#  Additivity of symmetric and subspace designs

Marco Buratti

Anamari Nakić Dipartimento di Scienze di Base e Applicate per l’Ingegneria
(S.B.A.I.), Sapienza Università di Roma, Via Antonio Scarpa, 10, Italy, email:
<EMAIL_ADDRESS>of Electrical Engineering and Computing,
University of Zagreb, Croatia, email<EMAIL_ADDRESS>
(August 28, 2024)

###### Abstract

A $2$-$(v,k,\lambda)$ design is additive (or strongly additive) if it is
possible to embed it in a suitable abelian group $G$ in such a way that its
block set is contained in (or coincides with) the set of all the zero-sum
$k$-subsets of $G$. Explicit results on the additivity or strong additivity of
symmetric designs and subspace 2-designs are presented. In particular, the
strong additivity of PG${}_{d}(n,q)$, which was known to be additive only for
$q=2$ or $d=n-1$, is always established.

Keywords: additive design; strongly additive design; symmetric design;
difference set; projective geometry; subspace design.

2010 MSC: 05B05, 05B10, 05B25.

##  1 Introduction

We assume familiarity with the very basic notions of design theory and finite
geometry. For the relevant background, we refer to [1] and to some chapters
[18, 19, 26] of the Handbook of Combinatorial Designs.

A design ${\mathscr{D}}=(V,{\mathscr{B}})$ is additive under an abelian group
$G$, or briefly $G$-additive, if there exists an injective map $f$ from $V$ to
$G$ such that $f(B)$ is zero-sum for every $B\in{\mathscr{B}}$. Such a map $f$
will be called an embedding of $\mathscr{D}$ in $G$. In the very special case
that $f$ maps $\mathscr{B}$ precisely onto the set of all zero-sum $k$-subsets
of $G$, the design $\mathscr{D}$ is said to be strongly $G$-additive and the
map $f$ is said to be a strong embedding of $\mathscr{D}$ in $G$. In order to
underline that an embedding $f$ is not strong we say that $f$ is a smooth
embedding.

By simply saying that ${\mathscr{D}}$ is additive or strongly additive one
means that ${\mathscr{D}}$ is $G$-additive or strongly $G$-additive for at
least one group $G$. In general, to determine whether a design $\mathscr{D}$
is additive seems to be difficult. To establish if it is strongly additive is
even more difficult. In the affirmative cases, it would be interesting to
determine the smallest $G$’s in which $\mathscr{D}$ can be smoothly or
strongly embeddable, respectively. Obviously, the smallest possible order for
$G$ is the number of points. This is why we say that a design $\mathscr{D}$
with $v$ points is strictly or almost strictly $G$-additive when it is
additive under a group $G$ of order $v$ or $v+1$, respectively.

In a recent paper [23] it is shown that there are designs which are additive
but not strongly additive.

The theory of additive designs was recently introduced by Caggegi, Falcone and
Pavone [12]. Other papers from these authors on the topic or related topics
are [10, 11, 13, 16]. Besides their intrinsic beauty, additive designs have
interesting connections with several branches of discrete mathematics such as
coding theory and additive combinatorics. We also note that the usage of zero-
sum blocks in the construction of designs is frequent (see, e.g., [4, 5, 15,
20, 28]), and that there are combinatorial designs different from the classic
ones which can be considered additive as, for instance, the so-called Heffter
arrays [22].

In their seminal paper Caggegi, Falcone and Pavone proved that every
$(v,k,\lambda)$ symmetric design is strongly additive but they were able to
indicate a concrete group where it is strongly embeddable, that is
$\mathbb{Z}_{k-\lambda}^{(v-1)/2}$, only in the hypothesis that the order
$k-\lambda$ is a prime not dividing $k$. Here we prove that every symmetric
$(v,k,\lambda)$ design is strongly additive under the (huge) group
$\mathbb{Z}_{k-\lambda}^{v}$ without any conditions on the parameters. We also
prove that a cyclic symmetric design is smoothly additive under a relatively
small group if suitable arithmetic conditions are met.

In order to explain our results on the “geometrical side”, we need to give
some notation and terminology. Given a prime power $q$, we denote by
$\mathbb{F}_{q}$, EA$(q)$ and $\mathbb{F}_{q}^{*}$ the finite field of order
$q$, its additive group (that is the elementary abelian group of order $q$)
and its multiplicative group, respectively. Also, AG$(n,q)$ and PG$(n,q)$ will
denote, respectively, the $n$-dimensional affine and projective geometries
over $\mathbb{F}_{q}$. The classical designs of points and $d$-dimensional
subspaces of AG$(n,q)$ and PG$(n,q)$ will be denoted by AG${}_{d}(n,q)$ and
PG${}_{d}(n,q)$, respectively. A $2$-$(n,k,\lambda)_{q}$ subspace design –
also called a $2$-$(n,k,\lambda)$ design over $\mathbb{F}_{q}$ or a $q$-analog
of a $2$-$(n,k,\lambda)$ design – is a classic 2-design of parameters
$({q^{n}-1\over q-1},{q^{k}-1\over q-1},\lambda)$ design whose points are
those of PG$(n-1,q)$ and whose blocks are suitable $(k-1)$-dimensional
subspaces of PG$(n-1,q)$. In particular, PG${}_{d}(n,q)$ is a
2-$(n+1,d+1,\lambda)_{q}$ design where $\lambda$ is the Gaussian coefficient
${n-1\brack d-1}_{q}$; we may call it the complete $n$-dimensional
$d$-subspace design. For general background on subspace designs we refer to
[3].

Note that any coset of a subgroup of EA$(q^{n})$ of order $q^{d}>2$ is zero-
sum. Hence the designs which are most obviously additive are AG${}_{d}(n,q)$
with $(d,q)\neq(1,2)$ and the designs over $\mathbb{F}_{2}$. Indeed the blocks
of AG${}_{d}(n,q)$ are all the cosets of all the subgroups of $EA({q^{n}})$ of
order $q^{d}$, and the blocks of a 2-$(v,k,\lambda)_{2}$ design are suitable
subgroups of $EA(2^{v})$ of order $2^{k}$ deprived of the identity element.
Thus, besides the additivity of PG${}_{d}(n,2)$, we have the existence of an
additive 2-$(v,7,7)$ design for every odd $v$ in view of the main results in
[6, 27], and of an additive 2-(8191,7,1) design, that is the well-celebrated
2-$(13,3,1)_{2}$ design found in [2] (see also Section 6.1 in [9]). It is also
evident the strict additivity of every 2-$(q^{n},kq,\lambda)$ design whose
blocks are union of $k$ parallel lines of AG$(n,q)$ (see [8, 11, 21] for some
examples). In all these cases the embedding map is simply the identity.

For $q>2$ the additivity of PG${}_{d}(n,q)$ and, more generally, of a 2-design
over $\mathbb{F}_{q}$ is not immediate. One can deduce it only for
PG${}_{n-1}(n,q)$ since this design – that of points and hyperplanes of
PG$(n,q)$ – is symmetric. In this paper we will prove that PG${}_{d}(n,q)$ is
always strongly additive under a huge group and smoothly additive under the
small EA$(q^{n+1})$. More generally, all subspace 2-designs are additive.
Unfortunately, their strong additivity remains in doubt.

Our four main results, one for each of the subsequent sections, can be
summarized as follows.

  * (1)

Every symmetric $(v,k,\lambda)$ design is strongly
$\mathbb{Z}_{k-\lambda}^{v}$-additive.

  * (2)

A cyclic symmetric $(v,k,\lambda)$ design is smoothly
$\mathbb{Z}_{p}^{ord_{v}(p)}$-additive for any prime $p$ dividing $k$ but not
$v$.

  * (3)

PG${}_{d}(n,q)$ is strongly $\mathbb{Z}_{q^{d}}^{(q^{n+1}-1)/(q-1)}$-additive.

  * (4)

Any $2$-$(n,k,\lambda)_{q}$ design is smoothly EA$(q^{n})$-additive.

Applying (3) with $d=1$ we get, in particular, that PG${}_{1}(n,q)$ is an
additive Steiner 2-design. In [7] we recently proved that for any $k$ which is
neither singly even nor of the form $2^{n}3$ there are infinitely many –
unfortunately huge – values of $v$ for which there exists a strictly additive
2-$(v,k,1)$ design. So the big challenge is to determine additive 2-$(v,k,1)$
designs with $v$ “reasonable” where $k$ is neither a prime power nor a prime
power plus one. Indeed for $k$ a prime power we have AG${}_{1}(n,k)$ whereas
for $k$ a prime power plus one we have PG${}_{1}(n,k)$.

It is important to note that the additive designs by Caggegi et al. bear no
known relation to the “additive BIB designs” considered in [24], in spite of
the misfortune of inadvertently similar terminology.

##  2 Strong additivity of the symmetric designs

We recall that a $(v,k,\lambda)$ symmetric design is a $2$-$(v,k,\lambda)$
design with as many points as blocks. As a consequence, the number of blocks
through a point is equal to $k$ and any two distinct blocks share exactly
$\lambda$ points. The trivial necessary condition for its existence is that
$\lambda(v-1)=k(k-1)$. Another necessary condition is given by the famous
theorem of Bruck, Ryser and Chowla [1, 18].

In this section we get the strong additivity of any symmetric design already
obtained by Caggegi et al. in [12]. Anyway, differently from them, we are
always able to indicate a group under which the strong additivity is
realizable.

We first need the following lemma.

######  Lemma 2.1.

If $X$ is a $k$-subset of a $(v,k,\lambda)$ symmetric design
${\mathscr{D}}=(V,{\mathscr{B}})$, then there exists a block
$B\in{\mathscr{B}}$ intersecting $X$ in at least $\lambda+1$ points.

###### Proof.

Consider the set $\Phi(X):=\\{(x,B)\ |\ x\in X;x\in B\in{\mathscr{B}}\\}$ of
all flags of $\mathscr{D}$ having the point in $X$. The number of pairs
$(\overline{x},B)$ belonging to $\Phi(X)$ with $\overline{x}$ fixed, is equal
to the number of blocks of $\mathscr{D}$ containing $\overline{x}$, that is
$k$. Thus we have $|\Phi(X)|=k^{2}$. The number of pairs $(x,\overline{B})$
belonging to $\Phi(X)$ with $\overline{B}$ fixed, is clearly equal to
$|\overline{B}\cap X|$. Thus we have $|\Phi(X)|=\sum_{B\in{\mathscr{B}}}|B\cap
X|$. Comparing the obtained equalities we get

$k^{2}=\sum_{B\in{\mathscr{B}}}|B\cap X|$

Assume for contradiction that $|B\cap X|\leq\lambda$ for every
$B\in{\mathscr{B}}$. In this case the above equality would give
$k^{2}\leq\lambda v$. This, together with the trivial identity
$\lambda(v-1)=k^{2}-k$, would imply that $\lambda\geq k$ which is absurd. ∎

######  Theorem 2.2.

A symmetric $(v,k,\lambda)$ design is strongly additive under
$\mathbb{Z}_{k-\lambda}^{v}$.

###### Proof.

Given a symmetric $(v,k,\lambda)$ design ${\mathscr{D}}=(V,{\mathscr{B}})$,
take an ordering $\\{x_{1},\dots,x_{v}\\}$ of $V$ and an ordering
$\\{\beta_{1},\dots,\beta_{v}\\}$ of $\mathscr{B}$. Consider the $v\times v$
matrix $M=(m_{i,j})$ with entries in $\mathbb{Z}_{k-\lambda}$ defined by

$m_{i,j}=\begin{cases}0&{\rm if}\ x_{i}\in\beta_{j}\cr 1&{\rm if}\
x_{i}\not\in\beta_{j}\end{cases}$

and consider the map

$f:x_{i}\in V\longrightarrow M_{i}\in\mathbb{Z}_{k-\lambda}^{v}$

where $M_{i}$ denotes the $i$-th row of $M$. For any two distinct points
$x_{i_{1}}$, $x_{i_{2}}$, there is a block $\beta_{j}$ containing $x_{i_{1}}$
but not $x_{i_{2}}$ since the number of blocks through $x_{i_{1}}$, that is
$k$, is strictly greater than the number of blocks containing both $x_{i_{1}}$
and $x_{i_{2}}$, that is $\lambda$. It follows that the $j$-th component of
$M_{i_{1}}$ is 0 whereas the $j$-th component of $M_{i_{2}}$ is 1 and then
$f(x_{i_{1}})\neq f(x_{i_{2}})$. Thus $f$ is injective. By definition of
strong additivity, it is enough to prove that $\beta$ is a block of
$\mathscr{D}$ if and only if $\beta$ is a $k$-subset of $V$ such that
$f(\beta)$ is zero-sum.

Let $\beta=\\{x_{i_{1}},x_{i_{2}},\dots,x_{i_{k}}\\}$ be a fixed block of
$\mathscr{D}$. Given any block $\beta_{j}$ we have either $\beta=\beta_{j}$ or
$|\beta_{j}\ \cap\ \beta|=\lambda$. In the first case we have $m_{i_{h},j}=0$
for $1\leq h\leq k$. In the second case we have $m_{i_{h},j}=1$ for exactly
$k-\lambda$ values of $h\in\\{1,\dots,k\\}$, that are the values of $h$ for
which $x_{i_{h}}\in\beta\setminus\beta_{j}$. In both cases we clearly have
$m_{i_{1},j}+m_{i_{2},j}+\ldots+m_{i_{k},j}=0$. This is true for any $j$,
hence $M_{i_{1}}+M_{i_{2}}+\dots+M_{i_{k}}=0$, i.e., $f(\beta)$ is zero-sum.

Now assume that $\beta=\\{x_{i_{1}},\dots,x_{i_{k}}\\}$ is a $k$-subset of $V$
such that $f(\beta)$ is zero-sum. Thus we have
$M_{i_{1}}+M_{i_{2}}+\dots+M_{i_{k}}=0$ and then

$m_{i_{1},j}+m_{i_{2},j}+\ldots+m_{i_{k},j}=0\quad\mbox{in
$\mathbb{Z}_{k-\lambda}$ for $1\leq j\leq v$}$ (2.1)

By Lemma 2.1 there is a block $\beta_{j}$ intersecting $\beta$ in at least
$\lambda+1$ points. Thus, up to a reordering of the indices, we can assume
that $\\{x_{i_{1}},x_{i_{2}},\dots,x_{i_{\lambda+1}}\\}$ is contained in
$\beta_{j}$ so that we have $m_{i_{h},j}=0$ for $1\leq h\leq\lambda+1$. It
follows that $m_{i_{h},j}=0$ also for $\lambda+2\leq h\leq k$ in view of (2.1)
and the fact that the entries of $M$ are only $0$’s and $1$’s. Hence every
point of $\beta$ is in $\beta_{j}$. This means that $\beta=\beta_{j}$ and the
assertion follows. ∎

##  3 Smooth additivity of cyclic symmetric designs

An incidence structure $(V,{\mathscr{B}})$ is said to be cyclic if there
exists a cyclic permutation on $V$ leaving $\mathscr{B}$ invariant. It is very
well-known that, up to isomorphism, every cyclic symmetric $(v,k,\lambda)$
design is of the form $(\mathbb{Z}_{v},\\{D+i\ |\ 0\leq i\leq v-1\\})$ where
$D$ is a cyclic $(v,k,\lambda)$ difference set. This means that $D$ is a
$k$-subset of $\mathbb{Z}_{v}$ such that its list of differences $\Delta
D=\\{d-d^{\prime}\ |\ d,d^{\prime}\in D;d\neq d^{\prime}\\}$ is exactly
$\lambda$ times $\mathbb{Z}_{v}\setminus\\{0\\}$.

From the previous section we already know that every symmetric $(v,k,\lambda)$
design $\mathscr{D}$ is additive under $\mathbb{Z}_{k-\lambda}^{v}$. It is
clear that the order of this group is in general huge. Here we prove that if
$\mathscr{D}$ is cyclic and the radical of $k-\lambda$ does not divide $v$,
then $\mathscr{D}$ is additive under a group of much smaller order.

In the following, assuming that $\gcd(u,v)=1$, we denote by $ord_{v}(u)$ the
multiplicative order of $u$ modulo $v$, i.e., the order of $u$ in the group of
units of $\mathbb{Z}_{v}$.

######  Theorem 3.1.

Let $\mathscr{D}$ be a cyclic symmetric $(v,k,\lambda)$ design and let $p$ be
a prime dividing $k-\lambda$ but not $v$. Then $\mathscr{D}$ is
$EA({p^{t}})$-additive with $t=ord_{v}(p)$.

###### Proof.

By definition of $t$, we have $p^{t}\equiv 1$ (mod $v$) so that $v$ is a
divisor of the order $p^{t}-1$ of $\mathbb{F}_{p^{t}}^{*}$. Let $g$ be a
generator of the subgroup of $\mathbb{F}_{p^{t}}^{*}$ of order $v$ and
consider the injective maps $f_{1}$ and $f_{-1}$ defined as follows:

$f_{1}:x\in\mathbb{Z}_{v}\longrightarrow
g^{x}\in\mathbb{F}_{p^{t}},\quad\quad\quad
f_{-1}:x\in\mathbb{Z}_{v}\longrightarrow g^{-x}\in\mathbb{F}_{p^{t}}.$

The assertion will be proved if we show that there is at least one
$i\in\\{1,-1\\}$ such that

$\sum_{b\in B}f_{i}(b)=0\quad\quad\mbox{for every block $B$ of $\mathscr{D}$}$
(3.1)

By assumption, the blocks of $\mathscr{D}$ are all the translates of a
$(v,k,\lambda)$ difference set $D$. Consider the two sums

$\displaystyle\sigma_{1}:=\sum_{d\in D}f_{1}(d)=\sum_{d\in
D}g^{d},\quad\quad\quad\displaystyle\sigma_{-1}:=\sum_{d\in
D}f_{-1}(d)=\sum_{d\in D}g^{-d}$

and let us calculate their product:

$\sigma_{1}\cdot\sigma_{-1}=\sum_{d\in D}g^{d-d}+\sum_{d\in D}\
\sum_{d^{\prime}\in D\setminus\\{d\\}}g^{d-d^{\prime}}=$

$=|D|+\sum_{\delta\in\Delta
D}g^{\delta}=k+\lambda\cdot(\sum_{i=1}^{v-1}g^{i})=$

$=(k-\lambda)+\lambda\cdot(\sum_{i=0}^{v-1}g^{i})=k-\lambda=0.$

The penultimate equality holds since the order of $g$ in
$\mathbb{F}_{p^{t}}^{*}$ is $v$ so that $g^{v}=1$ and then
$\displaystyle\sum_{i=0}^{v-1}g^{i}={g^{v}-1\over g-1}=0$. The last equality
holds since by assumption $k-\lambda$ is divisible by $p$ and hence
$k-\lambda$ is null in $\mathbb{F}_{p^{t}}$.

From the obtained equality $\sigma_{1}\cdot\sigma_{-1}=0$ we infer that one of
the two sums $\displaystyle\sigma_{1}$ and $\displaystyle\sigma_{-1}$ is equal
to 0, i.e., there is one element $i\in\\{1,-1\\}$ such that $\sigma_{i}=0$.
Take this element $i$, take any block $B=D+j$ of $\mathscr{D}$, and note that
the sum of the elements of $f_{i}(B)$ is $\displaystyle\sum_{d\in
D}g^{i(d+j)}=g^{ij}\cdot\displaystyle\sum_{d\in
D}g^{id}=g^{ij}\cdot\sigma_{i}=0$. Hence (3.1) holds and the assertion
follows. ∎

The above theorem gives many concrete results when $k-\lambda$ is coprime with
$v$. Anyway we may have applications of the theorem also for
$\gcd(k-\lambda,v)\neq 1$. For instance, according to [17] the existence of
cyclic difference sets of parameters $(465,145,45)$ and $(910,405,180)$ is
still open. If they exist, using Theorem 3.1 we can say that their associated
symmetric designs would be $\mathbb{Z}_{2}^{20}$-additive and
$\mathbb{Z}_{3}^{12}$-additive, respectively. Indeed in the first case 2 is a
divisor of $k-\lambda=100$ which does not divide $v=465$ and we have
$ord_{465}(2)=20$. In the second case 3 is a divisor of $k-\lambda=225$ which
does not divide $v=910$ and we have $ord_{910}(3)=12$.

We recall that the incidence structure points-hyperplanes of PG$(n,q)$ is a
cyclic symmetric design generated by the so-called Singer $({q^{n+1}-1\over
q-1},{q^{n}-1\over q-1},{q^{n-1}-1\over q-1})$ difference set. Here
$k-\lambda={q^{n}-1\over q-1}-{q^{n-1}-1\over q-1}=q^{n-1}$ is obviously
coprime with $v={q^{n+1}-1\over q-1}$. Assume that $q=p^{\alpha}$ with $p$
prime and note that we have

$p^{\alpha(n+1)}=q^{n+1}=(q-1)v+1\equiv 1\quad(mod\ v)$

This means that $ord_{v}(p)=\alpha(n+1)$. Thus, applying Theorem 3.1 we get
the following.

######  Corollary 3.2.

The point-hyperplane design of PG$(n,q)$ is $EA({q^{n+1}})$-additive. In
particular, the desarguesian projective plane of order $q$ is
$EA({q^{3}})$-additive.

######  Example 3.3.

Consider the Singer $(13,4,1)$ difference set $D=\\{0,1,3,9\\}$ generating
PG$(2,3)$, the projective plane of order 3. Let $r$ be a root of the primitive
polynomial $x^{3}+2x^{2}+1$ over $\mathbb{F}_{3}$. Taking $r$ as primitive
element of $\mathbb{F}_{3^{3}}$, a generator of the subgroup of
$\mathbb{F}_{3^{3}}^{*}$ of order 13 is $g=r^{2}$. Let us calculate the two
sums $\sigma_{1}$ and $\sigma_{-1}$.

$\sigma_{1}=g^{0}+g^{1}+g^{3}+g^{9}=r^{0}+r^{2}+r^{6}+r^{18}=$
$=(0,0,1)+(1,0,0)+(2,2,0)+(0,1,1)=(0,0,2);$

$\sigma_{-1}=g^{0}+g^{-1}+g^{-3}+g^{-9}=r^{0}+r^{-2}+r^{-6}+r^{-18}=$
$=(0,0,1)+(0,2,1)+(2,0,2)+(1,1,2)=(0,0,0).$

Thus a smooth embedding of PG$(2,3)$ in $\mathbb{Z}_{3}^{3}$ is given by the
map

$f^{-1}:x\in\mathbb{Z}_{13}\longrightarrow g^{-x}\in\mathbb{F}_{3^{3}}$

In other words PG$(2,3)$ can be seen as the design $(V,\mathscr{B})$ where

$V=\\{001,100,122,220,112,121,120,020,201,011,202,111,021\\}$

and where $\mathscr{B}$ consists of the following zero-sum blocks

$\\{001,021,202,112\\},\quad\\{021,111,011,220\\},\quad\\{111,202,201,122\\},$
$\\{202,011,020,100\\},\quad\\{011,201,120,001\\},\quad\\{201,020,121,021\\},$
$\\{020,120,112,111\\},\quad\\{120,121,220,202\\},\quad\\{121,112,122,011\\},$
$\\{112,220,100,201\\},\quad\\{220,122,001,020\\},\quad\\{122,100,021,120\\},$
$\\{100,001,111,121\\}.$

Another important class of symmetric designs is that of Paley. For any given
prime $v=4\lambda+3$, the set $\mathbb{Z}_{v}^{\Box}$ of non-zero squares of
$\mathbb{F}_{v}$ is the so-called Paley $(4\lambda+3,2\lambda+1,\lambda)$
difference set. Let Paley$(v)$ be its associated symmetric design. Here
$k-\lambda=(2\lambda+1)-\lambda=\lambda+1$ is clearly coprime with
$v=4\lambda+3$. Hence, applying Theorem 3.1 we get the following.

######  Corollary 3.4.

Let $v=4\lambda+3$ be a prime, let $p$ be any prime divisor of $\lambda+1$,
and let $t=ord_{v}(p)$. Then Paley$(v)$ is additive under
$\mathbb{Z}_{p}^{t}$.

Note, in particular, that if $v=4\lambda+3$ is a Mersenne prime, say
$v=2^{t}-1$, then 2 divides $k-\lambda$ but not $v$ and $ord_{v}(2)=t$. Thus
Theorem 3.1 allows to state the following.

######  Corollary 3.5.

If $v=2^{t}-1$ is a Mersenne prime, then Paley$(v)$ is almost strictly
$\mathbb{Z}_{2}^{t}$-additive.

##  4 Strong additivity of PG${}_{d}(n,q)$

Throughout this section, given a prime power $q$ and a positive integer $n$,
we denote by $[n]_{q}$ the number of points of PG$(n-1,q)$, hence

$[n]_{q}={q^{n}-1\over q-1}$

######  Theorem 4.1.

PG${}_{d}(n,q)$ is strongly additive under $\mathbb{Z}_{q^{d}}^{[n+1]_{q}}$.

###### Proof.

Set $v=[n+1]_{q}$ and $k=[d+1]_{q}$. Let
${\mathscr{P}}=\\{x_{1},\dots,x_{v}\\}$ be an ordering of the points of
PG$(n,q)$, and let ${\cal H}=\\{\pi_{1},\dots,\pi_{v}\\}$ be an ordering of
its hyperplanes.

Consider the $v\times v$ matrix $M=(m_{i,j})$ with entries in
$\mathbb{Z}_{q^{d}}$ defined by

$m_{i,j}=\begin{cases}0&{\rm if}\ x_{i}\in\pi_{j}\cr 1&{\rm if}\
x_{i}\not\in\pi_{j}\end{cases}$

and consider the map

$f:x_{i}\in{\mathscr{P}}\longrightarrow M_{i}\in\mathbb{Z}_{q^{d}}^{v}$

where $M_{i}$ denotes the $i$-th row of $M$. Given any two distinct points
$x_{i_{1}}$, $x_{i_{2}}$, we can take a hyperplane $\pi_{j}$ containing
$x_{i_{1}}$ but not $x_{i_{2}}$ so that the $j$-th component of $f(x_{i_{1}})$
is 0 whereas the $j$-th component of $f(x_{i_{2}})$ is 1. Hence
$f(x_{i_{1}})\neq f(x_{i_{2}})$ and $f$ is injective. The assertion will be
proved if we show that the following equivalence holds.

$S$ is a $d$-subspace of PG$(n,q)$ $\Longleftrightarrow$
$S\in{{\mathscr{P}}\choose k}$ and $f(S)$ is zero-sum.

$(\Longrightarrow).$ Let $S=\\{x_{i_{1}},\dots,x_{i_{k}}\\}$ be a $d$-subsapce
of PG$(n,q)$. For any hyperplane $\pi_{j}$ of $\cal{H}$ we have either
$S\subset\pi_{j}$ or $|S\cap\pi_{j}|=[d]_{q}$. In the first case we have
$m_{{i_{h}},j}=0$ for $1\leq h\leq k$ and then
$m_{i_{1},j}+m_{i_{2},j}+\dots+m_{i_{k},j}=0$. In the second case, up to a
reordering of the indices we can assume that
$S\cap\pi_{j}=\\{x_{i_{1}},\dots,x_{i_{[d]_{q}}}\\}$ so that we have

$m_{i_{h},j}=\begin{cases}0&{\rm for}\ 1\leq h\leq[d]_{q}\vskip 6.0pt plus
2.0pt minus 2.0pt\cr 1&{\rm for}\ [d]_{q}+1\leq h\leq k\end{cases}$

We get again that $m_{i_{1},j}+m_{i_{2},j}+\dots+m_{i_{k},j}$ is null since it
is the sum of $k-[d]_{q}=[d+1]_{q}-[d]_{q}=q^{d}$ ones and the sum has to be
performed in $\mathbb{Z}_{q^{d}}$. In view of the arbitrariness of $j$, it
follows that $M_{i_{1}}+M_{i_{2}}+\dots+M_{i_{k}}=0$, i.e., $f(S)$ is zero-
sum.

$(\Longleftarrow).$ Assume that $S=\\{x_{i_{1}},x_{i_{2}},\dots,x_{i_{k}}\\}$
is a $k$-subset of $\mathscr{P}$ such that $f(S)$ is zero-sum. Thus we have
$M_{i_{1}}+M_{i_{2}}+\dots+M_{i_{k}}=0$ and then

$m_{i_{1},j}+m_{i_{2},j}+\dots+m_{i_{k},j}=0\quad\mbox{in $\mathbb{Z}_{q^{d}}$
for $1\leq j\leq v$}$ (4.1)

Consider the set $\Phi$ of all pairs $(x_{i_{h}},\pi_{j})$ with $1\leq h\leq
k$ and $\pi_{j}\ni x_{i_{h}}$. By a very similar counting argument as that
used in the proof of Lemma 2.1 we get

$k\cdot[n]_{q}=\sum_{j=1}^{v}|\pi_{j}\cap S|$ (4.2)

Indeed the number of pairs $(x_{i_{h}},\pi_{j})\in\Phi$ with
$h\in\\{1,\dots,k\\}$ fixed, is equal to the number of hyperplanes through
$x_{i_{h}}$, that is $[n]_{q}$. Thus $\Phi$ has size equal to the left-hand
side of (4.2). Also, the pairs $(x_{i_{h}},\pi_{j})\in\Phi$ with $j$ fixed, is
equal to $|\pi_{j}\cap S|$ so that $\Phi$ has size equal to the right-hand
side of (4.2).

Let $J$ be the set of $j$’s for which $\pi_{j}$ intersects $S$ in more than
$[d]_{q}$ points and set $\overline{J}=\\{1,2,\dots,v\\}\setminus J$.

Take any $j$ in $J$. By definition of $J$, up to a reordering of the indices
$i_{1}$, $i_{2}$, …, $i_{k}$, we can assume that
$\\{x_{i_{1}},x_{i_{2}},\dots,x_{i_{[d]_{q}+1}}\\}$ is contained in $\pi_{j}\
\cap\ S$ so that we have $m_{i_{h},j}=0$ for $1\leq h\leq[d]_{q}+1$. It
follows that $m_{i_{1},j}+m_{i_{2},j}+\dots+m_{i_{k},j}$ is the sum of a
number of $1$s which is at most equal to
$k-([d]_{q}+1)=[d+1]_{q}-[d]_{q}-1=q^{d}-1$. Considering (2.1), this number is
necessarily zero, i.e., $m_{i_{h},j}=0$ for $1\leq h\leq k$. We conclude that
every point of $S$ is in $\pi_{j}$, i.e., $S\subset\pi_{j}$.

Thus the contribute of each $j\in J$ to the sum in (4.2) is exactly equal to
$|S|=k$ whereas the contribute of each $j\in\overline{J}$ is at most equal to
$[d]_{q}$ by definition of $\overline{J}$. Thus we can write

$\sum_{j=1}^{v}|\pi_{j}\cap S|\leq k|J|+(v-|J|)[d]_{q}$

and then, by (4.2),

$k\cdot[n]_{q}\leq k|J|+(v-|J|)[d]_{q}$

which is equivalent to $|J|\geq{k\cdot[n]_{q}-v[d]_{q}\over k-[d]_{q}}$. Now
we have

${k\cdot[n]_{q}-v[d]_{q}\over k-[d]_{q}}=$ $=\biggl{(}{q^{d+1}-1\over
q-1}\cdot{q^{n}-1\over q-1}-{q^{n+1}-1\over q-1}\cdot{q^{d}-1\over
q-1}\biggl{)}\cdot{1\over{q^{d+1}-1\over q-1}-{q^{d}-1\over q-1}}=$
$={-q^{d+1}-q^{n}+q^{n+1}+q^{d}\over(q-1)^{2}}\cdot{q-1\over q^{d+1}-q^{d}}=$
$={(q^{n}-q^{d})(q-1)\over(q-1)^{2}}\cdot{q-1\over q^{d}(q-1)}={q^{n-d}-1\over
q-1}=[n-d]_{q}$

so that we can write

$|J|\geq[n-d]_{q}$ (4.3)

Let $S^{\prime}$ be the subspace of PG$(n,q)$ which is the intersection of all
the hyperplanes $\pi_{j}$ with $j\in J$ and let $d^{\prime}$ be its dimension.
We have already seen that $S$ is contained in every $\pi_{j}$ with $j\in J$ so
that $S$ is also contained in $S^{\prime}$. It follows that
$[d+1]_{q}=|S|\leq|S^{\prime}|=[d^{\prime}+1]_{q}$ and then $d\leq
d^{\prime}$.

Now let $\Pi(S^{\prime})$ be the pencil of hyperplanes containing
$S^{\prime}$. By the principle of duality, $\Pi(S^{\prime})$ has size equal to
the number of points belonging to a $(n-d^{\prime}-1)$-subspace, that is
$[n-d^{\prime}]_{q}$. By definition of $S^{\prime}$, we have $\\{\pi_{j}\ |\
j\in J\\}\subset\Pi(S^{\prime})$ so that we have $|J|\leq[n-d^{\prime}]_{q}$.
This, together with (4.3), gives $[n-d]_{q}\leq[n-d^{\prime}]_{q}$ and hence
$d^{\prime}\leq d$. Anyway we already noted that $d\leq d^{\prime}$ so that
$d=d^{\prime}$ and then $|S|=[d]_{q}=[d^{\prime}_{q}]=|S^{\prime}|$.
Considering that $S$ is contained in $S^{\prime}$ we conclude that
$S=S^{\prime}$, i.e., $S$ is a $d$-subspace of PG$(n,q)$ and the assertion
follows. ∎

Applying the above theorem with $d=n-1$ we get that the point-hyperplane
design associated with PG$(n,q)$ is strongly additive under
$\mathbb{Z}_{q^{n-1}}^{[n+1]_{q}}$. Note that this is exactly the same result
obtainable by applying Theorem 2.2.

##  5 Additivity of subspace designs

In the previous section we established the strong additivity of
PG${}_{d}(n,q)$ under a quite huge group. Here we show that if we are content
with the smooth additivity, it is enough to suitably embed it in the
elementary abelian group of order $q^{n+1}$. This is a special case of the
following.

######  Theorem 5.1.

Every $2$-$(n,k,\lambda)_{q}$ design is EA$(q^{n})$-additive.

###### Proof.

Let us take, as it is standard, $\mathbb{F}_{q^{n}}^{*}/\mathbb{F}_{q}^{*}$ as
point set of PG$(n-1,q)$. Note that two points $\overline{x}$ and
$\overline{y}$ coincide if and only if ${x\over y}\in\mathbb{F}_{q}^{*}$,
which is equivalent to saying that $({x\over y})^{q-1}=1$ since
$\mathbb{F}_{q}^{*}$ is clearly the group of $(q-1)$-th roots of unity of
$\mathbb{F}_{q^{n}}$. Thus we have:

$\overline{x}=\overline{y}\Longleftrightarrow x^{q-1}=y^{q-1}.$

This means that the map

$f:\overline{x}\in\mathbb{F}_{q^{n}}^{*}/\mathbb{F}_{q}^{*}\longrightarrow
x^{q-1}\in\mathbb{F}_{q^{n}}$

is well-defined and injective.

Recalling that every block of a $2$-$(n,k,\lambda)_{q}$ design is a subspace
of PG$(n-1,q)$, the assertion will be proved if we show that we have

$\displaystyle\sum_{\overline{z}\in S}f(\overline{z})=0\quad\mbox{for every
subspace $S$ of PG$(n-1,q)$}$ (5.1)

Given any two distinct points $\overline{x}$ and $\overline{y}$ of a line $L$
we have

$L\setminus\\{\overline{x}\\}=\\{\overline{\lambda x+y}\ |\
\lambda\in\mathbb{F}_{q}\\}$ (5.2)

Using the binomial Newton’s formula we can write:

$\begin{array}[]{l}\displaystyle\sum_{\lambda\in\mathbb{F}_{q}}(\lambda
x+y)^{q-1}=\sum_{\lambda\in\mathbb{F}_{q}}\sum_{i=0}^{q-1}{q-1\choose
i}\lambda^{i}x^{i}y^{q-1-i}=\vskip 6.0pt plus 2.0pt minus 2.0pt\\\
=\displaystyle\sum_{i=0}^{q-1}{q-1\choose i}\mu_{i}x^{i}y^{q-1-i}\\\
\end{array}$ (5.3)

where $\mu_{i}=\displaystyle\sum_{\lambda\in\mathbb{F}_{q}}\lambda^{i}$. It is
well-known (see, e.g., the lemma on page 5 of [25]) that we have:

$\sum_{\lambda\in\mathbb{F}_{q}}\lambda^{i}=\begin{cases}0&{\rm for}\ 0\leq
i\leq q-2;\cr-1&{\rm for}\ i=q-1\end{cases}$

It follows that the only non-null addend in the last sum of (5.3) is the one
corresponding to $i=q-1$, that is $-x^{q-1}$. Then, considering (5.2), we have
$\displaystyle\sum_{\overline{z}\in L}f(\overline{z})=0$. This proves that
(5.1) holds for 1-dimensional subspaces. Now take any $d$-dimensional subspace
$S$ of PG$(n-1,q)$, take a point $\overline{x}$ of $S$ and let $\mathscr{L}$
be the set of lines of $S$ through $\overline{x}$. Consider the double sum

$\sigma=\sum_{L\in\mathscr{L}}\sum_{\overline{z}\in L}f(\overline{z})$

Given that (5.1) has been proved for the lines, $\sum_{\overline{z}\in
L}f(\overline{z})$ is null for every $L\in\mathscr{L}$ and hence $\sigma$ is
null as well. Also note that in the expansion of $\sigma$ the addend
$f(\overline{y})$ appears exactly once for every $\overline{y}\in
S\setminus\\{\overline{x}\\}$ whereas the addend $f(\overline{x})$ appears
exactly $|\mathscr{L}|$ times. Thus we can write

$\sigma=0=\sum_{\overline{z}\in
S}f(\overline{z})+(|\mathscr{L}|-1)f(\overline{x})$

Now note that $\mathscr{L}$ has size ${q^{d}-1\over q-1}$ so that
$|\mathscr{L}|-1=q\cdot{q^{d-1}-1\over q-1}$ which is zero in
$\mathbb{F}_{q}$. We conclude that $\sum_{\overline{z}\in
S}f(\overline{z})=0$, hence (5.1) is completely proved and the assertion
follows. ∎

As a consequence, recalling that PG${}_{d}(n,q)$ is a 2-$(n+1,d+1,{n-1\brack
d-1}_{q})_{q}$ design, we can state the following.

######  Corollary 5.2.

PG${}_{d}(n,q)$ is EA$(q^{n+1})$-additive.

######  Example 5.3.

Let us give an additive representation of PG${}_{1}(3,3)$. Let $g$ be a root
of the primitive polynomial $x^{4}+x+2$ over $\mathbb{F}_{3}$. A standard
presentation of PG${}_{1}(3,3)$ is the following:

point set $V=\mathbb{F}_{3^{4}}^{*}/\mathbb{F}_{3}^{*}$;

block set $\mathscr{B}=\\{B_{i}\cdot g^{j}\ |\ 1\leq i\leq 3;0\leq j\leq
39\\}\ \cup\ \\{B_{4}\cdot g^{j}\ |\ 0\leq j\leq 9\\}$ where

$B_{1}=\\{g^{0},g^{1},g^{4},g^{13}\\};\quad
B_{2}=\\{g^{0},g^{2},g^{17},g^{24}\\};$
$B_{3}=\\{g^{0},g^{5},g^{26},g^{34}\\};\quad
B_{4}=\\{g^{0},g^{10},g^{20},g^{30}\\}.$

The map

$f:g^{i}\in\mathbb{F}_{3^{4}}^{*}/\mathbb{F}_{3}^{*}\longrightarrow
g^{2i}\in\mathbb{F}_{3^{4}}$

turns our design $(V,{\mathscr{B}})$ into the isomorphic design
$(f(V),f({\mathscr{B}}))$ where the point set $f(V)$ is the set of non-zero
squares of $\mathbb{F}_{3^{4}}$ and where each block $f(B)$ is zero-sum. For
instance we have:

$f(B_{1})=\\{g^{0},g^{2},g^{8},g^{26}\\}=\\{(0,0,0,1),(0,1,0,0),(0,1,1,1),(0,1,2,1)\\};$
$f(B_{2})=\\{g^{0},g^{4},g^{34},g^{48}\\}=\\{(0,0,0,1),(0,0,2,1),(0,1,2,2),(0,2,2,2)\\};$
$f(B_{3})=\\{g^{0},g^{10},g^{52},g^{68}\\}=\\{(0,0,0,1),(1,1,2,1),(1,0,0,2),(1,2,1,2)\\};$
$f(B_{4})=\\{g^{0},g^{20},g^{40},g^{60}\\}=\\{(0,0,0,1),(2,2,1,0),(0,0,0,2),(1,1,2,0)\\}.$

Unfortunately, we are not able to answer the question on whether there are
strongly additive subspace 2-designs which are not complete. We think that
investigating this question is much worth of attention since, maybe, it could
give some answers also on the very hard existence problem for non-complete
designs over a finite field.

## Acknowledgements

This work has been performed under the auspices of the G.N.S.A.G.A. of the
C.N.R. (National Research Council) of Italy.

The second author is supported in part by the Croatian Science Foundation
under the project 9752.

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