Siberian Mathematical Journal, Vol. 43, No. 5, pp. 967–976, 2002 c 2002 Yushchenko A. V. Original Russian Text Copyright

THE FORMS AND REPRESENTATIONS OF THE LIE ALGEBRA sl2 (Z) A. V. Yushchenko

UDC 519.48

Abstract: We study the structure of integral p-adic forms of the splitting three-dimensional simple Lie algebra over the field of p-adic numbers. We discuss the questions of diagonalizability of such forms and description for maximal diagonal ideals. We consider torsion-free finite-dimensional modules over the splitting three-dimensional simple Lie algebra with integral and p-adic integral coefficients. We describe diagonal modules, demonstrate finiteness of the number of modules in each dimension, and prove a local-global principle for irreducible modules. Keywords: Lie algebra, form of an algebra, irreducible module, diagonal algebra

This article is devoted to the Zp -forms of the splitting three-dimensional simple Lie algebra sl2 (Qp ) and the sl2 (Z)-modules. In Section 1 we find a basis for an arbitrary Zp -form and study the structure of the form. In Section 2 we discuss the questions of diagonalizability of Zp -forms and description for maximal diagonal ideals. In Section 3 we consider torsion-free finite-dimensional sl2 (Z)-modules, describe diagonal modules, demonstrate that there are only finitely many torsion-free sl2 (Z)-modules in each dimension, and prove a local-global principle for irreducible modules. Henceforth we use the following
standard notations: Qp is the field of p-adic numbers, Zp is the ring of p-adic integral numbers, and ap is the Legendre symbol. 1. Structure of the Zp -forms of sl2 (Qp ). In this and subsequent sections we suppose that p is a prime number other than 2. Definition. A Lie algebra L over Zp is said to be a Zp -form of sl2 (Qp ) if L ⊗Zp Qp ∼ = sl2 (Qp ) and L is a finite-dimensional Zp -module. The condition L ⊗Zp Qp ∼ = sl2 (Qp ) implies that L is a free Zp -module. Definition. A Zp -form L of the Lie algebra sl2 (Qp ) is called diagonal if there exist a semisimple h ∈ L and eigenvectors e and f of h such that L = he, h, f i. The set {e, h, f } is referred to as a diagonal basis for L. If eh = pn e, ef = pd h, and f h = −pn f then we denote such an algebra by S(n, d). Theorem 1. Let L be a Zp -form of sl2 (Qp ). Then there exist a diagonal subalgebra S and an element A in L such that L = S + hAi. Proof. Let S be a maximal diagonal subalgebra in L. Since L is a finite-dimensional Zp -module, we have L = S + hA1 , . . . , Ak i and so it suffices to prove the theorem in the case when L = S + hA, Bi. Take a diagonal basis he, h, f i, eh = pn e, ef = pd h, f h = −pn f in S. If A or B lies in S then the claim of the theorem is obvious. Thus, A, B ∈ / S. In this case A=

αe + βh + γf , ps

B=

ae + bh + cf ; pm

moreover, at least one of the coefficients A and B is not divisible by p. Assume that s ≥ m. Observe that two of the three coefficients of the elements A and B are not divisible by p. Indeed, suppose that β and γ are divisible by p. Considering the element ps−1 A = αe+βh+γf , we then infer that p e p ∈ L; a contradiction with the maximality of S. In particular, we may always assume that α 6≡ 0(p). Omsk. Translated from Sibirski˘ı Matematicheski˘ı Zhurnal, Vol. 43, No. 5, pp. 1197–1207, September–October, 2002. Original article submitted December 1, 1999.

c 2002 Plenum Publishing Corporation 0037-4466/02/4305–0967 $27.00

967

Indeed, if α ≡ 0(p) and γ 6≡ 0(p) then we choose another basis in S: he0 , h0 , f 0 i, e0 = f , h0 = −h, f 0 = e. 0 0 +αf 0 . Then A = γe −βh ps Look at the element G = αB − aps−m A =

(αb − a β)h + (αc − a γ)f . pm

Suppose that the coefficients of G are divisible by pm . Then we have the system of comparisons a β ≡ αb(pm ),

γb ≡ cβ(pm ).

Since α 6≡ 0(p), it follows that b = αβ a(pm ) and c = αγ a(pm ); whence B=

ae + bh + cf a αe + βh + γf a ≡ ≡ ps−m A m m p α p α

(mod S).

Therefore, L = S + hAi. Now, suppose that at least one of the coefficients of G is not divisible by pm . Then they have the same order modulo p. Indeed, multiplying G by a suitable power of p, we conclude that hp or fp lies in L; a contradiction with the maximality of S. Put νp (αb − a β) = νp (αc − αc) = t. Here and below, νp (x) is the order of x modulo p. αc−a γ pt h+τ f Look at the element H = αb−a β G. We have H = pm−t , where τ = αb−aβ , τ 6≡ 0(p). This element is semisimple and its eigenvectors are F = f and  n−d for n > d, p e + τ h + 12 τ 2 f E= 1 2 d−n d−n e+p τh + p for n ≤ d. 2τ f Put J = hE, H, F i. If n ≤ d then obviously S ⊂ J; a contradiction with the maximality of S. Suppose that n > d. Express the elements e, h, f , A, and B through E, H, and F . We have 1 1 (E − τ pm−t H + τ 2 F ), h = pm−t H − τ F, f = F, 2 pn−d     1 2 1 n−d+s s n−d β − p ατ )H + p (γ − βτ ) + ατ F , A = n+d−s αE + (p 2 p     1 1 B = n+d−m a E + (pn−d+s b − ps aτ )H + pn−d (c − bτ ) + aτ 2 F . 2 p e=

Whence a ps−m A − αB = ps−m (βa − αb)H ≡ 0

(mod J),

ps A − αe ≡ 0

(mod J).

Since α 6≡ 0(p), it follows that B≡

a s−m p A α

(mod J),

e≡

1 s−m p A α

(mod J),

h, f ∈ S.

Hence, L = J + hAi, which completes the proof of the theorem. Theorem 2. Let L be a Zp -form of sl2 (Qp ), L = S + hAi, where S = S(n, d) is a maximal diagonal subalgebra and A = αe+βh+γf . Then n, d ≥ s and S is an ideal in L. ps Proof. As in Theorem 1, it can be shown that we may assume that α 6≡ 0(p). Consider two cases: n < d and n ≥ d. 968

Case 1: n < d. Suppose that n < s. Let β 6≡ 0(p). Look at the element 2aeA − β(pn A + Ah) = Then

h p

2pd αγ − pn β 2 ) 2pd−n αγ − β 2 h = h ∈ L. ps ps−n

∈ L; a contradiction with the maximality of S.

Now, let β ≡ 0(p). Then ps−1 A =

αe+γf p

+ th for some t ∈ Zp and

ps−n−1 Ah + ps−1 A =

α e + ps−1 th. p

Hence, pe ∈ L; a contradiction with the maximality of S. Case 2: n ≥ d. Suppose that d < s. Look at the element B=

1 pd h + pn (β/α)f h + pn−d (β/α)f Af = = . α ps ps−d n−d 2

This element is semisimple and its eigenvectors are E = e + αβ h + p 2α2β f and F = f . Obviously, S ⊂ hE, B, F i; a contradiction with the maximality of S. Thus, n, d ≥ s. It is straightforward that Ae, Ah, Af ∈ S, and so S is an ideal in L. The proof of the theorem is over. Corollary. Let L be a Zp -form of sl2 (Qp ), L = S + hAi, where S = S(n, d) = he, h, f i is a maximal diagonal subalgebra and A = αe+βh+γf , α 6≡ 0(p). Then {A, h, f } is a basis for L. ps 2. Diagonalizability conditions for Zp -forms and maximal diagonal ideals. In accordance with Theorems 1 and 2, we can further consider only the Zp -forms of sl2 (Qp ) defined as follows: L = S + hAi, where S = S(n, d) = he, h, f i, eh = pn e, ef = pd h, f h = −pn f,

A=

αe + βh + γf , ps

(1)

n, d ≥ s.

In this case S is a diagonal ideal in L. Lemma 1. Let L be a Zp -form of sl2 (Qp ) defined by (1) and let H =

ae+bh+cf pm

∈ L, 0 < m ≤ s.

1. If n ≥ d, n−d ≡ 0(2), and pn−d β 2 −2αγ 6≡ 0(p) then H is semisimple if and only if 2. If n ≥ d, n − d ≡ 1(2), and pn−d β 2 − 2αγ 6≡ 0(p) then H is not semisimple. 3. If n < d and β 6≡ 0(p) then H is semisimple.

pn−d β 2 −2αγ p

= 1.

Proof. First, observe that a ≡ µα(p), b ≡ µβ(p), c ≡ µγ(p), and µ 6≡ 0(p). 1. Let H = ae+bh+cf ∈ L be a semisimple element, let E be an eigenvector, E = xe + yh + zf , and pm 1 n EH = pm (p (xb − ya)e + pd (xc − za)h + pn (yc − zb)f ) = kh (xe + yh + zf ). We obtain the system of linear equations  n−m b − kh )x − pn−m ay = 0,  (p pd−m cx − kh y − pd−m az = 0, (2)  n−m n−m p cy − (p b + kh )z = 0. This system has a nonzero solution; therefore, its determinant equals zero. Calculate it: (p2(n−m) b2 − kh2 ) kh − 2pn+d−2m ackh = 0. Since kh 6= 0, it follows that p2(n−m) b2 − kh2 − 2pn+d−2m ac = 0. Dividing the last equality by pn+d−2m , we obtain pn−d b2 − 2ac = k 2 , with kh = p(n+d)/2−m k; i.e., pn−d b2 − 2ac is a square
n−d 2 in Zp and the condition pn−d b2 − 2ac ≡ µ2 (pn−d β 2 − 2αγ) 6≡ 0(p) implies that p βp −2αγ = 1. 969

Conversely, put H=

ae + bh + cf , ps

p2t b2 − 2ac 6≡ 0(p),

t=

n−d . 2

From the conditions pn−d β 2 − 2αγ = 1, µ2 (pn−d β 2 − 2αγ) ≡ pn−d b2 − 2ac(p) p we find that pn−d b2 − 2ac is a square in Zp ; i.e., pn−d b2 − 2ac = k 2 . Then   t   t pt c pa pt c pa l2 l1 e+h+ t f , F =p e+h+ t f (3) E=p pt b − k p b+k pt b + k p b−k are eigenvectors of H. Here l1 , l2 ∈ N are such that E, F ∈ L, EH = kh E, F H = −kh F , and kh = n+d p 2 −s k. ∈ L be a semisimple element. By analogy to item 1, calculating the determinant 2. Let H = ae+bh+cf pm n+d−1

of system (2), we obtain (pn−d b2 − 2ac)p = k 2 , with kh = p 2 −m k. Hence, pn−d b2 − 2ac ≡ µ2 (pn−d β 2 − 2αγ)(p) ≡ 0(p), and so pn−d β 2 − 2αγ ≡ 0(p); a contradiction. 3. Assume that n < d and H = ae+bh+cf ∈ L. Since n < d and b 6≡ 0(p), it follows that b2 − 2pn−d ac pm is a nonzero square in Zp . Put b2 − 2pn−d ac = k 2 . It is easy to verify that E = a(b + k)e + (b2 − k 2 )h + c(b − k)f, F = a(b − k)e + (b2 − k 2 )h + c(b + k)f are eigenvectors of H, EH = pn−m kE, F H = −pn−m kF . The proof of the lemma is over. Let J be a subalgebra of L and let hA, B, Ci be a basis for J. Denote by ∆(A, B, C) the determinant of the transition matrix from he, h, f i to hA, B, Ci. Then it is easy to verify that [L : J] = s+νp (∆(A, B, C)). Lemma 2. Let L be a Zp -form of sl2 (Qp ) defined by (1), n ≥ d, let H = ae + bh + cf ∈ S be a semisimple element, and let E and F be eigenvectors of H. Then hE, H, F i ⊂ S. Proof. If n − d ≡ 0(2) then take t = (n − d)/2 in the case of n − d ≡ 1(2), t = (n − d + 1)/2. Then  t   t  pa pt c pa pt c l1 l2 E=p e+h+ t f , F =p e+h+ t f pt b − k p b+k pt b + k p b−k are eigenvectors of H and l1 and l2 are minimal numbers for which E, F ∈ L. Then 4pt k 3 ∆(E, H, F ) = pl1 +l2 2t 2 . p b − k2 Put pt c pt a m1 0 = p a , = pm2 c0 , a0 , c0 6≡ 0(p). pt b − k pt b + k Now,   p2t ac m1 0 m2 0 = 2t. νp (p a p c ) = m1 + m2 = νp p2t b2 − k 2 Since 2t ≥ 0, we have m1 ≥ 0 or m2 ≥ 0. Examine the case of m1 ≥ 0, m2 < 0. Let m02 = −m2 . Then 0 0 pm1 +m2 a0 e + pm2 h + c0 f E= . 0 pm2 −l1 If m02 − l1 > 0 then the conditions E ∈ L and E ∈ / S yield the system of comparisons 0

0

0

0

0

pm1 +m2 a0 ≡ λα(pm2 −l1 ), pm2 ≡ λβ(pm2 −l1 ), c0 ≡ λγ(pm2 −l1 ). The last comparison and c0 6≡ 0(p) imply λ 6≡ 0(p). Since m1 + m02 > 0 and m02 > 0, it follows that α, β ≡ 0(p). We obtain a contradiction with the condition pn−d β 2 − 2αγ 6≡ 0(p). Hence, m02 − l1 ≤ 0 and E ∈ S. Similarly, we prove that F ∈ S. The case of m1 < 0, m2 ≥ 0 is settled in the same way. The proof of the lemma is over. 970

Theorem 3. Let L be a Zp -form of sl2 (Qp ) defined by (1). 1. If n ≥ d, n ≡ d(2), and pn−d β 2 − 2αγ 6≡ 0(p) then L is diagonal if and only if n = d and β 2 −2αγ
= 1. p 2. If n < d and β 6≡ 0(p) then L is diagonal. Proof. 1. In view of Lemma 2, it suffices to consider the case in which a semisimple element like H = ae+bh+cf belongs to a diagonal basis. From the condition H ∈ L we obtain ps a ≡ µα(ps ), b ≡ µβ(ps ),
2 By Lemma 1 we have b −2ac = 1, and so p 

c ≡ µγ(ps ),

µ 6≡ 0(p).

  2  µ2 (β 2 − 2αγ) β − 2αγ = = 1. p p

Assume that αγ 6≡ 0(p). Then ac 6≡ 0(p) and  t  1 pa pt c E = l1 e+h+ t f , pt b − k p b+k p

1 F = l2 p



pt a pt c e + h + f pt b + k pt b − k



are eigenvectors of H. Let l1 > 0. Then from the conditions E ∈ L and E ∈ / S we obtain the system of comparisons pt a pt c l1 l1 ≡ λα(p ), 1 ≡ λβ(p ), ≡ λγ(pl1 ). pt b − k pt b + k From the second comparison we have λ 6≡ 0(p), β 6≡ 0, λ ≡ 1/β(p), and if t > 0 then α/β ≡ γ/β ≡ 0(p); a contradiction. Hence, t = 0 or t 6= 0 and l1 = l2 = 0. Suppose that t = 0. Then the system of comparisons looks like a α ≡ , b−k β

c c ≡ . b+k β

Multiplying these comparisons and using the condition 2ac = b2 −k 2 , we obtain 2αγ ≡ 1(p), or β 2 −2ac ≡ β2 0(p); a contradiction. Thus, if αγ 6≡ 0(p) then l1 = l2 = 0. By the remark before Lemma 2, L = hE, H, F i if and only if νp (∆(E, H, F )) = −s. Calculating t 3 ∆(E, H, F ), we obtain ∆(E, H, F ) = − ppskac . Since ac 6≡ 0(p) and k 6≡ 0(p), we have νp (∆(E, H, F )) = t − s, and so t = 0 and n = d. Now, assume αγ ≡ 0(p). Then the condition pn−d β 2 − 2αγ 6≡ 0(p) implies that n = d; i.e., t = 0. Let H = ae+bh+cf be a semisimple element, 2ac = b2 − k 2 . Since p 6= 2, we have b − k ≡ 0(p), b + k 6≡ 0(p), ps or b − k 6≡ 0(p), b + k ≡ 0(p). Consider the case of b − k ≡ 0(p), b + k 6≡ 0(p). Then   a(b − k) 1 a2 E = l1 e+ h+f , ac − k(b − k) ac − k(b − k) p   1 c(b − k) c2 F = l2 e + h+ f ac − k(b − k) ac − k(b − k) p are eigenvectors of H. Note that ac − k(b − k) 6≡ 0(p). Let l1 6= 0. Then from the condition E ∈ L we obtain the system of comparisons a2 ≡ λα(p) ≡ λa (p), ac − k(b − k)

a(b − k) ≡ λβ(p) ≡ λb(p), ac − k(b − k)

1 ≡ λγ(p). 971

The third comparison implies that c 6≡ 0(p) and hence a ≡ 0(p). The second comparison yields a ≡ b(ac−k(b−k)) (p). Therefore, b ≡ 0(p) and b2 − ac ≡ β 2 − αγ ≡ 0(p); a contradiction. Thus, l1 = 0. c(b−k) Similarly, we prove that l2 = 0. Calculating ∆(E, H, F ), we obtain ∆(E, H, F ) =

k 3 (b − k)3 . ps (ac − k(b − k))2

Since k, b − k, and ac − k(b − k) are not divisible by p, it follows that νp (∆(E, H, F )) = −s and so L is diagonal. The case of b − k 6≡ 0(p), b + k ≡ 0(p) is settled in much the same way. 2. Let H = ae+bh+cf . By Lemma 1, H is a semisimple element. Suppose that b − k 6≡ 0(p), ps b + k ≡ 0(p), and the eigenvectors of H are defined by the following formulas:  pd−n a2 pd−n a(b − k) E= e + d−n h+f , pd−n ac − k(b − k) p ac − k(b − k)   pd−n c(b − k) pd−n c2 F = e + d−n h + d−n f . p ac − k(b − k) p ac − k(b − k) 

Let us check that E, H, F is a basis for L. Indeed, ∆(E, H, F ) =

k 3 (b − k)3 . ps (pd−n ac − k(b − k)2 )

Since k 6≡ 0(p) and b − k 6≡ 0(p), we have νp (∆(E, H, F )) = −s, and so E, H, F is a basis for L. In the case of b − k 6≡ 0(p), b + k ≡ 0(p), the eigenvectors of H are  pd−n a(b + k) pd−n a2 e + h + f , pd−n ac − k(b + k) pd−n ac − k(b + k)   pd−n c(b + k) pd−n c2 F = e + d−n h + d−n f . p ac − k(b + k) p ac − k(b + k) 

E=

By analogy to the previous case we show that νp (∆(E, H, F )) = −s, completing the proof of the theorem. Suppose that n ≥ d and pn−d β 2 −2αγ 6≡ 0(p). Clarify the conditions when the element u1 e+u2 h+u3 f belongs to the ideal hE, H, F i, where H = ae+bh+cf is a semisimple element with the corresponding ps eigenvectors E and F defined by formulas (3). In the proof of Theorem 3 we have demonstrated that l1 = l2 = 0. Since xE + yH + zF = u1 e + u2 h + u3 f , it follows that a pt a pt a x + y + z = u1 , pt b − k ps pt b + k

x+

b y + z = u2 , ps

pt c c pt c x + y + z = u3 . pt b + k ps pt b − k

Solving the system, we obtain 1 ((pt b + k)cu1 + (pt b − k)au3 − 2acpt u2 ), 2k 2 pt ps y = − 2 (cu1 + au3 − p2t bu2 ), k 1 z = 2 t ((pt b − k)cu1 + (pt b + k)au3 − 2acpt u2 ). 2k p

x=

972

(4)

Proposition 1. Let L be a nondiagonal Zp -form of the Lie algebra sl2 (Qp ) defined by relations (1), n ≥ d, n − d ≡ 0(2), pn−d β 2 − 2αγ 6≡ 0(p), t = (n − d)/2, H = ae+bh+cf , and let hE, H, F i be a maximal ps t t diagonal ideal. Then hp e, h, p f i ⊂ h E, H, F i and for every l < t the elements pl e and pl f do not belong to the ideal hE, H, F i. Proof. Let t = 0. The preceding theorem and Lemma 1 imply that L lacks diagonal ideals like hE, H, F i, where H = ae+bh+cf . Now, let t 6= 0. Then ac 6≡ 0(p). From formulas (4) with e = ps xE + yH + zF we have x=

1 (pt b + k)c, 2k 2 pt νp (x) = −t,

y=−

ps c, k2

νp (y) = s,

z=

1 (pt b − k)c, 2k 2 pt

νp (z) = −t.

Hence, pt e ∈ hE, H, F i and pl e ∈ / hE, H, F i for every l < t. The fact that h, pt f ∈ hE, H, F i is verified analogously. The proof of the proposition is over. Theorem 4. Suppose that L is a nondiagonal Zp -form of sl2 (Qp ) defined by relations (1), pn−d β 2 − 2αγ 6≡ 0(p). 1. Let n > d, n − d ≡ 1(2). Then S is the only maximal diagonal ideal in L.
n−d 2 2. Let n ≥ d, p bp −2ac = −1. Then S is the only maximal diagonal ideal in L.
3. Let n > d, n−d ≡ 0(2), −2ac = 1. Then all maximal diagonal ideals in L look like S or hE, H, F i, p ae+bh+cf is semisimple and E and F are eigenvectors for ps 0 0 0 0 Moreover, if H = a e+bpsh+c f is another semisimple element then ac0 ≡ a0 c(ps+(n−d)/2 ).

where H = if

H. hE, H, F i = hE 0 , H 0 , F 0 i if and only

Proof. The cases 1 and 2 are immediate from Lemmas 1 and 2. We consider the case 3. Take a semisimple element H = ae+bh+cf and the corresponding ideal ps hE, H, F i. From Proposition 1 we obtain he, h, f i 6⊂ hE, H, F i. Hence, he, h, f i is a maximal diagonal ideal in L. 0 0 0 Now, let H 0 = a e+bpsh+c f be another semisimple element and let hE 0 , H 0 , F 0 i be the corresponding ideal. Let us clarify when E 0 , H 0 , F 0 ∈ hE, H, F i. Assume that E 0 = xE + yH + zF . From (4) we obtain   1 pt a0 pt c0 t t t x = 2 t (p b + k)c t 0 + (p b − k)a t 0 − 2acp ∈ Zp , 2k p p b − k0 p b + k0 and similarly find out that z, y ∈ Zp ; i.e., E 0 ∈ hE, H, F i. We carry out the same arguments for 0 0 0 H 0 = a e+bpsh+c f . As a preliminary, note that a0 ≡ µa(ps ), b0 ≡ µb(ps ), c0 ≡ µc(ps ), and µ 6≡ 0(p). Let H 0 = xE + yH + zF . From (4) we obtain   k 1 0 0 x = 2 µ(bca2 + bac2 − 2acb2 ) − s+t (ca − c a) , 2k p and similarly calculate z. Thus, x, z ∈ Zp if and only if ca0 − c0 a ≡ 0(ps+t ). The coefficient  0  0 ps a c0 2t b y = 2 c s +a s −p b s 2k p p p always belongs to Zp . Hence, H 0 ∈ hE, H, F i if and only if ca0 − c0 a ≡ 0(ps+t ); by symmetry we conclude H ∈ hE 0 , H 0 , F 0 i if and only if ca0 − c0 a ≡ 0(ps+t ). This completes the proof of the theorem. 973

3. Representations of the Lie ring sl2 (Z). Put R = sl2 Z and let p be an arbitrary prime number. Denote by O the category of torsion-free sl2 Z-modules of finite type. Assume that M ∈ O. Then MQ = M ⊗Z Q is a finite-dimensional sl2 Q-module; moreover, dim M = dimQ (MQ ). Also, define the sl2 (Zp )-module Mp = M ⊗Z Zp . Fix a standard basis {e, h, f } for R, where eh = 2e, ef = h, and f h = −f. Introduce the notation: M = hw0 , . . . , wk i, provided that the elements w0 , . . . , wk constitute a basis for the module M . Let V be a sl2 Q-module. Choose some basis hw0 , . . . , wk i for it, so that the action of e, h, f be determined by means of integral coefficients. Now, we can consider the R-module VZ = hw0 , . . . , wk i. We call it the reduction of V with respect to the basis w0 , . . . , wk . An R-module M is said to be irreducible if the corresponding sl2 Q-module MQ = M ⊗Z Q is irreducible. It is well known that there exists a unique irreducible (m + 1)-dimensional sl2 Q-module VQ which can be determined, for example, as follows: VQ = hv0 , . . . , vm i, where vi e = vi+1 ,

vm e = 0,

vi h = (m − 2i)vi ,

vi f = −i(m − i + 1)vi−1 .

The reduction of the module VQ with respect to this basis is referred to as the standard R-module of dimension m + 1. Let M ∈ O be an arbitrary irreducible (m + 1)-dimensional module. Then MQ ∼ = VQ . Call a module D diagonal if D has a basis of eigenvectors of h; i.e., D = hd0 , . . . , ds | ∃ki ∈ Z di h = ki di i. Denote by Md the maximal diagonal submodule in M . Obviously, Md is generated by the set of all eigenvectors {v ∈ M | ∃k ∈ Z vh = kv}. Note that if M is irreducible then Md is irreducible too. The following proposition completely describes irreducible diagonal R-modules. Proposition 2. Let α = (α0 , . . . , αm−1 ) ∈ Zn+ be such that (i + 1)(m − i)/αi = βi+1 ∈ Z. Define the module D(α) = hd0 , . . . , dm i as follows: di e = αi di ,

dm e = 0,

di f = βi di−1 ,

di h = (m − 2i)di ,

i = 0, . . . , m.

Then D(a) is an irreducible diagonal R-module and all irreducible diagonal modules in O can be obtained in this way for some α. Moreover, D(α) ∼ = D(γ) if and only if α = γ. Proof. The fact that D(α) is an R-module is verified by straightforward calculations. Irreducibility follows from the definition, because D(α)Q ∼ = VQ . Let W ∈ O be an irreducible diagonal module. Then WQ ∼ = VQ ; we assume that WQ = VQ . Put W = hw0 , . . . , wm i and wi = xi vi , where xi ∈ Q. Here wi e = (xi vi )e = xi vi+1 = (xi /xi+1 )wi+1 , whence αi = (xi /xi+1 ) ∈ Z. Moreover, wi f = (xi vi )f = −xi i(m − i + 1)vi−1 = −(xi i(m − i + 1)/xi−1 )wi+1 and βi = −(xi i(m − i + 1)/xi−1 ) ∈ Z. Furthermore, αi βi+1 = −(xi /xi+1 )(xi+1 (i + 1)(m − i)/xi ) = −(i + 1)(m − i). Hence, W = D(α). Let M1 = D(α), M2 = D(γ), M1 ∼ = M2 , and let ϕ : M1 → M2 be an isomorphism. Look at M1 = hw0 , . . . , wm i and M2 = hu0 , . . . , um i, the corresponding diagonal bases. Then it is obvious that ϕ(wi ) = xi ui ; moreover, xi = pm 1. Look at the element ϕ(wi e) = ϕ(αi wi+1 ) = αi xi+1 ui+1 . On the other hand, ϕ(wi e) = ϕ(wi )e = xi ui e = xi γi ui+1 . We obtain αi = pm γi and since αi , γi > 0, it follows that αi = γi . Hence, α = γ, which completes the proof of the proposition. Lemma 3. Let M ∈ O be an irreducible R-module. Then there exists a constant c ∈ Z, depending only on dim M , such that cV ⊂ M ⊂ V , where V is the standard R-module. Proof. PutP M = hw0 , . . . , wm i. Since MQ ∼ = VQ , we can identify the module M with the submodule V . Indeed, wi = aij vj , where aij ∈ Q. Denote the common denominator of the elements aij by a, and 974

P take a1 V instead of the module V . Then wi = aaij vj , but now aaij ∈ Z. Similarly, we show that V can be chosen so that aij be mutually coprime; i.e., (aij ) = 1. Consider the equalities X  k m s wi f e f = aij vj f k em f s = aik βk . . . β1 βm . . . βm−s vm−s ∈ M, where βi = −i(m − i + 1). Put c = (β1 . . . βm )2 =

m−1 Y

!2 (i + 1)(m − i)

.

i=0

Then aik cvj ∈ M for all i, j, k ∈ 0, . . . , m, and since we have chosen (aij ) = 1, we obtain cvj ∈ M for every i ∈ 0, . . . , m. Hence, cV ⊂ M ⊂ V , which completes the proof of the lemma. Proposition 3. Let M ∈ O be a torsion-free finite-dimensional R-module. Then there exist a diagonal module D and a constant c, depending only on dimM , such that cD ⊂ M ⊂ D. Proof. Consider the module M = hw0 , . . . , wm i. The module MQ is completely reducible. Hence, MQ = V0 ⊕ · · · ⊕ Vk , where Vi is an irreducible sl2 Q-module. Choose diagonal bases for Vi so that wj be expressible with integral coefficients. Denote by Ni the reductions of the modules Vi with respect to these bases. Then Ni is an irreducible diagonal R-module and the module M embeds into the direct sum of Ni ; M ⊂ N1 ⊕ · · · ⊕ Nk . The module M ∩ Ni is irreducible; therefore, by the preceding lemma there exists a constant ci such that ci ViQ ⊂ M ∩ Ni ⊂ Vi , where Vi is the standard R-module of some dimension. Put D = V0 ⊕ · · · ⊕ Vk and c = ci . Then cD ⊂ M ⊂ D. The proof of the proposition is over. Corollary. There are only finitely many torsion-free finite-dimensional sl2 Z-modules in each dimension. In the following theorem we demonstrate a local-global principle for sl2 Z-modules. Theorem 5. Let M, N ∈ O be irreducible finite-dimensional sl2 Z-modules. Then M ∼ = N if and only if Mp ∼ = Np for every prime number p. Proof. We first prove the theorem in the case of diagonal modules. Let M = D(α), N = D(γ), M = hd0 , . . . , dm i, M = hc0 , . . . , cm i. Suppose that Mp ∼ = Np for every prime p and let ϕp : Mp → Np be the corresponding isomorphism. It is easy to verify that ϕp (di ) = xp,i ci ; moreover, xp,i is invertible in Zp ; i.e., p - xp,i . Then ϕp (di e) = ϕp (αi di+1 ) = αi xp,i+1 ci+1 . On the other hand, ϕp (di e) = ϕp (di )e = γi xp,i ci+1 , and so αi = yp,i γi in Zp , where yp,i = xp,i /xp,i+1 , p - yi . It follows that αi = yi γi in Z, and from the condition p - yp,i for every p we find out that p - yi in Z for every p. Hence, yi = pm 1 and αi = pm γi . Since αi , γi > 0, we have αi = γi and by Proposition 2 M = N . The theorem is thus proved in the case of diagonal modules. Let M, N ∈ O be arbitrary irreducible modules. Let C = Md and D = Nd be maximal diagonal submodules in M and N . Then Cp and Dp are maximal diagonal submodules in Mp and Np respectively. From the condition Mp ∼ = Dp for every p, and by the above this is = Np for every p we infer that Cp ∼ ∼ equivalent to C = D. Henceforth we can identify the maximal diagonal submodules in M and N ; Md = Nd = D = hd0 , . . . , dm i. Denote by ϕp : Mp → Np the corresponding isomorphisms over Zp . Then ϕP p (D) = D. Take m ∈ M . Since MQ = DQ , we can express m in terms of di , m = ai di , ai ∈ Q. Denote the common denominator of ai by a. Then am ∈ D and in consequence ϕp (am) = am. Hence, ϕp (m) = m ∈ Np . It remains to prove that if m ∈ Np for every prime p then m ∈ N . 975

P Put N = hn0 , . . . , nm i. Since MQ = NQ , we have m = xi ni , where xi ∈ Q. On the other hand, from the condition m ∈ Np we obtain X m= bp,i ni , bp,i ∈ Zp . Thus, xi ∈ Zp for every prime p; whence xi ∈ Z. Thus, m ∈ N , which completes the proof of the theorem. References 1. Jacobson N., Lie Algebras [Russian translation], Mir, Moscow (1964). 2. Borevich Z. I. and Shafarevich I. R., The Theory of Numbers [in Russian], Nauka, Moscow (1985). 3. Yushchenko A. V., “The Zp -forms of the Lie algebra sl2 (Qp ),” Vestnik Omsk Univ., 1, 19–21 (1999).

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