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

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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:

S^∗be 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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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 A^p^k is also a subfield ofF,

if A⊆Fis an additive subgroup, then A^p^k is also an additive subgroup.

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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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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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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 x^p^d +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 letP_mbe them-dimensional subspace of PG(n−1,q)corresponding toGF(q^m+1).

If 2m+2 dividesn, then letL_mbe them-dimensional subspaces ofPG(n−1,q)corresponding to the set of roots ofx^q^m+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.x^q−1=y^q−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⁺¹_q−1⁻¹. The points ofPG(n−1,q)can be represented by

{α⁰, α¹, α², . . . , α^v⁻¹}, wherev =θn−1.

A further identification withZ_v, that is the cyclic group of orderv, can be made byα^j ↔j.

LetD={a₁,a₂, . . . ,a_k} ⊂Z_v be the set of elements

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

ThenD+j ={a₁+j,a₂+j, . . . ,a_k +j} ⊂Z_v corresponds to the hyperplaneα^jH. 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 Z_θ_n, 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 →x^t is a permutation of the elements ofGF(qⁿ)iff gcd(t,qⁿ−1) =1,

the functionhxi → hx^tiis a permutation of the points of PG(n−1,q)iffgcd(t,^q_q−1ⁿ⁻¹) =1.

Suppose thatx →x^t permutes the elements ofGF(qⁿ). If there exist two(n−1)-dimensionalGF(q)-subspaceAandBsuch thatA^t =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 → hx^tiis a collineation of the associatedPG(n−1,q).

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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 groupZ_r, 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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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) = (θ_n, θ_n−1, θ_n−2). Then t is a multiplier of D if and only if t ≡p^k (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≡3^k (modulo 9³−1).

Corollary

Let A be an(n−1)-dimensionalGF(q)-subspace ofGF(qⁿ). Then A^t is an(n−1)-dimensionalGF(q)-subspace if and only ifgcd(t,qⁿ−1) =1 and t ≡p^k (mod ^q_q−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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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 rp^k(qⁱ+q^j)≡1(mod q²+q+1),

2 rp^k ≡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 pointsetH^r is a (possibly degenerate) quadric if there exist integers i,j,k such that:

1 rp^k(qⁱ+q^j)≡1(modθ_n).

The converse is true if q =2.

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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 someP_m, 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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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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Lemma

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

s_i(x₁,x₂, . . . ,x_r) := X

1≤j₁<j₂<...<j_i≤r

(x_j₁+x_j₂+. . .+x_j_i)^r.

Then r−1

X

i=0

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

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

LetA^r :={a₁a₂. . .a_r :a_i ∈A, fori =1,2, . . . ,r}.

With the previous lemma we have proved the following:

Corollary

Ifchar(F) =0orchar(F)>r , thenhA^ri=h{a^r :a∈A}i.

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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 hW^ri 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

dim_EAB≥dim_EA+dim_EB−dim_EH(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| ≥ ²_q|B| −1,then B^∗ =A⁻¹and both A and B are one-dimensionalGF(q^d)-subspaces.

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

|B^∗∩A⁻¹| ≥ ²_q|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(q^k −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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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 q^m+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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Ifqis a prime power andd +1 dividesn, then P_d ={hxi ∈PG(n−1,q) :x^q^d+1 =x}.

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

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

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Lemma

For a pointhxi ∈PG(3,q)we havehxi ∈Q if and only ifhx^q²i=hx⁻¹i.

Proof. The pointhxiis an element ofQif and only if

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

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

Proof. We havehxi ∈ P₁∪ L₁if and only ifx^2(q²⁻¹⁾=1.On the other handhxi ∈Qif and only if 1=x^(q²^+1)(q−1) =x^2(q²^{−1)(q−1)/2}x^2(q−1). This implies thathxi ∈Q∩(L₁∪ P₁)if and only ifx^2(q−1)=1 and hencex^q−1=1 andhxi=P₀orx^q−1=−1 andhxi=L₀.If

x^2(q−1) =1,thenx2(q−1)(q+1)/2=x^q²⁻¹=1 and henceP₀,L₀∈ P₁.

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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=hp_Hiand for eachhxi ∈ H \ {L₁∪ hp_Hi}let

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

−al_x+bpH=x^q² and hencehx^q²iis a point ofhl_x,pHi ⊂ H.Adding the first equation to the second yields 2bpH=x^q²+x. Hereb6=0 sincehxi∈ L/ ₁,thus we have:

hp_Hi=hx^q²+xi. (4) Since (4) holds also forhxi=hp_Hiwe have that it holds for each hxi ∈ H \ L₁.

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First we showH ∩ H⁻¹⊇(Q∩ H)∪ L₁.

Ifhxi ∈ L₁,thenhx⁻¹i ∈ L₁sinceL₁is inverse-closed. Ifhxi ∈Q∩ H, thenhx^q²i ∈ Handhx⁻¹i ∈Qbut sincehxi ∈Qwe have

hx^q²i=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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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 hy^q²+yi=hy^−q²+y⁻¹i,or equivalently:

(y^q²+y)^q−1= (y^−q² +y⁻¹)^q−1.

Multiply both sides by(y^q² +y)(y^−q² +y⁻¹)y^q³^+q 6=0 to obtain:

(y^q²+y)^q(y^−q²+y⁻¹)y^q²⁺¹y^q³^−q²^+q−1= (y^−q²+y⁻¹)^q(y^q²+y)y^q(q²⁺¹⁾, (y^q²+y)^q(y +y^q²)y^q³^−q²^+q−1= (y +y^q²)^q(y^q²+y),

(y^q²+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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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 someP_m, 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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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 ofP_m 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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Proof. Denote the cubic curveB ∩ A⁻¹byF. The condition in 1 impliesN_q(F)>q+1+b2√

qc, whereN_q(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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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 thatX_i 6=X fori =1,2. LetX be a projective algebraic set and denote withN_q(X)the number of rational points ofX, then the Lang-Weil inequailty says that ifX is a projective variety of dimensionr and degreed, then:

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

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LetAandBbek-dimensional subspaces ofPG(n,q)withk ≥3 such thatB^∗6=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(q^k−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(q^k−1)/(q−1) only ifk ≤3. (I have not seen the whole proof yet.)

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