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Inverses of the elements of a linear subspace and related problems


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Inverses of the elements of a linear subspace and related problems

Bence Csajbók

University of Basilicata, Italy

Szeged, June 10-14, 2013


Inverse of an additive subgroup Inverse-closed additive subgroups

Inverse-closed additive subgroups in rings


For a subsetS of a (unitary) ring let:

Sbe the set of invertible elements inS,

S−1be the inverse set ofS, that isS−1={s−1:s∈S}.

IfS−1=S, thenSis called inverse-closed.

Theorem (Kroll, 1992)

Let V be a commutative, unitary, associativeK-algebra with

char(K)6=2. If A is an inverse-closedK-subspace in V and1∈A, then A is a subalgebra of V .

Theorem (Goldstein, Guralnick, Small and Zelmanov, 2006) Characterised pairs A⊆D, where A is an inverse-closed additive subgroup, D is a division ring withchar(D)6=2.


Inverse of an additive subgroup Inverse-closed additive subgroups

The field case


For a subsetS of a fieldFwe use the following notation:

Fora∈FletaS ={as:s ∈S}, that is the elements ofS multiplied bya,

For an integern, letSn={sn:s ∈S}, that is the set ofn-th powers of the elements inS.


LetFbe a field of characteristic p>0and let k be a positive integer:

If A⊆Fis subfield, then Apk is also a subfield ofF,

if A⊆Fis an additive subgroup, then Apk is also an additive subgroup.


Inverse of an additive subgroup Inverse-closed additive subgroups

Theorem (Goldstein, Guralnick, Small and Zelmanov, 2006 and independently Mattarei, 2007)

Let A be an inverse-closed additive subgroup in a fieldF, with characteristic p≥0, then:

1 p6=2and A is a subfield ofFor A=K, whereKis a subfield of F, /∈Kand2∈K,

2 p=2and A is aK2-subspace ofKfor some subfieldKofF. These are indeed inverse-closed additive subgroups:

1 The inverse ofk is(2k)−1, where(2k)−1is inK, soA=Kis inverse-closed.

2 The inverse ofa∈Aisa(a2)−1, where(a2)−1∈K2thusAis inverse-closed.


Inverse of an additive subgroup Hua’s identity

Hua’s identity

Lemma (Hua’s identity)

For invertible a,b such that ab−1is also invertible:

a−(a−1+ (b−1−a)−1)−1=aba.

It follows suddenly that if 1,a,b∈A, that is an inverse-closed additive subgroup in a fieldF, thena2,b2,(a+b)2∈A, thus also


So ifchar(F)6=2, thenAis closed also to multiplication, hence it is a subfield ofF.


Inverse of an additive subgroup Hua’s identity

Theorem (BCs)

Let A be an infinite additive subgroup of the fieldF, wherechar(F)6=2.

If|A\A−1|<|A|, then A is inverse-closed.

Sketch of Proof.

Suppose, contrary to our claim, that there is an elementa∈A\A−1. It can be proved that there exists an elementx ∈A such that


SinceAis an additive subgroup, this implies

(x+x−1)2(2a)−1−x2(2a)−1−x−2(2a)−1=a−1∈A, which contradicts the choice ofa.


Inverse of an additive subgroup Hua’s identity

Again Hua’s identity can be used:

a−(a−1+ (b−1−a)−1)−1=a2b

x−(x−1+ (2a−x)−1)−1=x2(2a)−1, (1) x−1−(x+ (2a−x−1)−1)−1=x−2(2a)−1, (2) (x+x−1)−((x+x−1)−1+(2a−(x+x−1))−1)−1= (x+x−1)2(2a)−1, (3) hold for eachx ∈Fwhen the elements of the set

{x,x−1,x +x−1,(2a)−1x−1,(2a)−1x−1−1,(2a)−1(x+x−1)−1}

are defined and non-zeros. One has to find anx such that the left-hand sides in (1), (2), (3) are inA...

The same result holds ifchar(F) =2 and 1∈A.

I don’t known the answer if 1∈/ A.


Inverse of an additive subgroup Finite fields

The finite field case

Theorem (Goldstein, Guralnick, Small and Zelmanov, 2006 and independently Mattarei, 2007)

Let A be a non-trivial inverse-closed additive subgroup ofGF(q), with q =pn, p prime. Then

A is either a subfield ofGF(q)or

p6=2and A consists of all elements x ∈GF(q)such that xpd +x =0for some1≤d <n and2d|n.

ConsiderGF(qn)as ann-dimensional vector space overGF(q). We can identify thek-dimensionalGF(q)-subspaces ofGF(qn)with the (k−1)-dimensional projective subspaces ofPG(n−1,q).

Ifm+1 dividesn, then letPmbe them-dimensional subspace of PG(n−1,q)corresponding toGF(qm+1).

If 2m+2 dividesn, then letLmbe them-dimensional subspaces ofPG(n−1,q)corresponding to the set of roots ofxqm+1+x =0.


Inverse of an additive subgroup When the inverse is also an additive subgroup

When the inverse set is also an additive subgroup

Corollary (BCs)

Let A be a non-trivial additive subgroup of the fieldF. Suppose that A−1∪ {0}is also an additive subgroup.

1 Ifchar(F)6=2, then A is a one-dimensional subspace over a subfield ofF,

2 Ifchar(F) =2, then A=aB, for some a∈F, where B is a K2-subspace ofKfor some subfieldKofF.

Corollary (BCs)

Let A be a non-trivial additive subgroup ofGF(q). If A−1∪ {0}is also an additive subgroup, then A is a one-dimensional subspace over a subfield ofGF(q). (In Case 2 above the map x →x2is a field automorphism ifFis finite.)


The Singer group and difference sets The Singer group

The Singer group

For an elementx ∈GF(qn) denote byhxithe point of

PG(n−1,q)identified with theGF(q)-subspace generated byx. With this notationhxi=hyiiffx/y ∈GF(q), i.e.xq−1=yq−1. Letαbe a primitive element ofGF(qn). The collineation group generated byhxi → hαxiis a cyclic Singer groupGof

PG(n−1,q), that is a cyclic collineation group of order

(qn−1)/(q−1)permuting the points in one orbit. Moreover it also permutes the hyperplanes in one orbit.


The Singer group and difference sets The Singer group

Letθndenote the number of points ofPG(n,q), that is qn+1q−1−1. The points ofPG(n−1,q)can be represented by

0, α1, α2, . . . , αv−1}, wherev =θn−1.

A further identification withZv, that is the cyclic group of orderv, can be made byαj ↔j.

LetD={a1,a2, . . . ,ak} ⊂Zv be the set of elements

corresponding to a hyperplaneH, hencek = (qn−1−1)/(q−1).

ThenD+j ={a1+j,a2+j, . . . ,ak +j} ⊂Zv corresponds to the hyperplaneαjH. Thus the translates ofDcorrespond exactly to the hyperplanes ofPG(n−1,q).


The Singer group and difference sets Difference sets

Difference Sets


LetGbe a finite abelian group (written additively). We say thatD⊂G is a(v,k, λ)-difference set ofGif|G|=v,|D|=k, and for each 06=g ∈G, there are exactlyλpairsd,d0 ∈Dsuch thatd −d0=g.

The difference setD⊂Gis said to be cyclic ifGis cyclic.

Note that ifD⊂Gis a difference set, thenD+jis also a difference set.

Theorem (Singer – the classical Singer difference set)

In the cyclic group Zθn, the set of elements D corresponding to a hyperplane ofPG(n,q)is a(θn, θn−1, θn−2)-difference set.


The Singer group and difference sets Difference sets


A 2−(n,k, λ)block design is a set ofv points arranged intobblocks of sizek, where any 2 points are together on exactlyλblocks. A design is called symmetric (or square) ifv =b.

LetD⊂Gbe a(v,k, λ)difference set. We can define the points of a design as the elements ofG, and the blocks as the translates ofD, that are: {g+D:g ∈G}. This construction gives a symmetric design with an automorphism groupGacting regularly on the points (and blocks) ofG.


The Singer group and difference sets Multipliers

Definition (Hall)

LetD⊂Gbe an abelian difference set. A natural number,tis a (numerical) multiplier ofDifgcd(t,|G|) =1 andtD=D+j for somej.

The functionx →xt is a permutation of the elements ofGF(qn)iff gcd(t,qn−1) =1,

the functionhxi → hxtiis a permutation of the points of PG(n−1,q)iffgcd(t,qq−1n−1) =1.

Suppose thatx →xt permutes the elements ofGF(qn). If there exist two(n−1)-dimensionalGF(q)-subspaceAandBsuch thatAt =B, then:

t is a multiplier of the classical Singer difference set,

the image of any(n−1)-dimensionalGF(q)-subspace is again an (n−1)-dimensionalGF(q)-subspace,

hxi → hxtiis a collineation of the associatedPG(n−1,q).


The Singer group and difference sets Multipliers

It is widely studied when−1 is a multiplier of a difference set.

Theorem (Johnsen, 1963)

Let D be a non-trivial(v,k, λ)-difference set in an abelian group G. If

−1is a multiplier of D, then G is not cyclic. (Non-trivial means 0< λ <k <v −1)

The proof relies on the simple fact that in the cyclic groupZr, the map g → −gfixes one element ifr is odd and two ifr is even. While in the automorphism group of the associated design, if an involution fixes a point, then it fixes more than two points. This also implies that in PG(n,q)there is

one inverse-closed point (P0) iffθn is odd, i.e. nis even and there are two (P0andL0) iffnis odd.


If A is an(n−1)-dimensionalGF(q)-subspace ofGF(qn), then A−1is not an(n−1)-dimensionalGF(q)-subspace. (We have already known

n n−1


The Singer group and difference sets Multipliers

Conjecture(McFarland): If−1 is a multiplier of an abelian (v,k, λ)-difference setD(assumek <v/2), then either

(v,k, λ) = (4000,775,150)orDis a Hadamard difference set, i.e.

(v,k, λ) = (4N2,2N2−N,N2−N).

Theorem (Pott: Finite Geometry and Character Theory) Let D be a classical Singer difference set with parameters

(v,k, λ) = (θn, θn−1, θn−2). Then t is a multiplier of D if and only if t ≡pk (mod v ), for some k , where q is a power of the prime p.

InGF(93)the mapx →x61takes 2-dimensionalGF(9)-subspaces into 2-dimensionalGF(9)-subspaces, since 61≡35(modulo


9−1). But 616≡3k (modulo 93−1).


Let A be an(n−1)-dimensionalGF(q)-subspace ofGF(qn). Then At is an(n−1)-dimensionalGF(q)-subspace if and only ifgcd(t,qn−1) =1 and t ≡pk (mod qq−1n−1) where q is a power of the prime p.


The Singer group and difference sets Inverse of a line or hyperplane

What is the inverse of a line?

Theorem (Hall, 1947)

Let D ⊂G be a cyclic(n2+n+1,n+1,1)-difference set. Then 12D and2D are ovals if n is odd (and lines if n is even) in the associated design. (That is a projective plane with a Singer group, i.e. with a regular collineation group.)

If`is a line ofPG(2,q), then`−1is a conic (Hall, 1974).

Theorem (Bruck, 1973 and Jungnickel, Vedder, 1984)

Let D ⊂G be an abelian(n2+n+1,n+1,1)-difference set. Then−D is an oval, that is a set of n+1points, no three of them collinear (this is maximal if n is odd).

In general what is thet-th power of ak-dimensional subspace in PG(n,q)?


The Singer group and difference sets Inverse of a line or hyperplane

When the "power of a hyperplane" is a quadric

Theorem (Jackson, Quinn, Wild, 1996)

Let`be a line ofPG(2,q), q is a power of an odd prime p and let r be any integer. The pointset`r is a (possibly degenerate) conic if there exist integers i,j,k such that one of the following holds:

1 rpk(qi+qj)≡1(mod q2+q+1),

2 rpk ≡2(mod q2+q+1).

Theorem (Jackson, Quinn, Wild, 1996)

LetHbe hyperplane ofPG(n,q), q is a power of a prime p and let r be any integer. The pointsetHr is a (possibly degenerate) quadric if there exist integers i,j,k such that:

1 rpk(qi+qj)≡1(modθn).

The converse is true if q =2.


The Singer group and difference sets Inverse of a line or hyperplane

Theorem (Baker, Brown, Ebert, Fisher, 1994)

Let`be a line ofPG(2,q)and letgcd(r,q2+q+1) =1, then

`1/r is a curve of degree r ,

`−r is a curve of degree2r ,

`r is a curve of degree r .

Lemma (Faina, Kiss, Marcugini, Pambianco, 2002)

LetHbe a hyperplane ofPG(n,q), thenH−1is contained in a hypersurface of degree n.

Theorem (Faina, Kiss, Marcugini, Pambianco, 2002)

Let`be a line ofPG(n,q), then`−1is always an arc in somePm, where m+1|n+1. (A k -arc inPG(n,q)is a set of k points such that k ≥n+1and there are at most n points in each hyperplane. Note that PG(1,q)is an arc with this definition.)


The Singer group and difference sets Inverse of a line or hyperplane

Lemma (BCs)

IfAis a k -dimensional subspace ofPG(n,q), where0<k <n, then A−1⊆PG(n,q)is the intersection of k+1n

hypersurfaces of degree k +1.

(The equations come from the(k +1)×(k +1)subdeterminants of an n×(k +1)matrix.)

Typical application: If a line`intersectsA−1in more thank+1 points, then (from Bézout)`is contained inA−1, hence`−1is contained inA and we can use for example the previous result.


The Singer group and difference sets Te generated subspace

What is the GF(q)-subspace generated by the r -th powers of a GF(q)-subspace?

InPG(2,81)the maphxi → hx5905itakes lines to lines, but the mapx →x5905is not a bijection onGF(813).

If we ask about the generatedGF(q)-subspace, then there is no difference in the answers.


The Singer group and difference sets Te generated subspace


Let x1,x2, . . . ,xr be elements of a fieldFand define the following:

si(x1,x2, . . . ,xr) := X


(xj1+xj2+. . .+xji)r.

Then r−1



(−1)isr−i(x1,x2, . . . ,xr) =r!x1x2. . .xr. For example:


LetAr :={a1a2. . .ar :ai ∈A, fori =1,2, . . . ,r}.

With the previous lemma we have proved the following:


Ifchar(F) =0orchar(F)>r , thenhAri=h{ar :a∈A}i.


The Singer group and difference sets Te generated subspace

Theorem (Hou, Leung, Xiang, 2001)

Let s(r,n) be the maximal number such that there is an

s(r,n)-dimensionalGF(q)-subspace W inGF(qn)with the property hWri 6=GF(qn). Then


k|n k n

k −2 r



The above theorem is a corollary of the following vector space analogous of Kneser’s Addition Theorem in abelian groups:

Theorem (Hou, Leung, Xiang, 2001)

Let E ⊂K be fields and let A,B be finite-dimensional E -subspaces of K such that A6={0}, B 6={0}. Suppose that every algebraic element in K is separable over E . Then



Large inverse-closed subsets Motivation

Large inverse-closed subsets in subspaces

The classification of spatial equifocused arcs relays on a result on additive subgroups inGF(q), which are inverse-closed apart from at most two non-zero elements.

Lemma (Korchmáros, Lanzone, Sonnino, 2010)

If A is an additive subgroup ofGF(q), q =2n, with|A| ≥16and1∈A, then|A∩A−1| ≥ |A| −2implies that A =A−1and hence A is a subfield ofGF(q).

LetAbe a non-trivial additive subgroup ofGF(q). How big the positive integerdepending only on|A|can be chosen such that

|A∩A−1| ≥ |A| −impliesA =A−1?

In general: LetAandB be two non-trivialGF(q)-subspaces in GF(qn)with the same size. How big the positive integer depending only on|A|=|B|can be chosen such that

|B∩A−1| ≥ |B| −impliesB =A−1?


Large inverse-closed subsets Bounds

Theorem (BCs)

Let A and B be two non-trivial d -dimensionalGF(q)-subspaces in GF(qn). If|B∩A−1| ≥ 2q|B| −1,then B =A−1and both A and B are one-dimensionalGF(qd)-subspaces.

Ifq=2 thenB =A−1is a trivial consequence of

|B∩A−1| ≥ 2q|B| −1=|B|.In this case the following theorem gives same result under a weaker condition.

Theorem (BCs)

Let A and B be non-trivial additive subgroups ofGF(2n)with the same size which is greater than or equal to four. If|B∩A−1| ≥3|B|/4,then B =A−1and both A and B are one-dimensional subspaces over the same subfield ofGF(2n).


Large inverse-closed subsets Sharpness

When the bound is sharp (q odd)

BothA andB−1are union of one-dimensionalGF(q)-subspaces, hence the same holds forA∩B−1(andA∩B−1is divisible by(q−1)).

Thus we can study the problem inPG(n−1,q), and searching for k-dimensional subspacesAandBsuch that

|B ∩ A−1|=2(qk −1)/(q−1).


LetAandBbe two two-dimensionalGF(q)-subspaces ofGF(qn), that are lines`and`0 inPG(n−1,q). We have that`0∩`−1

contains at most 2(|A|q −1)/(q−1) =2 points or`=`0 and bothA andBare one-dimensionalGF(q2)-subspaces. The latter case cannot be whennis odd.

Ifn=3, then|`0∩`−1| ≤2 for each line`0. So`−1is a(q+1)-arc inPG(2,q), i.e. a conic ifqis odd (due to Segre).

This proves a previous result of Hall, and shows that the previous theorem is sharp for 2-dimensionalGF(q)-subspaces.


Large inverse-closed subsets Sharpness

PG(3, q)

Theorem (Ebert, 1985)

Let G denote the cyclic Singer group ofPG(n−1,q). If n=2m+2, then there is a subgroup H of order qm+1+1in G. If m is odd, then the orbits of H are caps (that is a set of points ofPG(n−1,q), no three of which are collinear).

Letm=1 andQ:={hxi ∈PG(3,q) :x(q2+1)(q−1)=1}, that is the orbit ofHcontainingh1i. Since{x ∈GF(q4) :x(q2+1)(q−1)=1}is a

subgroup of the multiplicative group ofGF(q4), we haveQ=Q−1. In PG(3,q), withqodd, Barlotti and Panella (1955) independently showed that each cap of sizeq2+1 is an elliptic quadric.

This means that ifqis odd, thenQis an inverse-closed elliptic quadric.


Large inverse-closed subsets Sharpness

Ifqis a prime power andd +1 dividesn, then Pd ={hxi ∈PG(n−1,q) :xqd+1 =x}.

Similarly ifq is odd and 2d+2 dividesn, then Ld ={hxi ∈PG(n−1,q) :xqd+1 =−x}.

LetAbe ad-dimensional subspace ofPG(n−1,q). ThenA−1is a d-dimensional subspace iffAcorresponds to a one-dimensional GF(qd+1)-subspace ofGF(qn). We call these subspaces the cosets of Pd. The one-dimensionalGF(qd+1)-subspaces ofGF(qn)are pairwise disjoint, so the cosets ofPd form a spread ofPG(n−1,q)byd-spaces.


Large inverse-closed subsets Sharpness


For a pointhxi ∈PG(3,q)we havehxi ∈Q if and only ifhxq2i=hx−1i.

Proof. The pointhxiis an element ofQif and only if

1=x(q2+1)(q−1)= (xq2)q−1xq−1or equivalently(x−1)q−1= (xq2)q−1. Lemma

If q is odd, then Q∩ P1={P0,L0}and Q∩ L1=∅.

Proof. We havehxi ∈ P1∪ L1if and only ifx2(q2−1)=1.On the other handhxi ∈Qif and only if 1=x(q2+1)(q−1) =x2(q2−1)(q−1)/2x2(q−1). This implies thathxi ∈Q∩(L1∪ P1)if and only ifx2(q−1)=1 and hencexq−1=1 andhxi=P0orxq−1=−1 andhxi=L0.If

x2(q−1) =1,thenx2(q−1)(q+1)/2=xq2−1=1 and henceP0,L0∈ P1.


Large inverse-closed subsets Sharpness

Example (BCs)

Ifqis odd andHis a plane ofPG(3,q),that containsL1,then H ∩ H−1= (Q∩ H)∪ L1.IfHcontainsP0orL0,then

|H ∩ H−1|=q+2,otherwise we have|H ∩ H−1|=2q+2.

Proof. SinceP1∩ L1=∅, the lineP1cannot be contained inH. Let H ∩ P1=hpHiand for eachhxi ∈ H \ {L1∪ hpHi}let

hlxi:=L1∩ hx,pHi. There exista,b∈GF(q), not both zero, such that alx +bpH=x.Takingq2-th powers on both sides yields

−alx+bpH=xq2 and hencehxq2iis a point ofhlx,pHi ⊂ H.Adding the first equation to the second yields 2bpH=xq2+x. Hereb6=0 sincehxi∈ L/ 1,thus we have:

hpHi=hxq2+xi. (4) Since (4) holds also forhxi=hpHiwe have that it holds for each hxi ∈ H \ L1.


Large inverse-closed subsets Sharpness

First we showH ∩ H−1⊇(Q∩ H)∪ L1.

Ifhxi ∈ L1,thenhx−1i ∈ L1sinceL1is inverse-closed. Ifhxi ∈Q∩ H, thenhxq2i ∈ Handhx−1i ∈Qbut sincehxi ∈Qwe have

hxq2i=hx−1iand hencehx−1i ∈Q∩ H, i.e.Q∩ H is inverse-closed.

What are the tangent planes ofQonL1?

The only inverse-closed points ofPG(3,q)areP0andL0thus ifSis an inverse-closed pointset ofPG(3,q), then|S\ {P0,L0}|has to be even.

Sinceqis odd, this implies that ifQ∩ His a non-singular conic, i.e. it hasq+1 points, then it cannot contain exactly one inverse-closed point. On the other hand we havehP0,L0i=P1andP1∩ L1=∅, henceHcontains at most one of the inverse-closed points. It follows now thatHis a tangent plane ofQif and only if it containsP0orL0.


Large inverse-closed subsets Sharpness

Now we proveH ∩ H−1⊆(Q∩ H)∪ L1.

Suppose that there is a pointhyi ∈(H ∩ H−1)\ L1,this implies hy−1i ∈(H ∩ H−1)\ L1. Applying (4) tohyiandhy−1iyields hyq2+yi=hy−q2+y−1i,or equivalently:

(yq2+y)q−1= (y−q2 +y−1)q−1.

Multiply both sides by(yq2 +y)(y−q2 +y−1)yq3+q 6=0 to obtain:

(yq2+y)q(y−q2+y−1)yq2+1yq3−q2+q−1= (y−q2+y−1)q(yq2+y)yq(q2+1), (yq2+y)q(y +yq2)yq3−q2+q−1= (y +yq2)q(yq2+y),

(yq2+y)q+1(y(q2+1)(q−1)−1) =0.

The first factor of the last equation cannot be zero sincehyi∈ L/ 1and this implieshyi ∈Q.

The size of(Q∩ H)∪ L1isq+2 ifHis a tangent ofQand 2q+2 otherwise, thus the same hold for|H ∩ H−1|.


Large inverse-closed subsets Sharpness

The same ideas work to prove the following:

Proposition (BCs)

Ifqis odd andHis a plane ofPG(3,q)that containsP1, then H ∩ H−1= (Q∩ H)∪ P1and|H ∩ H−1|=2q.

Lemma (BCs)

IfAis a k -dimensional subspace ofPG(n,q), where0<k <n, then A−1⊆PG(n,q)is the intersection of k+1n

hypersurfaces of degree k +1.

Theorem (Faina, Kiss, Marcugini, Pambianco, 2002)

Let`be a line ofPG(n,q), then`−1is always an arc in somePm, where m+1|n+1. (A k -arc inPG(n,q)is a set of k points such that k ≥n+1and there are at most n points in each hyperplane. SoP1is also an arc with this definition.)


Large inverse-closed subsets Sharpness

It follows now thatB ∩ A−1cannot contain two different lines since their inverses would be contained inA, hence they could not be arcs of PG(3,q) =P3, thus they would be arcs in a coset ofPm for somem, wherem+1 dividesn=4 andm<3. This could happen only ifm=1 but this would be a contradiction since two cosets ofP1cannot have a common point and hence they cannot be contained in the same plane.

Theorem (BCs)

LetAandBbe two planes ofPG(3,q).

1 If|B ∩ A−1|>q+1+b2√


|B ∩ A−1| ∈ {2q,2q+1,2q+2}.

2 We have|A ∩ A−1|=2q if and only ifP1⊂ A.

3 We have|A ∩ A−1|=2q+2if and only ifL1⊂ AandP0,L0∈ A./

4 There are no planesA,Bthat satisfies|B ∩ A−1|=2q+1. (We prove only forA=B.)


Large inverse-closed subsets Sharpness

Proof. Denote the cubic curveB ∩ A−1byF. The condition in 1 impliesNq(F)>q+1+b2√

qc, whereNq(F)is the number of rational points of the curveF. This means that the Hasse-Weil bound does not hold and henceF is reducible over some extension ofGF(q).F cannot contain more than one rational line and hence it has to be the union of a line`and a non-degenerate conicC. Depending on the mutual position of`andC,this implies|B ∩ A−1| ∈ {2q,2q+1,2q+2}.

If|A ∩ A−1| ∈ {2q,2q+1,2q+2}, thenAandA−1contain the same line denoted by`. The inverse of this line`−1is not an arc ofPG(3,q), hence it is a coset ofP1. NowA−1contains both lines`and`−1and hence`=`−1. This implies that`is inverse-closed and hence it isP1 orL1.

Our previous results describe the cubic curveA ∩ A−1, whenA contains an inverse-closed line and hence the remaining statements



Large inverse-closed subsets Sharpness

The Lang-Weil bound

A projective algebraic setX ⊆PG(n,q)is said to be geometrically irreducible (or projective variety) if there is no decomposition

X =X1∪X2, withX1andX2projective algebraic sets defined over the algebraic closure ofGF(q)such thatXi 6=X fori =1,2. LetX be a projective algebraic set and denote withNq(X)the number of rational points ofX, then the Lang-Weil inequailty says that ifX is a projective variety of dimensionr and degreed, then:

|Nq(X)−qr| ≤(d−1)(d−2)qr−1/2+C(n,r,d)qr−1, (5) whereC(n,r,d)depends only onn,r,d and not on the fieldGF(q).


Large inverse-closed subsets Sharpness

LetAandBbek-dimensional subspaces ofPG(n,q)withk ≥3 such thatB6=A−1. It follows now from the Lang-Weil inequailty and from one of our previous Lemma that ifB ∩ A−1is geometrically irreducible andq is large enough with respect tonandk, then|B ∩ A−1|cannot reach the bound 2(qk−1)/(q−1)in our theorem.

Sandro Mattarei told me that he has a proof for that the bound arising from the Lang-Weil inequality holds for|B ∩ A−1|when k >3 (andqis large enough) and|B ∩ A−1|=2(qk−1)/(q−1) only ifk ≤3. (I have not seen the whole proof yet.)






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