DOI 10.1007/s10469-017-9448-3

Algebra and Logic, Vol. 56, No. 4, September, 2017 (Russian Original Vol. 56, No. 4, July-August, 2017)

AUTOMORPHISM GROUPS OF DIAGONAL Zp -FORMS OF THE LIE ALGEBRA sl2 (Qp ), p > 2 A. N. Grishkov1∗ and M. N. Rasskazova2∗∗

UDC 512.554.3

Keywords: Lie algebra, diagonal Zp -form, automorphism group. It follows from [2] that two nondiagonal forms like S(n, d) + Zp A and S(n, d) + Zp A are isomorphic if the elements of A and A are conjugated via the group AutZp S(n, d). In the present paper, we settle just this question on conjugation. In other words, we describe the group AutZp S(n, d) and clarify under which conditions two elements of S(n, d) are conjugate under the action of this group on S(n, d), p > 2.

1. PRELIMINARIES For a given variety Vk of algebras over a commutative associative ring k and for a fixed extension k < K on Vk , we can introduce an equivalence relation ∼K such that for given algebras A, B ∈ Vk , the relation A ∼K B holds iff C  A ⊗k K  B ⊗k K. In this case we say that k-algebras A and B are k-forms of a K-algebra C. For a variety of Lie algebras, k-forms of K-simple Lie algebras under the condition that k < K is a finite extension of fields of characteristic 0 have been studied intensively [1]. For the case where k is not a field, we are unaware of any essential results. Difficulties that arise here are well characterized by the following: Problem. Classify k-forms of the Lie K-algebra sl2 (K) up to isomorphism as k-algebras for the following cases: ∗

Supported by RFBR, project No. 16-01-00577-a, by FAPESP and CNPq (Sections 1 and 2), and by Russian Science Foundation, project No. 16-11-10002 (Section 3). ∗∗ Supported by CNPq, project No. 456698/2014-0.

Instituto de Matem´atica e Estat´isitica, Universidade de S˜ao Paulo, Postal 66281, S˜ ao Paulo-SEP, Brasil, 05311-970. Dostoevskii Omsk State University, pr. Mira 55-A, Omsk, 644077 Russia; [email protected]. 2 Omsk State Technical University, pr. Mira 11, Omsk, 644050 Russia. Siberian State Automobile and Highway University, pr. Mira 5, Omsk, 644080 Russia; [email protected]. Translated from Algebra i Logika, Vol. 56, No. 4, pp. 406-420, July-August, 2017. Original article submitted March 4, 2016; revised November 14, 2016.

1

c 2017 Springer Science+Business Media New York 0002-5232/17/5604-0269 

269

k = Z < K = Q; k = Q(p) < K = Q; k = Zp < K = Q p . Here Zp and Qp stand for the ring of p-adic integers and the field of p-adic numbers, respectively, and Q(p) is the ring Q ∩ Zp . It seems that the first form is most complicated, whereas the last one is simplest. Among all Zp -forms of the Lie algebra sl2 (Qp ) we distinguish diagonal ones where, by definition, a Zp -form L is diagonal if L = S(n, d) = Zp e⊕Zh⊕Zf , [e, h] = pn e, [f, h] = −pn f , and [e, f ] = pd h. The first step toward classifying Zp -forms was taken in [2], in which the following theorem was proved. THEOREM 1.1 [2]. Let L be a nondiagonal Zp -form of sl2 (Qp ). Then there exists an ideal S(n, d) of the form L such that L = S(n, d) + Zp A and A = αe+βh+γf , where α, β, γ ∈ Zp , n ≥ s, ps and d ≥ s > 0. The theorem shows the importance of studying diagonal forms in classifying general Zp -forms. In particular, Theorem 1.1 implies that two nondiagonal forms like S(n, d)+Zp A and S(n, d)+Zp A are isomorphic if the elements of A and A are conjugated via the group AutZp S(n, d). In the present paper, we settle just this question on conjugation. In other words, we describe the group AutZp S(n, d) and clarify under which conditions two elements of S(n, d) are conjugate under the action of this group on S(n, d), p > 2. The main result of the paper is THEOREM 1.2. Let v = ae + bh + cf be an arbitrary element of a diagonal Zp -form of the Lie algebra sl2 (Qp ). Suppose also that at least one of the elements a, b, or c of the ring Zp is not divisible by p > 2. Then, for a suitable α ∈ Z∗p , an element αv is conjugate with respect to AutZp S(n, d) to one of the following elements: e, h, pq e + h, pr h + f , pq e − 12 f , e − p20 f , where p0 ∈ {1, . . . , p − 1} \ {1, 22 , . . . , (p − 1)2 }, E1 (q) = v φ = pq e + h + pn−d−q f /2,

(1)

E2 (r) = e + pr h + p2r+n−df /2,

(2)

β 2 − 1 n−d−q p f, 2 2q ≥ n − d > q, β 2 ≡ 1 (mod p),

H0 (q, β) = pq e + βh +

H1 (q, l, β) = pq e + (1 + βpd+q+l−n )h + (β + β 2 pd+q+l−n /2)pl f, β ∈ Z \ pZ, 0 < β < pn−d−q−l ,

(4)

β 2 − p2m−2n l p f, 2 β ∈ Z \ pZ, 0 < β < pl , m < n,

(5)

β 2 pd+q+n−2m ± 2β f, 2 β ∈ Z \ pZ, 0 < β < pl , m < n,

(6)

H2 (m, l, β) = pn−d−l e + βh +

H3 (m, q, β) = pq e + (βpd+q+n−2m ± 1)pr h +

270

(3)

β2 − 1 f, 2 β ∈ Z \ pZ, 0 < β < pl , m < n,

H4 (m, β) = p2m−n−d e + βpm−n h +

(7)

p2r+2n−2m β 2 − 1 f, 2 β ∈ Z \ pZ, 0 < β < pl , m < n,

(8)

β 2 − p2m−2n−2r − 1 f, 2 β ∈ Z \ pZ, 0 < β < pd−n−r ,

(9)

β 2 p2n−2m f, 2 β ∈ Z \ pZ, 0 < β < pl , m < n,

(10)

H5 (m, r, β) = p2m−n−d e + βpr h +

H6 (m, r, β) = p2r+n−de + pr βh +

H7 (m, l, β) = p2m−n−d−l e + βh +

β 2 pn−d−q−l − ξ l p f, 2 β ∈ Z \ pZ, 0 < β < pl , m < n, ξ ∈ {1, p0 }.

Tξ (q, l, β) = pq e + βh +

(11)

In this event all the above-listed elements are mutually nonconjugate. In fact, we prove a more precise result. First note that for any v = ae + bh + cf of S(n, d) there exists a number α ∈ Z∗p such that d(αv) = α2 (b2 p2n − 2acpn+d ) ∈ {pm , pm p0 }. Therefore, we may consider only those elements v in S(n, d) for which the norm d(v) lies in {pm , pm p0 }. In Tables 1-4 (see below), the first column enumerates all elements v = ae + bh + cf of S(n, d) with norm d(v) = b2 p2n − 2acpn+d ∈ {pm , pm p0 }, and the middle column specifies their corresponding conjugates from Theorem 1.2. The last column displays conditions under which this is possible. We hope that Theorem 1.2 will allow us to classify arbitrary Zp -forms for p > 2. It is also worth mentioning [3], in which representations of diagonal forms were studied. 2. AUTOMORPHISMS OF DIAGONAL FORMS FOR p > 2 As above, we denote by S = S(n, d) a diagonal Zp -form of the Lie algebra sl2 (Qp ) with diagonal basis e, h, f . The form S will be realized over a field K = Qp (p1/2 ) (an extension of the field Qp of degree 2). Let       0 0 −pn /2 0 0 −ps /2 , f= ; , h= e= ps 0 0 0 0 pn /2 here s = n+d 2 . This realization makes it natural to define the norm d(v) of an arbitrary element v = ae + bh + cf ∈ S(n, d) in the form   −bpn /2 −apn /2 = b2 p2n − 2acpn+d ; d(v) = −4det d n bp /2 cp

271

i.e., d(v) = b2 p2n − 2acpn+d . It is easy to see that S has second-order automorphisms φ1 and φ2 such that φ1 φ2 = φ2 φ1 , −1 φ φ1 φ2 φ2 φ2 e 1 = f , f φ1= e,  h = −h, e = p0 e, f = p0 f , and h = h. Here, as in Theorem 1.2, p0 ∈ Z, 0 < p0 < p, pp0 = −1; i.e., p0 is not a square modulo p. Denote by G˜0 the subgroup of matrices in sl2 (Zp ),  G˜0 =

x p|n−s|y p|s−n|z t

    2|n−s| yz = 1, x, y, z, t ∈ Zp .  xt − p  

Next we define a subgroup G0  G˜0 /{λE | λ ∈

Z∗p },

E=

1 0 0 1

 .

THEOREM 2.1. Let S(n, d) be a diagonal Zp -form of the Lie algebra sl2 (Qp ) and let G = AutZp S(n, d) be the automorphism group of S(n, d). Then, for n = d, the group G = G0 ∪ G0 φ1 ∪ G0 φ2 ∪ G0 φ1 φ2 has a normal subgroup G0 of index four. If n = d, then the normal subgroup G0 has index two in G, and G = G0 ∪ φ2 G0 . Proof. First note that the matrix   x p|n−s|y p|s−n|z t induces (by conjugation) on S(n, d) an automorphism g = g(x, y, z, t) such that eg = t2 e + tzpk+s−nh + p2k z 2 f /2, hg = 2pn−s+k yte + (xt + p2k yz)h + pn−s+k xzf,

(12)

f g = 2p2k y 2 e + 2xypk+s−n h + x2 f, where k = |s − n|, s = (n + d)/2, xt − p2|n−s| yz = 1, and x, y, z, t ∈ Zp . Then v g = (αe + βh + γf )g = ke e + kh h + kf f , where ke = αt2 + 2βpn−s+k yt + 2γp2k y 2 , kh = αtzpk+s−n + β(xt + p2k yz) + 2γxypk+s−n ,

(13)

kf = αp2k z 2 /2 + βpn−s+k xz + γx2 . For an element v = ae + bh + cf ∈ S(n, d), we define N OD(v) = N OD(a, b, c) as the greatest power pm such that pam , pbm , pcm ∈ Zp . It is easy to see that N OD(v) = N OD(v φ ), if φ ∈ AutZp S(n, d), and N OD(v) = 1 if the element v can be included in the Zp -basis for S(n, d). LEMMA 2.1. For an element v = ae + bh + cf ∈ S(n, d), there exists a Zp -automorphism φ of the Lie ring S(n, d) such that v φ = e if and only if N OD(v) ∈ Z∗p and one of the following conditions is satisfied: (i) a = b = 0; (ii) c = b = 0; 272

(iii) d(v) = 0, n ≥ d, and a or c is invertible in Zp ; (iv) d(v) = 0, n < d, 0 = b ∈ p(d−n) Zp . Proof. Let φ ∈ G and v = eφ = ae + bh + cf . Then d(e) = d(v) = p2n b2 − 2acpn+d = 0.

(14)

Suppose ac = 0. Applying φ1 if necessary, we may assume that c = 0. In this case (14) implies b = 0. Consider the case ac = 0. Suppose that a = a0 pq and c = c0 pl , q, l > 0, where a0 , c0 ∈ Z∗p . By virtue of N OD(v) = 1, we have b ∈ Z∗p . It follows from (14) that b2 pn = 2a0 c0 pq+l+d . Applying φ1 and φ2 if necessary, we may assume that a0 = 1. Therefore, n = d + r + s and b2 = 2c0 ∈ Z∗p . Let w = hφ = a1 e + b1 h + c1 f ; then [v, w] = pn v and [v, w + αv] = pn v. If α = −b1 /b then H = w − (b1 /b)v = xe + yf . The equality [ae + bh + cf, xe + yf ] = pn (ae + bh + cf ) implies that −pn bxe − pd cxh + pd ayh + pn byf = pn ae + pn bh + pn cf, whence a = pq = −bx, c = c0 pl = by, and b = pd−n (ay − cx). Since q, l > 0, we have x ≡ y ≡ 0 (mod p), which contradicts the fact that {v, w, f φ } and {v, H, f φ } are Zp -bases of S(n, d), and the N OD(u) of any element u in the basis is equal to 1. Suppose that q = 0, i.e., a = a0 = 1. Then (14) implies that c = b2 pn−d /2. Consequently, eg = eφ if g = g(1, 0, b, 1) for n ≥ d. Let n < d and b = b0 pr , with l ≥ d − n. Then eg = eφ if g = g(1, 0, b0 pr−d+n , 1). Now consider the case where n < d, abc = 0, d(v) = 0, a = 1, b = b0 pl , c = c0 pr , and l < d − n. It follows from (14) that 2r = d − n + l. Let v = eφ , φ ∈ G. Since [e, S(n, d)] ⊆ Zp h ⊕ Zp e, we have [eφ , S(n, d)] ⊆ Zp hφ ⊕ Zp eφ , and hence [v, f ] = pd h + pn bf = αhφ + βeφ . Then v = eφ , w = pd−n−l h + b0 f ∈ Zp hφ ⊕ Zp eφ , [v, w] = αp−l v, and [f, w] = −pd−l f . The trace of any operator of multiplication by w is equal to 0, so α = pd . This yields [v, f ] − βeφ = pd hφ = −βe + (pd − βb0 pl )h + (b0 pn+l − βc)f. d

n

bf Consequently, β = pd β0 . Then p h+p = h + b0 pn+l−d f ∈ S(n, d) and n + l ≥ d, which is a pd contradiction with l < d − n. The lemma is proved. Incidentally, we have proved that if v = eφ , where φ ∈ AutZp S(n, d), then there exists ψ ∈ G −1 for which eφψ = e. We come back to the proof of Theorem 2.1. Let φ ∈ AutZp S(n, d) and v = eφ . Taking into account the remark made above, we may assume that eφ = e. Since {t ∈ S(n, d) | [e, t] ∈ Zp e} = Zp h ⊕ Zp e, it follows that hφ = αh + βpm e, α, β ∈ Zp . The equality [eφ , hφ ] = [e, hφ ] = pn e implies that α = 1. Let f ψ = xe + yh + zf . In view of [f ψ , hψ ] = −pn f ψ , we have

[xe + yh + zf, h + βpm e] = xpn e − zpn f − βypm+n e − βpd+m zh = −pn (xe + yh + zf ). 273

Consequently, x = αpm y/2 and y = αzpd+m−n . By virtue of [eψ , f ψ ] = pd hψ , we obtain z = 1. Then ψ = g(1, αpd+m−n , 0, 1). The theorem is proved. 3. CLASSIFICATION OF ELEMENTS OF THE DIAGONAL FORM S(n, d) Now we are in a position to solve the main problem, that of classifying elements of the diagonal form S(n, d). First note that the determinant d(v) introduced above is a G-invariant, with G = AutZp S(n, d). We have d(αv) = α2 d(v) and the group Z∗p /(Zp ∗)2 is generated by two involutions p and p0 . Therefore, to classify elements of the diagonal form S(n, d), it suffices to consider the following cases: d(v) = 0; d(v) = p2m ; d(v) = p0 p2m ; d(v) = ξp2m+1 . Hereinafter, ξ ∈ {1, p0 }. We fix an element v = ae + bh + cf ∈ S(n, d) such that N OD(v) = 1. Let a = a0 pq , b = b0 pr , and c = c0 pl ; then qrl = 0. Applying, if necessary, an automorphism φ1 , e → a−1 0 e, h → h, f → a0 f , we may assume that a0 = 1 and q ≥ l, lr = 0. For elements v, w ∈ S(n, d), we write v ∼ w if there exist g ∈ G and α ∈ Z∗p such that v g = αv. Case d(v) = 0. In view of Lemma 2.1, the element v is conjugate to e iff one of conditions (i)-(iv) of that lemma is satisfied. Suppose that none of the conditions are met. Replacing v with v b0 , we may assume that b0 = 1. Let n ≥ d. Then 0 < l ≤ q and b ∈ Z∗p , and d(v) = 0 implies that n = d + q + l and c0 = 1/2. If g = g(x, y, z, t) ∈ G, then in view of (13) it is true that v g = ke e + kh h + kf f , where ke = pq t2 + 2bpn−d yt + 2c0 pn−d+l y 2 = pq (t2 + 2bpn−d−q + 2c0 pn−d+l−q ) = pq (t2 + 2bpl yt + b2 p2l y 2 ) = pq (t + bypl )2 . Put t = 1 − bypl . Then ke = pq and kh = pq tz + (xt + pn−d yz) + xypl = 1 + b1 pl , where b1 = pq−l tz + 2pq yz + xy by virtue of xt − pn−d yz = 1 and n = d + q + l. It is easy to see that there exist y and z such that b1 = 0, t = 1 − bypl , and xt − pn−d yz = 1. Consequently, there exists an automorphism φ ∈ G for which v φ = pq e + h + pn−d−q f /2 = E2 (q). Now let n < d. Then k = (d − n)/2 and b2 = 2acpd−n . Consequently, we can put a0 = 1, b = b0 pr , c = c0 , 2c = b20 , and 2r = d − n + q. If r ≥ d − n, then v g = e for some g ∈ G in view of Lemma 2.1. Let r < d − n; then 0 < 2r = d − n + l < 2d − 2n or d > n + l. Let g = g(x, y, z, t) ∈ G. Consider v g . By virtue of (13), we obtain ke = t2 + 2pr yt + pd−n+l y 2 , kh = tzpd−n + pr (xt + pd−n yz) + xypd−n+l 274

TABLE 1. d(v) = 0 N

v

1

e + bh + b2 pn−d f /2

e

n≥d

+

e

d>n

b2 pn−d−q f /2

E1 (q)

n > d> d−q

b2 p2r+n−d f /2

E2 (r)

d>n

e+

3

pq e

+ bh +

e+

bpr

4

Conditions

pd−n bh

2

+

b2 pd−n f /2

vg

= pr (tzpd−n−r + (1 + 2pd−n yz) + xypd−n+l−r ) = pr b0 , where b0 ≡ b0 (mod pd−n−r ). It is easy to prove that for any b0 ∈ Zp , there exist x, y, z, and t for which ke = 1, kh = pr , and xt − pd−n yz = 1. Consequently, the element v is conjugate to E2 (r). Summing up the consideration of the case d(v) = 0, we arrive at the following: PROPOSITION 3.1. If v = ae + bh + cf ∈ S(n, d), a = pq , N OD(v) = 1, and d(v) = 0, then v is conjugate to one of the elements e, E1 (q), or E2 (r) [see (1), (2)]. Case d(v) = p2m ≤ p2n . As above, we assume that v = ae + bh + cf , N OD(v) = 1, c = 0 or ac = 0 and a = pq , b = b0 pr , c = c0 pl , rl = 0, q ≥ l. Suppose first that m = n. The equality b2 p2n − 2acpn+d = p2n implies b20 p2r − 1 = 2c0 pd−n+q+l .

(15)

Version 1. Let c = 0. Then r = 0, and we may assume that b = 1, i.e., v = h + pq e. Consider v g(x,y,0,t) = ke e + kh h + kf f ; then kf = pq+2k z 2 /2 + pn−s+k xz = 0, kh = pq tzpk+s−n + (xt + p2k yz) = 1, ke = pq t2 + 2pn−s+k yt = t(pq t + 2pn−s+k y). If n < d then n − s + k = 0. Putting y = −pq t/2, we see that v g(x,y,0,t) = h. If n ≥ d then n − s + k = n − d. For q ≥ n − d, we can put y = −pq−n+d /2, and then again v g(x,y,0,t) = h. If q < n−d, then we cannot simplify the representation of an element v = h+pq e by conjugating it by automorphisms of G. Version 2. Let b0 c0 = 0 and r > 0. It follows from (15) that n = d + q, l = 0, and c0 = c = 2 (b0 p2r − 1)/2. Consider v g(x,y,z,t) = ke e + kh h + kf f . Then ke = pn−d t2 + 2b0 p2r+n−d yt + 2cpn−d y 2 = pn−d (t2 + 2b0 p2r yt + 2cy 2 )

275





b0 p2r − 1 1 + b0 p2r t y+ t . = 2cpn−d y + 2c 2c 2r

If we put y = 1−b2c0 p t = −t we obtain ke = 0. In this case kh ∈ Z∗p and kf = p2(n−d) z 2 /2 + b0 p2r+n−d xz + cx2 ∈ Z∗p if, for example, z = 0. We have thus proved the existence of g ∈ G such that v g = h + f . If n > d, then, as shown above, this element is not conjugate to h. If, however, n = d, then h + f and h are conjugate. Version 3. Let bc = 0 and r = 0. It follows from (15) that b2 − 1 = 2c0 pd−n+q+l . 2 Version 3a. Let n = d + q + l and c0 = b 2−1 ∈ Z∗p . If v g(x,y,z,t) = ke e + kh h + kf f , then



b−1 b+1 q 2 n−d n−d+l 2 n−d+l yt + 2c0 p y = 2c0 p t y+ t . ke = p t + 2bp y+ 2cpl 2cpl If l > 0, then ke = 0 in virtue of b2 − 1, t ∈ Z∗p . Consequently, ke = pq (t2 + 2bpn−d−q yt + 2c0 pn−d+l−q y 2 ) ∈ pq Z∗p . Thus the number q is invariant with respect to G. Since kh = tzpl + b(1 + 2pn−d yz) + 2c0 xypl ≡ b (mod pl ), it follows that v is conjugate to H1 (q, β) = pq e + βh + β 2 −1 n−d−q f , 2q ≥ n − d > q, β 2 ≡ 1 (mod p), β ≡ b (mod pl ). Moreover, we can ensure that 2 p β ∈ Z, 0 ≤ β < pl . And we admit the possibility that β = 0. If l = 0, then we can take y = 1−b 2c t, and hence ke = 0. It is easy to see that v is conjugate to h + f . Consequently, there exists an automorphism g ∈ G such that v g = h for q = 0. Otherwise h and h + f are not conjugate. Version 3b. Let b = 1 + b1 pi and b1 ∈ Z∗p , i > 0. The situation where b = −1 + b1 pi is treated similarly. It follows that 2b1 pi + b21 p2i = 2c0 pd+q+l−n and i = d + q + l − n. If v g(x,y,z,t) = ke e + kh h + kf f , then ke = pq t2 + 2bpn−d ty + 2c0 pn−d+l y 2



b+1 b−1 n−d+l = 2c0 p y+ y+ 2c0 pl 2c0 pl



b1 pd+q+l−n b+1 t y + t . = 2c0 pn−d+l y + 2c0 pl 2c0 pl Therefore, we can take ke = 0 if d + q ≥ n. In this case kh ∈ Z∗p and kf = pq+2k z 2 /2 + bpn−s+k xz + c0 pl x2 . If n − d > l, then v is conjugate to h + pl f and h + pl e. If l ≥ n − d ≥ 0 or d ≥ n, then v is conjugate to h.   d+q−n b+1 t . We Finally, if n > d + q, then ke = 0 in view of t ∈ Z∗p and y = − b1 p2c0 t or y = − 2c l 0p have l > 0 since 0 < m = d + q + l − n < l. In this case v is conjugate to H2 (q, l, β) = pq e + (1 + βpd+q+l−n )h + (β + β 2 pd+q+l−n /2)pl f, β ∈ Z \ pZ, 0 < β < pn−d−q−l . p2n+2r−2m b2 −1

0 . It follows that Now suppose that m < n; then 2m = n + d + q + l < 2n and c0 = 2 l q ke = 0 and v is conjugate to H7 (m, l, β), if r = 0, β ∈ {1, . . . , p } \ {pZ}, and to p e − 12 f if r > 0.

276

TABLE 2. d(v) = p2m ≤ p2n N

v

vg

Conditions

1

h + pq a0 e h + pq a0 f b2 p2r −1 pn−d e + b0 pr h + 0 2 f b2 p2r −1 e + b0 p r h + 0 2 f 2 pq e + bh + b 2−1 pn−d−q f

h

q ≥n+k−s

2 3 4 5

6 7 8 9 10 11 12

b2 −1 2 f b2 −1 e + bh + 2 f pq e + (1 + pd+q+l−n b0 )h d+q+l−n ) + b0 (2+b0 p2 pl f pq e + (1 + pd+q+l−n b0 )h d+q+l−n ) + b0 (2+b0 p2 pl f pq e + (1 + pd+q+l−n b0 )h d+q+l−n ) + b0 (2+b0 p2 pl f p2m−n−d e + b0 pr h (b2 p2n+2r−2m −1) + 0 f 2 p2m−n−d−l e + bh

pn−d e + bh +

h+

0 ≤q
pq e

h+e

0 < n − d, 0 < r

h

n = d, r > 0

H0 (q, β)

0
h+e

n − d > 0, b2 ≡ 0, 1 (mod p)

h

n = d, b2 ≡ 1 (mod p)

h

d ≥ n, q ≥l ≥n−d

h+

d+q ≥n>d+l

pl e

H(q, l, β)

0
pq e − 12 f

2n > 2m ≥ n + d

H7 (m, l, β)

2n > 2m ≥ n + d + l

2 2m−2n + b p 2 −1) pl f

Case d(v) = p2m > p2n . As above, we assume that v = pq e + b0 pr h + c0 pl f , rl = 0, b0 , c0 ∈ Z∗p , and lastly p2n+r b20 − 2c0 pn+d+q+l = p2m , p2r b20 − 2c0 pd+q+l−n = p2m−2n .

(16)

. Since n ≥ d, we Version 1. Let r = 0. Then b = b0 , n = d + q + l ≥ d + 2l, and c0 = b −p 2 have v g(x,y,z,t) = ke e + kh h + kf f and



b+1 1−b q 2 n−d n−d+l 2 n−d+l yt + 2c0 p y = 2c0 p ke = p t + 2bp y+ y+ . 2c0 pl 2c0 pl 2

2m−2n

If l > 0 and b2 ≡ 1 (mod pl ), then ke = 0 and v is conjugate to H2 (m, l, β) = pn−d−l e + βh +

β 2 − p2m−2n l p f. 2

If, however, b = ±1 + b1 pl , then we may assume that ke = 0 and v is conjugate to pm−n h + f . Version 2. Let r > 0. Then l = 0 and c = c0 . By virtue of (16), we have the following options: 2r = 2m−2n < d+q−n; 2r = d+q−n = 2m−2n; 2r > d+q−n = 2m−2n; 2r = d+q−n < 2m−2n. 277

TABLE 3. d(v) = p2m > p2n N 1 2

v pn−d−l e +

2

pl f

pq e + (b1 pd+q+n−2m ± 1)pm−n h +

3

bh +

b2 −p2m−2n

b21 pd+q+n−2m ±2b1 2

vg

Conditions

H2 (l, m, β)

n − d ≥ 2l > 0, b ≡ β ≡ 0 (mod p)

pm−n h + f

b1 ≡ 0 (mod p) 2n + q ≥ d + q + n > 2m or d + q + n > 2m, d > n ≥ m − q

f

pq e + (b1 pd+q+n−2m ± 1)pm−n h

H3 (q, m, β)

b2 pd+q+n−2m ±2b1 + 1 f 2

b1 ≡ 0 (mod p) d + q + n > 2m > 2q + 2n, b1 ≡ β (mod pm−n−q )

4

p2m−d−n e + b0 pm−n h +

b20 −1 2 f

pm−n h + f

2m ≥ n + d, n ≥ d, or m ≥ d, b20 ≡ 0, 1 (mod p)

5

p2m−d−n e + b0 pm−n h +

b20 −1 2 f

H4 (q, m, β)

d > m, b20 ≡  0, 1 (mod p), β ≡ b0 (mod pd−m )

pm−n h + f

2r + 2n > 2m, m ≥ d

H5 (m, r, β)

2r + 2n > 2m, m < d, β ≡ b0 (mod pr+n−d)

pm−n h + f

m > n + r, m ≥ d

H6 (m, r, β)

d > m > n + r, b0 ≡ β (mod pd−n−r )

6

p2m−n−d e + b0 pr h +

7

p2m−n−d e + b0 pr h +

8

p2r+n−d e + pr b0 h +

9

p2r+n−d e + pr b0 h +

b20 p2r+2n−2m −1 f 2 2 2r+2n−2m b0 p −1 f 2 b20 −p2m−2n−2r f 2 2 2m−2n−2r b0 −p f 2

Version 2a. Let r = m − n, b0 = ±1 + b1 pd+q+n−2m , and 2c0 = b21 pd+q+n−2m ± 2b1 . If n ≥ d, g(x,y,z,t) m−n h + f , since v g(x,y,z,t) = then there exist x, y, z, t ∈ Zp , xt − pn−dyz = 1, such   that vr m−n  = p r +pm−n b p b p −p ke e + kh h + kf f and ke = 2cpn−d y + 0 2c y + 0 2c . Below we use the designations ke , kh , and kf in just this sense.       If n < d, then ke = 2cpd−n y + pm−d b02c+1 y + pm−d b02c−1 .   Let n + q ≥ m. Then v is conjugate to pm−n h + f since we may put y = −pm−d b02c−1 =   +1 b1 if b0 = 1+b1 pd+q+n−2m or y = −pm−d b02c = −pn+q−m b2c1 if b0 = −1+b1 pd+q+n−2m . −pn+q−m 2c If m > q + n, then kl = 0 and kh = pq+d−n tze + pm−n b0 (xt + pd−n yz) + 2cxypd−n . If d > m > n + q, then kh = pm−n β0 and β0 ≡ b0 (mod pd−m ). Since b0 = 1 + b1 pd+q+n−2m , it follows that β0 = 1 + β1 pd+q+n−2m and β1 ≡ b1 (mod pm−n−q ). The element v is conjugate 2 d+q+n−2m ±2β f , β ≡ b1 (mod pm−n−q ), β ∈ to H3 (m, q, β) = pq e + βpd+q+n−2m ± 1pr h + β p 2 {1, . . . , pm−n−q } \ pZp . Let m ≥ d. For x = t = 1, there is y such that kh = 0 and v is conjugate to p2m−d−n e − f /2.

278

TABLE 4. d(v) ∈ {p0 , p2m+1 , p0 pm } N

v

1

pq e + b0 pr h +

2

pq e + bh +

b20 pn+2r−d−q −ξ f 2

b2 pn−d−q−l −ξ l pf 2

vg pq e −

Conditions ξ 2f

Tξ (q, l, β)

n − d + 2r > q, ξ = p0 or n + d + q ≡ 1 (mod 2) n − d ≥ q + l, q ≥ l > 0

2 Version  2b. Let 2r = d + q − n = 2m − 2n > 0, l = 0, c0 = c, b0 − 2c0 = 1, and ke = −1 m−d 2cpn−d y + b02c+1 pm−d t y + b02c p t . If n ≥ d, then we may assume that ke = 0 in virtue of m−n m > n ≥ d. Hence v is conjugate to p h + f. Let n < d. The element v is conjugate to pm−n h + f , for m ≥ d, and to H4 (m, β) = p2m−n−d e + 2 βpm−n h + β 2−1 f for m < d. Version 2c. Let 2r > d + q − n = 2m − p2r+2n−2m b20 − 2c = 1. If n ≥ d, then  2n. Then m−d m−d (b pr+n−m −1) (b0 pr+n−m +1) 0 ke = pq t2 + 2b0 pr+n−d yt + 2cpn−dy 2 = 2cpn−d y + p t y+p t . In 2cpn−d 2cpn−d m−n view of m > n, we may assume that ke = 0 and the element v is conjugate to p h + f .   pm−d (b0 pr+n−m +1) q 2 r d−n 2 d−n y = 2cp t y+ Let d > n; then ke = p t + 2b0 p yt + 2cp y+ 2cpn−d  m−d r+n−m p (b0 p −1) t . 2cpn−d If m ≥ d, then v is conjugate to pm−n h + f . In the case d > m, v is conjugate to H5 (m, r, β) = 2r+2n−2m β 2 −1 f . Note that in this case r + n − d > 2m − n − d ≥ 0 and β ≡ b0 p2m−n−d e + βpr h + p 2 (mod pr+n−d), β ∈ Z \ pZ, 0 < β < pr+n−d, since kh = p2m−2n tz + b0 pr (1 + 2pd−n yz) + 2cxypd−n ≡ pr b0 (mod pr+n−d ). Version 2d. Let 2r = d + q − n < 2m − 2n. The element v is conjugate to pm−n h + f , for m ≥ d, 2 2m−2n−2r and to H6 (m, r, β) = p2r+n−d e + pr βh + β −p 2 f , β ∈ Z \ pZ, 0 < β < pd−n−r , for d > m. Case d(v) ∈ {p0 , p2m+1 , p0 pm }. This case is simpler since ke = 0 for any x, y, z, t ∈ Zp , xy − p2k yz = 1. Hence the canonical form of v = ae + bh + cf shows up as pq e + βpr h + γpl f , where p2n+2r β 2 − 2γpn+d+l+q = d(v). 2r If d(v) = p0 , then n = d = l = q = 0 and γ = βp 2−p0 . Consequently, there exist x, y, z, and t such that xt − yz = 1 and kh = tz + b0 pr (1 + 2yz) + 2cxy = 0—for example, for x = t = 1, z = 0, r and y = − b02cp . Thus the canonical form appears as e − p20 f . Let d(v) = ξpm , m ≥ 0, ξ ∈ {1, p0 }, with ξ = p0 if m is an even number. There are two options: m = 2n + 2r < n + d + q + l, ξ = p0 , and m = n + d + q + l ≤ 2n + 2r. The first option is impossible since p0 ≡ b20 (mod p). b2 p2n+2r−m −ξ and For the second option, c0 = 0 2

kh = tzpq+k+s−n + b0 pr (1 + 2p2k yz) + 2c0 xypl+k+s−n. If l = 0, then it is not hard to ensure that kh = 0. If l > 0, then r = 0, b = b0 , and kh ≡ b (mod pl ) in view of n − d > q + l and k + s − n = 0. 279

Thus v is conjugate to T (q, l, β) = pq e + βh +

β 2 pn−d−q−l −ξ l p f. 2

REFERENCES 1. N. Jacobson, Lie Algebras, Interscience, New York (1962). 2. A. V. Yushchenko, “The forms and representations of the Lie algebra sl2 (Z),” Sib. Math. J., 43, No. 5, 967-976 (2002). 3. A. N. Grishkov, “Representations of the Lie ring sl2 (Z) over the ring of integers,” Sib. Math. J., 41, No. 6, 1105-1110 (2000).

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