## 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

Definition

For a subsetS of a (unitary) ring let:

S^{∗}be the set of invertible elements inS,

S^{−1}be 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

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, letS^{n}={s^{n}: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.

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, /∈Kand^{2}∈K,

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

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

2 The inverse ofa∈Aisa(a^{2})^{−1}, where(a^{2})^{−1}∈K^{2}thusAis
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, thena^{2},b^{2},(a+b)^{2}∈A, thus also

(a+b)^{2}−a^{2}−b^{2}=2ab∈A.

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

x^{2}(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}−x^{2}(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}=a^{2}b

⇓

x−(x^{−1}+ (2a−x)^{−1})^{−1}=x^{2}(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)^{−1}x−1,(2a)^{−1}x^{−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 =p^{n}, 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(q^{n})as ann-dimensional vector space overGF(q). We
can identify thek-dimensionalGF(q)-subspaces ofGF(q^{n})with the
(k−1)-dimensional projective subspaces ofPG(n−1,q).

Ifm+1 dividesn, then letP_{m}be them-dimensional subspace of
PG(n−1,q)corresponding toGF(q^{m+1}).

If 2m+2 dividesn, then letL_{m}be them-dimensional subspaces
ofPG(n−1,q)corresponding to the set of roots ofx^{q}^{m+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
K^{2}-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 →x^{2}is a field
automorphism ifFis finite.)

The Singer group and difference sets The Singer group

## The Singer group

For an elementx ∈GF(q^{n})^{∗} 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(q^{n}). The collineation group
generated byhxi → hαxiis a cyclic Singer groupGof

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

(q^{n}−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 ^{q}^{n+1}_{q−1}^{−1}.
The points ofPG(n−1,q)can be represented by

{α^{0}, α^{1}, α^{2}, . . . , 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={a_{1},a_{2}, . . . ,a_{k}} ⊂Z_{v} be the set of elements

corresponding to a hyperplaneH, hencek = (q^{n−1}−1)/(q−1).

ThenD+j ={a_{1}+j,a_{2}+j, . . . ,a_{k} +j} ⊂Z_{v} corresponds to the
hyperplaneα^{j}H. Thus the translates ofDcorrespond exactly to
the hyperplanes ofPG(n−1,q).

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,d^{0} ∈Dsuch thatd −d^{0}=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.

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.

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(q^{n})iff
gcd(t,q^{n}−1) =1,

the functionhxi → hx^{t}iis a permutation of the points of
PG(n−1,q)iffgcd(t,^{q}_{q−1}^{n}^{−1}) =1.

Suppose thatx →x^{t} permutes the elements ofGF(q^{n}). 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^{t}iis 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 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 (P_{0}) iffθn is odd, i.e. nis even
and there are two (P_{0}andL_{0}) iffnis odd.

Corollary

If A is an(n−1)-dimensionalGF(q)-subspace ofGF(q^{n}), then A^{−1}is
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) = (4N^{2},2N^{2}−N,N^{2}−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(9^{3})the mapx →x^{61}takes 2-dimensionalGF(9)-subspaces
into 2-dimensionalGF(9)-subspaces, since 61≡3^{5}(modulo

9^{3}−1

9−1). But 616≡3^{k} (modulo 9^{3}−1).

Corollary

Let A be an(n−1)-dimensionalGF(q)-subspace ofGF(q^{n}). Then A^{t} is
an(n−1)-dimensionalGF(q)-subspace if and only ifgcd(t,q^{n}−1) =1
and t ≡p^{k} (mod ^{q}_{q−1}^{n}^{−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(n^{2}+n+1,n+1,1)-difference set. Then ^{1}_{2}D
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`^{−1}is a conic (Hall, 1974).

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

Let D ⊂G be an abelian(n^{2}+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 rp^{k}(q^{i}+q^{j})≡1(mod q^{2}+q+1),

2 rp^{k} ≡2(mod q^{2}+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}(q^{i}+q^{j})≡1(modθ_{n}).

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,q^{2}+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^{−1}is contained in a
hypersurface of degree n.

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

Let`be a line ofPG(n,q), then`^{−1}is 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.)

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+1}^{n}

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^{−1}in more thank+1 points,
then (from Bézout)`is contained inA^{−1}, hence`^{−1}is 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 → hx^{5905}itakes lines to lines, but the
mapx →x^{5905}is not a bijection onGF(81^{3}).

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

Lemma

Let x_{1},x_{2}, . . . ,x_{r} be elements of a fieldFand define the following:

s_{i}(x_{1},x_{2}, . . . ,x_{r}) := X

1≤j_{1}<j_{2}<...<j_{i}≤r

(x_{j}_{1}+x_{j}_{2}+. . .+x_{j}_{i})^{r}.

Then r−1

X

i=0

(−1)^{i}s_{r}_{−i}(x_{1},x_{2}, . . . ,x_{r}) =r!x_{1}x_{2}. . .x_{r}.
For example:

(a+b+c)^{3}−(a+b)^{3}−(a+c)^{3}−(b+c)^{3}+a^{3}+b^{3}+c^{3}=6abc.

LetA^{r} :={a_{1}a_{2}. . .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^{r}i=h{a^{r} :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(q^{n})with the property
hW^{r}i 6=GF(q^{n}). 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_{E}AB≥dim_{E}A+dim_{E}B−dim_{E}H(AB),

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 =2^{n}, with|A| ≥16and1∈A,
then|A^{∗}∩A^{−1}| ≥ |A^{∗}| −2implies that A^{∗} =A^{−1}and 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(q^{n})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(q^{n}). If|B^{∗}∩A^{−1}| ≥ ^{2}_{q}|B| −1,then B^{∗} =A^{−1}and both A and B are
one-dimensionalGF(q^{d})-subspaces.

Ifq=2 thenB^{∗} =A^{−1}is 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(2^{n})with the same
size which is greater than or equal to four. If|B∩A^{−1}| ≥3|B|/4,then
B^{∗} =A^{−1}and both A and B are one-dimensional subspaces over the
same subfield ofGF(2^{n}).

Large inverse-closed subsets Sharpness

## When the bound is sharp (q odd)

BothA^{∗} andB^{−1}are union of one-dimensionalGF(q)-subspaces,
hence the same holds forA∩B^{−1}(andA∩B^{−1}is 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(q^{n}),
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(q^{2})-subspaces. The latter case
cannot be whennis odd.

Ifn=3, then|`^{0}∩`^{−1}| ≤2 for each line`^{0}. So`^{−1}is 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 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^{(q}^{2}^{+1)(q−1)}=1}, that is the orbit
ofHcontainingh1i. Since{x ∈GF(q^{4}) :x^{(q}^{2}^{+1)(q−1)}=1}is a

subgroup of the multiplicative group ofGF(q^{4}), we haveQ=Q^{−1}. In
PG(3,q), withqodd, Barlotti and Panella (1955) independently
showed that each cap of sizeq^{2}+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
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^{−1}is a
d-dimensional subspace iffAcorresponds to a one-dimensional
GF(q^{d+1})-subspace ofGF(q^{n}). We call these subspaces the cosets of
P_{d}. The one-dimensionalGF(q^{d+1})-subspaces ofGF(q^{n})are pairwise
disjoint, so the cosets ofP_{d} form a spread ofPG(n−1,q)byd-spaces.

Large inverse-closed subsets Sharpness

Lemma

For a pointhxi ∈PG(3,q)we havehxi ∈Q if and only ifhx^{q}^{2}i=hx^{−1}i.

**Proof.** The pointhxiis an element ofQif and only if

1=x^{(q}^{2}^{+1)(q−1)}= (x^{q}^{2})^{q−1}x^{q−1}or equivalently(x^{−1})^{q−1}= (x^{q}^{2})^{q−1}.
Lemma

If q is odd, then Q∩ P_{1}={P_{0},L_{0}}and Q∩ L_{1}=∅.

**Proof.** We havehxi ∈ P_{1}∪ L_{1}if and only ifx^{2(q}^{2}^{−1)}=1.On the other
handhxi ∈Qif and only if 1=x^{(q}^{2}^{+1)(q−1)} =x^{2(q}^{2}^{−1)(q−1)/2}x^{2(q−1)}.
This implies thathxi ∈Q∩(L_{1}∪ P_{1})if and only ifx^{2(q−1)}=1 and
hencex^{q−1}=1 andhxi=P_{0}orx^{q−1}=−1 andhxi=L_{0}.If

x^{2(q−1)} =1,thenx2(q−1)(q+1)/2=x^{q}^{2}^{−1}=1 and henceP_{0},L_{0}∈ P_{1}.

Large inverse-closed subsets Sharpness

Example (BCs)

Ifqis odd andHis a plane ofPG(3,q),that containsL_{1},then
H ∩ H^{−1}= (Q∩ H)∪ L_{1}.IfHcontainsP_{0}orL_{0},then

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

**Proof.** SinceP_{1}∩ L_{1}=∅, the lineP_{1}cannot be contained inH. Let
H ∩ P_{1}=hp_{H}iand for eachhxi ∈ H \ {L_{1}∪ hp_{H}i}let

hl_{x}i:=L_{1}∩ hx,p_{H}i. There exista,b∈GF(q), not both zero, such that
alx +bpH=x.Takingq^{2}-th powers on both sides yields

−al_{x}+bpH=x^{q}^{2} and hencehx^{q}^{2}iis a point ofhl_{x},pHi ⊂ H.Adding
the first equation to the second yields 2bpH=x^{q}^{2}+x. Hereb6=0
sincehxi∈ L/ _{1},thus we have:

hp_{H}i=hx^{q}^{2}+xi. (4)
Since (4) holds also forhxi=hp_{H}iwe have that it holds for each
hxi ∈ H \ L_{1}.

Large inverse-closed subsets Sharpness

First we showH ∩ H^{−1}⊇(Q∩ H)∪ L_{1}.

Ifhxi ∈ L_{1},thenhx^{−1}i ∈ L_{1}sinceL_{1}is inverse-closed. Ifhxi ∈Q∩ H,
thenhx^{q}^{2}i ∈ Handhx^{−1}i ∈Qbut sincehxi ∈Qwe have

hx^{q}^{2}i=hx^{−1}iand hencehx^{−1}i ∈Q∩ H, i.e.Q∩ H is inverse-closed.

What are the tangent planes ofQonL_{1}?

The only inverse-closed points ofPG(3,q)areP_{0}andL_{0}thus ifSis an
inverse-closed pointset ofPG(3,q), then|S\ {P_{0},L_{0}}|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 havehP_{0},L_{0}i=P_{1}andP_{1}∩ L_{1}=∅,
henceHcontains at most one of the inverse-closed points. It follows
now thatHis a tangent plane ofQif and only if it containsP_{0}orL_{0}.

Large inverse-closed subsets Sharpness

Now we proveH ∩ H^{−1}⊆(Q∩ H)∪ L_{1}.

Suppose that there is a pointhyi ∈(H ∩ H^{−1})\ L_{1},this implies
hy^{−1}i ∈(H ∩ H^{−1})\ L_{1}. Applying (4) tohyiandhy^{−1}iyields
hy^{q}^{2}+yi=hy^{−q}^{2}+y^{−1}i,or equivalently:

(y^{q}^{2}+y)^{q−1}= (y^{−q}^{2} +y^{−1})^{q−1}.

Multiply both sides by(y^{q}^{2} +y)(y^{−q}^{2} +y^{−1})y^{q}^{3}^{+q} 6=0 to obtain:

(y^{q}^{2}+y)^{q}(y^{−q}^{2}+y^{−1})y^{q}^{2}^{+1}y^{q}^{3}^{−q}^{2}^{+q−1}= (y^{−q}^{2}+y^{−1})^{q}(y^{q}^{2}+y)y^{q(q}^{2}^{+1)},
(y^{q}^{2}+y)^{q}(y +y^{q}^{2})y^{q}^{3}^{−q}^{2}^{+q−1}= (y +y^{q}^{2})^{q}(y^{q}^{2}+y),

(y^{q}^{2}+y)^{q+1}(y^{(q}^{2}^{+1)(q−1)}−1) =0.

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

The size of(Q∩ H)∪ L_{1}isq+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 containsP_{1}, then
H ∩ H^{−1}= (Q∩ H)∪ P_{1}and|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+1}^{n}

hypersurfaces of degree k +1.

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

Let`be a line ofPG(n,q), then`^{−1}is 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. SoP_{1}is
also an arc with this definition.)

Large inverse-closed subsets Sharpness

It follows now thatB ∩ A^{−1}cannot contain two different lines since their
inverses would be contained inA, hence they could not be arcs of
PG(3,q) =P_{3}, 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 ofP_{1}cannot 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 ifP_{1}⊂ A.

3 We have|A ∩ A^{−1}|=2q+2if and only ifL_{1}⊂ AandP_{0},L_{0}∈ 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^{−1}byF. 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^{−1}contain the same
line denoted by`. The inverse of this line`^{−1}is not an arc ofPG(3,q),
hence it is a coset ofP_{1}. NowA^{−1}contains both lines`and`^{−1}and
hence`=`^{−1}. This implies that`is inverse-closed and hence it isP_{1}
orL_{1}.

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

follow.

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 =X_{1}∪X_{2}, withX_{1}andX_{2}projective 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).

Large inverse-closed subsets Sharpness

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^{−1}is 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.)

## THANK YOU

THANK YOU FOR YOUR ATTENTION