2022 Aﬃne vector space partitions

An aﬃne vector space partition of AG( n, q ) is a set of proper aﬃne subspaces that partitions the set of points. Here we determine minimum sizes and enumerate equivalence classes of aﬃne vector space partitions for small parameters. We also give parametric constructions for arbitrary ﬁeld sizes.


Introduction
A vector space partition P of the projective space PG(n − 1, q) is a set of subspaces in PG(n − 1, q) which partitions the set of points.For a survey on known results we refer to [Hed12].We say that a vector space partition P has type (n − 1) mn−1 . . . 2 m2 1 m1 if precisely m i of its elements have dimension i, where 1 ≤ i ≤ n.The classification of the possible types of a vector space partition, given the parameters n and q, is an important and difficult problem.Based on [Hed86], the classification for the binary case q = 2 was completed for n ≤ 7 in [EZSS + 09].Under the assumption m 1 = 0 the case (q, n) = (2, 8) has been treated in [EZHS + 10].It seems quite natural to define a vector space partition A of the affine space AG(n, q) as a set of subspaces in AG(n, q) that partitions the set of points.However, it turns out that those partitions exist for all types which satisfy a very natural numerical condition.If we impose the additional condition of tightness, that is that the projective closures of the elements of A have an empty intersection, then the classification problem becomes interesting and challenging.This condition is natural in the context of hitting formulas as introduced in [Iwa89], that is for logical formulas in full disjunctive normal form (DNF) such that each truth assignment to the underlying variables satisfies precisely one term.For a more recent treatment and applications we refer to [PS22].Here we consider the geometrical and the combinatorial point of view.
Variants of vector space partitions of PG(n − 1, q) have been studied in the literature.In [EZSS + 11] the authors study (multi-)sets of subspaces covering each point exactly λ times.The problem of covering each k-space exactly once is considered in [HHKK19].A more general partition problem for groups is studied in [Hed86].However, we are not aware of any publication directly treating the introduced affine vector space partitions.
The paper is organized as follows.In Section 2 we formally introduce affine vector space partitions, state the preliminaries, and develop the first necessary existence conditions.Here we are guided by the published necessary conditions for vector space partitions.We also argue why tightness (see above) and irreducibility, that is there exists no proper subset A ′ A such that the union of all elements of A ′ is a subspace of AG(n, q), are necessary to obtain an interesting existence question.In Section 3 we classify affine vector space partitions for arbitrary field sizes but small dimensions.Section 4 is concerned with the binary case.We completely determine the possible dimension distributions of tight irreducible affine vector space partitions of PG(n − 1, 2) for all n ≤ 7.In a few cases we give theoretical or computational classifications of the corresponding equivalence classes of tight irreducible vector space partitions.A very nice example consists of eight solids in PG (6,2) whose parts at infinity live on the Klein quadric Q + (5,2).A generalization to arbitrary finite fields of characteristic 2 is given in Subsection 5.2.Parametric constructions of tight irreducible affine vector space partitions using spreads or hitting formulas complete Section 5.In Section 6 we determine the smallest possible size of an irreducible tight affine vector space partition of PG(7, 2) and give a parametric upper bound for PG(n − 1, 2) of size roughly 3n 2 , which is significantly smaller than the conjectured smallest size of an irreducible hitting formula mentioning all variables.We close with a conclusion and a list of open problems in Section 7. To keep the paper self-contained we present some additional material in an appendix.Section A contains details on integer linear programming formulations that we have utilized to obtain some computational results.Section B contains a few technical results that might be left to the reader or collected from the literature.Lists of hitting formulas that can be used to construct tight irreducible affine vector space partitions of the minimum possible size are given in Section C.

Preliminaries and necessary conditions
Definition 1.An affine vector space partition A of AG(n, q) is a set {A 1 , . . ., A r } of subspaces of AG(n, q) such that 0 ≤ dim(A i ) ≤ n − 1 for all 1 ≤ i ≤ r and every point (element of F n q \0) is contained in exactly one element A i .The integer r is called the size of the affine vector space partition.
We write #A for the size of A. For each affine subspace A ∈ AG(n, q) we write A for its projective closure.With this A := A : A ∈ A is the natural embedding of an affine vector space partition of AG(n, q) in PG(n, q).Denoting the hyperplane at infinity by H ∞ , we can directly define an affine vector space partition in PG(n, q): Definition 2. An affine vector space partition U of PG(n − 1, q) is a set {U 1 , . . ., U r } of subspaces of PG(n − 1, q) such that 1 ≤ dim(U i ) ≤ n − 1 for all 1 ≤ i ≤ r and there exists a hyperplane H ∞ such that every point (1-dimensional subspace) outside of H ∞ is contained in exactly one element U i and U i ≤ H ∞ for all 1 ≤ i ≤ r.The integer r is called the size of the affine vector space partition and also denoted by #U.
In the following we will mostly speak of an affine vector space partition, abbreviated as avsp, and will consider its embedding in PG(n − 1, q).The type of an avsp U = {U 1 , . . ., U r } is given by (n − 1) mn−1 . . . ( The analog of Equation (1) for vector space partitions of PG(n − 1, q) is called the packing condition.While the packing condition for vector space partitions of PG(n − 1, q) is just a necessary but not a sufficient condition for the existence with a given type, for avsps Equation ( 1) is both necessary and sufficient.
Proof.Consider a subspace K of H ∞ with dim(K) = n − 2. By H 1 , . . ., H q we denote the q hyperplanes containing K that are not equal to H ∞ .Clearly, we have 0 ≤ m n−1 ≤ q and we can choose H 1 , . . ., H mn−1 as the first elements of U. The remaining elements are constructed recursively.For each index m n−1 +1 ≤ j ≤ q we consider an avsp of type where the m (j) i ∈ N 0 are chosen such that the packing condition is satisfied for H j and for all 1 ≤ i ≤ n − 2. Such a decomposition can be easily constructed, see e.g.Algorithm 1 in Section B.
Definition 4. We call an avsp U = {U 1 , . . ., U r } reducible if there exists a subspace U and a subset S {1, . . ., r} such that dim(U ) < n, #S > 1 and {U i : i ∈ S} is an avsp of S. Otherwise U is called irreducible.
Lemma 5.The smallest size of an irreducible avsp U of PG(n − 1, q) is given by #U = q.
Proof.Let U be an avsp of PG(n − 1, q).Since there are q n−1 points to cover and each subspace covers at most q n−2 points, we have #U ≥ q.Now consider a hyperplane K of H ∞ .By H 1 , . . ., H q we denote the q hyperplanes containing K and not being equal to H ∞ .With this, {H 1 , . . ., H q } is an irreducible avsp of PG(n − 1, q).
For a vector space partition P of PG(n − 1, q) we have dim(A) + dim(B) ≤ n for each pair {A, B} of different elements of P, which is also called dimension condition.Using this it can be easily shown that #P ≥ q n −1 q n/2 −1 = q n/2 + 1 if n is even and #P ≥ q (n+1)/2 + 1 if n is odd.Both bounds can be attained by spreads, i.e., vector space partitions of type (n/2) q n/2 +1 , and lifted MRD codes of maximum possible rank distance, i.e., vector space partitions of type ((n + 1)/2) 1 ((n − 1)/2) q (n+1)/2 , respectively.In [NS11] the authors determine the minimum size σ q (n, t) of a vector space partition of PG(n, q) whose largest subspace has dimension t.Lemma 6.Let U be an irreducible avsp of PG(n − 1, q) and U 1 , . . ., U q ∈ U be q different elements with Proof.Let U := U 1 , . . ., U q and u := dim(U 1 ).Since U \H ∞ contains q u points and U i \H ∞ contains q u−1 points for each 1 ≤ i ≤ q, the set U\ {U 1 , . . ., U q } ∪ {U } is an avsp unless dim(U ) = u + 1 = n.
As an analog of the dimension condition for vector space partitions in PG(n − 1, q) we have: (3) Due to the following general construction for (irreducible) avsps we introduce a further condition.
Definition 10.Let U = {U 1 , . . ., U r } be an avsp of PG(n − 1, q).We call U tight iff the intersection of all U i does not contain a point, i.e. the intersection is trivial.
(1) U ∩ H ∞ ≤ K; (2) dim(U ∩ H i ) = dim(U ) for all 1 ≤ i ≤ q; (3) there are q (dim(U ) − 1)-spaces in U containing U ∩ K and not being contained in Assume that P is a vector space partition of PG(n − 1, q) with type k m1 1 . . .k m l l , where k 1 > • • • > k l and k i > 0 for all 1 ≤ i ≤ l.The so-called tail T of P is the set of all k l -spaces in P, i.e., the set of all elements with the smallest occurring dimension.In [Hed09] several conditions on #T have been obtained.In our situation we can also consider the tail The packing condition (1) directly implies that q k l−1 −k l divides #T = m l if l ≥ 2 and that q divides #T = m l if l = 1.In [Kur18] the results on the tail of a vector space partition of PG(n − 1, q) were refined using the notion of ∆-divisible sets of k-spaces.
for every hyperplane H, where H ∩ S denotes the (multi-)set of elements of S that are contained in H.
Proof.Clearly we have #T = #T ′ .From the packing condition (1) we directly conclude #T ≡ 0 (mod q k l−1 −k l ) if l ≥ 2 and #T ≡ 0 (mod q) if l = 1.Let K be an arbitrary hyperplane of H ∞ and H 1 , . . ., H q be the q hyperplanes of PG(n − 1, q) not being equal to H ∞ .Call the points outside of H ∞ that are contained in some element of U with dimension strictly larger than k l covered and all others outside of H ∞ uncovered.Since each k-space covers either q k−1 or q k−2 points of H i \H ∞ , the number of uncovered points in H i \H ∞ is divisible by q k l−1 −2 if l ≥ 2 and by q k l −1 if l = 1, where 1 ≤ i ≤ q is arbitrary.Let a be the number of k l -spaces in U that are completely contained in H i , so that the number of uncovered points in If l ≥ 2 we have x ≡ 0 (mod q k l−1 −2 ) and #T ≡ 0 (mod q k l−1 −k l ), so that (q − 1)a ≡ 0 (mod q k l−1 −k l ) and a ≡ 0 (mod q k l−1 −k l ).If l = 1 we have x ≡ 0 (mod q k l −1 ) and #T ≡ 0 (mod q), so that (q − 1)a ≡ 0 (mod q) and a ≡ 0 (mod q).
∆-divisible (multi-)sets S of k-spaces in PG(n − 1, q) have been studied in [Kur18].If we replace each k-space by its q k −1 q−1 points we obtain a ∆q k−1 -divisible multiset of #S • q k −1 q−1 points in PG(n − 1, q).The possible cardinalities, given the divisibility constant and the field size, have been completely characterized in [KK20, Theorem 1].Here we will use only a few results on the possible structure of the tail (or more precisely of T ′ ) which allow more direct proofs.
Lemma 16.Let U be an avsp of PG(n − 1, q) with tail T .If #T = q, then either U is reducible or we have U = T and n = 2.
Proof.Denote the dimension of the elements of T by k.Lemma 15 yields that T ′ := {T \H ∞ : T ∈ T } is a q-divisible multiset of (k − 1)-spaces.So, each hyperplane of H ∞ contains either all q or zero elements from T ′ , so that T ′ is a q-fold (k − 1)-space.With this, the stated results follows from Lemma 6.

The structure of the tail for small parameters
If #T is small, then we can also characterize the tail.To this end, let S denote a set of k-spaces in PG(n − 1, q).The corresponding spectrum (a i ) i∈N0 is given by the numbers a i of hyperplanes that contain exactly i elements from S, so that The condition that S is spanning, i.e. S : S ∈ S = PG(n − 1, q), is equivalent to a #S = 0. Double-counting the k-spaces gives Lemma 18.Let S be a 2-divisible set of four k-spaces in PG(n − 1, 2).Then there exists a (k − 1)-space B, a plane E, and a line L ≤ E with dim( E, B ) = k + 2, such that S = { P, B : P ∈ E\L}.
Proof.Assume that P is a point that is contained in at least one but not all elements from S. Let x denote the number of elements of S that contain P .Since all hyperplanes contain an even number of elements from S we have x = 3. Assume x = 2 for a moment and let S, S ′ ∈ S be the two elements not containing P .There are 2 n−k−1 hyperplanes that contain S but do not contain P , so that all of those hyperplanes contain S and S ′ .The intersection of these hyperplanes has dimension at most k and contains S as well as S ′ , so that S = S ′ , which is a contradiction.Thus, each point P in PG(n − 1, 2) is contained in 0, 1 or 4 elements of S.
By (a i ) i∈N0 we denote the spectrum of S. W.l.o.g.we assume that S is spanning, i.e., we have a 4 = 0. From the equations ( 4) and ( 5) we conclude If there is no point P that is contained in all four elements of S, then the elements of S are pairwise disjoint and double-counting pairs yields 2 2 which has the unique solution n = 3, k = 1.So, by recursively quotienting out points P that are contained in all elements of S we conclude the existence of a (k − 1)-space B that is contained in all four elements of S. Quotienting out B yields a spanning 2-divisible set of points in PG(2, 2) with a 0 = 1 and a 2 = 6.Choosing E as the ambient space and L as the empty hyperplane yields the stated characterization since in PG(2, 2) there are exactly four points outside a hyperplane.If k = 1, i.e., the k-spaces are points, the equations ( 4)-( 6) are also known as "standard equations" or the first three MacWilliams equations for the corresponding linear code.
We remark that Lemma 18 is based on the fact that each 2-divisible set of 4 points is an affine plane.For q > 2 there there further possibilities for q-divisible sets of q 2 points over F q , see [DBDMS19, KM21] on the so-called cylinder conjecture.
3 Classification of avsps in PG(n − 1, q) for small parameters By definition, there is no avsp in PG(1 − 1, q).In PG(2 − 1, q) there is a unique avsp.It has type 1 q and is irreducible and tight.
Lemma 19.Let U be an avsp of PG(n − 1, q), where n ≥ 3.If there exist pairwise different hyperplanes U 1 , . . ., U l ∈ U, then there exists an Proof.The statement is trivial for l ≤ 1, so that we assume l ≥ 2. Due to the dimensions we have dim Proposition 20.Let U be an irreducible avsp of PG(n − 1, q), where n ≥ 3.If U is of type (n − 1) mn−1 . . . 2 m2 1 m1 , then we have m n−1 ≤ q − 2 or m n−1 = q.In the latter case U is not tight.
Proof.We assume m n−1 = q − 1 ≥ 1 and let K ≤ H ∞ as in Lemma 19.With this, let H = H ∞ be the unique hyperplane with K ≤ H that is not contained as an element in U and U ′ arise from U by removing the q − 1 (n − 1)-dimensional elements.Thus, U ′ is an avsp of H, i.e., U is reducible. If ) is contained in all elements of U, i.e., U is not tight.
Let U = {U 1 , . . ., U r } be an avsp of PG(n − 1, q), I ⊆ {1, . . ., r}, and V be a proper subspace with V ≤ H ∞ .If #I ≥ 2 and {U i : i ∈ I} is an avsp of V , then we say that the spaces U i with i ∈ I can be joined to V .Note that this is exactly the situation when U is reducible.In PG(n − 1, 2) any two points outside of H ∞ can be joined to a line, so that: Lemma 22.Let U be an irreducible tight avsp of PG(n − 1, 2) of type (n − 1) mn−1 . . . 1 m1 , where n ≥ 3. Then,we have m 1 = 0.
Theorem 23.Let U be an avsp of PG(3 − 1, q) with type 2 m2 1 m1 .Then, we have 0 ≤ m 2 ≤ q, m 1 = q • (q − m 2 ), all lines in U contain a common point P ≤ H ∞ , and the 1-dimensional elements can be grouped into pairwise disjoint sets of size q that can be joint to a line each.
Proof.The existence of P follows from Lemma 19 and the parameterization of m 2 , m 1 follows from the packing condition (1).If m 2 = 0 then choose an arbitrary point P ≤ H ∞ .By L 1 , . . ., L q we denote the q lines containing P that are not equal to H ∞ .For each line L i that is not an element of U there exist q points in U that can be joined to L i .(Note that L i ∩ L j = P for all 1 ≤ i < j ≤ q.) We remark that all possibilities for 0 ≤ m 2 ≤ q can indeed by attained.In general there exist several non-isomorphic examples.
(1) Let U be an irreducible avsp of PG(3 − 1, q).Then U is of type 2 q and non-tight.
Let U be an avsp of PG(n − 1, q), where n ≥ 3, and K ≤ H ∞ be an arbitrary (n − 2)-space.We say that U (1) , . . ., U (q) is a K-decomposition of U if the q hyperplanes containing K and not being equal to H ∞ can be labeled as H 1 , . . ., H q such that for all 1 ≤ i ≤ q.Note that U (i) is an avsp of H i for each 1 ≤ i ≤ q (including the case Moreover, any labeling of the q hyperplanes H i induces a K-decomposition.Observe that for a fixed (n − 2)-space K ≤ H ∞ each pair of K-decompositions arises just by relabeling, so that we also speak of the K-decomposition of U since the actual labeling will not matter in our context.
Affine vector space partitions of PG(4 − 1, q) that contain at least one hyperplane as an element can be characterized easily.
4 Classification of tight irreducible avsps in PG(n − 1, 2) for small dimensions n The cases n ≤ 3 have already been treated in Section 3, so that we assume n ≥ 4 in the following.Our aim is to classify all possible types (n − 1) mn−1 . . . 1 m1 such that a tight irreducible avsp U exists in PG(n − 1, 2).We have m n−1 = 0 and m 1 = 0 due to Corollary 21 and Lemma 22. From Lemma 17 we conclude m l = 2 for the smallest index 1 ≤ l ≤ n − 1 with m l > 0. The possible vectors (m n−2 , . . ., m 2 ) ∈ N n−3 0 are quite restricted by the packing condition (1).For n = 4 the only remaining possibility is type 2 4 .From Lemma 6 we conclude that the four lines are pairwise disjoint, i.e., they form a partial line spread of cardinality 4. It is well known that each partial line spread of cardinality q 2 in PG(3, q) can be extended to a line spread, which has size q 2 + 1. 1 For q = 2 there is only the Desarguesian line spread and since it has a transitive automorphism group, there is only one equivalence class.The numbers of line spreads in PG(3, q) are 1, 2, 3, 21, 1347 for q = 2, 3, 4, 5, 7.
In the three subsequent subsections we will consider tight irreducible avsps in PG(n−1, 2) for n ∈ {5, 6, 7}.The possible types are completely determined in all cases, where realizations are computed using an integer linear programming (ILP) formulation, see Section A in the appendix for the details.If the sizes of the avsps are not too large we were able to also compute all equivalence classes of avsps using a slight modification of an algorithm from [Lin04], see also [K Ö06, Algorithm 4.5].A GAP implementation , based on the GAP package "FinInG" [BBC + 18] for computations in finite incidence geometry, can be obtained from the authors upon request.In the theoretical parts we will also use classification for 2-divisible sets points that can e.g.be found in [HHK + 17] or [Kur21].For the convenience of the reader we will also give a few selected proofs in Section B in the appendix.
Proof.First we assume that two elements of {U 1 , U 2 , U 3 } can be joined to an (n−1)-space H. Without loss of generality, we assume that U 1 and U 2 can be joined to H. Let K := H ∩ H ∞ , so that dim(K) = n − 2. By H ′ we denote the unique hyperplane containing K that is not equal to H or H ∞ .Observe that {U 3 , . . ., U r } is an avsp of H ′ and K is "the hyperplane at infinity" of H ′ .Next we set K ′ := K ∩ U 3 , so that dim(K ′ ) = n − 3. Let B denote the unique (n − 2)-space in H ′ that contains K ′ and is not equal to U 3 or K.With this, {U 4 , . . ., U r } is an avsp of B (including the case r = 4, U 4 = B).Note that the (n − 3)-space K ′ is contained in all elements of {H, U 3 , B}.Since {U 1 , U 2 } forms an avsp of H and dim(U 1 ) = dim ( U 2 ) = n − 2, there exists an (n − 4)-space C that is contained in all elements of {U 1 , U 2 , U 3 , B}.
Otherwise, we assume that no two elements of {U 1 , U 2 , U 3 } can be joined to an (n − 1)-space, so that dim(U i ∩ U j ) = n − 4 for all 1 ≤ i < j ≤ 3. We set If dim(K) = n − 2, then consider the K-decomposition U (1) , U (2) of U and let H 1 , H 2 be the corresponding hyperplanes.Since E 1 , E 2 , E 3 ≤ K, we have that either U i ≤ H 1 or U i ≤ H 2 for all indices 1 ≤ i ≤ 3.By the pigeonhole principle two of the three (n − 2)-spaces in U have to be contained in the same hyperplane, so that they can be joined, which is a contradiction.Thus, we have dim Let P 1 , P 2 be two different arbitrary points outside of H ∞ that or not covered by U 1 , U 2 , or U 3 .For pairwise different i, j, h ∈ {1, 2, 3} consider the (n − 2)-space K i,j,j := C, v i , v j + v h and let H i,j,j be the hyperplane that contains K i,j,h and U i .Since all points in H i,j,h \H ∞ are covered by U 1 , U 2 , U 3 the points P 1 , P 2 have to be contained in the other hyperplane containing K i,j,h not equal to H i,j,h and H ∞ , so that Together with the conditions m n−1 = m 1 = 0 and the packing condition (1) we obtain: Corollary 30.Let U be an irreducible tight avsp of PG(5 − 1, 2).Then the type of U is given by 3 2 2 4 , 3 1 2 6 , or 2 8 .
All types can indeed be realized and corresponding numbers of equivalence classes are given by 3, 4, and 2, respectively.I.e., for n = 4 we have 9 non-isomorphic examples in total.Below are representatives: We remark that the hypothetical type 3 3 2 2 is also excluded by Corollary 17.
Similarly as we have constructed T ′ from the tail T in Lemma 15, we can consider the set U ′ := {U ∩ H ∞ : U ∈ U} for an avsp U of PG(n − 1, q).If U is an irreducible tight avsp of PG(n − 1, q) of type 2 m2 3 m3 , where m 2 = q n−1 − qm 3 , then U ′ is a configuration of m 2 points and m 3 lines in H ∞ ∼ = PG(n − 2, q).The points are pairwise disjoint, so that Lemma 15 yields that they form a q-divisible set.Any two lines can meet in at most a point.If n = 5, then any two lines indeed intersect in a point.So, the maximum point multiplicity is at most m 3 + 1.In the following we will theoretically classify the possibilities for U ′ for tight irreducible avsps of PG(4, 2) of type 2 m2 3 m3 up to symmetry.Corollary 29 gives m 3 ∈ {0, 1, 2}.First we will deduce two general necessary criteria for U ′ .Lemma 31.Let U be an irreducible avsp of PG(n − 1, q) not of type (n − 1) q and U ′ := {U ∩ H ∞ : U ∈ U}.Then U ′ is spanning, i.e., U ′ spans H ∞ .
Proof.Assume that K is a hyperplane of H ∞ that contains all elements of U ′ .From Lemma 12 we can conclude that the K-decomposition U (1) , . . ., U (q) , with corresponding hyperplanes H 1 , . . ., H q , is a partition of U, i.e., the elements of U (i) can be joined to H i for all 1 ≤ i ≤ q.Since we have assumed that U is not of type (n − 1) q we obtain a contradiction.
Lemma 32.Let U be an avsp of PG(n − 1, q) of type (n − 1) mn−1 . . . 2 m2 and U Proof.For an arbitrary but fix hyperplane K of H ∞ let U (1) , . . ., U (q) be the K-decomposition of U with corresponding hyperplanes H 1 , . . ., H q .From Lemma 12 we conclude that for each element U ∈ U with U ∩ H ∞ ≤ K there exists an index 1 ≤ j ≤ q such that U ≤ H j .The c K i,j just count how many (i + 1)dimensional elements of that type are contained in H j .Since the hyperplanes H 1 , . . ., H q are pairwise disjoint, we obtain Equation (8).From Lemma 13 we conclude that for each element U ∈ U such that U ∩ H ∞ ≤ K we have # (U ∩ H j \H ∞ ) = q dim(U)−2 , so that the packing condition yields Equation (9).
Let us consider the case m 3 = 0 first.Here the 2 3 points in H ∞ ∼ = PG(3, 2) form a spanning 2-divisible set P of points.So, P is either an affine solid attained in Example E6 or given by the points of a plane and an intersecting line without the intersection point attained in Example E1, cf.Lemma 52.In the first case the geometrical object of the pairwise disjoint eight lines, that are disjoint from a special plane, in PG(4, 2) is also known under the name of a lifted MRD code or a vector space partition of PG(4, 2) of type 2 8 3 1 .The uniqueness up to symmetry is a well known result.
For m 3 = 1 the six points in H ∞ ∼ = PG(3, 2) form a 2-divisible set P of points, so that P is given by two disjoint lines L 1 , L 2 , cf.Lemma 51.Let us denote the unique line in H ∞ by L. Up to symmetry, the lines L 1 , L 2 , and L can be arranged as follows: • all three lines are pairwise disjoint, so that there are no multiple points, see Example E5; • L = L 1 , so that there are three double points forming a line, see Example E2; • L intersects both L 1 and L 2 in a point, so that we have two double points, see Example E3; • L intersects exactly one of the lines L 1 and L 2 in a point, so that we have a unique double points, see Example E4.
For m 3 = 2 let us denote the two lines by L 1 , L 2 , their intersection point by P , and their span by E, which is a plane.The four points in P form an affine plane, cf.Lemma 18. I.e., there exist a plane E ′ and a line L ′ ≤ E ′ with P = E ′ \L ′ .From Lemma 31 we conclude E = E ′ , so that we set L * := E ∩ E ′ with dim(L * ) = 2. From Lemma 32 we conclude that each hyperplane H of H ∞ that contains either the line L 1 or the line L 2 , but not both, has to intersect P in at least two points.(More technically, ) With this we obtain the following list of possible arrangements of L 1 , L 2 , and E ′ \L ′ : • P is outside of E ′ .Since L i ≤ P, L ′ for i = 1, 2, we have that |L i ∩ E ′ \L ′ | = 1 for i = 1, 2, i.e., we have three double points forming a basis, see Example E7; • P ∈ P = E ′ \L ′ .Since E = E ′ , one of lines L 1 , L 2 intersects E ′ exactly in the point P , so that the other line meets P in a second point, i.e., we have one triple and one double point, see Example E9; • P ∈ L ′ .W.l.o.g.we assume L 1 ∩ E ′ = P , so that L 2 ≤ E ′ .Since the plane π := L 1 , L ′ is disjoint to P, we conclude L 2 ∈ π, so that L 2 = L ′ .So, P is the unique double point, see Example E8.
Thus we have completed the classification of possibilities for U ′ that may correspond to tight irreducible avsps U of PG (4,2).We call the process of moving from U ′ to U the extension problem.An integer linear programming formulation is given in Section A in the appendix.Note that the extension problem comprises additional symmetry given by the pointwise stabilizer of H ∞ of order q n−1 .
Given a set U ′ satisfying all of the necessary conditions mentioned so far it is neither clear that an extension to a corresponding avsp U always exists nor that it is, in the case of existence, unique up to symmetry.Indeed, we will give counter examples later on.However, for the nine classified configurations U ′ in PG (3,2) it turns out that there always is an up to symmetry unique extension.
For n = 7 we conclude , and dim(L ′ ) = 3, so that L ′ = E and dim(E i ∩ B) ≥ 1 for i = 1, 2. Thus, we have U ′ ≤ E, B , i.e., U ′ is not spanning, which is a contradiction.
For n = 6 we have dim(B) = 1, i.e., B is a point.Since U is tight we have B ≤ L ′ .W.l.o.g.we assume B ≤ E 1 .Since E 1 intersects each of the lines L j in at least a point, we have E 1 = E. Since U is irreducible E 2 is not contained in the solid S := E, B .Since E 1 intersects each of the lines L j in at least a point, we have that the line L ′ ≤ E intersects each of the lines L j in at least a point.Since B ≤ L ′ this is impossible.
Proposition 34.Let U be a tight irreducible avsp of PG (5,2), then U has one of the following types: All types are realizable.
Proof.Let the type of U be 5 m5 . . .For small sizes we have enumerated the isomorphism types of tight irreducible avsps in PG (5,2), see Table 1.The last row concerns the parts U ′ at the hyperplane H ∞ at infinity w.r.t. the avsps U counted up to isomorphy in the second row.So, for e.g.types 4 1 3 4 2 4 and 4 2 2 8 there exist configurations U ′ that allow more than one extension up to symmetry.type 4 For the minimum possible size of a tight irreducible avsp in PG(5, 2) we can write down all implications of the stated necessary conditions for the part U ′ at infinity.So, for type 4 1 3 6 configuration U ′ consists of one plane E and six lines L = {L 1 , . . ., L 6 } satisfying the following conditions: (1) the configuration is spanning, i.e., E, L 1 , . . ., L 6 = PG(4, 2); (2) the configuration is tight, i.e., there does not exist a point P that is contained in E and all lines in L; (3) the lines in L form a 2-divisible set of lines, i.e., each hyperplane contains an even number of lines; (4) each line L i intersects E in at least a point; (5) hyperplanes that contain E also contain at least two lines.
Proof.All two possibilities are excluded using ILP computations, see Section A. They are also excluded using GAP computations.
All types are realizable.
Corollary 38.If U is a tight irreducible avsp of PG(6, 2) of minimum possible size, then #U = 8 and U has type 4 8 .
Here we describe all four isomorphism types of homogeneous irreducible tight avsps U of PG(6, 2) of type 4 8 .Geometrically each U is given by eight solids S 1 , . . ., S 8 in PG(6, 2) intersecting a hyperplane H ∞ in a plane (plus some extra conditions).Here we directly consider the part U ′ at infinity, i.e. the eight planes π 1 , . . ., π 8 ∈ H ∞ ∼ = PG (5,2) given by π i = S i ∩ H ∞ .The conditions for the pairwise intersections are Since the planes form a spanning 2-divisible set we have for every hyperplane H of H ∞ ∼ = PG (5,2).Let e i denote the ith unit vector, i.e., the vector with a 1 at the i-th position and zeros everywhere else.If the pairwise intersection of the planes π i is a line in all cases then they span a solid, which contradicts the condition that not all eight planes can be contained in a hyperplane.W.l.o.g.we assume π 1 = e 1 , e 2 , e 3 and π 2 = e 3 , e 4 , e 5 , i.e., the intersection point between π 1 and π 2 is e 3 .Since the intersection of all eight planes is empty we assume w.l.o.g. that π 3 does not contain π 1 ∩ π 2 = e 3 .Up to symmetry we have the following three cases for π 3 : e 4 , e 2 + e 5 ; and Starting from the three possibilities for π 1 , π 2 , π 3 we build up a graph whose vertices consist of the planes that have intersection dimension 1 or 2 with π i for 1 ≤ i ≤ 3, c.f. Condition (10).Two vertices π and π ′ are connected by an edge if 1 ≤ dim(π ∩ π ′ ) ≤ 2, c.f. Condition (10).For these graphs we determine all cliques of size five and check Condition (11) afterwards: The overall computation took just a few minutes.Note that the constructed 752 cases are just candidates for the extension problem to eight solids.Up to symmetry they decompose into just four non-isomorphic examples.It turns out that they can be distinguished by the maximum number γ 0 of incidences of a point and the eight planes, which has to lie between 2 and 5.In Table 2 we summarize incidence counts.
For γ 0 = 2 we consider an arbitrary plane π contained in the hyperbolic quadric Q = Q + (5, 2), which form a single orbit under its collineation group PGO + (6, 2) = C 2 × PGL(3, 2) = S 8 of order 40,320.From the 35 points on Q the points in π have no incidences with the eight planes while all other 28 points on Q have exactly two incidences.This example is obtained in 16 cases.The symmetry group of the eight planes has order 1344 and type C Choose seven planes π i := f j : j ∈ ℓ i for 1 ≤ i ≤ 7 and an eight plane.π 8 = K := j∈ℓi f j : 1 ≤ j ≤ 7 .Note that K itself is also a Fano plane (of course with a different embedding).The points with three incidences with the eight planes are the f i for 1 ≤ i ≤ 7 and the points with two incidences with the eight planes are the points of K.This example is obtained in 112 cases.The symmetry group of the eight planes has order 168 and type PGL (3,2). For With this, we construct the eight planes as where The points with four incidences with the eight planes are Q 1 , . . ., Q 4 .The lines with two incidences with the eight planes are Q i , Q i+1 for 1 ≤ i ≤ 4 (again setting Q 5 = Q 1 ; so this is some kind of a cyclic construction).This example is obtained in 192 cases.The symmetry group of the eight planes has order 128 and type D 2 8 : C 2 .For γ 0 = 5 let {Q 1 , Q 2 , R 1 , R 2 , S 1 , T 1 } be a basis of H ∞ .With this, we set S 2 := S R 2 , S i for i ∈ {1, 2}, and Q i , R i , T j for i, j ∈ {1, 2}, which also reflects the three orbits of the eight planes w.r.t. the action of their automorphism group.The points with five incidences with the eight planes are R 1 and R 2 .The points with four incidences with the eight planes are Q 1 and Q 2 .The points with three incidences with the eight planes are R 1 + Q 1 and R 2 + Q 2 .The lines with three incidences with the eight planes are R 1 , Q 1 and R 2 , Q 2 .The lines with two incidences with the eight planes are R 1 , R 2 and Q 1 , Q 2 .This example is obtained in 432 cases.The symmetry group of the eight planes has order 1024.

Constructions from projective spreads
A k-spread in PG(n − 1, q) is a disjoint set of k-spaces that partitions PG(n − 1, q).It is well known that k-spreads exist iff k divides n.
Proposition 39.For each positive even integer n there exists a tight irreducible avsp U of PG(n − 1, q) of type (n/2) m , where m = q n/2 .Proof.Let k = n/2 and P be a k-spread of PG(n − 1, q), which has size q k + 1.Now choose an arbitrary element K ∈ P and an arbitrary hyperplane H containing K. With this we set U = P\{K} where we choose H as the hyperplane at infinity.By construction U is an avsp of PG(n − 1, q).Since all elements are pairwise disjoint U is tight and since any two elements span PG(n − 1, q) U is irreducible.
We have seen that in PG(5, 2) there exist tight irreducible avsps of types 3 8 and 2 16 .Starting from a 2-spread of PG(5, q) we can clearly obtain a tight avsp U by removing all lines that are completely contained in an arbitrarily chosen hyperplane H.However, it may happen that U is reducible.This is indeed the case if we start with the Desarguesian line spread.In PG(5, 2) there exist 131,044 non-isomorphic line spreads [MT09].
Conjecture 40.For each integer 1 < k < n that divides n there exists a tight irreducible avsp U of PG(n − 1, q) of type k m , where m = q n−k .
If n is odd no ⌊(n − 1)/2⌋-spread exists, but we can construct tight irreducible avsps from some special large partial spreads.
Proposition 41.For each odd integer n ≥ 5 there exists a tight irreducible avsp U of PG(n − 1, q) of type ((n − 1)/2) m , where m = q (n+1)/2 .Proof.Let k = (n − 1)/2 and P be a vector space partition of PG(n − 1, q) of type (k + 1) 2 k m , where m = q k+1 .Now choose an arbitrary hyperplane H containing the unique (k + 1)-dimensional element K of P. With this we set U = P\{K} where we choose H as the hyperplane at infinity.By construction U is an avsp of PG(n − 1, q).Since all elements are pairwise disjoint U is tight.Any two elements of U span a hyperplane of PG(n − 1, q).Since the elements of U ′ := {U ∩ H ∞ : U ∈ U} span H ∞ , not all elements of U ′ can be contained in a hyperplane of H ∞ and U is irreducible.
Vector space partitions of the used type can be obtained from lifted MRD codes, see e.g.[SSW19] for a survey on MRD codes.They also occur as extendible partial k-spreads, where k = (n − 1)/2, of the second largest size q k+1 and are the main building block in the construction of partial k-spreads of size q k+1 + 1 as described by Beutelspacher [Beu75].For more details on the relations between these different geometrical objects we refer e.g. to [HKK19].
For each n ≥ 5 there also exist a vector space partition P of PG(n − 1, q) of type (n − 2) 1 2 m , where m = q n−2 .Choosing a hyperplane that contains the unique (n − 2)-space as the hyperplane at infinity we can obtain a tight avsp U of PG(n − 1, q) of type 2 q n−2 .The remaining question is whether we can choose P in such a way that U becomes irreducible.

Constructions from the Klein quadric
It seems very likely that the avsp of PG(6, 2) of type 4 8 with maximum point multiplicity 2, see Subsection 4.2, can be generalized to arbitrary field sizes.
Theorem 42.There exists a tight irreducible avsp of type 4 q 3 in AG(6, q) for q even.Proof.We will use the following finite field model of AG (6, q).Let V = F q 3 × F q 3 × F q and let H ∞ be the hyperplane X 3 = 0.So we identify AG(6, q) with the elements of V of the form (a, b, c), where c = 0. Consider the following quadratic form on H ∞ : Q(x, y, 0) := T r q 3 /q (xy).
Let P be the set of planes in the Klein quadric Q = Q + (5, q) that is disjoint to an arbitrary but fixed plane π in Q.One can verify that P is a spanning q-divisible set of q 3 planes in PG(5, q) such that the intersection of a pair of planes is a point, i.e., all known conditions for the part U ′ at infinity of a tight irreducible avsp of PG(6, q) of type 4 q 3 are satisfied.The remaining question is whether a solution of the extension problem for P exists.
Conjecture 43.The extension problem for P admits a solution for all prime powers q.
Theorem 42 shows the conjecture for q even.By computer we showed Conjecture 43 for q = 3, 5.

Conclusion
We have introduced the geometrical object of affine vector space partitions.To make their study interesting we need the additional conditions of tightness and irreducibility, which are natural in the context of hitting formulas.A very challenging problem is the determination of the minimum possible size of an irreducible tight avsp of PG(n − 1, q).To this end we have obtained some preliminary results for arbitrary field sizes but small dimensions and for the binary case with medium sized dimensions.We also gave a parametric construction that matches the known exact values in many cases.That irreducible tight avsps are nice geometric objects can be e.g.seen at their sometimes large automorphism groups as well as the mentioned connection to the hyperbolic quadric Q + (5, q).While we have obtained a few insights, many questions remain open.So, we would like to close with a list of a few open problems: Searching a tight irreducible avsp U in PG(n − 1, q) directly can be achieved by a similar model.Now let C be the set of subspaces of PG(n − 1, q) that are not incident with H ∞ .Again we use binary variables x C for all C ∈ C with the meaning x C = 1 iff C ∈ U. Partitioning the affine points is modeled by for all points P not contained in H ∞ .The condition that U is tight can be written as for all points Q ≤ H ∞ .In order to model the condition that U is irreducible we say that a subspace A escapes a subspace B if A has both points that are contained and points that are not contained in B. So, for each B ∈ C we require Of course we can fix the type of U by additional equations.Using a target function we can minimize or maximize #U as well as the number of i-dimensional elements.We have to mention that this ILP formulation comprises a lot of symmetry, so that it can be solved in reasonable time for small parameters n and q only.However, we can use the inherent symmetry to fix some of the x C variables.I.e. the symmetry group acts transitively on the set of a-spaces that are not contained in H ∞ .For pairs of an a-space A and a b-space B that both are not contained in H ∞ , the different orbits under the action of the symmetry group are characterized by the invariant dim(A ∩ B).

B Technical details
In order to keep the paper more readable, we have moved some technical details, that may also be left to the reader, to this section.The proof of Lemma 3 uses the numbers m (j) i satisfying certain constraints.For completeness we state how those number can be computed in Algorithm 1.
Lemma 50.Let P be a 2-divisible set of three points in PG(3, 2) then P forms a line.
Proof.Let P = {P 1 , P 2 , P 3 } and L := P 1 , P 2 .Since all hyperplanes containing L have to contain P, we have P 3 ∈ L.
Lemma 51.Let P be a 2-divisible set of six points in PG(3, 2) then P is the disjoint union of two lines.
Proof.If H is a hyperplane containing all points of P, then there is a unique point P ≤ H with P / ∈ P. Since every hyperplane H ′ that does not contain P intersects P in cardinality 3, so that this case cannot occur, i.e., P is spanning.From the standard equations we compute a 0 = 0, a 2 = 9, and a 4 = 6 for the spectrum.From the MacWilliams transform for the corresponding linear code we conclude the existence of a triple of points P ′ forming a line.Since P\P ′ is also 2-divisible the statement follows from Lemma 50.
We remark that there exists a second 2-divisible set of six points -a projective base of dimension 5, which clearly cannot be embedded in PG(3, 2).
Lemma 52.Let P be a 2-divisible set of eight points in PG(3, 2) then P is either an affine solid or given by the points of a plane and an intersecting line without the intersection point.For n = 7 there is a unique example: