A study of a permutation representation of P GL(4, q) via the Klein correspondence. May 20, 1999

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1 A study of a permutation representation of P GL4 q via the Klein correspondence May ntroduction nglis Liebeck and Saxl [8] showed that regarding the group GLn q as a subgroup of GLn q the character of GLn q induced by GLnq is multiplicity-free that is the multiplicity of each irreducible character is 0 or. For any multiplicity-free permutation representation one can define a character table of the associated association scheme. However very little is known for the permutation representation of GLn q on the right cosets GLn q/gln q. The case n = is related to the work of Bannai Hao and Song []. They investigated the character table of O m+ q acting on each of the positive- and negative-type of nonisotropic points which are controlled by the charcter table of the association scheme obtained from the action of P GL q acting on the cosets by dihedral subgroups D q and D q+ respectively. The association scheme obtained from the action of P GL q on the cosets of D q+ is a quotient of the association scheme obtained from the action of GL q on the cosets of GL q/gl q. Now consider the case n =. t is easier to investigate GL4 q/ GL q θ than GL4 q/gl q where θ is the Frobenius automorphism x x q because GL4 q/ GL q θ can be identified with the set of regular spreads in the projective space P G3 q. The aim of this paper is to determine the G-orbits on the set of regular spreads when q is an odd prime power where G is the image of the group GL q θ in P GL4 q. This in particular determines the number of double cosets G\P GL4 q/g. By using the Klein correspondence [9] we obtain a one-to-one correspondence between the set of regular spreads and the set of anisotropic lines with respect to a quadratic form on P G5 q. Moreover we also obtain an imbedding κg of the group G to P GL6 q so that we can identify the G-orbits on the set of regular spreads with the κg-orbits on the set of anisotropic lines.

2 Regular Spread n this paper let q be an odd prime power and V be a n + -dimensional vector space over F q n=3 or 5. Denote P Gn q := { v v V \ {0}}. where v means -dimensional vector subspace generated by v. the element of P Gn q is called point. Call T P Gn q is k-dimensional subspace if there exists a vector subspace U = u u k+ V with dimension k + such that T = { u u U \ {0}} and denote T = u u k+. n particular and - dimensional subspace are called line and plane respectively. Let L be the set of all lines in P G3 q. For a subset {L i } L consisting of mutually skew lines i j L i L j = φ define T {L i } := {L L L L i φ i}. n particular put R = RA B C := T T {A B C} where A B and C are mutually skew. Call the line set RA B C the regulus containing A B C. The regulus RA B C consists of q + lines including A B and C. On the other hand T {A B C} also forms a regulus. We call this opposite regulus and denote it by R. We call S L is a regular spread iff S forms a partition of the point set of P G3 q and RA B C S A B C S all distinct The following results are from Bruck s paper [4]: Proposition. The projective linear group P GL4 q acts transitively on the regular spreads of P G3 q. Proposition. Let R be a regulus of P G3 q and L a nonsecant line of the quadric Q = QR that is L Q = φ. Then there exists a unique regular spread S of P G3 q containing R and L. Let e e e 3 e 4 be a basis of 4-dimensional vector space then define L := e e LA := a e + a e + e 3 a e + a e + e 4 where A = a ij is a matrix. L and LA are lines of P G3 q. For a regular spread S there exists a basis and a irreducible matrix U GL q satisfying S = {LA A = a ij = a + bu a b F q } {L }. Conversely for any basis {e i } i= 4 and for any irreducible matrix U the line set forms a regular spread.

3 This fact is proved from the following properties: L is skew to L A : matrix L = LA. LX LY are skew X Y is nonsingular. Moreover we have the following. RL LO L = {La a GF q} {L }. F := {a + bu a b F q } forms a field isomorphic to F q. a b d b For A = F A c d q = A where A := = a + d A. c a QR 0 = {x e +x e +x 3 e 3 +x 4 e 4 x x 4 x x 3 = 0} where R 0 := RL LO L. Denote by S = SU the regular spread defined by a irreducible matrix U = u ij. Let G be the subgroup in P GL4 q fixing this regular spread. For σ GL4 q identify σ whth the element of P GL4 q = GL4 q/{k 4 k F q } with representative σ. Then we have G = G 0 θg 0 where R O θ := O R 0 R := where r satisfies u r r = u u and G 0 consists of the elements A B C D where A B C D are in F AD BC O. The group G has order G = q 4 q 4 q /q and G is 3-transitive on S. The subgroup of G fixing L LO and L is A O θ O A A F := F \ {O}. The element θ satisfies from RX = X R for X F θ is the identity in particutar L θ = L LAθ = LA n this study define another regular spread. For S = SU put S := {L } {LAR A R O F } which is the image of S by the element of P GL4 q. O 3

4 The goal of this paper is to classify regular spreads by the action of G G is the stabilizer defined above of the regular spread S = SU. We first compute the number of regular spreads. From Proposition. the number of regular spreads is P GL4 q G = q4 q 3 q. To investigate regular spreads we need the Klein correspondence. 3 The Klein correspondence Let x = x x 4 y = y y 4 be vectors of 4-dimensional vector space supposing that x y are linearly independent. Put v ij = x i y j y j x i. Denote the vector v v 3 v 4 v 3 v 4 v 34 of 6-dimensional vector space by κv w. This implies that for any L = v w we can define κl := κv w as the point of P G5 q. κl = v v 3 v 4 v 3 v 4 v 34 satisfies Define for a vector v = v v 6 and v v 34 v 3 v 4 + v 4 v 3 = 0 ϖv := v v 6 v v 5 + v 3 v 4. Q 6 := { v P G5 q ϖv = 0}. Remark that in general ϖv is not well-defined in case ϖv 0 but whether ϖv is a square or not is independent of the choice of a vector representing v. We can easily see that the correspondence κ : {line of P G3 q} Q 6 is -to- correspondence. We call this correspondence κ the Klein correspondence. Define ϖv w := ϖv+w ϖv ϖw = v w 6 v w 5 +v 3 w 4 +v 4 w 3 v 5 w +v 6 w v = v v 6 w = w w 6 :vectors. Remark that in generalϖv v ϖav bv for some a b F q. But we need only now that ϖv v = 0 which doesn t depend on choosing a b F q. Define for a subspace in P G5 q U U := {v P G5 q ϖu v = 0 u U}. The proof of following lemma is written in [7] Lemma. Two lines of P G3 q L L intersect two lines satisfies ϖκlκl = 0 The action of P GL4 q on P G5 q is also defined by the Klein correspondence as following: 4

5 v v v v 3 For σ = P GL4 q κσ := v 4 v 3 v 3 v 4 v 4 v 34 where v ij := κv i v j. For example if α σ = β α β then α κσ = α β α β. α αβ αβ β α We can easily see that this is a well-defined injective mapping from P GL4 q to P GL6 q. So we can identify P GL4 q with its image κp GL4 q. We can check the following properties easily κlσ = κlκσ κσκσ = κσσ L L σ σ P GL4 q. Now consider the image of regular spreads by the Klein correspondence. t is enough to consider only one fixed regular spread by Proposition.. 0 Now add assumptions : S 0 = SU is a regular spread where U = is a generator α β of F. Then α is a generator of F q and β + 4α is nonsquare. For S 0 Let κs 0 be the subspace of P G5 q generated by {κl L S 0 } which is a 3-dimensional subspace. Then κs 0 Q 6 = {κl L S 0 } and κs 0 = Moreover x κs 0 ϖx 0. 0 β α

6 Definition For a line l P G5 q we call l is an isotropic line. l Q 6. l is a secant line. l Q 6 =. l is a tangent line. l Q 6 =. l is an anisotropic line. l Q 6 = 0. From the definition κs 0 is an anisotropic line. κp GL4 q acts on Q 6 since P GL4 q acts on L. So κg acts on {point of P G5 q} \ Q 6 that is the set of nonsingular points. Since P GL4 q is transitive on the set of regular spreads κs forms an anisotropic line for any regular spread S. Hence we can define a mapping from regular spreads to anisotropic lines in P G5 q. By calculating the numbers of isotropic secant and tangent lines we have {anisotropic line in P G5 q} = q4 q q3. which is just the number of regular spreads in P G3 q and for two regular spreads S S S S κs κs κs κs. Hence the mapping from the set of regular spreads to the set of anisotropic lines is a bijection. This implies that we can identify the action of G on the set of regular spreads with that of κg on the set of anisotropic lines. From the definition of G κg fixes κs 0 and κs 0. 4 Classification of orbits n this section l denotes an anisotropic line and we denote κs 0 by l 0. Recall that S 0 = SU = {LA A F } {L } S = {LAR A F } {L } and U generates F. We can easily check that κ S = α β 0 0. is the anisotropic line in l 0. To calculate the number of orbits of G on regular spreads equivalently that of orbits of κg on anisotropic lines we first classify the orbits by the dimensions of intersections 6

7 l l 0 l l 0 l l 0 and l l 0. There are 6 possibilities: dim l l0 dim l l 0 dim l l0 = dim l l where means that the intersetion is empty. We denote this matrix by M. Each of the dimensions is preserved by the action of κg. Thus we can consider the action of κg by dim l l 0 separately. 4. dim l l 0 = 3 When M = 3 4. dim l l 0 = then l=l 0. So this corresponds to the orbit {l 0 }. When diml l 0 = by using a point v l l l 0 = v l. Next consider the cases whether v is in Q 6 or not. First consider when v is not in Q v is not in Q 6. When M = we need the following propositions. 0 Lemma 4. κg is transitive on the set of nonsingular points in l 0. n particular we have H := {σ G κσ fixes } = κ θ X O O X A A A X X F X = X A =. K := {σ G κσ fixes 0 α β β.0} R U a b = κ θ = R U c d a b c d P GL q. From this claim we can see that for each nonsingular point in l0 the number of anisotropic lines in l0 through this point is constant. Denote this number by m. Counting the number 7

8 of element in the set { p l p Q 6 l : anisotropic in l0 p l} we have m = qq +. For the aciton of K on the set of anisotropic lines {l l l 0 = 0 α β β.0 l 0 } This line set has q-q+/ lines we have following results: there exists only one orbit with order q + / but each regular spread S in this orbit 0 satisfies M = and other orbit has order q + which satisfies M = v is in Q 6. Suppose v is singular. Since G acts transitively on the set of singular points we can put without loss v = We have l l 0 = v l 0 = α β = v. The number of anisotropic lines such that l l0 = v l0 is q. Moreover these q lines forms an orbit of the subgroup of κg fixing above subspace v l0. Hence the set of anisotropic lines {l l l0 = v l0 for some v Q 6 } forms an orbit of κg and each line l in this set satisfies M = dim l l 0 = To classify the anisotropic lines which satisfies M = 0 0 or by the orbits of κg consider three cases : the intersection l l 0 0 and secant line. Now denote M 0 := M 0 :=. is anisotropic tangent Lemma 4.4 κg is transitive on the set of anisotropic lines tangent lines and secant lines in l 0 respectively. Proof By Klein correspondence a secant line in l0 = κs corresponds to a pair of lines in S. Since G is 3-transitive on S κg is transitive on the set of secant lines in l0. From l 0 = β α 0 0 0

9 each anisotropic line are orthogonal to just one secant line in l0. Hence κg is transitive on the set of anisotropic lines in l0. To prove that κg is transitive on the set of tangent lines in l0 show that κg v is transitive on the set {m l0 : tangent v m} where v = But this lines set is equal to { v v v l 0 } and the group κg v contains the subgroup κ which is transitive on l U 0. From this lemma we can consider for a fixed intersection line in each case separately. Lemma 4.5 For a given line m in l 0 put N := #{ p l 0 p l 0 \ Q 6 p l 0 m}. N := #{ p l 0 p l 0 Q 6 p l 0 m}. Then we have m : anisotropic N = q N = m : tangent N = q N = m : secant N = q + N = 0. Proof Since #{W l 0 dim W = m W } = #{-dim subspace in l 0 } #{line in a -dim subspace} #{line in l 0 } = q + q + q + q + q + q + q + = q + N = q+ N. And from previous lemma the number of -dimensional subspace p l0 containing m depends on whether p Q 6 or not and the line m is anisotropic tangent or secant. For p l0 Q 6 p l0 contains q anisotropic lines q + tangents and 0 secant. Hence N equals l0 Q 6 q #{anisotropic line in κs} = q + q q q + / = if l is anisotropic. l0 Q 6 q + = q + q + #{tangent in κs} q + q + = if l is tangent. l0 Q 6 0 #{secant in κs} = 0 if l is secant. 9

10 Lemma 4.6 For any anisotropic line l and for any point v l v l such that { v v } forms an orthogonal basis of l that is v v = l ϖv v = 0. Proof Choose a point w satisfying v w = l. When ϖv w = 0 this lemma is proved by putting v = w. Otherwise by ϖv av + bw = aϖv v + bϖv w = aϖv + bϖv w ϖv av +bw = 0 iff a = kϖv w b = kϖv for some k F q. Since ϖv 0 av + bw = v. The uniqueness is clear from above equations. Lemma 4.7 Let V = l l 0 where l is a line in l 0 dim V = 3. f a line in V l satisfies M = M then these two point sets are just l and l 0 respectively. {v l v v + v l} {v l 0 v v + v l} Proof For the mapping l v + v v v l v l 0 is injection otherwise there exist two distinct points v + v v + v in l for some v so v v is in l and also in l 0. This is contradiction. Since l and l are lines this mapping is bijection hence the image set of this mapping is just l. Similarly the image set of l v + v v is just l 0. Now consider the action of κg in each cases l l 0 is an anisotropic When m = l l0 is an anisotropic line suppose that m = κ S. By using l m m l the number of {l l l0 = m} is and #{l l κs = m} = q q + q + q = qq q +. #{l l l 0 = m M = M 0 } and #{l l l 0 = m M = M } are following #{l l l 0 = m M = M 0 } = q q #{l l l 0 = m M = M } = qq q + q q = q 3 q + 0

11 From m = 0 β α m = v := = κl v := = κlo v3 := 0 β 0 0 v 4 := 0 α Constructing a subgroup of κg which consists of an element fixing m equivalently fixing m this group is A A κ θ B B A B F. This group is generated by θ φ := φ U := To classify by orbits for each σ in above group define A σ P GL4 q such that v v v 3 v 4 κσ = A σ where v... v 4 are from. By using these A σ s and Lemmas the orbits in this case is following Then we have A θ = A R φ = α A φ = U By using these A σ s and Lemmas the orbits in this case is following v v v 3 v 4 n {l l l 0 = m M = M 0 } each orbit has length q. The number of orbits is q q q + = q. n {l l l 0 = m M = M } each orbit has length q. The number of orbits is q 3 q + q = 4 q l l 0 is tangent When m = l κs is a tangent line we can suppose that m = v 0 α β β 0. Then m v = = l u :=

12 Then we have #{l l l 0 = m} = q3 q q q + = qq q. #{l l l 0 = m M = M 0 } = qq #{l l l 0 = m M = M } = qq q qq = qq q 3. The subgroup {σ κg κσ fixes m = v 0 α β β 0} is written as A κ B aa A F a 0 B F U This group is generated by θ τ := τ U a := τ3 a X := X element of above subgroup σ define B σ P GL4 q such that v u v 3 v 4 κσ = B σ v u v 3 v 4.. For each where v 3 v 4 is from. Then B θ = R B τ = U q B τ a = a a B τ X 3 = By using this B σ we conclude that: The line set {l l l0 = m M = M 0 } forms an orbit of above group. n {l l l0 = m M = M } each orbit has length qq. And the number of orbits is qq q 3 qq = q 3.

13 4.3.3 l l 0 is secant When l l 0 is a secant line we can suppose that m = is generated by And we have We can see that m = α β 0 0 v = v = l 0 #{l l l 0 = m M = M 0 or M } = q q + q.then m #{l l l 0 = m M = M 0 } = q + q 3. #{l l l 0 = m M = M } = q + q3 5q + 7q +. The subgroup of G {σ G κσ fixes m} is just θ H 0 where H 0 is from 3.. anisotropic case. H 0 = θ U U From this we have that n {l l l0 = m M = M 0 } each orbit has length q +. And the number of orbits is U q + q 3 = q 3. q + n {l l l0 = m M = M 0 } there are q orbits with length q + and there are q 4 3 orbits with length q +. The next table is the results of this paper. From this table the total of orbits is Put N = N + = q + q. N = q + q +. q +. 3

14 Table l l 0 M size length in a orbit the number of orbits l 0 < p > l q + qq + q 3q p l0 + qq + Q 6 0 q + qq + q + qq + q 3 < p > l0 p l0 \ Q 6 anisotropic tangent secant q q + q q + q q N q N q q 3 q + N q N q 4 qq N qq q 3N qq N qq N q 3 N + N + q + q 3N + q + N + q 3 q q + N + q + N + q q + q 3 N + q + N + q 3 4 4

15 References [] Bannai Eiichi Character tables of commutative association schemes. Finite geometries buildings and related topics Pingree Park CO [] Bannai Eiichi; Hao Shen; Song Sung-Yell Character tables of the association schemes of finite orthogonal groups acting on the nonisotropic points. J. Combin.Theory Ser. A no [3] Bannai Eiichi; to Tatsuro Algebraic combinatorics.. Association schemes. Benjamin/Cummings Publisher Co. nc. Menlo Park Calif [4] Bruck R. H. Construction problems of finite projective planes. 969 Combinatorial Mathematics and its Applications. pp Univ. North Carolina Press Chapel Hill N.C. [5] Brouwer A. E.; Cohen A. M.; Neumaier A. Distance-regular graphs. Ergebnisse der Mathematik und ihrer Grenzgebiete 3 [Results in Mathematics and Related Areas3] 8. Springer-Verlag Berlin-New York 989. [6] Bruck R. H.; Bose R. C. The construction of translation planes from projective spaces. J. Algebra [7] Hirschfeld J. W. P. Finite projective spaces of three dimensions. Oxford Mathematical Monographs.Oxford Science Publications. The Clarendon Press Oxford University Press New York 985. [8] nglis Nicholas F. J.; Liebeck Martin W.; Saxl Jan Multiplicity-free permutation representations of finite linear groups. Math. Z no [9] Taylor Donald E. The geometry of the classical groups. Sigma Series in Pure Mathematics 9.Heldermann Verlag Berlin 99. 5