VOL. 55, 1966 MATHEMATICS: N. JACOBSON 243

We reduce this to a normal form x ax or x -- x + b by means of an appropriatelinear transformation of x. Then we obtain a d-closed meromorphic 1-formx-'dx or dx defined on S. Since S contains no curve, x-ldx or dx is holomorphic.This contradicts that the first Betti number of S is equal to 1, q.e.d.

* This paper was written while the author was engaged part-time on a research project at Stan-ford University sponsored by the Army Research Office (Durham).

1 Kodaira, K., and D. C. Spencer, "On deformations of complex analytic structures, II," Ann.Math., 67, 403-466 (1958), Theorem 15.1.

2 Kodaira, K., "On the structure of compact complex analytic surfaces, II," to appear in Am.J. Math.

3 Kodaira, K., "On the structure of compact complex analytic surfaces, I," Am. J. Math., 86,751-798 (1964).

4 Ibid., Section 1.5 Ibid., Theorem 19.6 Compare Gunning, R. C., "Connections for a class of pseudogroup structures," in Proceedings

of the Conference on Complex Analysis, (Minneapolis 1964), pp. 186-194.

STRUCTURE THEORY FOR A CLASS OF JORDAN ALGEBRAS

BY N. JACOBSON

DEPARTMENT OF MATHEMATICS, YALE UNIVERSITY

Communicated December 10, 1965

In this note, we shall give a structure theory for a class of Jordan algebras whichcorresponds rather closely to the class of semisimple Artinian rings in the associativetheory. Throughout the paper, "algebra" will mean algebra over a field 41 ofcharacteristic not two, which is not necessarily associative or of finite dimension-ality. For Jordan algebras we write a- b for the product, so the defining identitiesare: a-b = bra, (a2-b).a = a'.(b a), where a-2 = ala. If 21 is an arbitraryalgebra with product ab, then we define a * b = 1/2(ab + ba) and denote the algebrawith this product as 21+. If 2 is associative, then 21+ is Jordan. Such Jordanalgebras and their subalgebras are called special. An important composition inJordan algebras is the trilinear product { abc } (a. b) * c + (b - c) a - (a. c) * b.We write Ua for the linear mapping x -- axa}, so Ua = 2Ra2 - Ras, where Ra isx -xx a. We recall the following important identities:

{aba}-2 = {a{ba 2b}a}). (1)Ubn = Ub.., n = 1,2,3, . . .. (2)

UaUbUa Ubuarj1 (3)

If Ua,b denotes the mapping x - Iaxb} = {bxa}, then Uafb = Ua + Ub + 2Uab.If e is an idempotent in the Jordan algebra a, then it is easily seen that the range,JUe of Ue coincides with the Peirce space 31(e) = lxixe = x}. We recall alsothat if a has an identity element 1, then the element aEa is invertible with inverse bif a-b = 1, a2.-b = a.2 This relation is symmetric and b is unique. Also, a is in-vertible if and only if 1E a3Ua and if a is invertible with inverse b, then UaU, = 1 =

244 MATHEMATICS: N. JACOBSON PROC. N. A. S.

UbUa. If a is invertible, so are all of its powers and the criterion for invertibilityin terms of Ua shows that if a", n a positive integer, is invertible, then a is invertible.A Jordan algebra 3 is called a division algebra if 3 has a 1 X 0 and every a $ 0 in 3is invertible.

1. Quadratic Ideals.-We shall base our structure theory on a notion which hasbeen introduced by Topping3 for Jordan algebras of operators in Hilbert space andwhich is defined for arbitrary Jordan algebras as follows.

Definition: A subset Q3 of a Jordan algebra 3 is called a quadratic ideal in 3 if e3is a subspace and p3D3Ub for every b&Q3.

It is clear that a subspace is a quadratic ideal if and only if Iblab2}C3for allbi(E3, aE3. In particular, if Q3 is a quadratic ideal and b,, b2, b3,C, then { bib2b3}&E 3. Hence, (b, b2) b3 = 1/21 bib2b3} + 1/21 b2bib3}& 3. If 3 has an identityelement, then l is a subalgebra of 3. Clearly, any ideal in 3 is a Jordan ideal. If3 = 2 + where 2[ is associative, then any left or right ideal of W is a quadratic ideal of2tV+, since aUb = bab. If e is a quadratic ideal in 3 and aE, then 513Ua is a quad-ratic ideal since if bEQ3, then 3UbUa = 3UaUbUa C3UbUa CqUa. In particular,3Ua is a quadratic ideal for any ae3. We shall call this the principal quadraticideal determined by a. If e is an idempotent, then 3Ue is the Peirce space 31(e) andthis is a subalgebra of 3 with e as identity element. MIoreover, if 53 is a quadraticideal of 3l(e), theii it is a quadratic ideal of 3 since for bC-, b = bUe, and aUb =

3Ubua = ((3 Ue) Ub) Ue CQUe = 3. An element b of 3 will be called an absolute zero

divisor in 3 if 3 Ub = 0 (that is, Ub = 0). The equations just noted show also thatif bE3Ue is an absolute zero divisor in 3Ue, then b is an absolute zero divisor in 3.For the idempotent e one has the Peirce decomposition 3 = 3,(e) ( 3,/,(e) ®3

3o(e), where 3i(e) = xxi~xi e = ixi}. Then 3o(e) is a subalgebra and it is well knownthat 3o(e) - 31(e) = 0, 3o(e) * 31/2(e) C3,/,(e) and if xC3l/,(e), b&3o(e), then x*b2 =2(x*b) *b. Hence, al(e)Ub = Oand I/,(e)Ub = Oifbeao(e). Then UUb = 3o(e) Ub.This implies that any quadratic ideal in 3o(e) is a quadratic ideal of 3. Also, ifbE 30(e) is an absolute zero divisor in 3o(e), then it is an absolute zero divisor in 3.A quadratic ideal 53 of 3 is minimal if 3 $ 0 and there exists no quadratic ideal (E

such that 53D)S:0.THEOREM 1. If 53 is a minimal quadratic ideal in a Jordan algebra 3, then one has

one of the following possibilities: I. 3 = 4b, where b is an absolute zero divisor.II. 53 = 3Ub for every nonzero bE=3 and b b2, b3 = 0 for all bi 3. Moreover, ifeither 3 has an identity element or 3 contains no absolute zero divisor X 0, then 53 2 = 0.III. 53 = 3Ue for an, idempotent e and 3 is division subalgebra of 3.

Proof: Suppose 5 contains an element b $ 0 such that Ub = 0. Then 4b is anonzero quadratic ideal contained in 593 so 53 = 4b and we have case I. Nowassume that for every bE3 we have 3Ub -$ 0. Then 3U, is a nonzero quadraticideal contained in 53, so 3Ub = 53 for every b X 0 in 3. Also, QUb is a quadraticideal contained in 3, so either Q3Ub = 0 or 03Ub = 3-. Suppose there exists a b $ 0in 53 such that 51Ub = 0, and letb'C3. Since = 3Ub, b'= aUbfor aE3. Then53Ub' = 13UaUb = 53UbUaUb = 0. Thus, 53Ub = 0 for all bEz13 and 51Ub,,b2 = 0for all bb2&EO. This implies that b,-b2 b3 = 0 for all biEz&0 so we have case II.If 3 has a 1, then 53 is a subalgebra. Then in case II, b'3 = 0 for every bEzQ, so ifb-2 X 0, then b 2 is a nonzero element of 3. Hence, there exists an a in 3 such thatb = {b2ab 2} . Then b 2 = I b 21ab-4a} b 2} = 0, by (1). This contradiction

VOL. 55, 1966 MATHEMATICS: N. JACOBSON 245

shows that b 2 = 0 for all b)3. Since 3 is a commutative algebra of character-istic $ 2, this implies that 2 = 0. Again suppose we have case II and 3 containsno absolute zero divisor $ 0. If b $ 0 in A, then e = aUb :(3Ub)Ub = llt.2,by (2). If 3Ub.2 = 0, we have b 2 = 0. Otherwise, SUb.2 = i3 by the miinimalityof 513. Then b = { b2ab2}, a(-, and we obtain b2 = 0 contrary to Ut.2=2Hence, again we have e 2 = 0. It remains to consider the remaining alternativefor QA: e3 = S3UUb for every b $ 0 in A3. Take b $ 0 in A3. Then there exists ac(EQ such that b = CUb. This implies that Ub = UbUCUb, so E = UbUC and FUCUb are idempotent linear transformations in 3 which map e into itself -and havesurjective restrictions to Q3. Hence, these restrictions are the identity mapping1, on A, so UbUC, = 1Z = UcUb, where U denotes the restriction of U to -A. Con-sider f = Icb2c})EZ . We have f-2 = { cb2c} 2 = {c b2cb2}1c} = { c{ b{ bc2b}b}c} = {bc.2b}, since {bc2b}eI3 and UaU, = 1e. Similarly, {bc2b} 2 = {cb2c}.Hence, f4 = (f2)2 = {bc2b}b = {cb2c = f. Put e = f3. Then eC 3 ande2= 6= 3 = e, so e is idempotent. Now Uf = UcUb2UC = 1Z and U.Uf3= 1. Hence, e $ 0. ThenZ = 3U6 = 31(e) the Peirce 1-space of e. Hence,Q3 i§ a subalgebra of S. If dEz53 and d $ 0, then 53Ud = Z3 contains e. This im-plies that d is invertible in 513 and so Z is a division algebra. Hence, we have caseIII.We next obtain sufficient conditions that a Jordan algebra contain idempotents

$ 0, 1. For this we requireLEMMA 1. Let i be a Jordan algebra with identity element 1 and elements a,b such

that a 2 = 0 = b 2 and 2a*b = 1. Then a contains an idempotent element e$ 0,1.Proof: One sees directly that 1, a,b are linearly independent. Choose a43 in 4),

so that ca3 = 1. Put f = 1 + aa + fOb. Then f 2 = 2f. Hence, e = 1/2f is alnidempotent $ 0,1.THEOREM 2. Let 3 be a Jordan algebra such that: (1) a has an identity element

1 ($ 0). (2) a contains no absolute zero divisors $ 0. (3) 3 contains minimal quad-ratic ideals. Then either a is a division algebra or a contains an idempotent elemente $ 0,1.

Proof: Let e3 be a minimal quadratic ideal in S. Then Z is not of type I bycondition (2). If e3 is of type III, either a = e3 is a division algebra or we have anidempotent e $ 0,1. Hence, we may assume Ql3 is of type II. Also, we have-2 = 0. Now choose b $ 0 in e and aCa so that b = {bab}. Let c = as b. Wehave the identity

(a b)2 = 1/2a-{bab} + 1/4{ba.2b} + 1/4{ab2a} (4)in any Jordan algebra, which is clear by Shirshov's theorem. In the presentsituation this gives c-2 = 1/2c + 1/4{ ba 2b}. Since I ba2b}EE, we have (2c2 -c) 2 = 0. This implies that either c 2 = 0, or there exists an idempotent e $ 0 ofthe form ac + 6c 2 + 'yC3, a3, y(E 4). In the first case we have -2c = ba 2b}, soceQ3. Then b = aUb = 2(a.b)-b -a-ba2 = 2c b = 0, since t32 = 0. Hence,we have the second case: e = ac + ACt2 + yCc3 is a nonzero idempotent. Supposee = 1. Then c is a root of a cubic equation with nonzero constant term. Since(2c2 - c)2 = 0, this implies that (2c - 1),2 = 4C 2 - 4c + 1 = 0. Then I ba 2b}= 4c2 - 2c = 2a b - 1. Since b 2 = 0, XUb = 2(x - b) - b, so the foregoing equa-tion gives 2(a 2*b) *b = 2a-b - 1 or 2a'*b = 1 for a' = a - a-2*b. Then 2(a'*b)*b

246 MATHEMATICS: N. JACOBSON PROC. N. A. S.

= b, so a'Ub = b and replacing a by a' permits us to assume that 2a b = 1. Thenc = 1/2 and {ba 2b} = 0. Hence, {aba}*2 = 0 by (1). Also, 2{ aba} *b =2a-{bab} = 2arb = 1 and {aba}Ub = 2({abaj b)-b = b. Hence, replacing aby I aba}, we have a 2 = 0 = b 2 and 2a - b = 1. Then we obtain an idempotentF0, 1 by Lemma 1.

2. Axioms and Structure Theory.-We shall now investigate the structure ofJordan algebras satisfying the following axioms:

(i) 3 has an identity element 1.(ii) 3 contains no absolute zero divisors 5 0.(iii) The minimum condition holds for quadratic ideals determined by idem-

potent elements, and every quadratic ideal determined by an idempotent $ 0 con-tains a minimal quadratic ideal.We shall call a Jordan algebra 3 regular if aC3Ua for every aC3. Since XUa =

axa in 21+, 21 associative, it is clear that this is analogous to the well-known definitionof regularity for associative rings due to von Neumann. Clearly, regularity im-plies (ii). If 21 = Homa(9),91), where 9N is a finite dimensional vector space overthe division algebra, then it is well known that 21 is regular. Hence, 2[+ is regular.It is easy to see that W+ satisfies axioms (i)-(iii). It is easy to see also that if21 = Homa(9,92) has an involution J (anti-automorphism of period two), thenthe subalgebra St(2[,J) of 21+ of J-symmetric elements satisfies the axioms.THEOREM 3. Let 3 be a Jordan algebra satisfying axioms (i)-(iii). Then 1 =

n

ej in 3, where the ej are orthogonal idempotent elements in 3 such that 3U, is a1

division algebra, i = 1,2,...,n.Proof: It is clear from the remarks at the beginning of §1 that if e is a nonzero

idempotent in 3, then 3 U, satisfies the axioms. Let 3 U, be a minimal element inthe collection of quadratic ideals determined by idempotents e $ 0. We claim that3Ue is a division algebra. Otherwise, Theorem 2 implies that 3Ue contains anidempotent f $ O,e. Then aUe3aUf contrary to the minimality of 3 U,. There-fore, we see that there exist idempotents e in 3 such that 3 Ue is a division algebra.Suppose e1,e2,...,e, are orthogonal idempotents in 3 such that 3 U, is a division

algebra. Set fi = 1 - ej. Then f, is an idempotent and 3 Up, D3U . ...1 r

D3Ufr. If fr = 0, then 1 = ej and the theorem is proved. If fr 5 0, then 3 Uf1satisfies our conditions and so contains an idempotent e,+1 such that 3 Ue,+ is adivision algebra. Then e1,e2,... ,er+l is set of orthogonal idempotents and 3.Uf,

r+1v ... D3 Ur, D3 U17 + I if fT+1 = 1- ej. By (iii) this process must terminate

with an f,, = 0 and this proves the theorem.n

If 1 = ei, where the ei are orthogonal idempotents in 3, then we have the Peirce

decomposition 3 = E ij, where 3ii = jUei, jij = lUei,ej = {xijxij.ei =i<j

12Xij = xij ejl for i $ j.LEMMA 2. Let 3 = 311 G 312 E 322 be the Peirce decomposition of a Jordan

algebra 3 with 1 relative to the orthogonal idempotent elements el, e2 such that el + e2 = 1

VOL. 55, 1966 MATHEMATICS: N. JACOBSON 247

and assume 3a is a division algebra for i = 1,2. Then any a12e312 is either invertibleor a12-2 = 0.

Proof: It is well known that & 2C& + ajj. Hence, a12 2 = a, + a2, ajS_3jThen (al22.el) -a12 = a122, (ela12) gives a, a12 = (al + a2) (1/2al2), so a, -

a2- a12. Suppose a,1 a12 = 0. Then a2-a12 = 0 and al2 =3 0. Then a, 2 + a22 =a12 4 = 0. Since 3& is a division algebra, this implies that a, = 0 so a12 2 = 0.If al. al2 4 0, then a2 *a12 $ 0 so the a, $ 0 and a12 2 = a, + a2 is invertible. Thisimplies that a12 is invertible.

If el, e2, . . ., en is a set of orthogonal idempotents such thatE e; = 1 and 3 =E aij the corresponding Peirce decomposition, then we say that e, and el are con-i<jnected if there exists an element in 3kl which is invertible in the subalgebra 3' +3kl + 3It of 3, where we take Ik = 3a if k > 1.Lemma 2 implies that, under the indicated hypotheses, either el and e2 are con-

nected or 312 is an ideal such that 312 2 = 0.LEMMA 3. If 3 is as in Lemma 2, then either 3 is simple, 3 = 311 (3 322, and the

aii are ideals, or 3 contains an ideal 9 such that 9-2 = 0.Proof: If 3 is simple or 3122 = 0, we have the desired result. Hence,

assume 3 not simple and 312 2 $ 0. Let 9 be an ideal $ 0, 3. We have 91 =911 + 912 + 922, where %ij =%n9 ij. By Lemma 2 and 91 5 3 we have 912 2 = 0.Also, since 3ii is a division algebra and 9ii is an ideal in 3ii, either 9ii = 0 or9Rii = 3ii. In the latter case 91:312 = ei-312. Then 9112 = 312 $ 0 contrary to91122 = 0, 312 2 $ 0. Hence, 9ii = 0 and W = 912 satisfies 9112 2 = 0.

If 91 is an ideal in a Jordan algebra such that 912, then any bEE9 is an absolutezero divisor since xUb = 2(x - b) b - x b2 = 0 since x *be9. Hence, if the algebra3 of Lemma 2 has no absolute zero divisors = 0, then either 3 is simple and 3 =all @ a22-We can now prove theFIRST STRUCTURE THEOREM. Let 3 be a Jordan algebra satisfying the axioms

(i)-(iii). Then 3 is a direct sum of a finite number of ideals which are simple algebrassatisfying (i)-(iii).

Proof: Let 3 = E 3aij, where 3ii is a division algebra as in Theorem 3. If 3i<j

is simple, there is nothing to prove, so we assume 3 has an ideal S $ 0, 3. ThenS = RSip, where Sij = S nfl j. Then Rij + Rij + Sji is an ideal in 3ij + aij + ajjand the latter has the form 3 Up, f an idempotent, so it satisfies our axioms. Hence,by Lemma 3 and the remark following it, either 3ii + aij + 3fj is simple or aij = 0.In the first case, either SIi + Sij + Sjj = 3ii + 3ij + 3jj or an + Rij + StIj = 0.In the second case ftij = 0 and SIi + SI, = 0, 3i", 3jj, or 3ii + 3,j. This impliesthat = Z'3kl a sum of Peirce spaces 3ij such that if 3kl $ 0 CQ, then 3kk,31l CS. Let S' = E~'aj the sum of Peirce spaces not contained in R. Then3 = S 3 SI' and we proceed to show that S' is an ideal. If we define 3ji = aij forj > i, then it is well known that we have the following multiplication table for thePeirce spaces: s ake = O if {i,j} flnkl = 0, 3ii3jk C3ik if i, j, k are distinct,Si-aij = 3ij and 3ij2 C3., + 3jj. Hence, the fact that SI' is an ideal willfollow by showing that there is no index k such that there exists a nonzero 3kl CRand a nonzero 31Cft'. Suppose the contrary. Then 3kkCR and 3ki = 3kk-akiCIt which contradicts 3&Cft'. Thus, we have shown that every ideal in 3 has a

248 MATHEMATICS: N. JACOBSON PROC. N. A. S.

complementary ideal. This complete reducibility of the lattice of ideals impliesthat 3 is a direct sum of minimal ideals. Since a has a 1, a is a direct sum of afinite number of minimal ideals. Hence, 3 = al (E 32@ ... @ 2}& where 3i is aminimal ideal in $. Then a, as algebra satisfies the axioms. Hence, if it is not sim-ple, then 2i = Sit 3iare nonzero ideals in i. Sinceea 3, =0if i 5 j, ai',2i"are ideals in 3. This contradicts the minimality of 3i. Hence every 3a is a simplealgebra.

Let 3 = E ij3 be the Peirce decomposition of a Jordan algebra 3 with 1i<j

relative to the orthogonal idempotents el, e2, ..., en such that z ei = 1. Let k $ 1and let akIC$kz (= 31k) be invertible in 3kk + 3k1 + 3al. Then one sees easilythat the inverse bkl of akl in kk + Sk + 31 is in 3kl. We can now proveLEMMA 4. Connectedness is an equivalence relation for the set of idempotents

e1,e2, ... ,en (orthogonal with E ei = 1).Proof: The relation is clearly reflexive and symmetric. Now assume i, j, k are

distinct, ei and ej are connected? and ej and ek are connected. Then we have a1j,bij2aij which are inverses in bii + .ij + 2jj and ak, bik1E2j} which are inverses in3jj + ajk + 3kk1 Put aik = 2aij * a3k, bik = 2bij bjk. We shall show that aik andbikare inverses in 2ii + 3ik + 1kk. For the proof we need the following identitiesfor elements of the Peirce components: (Xi; * Yij) Zjk = Xi; (yij * Zjk) + Yij * (Xi; *Zjk).((Xij Y;k) -Zki) e, = (xij (Yjk1Zki)) vei, (xii *yij) *Zjk = Xii* (YijZjk), where i, j, kare distinct xij, yYijC3ij Zje3jk, etc. These are obtained by making suitablesubstitutions in the basic multilinear identities for Jordan algebras. We shall alsoneed the fact that if a and b are inverses in a subalgebra of 3, then their multiplica-tion Ra and Rb commute in 3. This is well known if the subalgebra has the sameidentity as 3 and the general case can be established using the Peirce decompositionrelative to the identity element of the subalgebra. Now we have ak bik = 2ajk(bk . bij) = ajk * (bjk * bij) + bjk * (ak bij) = (ajk . bjk) * bij = (ej + ek) * bij = 1/2bij and(aik*bik) *ei = 2((aij.a*k) .bik) *ei = 2(aij * (ajk * bik)) - ej = (aij * bij) * ej = ei. By sym-metry, (a11. bi,) eek = ek. Hence, aik * bi, = ei + e.,. Next we have bij * aik = 1/2ajkby the same type of argument used to prove ak,, bik = 1/2bij. Hence, (aik 2 ei) bi= 2(aik-2-ee)*(bij-bjk) = 2((aik 2ei) -bij) -bk = 2(aik 2bij) -bjk = 4(aik1 (aikb1j))b k = 2(aik* ajk) *bjk = (aik * ak) * bk + (aik * bjk) * ajk = aik * (bk * ajk) = aik (ej + ek) =

'/2aik- Similarly, (aik 2 * ek) - bik = 1/2aik. Hence, aik 2 - bik = aik and aikbik areinverses relative to ei + ek. This proves that connectedness is a transitive relation.

Let Z be an algebra with an identity element and an involution d -- d, Z., thealgebra of n X n matrices with entries, y a diagonal matrix with diagonal entries yjwhich are symmetric elements of the nucleus of Z having inverses in this nucleus.Let 'QX,'vy) be the set elements ATDCZ such that y-'A'-y = A, where the primedenotes the transpose and if A = (aij), then A = (at,). Then t)(Z.,{y) is a sub-algebra of , and it is known that for n = 3, ,)(Tn)nY) is Jordan if and only if Zis alternative and the symmetric elements of Z are in the nucleus, and for n > 4,!(Tn),y) is Jordan if and only if Z is associative.4 In these cases we call n Y)a Jordan matrix algebra and Z and its involution a coordinate algebra for !D(Zn, 'y).We can now state theSECOND STRUCTURE THEOREM. Let 3 be a simple Jordan algebra satisfying

axioms (i)-(iii). Then 3 is one of the following types: (1) a division algebra, (2) a

VOL. 55, 1966 MATHEMATICS: A. JACOBSON 249

Jordan algebra whose identity is a sum of two connected orthogonal idempotents ejsuch that 3Ue. is a division algebra, (3) a Jordan matrix algebra whose coordinatealgebra is either a Cayley algebra over its center with standard involution, a splitquaternion algebra over the center with standard involution, a direct sum of an associativedivision algebra and its opposite with involution exchanging the two factors, or an associ-ative division algebra with involution.

Proof: Let a = E 2ij be the Peirce decomposition of a relative to a set ofis}

n

orthogonal idempotents ei such that E ei = 1 and $ii is a division algebra (Theorem1

3). Let i $ j and assume $ij $ 0. Then Lemma 3 and the fact that aii +3ij + ajj satisfies the axioms imply that this is a simple algebra. Hence, a j 2 $ 0

and ei and ej are connected by Lemma 2. Let el,. . . ,em be the complete set of eiconnected with e1 and set E'= zm akl- Since connectedness is a transitive rela-

k,l =1

tion, every (ki with k < m and i > m is 0. This implies that a' is an ideal. Hence,3' = 3 and all the ei are connected. If n = 1, a is a division algebra and if n = 2,we have the second alternative. If n > 3, then by the Coordinatization The-orem for Jordan algebras, a is a Jordan matrix algebra.5 Also, since 3 is simple, thecoordinate algebra Z is a simple algebra with involution, that is, it has no ideals

0, Z which are invariant under the involution. It follows that either Z = A (3A' where A is simple with involution exchanging the two factors, or Z itself issimple. In the first case, .t(Z) the subalgebra of Z+ of symmetric elements is iso-morphic to Ai+ and in both cases t(Z) is isomorphic to 3Uei which is a Jordandivision algebra. If Z is not associative, then the condition that the symmetricelements are in the nucleus rules out Z = A /A'. Hence, in this case Z is simple,so by a theorem of Kleinfeld's, T is a Cayley algebra.6 Moreover, the involution is astandard one since the symmetric elements are in the nucleus which, for Cayleyalgebras, coincides with the center. If Z is associative, then it is well known that anelement a has an inverse in the Jordan algebra Z+ if and only if it has an inverse inZ in the usual sense. Hence, every nonzero element of St(Z) has an inverse in Z.If Z = A G) A', this implies that A is a division algebra. The remaining case:Z simple associative is settled by the following lemma which is due to J. i\I. Osborn,Jr. (oral communication).LEMMA 4. Let Z be a simple associative ring of characteristic $ 2 containing an

identity element and having an involution 3 such that every nonzero symmetric elementofZ has an inverse. Then either Z is a quaternion algebra over its center with standardinvolution or Z is a division algebra.

Proof: We assume that T is not a quaternion algebra with standard involution.Then, by a theorem of Herstein's, the symmetric elements ofZ generate ZY Let &and S denote the sets of symmetric and skew elements, respectively, of Z. Leta(- and write a = h + s, where hEt, see. If a is not invertible, then aa-' = 0and a-a = 0. Otherwise, since aaJ, aJaCt, the hypothesis on £& implies that one ofaaJ, aia is 0 and the other is invertible. By symmetry it is enough to suppose thataia = 0 and there exists an x such that aaJx = 1 = xaaJ. Then 0 = aJaaJx = aJso aaJ = 0 = aJa, a contradiction. We now prove that if h $ 0, then a = h + s

2.50 MATHEMATICS: N. JACOBSON PROC. N. A. S.

is invertible. Otherwise,

0 = '/2(aaJ + ala) = 1/2[(h + s)(h- s) + (h- s)(h + s)I = h2- 82,

0 = '/2(aaJ- ala) = 1/2[(h + s)(h -s) - (h -s)(h + s)] = sh -hs.

Hence, sh = hs and h2 = S2. Since h X 0 is in So, h-1 exists. Hence, if c = h-1s,then c2 = h-1sh-'s = h-2s2 = 1. Also, cJ = -h-'s = -c. Let k be any elementof A. Then (1 + c)k(l + c)J = (1 + c)k(l- c) and (1 + c)Jk(l + c) = (1 - c)k(l + c). Also, if b = (1 + c)k(l + c)J, then b2 = (1 + c)k(l- c)(1 + c)k(l- c)= 0 since (1 + c)(1 -c) = 1-c2 = 0. Since b Sthis gives (1 + c)k(1 - c) =0. Similarly, (1- c)k(l + c) = 0, so we have k + ck - kc - ckc = 0, k- ck +kc - ckc = 0. Hence, kc = ck. By Herstein's theorem the symmetric elementsgenerate Z. Hence, c is in the center. Since c2 = 1, e = 1/2(1 - c) is a centralidempotent. Since c is skew, c $ 1 so e $ 0, 1. This contradicts the simplicityof Z. Hence, we have shown that if a = h + s, he,?, sCS, and h $ 0, then a isinvertible. Next, suppose a = s is skew. If hs is not skew for some h&.i, thenthere exists a u such that uhs = 1 = - shuJ, so s is invertible. Hence, we mayassume (hs)J = -hs for all hCD. This gives sh = hs, so s commutes with everysymmetric element. Then s is in the center and since the center is a field, eithers = 0 or it has an inverse.

In the case3 of the theorem it is easy to see that a is either a reduced exceptionalsimple Jordan algebra over the center, or 3 has the form W+, 2 simple Artinian, ora = ,i(W,J), where 2t is simple Artinian with involution J. Conversely, one sees

easily that these algebras satisfy the axioms. Clearly, any Jordan division algebrasatisfies the axioms also. Further analysis of the remaining case2 is desirable andwe hope to undertake this in a later communication.

3. Alternative Set of Axioms.-Let a be a Jordan algebra, 3' = 44 3 3 theJordan algebra obtained by adjoining an identity element 1 to 3. If b' = 11 + b,bE3, then b'2 - 1321 + (213b + b 2), so b"2 = 0 implies 13 = 0 and b' = bc3. Itfollows that 3' has no absolute zero divisors $ 0 if 3 has none. Next, assumeb' 2 = b'. Then 13 = 0 or 1. In the first case, b' = bC3 and in the second b 2 =-b, so f = -b is an idempotent and b' = 1-f. If e-2 = e-3, then 31'(e) =31(e) since a is an ideal in 3'. If e' = 1 - f, f 2 = fEa, then 31'(e') = 3o(f) + be'Clearly, the minimum condition holds for quadratic ideals of this form if and onlyif it holds for quadratic ideals of 3 of the form So(f), f 2 = f.Assume 3 has no absolute zero divisors, so 3' has none. Let Q3 be a minimal

quadratic ideal of 3. By Theorem 1, e = 3Ub where b ZB and ?3 is a subalgebraof 3. Then 3'Ub = 3Ub + a1bb = A, so Q3 is a quadratic ideal of 3', which isclearly minimal. If e' = 1 _ f, f,2 = feC, then 31'(e') = So(f) + be'. Hence,if 3o(f) contains a minimal quadratic ideal of 3, then 3i'(e') contains a minimalquadratic ideal of 3'. If 30(f) = 0, then 31'(e') = be' is a minimal quadraticideal of 3'.We now state the following axiom on 3.(iv)- The minimum condition holds for quadratic ideals of the form 3o(e), e 2 = e,

and if JO(e) X 0, then So(e) contains a minimal quadratic ideal.THEOREM 4. If 3 satisfies (ii), (iii), and (iv), then 3 has an identity element, so the

structure theorems hold for 3.

VOL. 55, 1966 MATHEMATICS: L. V. AHLFORS 251

Proof: The foregoing results show that S' = A1 @ a satisfies (i), (ii), (iii).Hence, by the proof of the first structure theorem, every ideal in a' has a comple-mentary ideal. In particular, 3'= 03, where e is an ideal. Then a hasan identity element.Axioms (iii) and (iv) are evidently satisfied if 3 is finite dimensional over 4.

Also, it is quite easy to show in this case that (ii) is equivalent to the assumptionthat a has no nil ideals. Therefore, our results give a new and improved deriva-tion of Albert's structure theorems for finite dimensional semisimple Jordanalgebras.

1 See author's paper, "A coordinatization theorem for Jordan algebras," these PROCEEDINGS, 48,1154-1160 (1962); and the references in this paper.2See ref. 1; also Jacobson, N., "A theorem on the structure of Jordan algebras," these

PROCEEDINGS, 42, 140-147 (1956); and McCrimmon, K., "Norms and non-commutative Jordanalgebras," forthcoming in Pacific J. Math.

3 "Jordan algebras of self-adjoint operators," Mem. Am. Math. Soc., 53 (1965).4See ref. 1, p. 1154, and the reference given there to a dissertation by Dallas Sasser.See ref. 1, p. 1158.

6 "Simple alternative rings," Ann. Math., 58, 544-547 (1953).7 "Lie and Jordan systems in simple rings with involution," Am. J. Math., 78, 629-649 (1956).

FUNDAMENTAL POLYHEDRONS AND LIMIT POINT SETSOF KLEINIAN GROUPS*

BY LARS V. AHLFORS

DEPARTMENT OF MATHEMATICS, HARVARD UNIVERSITY

Communicated December 16, 1965

1. We denote by Q the group of orientation-preserving M6bius transformationsin R3 which leave B = {x/IxI < 1 } invariant. In other words, transformations inQ are products of an even number of reflections in spheres orthogonal to the unitsphere S = OB.Any discrete subgroup of Q is called a Kleinian group. Poincar6l has shown that

any such group G is discontinuous in B. In its action on S it is discontinuous on anopen set D, which may be empty. The complement L = S - D is the set of limitpoints. If L is all of S, we say that G is of the first kind. If not, L is nowhere denseon S and G is of the second kind. This terminology stresses the analogy withFuchsian groups.

2. The orbit space M = (B U D)/G is a connected orientable 3-manifold withboundary. The latter can be identified with MO = D/G which need not be con-nected, but whose components carry a structure of Riemann surface.The immediate problem is to study the structure of M and Mo as well as the

properties of D and L, particularly when G is a finitely generated group. So faronly MO has been investigated with some degree of success.2 It is hoped that asystematic study of M will lead to more complete results, also as far as Mo and Lare concerned.

3. For the study of M it is advantageous to introduce the isometric funda-