ASYMPTOTICS OF THE NUMBER OF INVOLUTIONS IN FINITE CLASSICAL GROUPS

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1 ASYMPTOTICS OF THE NUMBER OF INVOLUTIONS IN FINITE CLASSICAL GROUPS JASON FULMAN, ROBERT GURALNICK, AND DENNIS STANTON Abstract Answering a question of Geoff Robinson, we compute the large n iting proportion of i GL(n, q/q n /, where i GL(n, q denotes the number of involutions in GL(n, q We give similar results for the finite unitary, symplectic, and orthogonal groups, in both odd and even characteristic At the heart of this work are certain new sum=product identities Our self-contained treatment of the enumeration of involutions in even characteristic symplectic and orthogonal groups may also be of interest Introduction On math overflow, Dima Pasechnik asked about the asymptotic behavior of the number of involutions in GL(n,, as n Again on math overflow, Geoff Robinson observed that for n even, the number of involutions in GL(n, is unreasonably close to the estimate n / In an to us in October 05, Robinson asked whether for n even we could compute the iting value of i GL (n, / n /, where i GL (n, is the number of involutions in GL(n,, and asserted that he could show that this iting value should be between 3/4 and something probably larger than Letting i GL (n, q denote the number of involutions in GL(n, q, we are able to compute the more general iting proportion i GL (n, q/q n / The answer depends on whether n is even or odd and on whether q is even or odd For example if n is even, and if q is even and fixed, it is shown in Section 3 that i GL (n, q n q n / = ( + q/q i + i i ( q/q i = i ( + q 5 8i ( + q 3 8i ( q 8i ( q i When q = this is 6793, which is indeed between 3/4 and something probably larger than The reason for dividing by q n / is that for n even, Date: Submitted February 3, 06; revised February 8, 07 Key words and phrases involution, finite classical group 00 AMS Subject Classification: 05E99, 0G40

2 JASON FULMAN, ROBERT GURALNICK, AND DENNIS STANTON n / is the dimension (in the algebraic group of the largest conjugacy class of involutions Similarly when n is odd, one divides by q (n / Our asymptotic results are cited by Ginzburg and Pasechnik in their recent work on random chain complexes [7] At the heart of these results for GL(n, q are the following sum=product identities, proved in [5] These identities state that: u n q (n n/ q r(n 3r GL(r, q GL(n r, q = i ( + u/qi i ( u /q i u n q (n n i GL(r, q GL(n r, q = ( + u/qi i ( u /q i The first identity is useful for even characteristic, and the second identity is useful for odd characteristic We give new proofs of these identities, and derive analogous identities for the other finite classical groups, which we believe to be interesting and new The proof of the even characteristic symplectic identity is tricky, as are the orthogonal group identities The point of these identities is that it allows us to obtain product formulae for the generating functions for proportions of involutions in finite classical groups These product formulae are perfectly suited for asymptotic analysis, using the following elementary lemma of Darboux (given an exposition in [0] Lemma Suppose that f(u is analytic for u < r, r > 0 and has a finite number of simple poles on u = r Let {w j } denote the poles, and suppose g j (u u/w j that f(u = j with g j (u analytic near w j Then the coefficient of u n in the Taylor expansion of f(u around 0 is g j (w j wj n + o(/r n j For each of the families of groups we consider, we get an infinite product formula for the iting value of i(n, q/q d(n,q for q fixed and n Here d(n, q is the dimension of the variety of involutions in the corresponding algebraic group It turns out that the it of this formula as q is precisely the number of components of maximal dimension in the variety (in every case this is either or This paper is organized as follows Section derives many sum=product identities which are crucial to our work The remaining sections study asymptotics of the number of involutions Section 3 treats the general linear groups, and as an easy corollary, the unitary groups Section 4 treats the symplectic groups, and Section 5 treats the orthogonal groups We mention that Sections 4 and 5 give self-contained derivations of the number of involutions in the symplectic and orthogonal groups While this is easy in odd

3 THE NUMBER OF INVOLUTIONS 3 characteristic, the enumeration of involutions in even characteristic is subtle (see [], [3], [4], [8] for related results Our treatment of involutions is used in a recent article of Garibaldi and Guralnick [6] To close the introduction, we make three remarks about future work First, it would be interesting to find proofs of our results which do not require generating functions Second, our methods will work for the subgroups SO and Ω of the orthogonal groups, and should work for the spin groups Third, enumeration of involutions is connected to representation theory via Frobenius-Schur indicators This is developed for GL and U in [5], but much remains to be done for other finite classical groups For some results for Sp and O (in odd characteristic, see [] and the references therein Identities This section derives some sum=product identities which will be crucially applied later in the paper Throughout this section (and the rest of this paper we use the following notation: n (A; q n = ( Aq k, (A; q = ( Aq k if < q < k=0 Theorem (the q-binomial theorem will be used throughout See page 7 of [] for a proof k=0 Theorem If < q < and x <, then n=0 (A; q n (q; q n x n = (Ax; q (x; q The following two well known special cases are used Corollary If < q < and x <, then n=0 Proof Set A = 0 in Theorem x n (q; q n = Corollary 3 If < q <, then n=0 (x; q x n q (n (q; q n = ( x; q Proof Replace x by x/a in Theorem, and let A Lemma 4 is one of our main lemmas

4 4 JASON FULMAN, ROBERT GURALNICK, AND DENNIS STANTON Lemma 4 If < q, and ab/q <, then q (m m ( a m k ( b k m=0 k=0 (q; q k q (k (q; qm k q (m k = ( a/q; /q ( b/q; /q (ab/q; /q = H(a, b, q Proof Let Q = /q Replacing m by m + k, the double sum is a m b k D = Q (k+ +( m+ mk (Q; Q k (Q; Q m k=0 m=0 The m-sum is evaluable by Corollary 3 to So ( aq k ; Q = ( aq k ; Q k ( aq; Q D = ( aq; Q by Theorem k=0 = ( /a; Q k ( aq; Q a k Q (k ( /a; Q k (abq k = ( aq; Q ( bq; Q (Q; Q k (abq; Q There is a companion lemma which is also useful Lemma 5 Suppose that q >, that s is a non-negative even integer, and that t is a non-negative integer Let a, b be complex and suppose that bq s/ < Then n=0 q (n (n t/s a n sr b r q r (/q; /q r q (n sr t (/q; /q n rs t q r(sn (+s /r =a t q (t ( aqt ; /q (bq s/ ; /q = G(a, b, q, s, t Proof Replacing n by n + rs + t one obtains G(a, b, q, s, t = a t q (t ( bq s/ r (/q; /q r Then apply Corollaries and 3 n=0 q (n ( aq t n (/q; /q n Proposition 6 will be useful for writing certain sums and differences of infinite products as a single infinite product Proposition 6 Let < R < and F (X, R = (R; R ( X; R ( R/X; R

5 THE NUMBER OF INVOLUTIONS 5 Then (F (X, R + F ( X, R =F (RX, R 4, (F (X, R F ( X, R =XF (R3 X, R 4 Proof Use the Jacobi triple product identity ([], p F (X, R = R (n X n n= to find the even and odd parts of F (X, R Theorem 7 will be useful in treating even characteristic general linear and unitary groups Recall that in any characteristic, GL(j, q = q (j (q j (q Theorem 7 appeared as Corollary 35 of [5], but for completeness we give a different proof here Theorem 7 For q > and u < q, u n q (n n/ Proof Because for any r 0, q r(n 3r GL(r, q GL(n r, q = ( GL(r, q = q r (/q; /q r, i ( + u/qi i ( u /q i we use Lemma 5 to evaluate the double sum, which is G(u, u, q,, 0 Theorem 8 will be helpful in treating odd characteristic general linear and unitary groups This result was Corollary 37 of [5], but we give a different (and simpler proof here Theorem 8 For q > and u < q, u n q (n n GL(r, q GL(n r, q = i ( + u/qi i ( u /q i Proof Using (, we can apply Lemma 4 to evaluate the double sum, which is H(u, u, q Theorem 9 gives a useful sum=product formula for the symplectic groups in odd characteristic This can also be obtained by replacing q by q in Theorem 8 Recall that in any characteristic, for n 0, n ( Sp(n, q = q n (q i = q n +n (/q ; /q n i= Theorem 9 For q > and u < q, n u n q n Sp(r, q Sp(n r, q = i ( + u/qi i ( u /q i

6 6 JASON FULMAN, ROBERT GURALNICK, AND DENNIS STANTON Proof We can apply ( and Lemma 4 to evaluate the double sum, which is H(u, u, q The following rather tricky identity will be useful for treating even characteristic symplectic groups Theorem 0 For q > and u <, u n n n n q n + + A r B r where r even r= r even r= r odd C r A r =q r(r+/+r(n r Sp(r, q Sp(n r, q, = i ( + u/qi u i ( u /q i B r =q r(r+/+r(n r q r Sp(r, q Sp(n r, q, C r =q r(r+/+r(n r Sp(r, q Sp(n r, q Proof Each of the three double sums may be computed using Lemma 5 Replacing r by r, r +, and r + respectively in these three double sums, we see that the three sums become G(u, u,q,, 0 + q 6 G(u/q 4, u q, q,, + q G(u/q, u q, q,, ( =( u/q ; /q u u (u /q ; /q + (u ; /q + (u ; /q = ( u/q ; /q u (u /q ; /q Next we prove three sum=product identities which will be useful for studying odd characteristic orthogonal groups Recall that in any characteristic, (3 n O + (n +, q = O (n +, q = q n (q i = q n +n (/q ; /q n and that O + (, q = O (, q = Recall also that in any characteristic, for n, (4 O ± (n, q = qn n n q n (q i = qn ± q n ± (/q ; /q n i= It is also helpful to use (4 to define O ± (n, q when n = 0 Thus i= O + (0, q =, / O (0, q = 0 Our first identity is useful in treating odd characteristic even dimensional orthogonal groups of positive type

7 Theorem For q > and u < /q, u n q n = [ n THE NUMBER OF INVOLUTIONS 7 n O + (r, q O + (n r, q + r= ] O (r, q O (n r, q i ( + u/q(i ( uq i ( u /q (i + i ( + u/qi i ( u /q (i Proof It suffices to show that the left-hand side of the statement of Theorem is equal to [ ( u/q; /q (u ; /q + ( + uq ( u; /q ] (q u ; /q Because the value of O + (r, q depends on the parity of r, we split each sum into two sums, depending on the parity of r CASE : r = R, first sum: S = n ( u n ( + q R ( + q n R q n 4 q R R (q ; q R q (n R (n R (q ; q n R n=0 R=0 To write this in the form of Lemma 4 with q replaced by q, note that q R R = q (R, q (n R (n R = q (n R q n ( + q R ( + q n R = q (n q n ( + q R ( + q n R The numerator has 4 terms q n, q n+r, q n R, q n We pick (a, b in Lemma 4, so that these four terms appear The choices are (a, b = (uq, uq, (a, b = (uq, uq, (a, b = (uq, uq, and (a, b = (uq, uq respectively to obtain S = [ ( u/q; /q 4 (u ; /q + ( u/q; /q ( u; /q (qu ; /q + ( u; /q ] (q u ; /q CASE : r = R +, first sum: S = n u n ( R ( n R q n 4 q R (q ; q R q (n R (q ; q n R n= R=0 We use Lemma 4 with q replaced by q and (a, b = (uq, uq, S = uq 4 ( u; /q (u q ; /q CASE 3: r = R, second sum: S = n ( u n (q R (q n R q n 4 q R R (q ; q R q (n R (n R (q ; q n R n=0 R=0

8 8 JASON FULMAN, ROBERT GURALNICK, AND DENNIS STANTON The numerator has 4 terms q n, q n+r, q n R, q n As in CASE we obtain S 3 = [ ( u/q; /q 4 (u ; /q ( u/q; /q ( u; /q (qu ; /q + ( u; /q ] (q u ; /q CASE 4: r = R +, second sum: S 4 = 4 n= n u n q n R=0 This is the same as CASE, So S + S + S 3 + S 4 = S 4 = uq 4 ( R ( n R q R (q ; q R q (n R (q ; q n R ( u; /q (u q ; /q [ ( u/q; /q (u ; /q + ( + uq ( u; /q ] (q u ; /q Our second identity is useful in treating odd characteristic even dimensional orthogonal groups of negative type Theorem For q > and u < /q, u n q n = [ n n O + (r, q O (n r, q + r= ] O (r, q O + (n r, q i ( + u/q(i ( uq i ( u /q (i i ( + u/qi i ( u /q (i Proof It suffices to prove that the left hand side of Theorem is equal to [ ( u/q; /q (u ; /q + ( + uq ( u; /q ] (u q ; /q Note that the two sums on the left hand side of the theorem are equal, so we focus on the first sum Because the value of O + (r, q depends on the parity of r, we split the first sum into two cases, depending on the parity of r CASE : r = R: S = ( u n q n 4 n=0 n R=0 ( + q R (q n R q R R (q ; q R q (n R (n R (q ; q n R

9 THE NUMBER OF INVOLUTIONS 9 The numerator has 4 terms q n, q n+r, q n R, q n The middle two terms cancel, so as in CASE of Theorem S = [ ( u/q; /q 4 (u ; /q + ( u; /q ] (q u ; /q CASE : r = R + : As in CASE of Theorem S = 4 So we have = uq 4 n= (S + S = n u n q n R=0 ( u; /q (u q ; /q ( R ( n R q R (q ; q R q (n R (q ; q n R [ ( u/q; /q (u ; /q + ( + uq ( u; /q ] (u q ; /q The following sum=product formula will be helpful in treating odd dimensional orthogonal groups in odd characteristic Theorem 3 For q > and u <, u n q n [ n+ is equal to n O + (r, q O + (n + r, q + r= i ( + u/qi u i ( u /q i Proof It suffices to prove that the left hand side is equal to ( u; /q ( u/q ; /q (u ; /q ] O (r, q O (n + r, q We split each sum into two sums, depending on the parity of r CASE : r = R, first sum: S = n ( u n q (n R ( + q R q n 4 q R R (q ; q R q (n R (n R (q ; q n R n=0 R=0 This is nearly CASE of Theorem, the numerator has terms q R n and q R n We apply Lemma 4 with (a, b = (u, uq, (a, b = (u, uq, and q replaced by q to obtain S = [ ( u/q ; /q ( u/q; /q 4 (u /q; /q + ( u/q ; /q ( u; /q ] (u ; /q

10 0 JASON FULMAN, ROBERT GURALNICK, AND DENNIS STANTON CASE : r = R +, first sum: S = n ( u n ( + q n R q n 4 q R (q ; q n=0 R=0 R q (n R (n R (q ; q n R = [ ( u; /q ( u/q ; /q 4 (u ; /q + ( u/q; /q ( u/q ; /q ] (u /q; /q where we used Lemma 4 with (a, b = (uq, u, (a, b = (uq, u and q replaced by q CASE 3: r = R, second sum: S 3 = n ( u n (q R q n 4 q R R (q ; q R q (n R (q ; q n R n=0 R=0 The numerator has terms q n q (n R (q R = q R q R Using (a, b = (u, uq, (a, b = (u, uq and q replaced by q in Lemma 4, we obtain S 3 = [ ( u/q ; /q ( u; /q 4 (u ; /q ( u/q ; /q ( u/q; /q ] (u /q; /q CASE 4: r = R +, second sum: S 4 = n ( u n (q n R q n 4 q R (q ; q R q (n R (n R (q ; q n R n=0 R=0 This is basically the same as CASE, S 4 = [ ( u; /q ( u/q ; /q 4 (u ; /q ( u/q; /q ( u/q ; /q ] (u /q; /q So we have S + S + S 3 + S 4 = ( u; /q ( u/q ; /q (u ; /q Finally, we give two identities which will be useful for analyzing the number of involutions in even characteristic orthogonal groups Theorem 4 will be useful for positive type orthogonal groups in even characteristic Theorem 4 For q > and u < /q, q n u n n n + A r where = r even r= r even B r + n r= r odd i ( + u/q(i ( uq i ( u /q (i + i ( + u/qi i ( u /q (i, C r

11 THE NUMBER OF INVOLUTIONS A r = q r(r /+r(n r Sp(r, q O + (n r, q B r = q r(r+/+(r (n r Sp(r, q Sp(n r, q C r = q r(r /+(r (n r Sp(r, q Sp(n r, q Proof The first double sum, after replacing r by r, may be rewritten, after using (, (4, and Lemma 5, as ( G(uq, u q, q,, 0 + G(uq, u q, q,, 0 The last two double sums, after replacing r by r + and r + respectively, are ( q 4 G(u/q, u q 4, q,, + q G(u, u q 4, q,, The identity ( G(uq, u q, q,, 0 + q 4 G(u/q, u q 4, q,, + q G(u, u q 4, q,, ( u; /q = ( uq (u ; /q completes the proof Theorem 5 will be useful for negative type orthogonal groups in even characteristic Theorem 5 For q > and u < /q, n q n u n n + A r where = r even r= r even B r + n r= r odd i ( + u/q(i ( uq i ( u /q (i i ( + u/qi i ( u /q (i, C r A r = q r(r /+r(n r Sp(r, q O (n r, q B r = q r(r+/+(r (n r Sp(r, q Sp(n r, q C r = q r(r /+(r (n r Sp(r, q Sp(n r, q Proof The only change to the proof of Theorem 4, is that the first double sum has a minus sign in the numerator factor for O (n r, q The sum for /A r can be extended to r = n since / O (0, q = 0 by (4

12 JASON FULMAN, ROBERT GURALNICK, AND DENNIS STANTON 3 General linear and unitary groups To begin we study asymptotics of the number of involutions in GL(n, q, when q is fixed and n Throughout we let i GL (n, q denote the number of involutions in GL(n, q The following explicit formula for i GL (n, q can be found in Section of [9] Proposition 3 The number of involutions in GL(n, q is i GL (n, q = { n/ GL(n,q q r(n 3r GL(r,q GL(n r,q n GL(n,q GL(r,q GL(n r,q, for q odd To begin we treat the case of even characteristic, for q even, Theorem 3 ( Let n be even and let q be even and fixed Then i GL (n, q = ( + q/q i + n q n / i i ( q/q i = i ( + q 5 8i ( + q 3 8i ( q 8i ( q i ( Let n be odd and let q be even and fixed Then i GL (n, q q = ( + q/q i n q (n / i i ( q/q i = i ( + q 7 8i ( + q 8i ( q 8i ( q i Proof Replacing u by u q in Theorem 7 gives that (5 is equal to (6 u n q n/ q (n n/ q r(n 3r GL(r, q GL(n r, q i ( + u q/q i ( u i ( u /q i First consider the n it of the coefficient of u n in (5 This is equal to n q n/ n/ (q n (q q n/ = n (q n (q i GL(n, q = i ( /qi i GL (n, q n q n / GL(n, q q r(n 3r GL(r, q GL(n r, q

13 THE NUMBER OF INVOLUTIONS 3 The first equality is from Proposition 3 Now consider (6 Except for poles at u = ±, it is analytic in a disc of radius greater than Hence Lemma gives that for n even, the n it of the coefficient of u n in (6 is (7 It follows that for n even, i GL (n, q = n q n / [ i( + q/q i i ( /qi + Letting Q = /q, this is equal to i( q/q i i ( /qi ] ( + q/q i + i i ( q/q i (( Q / ; Q + (Q / ; Q = (( Q / ; Q ( Q 3/ ; Q + (Q / ; Q (Q 3/ ; Q = ( (Q ; Q F (Q /, Q + F ( Q /, Q = (Q ; Q F (Q 3, Q 8, where F is defined in Proposition 6 and where in the last step we have applied Proposition 6 with X = Q / and R = Q Similarly, if n is odd, Lemma gives that the n it of the coefficient of u n in (6 is (8 It follows that for n odd, i GL (n, q = n q n / [ i( + q/q i i ( /qi i( q/q i ] i ( /qi ( + q/q i i i ( q/q i Letting Q = /q and arguing as in the n even case gives that this is equal to ( (Q ; Q F (Q /, Q F ( Q /, Q = q (Q ; Q F (Q 7, Q 8 This proves the theorem Remarks: ( The expression in part of Theorem 3 is equal to 6793 when q =, and tends to as q The expression in part of Theorem 3 is equal to 9 when q =, and tends to as q

14 4 JASON FULMAN, ROBERT GURALNICK, AND DENNIS STANTON ( In parts and of Theorem 3, one can rewrite the ( q 8i /( q i as ( + q 4i ( + q i This makes it clear that the its in the theorem decrease monotonically to as q Our reason for not doing this in the statement of Theorem 3 is to emphasize the role in the numerator played by the base 8 Next we treat odd characteristic Our main result is the following theorem Theorem 33 i GL (n, q n q n / ( Let n be even and let q be odd and fixed Then = ( + q/q i + ( q/q i i i = i ( + q 4i ( q 4i ( q i ( Let n be odd and let q be odd and fixed Then i GL (n, q q = ( + q/q i ( q/q i n q (n / i i = i ( + q 4i ( q 8i ( q i Proof Replacing u by u q in Theorem 8 gives the equation (9 u n q n/ q (n n GL(r, q GL(n r, q = i ( + u q/q i ( u i ( u /q i Consider the n it of the coefficient of u n in the left hand side of (9 This is equal to q n/ n GL(n, q n (q n (q GL(r, q GL(n r, q q n/ = n (q n (q i GL(n, q = i ( /qi i GL (n, q n q n / The first equality is from Proposition 3 Now consider the right hand side of (9 Except for poles at u = ±, it is analytic in a disc of radius greater than Hence Lemma gives that for n even, the n it of the coefficient of u n in the right hand side of (9 is [i ( + q/q i i ( + /qi i ( q/q i ] i ( /qi

15 It follows that for n even, i GL (n, q = n q n / THE NUMBER OF INVOLUTIONS 5 ( + q/q i + i i ( q/q i Letting Q = /q and F be as in Proposition 6, this is equal to ( F (Q /, Q + F ( Q /, Q = F (Q, Q 4, (Q; Q (Q; Q where we have used Proposition 6 Similarly, if n is odd, Lemma gives that the n it of the coefficient of u n in the right hand side of (9 is It follows that for n odd, [i ( + q/q i i ( /qi i GL (n, q = n q n / i ( q/q i i ( /qi ( + q/q i i i ] ( q/q i Letting Q = /q, this is equal to ( F (Q /, Q F ( Q /, Q (Q; Q This proves the theorem = q (Q; Q F (Q 4, Q 4 Remark: The expression in part of Theorem 33 is equal to 85 when q = 3, and tends to as q The expression in part of Theorem 33 is equal to 3647 when q = 3, and tends to as q Next we study asymptotics of the number of involutions in the finite unitary groups U(n, q Recall that the (general unitary group is defined to be the subgroup of GL(n, q stabilizing a non degenerate Hermitian form (any two such forms are equivalent and so the group is unique up to conjugacy typically we assume the standard basis is an orthonormal basis with respect to the Hermitian form The results will be an easy consequence of the general linear case Throughout we let i U (n, q denote the number of involutions in U(n, q Recall that in any characteristic n (0 U(n, q = q n(n / (q i ( i = ( n GL(n, q i= The following elementary fact was Proposition 44 of [5] in even characteristic and was Proposition 49 of [5] in odd characteristic

16 6 JASON FULMAN, ROBERT GURALNICK, AND DENNIS STANTON Proposition 34 The number of involutions in U(n, q is i U (n, q = { n/ U(n,q q r(n 3r U(r,q U(n r,q n U(n,q U(r,q U(n r,q, for q odd, for q even, Theorem 35 treats the asymptotics of involutions in U(n, q with q even Theorem 35 ( Let n be even and let q be even and fixed Then i U (n, q = n q n / k ( q 3 8k ( q 5 8k ( q 8k ( q k ( Let n be odd and let q be even and fixed Then n i U (n, q q (n / = k ( q 8k ( q 7 8k ( q 8k ( q k Proof For part, it follows from Propositions 3 and 34 that for n and q even, i U (n,q is obtained by replacing q by q in i GL(n,q The result now q n / q n / follows from part of Theorem 3 For part, it follows from Propositions 3 and 34 that for n odd and q i even, U (n,q is obtained by replacing q by q in i GL(n,q The result now q (n / follows from part of Theorem 3 q (n / Next we treat the case of odd characteristic unitary groups Theorem 36 ( Let n be even and let q be odd and fixed Then i U (n, q = n q n / k ( + q 4k ( q 4k ( ( /q k ( Let n be odd and let q be odd and fixed Then n i U (n, q q (n / = k ( + q 4k ( q 4k ( ( /q k Proof For part, it follows from Propositions 3 and 34 that for n even and q odd, i U (n,q is obtained by replacing q by q in i GL(n,q The result q n / q n / now follows from part of Theorem 33 For part, it follows from Propositions 3 and 34 that for n odd and q i odd, U (n,q is obtained by replacing q by q in i GL(n,q The result now q (n / follows from part of Theorem 33 4 Symplectic groups q (n / This section studies the asymptotics of the number of involutions in the finite symplectic groups We begin with the case of odd characteristic Lemma 4 gives a formula for the number of involutions (see also the proof of Theorem 3 of []

17 THE NUMBER OF INVOLUTIONS 7 Lemma 4 Suppose that q is odd Sp(n, q is equal to Then the number of involutions in n Sp(n, q Sp(r, q Sp(n r, q Proof Let g Sp(n, q be an involution Let V denote the natural ndimensional module Then V = V V where V a is the eigenspace of g with eigenvalue a Note that each V is a nondegenerate space of even dimension Set r = dim V Since any two nondegenerate spaces of the same dimension are in the same Sp(n, q orbit, we see the conjugacy classes of involutions are determined by r Moreover, the centralizer of g is obviously isomorphic to Sp(r, q Sp(n r, q and so the size of the conjugacy class of g is Sp(n, q Sp(r, q Sp(n r, q Since r can take on any value between 0 and n, this proves the theorem The following theorem is our main result Theorem 4 Let i Sp (n, q be the number of involutions in Sp(n, q ( Let n be even and q be odd and fixed Then i Sp (n, q = ( + /q i + ( /q i n q n i i = i ( + q 4 8i ( q 8i ( q i ( Let n be odd and q be odd and fixed Then i Sp (n, q n q n = q ( + /q i ( /q i i i = i ( + q 8i ( q 6i ( q i Proof Replacing u by uq in Theorem 9 gives ( n u n q n +n i Sp(r, q Sp(n r, q = ( + u/qi ( u i ( u /q i

18 8 JASON FULMAN, ROBERT GURALNICK, AND DENNIS STANTON Consider the n it of the coefficient of u n in the left hand side of ( This is equal to n Sp(n, q n q n ( /q ( /q n Sp(r, q Sp(n r, q = n = q n ( /q ( /q n i Sp(n, q i ( /qi n i Sp (n, q q n The first equality was Lemma 4 Now consider the right hand side of ( Except for poles at u = ±, it is analytic in a disc of radius greater than Hence Lemma gives that for n even, the n it of the coefficient of u n in the right hand side of ( is It follows that for n even, [i ( + /qi i ( /qi i Sp (n, q = n q n + i ( /qi ] i ( /qi ( + /q i + i i ( /q i Letting Q = /q and using Proposition 6 gives that this is equal to ( F (Q, Q (Q ; Q + F ( Q, Q = (Q ; Q F (Q 4, Q 8 Similarly, if n is odd, Lemma gives that the n it of the coefficient of u n in the right hand side of ( is [i ( + /qi i i ( ( /qi ] /qi i ( /qi It follows that for n odd, i Sp (n, q = n q n ( + /q i i i ( /q i Letting Q = /q and using Proposition 6 gives that this is equal to ( F (Q, Q (Q ; Q F ( Q, Q = This proves the theorem (Q ; Q QF (Q 8, Q 8 Remark: The expression in part of Theorem 4 is equal to 689 when q = 3, and tends to as q The expression in part of Theorem 4 is equal to 89 when q = 3, and tends to as q Next we consider the case that q is even

19 THE NUMBER OF INVOLUTIONS 9 To begin we use elementary means to derive a formula for the number of involutions in Sp(n, q when q is even For related results, see [], [3], [4], [8] Theorem 43 When q is even, the number of involutions in Sp(n, q is equal to n Sp(n, q n Sp(n, q n Sp(n, q + + A r B r C r where r even r= r even r= r odd A r =q r(r+/+r(n r Sp(r, q Sp(n r, q, B r =q r(r+/+r(n r q r Sp(r, q Sp(n r, q, C r =q r(r+/+r(n r Sp(r, q Sp(n r, q The following lemma will be helpful for proving Theorem 43 Lemma 44 Suppose q is even, and let g Sp(n, q be an involution Let V be the natural module of dimension n for Sp(n, q Then ( W = (g V is totally singular of dimension r, where r n is the rank of g ( W = V g is the fixed space of g on V Proof If U is a nontrivial nondegenerate subspace of V g, then clearly, V = U U is a g-stable decomposition whence the result follows by induction So we may assume that V g is totally singular Since dim V g n, it follows that V g is n-dimensional and is a maximal isotropic space Since g =, W V g and since dim W + dim V g = n, we see in this case that r = n and V g = W and the result holds Now we prove Theorem 43 Proof (Of Theorem 43 Let G = Sp(n, q, and let g G be an involution With respect to the flag 0 < (g V < V g < V, and an appropriate basis, g can be written as: I 0 h 0 I I where h is a symmetric matrix of rank r Let P be the parabolic subgroup stabilizing the flag Then P acts on the set of all such elements with any symmetric element in the upper corner by congruence (via GL(r, q We use the fact that two symmetric m m matrices over a perfect field of characteristic are congruent if and only if they have the same rank and are both skew or are both nonskew (in particular if the rank is odd, then they are all congruent

20 0 JASON FULMAN, ROBERT GURALNICK, AND DENNIS STANTON Thus, we see that the conjugacy classes of involutions in G are determined by the rank r and if r > 0 is even on whether h is skew or not Clearly, two involutions written as above are conjugate in G if and only if they are conjugate in P if and only if the right hand corner terms are congruent In particular, there are n + + n/ classes if n is even and n + + (n / classes if n is odd Now we can write down the centralizers of such elements Clearly any element of C G (g preserves the flag and so C G (g is contained in P Note that P = LQ where Q is the unipotent radical with Q = q r(r+/+r(n r Clearly Q centralizes g Note that L = GL(r, q Sp(n r, q and that Sp(n r, q also centralizes g So we just need to compute the centralizer of g in K := GL(r, q Now A K acts on g by sending h to AhA Thus, C K (g = Sp(r, q if r is even and h is an alternating form If h is not alternating, by the remarks above, we can take h = I and so C K (g = {A K AA = I} Of course, if q were odd, this would just be an orthogonal group However, with q even, this is not the case We now compute the order of C K (g Let U be an r-dimensional space with a nondegenerate symmetric bilinear pairing (, on U that is not alternating (so (u, u = for some u U Let U be the set of all u with (u, u = 0 Since we are in a field of characteristic, this is a hyperplane in U Note that the form restricted to U is alternating Set U 0 = U Note that U 0 is -dimensional (because our form is nondegenerate If U 0 U = 0, then r is odd and L preserves this decomposition whence C K (g = Sp(r, q (the action is determined precisely by the action on U since it must be trivial on U 0 and preserve the splitting If U 0 < U, then U /U 0 is a nonsingular space of dimension r with the corresponding alternating form and so r is even In this case any element of C K (g is trivial on U 0 Let v U be a vector with v outside of U 0 with (v, v = Then there exists x C K (g with xv = v + u for any u U Moreover, C K (g induces Sp(r, q on U 0 /U and it is an easy exercise to see that C K (g = Sp(r, q q r Summarizing, we can write down the orders of centralizers for each class representative: For each even r with 0 r n, we have a class C(r with the centralizer having order A r = q r(r+/+r(n r Sp(r, q Sp(n r, q For each r with r n, we have a class D(r with centralizer having order for r even:

21 THE NUMBER OF INVOLUTIONS and for r odd: B r = q r(r+/+r(n r q r Sp(r, q Sp(n r, q C r = q r(r+/+r(n r Sp(r, q Sp(n r, q Adding these contributions completes the proof This leads to the following result Theorem 45 Let i Sp (n, q be the number of involutions in Sp(n, q Let q be even and fixed Then i Sp (n, q = + /q n q n +n i ( i Proof Combining Theorem 43 and Theorem 0 gives that ( u n q i Sp(n, q n Sp(n, q = i ( + u/qi u i ( u /q i Consider the n it of the coefficient of u n in the left hand side of ( This is equal to = i Sp (n, q n q n +n n i= ( /qi i ( /qi i Sp (n, q n q n +n Now consider the right hand side of ( Except for a pole at u =, it is analytic in a disc of radius greater than Hence Lemma gives that the n of the coefficient of u n in the right hand side of ( is i ( + /qi i ( /qi The theorem follows Remark: The expression in Theorem 45 is equal to 3559 when q =, and tends to as q 5 Orthogonal groups This section studies the asymptotics of the number of involutions in orthogonal groups over a finite field of size q Recall that the orthogonal group is the stabilizer of a nondegenerate quadratic form on a d-dimensional space If d = n is even, there are two types of nondegenerate quadratic forms and we denote their stabilizers by O ± (n, q Note that O + (n, q is the stabilizer of a nondegenerate form that contains totally singular subspaces of dimension n (in the other case, the totally singular subspaces have dimension at most n If d is odd, we only consider q odd (as the orthogonal

22 JASON FULMAN, ROBERT GURALNICK, AND DENNIS STANTON group in characteristic in odd dimension is isomorphic to the symplectic group We assume first that the characteristic is odd Lemma 5 gives an exact expression for the number of involutions (see also [] We note that the two terms in part of Lemma 5 are equal Lemma 5 Suppose that the characteristic is odd ( The number of involutions in O + (n, q is equal to n O + n (n, q O + (r, q O + (n r, q + O + (n, q O (r, q O (n r, q r= ( The number of involutions in O (n, q is equal to n O (n, q n O + (r, q O (n r, q + r= O (n, q O (r, q O + (n r, q (3 The number of involutions in O + (n +, q (or in O (n +, q is equal to n+ O + (n +, q n O + (r, q O + (n + r, q + r= O + (n +, q O (r, q O (n + r, q Proof Let g O ± (n, q be an involution and let V denote the natural orthogonal module Then V = V V where V is the fixed space of g and V is the eigenspace of g In particular, these eigenspaces are nondegenerate Let r = dim V The conjugacy class of g is determined by the orbit of V under the orthogonal group Given a fixed r with 0 < r < n, there will be two orbits depending upon the type of V (note that the type of V and V determines the type of V If r = 0 or n, then g is a scalar Clearly the centralizer of g is O(V O(V and so the size of the conjugacy class of g is O ± (n, q O(V O(V, whence the result This leads to the following result Theorem 5 Let i ± O (n, q denote the number of involutions in O± (n, q Let q be odd and fixed Then i ± O (n, q = ( + /q i n q n i Proof We prove the assertion for i + O (n, q as the proof for i O (n, q is almost identical (using part of Lemma 5 instead of part of Lemma 5, and Theorem instead of Theorem

23 THE NUMBER OF INVOLUTIONS 3 Replacing u by u/q in Theorem, and applying part of Lemma 5 gives that i (3 ( + u/qi ( u i ( u /q i + i ( + u/qi i ( u /q i is equal to (4 u n i + O (n, q ( /q n ( /q ( /q (n q n Taking the n it of the coefficient of u n in (4 gives i ( /qi i + O (n, q n q n Consider the n it of the coefficient of u n in the first infinite product in (3; by Lemma, it is equal to i ( + /qi i ( /qi Since the second term of (3 is analytic in a disc of radius bigger than, it follows that the n it of the coefficient of u n in the second infinite product of (3 is equal to 0 Summarizing, we have shown that i ( /qi i + O (n, q = n q n which implies the theorem i ( + /qi i (, /qi Remark: The expression in Theorem 5 is equal to 996 when q = 3, and tends to as q We next treat the case of odd dimensional orthogonal groups in odd characteristic Theorem 53 Let i O (n+, q be the number of involutions in O(n+, q Let q be odd and fixed Then i O (n +, q = + /q n q n +n i ( i Proof It follows from Theorem 3 and part 3 of Lemma 5 that (5 u n q i O(n +, q n O(n +, q = i ( + u/qi u i ( u /q i The n it of the coefficient of u n on the left hand side of (5 is equal to i ( /qi i O (n +, q n q n +n

24 4 JASON FULMAN, ROBERT GURALNICK, AND DENNIS STANTON By Lemma, the n it of the coefficient of u n on the right hand side of (5 is equal to i ( + /qi i ( /qi Comparing these two expressions proves the theorem Remark: The expression in Theorem 53 is equal to 538 when q = 3, and tends to as q Next we give results for even characteristic orthogonal groups Note that it is not necessary to treat odd dimensional orthogonal groups in even characteristic, as these are isomorphic to symplectic groups To begin, we derive a formula for the number of involutions in even characteristic orthogonal groups Let G = O ɛ (n, q with q a power of Let ν denote the quadratic form preserved by G on the natural module V of dimension n Recall that ν defines by a nondegenerate G-invariant alternating form on V defined by (v, w := ν(v + w + ν(v + ν(w If g G is an involution, then g X = Sp(n, q with respect to the alternating form defined above So by Lemma 44, W := (g V is totally singular with respect to the alternating form of rank r We next note: Lemma 54 If x, g G are involutions and are conjugate in X, then they are conjugate in G Proof If n = or 4, this is clear So assume that n 6 If g is a transvection in X, then we see that g is trivial on nondegenerate -spaces of either type (and so is x, whence the result follows by induction Suppose next that g corresponds to a symmetric element in X of rank at least Then g leaves invariant a nondegenerate 4-space where it has Jordan blocks of size with each nondegenerate Working in Sp(4, q, we see that g preserves nondegenerate -spaces of either type (where it acts nontrivially and so the result follows by induction Suppose that g corresponds to a skew element in X We see from the section on even characteristic symplectic groups that V is an orthogonal direct sum of 4-dimensional subspaces and one space where g acts trivially If the last space is actually there, we see by reducing to the case that n = 6, that g acts trivially on -dimensional invariant subspaces of both types, whence the result follows by induction So the remaining case is that all Jordan blocks have size It follows that on each 4-dimensional block above, the type must be + (such elements do not live in O (4, q, which is isomorphic to SL(, q, the semidirect product of SL(, q with a group of order whose generator induces the field automorphism on SL(, q given by the qth power map on the field of size q, whence the type of G must be + as well and the result is clear in this case

25 THE NUMBER OF INVOLUTIONS 5 It also follows from the argument above that each class in X occurs in G unless g corresponds to an alternating element in X of rank n (and so n is even In the latter case, g can only be in O + (n, q In the proof of Theorem 43, the conjugacy classes of involutions in Sp were called C(r and D(r Let C ɛ (r and D ɛ (r denote the intersections with G (of type ɛ By the previous lemma, this is either empty or a conjugacy class in G Now we have to compute centralizers As noted if g is in C(r with r even, then g is conjugate to an element of G unless r = n and ɛ = Then g may be written as: I 0 h 0 I 0, 0 0 I where h is skew of rank r We see that g G 0 := SO ɛ (n, q and C G (g = Sp(r, q O ɛ (n r, q q r(n r q r(r / Note that if r = n, then the O part of the centralizer is not there and ɛ = + Note also that if r < n, the G and G 0 classes are the same (because g commutes with a transvection in G while if r = n, there is a single G-class that splits into G 0 classes (corresponding to the classes of maximal totally isotropic subspaces We can compare C ɛ (r and C(r We have C ɛ (r = [G : C G (g] = G C X(g X C G (g C(r = qn r + ɛ q n + ɛ C(r The above formula again shows that C (n = 0 Next we consider the classes D(r with r n Note that for any such element we can write V as an orthogonal sum of nondegenerate - dimensional invariant spaces This implies that the G-classes and G 0 classes coincide (since any such element commutes with a transvection which is in G but not in G 0 Consider X acting on the orthogonal module of dimension n + (this is an indecomposable X-module with a -dimensional socle The stabilizers of the complements to the -dimensional socle are precisely the orthogonal groups contained in X Of course, the number of complements is q n If g D(r, then the number of g-invariant complements in q n r Thus, g lives in a total of q n r different orthogonal subgroups of X Lemma 55 Suppose that g D(r X Then g is in q n r / orthogonal subgroups of X of each type Proof We can decompose V = V 0 V V r where each V i, i > 0 is a nondegenerate -dimensional space on which g acts nontrivially and g acts trivially on V 0 Any quadratic form giving rise to the given alternating

26 6 JASON FULMAN, ROBERT GURALNICK, AND DENNIS STANTON form is determined by its restrictions to each V i Clearly, there are the same number of quadratic forms of each type Now we can compute D ɛ (r Let g D ɛ (r Let Ω be the set of orthogonal subgroups of X of type ɛ Since any two orthogonal subgroups of the same type are conjugate in X and their own normalizers (indeed, they are almost always maximal, we see that Ω = [X : O ɛ (n, q] = q n (q n + ɛ/ Let Ω(g denote those elements of Ω containing g (so this is the fixed point set of g acting on Ω By the previous lemma Ω(g = q n r / An elementary formula (essentially coming from Burnside s Lemma shows that Thus, and so Ω(g D(r = Ω D ɛ (r D ɛ (r = (/qn r D(r = Ω q n r q n (q n + ɛ D(r, D ɛ (r = qn r q n + ɛ D(r Of course D(r was computed in the proof of Theorem 43 Summarizing, we have proved the following theorem Theorem 56 Let q be even ( The number of involutions in O + (n, q is equal to n r even where O + (n, q A r + n r= r even O + (n, q B r + n r= r odd O + (n, q C r A r = q r(r /+r(n r Sp(r, q O + (n r, q B r = q r(r+/+(r (n r Sp(r, q Sp(n r, q C r = q r(r /+(r (n r Sp(r, q Sp(n r, q ( The number of involutions in O (n, q is equal to n r even where O (n, q A r + n r= r even O (n, q B r + n r= r odd O (n, q C r

27 THE NUMBER OF INVOLUTIONS 7 A r = q r(r /+r(n r Sp(r, q O (n r, q B r = q r(r+/+(r (n r Sp(r, q Sp(n r, q C r = q r(r /+(r (n r Sp(r, q Sp(n r, q Note that in the A r sum in part, r only ranges from 0 to n, and that the values of B r and C r are the same in parts and of the theorem This leads to the following result Theorem 57 Let i ± O (n, q denote the number of involutions in O± (n, q Let q be even and fixed Then i ± O (n, q = ( + /q i n q n i Proof We prove the assertion for i + O (n, q as the proof for i O (n, q is almost identical (using part of Theorem 56 instead of part of Theorem 56, and Theorem 5 instead of Theorem 4 Replacing u by u/q in Theorem 4 and applying Theorem 56 gives that (6 is equal to (7 ( u i( + u/qi i ( u /q i + i( + u/qi i ( u /q i u n i + O (n, q ( /q n ( /q ( /q (n q n Taking the n it of the coefficient of u n in (7 gives i ( /qi i + O (n, q n q n Consider the n it of the coefficient of u n in the first infinite product in (6 By Lemma, it is equal to i ( + /qi i ( /qi Since the second term of (6 is analytic in a disc of radius bigger than, it follows that the n it of the coefficient of u n in the second infinite product of (6 is equal to 0 Summarizing, we have shown that i ( /qi i + O (n, q = n q n which implies the theorem i ( + /qi i ( /qi,

28 8 JASON FULMAN, ROBERT GURALNICK, AND DENNIS STANTON Remark: The expression in Theorem 57 is equal to 7583 when q =, and tends to as q 6 Acknowledgements Fulman was partially supported by Simons Foundation Grant Guralnick was partially supported by NSF grants DMS and DMS Stanton was partially supported by NSF grant DMS We are very grateful to Geoff Robinson for helpful correspondence, and thank Ryan Vinroot for some references We thank the referee for comments References [] Andrews, G, The theory of partitions, Encyclopedia of Mathematics and its Applications, Vol Addison-Wesley Publishing Co, Reading, Mass-London-Amsterdam, 976 [] Aschbacher, M and Seitz, G, Involutions in Chevalley groups over fields of even order, Nagoya Math J 63 (976, -9 [3] Dye, R H, On the conjugacy classes of involutions of the orthogonal groups over perfect fields of characteristic, Bull London Math Soc 3 (97, 6-66 [4] Dye, R H, On the conjugacy classes of involutions of the simple orthogonal groups over perfect fields of characteristic two, J Algebra 8 (97, [5] Fulman, J and Vinroot, R, Generating functions for real character degree sums of finite general linear and unitary groups, J Algebr Comb 40 (04, [6] Garibaldi, S and Guralnick, R, Spinors and essential dimension, Compos Math, to appear, arxiv: (06 [7] Ginzburg, V and Pasechnik D, Random chain complexes, arxiv: (06 [8] Liebeck, M and Seitz, G, Unipotent and nilpotent classes in simple algebraic groups and Lie algebras, Mathematical Surveys and Monographs, 80 American Math Society, Providence, RI, 0 [9] Morrison, K, Integer sequences and matrices over finite fields, J Integer Seq 9 (006, Article 06, 8 pp [0] Odlyzko, A, Asymptotic enumeration methods, Chapter in Handbook of Combinatorics, Volume MIT Press and Elsevier, 995 [] Vinroot, R, Character degree sums and real representations of finite classical groups of odd characteristic, J Algebra Appl 9 (00, Department of Mathematics, University of Southern California, Los Angeles, CA address: fulman@uscedu Department of Mathematics, University of Southern California, Los Angeles, CA address: guralnic@uscedu School of Mathematics, University of Minnesota, Minneapolis, MN address: stanton@mathumnedu