{2})$ design is resolvable. For each $1\leq i\leq 3^{r-1}$, STS$(G_{i})={G_{i}}^{1}\cup{G_{i}}^{2}\cup{G_{i}}^{3}\cup{G_{i}}^{4}\cup\cdots\cup{G_{i}}^{\frac{3^{k-r+1}-1}{2}}$, then $\cup_{i}{G_{i}}^{1}$, $\cup_{i}{G_{i}}^{2}$, $\cdots$, $\cup_{i}{G_{i}}^{\frac{3^{k-r+1}-1}{2}-1}$, and $\cup_{i}{G_{i}}^{\frac{3^{k-r+1}-1}{2}}$ are parallel classes. So the first part of the decomposition produces $\frac{3^{k-r+1}-1}{2}$ parallel classes.

Let $C_{p}$ be $p$-th column of the check matrix of $S$, there exist $i,j,k$ such that $C_{i}+C_{j}+C_{k}=0$. Let $J_{l}=\{(i,j,k)\mid C_{i}+C_{j}+C_{k}=0\}$, it is easy to verify that $J_{l}$ is $S$. Thus $g=\frac{3^{r-1}(3^{r-1}-1)}{6}$ Latin squares of order $3^{k-r+1}$ can be denoted by

[TABLE]

$\cup_{(i,j,k)\in J_{l}}LS(H_{i},H_{j},H_{k})$ is $1$-$(v,3,3^{k-r+1})$ design and it produces $3^{k-r+1}$ parallel classes, the other part of the decomposition produces $\frac{3^{r-1}(3^{r-1}-1)}{6}/3^{r-2}\times 3^{k-r+1}=\frac{3^{k}-3^{k-r+1}}{2}$ parallel classes.

It means that the block of $S$ is partitioned into $\frac{3^{k}-1}{2}$ parallel classes. Therefore, we obtain that $S$ is resolvable. ∎

Corollary 3.8**.**

The number $(k,t)$ of isomorphism classes of resolvable Steiner triple systems on $3^{k}$ points with $3$-rank exactly $3^{k}-k-1+t$, where $2\leq t\leq k-1$, satisfies

[TABLE]

where $T=3^{t}$, $T^{\prime}=3^{t-1}$, $M=3^{k-t}$ and $M^{\prime}=3^{k-t+1}$; $N_{1}(T^{\prime})$ is the number of $STS(T^{\prime})$; $\widetilde{N_{1}}(T)$ is the number of resolvable $STS(T)$; $N_{3}(T^{\prime})$ is the number of Latin squares of order $T^{\prime}$; $\widetilde{N_{3}}(T^{\prime})$ is the number of Latin squares of order $T^{\prime}$ having an orthogonal mate.

Proof.

The proof is essentially the same as for Theorem $4.4$ in [7], with the only difference that we count only resolvable $STS$. ∎

4 Acknowledgement

This research is supported by National Natural Science Foundation of China (61672036) and Excellent Youth Foundation of Natural Science Foundation of Anhui Province (1808085J20).