Partial covers of PG(n,q)
aa r X i v : . [ m a t h . C O ] O c t Partial covers of PG( n, q ) S. Dodunekov L. Storme G. Van de Voorde ∗ Abstract
In this paper, we show that a set of q + a hyperplanes, q > a ≤ ( q − /
4, that does not cover PG( n, q ), does not cover at least q n − − aq n − points, and show that this lower bound is sharp. If the number of non-covered points is at most q n − , then we show that all non-covered pointsare contained in one hyperplane. Finally, using a recent result of Blokhuis,Brouwer, and Sz˝onyi [3], we remark that the bound on a for which theseresults are valid can be improved to a < ( q − / a is sharp. Let PG( n, q ) denote the n -dimensional projective space over the finite field F q with q elements, where q = p h , p prime, h ≥
1. We denote the number of pointsin PG( n, q ) by θ n , i.e., θ n = q n +1 − q − .Let C be a family of q + a hyperplanes of PG( n, q ). Denote by C ( P ) the setof hyperplanes of C containing P . A ( q + a )-cover C of PG( n, q ) is a family C of q + a different hyperplanes in PG( n, q ) such that |C ( P ) | ≥ , ∀ P ∈ PG( n, q ). A partial ( q + a )-cover S is a set of q + a hyperplanes such that there is at leastone point Q in PG( n, q ) such that |S ( Q ) | = 0. A point H for which |S ( H ) | = 0,is called a hole of S . We denote the set of holes of S by H S .A blocking set of PG( n, q ) is a set B of points such that each hyperplane ofPG( n, q ) contains at least one point of B . A blocking set B is called trivial ifit contains a line of PG( n, q ). If a hyperplane contains exactly one point of ablocking set B in PG( n, q ), it is called a tangent hyperplane to B , and a point P of B is called essential when it belongs to a tangent hyperplane to B . Ablocking set B is called minimal when no proper subset of B is also a blockingset, i.e., when each point of B is essential.It is clear that a cover of PG( n, q ) is a dual blocking set. Dualizing theabove definitions yields that a cover C is called trivial if it contains all hyper-planes through a certain ( n − minimal if no proper subset of C isa cover. A hyperplane π is essential to a cover C if there is a point P ∈ π suchthat C ( P ) = { π } . The following reducibility results will be used throughout this article. ∗ This author is supported by the Fund for Scientific Research Flanders (FWO - Vlaan-deren). esult 1. [7, Remark 3.3] A blocking set of size at most q in PG(2 , q ) isuniquely reducible to a minimal blocking set. Result 2. [6, Corollary 1] A blocking set of size smaller than q in PG( n, q ) isuniquely reducible to a minimal blocking set. In Theorem 7, we extend the following result of Blokhuis and Brouwer togeneral dimension.
Result 3. [2] Let B be a blocking set in PG(2 , q ) . If | B | = 2 q − s , then thereare at least s + 1 tangent lines through each essential point of B . Finally, for q a prime, we use the following result, proven by Blokhuis [1] for n = 2. Result 4. [5] Let B be a non-trivial blocking set in PG( n, p ) , where p is an oddprime. Then | B | ≥ p + 1) / . PG(2 , q ) Throughout this section, S will denote a partial ( q + a )-cover of PG(2 , q ), with0 ≤ a ≤ ( q − / q > Theorem 5. If |H S | ≤ q + a , then |H S | ≤ q , and the holes are collinear.Proof. Let |H S | = x . Suppose that there are three non-collinear points in H S ,otherwise the theorem is proven. The set H S can be covered by at most ( x +1) / L . Let L ′ be the minimal cover of H S contained in L . The set S ∪ L ′ is a cover in PG(2 , q ). Since the size of S ∪ L ′ isat most q + a + ( q + a + 1) / ≤ q , there is a unique minimal cover C containedin S ∪ L ′ (Result 1).Let ℓ y ∈ L ′ be a y -secant to H S with y ≤ ( q − a − /
2. Interchanging ℓ y by y other lines gives, together with the lines of S , another cover C ′ , with |C ∪ C ′ | ≤ q + a + ( q + a + 1) / q − a − / ≤ q . Hence, by the uniquereducibility property, there is a unique minimal cover contained in C ∪ C ′ , hencein C ∩ C ′ . This minimal cover does not contain ℓ y , hence L ′ contains only lineswith at least ( q − a − / |L ′ | = 1, then the theorem is proven. Remark that if there is only onelong secant and there are q holes, then a = 0.Suppose that |L ′ | = z . These z secants, together with the q + a lines of S ,form a cover C ′′ . Then there is a line L in L ′ with less than ( q + a +1+ (cid:0) z (cid:1) ) /z holes.Suppose to the contrary that any line in L ′ contains at least ( q + a + 1 + (cid:0) z (cid:1) ) /z holes, then there are at least z ( q + a + 1 + (cid:0) z (cid:1) ) /z − (cid:0) z (cid:1) = q + a + 1 > q + a holes, a contradiction. We construct a new cover by replacing this line L withless than ( q + a + 1 + (cid:0) z (cid:1) ) /z lines, one through each hole on L . In total, withthe z secants and the lines of S , this set of lines constitutes a cover C ′′′ of sizeat most q + a + z + ( q + a + 1 + (cid:0) z (cid:1) ) /z . If q + a + z + ( q + a + 1 + (cid:18) z (cid:19) ) /z ≤ q, (1)2he unique reducibility property (Result 1) shows that there is a minimal covercontained in C ′′ ∩ C ′′′ , which does not contain the line L . This implies that theline L was not essential to the cover C ′′ , a contradiction. It is easy to checkthat for z ≥ z <
9, inequality (1) holds for a ≤ ( q − / q > C ′′ . Oneach of these secants, there are at least ( q − a − / q − a − / − · / q − a − / − > q + a if a < (7 q − /
25. Since a ≤ ( q − /
4, and ( q − / < (7 q − /
25, thetheorem follows.
Corollary 6.
Let q be a prime. If |H S | ≤ q + a , then S consists of q linesthrough the same point R and a lines l , . . . , l a , not through R .Proof. It follows from Theorem 5 that the holes are contained in one line, say M . Then the lines of S , together with M , constitute a cover C of size q + a + 1 < q + 1) /
2. Result 4, together with Result 1, shows that the unique minimalcover contained in C is the set of all lines through a point R . It is clear that theline M is one of the lines through R . The other a lines are random, but do notcontain R . PG( n, q ) Before extending the results of Section 2 to general dimension, we need theextension of Result 3.
Theorem 7.
The number of tangent hyperplanes through an essential point ofa blocking set B of size q + a + 1 , | B | ≤ q , in PG( n, q ) is at least q n − − aq n − .Proof. The arguments of this proof are based on the proof of Proposition 2.5 in[8]. For n = 2, Result 3 proves this theorem. Assume by induction that thetheorem holds for all dimensions i ≤ n −
1. Let B be a blocking set in π =PG( n, q ). Since | B | ≤ q , there is an ( n − L in π that is skew to B .Let H be a hyperplane through L . Embed π in PG(2 n − , q ). Let P be aPG( n − , q ), skew to π , in PG(2 n − , q ). Then h B, P i , the cone with vertex P and base B , is a blocking set with respect to the ( n − n − , q ).Let H ∗ = H be a hyperplane through L only sharing one point Q with B . Since | B | is at most 2 q , there are at least 2 tangent hyperplanes through L , hence H ∗ can be chosen different from H .Let S be a regular ( n − L and h Q, P i in W , the (2 n − L and h Q, P i . Using the Andr´e-Bruck-Boseconstruction (see [4]), this yields a projective plane PG(2 , q n − ) = Π W . Thearguments of [8, Proposition 2.5] show that H defines a line ℓ in Π W , only havingessential points of the blocking set ¯ B of size 1 + ( q + a ) q n − = q n − + aq n − + 1,where ¯ B is the blocking set in PG(2 , q n − ), corresponding to h B, P i . Thisnumber of points comes from h Q, P i at infinity, which is one point of the blockingset, and the q + a affine points R i of B , all on a cone h R i , P i with q n − affinepoints. Result 3 shows that any essential point of ¯ B lies on at least q n − − aq n − B in Π W . We will show that the number oftangent lines through an essential point of the blocking set ¯ B in Π W is a lowerbound on the number of tangent hyperplanes through an essential point of B in PG( n, q ).A tangent line through an affine essential point R of ¯ B corresponds to an( n − h R, Ω i , with Ω a spread element of S . The space h R, Ω i is notnecessarily a tangent hyperplane to B in PG( n, q ). Note that Ω = h Q, P i , sinceboth are spread elements and cannot coincide since h Q, P i is an element of theblocking set, hence h R, Q, P i cannot be a tangent space.The projection of h R, Ω i from P onto PG( n, q ) is an ( n − R in PG( n, q ) which is skew to Q since Ω ∩ h Q, P i = ∅ , andwhich only has R in common with B since h Ω , R i ∩ h B, P i = { R } . Hence, thisprojection is a tangent ( n − R to B in PG( n, q ). So we haveshown that any tangent line in R to ¯ B gives rise to a tangent hyperplane to B in R . If any tangent line to ¯ B in R gives rise to a different tangent hyperplaneto B , the theorem is proven.Let η be a tangent hyperplane to B in R which is the projection of twotangent lines h Ω , R i and h Ω ′ , R i . The dimension of h η, P i is 2 n −
3, anddim( h η, P i ∩ W ) = 2 n −
4. A hyperplane of PG(2 n − , q ) contains exactlyone element of a regular ( n − ′ , Ω = Ω ′ . So η is the projection of at most one such ( n − Q of B , it is possible to select a tangent hyperplane H through Q , and to let this tangent hyperplane H play the role described inthe preceding paragraph. Since Q is an affine essential point, this implies that Q lies in at least q n − − aq n − tangent hyperplanes to B . Lemma 8.
Let S be a partial ( q + a ) -cover of PG( n, q ) , a < q . If all holes of S are contained in a hyperplane π of PG( n, q ) , then |H S | ≥ q n − − aq n − .Proof. The hyperplanes of S , together with the hyperplane π that containsall holes, form a cover of size q + a + 1, in which π is an essential hyperplane.Dualizing gives a blocking set B of size q + a +1, where the dual of π is an essentialpoint. Theorem 7 shows that the dual of π lies on at least q n − − aq n − tangenthyperplanes to B . Dualizing again shows that π contains at least q n − − aq n − points that are only covered by π . Removing π shows that there are at least q n − − aq n − holes. Remark 9.
The lower bound in Lemma 8 is sharp. Let S be the set of q hyperplanes through a fixed ( n − -space π n − . Let H be the hyperplane through π n − , which is not chosen. Take a hyperplanes for which the ( n − -dimensionalintersections with H are all distinct and go through a common ( n − -space of π n − , then there are exactly q n − − aq n − holes. From now on, S denotes a partial ( q + a )-cover of PG( n, q ), n ≥
3. Wedenote the following property by ( A x ):( A x ) If S is a partial ( q + b )-cover in PG(2 , q ), b ≤ x < ( q − / , with at most q + b holes, then q − b ≤ |H S | ≤ q and the holes are collinear.Note that we have shown in Theorem 5 and Lemma 8 that the property ( A x )holds for x ≤ ( q − / q >
13. 4 emma 10.
Assume ( A x ) for all x ≤ a . If a partial cover S ′ of PG(2 , q ) contains 3 non-collinear holes, then |S ′ | > q + a .Proof. If |S ′ | = q + a , this follows immediately from property ( A x ). So sup-pose that |S ′ | = q + x ′ , x ′ < a , and that there are 3 non-collinear holes, say H , H , H . Let P be a point not on H H , H H , or H H . Adding a − x ′ lines through P , different from P H , P H , P H , to the partial cover S ′ givesa cover S ′′ , with |S ′′ | = q + a . Applying property ( A x ) to S ′′ , the corollaryfollows. Lemma 11.
Assume ( A x ) for all x ≤ a , and |H S | ≤ q n − . A line that contains holes of S , contains at least a + 3 holes of S .Proof. Let L be a line with t holes, t < q − a , and let π be a plane through L .Assumption ( A x ) shows that if π contains at most q + a holes, there are at least q − a holes, which are all collinear, a contradiction. Hence, every plane through L contains at least q + a + 1 holes, which implies that there are at least θ n − ( q + a + 1 − t ) + t holes in PG( n, q ), which has to be at most q n − . If t = a + 2, θ n − ( q + a + 1 − a −
2) + a + 2 > q n − , a contradiction. Hence, t is at least a + 3. Lemma 12.
Assume ( A x ) for all x ≤ a , and |H S | ≤ q n − . Every hole of S lieson more than q n − / lines with at least q − a holes.Proof. Let R be a hole. There is a line L through R containing only coveredpoints and R , otherwise there would be at least θ n − +1 holes. Using assumption( A x ) and Lemma 10, we see that a plane through L contains either at most q − R , different from L , or it contains at least q + x holesdifferent from R .Suppose that there are X planes through L with at most q − R . Using assumption ( A x ) and Lemma 10, we see that the number of holesis at least X ( q − a −
1) + ( θ n − − X )( q + a ) + 1 , which has to be at most q n − . Putting X = q n − / q n − / q holes. Again usingassumption ( A x ), we see that in each of these planes, there is a line through R containing at least q − a − Theorem 13.
Assume ( A x ) for all x ≤ a , and |H S | ≤ q n − . Then the holes of S are contained in one hyperplane of PG( n, q ) .Proof. For n = 2, this is assumption ( A x ) with x = a . Suppose by inductionthat this theorem holds for any dimension i ≤ n − π of PG( n, q ) with at most q n − holes. Let R be a hole. There is a line L through R containing only coveredpoints and R . Suppose that all planes through L contain more than q holes,then there would be at least θ n − q + 1 holes, a contradiction. Suppose thatthere is a d -dimensional space π d with at most q d − holes. Then there is a( d + 1)-dimensional space containing π d with at most q d holes. Otherwise, the5umber of holes would be at least θ n − d − ( q d + 1 − q d − ) + q d − , a contradictionif d ≤ n −
1. Hence, by induction, there is a hyperplane π of PG( n, q ) with atmost q n − holes.Using the induction hypothesis, all holes in π are contained in an ( n − π n − of π . Moreover, Lemma 8 shows that the number ofholes in π n − is at least q n − − aq n − .There are at least θ n − ( q − a −
1) + 1 holes in PG( n, q ) since every planethrough L contains at least q − a − R ′ that is not contained in π n − .Now we distinguish between two cases. Case 1:
All lines through R ′ with at least q − a holes are lines which intersect π n − . Lemma 12 shows that there are at least q n − / a + 3 holes (see Lemma 11), counting theholes in h R ′ , π n − i yields that this number is at least q n − ( q − a − / q n − − aq n − − q n − / a + 2) + 1 . If all holes are contained in h R ′ , π n − i , the theorem is proven. Suppose nowthat not all holes are contained in the hyperplane h R ′ , π n − i . Let R ′′ be a holenot in h R ′ , π n − i . Connecting R ′′ with all the holes in h R ′ , π n − i yields at least( a + 2)( q n − ( q − a − / q n − − aq n − − q n − / a + 2) + 1) + 1 holes, whichis more than q n − , a contradiction. Case 2:
There is a line through R ′ , skew to π n − , with at least q − a holes.This yields at least( q − a )( q n − − aq n − )( a + 1) + q n − − aq n − + q − a > q n − holes, a contradiction. Theorem 14.
Assume ( A x ) for all x ≤ a , then the number of holes of S is atleast q n − − aq n − .Proof. This follows immediately from Theorem 13 and Lemma 8.
Corollary 15.
Assume ( A x ) for all x ≤ a , and |H S | ≤ q n − . If q is a prime, S consists of q hyperplanes through a common ( n − -space π and a otherhyperplanes, not through π .Proof. It follows from Theorem 13 that the holes are contained in one hyper-plane, say µ . Then the hyperplanes of S , together with µ , constitute a cover C of size q + a + 1 < q + 1) /
2. Result 4, together with Result 2, shows that theunique minimal cover contained in C is the set of all hyperplanes through an( n − π . Since this set covers PG( n, q ) entirely, the hyperplane µ is oneof the hyperplanes through π . The other a hyperplanes are random, but do notcontain π .As remarked before, assumption ( A x ) holds for all partial ( q + a )-covers ofPG( n, q ), x ≤ ( q − / q > A x ) is valid for all x , where x < ( q − /
3. Moreover, the followingexample shows that the upper bound a < ( q − / xample 16. Let a = ( q − / and let S be a set of q − lines L i through apoint P , and a + 1 other lines through a fixed point, lying on one of the lines L i . Then there are q − a −
1) = q + a holes, lying on two lines. Combining Theorems 13, 14 and Corollary 15 with the result of Brouwer,Blokhuis and Sz˝onyi, yields the following theorem.
Theorem 17. If S is a partial ( q + a ) -cover of PG( n, q ) , a < ( q − / , withat most q n − holes, then there are at least q n − − aq n − holes and the holesare contained in one hyperplane. If q is a prime, S consists of q hyperplanesthrough a common ( n − -space π and a other hyperplanes, not through π . Acknowledgements:
This research was supported by the Project
Com-bined algorithmic and theoretical study of combinatorial structures between theFund for Scientific Research Flanders-Belgium (FWO-Flanders) and the Bul-garian Academy of Sciences. The third author wants to thank the people at theBulgarian Academy of Sciences in Veliko Tarnovo and Sofia for their hospitalityduring her research visit in the framework of this project.The authors thank the referee and A.E. Brouwer for their valuable comments.
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FiniteFields Appl. (1997), 187–202.[8] T. Sz˝onyi and Zs. Weiner, Small blocking sets in higher dimensions. J.Combin. Theory, Ser. A (2001), 88–101.Address of the authors:Stefan Dodunekov:IMI / MFI8, Acad. G. Bonchev 7113 Sofia (Bulgaria)[email protected] Storme, Geertrui Van de Voorde:Department of pure mathematics and computer algebraGhent UniversityKrijgslaan 281-S229000 Ghent (Belgium) { ls,gvdvoorde } @cage.ugent.behttp://cage.ugent.be/ ∼ { ls,gvdvoorde }}