Optimal subspace codes in ${\rm PG}(4,q)$

We investigate subspace codes whose codewords are subspaces of ${\rm PG}(4,q)$ having non-constant dimension. In particular, examples of optimal mixed-dimension subspace codes are provided, showing that ${\cal A}_q(5,3) = 2(q^3+1)$.


Introduction
Let V be an n-dimensional vector space over GF(q), q any prime power.The set S(V ) of all subspaces of V , or subspaces of the projective space PG(V ), forms a metric space with respect to the subspace distance defined by d s (U, U ′ ) = dim(U + U ′ ) − dim(U ∩ U ′ ).In the context of subspace codes, the main problem is to determine the largest possible size of codes in the space (S(V ), d s ) with a given minimum distance, and to classify the corresponding optimal codes.The interest in these codes is a consequence of the fact that codes in the projective space and codes in the Grassmannian over a finite field referred to as subspace codes and constantdimension codes (CDC), respectively, have been proposed for error control in random linear network coding.An (n, M, d) q mixed-dimension subspace code is a set C of subspaces of V with |C| = M and minimum subspace distance The maximum size of an (n, M, d) q mixed-dimension subspace code is denoted by A q (n, d).In this paper we discuss the smallest open mixed-dimension case, which occurs when n = 5 and d = 3.In particular, examples of optimal mixed-dimension subspace codes are provided, showing that A q (5, 3) = 2(q 3 + 1).
We wish to remark that, independently, also Honold, Kiermaier and Kurz proved that A q (5, 3) = 2(q 3 + 1).We refer to their article [5] for their construction method, and for many other results on subspace codes.

Preliminaries
A partial line spread of PG(4, q) is a set of pairwise disjoint lines of PG (4, q).From [1], the largest partial line spread of PG(4, q) has size q 3 + 1.Let C be an optimal (5, 3) q subspace code.Since the lines contained in C are pairwise disjoint, it follows that C contains at most q 3 + 1 lines.A dual argument shows that C contains at most q 3 + 1 planes.Hence, if C consists of lines and planes, we have that |C| ≤ 2(q 3 + 1) and, in order to construct such a code one needs to find a set L of q 3 + 1 pairwise skew lines and a set P of q 3 + 1 planes mutually intersecting in exactly a point such that no line of L is contained in a plane of P.
Notice that C contains at most one point and, dually, C contains at most one solid.The next result gives an upper bound on the size of C. Lemma 1.1.If C contains a point, then C contains at most q 3 planes.Dually, if C contains a solid, then C contains at most q 3 lines.
Proof.We only need to prove the second assertion.Assume that C contains more than q 3 lines.From the discussion above, C contains a set L consisting of q 3 + 1 lines.Let L ′ be the set of points of PG(4, q) covered by the lines of L. Then every solid is covered by at least a member of L. Indeed, if S is a solid of PG(4, q) and s is the number of lines of L contained in S, we have that From the previous result, it follows that A q (5, 3) ≤ 2(q 3 + 1) and there are four possibilities for the code C: I) C consists of one point, q 3 + 1 lines and q 3 planes; II) C consists of q 3 lines, q 3 + 1 planes and one solid; III) C consists of one point, q 3 lines, q 3 planes and one solid; IV) C consists of q 3 + 1 lines and q 3 + 1 planes.
In the remaining part of the paper, we exhibit examples of (5, 2(q 3 + 1), 3) q codes showing that A q (5, 3) = 2(q 3 + 1).These results are achieved by using a suitable subgroup G of PGL(5, q) of order q 3 , if q is odd, or of order q 3 − q, if q is even.We shall find it helpful to work with the elements of G as matrices in GL (5, q).We shall consider the points of PG(4, q) as column vectors, with matrices in G acting on the left.

The odd characteristic case
Let q = p h , where p is an odd prime.Let PG(4, q) be the four dimensional projective space over GF(q) equipped with homogeneous coordinates (X 1 , X 2 , X 3 , X 4 , X 5 ), let π be the projective plane with equations X 4 = X 5 = 0 and let ℓ be the line of π with equations X 3 = X 4 = X 5 = 0. Let ω be a primitive element of GF(q) and denote by Π i the solid of PG(4, q) passing through π with equation Lemma 2.1.Let a, b, c be fixed elements of GF(q) such that the polynomial X 3 +aX 2 +bX+c = 0 is irreducible over GF(q) and . On the other hand, since M n r,s,t = M nr,ns,nt+n(n−1)rs , we have that each element of G distinct from the identity has order p.
Remark 2.2.The group G is not abelian, in particular the map M r,s,t →   an isomorphism between G and the so called Heisenberg group.
We need the following preliminary results.
Lemma 2.3.The group G has 2q + 3 orbits on the points of PG(4, q): a) q + 1 orbits of size one, each consisting of a point of ℓ, b) one orbit of size q 2 , consisting of the points of π \ ℓ, c) q + 1 orbits of size q 3 , each consisting of the points of the set Π i \ π.
Lemma 2.4.The group G has the following orbits on the lines of PG(4, q) distinct from ℓ: a) q + 1 orbits of size q, each consisting of the lines of π passing through a point of ℓ, b) (q + 1) 2 orbits of size q 2 , each consisting of lines of Π i , not in π, passing through a point of ℓ, 1 ≤ i ≤ q + 1, c) q(q + 1) orbits of size q 3 , each consisting of lines of Π i skew to ℓ, 1 ≤ i ≤ q + 1, but intersecting π in a point, d) q 3 orbits of size q 3 , each consisting of lines that are disjoint from π.
In particular, every line-orbit of type d) is a partial line spread.
Proof.To describe orbits of type a) or b) is trivial.
Line-orbits of type c) Let x be a line of Π i , for some i, such that x is skew to ℓ and let X = x ∩ π.The stabilizer of X in G, say G X , is the group of order q consisting of elements of type M 0,0,t .In particular, G X is an elation group having as axis the plane π and as center the line ℓ, and every orbit of G X on points not in π consists of q points of a line meeting ℓ in one of its points.It follows that x G X consists of q lines through X in a plane.Since the group G permutes the q 2 points of π \ ℓ in a single orbit, we have that the orbit x G contains q 3 lines.

Line-orbits of type d)
Let y be a line skew to π.Since y cannot be contained in a hyperplane Π i , 1 The stabilizer of y in G is trivial.Indeed, assume on the contrary that id = g ∈ G fixes the line y.Then, since G fixes Π i , we would have that g fixes y ∩ Π i = Y i , which is a contradiction, since, from Lemma 2.3, G acts sharply transitive on the points of Π i \ π.It follows that the orbit y G consists of q 3 lines such that the points covered by its members are all the q 4 + q 3 points of PG(4, q) \ π.As a consequence, lines in y G are pairwise skew.
We are ready to prove our main result of this section.
Theorem 2.5.There exists a set L consisting of q 3 +1 pairwise skew lines and a set P consisting of q 3 + 1 planes mutually intersecting in exactly a point, such that no line of L is contained in a plane of P.
Proof.Let α be a plane of PG(4, q) such that α ∩ π is a point A not belonging to ℓ.Then α ∩ Π i , 1 ≤ i ≤ q + 1, is a line passing through the point A. Let P ′ = α G .We claim that P ′ is a set consisting of q 3 planes pairwise intersecting in a point.Indeed, assume on the contrary that there exist two distinct planes α, α ′ in P ′ such that α ∩ α ′ is a line, say b.Let B be the line G-orbit containing b. Two possibilities occur: either b meets π in a point not on ℓ, or b is disjoint from π.
If the former case occurs, there exists a solid Π i such that b ∈ Π i , for some i, and B is a line-orbit of type c).Consider the tactical configuration whose points are the elements of B and whose blocks are the elements of P ′ .Since |B| = |P ′ | = q 3 and through b, there pass at least two distinct elements of P ′ , it follows that α contains at least two distinct elements of B. Hence α is a plane of Π i and so α ∩ π is a line, a contradiction.
If the latter case occurs, then B is a line-orbit of type d).Consider again the tactical configuration whose points are the elements of B and whose blocks are the elements of P ′ .Arguing as in the previous case, α contains at least two distinct elements of B, say b 1 and b 2 , but then b 1 meets b 2 in a point, contradicting the fact that two distinct lines in B are skew.The plane α contains q 2 lines that are disjoint from π and each one of them belongs to a distinct line-orbit of type d).Since there are q 3 line-orbits of type d), it follows that we can find a line-orbit of type d), say L ′ , such that no line in L ′ is contained in α.
Let r be a line of π distinct from ℓ and let L := L ′ ∪ {r}.Then L is a set consisting of q 3 + 1 pairwise skew lines.Let ξ be a plane passing through the line ℓ distinct from π and let P := P ′ ∪ {ξ}.Then, every plane in P ′ shares a point with ξ.Indeed, if there was a plane β ∈ P ′ meeting ξ in a line, then ξ ∩ π ∈ ℓ, a contradiction.Hence, P is a set consisting of q 3 + 1 planes mutually intersecting in a point.In particular, no line of L is contained in a plane of P, as required.Remark 2.6.With the same notation as in Theorem 2.5, the code C = P ∪ L is an optimal code of type IV ).Let X be a point of ℓ \ r, then the code C = P ′ ∪ L ∪ {X} is an optimal code of type I).Let j be such that Π j = π, ξ , then the code C = P ∪ L ′ ∪ {Π k }, k = j, is an optimal code of type II).Finally, the code C = P ′ ∪ L ′ ∪ {X, Π i } is an optimal code of type III).

The even characteristic case
Let q = 2 h .Let PG(4, q) be the four dimensional projective space over GF(q) equipped with homogeneous coordinates (X 1 , X 2 , X 3 , X 4 , X 5 ), let π be the projective plane with equations X 4 = X 5 = 0 and let ℓ be the line of π with equations X 1 = X 4 = X 5 = 0. Denote by Σ the solid of PG(4, q) with equation X 1 = 0. Then Σ ∩ π = ℓ.Let H be the hyperbolic quadric of Σ having equation X 2 X 5 + X 3 X 4 = 0.
Let α ∈ GF(q) be such that the polynomial X 2 + X + α = 0 is irreducible over GF(q) and let Then G is a group of order q 3 − q with structure C q+1 × (E q × C q−1 ).In particular the subgroup } is an elementary abelian group of order q and the subgroup As we will see in the next lemma, the group G fixes a pencil of quadrics, say F. The pencil F comprises q − 1 parabolic quadrics Q i , 1 ≤ i ≤ q − 1, the solid Σ and the cone C having as vertex the point N = (1, 0, 0, 0, 0) and as base H.The base locus of F is H and F is generated by the quadrics having equation X 2 1 = 0 and X 2 X 5 + X 3 X 4 = 0.The nucleus of the parabolic quadric Q i is the point N , 1 ≤ i ≤ q − 1.
We need the following preliminary results.
Lemma 3.1.The group G has q + 5 orbits on points of PG(4, q): a) the point N , b) an orbit of size q + 1, consisting of the points of ℓ, c) an orbit of size q 2 − 1, consisting of the points of π \ (ℓ ∪ {N }), d) an orbit of size q 2 + q, consisting of the points of H \ ℓ, e) an orbit of size q 3 − q, consisting of the points of Σ \ H, f ) an orbit of size q 3 − q, consisting of the points of C \ (π ∪ H), g) q − 1 orbits of size q 3 − q, each consisting of the points of Proof.The group G fixes the point N and the line ℓ.On the other hand, the subgroup G 1 acts transitively on the points of ℓ.If P ∈ π \ (ℓ ∪ {N }), then the stabilizer of P in G is the subgroup G 2 .Whereas, if P = (0, 0, 0, 0, 1, 0) ∈ H \ ℓ, then the stabilizer of P in G is the subgroup G 3 .Finally, straightforward computations show that if g ∈ G and P belongs either to Σ \ H, or to , respectively and P g = P if and only if g is the identity.
Let R 1 be the regulus of H containing ℓ and let R 2 be its opposite regulus.Since q is even, the lines of Σ that are tangent to H, together with the 2(q +1) lines of R 1 ∪R 2 , are the (q +1)(q 2 +1) lines of a symplectic polar space W(3, q) of Σ. See [4] for more details.Since the group G fixes H, it follows that G fixes W(3, q), also.We denote by T the set of q 3 − q lines of W(3, q) having exactly one point in common with H \ ℓ.In the next results we summarize useful information regarding the action of the group G and its subgroups.Lemma 3.2.Let t be a line of T , then t G 2 ∪ {ℓ} is a regulus of Σ.
Proof.The group G 2 fixes pointwise the plane π : X 4 = X 5 = 0 and in particular the line ℓ.Since G 2 fixes W(3, q) it follows that the parabolic congruence ρ of W(3, q) having as axis the line ℓ is fixed by G 2 .A straightforward computation shows that every line of ρ, distinct from ℓ, is fixed by G 2 .In particular, if r is a line of ρ \ {ℓ}, then G 2 permutes the q points of r \ {r ∩ ℓ} in a single orbit.Let t ∈ T .Then t is disjoint from ℓ. Since through every point of Σ \ {ℓ} there is exactly one line of the parabolic congruence ρ containing it, we have that there are q + 1 lines of ρ, say r 1 , . . ., r q+1 , each meeting t in a distinct point.The lines r 1 , . . ., r q+1 are pairwise disjoint, otherwise there would exist a triangle consisting of lines of W(3, q), contradicting the fact that W(3, q) is a generalized quadrangle.The assertion, now, follows from the fact that t G 2 ∪ {ℓ} consists of q + 1 lines covering (q + 1) 2 points of Σ and meeting three pairwise disjoint lines.

Lemma 3.3. The group G acts transitively on the elements of T .
Proof.Since G acts transitively on the points of H \ {ℓ}, we only need to prove that G permutes the q − 1 tangent lines that are concurrent at some point of The lines of T containing the point U 4 lie in the plane r 1 , r 2 .The stabilizer of U 4 in G, i.e.G 3 , fixes the plane r 1 , r 2 and induces a homology group of order q − 1 having as axis the line r 1 and as center the point U 2 .Hence, the q − 1 lines of T passing through U 4 are permuted in a single orbit under the action of G 3 .
Figure 2 illustrates the initial elements required in the code construction in PG(4, q), q even.It shows the plane π, the line ℓ, and the nucleus N , lying in the plane π, of all the quadrics Q i , 1 ≤ i ≤ q − 1.It shows the hyperbolic quadric H contained in the solid Σ.It shows the cone C with base H and with vertex the point N .The figure also includes the tangent line t to the hyperbolic quadric H, sharing the tangent point T with H. Proposition 3.4.Under the action of G, there are q 2 − q line-orbits of size q 3 − q consisting of mutually disjoint lines.
Proof.First of all notice that if r is a line meeting each of the sets Σ \ H, C \ (π ∪ H), Q i \ H, 1 ≤ i ≤ q −1, in exactly one point, then r G is a set consisting of q 3 −q lines covering (q +1)(q 3 −q) points and hence forming a partial spread of the ambient projective space.In the following we prove that there is a plane containing q 2 − q lines having the desired property.Since two lines in a projective plane have a point in common, no two of them are in the same G-orbit.Hence, they are representatives of q 2 − q distinct line-orbits of size q 3 − q forming a partial spread.Let t be a line of T , where T = t ∩ H, and consider the plane γ = N, t .It follows that γ meets Σ and C in the line t and N T , respectively.On the other hand, a solid Γ containing the plane γ has to meet the parabolic quadric Q i in a quadratic cone, since N ∈ Γ and N is the nucleus of Q i .Notice that every line passing through N meets Q i in a point.Hence, it turns out that two possibilities occur: either γ ∩ Q i is a conic, whose nucleus is N , or γ ∩ Q i is a line passing through T .Since t and N T are two lines that are tangent to Q i at the point T ∈ Q i , the former case cannot occur.Hence, γ meets Q i in a line passing through T , 1 ≤ i ≤ q − 1.It follows that every line of γ containing neither T nor N is a line meeting each of the sets Σ \ H, C \ (π ∪ H), Q i , 1 ≤ i ≤ q − 1, in exactly one point.
In order to prove the next result, we embed S = PG(4, q) as a hyperplane section of PG(5, q), the five-dimensional projective space over GF(q) equipped with homogeneous coordinates (X 1 , X 2 , X 3 , X 4 , X 5 , X 6 ).Let K be the Klein quadric of PG(5, q) defined by the equation X 1 X 6 + X 2 X 5 + X 3 X 4 = 0. Notice that K ∩ S = C. Also, the group G is extended canonically to a subgroup Ḡ of the stabilizer of K in PGL(6, q).In particular, Ḡ We will use the polarity ⊥ of PG(5, q) associated to the Klein quadric K, to show that the existence of line-orbits consisting of mutually disjoint lines implies the existence of plane-orbits consisting of planes mutually intersecting in one point.
Proposition 3.5.Under the action of G, there are q 2 − q plane-orbits of size q 3 − q consisting of planes mutually intersecting in one point.
Proof.Let S ′ be the hyperplane of PG(5, q) with equation X 1 = 0.Then, S ∩ S ′ = Σ and the Klein quadric K meets S ′ in a cone, say C ′ , having as vertex the point N ′ = (0, 0, 0, 0, 0, 1) and as base H.The group Ḡ acts identically on the line N N ′ , hence every hyperplane through Σ = (N N ′ ) ⊥ is fixed by Ḡ.Also, it is clear that the action of Ḡ on S ′ is the same as the action of G on S. Therefore, by Proposition 3.4, under the action of Ḡ, there are q 2 − q line-orbits of size q 3 − q consisting of mutually disjoint lines of S ′ .Let R be one of these q 2 − q orbits.Then, each of the lines in R is disjoint from π ′ = N ′ , ℓ , meets Σ in a point that is not in H, and has only one point in common with C ′ .Consider the plane σ = N, r , where r is a line of R.
Consider the set S = {x ⊥ | x ∈ σ Ḡ}.Since every plane in σ Ḡ contains the point N , it follows that every plane in S lies in S. On the other hand, since two distinct planes in σ Ḡ have exactly the point N in common, it follows that two distinct planes in S generate the hyperplane S and hence they meet exactly in a point.
With the same notation as in Proposition 3.5, we want to provide an alternative description of the q 2 − q plane-orbits of size q 3 − q consisting of planes mutually intersecting in one point.Firstly, we prove the following lemma.Lemma 3.6.σ ∩ K is a conic Proof.Notice that N ⊥ = S, N ′⊥ = S ′ , σ ∩S is a line tangent to K at the point N and σ contains at least two points of K, namely the points N and r ∩ C ′ .It follows that σ ∩ K is either a line, or two lines or a conic.If one of the first two cases occurs, then σ would contain a line of K through N , but all the lines of K through N lie in S, a contradiction.Hence, σ ∩ K is a conic.Remark 3.7.From Lemma 3.6, it follows that every plane in S meets C in a conic.Indeed, σ ∩ K is a conic and Ḡ stabilizes K. Consider the solid Σ ′ = σ, N ′ .Since Σ, N, N ′ is the whole ambient projective space PG(5, q), and Σ ′ contains both N , N ′ , we have that Σ ∩ Σ ′ is a line.From the proof of Proposition 3.4, if r is the line of R such that σ = N, r , then the plane N ′ , r meets Σ in a line t that is tangent to H at the point U .Hence, Σ ∩ Σ ′ = t.Also, since Σ ⊥ ⊂ Σ ′ , we have that Σ ′⊥ ⊂ Σ and, hence, Notice that, since q is even, ⊥ |Σ coincides with the polarity of Σ associated with W(3, q).We have that σ ⊥ ∩ Σ = (σ ∪ Σ ⊥ ) ⊥ = Σ ′⊥ = t and every plane in S meets Σ in a line of T .Let U = t ∩ (H \ ℓ).Let R be the point in common between the unique line in R 2 containing U and ℓ.Let U ′ = σ ⊥ ∩ π.Since σ ⊥ ∩ Σ = t, it follows that U ′ / ∈ ℓ.On the other hand, if U ′ ∈ RN , then the line U U ′ would be contained in σ ⊥ ∩ C, contradicting the fact that σ ⊥ ∩ C is a conic.In other words, let t be a line of T and let U = t ∩ (H \ ℓ).Let R be the point in common between the unique line in R 2 containing U and ℓ.Let U ′ be a point of π \ (ℓ ∪ RN ).Let γ be the plane containing the line t and the point U ′ .Then γ G consists of q 3 − q planes mutually intersecting in one point.There are q 2 − q choices for the point U ′ , and taking into account Lemma 3.3, each of these points gives rise to a representative for a plane-orbit of size q 3 − q consisting of planes mutually intersecting in one point.
Figure 3: Construction in PG(5, q), q even Figure 3 aims at giving a global overview of the setting in PG(5, q), q even, regarding the construction of the subspace code.We see the Klein quadric K, and the two hyperplanes S and S ′ .We see the 3-dimensional hyperbolic quadric H, contained in the intersection of S and S ′ .We see the two cones C and C ′ with base H and with respective vertices N and N ′ .Figure 3 also illustrates the plane σ = r, N .
With the same notation as in Remark 3.7, let γ be the plane t, U ′ such that γ G consists of q 3 − q planes mutually intersecting in one point.
Figure 4 illustrates the construction in the hyperplane S.
We say that γ G is a good plane-orbit.Also, we say that a line-orbit of size q 3 − q consisting of mutually disjoint lines as described in Proposition 3.4 is a good line-orbit.Proposition 3.8.A plane in a good plane-orbit contains exactly q − 1 lines of distinct good line-orbits.
Proof.We need to show that there are exactly q − 1 lines of γ meeting each of the sets Σ \ H, C \ (π ∪ H), Q i , 1 ≤ i ≤ q − 1, in exactly one point.Notice that F induces a pencil of plane quadrics F ′ on γ containing the line t = γ ∩ Σ and the conic γ ∩ C, see Remark 3.7.Furthermore, the base locus of F ′ is the point : Construction in the hyperplane S of PG(5, q), q even line would contain the point U and hence would meet the conic γ ∩ C in a further point, which is a contradiction.Also, γ ∩ Q i is a conic, 1 ≤ i ≤ q − 1, and i (γ ∩ Q i ) consists of q 2 − q + 1 points of γ.Each of the conics in F ′ admits as a nucleus the same point, say C, where C is a point of t distinct from U , see [3,Table 7.7].It turns out that the lines of γ meeting each of the sets Σ \ H, C \ (π ∪ H), Q i , 1 ≤ i ≤ q − 1, in exactly one point are the q − 1 lines passing through C and not containing U and U ′ .
We are ready to prove our main result of this section.Theorem 3.9.There exists a set L consisting of q 3 +1 pairwise skew lines and a set P consisting of q 3 + 1 planes mutually intersecting in exactly a point, such that no line of L is contained in a plane of P.
Proof.Let P 1 be a good plane-orbit.Let P 2 be the set of q + 1 planes generated by a line of R 1 and the point N .Of course, two distinct planes in P 2 share exactly the point N .We claim that a plane of P 1 , say γ, and a plane of P 2 , say τ , meet in a point.
Indeed, if t = γ ∩ Σ is disjoint from ℓ ′ = τ ∩ Σ or ℓ ′ = ℓ, then the assertion is trivial.If t ∩ ℓ ′ is the point U , with ℓ ′ = ℓ, then let R be the point in common between the unique line in R 2 containing U and ℓ.Let U ′ = γ ∩ π.By Remark 3.7, we have that U ′ / ∈ RN and, hence, γ, τ is the whole hyperplane S. Hence, γ ∩τ is a point and P = P 1 ∪P 2 is a set of q 3 +1 planes mutually intersecting in exactly a point.From Proposition 3.8, among the q 2 − q good line-orbits, there are exactly q − 1 of them covered by members of P 1 .Hence, there exists a good line-orbit, say L 1 , such that no line of L 1 is contained in a plane of P 1 .Notice that the (q + 1)(q 3 − q) points covered by the lines of L 1 are those in S \ (π ∪ H).Therefore, a line in L 1 cannot be contained in a plane of P 2 .Moreover, if L = L 1 ∪ R 2 , we have that L is a set of size q 3 + 1 consisting of mutually disjoint lines.In particular, no line of L is contained in a plane of P, as required.
Remark 3.10.With the same notation as in Theorem 3.9, the code C = P 1 ∪ P 2 ∪ L 1 ∪ R 2 is an optimal code of type IV ).In the same fashion, it can be seen that, if π ′ is a plane containing ℓ and not contained in Σ, then the code C = P 1 ∪ (P 2 \ {π}) ∪ {π ′ } ∪ L 1 ∪ R 2 is an optimal code of type IV ).Let π ′ = π and let S 1 be the solid generated by π and π ′ .If S 2 is a solid

Figure 1 :
Figure 1: Construction for q odd

Figure 2 :
Figure 2: Construction for q even