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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

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Inverse of an additive subgroup Inverse-closed additive subgroups

Inverse-closed additive subgroups in rings

Definition

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.

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Inverse of an additive subgroup Inverse-closed additive subgroups

The field case

Definition

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.

Lemma

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.

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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.

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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

(a+b)2−a2−b2=2ab∈A.

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

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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

x2(2a)−1,x−2(2a)−1,(x+x−1)2(2a)−1∈A.

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.

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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.

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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.

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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.)

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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.

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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).

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The Singer group and difference sets Difference sets

Difference Sets

Definition

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.

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The Singer group and difference sets Difference sets

Definition

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.

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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).

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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.

Corollary

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

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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

93−1

9−1). But 616≡3k (modulo 93−1).

Corollary

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.

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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)?

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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.

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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.)

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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.

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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.

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The Singer group and difference sets Te generated subspace

Lemma

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

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

1≤j1<j2<...<ji≤r

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

Then r−1

X

i=0

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

(a+b+c)3−(a+b)3−(a+c)3−(b+c)3+a3+b3+c3=6abc.

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

With the previous lemma we have proved the following:

Corollary

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

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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

s(r,n)=max

k|n k n

k −2 r

+1

.

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

dimEAB≥dimEA+dimEB−dimEH(AB),

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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?

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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).

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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).

Example

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.

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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.

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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.

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Large inverse-closed subsets Sharpness

Lemma

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.

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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.

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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.

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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|.

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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.)

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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√

qc,then

|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.)

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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

follow.

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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).

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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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