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\hypersetup{pdftitle={Subplanes of order 3 and maximal arcs in projective planes of order 12},
  pdfauthor={Tejasvi Singh Tomar}}

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\begin{document}

\title{Subplanes of order 3 and maximal arcs in projective planes of order 12}
\author{Tejasvi Singh Tomar}
\renewcommand{\urladdrname}{{\itshape ORCID}}
\urladdr{\href{https://orcid.org/0000-0003-4668-3639}{https://orcid.org/0000-0003-4668-3639}}
\subjclass[2020]{51E15; 05B07, 05B15, 51E21}
\keywords{Projective plane of order 12, subplane, maximal arc, Kirkman triple system, covering set of spreads, mutually orthogonal Latin squares, SAT solving, formal verification}

\begin{abstract}
Whether a projective plane of order 12 exists is open. A subplane of order 3 of such a plane is a case of equality in Bruck's bound $n\ge m^2+m$; the 27 points on no line of the subplane then form a maximal arc of degree 3, and together with the subplane a maximal arc $K$ of degree 4. We show that every line of a subplane of order 3 contained in $K$ lies in at most two such subplanes, whereas in $\PG(3,3)$, which like $K$ has 40 points, a line lies in four planes; the proof uses Egan and Wanless's classification of triples of mutually orthogonal Latin squares of order 9. We also show that $\AG(3,3)$ is not the design of any maximal arc of degree 3 in a plane of order 12, and that a permutation of the points that maps a subplane of order 3 onto itself and each line meeting it onto a line cannot have prime order other than 3. The proofs are computer-assisted, and their SAT refutations are certified.
\end{abstract}

\maketitle

\section{Introduction}\label{sec:intro}

Projective planes of order $n$ exist for all prime powers $n$, and no plane of any other order is known. The Bruck--Ryser theorem \cite{BR49} excludes infinitely many orders, and a large computer search excluded order 10 \cite{Lam91}. The smallest open order is 12. Most of the work on order 12 concerns collineations. A series of papers by Janko and van Trung, by Horvati\'c-Baldasar, Kramer and Matuli\'c-Bedeni\'c, by Suetake, and by Akiyama, Suetake and Tanaka shows that a collineation group of a plane of order 12 has order 1, 2 or 3 \cite{AST23}. The introduction of \cite{AST19} surveys the earlier steps. Hall and Wilkinson \cite{HW84} studied the ternary code of such a plane.

This paper concerns subplanes. Bruck \cite{Bru55} proved that a proper subplane of order $m$ of a projective plane of order $n$ satisfies $n=m^2$ or $n\ge m^2+m$. For $n=12$ this leaves $m\in\{2,3\}$, and $m=3$ is a case of equality: $12=3^2+3$. Janko and van Trung \cite{JvT80} showed that a plane of order 12 has no collineation of order 13 that leaves a subplane of order 3 invariant. Their proof uses their classification of the Kirkman triple systems of order 27 with an automorphism of order 13 \cite{JvT81}, and they later excluded collineations of order 13 altogether \cite{JvT82a}. We do not assume any collineations.

Throughout, $\pi$ is a projective plane of order 12 and $\pi_0$ is a subplane of order 3. A point outside $\pi_0$ is \emph{free} if it lies on no line of $\pi_0$, and \emph{non-free} otherwise. A line that meets $\pi_0$ in exactly one point is a \emph{tangent}. We write $F$ for the set of free points and $K=\pi_0\cup F$.

A \emph{maximal arc of degree $d$} in a projective plane of order $n$ is a set of points that meets each line in $0$ or $d$ points. Such an arc has $(d-1)(n+1)+1$ points, and $d$ divides $n$ unless $d=n+1$. Denniston \cite{Den69} constructed maximal arcs of all degrees $d$ dividing $q$ in $\PG(2,q)$ for $q$ even, and Ball, Blokhuis and Mazzocca \cite{BBM97} proved that $\PG(2,q)$ has none with $1<d<q$ for $q$ odd. Bruck's counts in the case of equality \cite[Lemma 3.1]{Bru55} imply that $F$ has $27$ points and is a maximal arc of degree $3$, and that $K$ is a maximal arc of degree $4$ (Section~\ref{sec:structure}). By the construction of Thas \cite{Tha74} and Wallis \cite{Wal73}, $K$ gives a partial geometry $\mathrm{pg}(8,9,6)$ and its dual $\mathrm{pg}(9,8,6)$. Brouwer's tables of strongly regular graphs list both with a question mark \cite{BroTables}, so a plane of order 12 with a subplane of order 3 would settle these two entries.

Each tangent carries exactly three free points, and the free points with these triples form a Kirkman triple system of order $27$; the $13$ classes $T_P$ of its resolution are formed by the triples on the tangents at the points $P$ of $\pi_0$. The most symmetric candidate for this triple system is the affine space $\AG(3,3)$, with the classes $T_P$ as the classes of parallel lines. We call this the \emph{affine case}. Then $\pi_0$ plays the role of the plane at infinity, and $\pi$ extends the points and lines of $\PG(3,3)$.

For a line $L$ of $\pi_0$, let $\Gamma_L$ be the graph on $F$ in which two free points are adjacent if the line joining them meets $L$ in a point of $\pi_0$. In the affine case each $\Gamma_L$ is the disjoint union of three copies of $K_9$, formed by three parallel planes of $\AG(3,3)$. Conversely, if this holds for every line $L$, the free points form a Hall triple system, hence $\AG(3,3)$ by the theorem of Hall and Bruck \cite{Hal60,Bru58}, and we are in the affine case. A component of $\Gamma_L$ with $9$ points, together with the four points of $\pi_0$ on $L$, is a subplane of order $3$ contained in $K$. So $\Gamma_L\cong3K_9$ means that $L$ lies in four subplanes of order $3$ contained in $K$, just as a line of $\PG(3,3)$ lies in four planes. Our main result shows that in a plane of order 12 the number is at most two.

\begin{introthm}[Section~\ref{sec:reduction}]\label{thm:lines}
Let $\pi$ be a projective plane of order $12$ with a subplane $\pi_0$ of order $3$, and let $K$ be the union of $\pi_0$ and its free points. Then every line of a subplane of order $3$ contained in $K$ lies in at most two such subplanes, and $K$ contains at most $14$ subplanes of order $3$.
\end{introthm}

If $\Gamma_L\cong3K_9$, the $81$ tangents at the points of $\pi_0$ off $L$ form a transversal design $\mathrm{TD}(5,9)$, that is, three mutually orthogonal Latin squares of order $9$. We go through the classification of these designs by Egan and Wanless \cite{EW16} and show with a SAT solver that none of them extends to the configuration of free points and tangents that a plane would give. Together with a count showing that every component of $\Gamma_L$ has $9$, $18$ or $27$ points, this leaves two possibilities for $\Gamma_L$ in a plane: it is connected, or it is the disjoint union of $K_9$ and a connected graph on $18$ points.

In the affine case, the $117$ points off $K$ give a \emph{covering set of spreads} of $\PG(3,3)$, that is, a family of spreads such that every pair of skew lines lies in exactly one of them. Bruck \cite{Bru69,Bru73} asked whether the points and lines of $\PG(3,q)$ extend to a projective plane of order $q(q+1)$, and proposed to build such planes from covering sets of spreads. For $q\ge4$ the existence of covering sets of spreads is open, with a partial result for $q=4$ \cite{Pri01}. For $q=3$, a search of Hall and Roth \cite{HR84}, which omitted some cases, found no extension, and Prince \cite{Pri99} gave a complete proof by showing that $\PG(3,3)$ has no covering set of spreads; we recheck his theorem by a certified computation (Theorem~\ref{thm:spreads}). We strengthen the exclusion of the affine case in two ways.

\begin{introthm}[Sections~\ref{sec:affine} and~\ref{sec:agarc}]\label{thm:affinecase}
\begin{enumerate}[label=\textup{(\alph*)}]
\item In the affine case, already the incidences of the free points and of the non-free points on three lines of $\pi_0$ with the tangents are impossible, whether the three lines are concurrent or form a triangle.
\item In no projective plane of order $12$ is there a maximal arc of degree $3$ whose points and secants form $\AG(3,3)$.
\end{enumerate}
\end{introthm}

Since the free points in the affine case form an arc as in (b), each part gives a proof, independent of Prince's theorem, that the affine case does not occur.

A collineation of $\pi$ that maps $\pi_0$ onto itself has order $1$ or $3$ (Proposition~\ref{prop:stab}). We therefore consider a weaker kind of symmetry. The \emph{core} of $\pi$ relative to $\pi_0$ is the incidence structure formed by all points of $\pi$ and the $130$ lines that meet $\pi_0$; it omits the $27$ lines that miss $\pi_0$. Similarly, the core of a maximal arc consists of all points and the secants of the arc. An automorphism of a core is a permutation of the points that maps each line of the core onto a line of the core; for the core relative to $\pi_0$ we also require that it maps $\pi_0$ onto itself.

\begin{introthm}[Section~\ref{sec:sym}]\label{thm:cores}
Let $\pi$ be a projective plane of order $12$.
\begin{enumerate}[label=\textup{(\alph*)}]
\item The core of a maximal arc of degree $3$ or $4$ in $\pi$ has no automorphism of order $13$.
\item If $\pi_0$ is a subplane of order $3$, every automorphism of prime order of the core relative to $\pi_0$ has order $3$.
\end{enumerate}
\end{introthm}

A core automorphism need not act on the lines outside the core, so the results on collineations cited above do not apply to it. Section~\ref{sec:sym} also excludes automorphisms of order $13$ of the triple system that the secants cut out on a maximal arc of degree $3$, and of the triple system of the free points with its labelled resolution; for the latter we recount the classification of \cite{JvT81}, whose list has four repetitions, so that there are $657$ isomorphism classes rather than $661$ (Remark~\ref{rem:jvt}). The same section shows that an automorphism of order $3$ as in (b) induces on $\pi_0$ an elation or a collineation with exactly one fixed point, and only the latter if it fixes a free point.

Section~\ref{sec:open} states open problems, among them the conjecture that no Steiner triple system of order $27$ has a family of parallel classes covering every pair of disjoint triples exactly once; it would imply that no projective plane of order $12$ contains a maximal arc of degree $3$ or $4$, and hence no subplane of order $3$.

Several proofs depend on exhaustive computer searches \cite{Knu00,Kissat,CaDiCaL}. Their SAT refutations have LRAT certificates \cite{LRAT17}, checked by the verified checker of Lean~4 \cite{Lean4}; Section~\ref{sec:verification} describes what else is checked, including the formalization of the results proved by hand other than Corollary~\ref{cor:twoplanes} (the last step of Theorem~\ref{thm:lines}), and what is not formalized, notably the translation of the geometric statements into formulas and searches.

Section~\ref{sec:prelim} collects definitions, and Section~\ref{sec:structure} derives the structure described above; the proofs of Theorems~\ref{thm:lines}--\ref{thm:cores} are in the sections named in their headings.

\section{Preliminaries}\label{sec:prelim}

\subsection{Planes}
A projective plane of order $n$ has $n^2+n+1$ points and as many lines. Each line has $n+1$ points, each point lies on $n+1$ lines, two points lie on exactly one line, and two lines meet in exactly one point. A subplane of order $m$ is a set of points and lines that forms a projective plane of order $m$ under the induced incidence. For $n=12$ there are $157$ points and $157$ lines, each line has $13$ points, and each point lies on $13$ lines. Since the projective plane of order 3 is unique, every subplane of order 3 is isomorphic to $\PG(2,3)$ \cite{HP73}.

The group $\PGL(3,3)$ has order $5616=2^4\cdot3^3\cdot13$. We use three facts about it.
\begin{itemize}
\item Every involution of $\PGL(3,3)$ is a homology: it fixes a line pointwise and a point off that line. Indeed, a matrix $A$ with $A^2=-I$ does not exist over $\F_3$ in dimension 3, since $\det(A)^2=-1$ is not a square in $\F_3$. Hence $A^2=I$, and $A$ is diagonalizable with eigenvalues $\pm1$.
\item An element of order 3 is either an elation, which fixes a line (its axis) pointwise and all lines through a point of the axis (its center), or it fixes exactly one point and one line, which are incident.
\item The subgroups of order 13 are Sylow 13-subgroups and therefore conjugate. They are generated by Singer cycles, which act regularly on the points and on the lines.
\end{itemize}

\subsection{Maximal arcs}
Let $K$ be a maximal arc of degree $d$ in a plane of order $n$, with $1<d\le n$. Lines meeting $K$ are \emph{secants}, the others are \emph{external lines}. A point of $K$ lies on $(|K|-1)/(d-1)=n+1$ secants. A point off $K$ lies on $|K|/d$ secants. The \emph{design} of $K$ is the set $K$ with the secants restricted to $K$ as blocks. It is a Steiner system $S(2,d,|K|)$.

\subsection{Designs}
A Steiner system $S(2,k,v)$ is a set of $v$ points with blocks of size $k$ such that two points lie in exactly one block. A Steiner triple system $\mathrm{STS}(v)$ is an $S(2,3,v)$. A \emph{parallel class} is a set of blocks that partitions the points, and a \emph{resolution} is a partition of the blocks into parallel classes. A Kirkman triple system $\mathrm{KTS}(v)$ is an $\mathrm{STS}(v)$ with a resolution. An $\mathrm{STS}(27)$ has $117$ triples, a parallel class consists of $9$ triples, and a resolution has $13$ classes.

The affine space $\AG(3,3)$ is the $\mathrm{STS}(27)$ on $\F_3^3$ whose triples are the lines $\{p,p+d,p+2d\}$. Three distinct points form a line if and only if their sum is $0$. The lines with a given direction $\langle d\rangle$ form a parallel class, and the $13$ directions form the plane at infinity $\PG(2,3)$. The projective space $\PG(3,3)$ has $40$ points and $130$ lines. A \emph{spread} of $\PG(3,3)$ is a set of $10$ lines that partitions the points.

A Steiner triple system is a \emph{Hall triple system} if any three points not in a common triple lie in a subsystem isomorphic to $\AG(2,3)$, the $\mathrm{STS}(9)$. Hall \cite{Hal60} showed that Hall triple systems correspond to commutative Moufang loops of exponent 3. Such a loop of order at most 27 is an elementary abelian group, since the smallest non-associative one has order 81 \cite{Bru58}; see also \cite{BL85}. Hence a Hall triple system on 27 points is isomorphic to $\AG(3,3)$.

A \emph{partial geometry} $\mathrm{pg}(s,t,\alpha)$ is a set of points and lines such that each line has $s+1$ points, each point lies on $t+1$ lines, two points lie on at most one line, and for each point $x$ and line $\ell\not\ni x$ exactly $\alpha$ points of $\ell$ are collinear with $x$. Its collinearity graph is strongly regular with parameters $v=(s+1)(st+\alpha)/\alpha$, $k=s(t+1)$, $\lambda=s-1+t(\alpha-1)$ and $\mu=\alpha(t+1)$. The dual structure is a $\mathrm{pg}(t,s,\alpha)$.

\section{Structure}\label{sec:structure}

The structure described in this section follows from the counts in Bruck's equality case \cite[Lemma 3.1]{Bru55}; Dembowski \cite[pp.~76--77]{Dem58} gives the incidence numbers of the three classes of points and lines. We give the short proofs to fix notation. We fix $\pi$ and $\pi_0$ and call the $13$ lines of $\pi_0$ the \emph{$\pi_0$-lines}. A tangent that meets $\pi_0$ in $P$ is a \emph{tangent at $P$}, and a line that misses $\pi_0$ is \emph{external}.

\begin{theorem}\label{thm:incidence}
The points of $\pi$ are the $13$ points of $\pi_0$, $117$ non-free points ($9$ on each $\pi_0$-line $L$, forming the set $N_L$), and $27$ free points. The lines of $\pi$ are the $13$ $\pi_0$-lines, $117$ tangents ($9$ at each point of $\pi_0$), and $27$ external lines. Moreover:
\begin{enumerate}[label=\textup{(\roman*)}]
\item a tangent at $P$ carries $P$, one non-free point of each $\pi_0$-line not through $P$, and $3$ free points;
\item an external line carries exactly one non-free point of each $\pi_0$-line;
\item a non-free point $X\in N_L$ lies on $L$, on one tangent at each point of $\pi_0$ off $L$, and on $3$ external lines;
\item a free point lies on one tangent at each point of $\pi_0$, and on no other line.
\end{enumerate}
\end{theorem}

\begin{proof}
A point $Q$ on no $\pi_0$-line is joined to the $13$ points of $\pi_0$ by $13$ distinct lines, because a line through $Q$ with two points of $\pi_0$ would be a $\pi_0$-line. These are all the lines through $Q$, which proves (iv). A tangent at $P$ meets each of the $9$ $\pi_0$-lines not through $P$ in a point outside $\pi_0$, which is non-free, and these $9$ points are distinct, since two $\pi_0$-lines meet in a point of $\pi_0$. The remaining $13-1-9=3$ points lie on no $\pi_0$-line, which proves (i). A non-free point $X\in N_L$ lies on $L$ and on the $9$ lines $XP$ with $P\notin L$. These lines are distinct tangents, since a line through $X$ with two points of $\pi_0$ is a $\pi_0$-line and hence equal to $L$. The remaining $3$ lines through $X$ miss $\pi_0$, which proves (iii). Each $\pi_0$-line has $13-4=9$ non-free points, and a non-free point lies on only one $\pi_0$-line. Hence there are $117$ non-free points and $157-13-117=27$ free points. By (iv) no external line contains a free point. An external line meets each $\pi_0$-line in a point outside $\pi_0$, and these $13$ points are distinct, which proves (ii). The line counts follow by duality, since the dual of $\pi_0$ is a subplane of order 3 of the dual plane.
\end{proof}

Two free points lie on a unique line, which is a tangent at a unique point $P$ of $\pi_0$. We call them \emph{partners at $P$} if they lie on a common tangent at $P$. For $P\in\pi_0$, let $T_P$ be the set of $9$ triples of free points cut out by the tangents at $P$. For a $\pi_0$-line $L$, the graph $\Gamma_L$ of the introduction joins two free points if they are partners at a point of $L$.

\begin{corollary}\label{cor:kts}
\begin{enumerate}[label=\textup{(\roman*)}]
\item The free points with the $117$ triples of $\bigcup_P T_P$ form a Kirkman triple system $\cK$ of order $27$ with resolution $\{T_P : P\in\pi_0\}$.
\item For a $\pi_0$-line $L$ and $X\in N_L$, let $C_X$ be the set of free triples of the $9$ tangents through $X$. Then $C_X$ is a parallel class of $\cK$ that contains one triple of each $T_P$ with $P\notin L$, and the $9$ classes $C_X$ with $X\in N_L$ partition $\bigcup_{P\notin L}T_P$.
\item If $X\in N_L$ and $X'\in N_{L'}$ with $L\ne L'$, then $|C_X\cap C_{X'}|\le1$.
\item Each external line $e$ gives a resolution $\{C_{e\cap L} : L \text{ a } \pi_0\text{-line}\}$ of $\cK$.
\item The graph $\Gamma_L$ is $8$-regular and is the union of the $4$ triangle factors $T_P$ with $P\in L$.
\end{enumerate}
\end{corollary}

\begin{proof}
(i) Two free points lie in exactly one triple, namely the free triple of their joining tangent. By Theorem~\ref{thm:incidence}(iv) each free point lies on exactly one tangent at $P$, so $T_P$ is a parallel class.

(ii) The $9$ tangents through $X$ are tangents at the $9$ points of $\pi_0$ off $L$, and they meet only in $X$, which is not free. Hence their free triples are pairwise disjoint and cover all $27$ free points. A tangent at $P\notin L$ meets $L$ in exactly one point of $N_L$, so each triple of $T_P$ lies in exactly one class $C_X$ with $X\in N_L$.

(iii) The points $X$ and $X'$ are distinct and lie on at most one common line. A common triple of $C_X$ and $C_{X'}$ lies on a common tangent.

(iv) A triple $t\in T_P$ lies on a tangent at $P$, which meets $e$ in a non-free point $X$. If $L$ is the $\pi_0$-line through $X$, then $P\notin L$ and $t\in C_X$. If $t$ lay in two of the classes, its tangent would contain two points of $e$ and would equal $e$.

(v) Each free point has $2$ partners at each of the $4$ points of $L$.
\end{proof}

\begin{definition}\label{def:orthres}
Let $L$ be a $\pi_0$-line. A parallel class of $\cK$ that contains one triple of $T_P$ for each $P\notin L$ is a \emph{transversal class} for $L$. A set of $9$ transversal classes for $L$ that partitions $\bigcup_{P\notin L}T_P$ is an \emph{orthogonal resolution} for $L$. Orthogonal resolutions for distinct lines $L$ and $L'$ are \emph{compatible} if a class of one and a class of the other share at most one triple.
\end{definition}

An orthogonal resolution for $L$ is a partial form of the usual notion of orthogonal resolutions of a design: its $9$ classes are orthogonal to the $9$ classes $T_P$ with $P\notin L$. Our use of ``compatible'' differs from that of Tonchev \cite{Ton17}, where it refers to resolutions of the whole design. By Corollary~\ref{cor:kts}, the classes $C_X$ ($X\in N_L$) form an orthogonal resolution for $L$, and these resolutions are pairwise compatible. Throughout, the labels $P$ of the classes $T_P$ carry the incidence structure of $\pi_0$.

\begin{proposition}\label{prop:stab}
The stabilizer of $\pi_0$ in the collineation group of $\pi$ has order $1$ or $3$.
\end{proposition}

\begin{proof}
By \cite{AST23} the collineation group of $\pi$ has order $1$, $2$ or $3$. An involution of a projective plane of order $n$ is an elation, a homology or a Baer involution \cite{HP73}. A Baer involution requires $n$ to be a square, and a homology of order 2 requires $n$ to be odd. So the involutions of $\pi$ are elations. Suppose an involution $\sigma$ maps $\pi_0$ onto itself. It does not fix $\pi_0$ pointwise, since the fixed points of an elation lie on its axis and $\pi_0$ contains a quadrangle. Hence $\sigma$ induces an involution of $\PGL(3,3)$, which is a homology. It fixes a line and a point off that line, hence five points of $\pi_0$, not all collinear. But the fixed points of the elation $\sigma$ lie on its axis, which meets $\pi_0$ in at most four points.
\end{proof}

\begin{proposition}\label{prop:arcs}
\begin{enumerate}[label=\textup{(\roman*)}]
\item Every line meets $F$ in $0$ or $3$ points, so $F$ is a maximal arc of degree $3$. Its secants are the $117$ tangents.
\item Every line meets $K$ in $0$ or $4$ points, so $K$ is a maximal arc of degree $4$ with $(4-1)\cdot13+1=40$ points. Its secants are the $130$ lines meeting $\pi_0$, and its external lines are the $27$ external lines of $\pi$. These form a maximal arc of degree $3$ in the dual plane.
\item The secants of $K$, restricted to $K$, form a Steiner system $\cD=S(2,4,40)$ containing $\pi_0$ as a subdesign. For $X\notin K$, the $10$ secants through $X$ form a parallel class $S_X$ of $\cD$, and $S_X$ contains exactly one block inside $\pi_0$. Two disjoint blocks of $\cD$ lie in exactly one class $S_X$.
\item If $\pi_1\subseteq K$ is a subplane of order $3$, then $K$ is the union of $\pi_1$ and its free points.
\item The $117$ points off $K$ and the $130$ secants form a partial geometry $\mathrm{pg}(8,9,6)$. Its dual is a $\mathrm{pg}(9,8,6)$ whose points are the blocks of $\cD$ and whose lines are the classes $S_X$.
\end{enumerate}
\end{proposition}

\begin{proof}
(i) and (ii) follow from Theorem~\ref{thm:incidence}: a $\pi_0$-line meets $K$ in $4+0$ points, a tangent in $1+3$, and an external line in none. A point off $K$ is non-free and lies on $3$ external lines, while a point of $K$ lies on none. Thus the external lines form a dual maximal arc of degree $3$.

(iii) Two points of $K$ lie on one line, which is a secant. A point $X\notin K$ lies on $13-3=10$ secants, which cover the $40$ points of $K$, four on each, and therefore cover each point once. A block of $\cD$ meets $\pi_0$ in $1$ or $4$ points. If $S_X$ contains $k$ blocks inside $\pi_0$, then $4k+(10-k)=13$, so $k=1$. Two disjoint blocks meet in $\pi$ in a point off $K$, which is the unique $X$ with both blocks in $S_X$.

(iv) A $\pi_1$-line is a secant of $K$, and its four points in $K$ are the points of $\pi_1$. Hence no point of $K\setminus\pi_1$ lies on a $\pi_1$-line, and $K\setminus\pi_1$ is contained in the set of free points of $\pi_1$. Both sets have $27$ points.

(v) A secant has $13-4=9$ points off $K$, and a point off $K$ lies on $10$ secants, so $s=8$ and $t=9$. Let $X\notin K$ and let $B$ be a secant not through $X$. Four of the secants through $X$ meet $B$ in a point of $K$, and the other $6$ meet $B$ in a point off $K$. Hence $\alpha=6$. The parameters of the collinearity graphs are $(117,80,52,60)$ and $(130,81,48,54)$.
\end{proof}

By Proposition~\ref{prop:arcs}(ii), the lines that meet $\pi_0$ are the secants of $K$, so the core of $\pi$ relative to $\pi_0$, defined in the introduction, is the core of the arc $K$. The core determines $\cK$, the labels of its resolution, and the orthogonal resolutions of all $\pi_0$-lines.

\begin{remark}\label{rem:converse}
Conversely, let $\cD$ be an $S(2,4,40)$ with a family $\cS$ of parallel classes such that any two disjoint blocks lie in exactly one class of $\cS$. A block is disjoint from $130-1-4\cdot12=81$ blocks, so there are $5265$ disjoint pairs. Since a class covers $\binom{10}{2}=45$ pairs, $|\cS|=117$. Each block lies in $81/9=9$ classes, and two classes share at most one block. Adding one point for each class, incident with the blocks of the class, gives $157$ points and $130$ lines of size $13$, in which two points lie on at most one line. This is the core of a putative plane. Of the $\binom{117}{2}=6786$ pairs of classes, $130\binom{9}{2}=4680$ share a block and $2106=27\binom{13}{2}$ do not. A projective plane of order $12$ arises exactly when one can add $27$ resolutions of $\cD$ by classes of $\cS$ such that two classes with no common block lie in exactly one of them. In the subplane case they correspond to the resolutions of Corollary~\ref{cor:kts}(iv).
\end{remark}

\section{The affine case}\label{sec:affine}

In this section we assume the affine case: there is a bijection from $F$ to $\F_3^3$ that maps each class $T_P$ onto the set of lines of $\AG(3,3)$ with one direction $d_P$. The $13$ classes $T_P$ partition the $117$ lines of $\AG(3,3)$, and $P\mapsto d_P$ is therefore a bijection from $\pi_0$ onto the set of directions. The first lemma shows that it is a collineation.

\begin{lemma}\label{lem:forced}
Let $\Delta$ be a set of four directions of $\AG(3,3)$ that are not collinear in the plane at infinity. Then $\AG(3,3)$ has no parallel class consisting of one line of each of the $9$ directions outside $\Delta$.
\end{lemma}

\begin{proof}
Let $\varphi$ be a linear form on $\F_3^3$. Over the $27$ points, $\sum_x\varphi(x)^2\equiv0\pmod 3$. Over a line $\{p+td : t\in\F_3\}$,
\[
\sum_{t}\bigl(\varphi(p)+t\varphi(d)\bigr)^2=3\varphi(p)^2+6\varphi(p)\varphi(d)+5\varphi(d)^2\equiv2\varphi(d)^2 \pmod 3,
\]
and $\varphi(d)^2$ depends only on the direction $\langle d\rangle$. If the parallel class exists, summing over its lines gives $\sum_{d\notin\Delta}\varphi(d)^2\equiv0$ for every $\varphi$. If $\varphi\neq0$, then $\varphi$ vanishes on $4$ directions, so the sum over all $13$ directions is $9\equiv0$. Hence $\#\{d\in\Delta:\varphi(d)\neq0\}\equiv0\pmod3$ for every $\varphi\neq0$. If $\Delta$ contains no three collinear directions, the line through two of them contains no other. If exactly three are collinear, the line through the fourth and one of the three contains no other. In both cases some $\varphi\neq0$ vanishes on exactly two directions of $\Delta$, and the count is $2$.
\end{proof}

\begin{corollary}\label{cor:natural}
In the affine case, $P\mapsto\langle d_P\rangle$ is a collineation from $\pi_0$ onto the plane at infinity of $\AG(3,3)$.
\end{corollary}

\begin{proof}
Let $L$ be a $\pi_0$-line and $X\in N_L$. By Corollary~\ref{cor:kts}(ii), $C_X$ is a parallel class with one line of each direction $d_P$, $P\notin L$. By Lemma~\ref{lem:forced} the four directions $d_P$, $P\in L$, are collinear.
\end{proof}

We identify $\pi_0$ with the plane at infinity accordingly. Then $K=F\cup\pi_0$ is the point set of $\PG(3,3)$.

\begin{lemma}\label{lem:spreadfamily}
In the affine case, the design $\cD$ of $K$ is the design of points and lines of $\PG(3,3)$, and the classes $S_X$ ($X\notin K$) form a covering set of spreads: a family of $117$ spreads such that every pair of skew lines of $\PG(3,3)$ lies in exactly one of them.
\end{lemma}

\begin{proof}
The blocks of $\cD$ inside $\pi_0$ are the lines at infinity, by Corollary~\ref{cor:natural}. The other blocks have the form $\{P\}\cup t$ with $t\in T_P$, that is, an affine line together with its point at infinity. These are the $130$ lines of $\PG(3,3)$. A parallel class of $\cD$ is a spread, and Proposition~\ref{prop:arcs}(iii) gives the covering property.
\end{proof}

Conversely, a covering set of spreads of $\PG(3,3)$ gives the core of a putative plane of order 12 in the affine case, as in Remark~\ref{rem:converse}.

\begin{theorem}[Prince \cite{Pri99}]\label{thm:spreads}
The lines of $\PG(3,3)$ admit no covering set of spreads.
\end{theorem}

\begin{proof}
Suppose $\cS$ is a covering set of spreads. The following data come from exhaustive computation. $\PG(3,3)$ has $8424$ spreads and $5265$ pairs of skew lines, and each pair of skew lines lies in exactly $72$ spreads. Let $\ell_1=\langle e_1,e_2\rangle$ and $\ell_2=\langle e_3,e_4\rangle$. Exactly one member $S$ of $\cS$ contains both lines. The stabilizer of $\{\ell_1,\ell_2\}$ in $\PGL(4,3)$ consists of the block-diagonal and block-antidiagonal matrices. It has order $2304$ and three orbits on the $72$ spreads through $\ell_1$ and $\ell_2$, of sizes $18$, $6$ and $48$. It maps $\cS$ to a family with the same property, so we may assume that $S$ is one of three orbit representatives $R_0,R_1,R_2$. The other members of $\cS$ share at most one line with $S$, and they cover the remaining $5220$ skew pairs exactly once. For each $i$, a dancing-links search over the spreads that share at most one line with $R_i$ finds no such exact cover. The same three problems, encoded as propositional formulas, are unsatisfiable.
\end{proof}

Under the Klein correspondence, lines of $\PG(3,3)$ become points of the hyperbolic quadric $Q^+(5,3)$, skew pairs become non-collinear pairs, and spreads become ovoids. Theorem~\ref{thm:spreads} therefore says that the non-collinearity graph of $Q^+(5,3)$ has no edge decomposition into ovoids.

\begin{theorem}[Hall and Roth \cite{HR84}, Prince \cite{Pri99}, with Corollary~\ref{cor:natural}]\label{thm:affine}
The affine case does not occur.
\end{theorem}

\begin{proof}
By Lemma~\ref{lem:spreadfamily} the affine case gives a covering set of spreads of $\PG(3,3)$, and by Theorem~\ref{thm:spreads} there is none.
\end{proof}

The next proposition strengthens Theorem~\ref{thm:affine}: it concerns only the free points and the classes $C_X$ on three lines. Let $\cK=\AG(3,3)$ with $T_P$ the direction class of $P$, where $\pi_0$ is the plane at infinity.

\begin{proposition}\label{prop:threelines}
For each line $L$ at infinity there are exactly $648$ transversal classes and $63{,}666$ orthogonal resolutions. For two distinct lines there are exactly $2304$ compatible pairs of orthogonal resolutions. No three lines admit pairwise compatible orthogonal resolutions, whether the three lines are concurrent or form a triangle.
\end{proposition}

\begin{proof}
The counts come from an exhaustive enumeration. For three lines, the conditions form a propositional formula that encodes the classes $T_P$, an orthogonal resolution for each of the three lines and the compatibility condition. A SAT solver shows that the formula is unsatisfiable, both for a concurrent triple and for a triangle. Since $\PGL(3,3)$ acts transitively on each of these two kinds of triples, this covers all triples.
\end{proof}

In the language of Theorem~\ref{thm:spreads}: the members of a covering set of spreads that contain a given line at infinity give an orthogonal resolution for that line, so the spreads through three lines of one plane of $\PG(3,3)$ cannot be chosen consistently.

\section{Maximal arcs whose design is \texorpdfstring{$\AG(3,3)$}{AG(3,3)}}\label{sec:agarc}

We now drop the subplane. Let $F$ be a maximal arc of degree $3$ in a projective plane of order $12$. It has $27$ points, and its design $\mathcal{T}$ is an $\mathrm{STS}(27)$.

\begin{definition}\label{def:classcover}
A \emph{class cover} of an $\mathrm{STS}(27)$ is a family of parallel classes such that any two disjoint triples lie in exactly one class of the family.
\end{definition}

In an $\mathrm{STS}(27)$ each triple is disjoint from $117-1-3\cdot12=80$ triples, so there are $4680$ disjoint pairs. Since a parallel class covers $\binom92=36$ of them, a class cover has exactly $130$ classes.

\begin{lemma}\label{lem:arccover}
The $130$ points off $F$ give a class cover of $\mathcal{T}$: the class of $X$ consists of the $9$ secants through $X$.
\end{lemma}

\begin{proof}
A point $X\notin F$ lies on $27/3=9$ secants, and they partition $F$. Two disjoint secants meet in exactly one point, which lies off $F$. Distinct points $X\ne X'$ give distinct classes, since at most one secant passes through both.
\end{proof}

Tonchev \cite[Theorem 2.1]{Ton17} characterizes the designs of maximal arcs by families of resolutions, and the class cover of Lemma~\ref{lem:arccover} is the condition that appears in his proof. A class cover need not extend to a plane: the $40$ external lines of $F$ are still missing. The following theorem is therefore stronger than Theorem~\ref{thm:affinecase}(b).

\begin{theorem}\label{thm:agcover}
$\AG(3,3)$ has no class cover.
\end{theorem}

\begin{proof}
By exhaustive search, $\AG(3,3)$ has $17{,}641$ parallel classes, in agreement with \cite[p.~444]{CMM92}. Its $4680$ disjoint pairs of lines are $468$ parallel and $4212$ skew pairs, and each pair lies in between $79$ and $142$ classes. Suppose $\cC$ is a class cover. It contains exactly one class through the parallel lines $\ell_0=\{(0,0,t)\}$ and $\ell_1=\{(1,0,t)\}$, which is one of the $79$ classes containing both lines. The stabilizer of $\{\ell_0,\ell_1\}$ in $\AGL(3,3)$ maps class covers to class covers. It has order $648$ and $6$ orbits on these $79$ classes, of sizes $9$, $1$, $6$, $3$, $6$ and $54$. So we may assume that the class of $\cC$ through $\ell_0$ and $\ell_1$ is one of six orbit representatives. This gives six cases, numbered $0$ to $5$ in this order, in each of which one class is fixed. In case $1$ it is the class of all lines parallel to $\ell_0$, which the whole stabilizer fixes.

We split further by the same argument. Suppose classes $C_1,\dots,C_m$ are fixed. Choose a disjoint pair $p$ covered by none of them, and let $H$ be a group of affine maps that fix each $C_i$ as a set of lines. In a class cover containing $C_1,\dots,C_m$, the member covering $p$ contains both lines of $p$ and shares no disjoint pair with any $C_i$. We call such classes \emph{admissible}. The stabilizer $H_p$ of $p$ in $H$ maps the cover to another cover containing $C_1,\dots,C_m$, so we may assume that the member covering $p$ is a representative of an $H_p$-orbit of admissible classes. Case $1$ is split in four levels, as in Table~\ref{tab:split}. This gives $23$ leaves: cases $0$, $2$, $3$, $4$, $5$ and the $18$ leaves of Table~\ref{tab:split}.

For each leaf, a SAT solver shows that no class cover contains its fixed classes. The formula has a variable for each of the $17{,}641$ classes, a unit clause for each fixed class, and for each disjoint pair a clause requiring at least one of its classes and a cardinality constraint allowing at most one.
\end{proof}

\begin{table}[ht]
\caption{The split of case 1 in the proof of Theorem~\ref{thm:agcover}. At each node, $H$ is the group used for the split, $H_p$ the stabilizer of the chosen pair, and the children are the representatives of the $H_p$-orbits of admissible classes. At the first split $H$ also fixes $\{\ell_0,\ell_1\}$.}\label{tab:split}
\begin{tabular}{lcccl}
\toprule
node & admissible classes & $|H|$ & $|H_p|$ & children \\
\midrule
case 1 & 36 & 648 & 72 & b0--b5 \\
b2 & 13 & 1944 & 108 & c0--c3 \\
c1 & 10 & 972 & 108 & d0, d1 \\
d1 & 33 & 972 & 12 & e0--e8 \\
\midrule
\multicolumn{5}{l}{leaves: b0, b1, b3, b4, b5, c0, c2, c3, d0, e0--e8 (18 in all)}\\
\bottomrule
\end{tabular}
\end{table}

\begin{proof}[Proof of Theorem~\ref{thm:affinecase}]
Part (a) is Proposition~\ref{prop:threelines}. For (b), if $F$ were such an arc, Lemma~\ref{lem:arccover} would give a class cover of $\AG(3,3)$, contradicting Theorem~\ref{thm:agcover}.
\end{proof}

\section{Symmetric cores}\label{sec:sym}

\subsection{Definitions}
An \emph{automorphism of the core relative to $\pi_0$} is a permutation $\sigma$ of the points of $\pi$ such that $\sigma(\pi_0)=\pi_0$ and $\sigma$ maps each line meeting $\pi_0$ onto a line meeting $\pi_0$. It maps $\pi_0$-lines to $\pi_0$-lines and tangents to tangents, so it permutes the free points and the non-free points. It induces a collineation $g$ of $\pi_0$, maps $T_P$ to $T_{gP}$, and maps $C_X$ to $C_{\sigma X}$. An \emph{automorphism of the core of a maximal arc $K$} is a permutation of the points of $\pi$ that maps each secant of $K$ onto a secant. It maps $K$ onto itself, because the points of $K$ lie on $13$ secants and the other points on fewer. It therefore induces an automorphism of the design of $K$. By Proposition~\ref{prop:arcs}(ii), an automorphism of the core relative to $\pi_0$ is an automorphism of the core of $K=\pi_0\cup F$.

For collineations, Roth \cite[Theorem 7]{Rot64} showed that a collineation of prime order whose fixed structure is a subplane of order $m$ with $m^2\ne n$ forces $n\ge m^2+m+2$. His argument uses a line that misses the subplane, so it does not apply to automorphisms of the core. In Lemma~\ref{lem:basic}, $\sigma$ is an automorphism of the core relative to $\pi_0$. Throughout this section, $g$ denotes the collineation of $\pi_0$ induced by such an automorphism $\sigma$.

\begin{lemma}\label{lem:basic}
\begin{enumerate}[label=\textup{(\roman*)}]
\item The fixed free points of $\sigma$ form a subsystem of $\cK$. Each triple inside it lies in a class $T_P$ with $gP=P$.
\item If $q$ is a fixed free point and $gP=P$, the tangent at $P$ through $q$ is fixed, and $\sigma$ fixes or swaps the two partners of $q$ at $P$.
\item If $g$ is the identity and $\sigma$ fixes a free point, then $\sigma$ is the identity.
\end{enumerate}
\end{lemma}

\begin{proof}
(i) Two fixed free points lie on a unique tangent, which is fixed. Its point $P\in\pi_0$ is fixed, and its third free point is fixed. (ii) is clear. (iii) Let $q$ be the fixed free point. The $13$ tangents through $q$ are fixed. For a $\pi_0$-line $L$, the $9$ tangents through $q$ at the points off $L$ meet $L$ in the $9$ points of $N_L$, so all non-free points are fixed. Then all tangents are fixed, since each of them joins a point of $\pi_0$ to a non-free point. If a free point $x$ were moved, the $13$ tangents through $x$ would be fixed, so all of them would contain $\sigma x$ as well, which is impossible.
\end{proof}

\subsection{Automorphisms of order 13}
Janko and van Trung \cite[Theorems A and B]{JvT81} classified the Kirkman triple systems of order $27$ whose $13$ parallel classes are permuted by an automorphism $\rho$ of order $13$. They showed that none of them has $9$ pairwise disjoint triples, in distinct $\rho$-orbits and in distinct parallel classes, such that the $13$ images of the set of the $4$ omitted parallel classes form the lines of a projective plane of order $3$. This excludes a collineation of order $13$ that leaves $\pi_0$ invariant \cite{JvT80}. Such a collineation fixes one of the $27$ external lines, since $27\equiv1\pmod{13}$. Let $X$ be a non-free point on that line. The images of $C_X$ are the classes of the $13$ non-free points on the line, and they partition the triples, so $C_X$ meets every $\rho$-orbit of triples once. Its omitted classes are those at the points of a $\pi_0$-line. This argument uses an external line, which the core does not contain, and a plane of order $12$ has no collineation of order $13$ at all \cite{JvT82a}. So the next theorem concerns exactly the case that these papers leave open.

\begin{theorem}\label{thm:z13sub}
If $\pi_0$ is a subplane of order $3$ of a projective plane of order $12$, the core relative to $\pi_0$ has no automorphism of order $13$.
\end{theorem}

\begin{proof}
Let $\sigma$ be such an automorphism. If $g$ were the identity, $\sigma$ would permute the $9$ tangents at each point of $\pi_0$ in orbits of size $1$ or $13$, so it would fix all core lines, and all points would be fixed. Hence $g$ has order $13$. All subgroups of order $13$ of $\PGL(3,3)$ are conjugate. So after replacing $\sigma$ by a suitable power and relabelling $\pi_0$, we may assume that $g$ is a fixed Singer cycle. Since $27\equiv1\pmod{13}$, $\sigma$ fixes a free point $q^*$. A second fixed free point would lie on a tangent through $q^*$ at a fixed point of $\pi_0$, and $g$ fixes no point. The other $26$ free points therefore form two orbits of size $13$. If one orbit contained both partners of $q^*$ at a point $P$, say $x$ and $\sigma^kx$, then $\sigma^k$ would map the tangent $q^*x$ onto itself and would fix $P$, so $13\mid k$ and $\sigma^kx=x$, a contradiction. Hence we may label the partners of $q^*$ at $P$ as $a_P$ and $b_P$ with $\sigma(a_P)=a_{gP}$ and $\sigma(b_P)=b_{gP}$. A non-free point $X\in N_L$ lies on exactly one tangent through $q^*$. We label $X$ by $L$ and by the point $P\notin L$ of this tangent. Then $\sigma$ acts on these labels by $(L,P)\mapsto(gL,gP)$.

With these labels, the core gives a satisfying assignment of a propositional formula. Its variables describe the classes $T_P$, that is, which free points are partners at $P$, and, for each $\pi_0$-line $L$, which non-free point of $L$ lies on the tangent at $P$ through a free point $q$. Its clauses say that these data form a Kirkman triple system with pairwise compatible orthogonal resolutions for all $13$ lines, that they respect the labels above, and that they are invariant under $\sigma$. A SAT solver shows that this formula is unsatisfiable.
\end{proof}

\begin{remark}\label{rem:jvt}
We recounted the classification of \cite{JvT81} with an independent program: an exact cover over the possible first parallel classes, and orbits under the normalizer of $\langle\rho\rangle$, which has order $4056$. There are $657$ isomorphism classes, $629$ of affine type and $28$ of antiaffine type in the sense of \cite{JvT81}. Exactly three are diagonal and none is projective, as in \cite[Theorem B]{JvT81}. Each of the $661$ entries of the list in \cite{JvT81} belongs to one of the $657$ classes, and each class occurs. The entries (74) and (77) are the images of (72), and the entries (75) and (76) the images of (73), under the multiplier $\mu$ of order $3$ of \cite[Section 1]{JvT81}. So the list contains four repetitions, and the count in \cite[Theorem A]{JvT81} should read $657$, of which $629$ are of affine type. The count $661$ is also quoted in later tables of Kirkman triple systems of order $27$ \cite{MR90,CMM92}. Four entries, (230), (339), (395) and (412), repeat a point as printed; each has exactly one single-symbol correction that gives a valid first parallel class, and we used it. None of this affects the conclusions of \cite{JvT80,JvT82a}.
\end{remark}

The classification also handles the case in which only the triple system of the free points, with its labelled resolution, is symmetric.

\begin{proposition}\label{prop:z13kts}
Let $\pi_0$ be a subplane of order $3$ of a projective plane of order $12$. Suppose that a permutation $\sigma$ of order $13$ of the free points maps each class $T_P$ onto a class $T_{gP}$, where $g$ is a collineation of $\pi_0$. Then no $\pi_0$-line has an orthogonal resolution. Such a plane therefore does not exist.
\end{proposition}

\begin{proof}
Since $27\equiv1\pmod{13}$, $\sigma$ fixes a free point $x$. If $g$ were the identity, $\sigma$ would fix the triple of each class through $x$, hence each of its two other points, since $\sigma$ has odd order; then $\sigma$ would fix all $27$ points. So $g$ has order $13$, it is a Singer cycle, and $\cK$ with the resolution $\{T_P\}$ is one of the systems of Remark~\ref{rem:jvt}. Fix $P_0$. Then $T_{g^jP_0}=\sigma^j(T_{P_0})$ for all $j$. All subgroups of order $13$ of $\PGL(3,3)$ are conjugate, and the collineations of $\pi_0$ that normalize $\langle g\rangle$ induce the multipliers $1$, $3$, $9$ on it. So, up to isomorphism and collineations of $\pi_0$, each of the $657$ systems carries four labellings, one for each coset of $\{1,3,9\}$ in $\Z_{13}^*$. By the action of $g$, all $\pi_0$-lines behave alike. An exhaustive search over the $657\cdot4$ labelled systems finds that $2594$ of them have no transversal class for a line, $33$ have between $1$ and $24$ and no orthogonal resolution, and the remaining one, $\AG(3,3)$ with the natural labelling, has $648$. A plane would give an orthogonal resolution for every line by Corollary~\ref{cor:kts}(ii), so $\cK$ would be $\AG(3,3)$ in the affine case, which does not occur by Theorem~\ref{thm:affine}.
\end{proof}

An automorphism of order $13$ of the core induces such a permutation of the free points, so Proposition~\ref{prop:z13kts} contains Theorem~\ref{thm:z13sub}. So does the next theorem, which assumes no subplane, in the special case $K=\pi_0\cup F$. We keep all three, because their proofs rest on different computations (a SAT formula, the classification of Remark~\ref{rem:jvt} with a search, and an enumeration of $\Z_{13}$-invariant designs with a search) and so check one another.

\begin{theorem}\label{thm:z13arc}
Let $K$ be a maximal arc of degree $4$ or $3$ in a projective plane of order $12$. The core of $K$ has no automorphism of order $13$.
\end{theorem}

\begin{proof}
We treat degree $4$ first. Let $\sigma$ be an automorphism of order $13$ of the core, and let $\cD$ be the design of $K$, an $S(2,4,40)$. If $\sigma$ fixed all points of $K$, it would fix all secants and then all points. The fixed points of $\sigma$ in $K$ form a closed set: the block through two fixed points is fixed, and its other two points are fixed because $\sigma$ has order 13. A closed set with $w\ge2$ points is a block or an $S(2,4,w)$, so $w\equiv1$ or $4\pmod{12}$. The number of fixed points is $\equiv40\equiv1\pmod{13}$, so it is $1$, $14$, $27$ or $40$. The values $14$ and $27$ are not $\equiv1$ or $4\pmod{12}$, and $40$ is excluded by the first observation. Hence it is $1$. Call the fixed point $\infty$ and write the other points as $(o,i)$ with $o\in\{0,1,2\}$, $i\in\Z_{13}$ and $\sigma(o,i)=(o,i+1)$.

No block is fixed, since a fixed block would consist of fixed points. Hence the $13$ blocks through $\infty$ form one orbit. Its base block meets each point orbit once, since otherwise two blocks through $\infty$ would share a second point. Translating each point orbit separately commutes with $\sigma$, so we may take the base block $B_\infty=\{\infty,(0,0),(1,0),(2,0)\}$. The pairs of points fall into $60$ orbits under $\sigma$: $6$ inside each point orbit, $13$ between any two point orbits, and $3$ through $\infty$. The orbit of $B_\infty$ covers $6$ of them. The other $117$ blocks form $9$ orbits, whose base blocks cover each of the remaining $54$ pair orbits exactly once. An exact-cover search over the $3762$ candidate base blocks gives $3{,}291{,}704$ labelled designs.

Now consider the parallel classes $S_X$ for $X\notin K$. A fixed $X$ would give a fixed class, whose block through $\infty$ would be fixed. Hence the $117$ points off $K$ form $9$ orbits of size $13$. Within an orbit, the $13$ classes contain the $13$ blocks through $\infty$, one each, so exactly one class of each orbit contains $B_\infty$. Call these classes $S_0,\dots,S_8$. They partition the $81$ blocks disjoint from $B_\infty$, and since two classes share at most one block,
\begin{equation}\label{eq:compat13}
|S_k\cap\sigma^dS_l|\le1\qquad\text{for all }k,l\text{ and }d\ne0.
\end{equation}
Conversely, if parallel classes $S_0,\dots,S_8$ through $B_\infty$ partition the blocks disjoint from $B_\infty$ and satisfy \eqref{eq:compat13}, then the $117$ classes $\sigma^dS_k$ cover every pair of disjoint blocks exactly once. Indeed, each point of a block $B$ not through $\infty$ lies in $B_\infty$ after exactly one translation, and these $4$ translations are distinct, since a translate with two points of $B_\infty$ would equal $B_\infty$, which contains $\infty$. So $B$ has exactly $4$ translates that meet $B_\infty$, and it lies in $9$ of the classes, which pairwise share only this block and so cover its $81$ disjoint blocks once. A dancing-links search over all $3{,}291{,}704$ designs, with condition \eqref{eq:compat13} checked during the search, finds no design that admits such classes.

For degree $3$ the argument is the same, with an $\mathrm{STS}(27)$. The number of fixed points is $1$, $14$ or $27$, a closed set with $w\ge2$ points has $w\equiv1$ or $3\pmod 6$, and $27$ is excluded as before, so it is $1$. There are two point orbits, the base triple $T_\infty=\{\infty,(0,0),(1,0)\}$ and $130=10\cdot13$ points off $K$. There are $164$ candidate base triples and $38{,}288$ labelled $\mathrm{STS}(27)$, and none admits $10$ classes through $T_\infty$ with the analogous properties.
\end{proof}

For degree $3$ a symmetry of the design alone suffices.

\begin{theorem}\label{thm:z13design}
In no projective plane of order $12$ is there a maximal arc of degree $3$ whose design has an automorphism of order $13$.
\end{theorem}

\begin{proof}
Let $F$ be such an arc and $\mathcal{T}$ its design. The degree-$3$ version of the first two paragraphs of the proof of Theorem~\ref{thm:z13arc} uses only the design and its automorphism of order $13$. Hence $\mathcal{T}$ is isomorphic to one of the $38{,}288$ labelled systems. By Lemma~\ref{lem:arccover}, $\mathcal{T}$ has a class cover. Four of the systems are $\AG(3,3)$, which has none by Theorem~\ref{thm:agcover}. For each of the other $38{,}284$ systems we list all parallel classes and search for a class cover by dancing links, with no symmetry assumed. No system has one, and each search ends after at most $5$ nodes. In $35{,}540$ of these systems some disjoint pair lies in no parallel class at all.
\end{proof}

\subsection{Automorphisms of order 2 and 3}

\begin{lemma}\label{lem:elation}
No automorphism of order $3$ of the core relative to $\pi_0$ fixes a free point and induces an elation on $\pi_0$.
\end{lemma}

\begin{proof}
Let $\sigma$ be such an automorphism, $q$ a fixed free point, and $g$ the induced elation with axis $a$ and center $c$. The fixed points of $g$ are the four points of $a$.

\emph{The fixed free points.} For $P\in a$, $\sigma$ fixes the two partners of $q$ at $P$ by Lemma~\ref{lem:basic}(ii), since an element of order $3$ cannot swap two points. A fixed free point $x\ne q$ lies on a fixed tangent with $q$, at a fixed point of $\pi_0$, that is, at a point of $a$. Thus the fixed free points are $q$ and its $8$ partners at the points of $a$. By Lemma~\ref{lem:basic}(i) they form a subsystem $C_0\cong\AG(2,3)$. For $P\in a$, the tangent at $P$ through a fixed free point is fixed, so its other two free points are fixed, again because $\sigma$ has order $3$. Hence the classes $T_P$, $P\in a$, restrict to the four parallel classes of $C_0$.

\emph{A fixed class $C_X$.} Let $\ell\ne a$ be a $\pi_0$-line through $c$, so $g\ell=\ell$, and let $P\in a\setminus\{c\}$, so $P\notin\ell$. The tangent at $P$ through $q$ is fixed and meets $\ell$ in a fixed point $X\in N_\ell$. Hence $C_X$ is invariant under $\sigma$.

\emph{Contradiction.} Let $t\in C_X$ lie in a class $T_{P'}$ with $gP'\ne P'$. Then $\sigma t\in C_X$ lies in a different class, so $\sigma t\ne t$ and $t\cap\sigma t=\emptyset$, and $t$ contains no fixed point. The fixed points of $g$ off $\ell$ are the three points of $a\setminus\{c\}$. So $C_0$ is covered by the three triples of $C_X$ in the classes $T_P$ with $P\in a\setminus\{c\}$, and each of these triples lies in $C_0$. They are pairwise disjoint lines of the affine plane $C_0$ from three different parallel classes. But lines from different parallel classes meet.
\end{proof}

\begin{proposition}\label{prop:involution}
The core relative to $\pi_0$ has no automorphism of order $2$.
\end{proposition}

\begin{proof}
Let $\sigma$ be an involution of the core. Since $27$ is odd, $\sigma$ fixes a free point, which we call $0$. The induced collineation $g$ is the identity or a homology. Let $s$ be the set of fixed points of $g$ at which $\sigma$ swaps the two partners of $0$. By Lemma~\ref{lem:basic}, the fixed free points other than $0$ are the partners of $0$ at the fixed points of $g$ outside $s$, and they form, with $0$, a subsystem of size $1+2(|\mathrm{Fix}\,g|-|s|)$. A proper subsystem of an $\mathrm{STS}(27)$ has $1$, $3$, $7$, $9$ or $13$ points.

If $g$ is the identity, $\sigma$ is the identity by Lemma~\ref{lem:basic}(iii). Hence $g$ is a homology with axis $a$ and center $c\notin a$, and $|\mathrm{Fix}\,g|=5$. The subsystem size forces $|s|\in\{1,2,4,5\}$. Up to the centralizer of $g$ there are seven cases: $s=\{c\}$ or $s=\{P_1\}$ with $P_1\in a$; two cases with $|s|=2$; two cases with $|s|=4$; and $s=a\cup\{c\}$.

\emph{$|s|=2$.} The $7$ fixed free points form a Fano plane. By Lemma~\ref{lem:basic}(i) its $7$ lines lie in the $5$ classes $T_P$ with $gP=P$. Two triples of one class are disjoint, while two Fano lines meet, so each class contains at most one of them, which is impossible for $7$ lines and $5$ classes.

\emph{$|s|=5$.} The only fixed free point is $0$. For a free point $x\ne0$, the tangent through $x$ and $\sigma x$ is fixed, so its point in $\pi_0$ is fixed, and its third free point is fixed, so it is $0$. Then $x$ is a partner of $0$ at a fixed point of $g$. There are only $10$ such partners, but $26$ free points $x\ne0$.

\emph{$|s|=1$.} The $9$ fixed free points form a subsystem $C_0\cong\AG(2,3)$. As in the proof of Lemma~\ref{lem:elation}, a triple of a class $C_X$ with $\sigma X=X$ that lies in a class $T_{P'}$ with $gP'\ne P'$ contains no fixed point.
\begin{itemize}
\item Let $s=\{c\}$. For $P\in a$, the tangent at $P$ through $0$ is a line of $C_0$ in $T_P$. These four lines pass through $0$, so they have four different directions in $C_0$, and the lines of $C_0$ in $T_P$ are pairwise disjoint, hence all of the direction $\delta_P$ of the first one. Let $\ell$ be a $\pi_0$-line through $c$, let $P_0=\ell\cap a$, and let $X$ be the point where the tangent at some $P\in a\setminus\{P_0\}$ through $0$ meets $\ell$. Then $X$ is fixed. The classes $T_c$ and $T_{P_0}$ do not occur in $C_X$, since $c,P_0\in\ell$. So $C_0$ is covered by the three triples of $C_X$ in the classes $T_P$ with $P\in a\setminus\{P_0\}$, and each of them lies in $C_0$. These are three pairwise disjoint lines of $C_0$ with three different directions, which is impossible.
\item Let $s=\{P_1\}$ with $P_1\in a$. Let $\ell$ be the $\pi_0$-line through $c$ and a point $P_0\in a\setminus\{P_1\}$, and let $X$ be the point where the tangent at $P_1$ through $0$ meets $\ell$. Then $X$ is fixed, and the fixed points of $g$ off $\ell$ are the three points of $a\setminus\{P_0\}$. The triple of $C_X$ in $T_{P_1}$ consists of $0$ and its two partners at $P_1$, which are swapped, so it contains one fixed point. The two other triples of $C_X$ in fixed classes contain at most $3$ fixed points each. Hence $C_X$ covers at most $1+3+3=7$ of the $9$ fixed free points, although it is a parallel class.
\end{itemize}

\emph{$|s|=4$.} All homologies of $\PGL(3,3)$ are conjugate, so we may fix $g$. Along each $2$-cycle of $g$ we label the partners of $0$ so that $\sigma$ maps the first partner at $P$ to the first partner at $gP$. At the fixed points of $g$, $\sigma$ fixes the two partners outside $s$ and swaps them on $s$. With these labels, and with the non-free points labelled through $0$ as in the proof of Theorem~\ref{thm:z13sub}, the action of $\sigma$ is determined by $s$. In each of the two cases, a SAT solver shows that the core encoding of Theorem~\ref{thm:z13sub}, with all $13$ lines and with the clauses for invariance under $\sigma$ instead of those for the Singer cycle, is unsatisfiable.
\end{proof}

\begin{corollary}\label{cor:primes}
Let $\sigma$ be an automorphism of prime order $p$ of the core relative to $\pi_0$. Then $p=3$. Moreover $\sigma$ either fixes no free point and induces an elation or a collineation with exactly one fixed point on $\pi_0$, or it fixes a free point and induces a collineation with exactly one fixed point.
\end{corollary}

\begin{proof}
If $p\ge5$ and $p\ne13$, then $p$ does not divide $|\PGL(3,3)|=2^4\cdot3^3\cdot13$, so $g$ is the identity. Since $p\nmid27$, $\sigma$ fixes a free point, and $\sigma$ is the identity by Lemma~\ref{lem:basic}(iii). The cases $p=13$ and $p=2$ are Theorem~\ref{thm:z13sub} and Proposition~\ref{prop:involution}. Let $p=3$. If $g$ is the identity, then $\sigma$ fixes no free point by Lemma~\ref{lem:basic}(iii). Then for a free point $q$, the tangent through $q$ and $\sigma q$ at some point $P$ is mapped to the tangent through $\sigma q$ and $\sigma^2q$ at $P$. These two tangents share $\sigma q$, so they coincide, and each orbit of $\sigma$ on $F$ is a triple of $\cK$ whose tangent is fixed. Such a fixed tangent at $P$ meets each $\pi_0$-line $L\not\ni P$ in a fixed point $X$. The $9$ tangents through a fixed $X\in N_L$ are fixed, so $C_X$ consists of $9$ orbits of $\sigma$, one in each class $T_{P'}$ with $P'\notin L$. As there are only $9$ orbits, the points $P$ of the fixed tangents are exactly the $9$ points off $L$. This holds for each $\pi_0$-line $L$ that carries a fixed non-free point, and there are at least two such lines. But only $6$ points of $\pi_0$ lie off two distinct lines. Therefore $g$ is not the identity. It is an elation or has exactly one fixed point. Lemma~\ref{lem:elation} excludes an elation together with a fixed free point.
\end{proof}

The cases left open by Corollary~\ref{cor:primes} have a rigid fixed structure.

\begin{lemma}\label{lem:fixedtriples}
Let $\sigma$ be an automorphism of order $3$ of the core relative to $\pi_0$ that fixes no free point, and let $g$ be the induced collineation of $\pi_0$. For a point $P$ of $\pi_0$ with $gP=P$, let $k_P$ be the number of triples of $T_P$ fixed by $\sigma$.
\begin{enumerate}
\item[(i)] A triple fixed by $\sigma$ lies in a class $T_P$ with $gP=P$, and it is an orbit of $\sigma$. Each $k_P$ is $0$, $3$, $6$ or $9$, and the sum of the $k_P$ is at most $9$.
\item[(ii)] If $gP=P$ and $g\ell=\ell$ for a $\pi_0$-line $\ell$ not through $P$, then $k_P$ is the number of fixed points of $\sigma$ in $N_\ell$.
\item[(iii)] If $g$ is an elation with axis $a$ and center $c$, then $k_P$ has the same value $k\in\{0,3\}$ for the three points $P\in a\setminus\{c\}$, and $k=3$ implies $k_c=0$.
\end{enumerate}
\end{lemma}

\begin{proof}
(i) Since $\sigma$ maps $T_P$ onto $T_{gP}$, a fixed triple lies in a class with $gP=P$. It contains no fixed point, so it is a union of orbits of size $3$, that is, one orbit. The automorphism permutes the $9$ triples of $T_P$ in orbits of size $1$ or $3$, so $k_P\equiv9\equiv0\pmod3$. An orbit is a single triple, so it lies in one class, and there are only $9$ orbits.

(ii) The classes $C_X$, $X\in N_\ell$, partition the triples of the classes $T_{P'}$ with $P'\notin\ell$. Hence $X\mapsto$ (the triple of $C_X$ in $T_P$) is a bijection from $N_\ell$ onto $T_P$. It commutes with $\sigma$, because $\sigma C_X=C_{\sigma X}$ and $gP=P$. So it maps the fixed points onto the fixed triples.

(iii) The elation fixes every line through $c$. For $P\in a\setminus\{c\}$ and a $\pi_0$-line $\ell\ne a$ through $c$ we have $P\notin\ell$, so by (ii) $k_P$ is the number of fixed points in $N_\ell$, which does not depend on $P$. By (i), $k_c+3k\le9$, so $k\in\{0,3\}$, and $k=3$ forces $k_c=0$.
\end{proof}

By part (i), the elation case without fixed free points splits into $35$ cases, one for each choice of the numbers $k_P$ with $P\in a$. Part (iii) leaves $5$ of them. A SAT solver refutes each of the other $30$ cases. The $5$ remaining cases, the $4$ cases (by $k_P$ at the unique fixed point) of the other type without fixed free points, and the type with a fixed free point remain open.

\begin{proof}[Proof of Theorem~\ref{thm:cores}]
Part (a) is Theorem~\ref{thm:z13arc}, and part (b) is Corollary~\ref{cor:primes}.
\end{proof}

\section{A reduction of the general case}\label{sec:reduction}

In this section $\cK$ is a Kirkman triple system of order $27$ whose resolution $\{T_P\}$ is labelled by the points $P$ of $\PG(2,3)$. For a line $L$ of $\PG(2,3)$, the graph $\Gamma_L$ joins two points if their triple lies in $T_P$ for some $P\in L$. In the situation of Corollary~\ref{cor:kts} this is the graph $\Gamma_L$ of Section~\ref{sec:structure}.

\begin{theorem}\label{thm:reduction}
Suppose that $\Gamma_L$ is a disjoint union of three copies of $K_9$ for every line $L$. Then there is an isomorphism from $\cK$ to $\AG(3,3)$ that maps each class $T_P$ onto the set of lines of one direction $d_P$, and $P\mapsto d_P$ is a collineation onto the plane at infinity.
\end{theorem}

\begin{proof}
\emph{Step 1: the classes of a triangle are collinear.} Let $q_1,q_2,q_3$ be points not in a common triple, and let $P_{ij}$ be the class of the triple through $q_i$ and $q_j$. These three classes are distinct, since each point lies in only one triple of each class. Let $L$ be the line through $P_{12}$ and $P_{13}$. In $\Gamma_L$ we have $q_1\sim q_2$ and $q_1\sim q_3$. Since $\Gamma_L$ is a union of cliques, $q_2\sim q_3$, so $P_{23}\in L$.

\emph{Step 2: $\cK$ is a Hall triple system.} With $L$ as in Step 1, let $C$ be the component of $\Gamma_L$ containing $q_1$, $q_2$ and $q_3$. If two points of a triple of $T_P$ with $P\in L$ lie in $C$, so does the third. Hence the triples of the classes $T_P$, $P\in L$, inside $C$ cover each pair of points of $C$ exactly once, and $C$ is a closed subsystem $\mathrm{STS}(9)\cong\AG(2,3)$. Thus any three points not in a common triple lie in a subsystem $\AG(2,3)$.

\emph{Step 3: $\cK\cong\AG(3,3)$.} This follows from Step 2 and the theorem of Hall and Bruck (Section~\ref{sec:prelim}).

\emph{Step 4: the labelling.} A closed set of $9$ points of $\AG(3,3)$ is a plane. By the argument of Step 2, every component of $\Gamma_L$ is closed. So the three components of $\Gamma_L$ are pairwise disjoint planes, and they are the cosets of a $2$-dimensional subspace $W_L$. For $P\in L$, each triple of $T_P$ lies in a component of $\Gamma_L$, so its direction lies in $W_L$. For distinct lines $L\ne L'$ we have $W_L\ne W_{L'}$. Otherwise the $7$ classes $T_P$ with $P\in L\cup L'$ would contribute $63$ triples inside the three cosets, which contain only $3\cdot12=36$ triples. Now let $P$ be a point and $L\ne L'$ two lines through $P$. Every triple of $T_P$ has its direction in the $1$-dimensional space $W_L\cap W_{L'}$. Hence $T_P$ is the direction class of $d_P=W_L\cap W_{L'}$. Distinct classes have distinct directions, since a direction class consists of $9$ lines. Finally $d_P\subseteq W_L$ for all $P\in L$, so the injective map $P\mapsto d_P$ maps lines to lines.
\end{proof}


The components of $\Gamma_L$ have a geometric meaning. Let $K=\pi_0\cup F$ be the maximal arc of degree $4$ of Proposition~\ref{prop:arcs}.

\begin{lemma}\label{lem:k9sub}
In a plane of order $12$ with a subplane $\pi_0$ of order $3$, let $L$ be a $\pi_0$-line and $C$ a component of $\Gamma_L$ with $9$ points. Then $C$ and the four points of $\pi_0$ on $L$ are the points of a subplane $\pi_1$ of order $3$, and $\pi_0\cap\pi_1$ consists of the points of $L$. Conversely, every subplane of order $3$ contained in $K$ that meets $\pi_0$ in the points of a line $L$ arises in this way.
\end{lemma}

\begin{proof}
Since $\Gamma_L$ is $8$-regular, $C$ is a $9$-clique. Two points of $C$ lie on a tangent at a point of $L$, whose third free point is adjacent to both, so it lies in $C$. Take as lines $L$ and the $12$ tangents at points of $L$ that carry a triple inside $C$. Two points of $C$ lie on one of these tangents, and two points of $L$ lie on $L$. A point of $C$ and a point $P\in L$ lie on the tangent at $P$ through that point, which carries a triple inside $C$ because $C$ is a component. Each of the $13$ lines contains $4$ of the $13$ points, so these points and lines form a projective plane of order $3$. Conversely, let $\pi_1\subseteq K$ meet $\pi_0$ in the points of $L$. Its other $9$ points are free, and its lines other than $L$ meet $\pi_0$ in one point of $L$, so they are tangents at points of $L$. Hence these $9$ points are pairwise adjacent in $\Gamma_L$ and form a component.
\end{proof}

\begin{lemma}\label{lem:subint}
In a plane of order $12$ with a subplane $\pi_0$ of order $3$, two distinct subplanes of order $3$ contained in $K$ meet in the four points of a line of both.
\end{lemma}

\begin{proof}
Let $\pi'\subseteq K$ be a subplane of order $3$. A line of $\pi'$ meets $K$ only in its $4$ points of $\pi'$, since $K$ is an arc of degree $4$. So no point of $K\setminus\pi'$ lies on a line of $\pi'$, and since both sets have $27$ points, $K\setminus\pi'$ is the set of free points of $\pi'$. By Proposition~\ref{prop:arcs} applied to $\pi'$, every line meets $K\setminus\pi'$ in $0$ or $3$ points. A line that meets $K$ meets it in $4$ points, so it meets $\pi'$ in $1$ or $4$ points. Now let $\pi''\subseteq K$ be another subplane of order $3$. Each line of $\pi'$ meets $K$, hence meets $\pi''$, in a point of $\pi'$. So $\pi'\cap\pi''$ meets every line of $\pi'$. It is also closed: the line through two common points is a line of both, and its $4$ points in $K$ lie in both. A closed set of points of $\PG(2,3)$ that meets every line is a line or the whole plane, and $\pi'\ne\pi''$.
\end{proof}

So $\Gamma_L$ is a disjoint union of three copies of $K_9$ if and only if $L$ lies in four subplanes of order $3$ contained in $K$: three components give three subplanes besides $\pi_0$, and conversely every subplane through $L$ other than $\pi_0$ meets $\pi_0$ in $L$ by Lemma~\ref{lem:subint}, so it comes from a component. By Corollary~\ref{cor:kts}, Theorem~\ref{thm:reduction} applies to the free points of a plane, so Theorems~\ref{thm:reduction} and~\ref{thm:affine} exclude planes in which every line of $\pi_0$ lies in four subplanes of order $3$ contained in $K$; Theorem~\ref{thm:oneline} below excludes a single such line. By Lemma~\ref{lem:subint} the subplanes of order $3$ in $K$ meet pairwise in lines.

The sizes of the components are restricted by counting. Here $\cK$ is again a Kirkman triple system with a labelled resolution, as at the start of this section.

\begin{lemma}\label{lem:sizes}
Let $L$ be a line of $\PG(2,3)$ and $C$ a union of components of $\Gamma_L$ with $c$ points. Then the number of triples of the classes $T_P$ with $P\notin L$ that lie in $C$ or in its complement is $(c-9)(c-18)/2$. Hence every component of $\Gamma_L$ has $9$, $18$ or $27$ points, and if $C$ is a component with $9$ points, every triple of a class $T_P$ with $P\notin L$ has exactly one point in $C$.
\end{lemma}

\begin{proof}
A point lies in one triple of each class, so it has $8$ neighbours in $\Gamma_L$, and two points that are not adjacent lie in a triple of a class $T_P$ with $P\notin L$. Since $C$ is a union of components, $C$ contains $\binom{c}{2}-4c=c(c-9)/2$ pairs of points that are not adjacent, and its complement contains $(27-c)(18-c)/2$. A triple of a class off $L$ with $k$ points in $C$ contains $\binom{k}{2}+\binom{3-k}{2}$ of these pairs, that is, $3$ if $k\in\{0,3\}$ and $1$ otherwise. Let $N$ be the number of these $81$ triples with $k\in\{0,3\}$. Summing over them gives $81+2N=c(c-9)/2+(27-c)(18-c)/2=c^2-27c+243$, so $2N=(c-9)(c-18)$. Since $N\ge0$, no union of components has between $10$ and $17$ points. A component has at least $9$ points, so its complement, if not empty, has at least $9$ as well. Hence every component has $9$, $18$ or $27$ points. For a component with $9$ points, $N=0$.
\end{proof}

So $\Gamma_L$ is $3K_9$, or the disjoint union of $K_9$ and a connected graph on $18$ points, or connected. In planes the first case does not occur.

\begin{lemma}\label{lem:td}
In a plane of order $12$ with a subplane $\pi_0$ of order $3$, let $L$ be a $\pi_0$-line such that $\Gamma_L$ has three components $A_1$, $A_2$, $A_3$ with $9$ points. Let $A_0$ be the set of points of $\pi_0$ off $L$, and $N_L$ the set of points of $L$ off $\pi_0$. Then each tangent at a point $Q\in A_0$ meets $L$ in a point of $N_L$ and each $A_i$ in one point, and the $81$ sets of five points obtained in this way are the blocks of a transversal design $\mathrm{TD}(5,9)$ with groups $A_0$, $A_1$, $A_2$, $A_3$ and $N_L$.
\end{lemma}

\begin{proof}
A tangent at $Q$ meets $\pi_0$ only in $Q\notin L$, so it meets $L$ in a point of $N_L$. Its three free points are pairwise non-adjacent in $\Gamma_L$, so they lie in different components. Two points from different groups lie on exactly one of these tangents. A point $Q\in A_0$ and a free point or a point of $N_L$ are joined by a line that is not a line of $\pi_0$, since a line of $\pi_0$ through $Q$ has no free point and meets $L$ in a point of $\pi_0$; so it is a tangent at $Q$. Two free points in different components are joined by a tangent at a point of $\pi_0$ that is not on $L$, by the definition of $\Gamma_L$. A free point and a point $X\in N_L$ are joined by a tangent at a point of $\pi_0$, which is not on $L$ because the line is not $L$.
\end{proof}

By Lemma~\ref{lem:k9sub}, each $A_i$ with the points of $\pi_0$ on $L$ is a subplane of order $3$. Taking it in place of $\pi_0$ gives the same $81$ lines, with $A_i$ in the role of $A_0$. Conversely, the design determines the classes $T_P$ for $P\notin L$: two free points in different components lie in a triple of $T_P$, where $P$ is the point of $A_0$ on their block.

\begin{theorem}\label{thm:oneline}
In a projective plane of order $12$ with a subplane $\pi_0$ of order $3$, no graph $\Gamma_L$ is a disjoint union of three copies of $K_9$. Equivalently, no line of $\pi_0$ lies in four subplanes of order $3$ contained in $K$.
\end{theorem}

\begin{proof}
Suppose that $\Gamma_L$ is, and let $A_0,\dots,A_3$, $N_L$ be as in Lemma~\ref{lem:td}. A $\mathrm{TD}(5,9)$ is the same as an orthogonal array with $5$ columns and $81$ rows over $9$ symbols, or three mutually orthogonal Latin squares of order $9$. Egan and Wanless \cite{EW16} classified these up to permutations of the columns and of the symbols in each column: there are $371$ classes, $232$ of them maximal. So the design of Lemma~\ref{lem:td} is isomorphic to one of their representatives, by an isomorphism that maps $N_L$ to one of the five columns. By the remark after Lemma~\ref{lem:td}, any of the other four columns may then play the role of $A_0$. For each representative and each choice of the column for $N_L$, we take the other four columns, in their order, as $A_0$, $A_1$, $A_2$, $A_3$, and write a formula for the following: a bijection from the symbols of $A_0$ to the points of $\pi_0$ off $L$, an affine plane on each of $A_1$, $A_2$, $A_3$ whose parallel classes are labelled by the points of $L$, and pairwise compatible orthogonal resolutions for all $13$ lines of $\pi_0$, in which the classes $T_P$ with $P\notin L$ and the orthogonal resolution for $L$ are those given by the design. The stabilizer of $L$ in $\PGL(3,3)$ acts on the points off $L$ as $\AGL(2,3)$, so we may fix the images of two symbols of $A_0$ and choose the image of a third among two points. All $371\cdot5=1855$ formulas are unsatisfiable.
\end{proof}

\begin{corollary}\label{cor:twoplanes}
In a projective plane of order $12$ with a subplane of order $3$, every line of a subplane of order $3$ contained in $K$ lies in at most two such subplanes, and $K$ contains at most $14$ subplanes of order $3$.
\end{corollary}

\begin{proof}
Every subplane $\pi'$ of order $3$ contained in $K$ can play the role of $\pi_0$, with the same arc $K$ (proof of Lemma~\ref{lem:subint}). If a line $\ell$ of $\pi'$ lay in three subplanes in $K$, the graph $\Gamma_\ell$ relative to $\pi'$ would have two components with $9$ points by Lemmas~\ref{lem:k9sub} and~\ref{lem:subint}, hence three by Lemma~\ref{lem:sizes}, against Theorem~\ref{thm:oneline}. By Lemma~\ref{lem:subint}, every other subplane in $K$ meets $\pi'$ in a line, and distinct subplanes meet $\pi'$ in distinct lines, so there are at most $13$ of them.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:lines}]
This is Corollary~\ref{cor:twoplanes}.
\end{proof}

The hypothesis of Theorem~\ref{thm:reduction} is not a consequence of a single orthogonal resolution.

\begin{proposition}\label{prop:single}
There is a Kirkman triple system of order $27$ with a resolution labelled by the points of $\PG(2,3)$ and an orthogonal resolution for one line $L$, in which $\Gamma_L$ is connected.
\end{proposition}

\begin{proof}
In a plane, a free point $q$ determines the map from the points off $L$ to $N_L$ that sends $P$ to the point where the tangent at $P$ through $q$ meets $L$. We describe the example by such maps. Identify the $9$ points of $\PG(2,3)$ off $L$, and a set $N$ of $9$ further symbols, with $\Z_3^2$, where $(x,y)$ has index $3x+y$. Let $f_1,f_2,f_3$ be the maps on $\{0,\dots,8\}$ whose lists of values are
\[
(0,2,6,8,5,3,1,4,7),\quad(0,6,7,2,8,4,3,5,1),\quad(5,1,3,0,7,8,4,6,2),
\]
and for $t\in\Z_3^2$ let $f_{i,t}(P)=f_i(P-t)+t$. The points of the system are the $27$ maps $f_{i,t}$. These maps are permutations, each pair $(P,X)$ satisfies $f(P)=X$ for exactly three of them, and two of them agree in at most one argument. For $P\notin L$, the class $T_P$ consists of the triples $\{f : f(P)=X\}$ with $X\in N$. The graph $\Gamma$ that joins two maps agreeing nowhere is $8$-regular and connected, and it has $81$ triangles, whereas $3K_9$ has $252$. A search finds a decomposition of $\Gamma$ into four triangle factors, and these are the classes $T_P$ with $P\in L$. Then two maps lie in exactly one triple, $\Gamma_L=\Gamma$, and the classes $C_X=\{\{f:f(P)=X\}: P\notin L\}$, $X\in N$, form an orthogonal resolution for $L$.
\end{proof}

\section{Open problems}\label{sec:open}

We keep the setting of Section~\ref{sec:reduction}: $\cK$ is a Kirkman triple system of order $27$ with a resolution labelled by the points of $\PG(2,3)$. In a plane of order 12 with a subplane $\pi_0$ of order 3, the free points form such a system, and all $13$ lines of $\pi_0$ have pairwise compatible orthogonal resolutions (Corollary~\ref{cor:kts}). Hence a positive answer to Problem~\ref{prob:TL}, together with Theorem~\ref{thm:oneline}, or a negative answer to Problem~\ref{prob:three} would show that no projective plane of order $12$ contains a subplane of order $3$. So would a proof of Conjecture~\ref{conj:EC} or of Conjecture~\ref{conj:LR} below.

\begin{problem}[two-line problem]\label{prob:TL}
If two distinct lines $L$ and $L'$ admit compatible orthogonal resolutions, is $\Gamma_L$ a disjoint union of three copies of $K_9$?
\end{problem}

\begin{problem}[three-line problem]\label{prob:three}
Is there a $\cK$ with pairwise compatible orthogonal resolutions for three concurrent lines?
\end{problem}

By Proposition~\ref{prop:threelines}, $\AG(3,3)$ with the natural labelling is not an example. Without the word ``compatible'', Problem~\ref{prob:TL} has a negative answer, even if every line admits an orthogonal resolution.

\begin{example}\label{ex:heis}
Let $H$ be the Heisenberg group of order $27$: its elements are the triples $(a,b,c)\in\F_3^3$, with product $(a,b,c)(a',b',c')=(a+a',b+b',c+c'+ab')$. Each element $(a,b,c)\ne(0,0,0)$ has order $3$, and we label the subgroup it generates by the point $[a:b:c+ab]$ of $\PG(2,3)$, which does not depend on the choice of the generator. Let $T_P$ be the set of left cosets of the subgroup labelled $P$. Two elements $x\ne y$ lie in a common left coset only of the subgroup generated by $x^{-1}y$, so these $13$ classes form a resolution of a Steiner triple system on $H$, labelled by the points of $\PG(2,3)$. The bijection $(a,b,c)\mapsto(a,b,c+ab)$ from $H$ to $\F_3^3$ maps every triple onto a line of $\AG(3,3)$, so the system is $\AG(3,3)$, but only $T_{[0:0:1]}$ is mapped onto a direction class; as the automorphisms of $\AG(3,3)$ are affine, the same holds for every isomorphism. A computer search shows that every line of $\PG(2,3)$ admits an orthogonal resolution, with $135$ transversal classes for each line, but that $\Gamma_L$ is a disjoint union of three copies of $K_9$ only for the four lines through $[0:0:1]$, the label of the centre of $H$; for the other nine lines it is connected.
\end{example}

Among the other systems we examined, orthogonal resolutions are rare: twelve single-line structures as in Proposition~\ref{prop:single}, all invariant under $\Z_3^2$, admit no orthogonal resolution for any second line. If $\cK$ has an orthogonal resolution for every line, and a permutation of order $13$ of its points maps $T_P$ to $T_{gP}$ for a Singer cycle $g$ and the resolution of $L$ to that of $gL$, for all $P$ and $L$, then a SAT computation shows that the component of its fixed point in each graph $\Gamma_L$ is a $K_9$.

The question for all $13$ lines at once, whether pairwise compatible orthogonal resolutions for all lines force each $\Gamma_L$ to be a disjoint union of three copies of $K_9$, is weaker and has the same consequence. A symmetry reduction turns it into four explicit SAT instances, which general-purpose solvers have not decided; nor have they decided the cases of order $3$ left open by Corollary~\ref{cor:primes}. In the proof of Theorem~\ref{thm:oneline}, by contrast, the transversal design fixes most of the triple system; for the two cases that Theorem~\ref{thm:oneline} and Lemma~\ref{lem:sizes} leave, we know no classification that would play this role.

\subsection*{Class covers}
The next conjecture concerns maximal arcs directly. By Lemma~\ref{lem:arccover} it would rule out maximal arcs of degree $3$ in planes of order 12, and by duality also those of degree $4$: the external lines of a maximal arc of degree $3$ form a maximal arc of degree $4$ in the dual plane, and conversely, as in the proof of Proposition~\ref{prop:arcs}(ii). It is stronger than the nonexistence of these arcs, since a class cover need not come from a plane.

\begin{conjecture}\label{conj:EC}
No Steiner triple system of order $27$ has a class cover.
\end{conjecture}

The lines of a partial geometry whose collinearity graph is the disjointness graph of the triples of an $\mathrm{STS}(27)$ are sets of $9$ pairwise disjoint triples, that is, parallel classes. So Conjecture~\ref{conj:EC} says that this graph, which is strongly regular with parameters $(117,80,52,60)$, is never the collinearity graph of a $\mathrm{pg}(8,9,6)$.

$\AG(3,3)$ has no class cover by Theorem~\ref{thm:agcover}, which needs a deep search, while every other system we tested fails after at most $5$ search nodes: the $38{,}284$ non-affine systems with an automorphism of order $13$ (Theorem~\ref{thm:z13design}) and several hundred tripled, cyclic, random and nearly affine systems.

A weaker property is \emph{(DP)}: any two disjoint triples lie in a common parallel class. A system with a class cover satisfies (DP), but (DP) does not characterize $\AG(3,3)$. The \emph{tripling} construction gives an $\mathrm{STS}(27)$ from a Latin square $Q$ of order $9$: take three copies $\AG(2,3)\times\{i\}$, $i\in\Z_3$, with their $36$ lines, and the $81$ triples $\{(x,0),(y,1),(Q(x,y),2)\}$. Some of these systems satisfy (DP) although they have far fewer parallel classes than $\AG(3,3)$.

A system is \emph{residually resolvable} at a triple $t$ if the $80$ triples disjoint from $t$ split into $10$ parallel classes of the $24$ points off $t$. A class cover gives such a splitting at every triple $t$, by the members of the cover that contain $t$; for the design of a maximal arc of degree $3$, these are the classes of the $10$ points off the arc on the secant $t$. By Theorem~\ref{thm:agcover}, the following conjecture therefore implies Conjecture~\ref{conj:EC}.

\begin{conjecture}\label{conj:LR}
An $\mathrm{STS}(27)$ that is residually resolvable at every triple is $\AG(3,3)$.
\end{conjecture}

$\AG(3,3)$ is residually resolvable at all $117$ triples, while each tripled system with (DP) that we found fails at most of its triples.

\section{Verification}\label{sec:verification}

All searches are exhaustive. Exact-cover problems are solved by Algorithm~X with dancing links and the minimum-column rule \cite{Knu00}. In the SAT encodings, a constraint ``exactly one of $x_1,\dots,x_k$'' is the clause $x_1\vee\dots\vee x_k$ together with an encoding of ``at most one'', by pairwise clauses if $k\le6$ and by a sequential counter otherwise; constraints ``exactly $2$'' and ``exactly $3$'' are encoded by sequential counters. Symmetry enters only through orbit representatives under explicit groups, as described in the proofs. The SAT solvers are kissat \cite{Kissat} and CaDiCaL \cite{CaDiCaL}. For every formula used in a proof, CaDiCaL wrote a certificate in the LRAT format \cite{LRAT17}, which \file{lrat-check} \cite{DRATtrim} and then the verified LRAT checker of the standard library of Lean~4 \cite{Lean4} checked.

For all formulas except the $1855$ of Theorem~\ref{thm:oneline}, the check by Lean is a theorem, proved without Mathlib and evaluated by \texttt{native\_decide}, which trusts the Lean compiler and runtime as well as the kernel. Since the checker is verified, neither the SAT solver nor the program that trims the certificates is trusted. For the $1855$ formulas, the same checker ran as a compiled Lean program.

For Theorem~\ref{thm:oneline}, the $371$ designs are the representatives published with \cite{EW16} (\url{https://users.monash.edu.au/~iwanless/data/MOLS/}), and our program checks that each consists of three pairwise orthogonal Latin squares. As a control, with a design obtained from $\AG(3,3)$ with the natural labelling, the formulas are satisfiable when only $L$ and one further line are required. A second implementation of the encoding, written from the mathematical description alone and with different variables, refutes $24$ of the formulas, among them the four slowest, and also finds the control satisfiable.

In each geometric Lean theorem, the passage from a geometric statement to a formula enters as a hypothesis, that is, a proposition defined in the source and passed as an argument, not an axiom; for Theorem~\ref{thm:z13sub}, for example, it says that a plane of order 12 with a subplane of order 3 whose core has an automorphism of order 13 yields a model of the formula. These hypotheses comprise the soundness of the encodings, the uniqueness of $\PG(2,3)$, the conjugacy arguments that fix the action on $\pi_0$, and the symmetry splits. Lean checks the computational parts of some of them: the lists of spreads and parallel classes, the formulas derived from these lists, and the orbit decompositions of the first symmetry step.

The searches in the proofs of Proposition~\ref{prop:z13kts} and Theorems~\ref{thm:z13arc} and~\ref{thm:z13design} are run in Lean as well. A search written in Lean enumerates the objects in question (the first parallel classes through $\{\infty,0,13\}$, and the $\Z_{13}$-invariant $S(2,4,40)$ and $\mathrm{STS}(27)$), a theorem proves that every such object is among its results, and a second procedure, proved sound, rejects every object except the affine ones; \texttt{native\_decide} evaluates both. In this way Lean proves that for every first parallel class and each of the four labellings, an orthogonal resolution for a line forces exactly $648$ transversal classes for this line; that no $\Z_{13}$-invariant $S(2,4,40)$ or $\mathrm{STS}(27)$ admits classes as in the proof of Theorem~\ref{thm:z13arc}; and that every $\Z_{13}$-invariant $\mathrm{STS}(27)$ with a class cover is a translate of one of four systems, which the kernel shows to be copies of $\AG(3,3)$. These theorems need no hypothesis, since they are stated for the objects themselves. The rest of the proofs of Proposition~\ref{prop:z13kts} and Theorems~\ref{thm:z13arc} and~\ref{thm:z13design} lies outside them: the reductions from a plane to these objects, the identification of the system with $648$ transversal classes as $\AG(3,3)$, and the use of Theorems~\ref{thm:affine} and~\ref{thm:agcover}.

The results proved by hand that are marked ``kernel'' in Table~\ref{tab:verify} are formalized in Lean~4 with Mathlib \cite{Mathlib} and checked by the kernel, with only the standard axioms; the statements are made for an arbitrary projective plane of order 12, with a subplane of order 3 where needed, or, in Lemma~\ref{lem:forced}, Theorem~\ref{thm:reduction} and Lemma~\ref{lem:sizes}, for the abstract objects of those statements. Table~\ref{tab:verify}, in the appendix, lists for each result that uses a computation or is formalized how it is established. The evidence in Section~\ref{sec:open} other than Example~\ref{ex:heis} rests on searches without certificates, except the Singer-symmetric formula, whose certificate \file{lrat-check} verified. Independent programs reproduce most counts and conclusions. The programs, the Lean sources and the formulas other than the $1855$ of Theorem~\ref{thm:oneline} are available on Zenodo at \url{https://doi.org/10.5281/zenodo.23687637}. The certificates are too large to store; the README there explains how to regenerate them and the remaining formulas, lists for each result the programs, the logs, the formulas and the Lean theorems, names the programs of the independent checks, and records the versions of the solvers and of Lean.

\appendix
\section{Verification status of the results}\label{app:verify}

{\footnotesize
\setlength{\tabcolsep}{3pt}
\begin{longtable}{@{}>{\raggedright}p{2.3cm}>{\raggedright}p{5.3cm}>{\raggedright}p{1.9cm}>{\raggedright\arraybackslash}p{5.9cm}@{}}
\caption{How the results are established. Lean theorems about formulas, and searches proved complete, are evaluated by \texttt{native\_decide}; the theorems about formulas have the hypotheses described in Section~\ref{sec:verification}.}\label{tab:verify}\\
\toprule
Result & Method & Certificate & Lean \\
\midrule
\endfirsthead
\toprule
Result & Method & Certificate & Lean \\
\midrule
\endhead
\bottomrule
\endfoot
Thms.~\ref{thm:spreads}, \ref{thm:affine} & dancing links and SAT, three cases & LRAT & theorems \\
Prop.~\ref{prop:threelines} & enumeration; SAT, two formulas & LRAT & theorems for the SAT part \\
Thm.~\ref{thm:agcover} & SAT, $23$ leaves & LRAT & theorems \\
Thm.~\ref{thm:z13sub} & SAT, one formula & LRAT & theorem \\
Rem.~\ref{rem:jvt} & exact cover and orbit count & --- & the $14{,}520$ classes recounted \\
Prop.~\ref{prop:z13kts} & search over $657\cdot4$ labelled systems & --- & search proved complete \\
Thm.~\ref{thm:z13arc} & dancing links over $3{,}291{,}704$ designs and $38{,}288$ systems & --- & searches proved complete \\
Thm.~\ref{thm:z13design} & dancing links over $38{,}284$ systems & --- & search proved complete \\
Lem.~\ref{lem:elation} & SAT form of the lemma, one formula & LRAT & theorem \\
Prop.~\ref{prop:involution} & hand proof and SAT; every case also by SAT & LRAT & theorems \\
Cor.~\ref{cor:primes} & hand proof from Thm.~\ref{thm:z13sub} and Prop.~\ref{prop:involution} & --- & kernel, with these two results as hypotheses \\
Lem.~\ref{lem:fixedtriples} & SAT for the $30$ cases excluded by hand & LRAT, \texttt{lrat-check} only & --- \\
Thm.~\ref{thm:oneline} & SAT, $1855$ formulas & LRAT & compiled checker (Section~\ref{sec:verification}) \\
Cor.~\ref{cor:twoplanes} & hand proof from Thm.~\ref{thm:oneline} and Lems.~\ref{lem:k9sub}--\ref{lem:sizes} & --- & --- \\
Prop.~\ref{prop:single} & explicit example & --- & kernel \\
Ex.~\ref{ex:heis} & explicit example; search & --- & kernel: the isomorphism, the resolutions and the lines with $\Gamma_L\cong3K_9$ \\
\addlinespace
Thms.~\ref{thm:incidence}, \ref{thm:reduction}; Cors.~\ref{cor:kts}, \ref{cor:natural}; Props.~\ref{prop:stab}, \ref{prop:arcs}; Lems.~\ref{lem:forced}, \ref{lem:spreadfamily}, \ref{lem:arccover}, \ref{lem:basic}, \ref{lem:elation}, \ref{lem:fixedtriples}, \ref{lem:k9sub}--\ref{lem:td} & hand proofs & --- & kernel; hypotheses: the theorem of Hall and Bruck for Thm.~\ref{thm:reduction}, \cite{AST23} and the classification of involutions \cite{HP73} for Prop.~\ref{prop:stab} \\
\end{longtable}
}

\section*{Acknowledgements}
Generative AI tools were used to write code, run analyses and draft text; the author directed the work and takes full responsibility for its content.
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