We present the basic properties of the octonions and construct the five exceptional simple Lie algebras

Fitzgerald, Robert W.

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Southern Illinois University CarbondaleOpenSIUCMiscellaneous (presentations, translations,interviews, etc) Department of Mathematics2009OctonionsRobert W. FitzgeraldSouthern Illinois University Carbondale, rfitzg@math.siu.eduFollow this and additional works at: http://opensiuc.lib.siu.edu/math_miscRepresentation Theory Seminar lecturesThis Article is brought to you for free and open access by the Department of Mathematics at OpenSIUC. It has been accepted for inclusion inMiscellaneous (presentations, translations, interviews, etc) by an authorized administrator of OpenSIUC. For more information, please contactopensiuc@lib.siu.edu.Recommended CitationFitzgerald, Robert W., "Octonions" (2009). Miscellaneous (presentations, translations, interviews, etc). Paper 40.http://opensiuc.lib.siu.edu/math_misc/40OctonionsThe simple Lie groups over C come in four infinite families plus five exceptional groups.All five of the exceptional groups, G2, F4 and E6, E7, E8, are constructed from the octo-nions. That is our goal.Here an algebra is a vector space over R with a bilinear multiplication (it need not becommutative or associative) and a unity 1. An involution on an algebra A is a linear map¯: A→ A such thatx¯ = x and xy = y¯x¯.We further always assume that x + x¯, xx¯ ∈ R · 1. The associated norm is N(x) = xx¯. Ais a composition algebra if N(xy) = N(x)N(y) and N is anisotropic (that is, N(x) = 0implies x = 0). For x in an algebra with an involution, we set Re(x) = 12 (x + x¯), the realpart of x. We call x purely imaginary if Re(x) = 0, or equivalently, x¯ = −x.1. The Cayley-Dickson Process.Let A be an algebra with an involution. Set CD(A) = A+Av, where v is a new symboland(a + bv)(c + dv) = (ac − d¯b) + (da + bc¯)v.Note that v2 = −1. Define a+ vb = a¯− bv.We first check that this is an involution. Clearly x¯ = x. For x = a+ bv and y = c+ dvwe have:y¯x¯ = (c¯− dv)(a¯ − bv)= (c¯a¯ − (−b)(−d)) + ((−b)c¯ + (−d)a¯)v= (c¯a¯ − b¯d)− (bc¯+ da)vxy = (ac − d¯b) + (da + bc¯)v= (ac − d¯b)− (da + bc¯)v= (c¯a¯ − b¯d)− (da + bc¯)v.Note that Re(a + bv) = Re(a) ∈ R and N(a + bv) = N(a) +N(b) ∈ R.Start with A = R and the trivial involution x¯ = x. Then CD(R) = C, namely replacingv by i, we have CD(R) = R + iR with (a + ib)(c + id) = (ac − db) + i(ad + bc). Theinvolution on CD(R) is the usual complex conjugation and N(a + ib) = a2 + b2. NextCD(C) = H, the quaternions, since if we replace v by j we have:j2 = −1 ji = −ij,(taking a = 0, b = 1, c = i, d = 0 in the formula). Again we get the usual involution onH and the norm is the sum of four squares. One can prove H is associative (because C iscommutative).The octonions are defined to be O = CD(H). We want to prove the basic properties ofO, in particular that it is a composition algebra, and to indicate why the process is notrepeated.Typeset by AMS-TEX12First, we will use ` in place of v. Thus O is 8-dimensional, with a basis of 1, i, j, k =ij, `, i`, j`, and k`. The real part is the coefficient of 1 and the norm is the sum of thecoefficients squared. Let a, b ∈ H and set x = a, y = b`. Thenx(y`) = a(b` · `) = a(−b) = −ab(xy)` = (a · b`)` = (ba`)` = −ba.Thus, since H is not commutative, O is not associative.Lemma 1. Let A be an algebra with an involution. Let x = a+vb, y = c+vd be elementsof CD(A).(1) Re(xy) = Re(yx).(2) xx¯ = N(x) = x¯x.Proof. (1) For purely imaginary x, y:xy = y¯x¯ = (−y)(−x) = yx.So Re(xy) = Re(xy) = Re(yx). Still assuming x, y are purely imaginary and lettinge, f ∈ R gives:Re(e + x)(f + y)) = Re(ef + ey + xf + xy)= ef +Re(xy)= fe +Re(yx)= Re(f + y)(e + x).(2) x¯x = (a¯ − vb)(a + vb) = a¯a+ b¯b = N(x) = xx¯. Lemma 2. Let A be an algebra with an involution. Then A is associative iffx¯(xy) = (x¯x)y and (xy)y¯ = x(yy¯)holds for all x, y ∈ CD(A).Proof. Let x = a+ bv and y = c+ dv. Thenx¯(xy) = (a¯− bv)[(ac − vd¯b) + (da + bc¯)v]= [a¯(ac− d¯b) − (da+ bc¯)(−b)] + [(da + bc¯)a¯ − b(ac − db¯)]v= [a¯(ac) − a¯(d¯b) + (a¯d¯)b + (cb¯)b] + [(da)a¯ + (bc¯)a¯ − b(c¯a¯) + b(b¯d)]v.Hence if A is associative, x¯(xy) = N(x)(c+dv) = N(x)y = (x¯x)y, by Lemma 1 (2). Forthe second equation, replace x by y¯ and y by x¯ and conjugate. Conversely, if x¯(xy) = (x¯x)ythen (a¯d¯)b = a¯(d¯b) and A is associative. The associator of x, y, z is [x, y, z] = (xy)z − x(yz). An algebra is alternative if [x, y, z]is an alternating function. That is, if σ ∈ S3 then we require that[xσ(1), xσ(2), xσ(3)] = sgn(σ)[x1, x2, x3],for all x1, x2, x3 ∈ A.3Lemma 3. Let x, y, z ∈ CD(A).(1) [x, y, z] changes sign when the involution is applied to one variable.(2) If CD(A) is alternative then [x, y, z] is purely imaginary.Proof. (1) Write x = e + x0 with e = Re(x) and x0 purely imaginary. Then:[x, y, z] = ((e + x0)y)z − (e + x0)(yz)= (ey)z + (x0y)z − e(yz) − x0(yx)= [x0, y, z].[x¯, y, z] = [−x0, y, z] = −[x0, y, z] = −[x, y, z].(2) We have:[x, y, z] = (xy)z − x(yz)= z¯(y¯x¯) − (z¯y¯)x¯= −[z¯, y¯, x¯]= [z, y, x] by (1)= −[x, y, z],by the alternative property. Hence [x, y, z] is purely imaginary. Theorem 4. Let A be an algebra with involution. Then CD(A) is alternative iff A isassociative. In particular, O is alternative.Proof. We only prove (⇐). Let x, y, z, w ∈ CD(A). By Lemma 2, [x¯, x, y] = 0 = [x, y, y¯].Lemma 3 (1) gives [x, x, y] = 0 = [x, y, y]. Hence:[w + x,w + x, y] = 0 = [x, y + z, y + z].In particular:(w + x)2y − (w + x)((w + x)y) = 0w2y + (wx)y + (xw)y + x2y = w(wy) +w(xy) + x(wy) + x(xy)[w,w, y] + [w, x, y] + [x,w, y] + [x, x, y] = 0[w, x, y] = −[x,w, y].Similarly, [x, y, z] = −[x, z, y]. Since (1 2) and (2 3) generate S3, we have that CD(A)is alternative. This is why the Cayley-Dickson process is not repeated. CD(O) is an algebra with aninvolution but it is not alternative, that is, there is no useful version of an associative law.Corollary 5. Let x, y, z ∈ O. The real parts of (xy)z, (yz)x, (zx)y, x(yz), y(zx) and z(xy)are the same.Proof. Theorem 4 and Lemma 3 (2) give that the real part of [x, y, z] is zero. HenceRe(x(yz)) = Re((xy)z). Lemma 1 (1) gives Re(x(yz)) = Re((yz)x). Repeat the process:Re([y, z, x]) = 0 so Re((yz)x) = Re(y(zx)) and so on. 4Corollary 6. O is a composition algebra.Proof. We have for x, y ∈ O:N(xy) = Re((xy)(xy))= Re((y¯ x¯)(xy))= Re(y¯(x¯(xy))) by Corollary 5= Re(y¯((x¯x)y)) by Lemma 2= N(x)Re(y¯y) = N(x)N(y).The Cayley-Dickson process will not yield more composition algebras. In fact, there areno others. Hurwitz (1898) proved that the only composition algebras over R are: R, C,H and O. His proof was via quadratic forms. First, as N(1) = 1 and N(x) = 0 only forx = 0, there is a basis {vi} of a composition algebra A such thatN(∑xivi) =∑x2i .The formula N(x)N(y) = N(xy) then gives a sum of squares identity of the form:(x21 + x22 + · · ·+ x2n)(y21 + y22 + · · ·+ y2n) = z21 + z22 + · · ·+ z2n zi ∈ R[X,Y ]Hurwitz proved that there is such an identity iff n = 1, 2, 4 or 8.We can now give the first construction: G2 = Aut(O). Since O is 8-dimensional, eachelement of G2 can be viewed as an 8 × 8 matrix (an 8-dimensional representation). Anautomorphism will fix the purely imaginary octonions, yielding a 7-dimensional represen-tation.2. Digression: Arbitrary fields.Let F be a field, A an algebra over F with an involution, and let λ ∈ F ∗. ThenCD(A,λ) = A+Av where(a + bv)(c+ dv) = (ac+ λdb¯) + (a¯d+ cb)v.The previous construction was with λ = −1.Then CD(F,−λ1) = F (√−λ1) and the norm of x1 + x2v is:N1(x1, x2) = x21 + λ1x22.CD(F (√−λ1),−λ2) = (−λ1,−λ2)F . This has a basis 1, e2, e3, e4 withe22 = −λ1 e23 = −λ2 e2e3 = e4 = −e3e2.The norm of∑xiei is:N2(x1, x2, x3, x4) = N1(x1, x2) + λ2N1(x3, x4) = x21 + λ1x22 + λ2x23 + λ1λ2x24.5Lastly, CD((−λ1,−λ2)F ,−λ3) is a generalized octonion with norm formN2(x1, x2, x3, x4) + λ3N2(x5, x6, x7, x8).Quadratic forms of these shapes are known as 1-fold, 2-fold, 3-fold Pfister forms, respec-tively. Then Hurwitz’s Theorem extends: Let q be a quadratic form in n variables. Thereis an identityq(x1, x2, . . . , xn)q(y1, y2, . . . , yn) = q(z1, z2, . . . , zn) zi ∈ F [X,Y ]iff q is equivalent to a k-fold Pfister form for k = 0, 1, 2 or 3. In particular, the only com-position algebras over F are: F , F (√−λ) and the generalized quaternions and octonions.Pfister’s connection is by his result (1966): There is such an identity but with zi ∈F (X,Y ) iff q is equivalent to a k-fold Pfister form for some k. In particular, n is a powerof 2.3. Jordan algebras.A Jordan algebra is an algebra where the multiplication satisfiesab = ba(JA1)a2(ab) = a(a2b).(JA2)Jordan algebras were introduced in 1932 as part of an algebraic formalism for quantummechanics. They can be thought of as similar to Lie algebras, with (JA1) in place ofanti-commuting and (JA2) in place of Jacobi’s identity. Further, Lie algebras can beconstructed from an associative algebra A by setting [a, b] = ab− ba. Jordan algebras canbe constructed from the anticommutatora ◦ b = 12 (ab + ba).Then clearly (JA1) holds. Note that a ◦ a = a2. For (JA2):a2 ◦ (a ◦ b) = a2 ◦ 12 (ab + ba)= 14 [a2(ab + ba) + (ab + ba)a2]= 14(a3b + a2ba + aba2 + ba3)a ◦ (a2 ◦ b) = a ◦ 12 (a2b+ ba2)= 14 [a(a2b+ ba2) + (a2b + ba2)a]= 14(a3b + aba2 + a2ba+ ba3)= a2 ◦ (a ◦ b).Thus (A, ◦) is a Jordan algebra. Note that (A, ◦) is not associative in general: a ◦ (b ◦ b)−(a ◦ b) ◦ b = (ab − ba)b − b(ab − ba), which fails unless A is commutative.6There is one important difference between Lie algebras and Jordan algebras. EveryLie algebra is a Lie subalgebra of an associative algebra with the standard Lie bracket–the Poincare-Birkoff-Witt theorem. However, not every Jordan algebra comes from theanticommutator. Call a Jordan algebra special if it is a Jordan subalgebra of an asso-ciative algebra with the circle product. Non-special Jordan algebras exist and are calledexceptional. An example is:H3(O) = {A ∈M3(O) : A¯t = A}=λ1 x2 x3x¯2 λ2 x4x¯3 x¯4 λ3 : λi ∈ R, xi ∈ O ,H3(O) is thus 27-dimensional. The operation is again the anticommutator (but not on anassociative algebra). The basic facts are:Theorem 7. H3(O) is an exceptional Jordan algebra. Any irreducible, exceptional Jordanalgebra (over R) is isomorphic to H3(O).This is exceedingly difficult to prove. H3(O) was shown to be a Jordan algebra byJordan-vonNeumann-Wigner in 1934. Albert, still in 1934, showedH3(O) was exceptional.Zelmanov (1978) proved the rest. Here we will show why H3(O) is a Jordan algebra andwhy Hn(O), n ≥ 4, is not. (When n = 1, 2 the hermitian matrices do form a Jordanalgebra but they are special.)Lemma 8. Hn(O) is closed under the anticommutator ◦.Proof. Let A = (xij ) and B = (yij ) be in Hn(O). Note xij = xji. Then:2(A ◦B)ij =n∑k=1xikykj + yikxkj=n∑k=1ykj xik + xkj yik=n∑k=1yjkxki + xjkyki= (BA +AB)ji = 2(A ◦B)ji.Thus A ◦Bt = A ◦B. Proposition 8. Hn(O), n ≥ 3 is a Jordan algebra iff n = 3.Proof. (JA1) clearly holds. (JA2), a2(ab) = a(a2b), is linear in b and so suffices to checkthis on vectors in the obvious basis, namely Eii and B = yEij + y¯Eji, y ∈ O. Weonly do the more difficult second case. To save on notation, we take i = 1, j = 2. LetA = (xij ) ∈ Hn(O) and write A2 = (zij ). We want to check ifA2(AB +BA) + (AB +BA)A2 = A(A2B +BA2) + (A2B +BA2)A.7NowAB +BA =x12y¯ x11y 0 . . . 0x22y¯ x21y 0 . . . 0x32y¯ x31y 0 . . . 0...... 0 . . . 0+yx21 yx22 yx23 . . .y¯x11 y¯x12 y¯x13 . . .0 0 0 . . ..........=x12y¯ + yx21 x11y + yx22 yx23 yx24 . . ..x22y¯ + y¯x11 x21y + y¯x12 y¯x13 y¯x14 . . .x32y¯ x31y 0 0 . . .x42y¯ x41y 0 0 . . ....... 0 0 . . .The identity holds at the entries (ij) with i ≤ 2 or j ≤ 2. For i, j ≥ 3, the entries are:[A2(AB +BA)]ij = zi1(yx2j ) + zi2(y¯x1j)[(AB +BA)A2]ij = (xi2y¯)z1j + (x1iy)z2j[A(A2B +BA2)]ij = xi1(yz2j) + xi2(y¯z1j)[(A2B +BA2)A]ij = (zi2y¯)x1j + (zi1y)x2j .Subtracting the second two from the sum of the first two gives that (JA2) holds iff−[zi1, y, x2j ]− [zi2, y¯, x1j ] + [xi2, y¯, z1j ] + [xi1, y, z2j ] = 0−[z1i, y, x2j ]− [z2i, y¯, x1j ] + [x2i, y¯, z1j ] + [x1i, y, z2j ] = 0 (hermitian)[z1i, y, x2j ]− [z2i, y¯, x1j ] + [x2i, y, z1j ]− [x1i, y, z2j ] = 0 Lemma 3.First suppose that n = 3. Then i = j = 3. Then H3(O) is a Jordan algebra iff[z13, y, x23]− [z23, y, x13] + [x23, y, z13]− [x13, y, z23] = 0,which holds by the alternative property of O.Next suppose n ≥ 4. Take i = 3 and j = 4. Then Hn(O) is a Jordan algebra iff:[z13, y, x24]− [z23, y, x14] + [x23, y, z14]− [x13, y, z24] = 0,But takeA =0 0 a 00 0 c 0a¯ c¯ 1 10 0 1 1 .We have:x14 = x24 = 0 x13 = z14 = a x23 = z24 = c.So (JA2) holds iff [c, y, a]− [a, y, c] = 0 iff 2[c, y, a] = 0 by the alternative property. But ifthis held for all c, y, a then O would be associative, which it is not. We can give the second construction: F4 = Aut(H3(O)). These can be viewed as27× 27 matrices. As automorphisms fix the matrices of trace 0, there is a 26-dimensionalrepresentation.84. Constructing the exceptional Lie algebras.A derivation of an algebra A is a linear map D : A→ A such thatD(xy) = xD(y) +D(x)y,for all x, y ∈ A (i.e. it satisfies the product rule). For an example, if A is associative anda ∈ A then Da(x) = ax− xa is a derivation:xDa(y) +Da(x)y = x(ay − ya) + (ax − xa)y= axy − xya = Da(xy).For another example, when A is only assumed to be alternative, define `a(x) = ax andra(x) = xa. ThenDa,b = [`a, `b] + [`a, rb] + [ra, rb]is a derivation (with usual Lie brackets). It is extremely tedious to check this.Our construction is due to Tits (1966). Let A be one of the composition algebras andlet B = H3(O), the exceptional Jordan algebra. Let A0 be the elements of A with realpart zero. Let B0 be the elements of B of trace zero. Define new products:for A0 a ? b = ab −Re(ab)for B0 x ? y = xy − 13Tr(xy).SetL(A,B) = Der(A) ⊕ (A0 ⊗B0)⊕Der(B),with the product, for a, b ∈ A0, x, y ∈ B0, D1 ∈ Der(A), D2 ∈ Der(B),usual Lie bracket on Der(A) and Der(B)[a⊗ x,D1 +D2] = D1(a) ⊗ x+ a⊗D2(x)[a⊗ x, b ⊗ y] = 112Tr(xy)Da,b + (a ? b) ⊗ (x ? y) + Re(ab)[rx, ry ]Then L(A,B) is a Lie algebra. We have:g2 = Der(O)f4 = Der(H3(O)) = L(R,H3(O))e6 = L(C,H3(O))e7 = L(H,H3(O))e8 = L(O,H3(O)).One can also do this by fixing A = O and letting B vary. The full table is known asFreudenthal’s Magic Square:L(A,H3(B)) =A\B | R C H OR | a1 a2 c3 f4C | a2 a2 ⊕ a2 a5 e6H | c3 a5 d6 e7O | f4 e6 e7 e8

Octonions

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Octonions

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