The character table of a split extension of the Heisenberg group H 1 (q) by Sp(2, q), q odd

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1 Università di Milano Bicocca Quaderni di Matematica The character table of a split extension of the Heisenberg group H (q by Sp(, q, q odd Marco Antonio Pellegrini Quaderno n. 7/8 (arxiv:math/85.48

2 Stampato nel mese di maggio 8 presso il Dipartimento di Matematica e Applicazioni, Università degli Studi di Milano-Bicocca, via R. Cozzi 53, 5 Milano, ITALIA. Disponibile in formato elettronico sul sito Segreteria di redazione: Ada Osmetti - Giuseppina Cogliandro tel.: fax: Esemplare fuori commercio per il deposito legale agli effetti della Legge 5 aprile 4 n.6.

3 The character table of a split extension of the Heisenberg group H (q by Sp(,q, q odd Marco Antonio Pellegrini Abstract In this paper we determine the full character table of a certain split extension H (q Sp(,q of the Heisenberg group H by the odd-characteristic symplectic group Sp(,q. Keywords: Character table, Heisenberg group, Symplectic group. AMS subject classification: C5. Introduction In his paper ([Gér] P. Gérardin constructed the Weil representations of the oddcharacteristic symplectic groups using the properties of a certain split extension H t (q Sp(t, q of the Heisenberg group H t (q of order q t+ by the symplectic group Sp(t, q. In this paper we explicitly determine the character table of this extension, in the case where t =. A motivation lies in the fact that knowledge of this character table seems to be useful in the study of the restrictions to parabolic subgroups of certain unipotent characters of odd-dimensional orthogonal groups (see [DPW]. Let V be the column vector space of dimension t over a finite field F of order q, where q is odd, and V is provided with a non-degenerate symplectic form j. Given w V, we denote by w the element of the dual space (we think at w as a row such that w w = j(w, w /. Let H t (q be the group consisting of the matrices h = h (w,z = w z w Mat(t +, F, where w V and z F. We call this group the Heisenberg group of V. H t (q is obviously a central extension of (V, + by (F, +. Furthermore, H t (q is a two-step nilpotent group of order q t+ whose center is isomorphic to F (cf. [Gér, Lemma.]. Let S be the symplectic group associated to the form j and, for each s S, denote by sw the image of w under the natural action of S on V. Then, the map 3

4 h (w,z h (sw,z defines an automorphism of H t (q fixing pointwise Z(H t (q. Viewed as acting on matrices, this map is the conjugation by the element s = diag(, s, Let us denote by G the semidirect product H t (q Sp(t, q defined by the above action of S. We want to construct the character table of G in the case where t =. So, G = H (q Sp(, q. In this case, we can write in a unique way a generic element g of G as w z g = g (s,w,z = sh (w,z = s sw, where s( S = Sp(, q (here we identify s S with s G, w V and z F. x If w = V, then we can take as w y the row ( y, x. Note that G = q 4 (q. The conjugacy classes In the sequel, we denote by (g the conjugacy class of G containing the element g, and by (g the size of the conjugacy class (g. The following lemma lists the conjugacy classes of G. Lemma.. Let F = GF(q, q odd, and let F = ν be the multiplicative group of F. Set z A (z =, B =, C (z = E (z = G m (z = I (z = z z z b m z ν, D k(z =, F(z =, H (z =, L m = 4 z ν k ν k z ν z νm ν m ν m,,,,

5 M m = νm ν m ν ν m+ where z F, k q 3, m q and b is an element of order q + (a Singer cycle of Sp(, q. These are elements of G, and G admits exactly q + 5q conjugacy classes (g with representative g, as listed in the Table below., g (g Parameters A (z z F B q(q C (z q z F D k (z q 3 (q + z F, k q 3 E (z q (q z F F(z q (q z F G m (z q 3 (q z F, m q H (z q(q z F I (z q(q z F L m q (q m q M m q (q m q Proof. Let g = g (s,w,z and g = g (s,w,z be two generic elements of G. Then g g g = g (s s s,s (w w +s w,z (w +s w +s w w. It easily follows that if g is conjugate to g in G, then s is conjugate to s in S. Moreover, if z z, then the elements g (s,,z and g (s,,z cannot be conjugate in G. Observe that g C G (g if and only if s C S (s w + s w = w + s w. w (s w = w (s w Let us consider the elements A (z = g (,,z, z F. It is straightforward to see that Z(G = Z(H (q = {A (z : z F } = (F, +. Therefore, each of these q elements of G forms a central class of order. In particular, A ( is the identity of G. Now, let us consider the element g (,w, = B H (q \ Z(H (q. Then, C G (B = q 3, i.e. (B = q(q. Since g (s,w,z Bg (s,w,z = g (,s w, w w, it turns out that the elements of H (q \Z(H (q form a single conjugacy class (B of G. Set g = g (s,,z {C (z, D k (z, E (z, F(z, G m (z}. Recall (e.g., see [Dor, 38] that S admits elements b of order q +, the so-called Singer cycles. As observed 5

6 before, for different values of z and s the elements g (s,,z belong to q + q distinct conjugacy classes of G. Now, an element g (s,w,z belongs to C G (g if and only if { s C S (s sw = w. ( Since s does not have eigenvalue, the condition sw = w implies w =. It follows that C G (g = q C S (s, and using the information about the centralizers of elements of S contained in [Dor, 38], we obtain the results listed in the statement of the lemma. Next, let us consider elements g = g (s,,z {H (z, I (z}. We argue as above, but note ( that this time s does admit the eigenvalue. This implies that in ( w =, where y F. So C y G (g = q C S (s = q 3, i.e. (g = q(q. Finally, let us consider elements g = g (s,w, {L m, M m }, where s = ( ǫ ( ν m, w = and ǫ {, ν}. Easy calculations show that if m q the elements g belong to distinct conjugacy classes of G. An element g (s,w,z belongs to C G (g if and only if { ( } a s C S (s = : a = ±, c F c a w + s w = w + s w. w(sw = w (s w Since the condition w + s w = w + s w implies a =, it follows that g (s,w,z can be chosen in q different ways. Thus, C G (g = q, i.e. (L m = (M m = q (q. So far, we have found q +5q distinct conjugacy classes, adding up to G elements. Thus, we are done., 3 The character table First of all, we observe that the character table of SL(, q = Sp(, q = G/H (q is well-known, e.g., see [Dor, 38], to which we refer for notation and all the information needed in the sequel. Next, note that, as Z(G = {A (z : z F }, for any irreducible character χ of G χ(c (z = χ(a (z χ(c ( χ( 6

7 for all z F. The same holds for the classes (D k (z, (E (z, (F(z, (G m (z, (H (z and (I (z. So, in the character table we only report the values of a character on C (, D k ( and so on. Since G/H (q = SL(, q, knowledge of the character table of SL(, q gives us by inflation q + 4 characters: namely G, η, η, ξ, ξ, θ j ( j q, ψ and χ i ( i q 3. Next, we construct q distinct irreducible characters of G having degree q. Denote by λ a fixed non-trivial character of Z(G = (F, +. Clearly, each of the q linear characters of Z(G can be parametrised as λ u (u F, where λ u (z = λ(uz for all z F. In particular, λ = Z(G. We know by [Gér, Lemma.] that H (q has exactly q non-linear irreducible characters λ u, defined as λ u (h = { qλu (h if h Z(H (q if h Z(H (q (u F. Furthermore, by [Gér, Theorem.4] the characters λ u can be extended to G. We denote such extensions by ω u (u F. The values taken by the characters ω u on the elements of S can be found in [Sze, Proposition ]. Namely: g A (z B C( D k ( E ( F( G m ( H ( I ( ω u (g q qλ u (z δ ( k δ δ ( m+ Q(λ u Q(λ u where Q(λ = t F λ( t /, Q(λ u = t F λ u ( t / = ( u F Q(λ and ( u { + if u is a square in F = F if u is not a square in F (it turns out that Q(λ = q. We are left to compute the values of the ω u s on the classes (L m and (M m. To this purpose, we compute = (ω u, ω u G = q4 (q + q (q q m=( ω u (L m + ω u (M m. q 4 (q This implies that ω u (L m = ω u (M m =, for all m q. It is easy to verify that the characters ω u η, ω u η, ω u ξ, ω u ξ, ω u θ j, ω u ψ and ω u χ i (u F are pairwise distinct irreducible characters of G. At this stage, q irreducible characters of G are still missing. We construct them as follows. 7

8 Let us consider the Sylow p-subgroup K of G consisting of the matrices of shape y/ x/ z k (a,x,y,z = a x + ay y, where a, x, y F. Define the linear characters µ u,u (u, u F of K setting µ u,u (k (a,x,y,z = λ u (aλ u (y = λ(u a + u y, where, as above, λ u denotes the non-trivial linear character of Z(G associated to u F (in particular, µ, = K. We consider the induced characters µ G u,u. First of all, note that (C (z K = and that the same holds also for (D k (z, (E (z, (F(z and (G m (z. So, the value of µ G u,u on these classes is, whereas the value on A (z is µ G u,u (A (z = q4 (q = q. q 4 ( To compute µ G s s u,u (B, we observe that if s = S and g = g s s (s,w,z, then µ u,u (gbg = λ u (λ u (s = λ u (s. So, if s = the matrix s can be chosen in q(q ways, whereas if we fix s, s can be chosen in q different ways. For u we obtain µ G u,u (B = q3 [q(q + q s λ u (s ] q 4 = q3 [q(q q ] q 4 = Next, we( look at the classes (H (z. The matrices s such that gh (zg K are s s of shape. It follows that /s and therefore µ u,u (gh (zg = λ u ( s λ u ( µ G u,u (H (z = q3 q t λ u ( t q 4 = + Q(λ u = + ( F Q(λ u In a similar way, one also obtains that µ G u,u (I (z = t λ u ( νt. 8

9 In particular, for u = we get µ G,u (H (z = µ G,u (I (z = q, whereas for u, we get µ G u,u (I (z = Q(λ u. The value of µ G u,u on L m is obtained in the same way as above: s has the same shape as in the case H (z, but µ u,u (gl m g = λ u ( s λ u ( νm s. Thus, µ G u,u (L m = q3 q t λ u ( t λ u ( ν m /t Similarly, in the case of M m, we obtain q 4 = ( λ u t 3 + u ν m. t t µ G u,u (M m = q3 q t λ u ( νt λ u ( ν m /t q 4 = ( λ u νt 3 + u ν m. t t In particular, for u = we get µ G,u (L m = µ G,u (M m =. Set κ = µ G,. Computing (κ, κ G, one sees that κ is irreducible. Furthermore, for all u, u F, κ is different from any of the µ G u,u s because κ (H ( + κ (I ( = q µ G u,u (H ( + µ G u,u (I ( =. Next, we show that we can always pick q pairwise distinct irreducible characters among the µ G u,u s. For instance, we can take as (u, u the pairs (, ν n and (ν, ν n, where n q. Set κ,n = µ G,ν n, κ ν,n = µ G ν,νn. We start showing that these characters are irreducible. Use of Mackey s formula implies that (κ,n, κ,n G = r R(µ,ν n, r µ,ν n K r K, where R is a complete set of representatives for the double cosets of K in G. As R we can choose the set {s(α, s(β α, β F }, where ( ( α /β s = s(α =, s = s(β = S. /α β Note that KsK = q 4 and KsK = q 5. Since the µ,ν n s are linear characters, it suffices to show that for r s(, the restrictions of µ,ν n and r µ,ν n to K r K are distinct. First, we look at the double cosets Ks(βK. For all s k K s K, we have µ,ν n( s k = λ ν n(βx and s µ,ν n( s k = µ,ν n(k = λ ν n(y. It follows that, if µ,ν n = s µ,ν n, then λ ν n(βx = λ ν n(y, for all x, y F. In particular, for x =, we have λ ν n(y = for all y F, i.e. Ker(λ ν n = Z(H (q, forcing ν n =, a contradiction. 9

10 Next, we look at the double cosets Ks(αK. For all s k K s K, we have µ,ν n( s k = λ (aα λ ν n( y and s µ α,ν n( s k = µ,ν n(k = λ (aλ ν n(y. It follows that, if µ,ν n = s µ,ν n, then λ (aα λ ν n( y = λ α (aλ ν n(y, for all a, y F. In particular, for y =, we get λ α = λ, and so α =. Clearly, for α =, the two restrictions are the same character. This proves that the characters κ,n are irreducible. In the same way, we can prove that the characters κ ν,n are also irreducible. To conclude, we are left to show that the characters κ,n and κ ν,n are pairwise distinct. This can be obtained proving that (κ d,n, κ d,n G =, for d, d {, ν}, n q and (d, n (d, n. As above, we exploit Mackey s formula. The double cosets Ks(βK are dealt with in the same way as before. In the case of the double cosets Ks(αK, for d = d we can argue as before. In the case (d, d = (, ν, if the restrictions of µ,ν n and µ ν,ν n are the same, then λ (aα λ ν n( y α = λ ν(aλ ν n (y for all a, y F. In particular, for y =, we get λ α = λ ν, a contradiction, since ν is not a square in F. In conclusion, the desired character table of G can be described as follows: A (z B C ( D k ( G q η q η q+ ξ q+ ξ q q q+ q+ q q q+ q+ δ(q δ(q δ(q+ ( k δ(q+ ( k θ j q q q ( j (q ψ q q q q χ i q + q + q + ( i (q + ρ ik + ρ ik κ q q κ,n q q κ ν,n q q ω u q qλ u (z δ ( k q(q q(q q(q+ q(q+ q(q λ u(z ω u η q(q λ ω u η u(z (q q(q+λ ω u ξ u(z (q+ q(q+λ ω u ξ u(z (q+ ω u θ j q(q q(q λ u (z ( j δ(q ω u ψ q q λ u (z δq ( k ω u χ i q(q + q(q + λ u (z ( i δ(q + ( k (ρ ik + ρ ik (q

11 E ( F( G m ( H ( G η η ξ ξ δ( + δq δ( δq δ(+ δq δ( δq δ( δq ( m+ ( + δq δ( + δq ( m+ ( δq δ( δq (+ δq δ(+ δq ( δq θ j ( j+ ( j+ (σ jm + σ jm ψ χ i ( i ( i κ q κ,n + ( F Q(λ κ ν,n ( F Q(λ ω u δ δ ( m+ Q(λ u ω u η ω u η ω u ξ δq + δq + δq δq + δq δq δq + δq ( + δqq(λ u ( δqq(λ u (+ δqq(λ u ( δqq(λ u ω u ξ ω u θ j ( j+ δ ( j+ δ ( m (σ jm + σ jm Q(λ u ω u ψ ( m ω u χ i ( i δ ( i δ Q(λ u I ( L m M m G η η ξ ξ ( δq ( + δq ( δq (+ δq ( + δq ( δq (+ δq ( δq ( δq ( + δq ( δq (+ δq θ j ψ χ i κ q κ,n ( F Q(λ t F λ( t3 +ν n+m t t κ ν,n + ( F Q(λ t F λ( νt3 +ν n+m t F λ( νt3 +ν n+m t t t F λ( ν t 3 +ν n+m ω u Q(λ u ω u η (+ δqq(λ u ω u η ( δqq(λ u ω u ξ ( + δqq(λ u ( δqq(λ ω u ξ u ω u θ j Q(λ u ω u ψ ω u χ i Q(λ u Notations. i, k q 3 q q, j, m, n. δ = (, ρ = e πı πı q, σ = e q+. F = GF(q, F = ν, u F. λ is a (fixed non-trivial character of Z(G. λ u is

12 the linear character of Z(G defined by λ u (z = λ(uz for all z Z(G. Q(λ = ( u λ( t /, Q(λ u = Q(λ. F t F For all χ Irr(G, we have (but this is omitted from the Table χ(c (z = and likewise for the other conjugacy classes. χ(a (z χ(c (, χ(a ( References [DPW] L. Di Martino, M. A. Pellegrini, T. Weigel, Minimal irreducibility and the unipotent characters of groups of type B m and C m, In Preparation [Dor] L. Dornhoff, Group representation theory. Part A: Ordinary representation theory, Pure and Applied Mathematics, 7, Marcel Dekker, Inc., New York 97 [Gér] P. Gérardin, Weil representations associated to finite fields, J. Algebra 46 ( [Sze] F. Szechtman, Weil representations of the symplectic group, J. Algebra 8 ( Marco Antonio Pellegrini, Dipartimento di Matematica e Applicazioni, Università degli Studi di Milano-Bicocca, Via R. Cozzi, 53, 5 Milano (Italy marco.pellegrini@unimib.it

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