The Octonions and G 2

Transcription

1 Lie Groups The Octonions and G 2 1 Introduction Let A be an algebra with unit. A is a composition algebra if its quadratic form, N, is multiplicative. One can show that the dimension of any composition algebra is a power of two. Proposition 1.1. Let H be a composition subalgebra of A and x A with x orthogonal to H. Then H xh is a composition algebra if H is associative. Theorem 1.2. (Hurwitz) The only composition algebras are R, C, H, and O. Proof. The smallest composition algebra is R. Since R is associative and i is orthogonal to R, C = R ir is a composition algebra. Now as C is associative and j is orthogonal to {1,i}, hencetoc, H = C jc is a composition algebra. Finally, since H is associative and is orthogonal to {1,i,j,k}, hencetoh, O = H H is a composition algebra. The process of doubling the composition algebra to obtain another composition algebra ends with the octonions because O is not associative. Since the dimension of a composition algebra is a power of two, these are the only composition algebras. Let H be a subalgebra of the composition algebra A and x A orthogonal to H. Then for a, b, c, d H, a(xd) =x(ād) (1) (xb)c = x(cb) (2) (xb)(xd) = d b (3) It now follows that H xh is closed under product and for a, b, c, d H we have (a + xb)(c + xd) =(ac d b)+x(cb +ād) (4) Proposition 1.3. Let A = H xh be a composition algebra, (x A is orthogonal to H A). Let J be another subalgebra of A, τ : H J an algebra isomorphism, and y A orthogonal to J. Then τ extends to σ Aut(A) with σ(a + xb) =τ(a) +yτ(b), hence σ(x) =y Proof. This is the only way one can extend τ to A. Sinceτ is R-linear, we have that σ is as well. Furthermore, τ is bijective and H xh = J yj, soσ is bijective as well. We have 1

2 only to check that σ(gh) =σ(g)σ(h) for g, h A. Using (4) on both (H, x) and (J, y), we have that σ((a + xb)(c + xd)) = σ (ac d b)+x(cb +ād) = τ(ac d b)+yτ(cb +ād) = τ(a)τ(c) τ(d)τ( b)+y()τ(b)+τ(ā)τ(d)) = τ(a)τ(c) τ(d)τ(b)+y()τ(b)+τ(a)τ(d)) =(τ(a)+yτ(b))(τ(c)+yτ(d)) = σ(a + xb)σ(c + xd). An example using this proposition is taking O = H H and H = J = H. Clearly, id : H H is an algebra isomorphism and is orthogonal J = H, so we can extend the identity map to σ Aut(O) such that σ() =. 2 The Octonions The octonions are an 8-diminensional vector space over R. O = R Ri Rj Rk Rk Rj Ri R, where {i, j, k, k, j, i, } all square to 1. Multiplication of the octonions is given by the triangle below. You can think of each line as circle, with multiplication of elements on the circle the same as for the circle with {i, j, k} illustrating multiplication within the quaternions. Each circle anticommutes as the quaternions do. For example, k j = i, butj k = i. 2

3 As previoulsy mentioned, the octonions have an additional failed property: they are not associative. For example, (i j) j = h j = i i(j j)=i ( ) = i Actually, very composition algebra is alternative, so O is an alternative algebra. An algebra A is alternative if x(xy) =(xx)y and (yx)x = y(xx) for all x, y A. Ifu, v are orthogonal, imaginery unit vectors, then the subalgebra generated by {1,u,v,uv} is isomorphic to H. Thus, products involving no more than two independent octonions are associative. The commutator for the octonions is defined as [a, b] =ab ba for a, b O. Wedefinethe associator for a, b, c O, tobe[a, b, c] =(ab)c a(bc). The associator measures how badly a product fails to be associative. Both the commutator and associator are antisymmetric. For example, [i, j, j] =i ( i) =2i [j, i, j] = i i = 2i. Conjugation is defined the same as for the quaternions, by negating the imaginery terms. If a = x 0 + x 1 i + x 2 j + + x 7 O, thenā = x 0 x 1 i x 2 j x 7 O. Furthermore, the norm N : O R is defined as N(a) = a 2 = aā = x x x2 7, and for a, b O, ab = bā. It follows that N(ab) =N(a)N(b) and O has no zero divisors. However, O is not a domain as multiplication is not associative. Let V be the orthogonal complement of 1 in O and S 6 = {x V : N(x) =1} R 7. Define G 2 = Aut(O). The next two sections will show that SO 4 and SU(3) are subgroups of G 2. 3 SU(3) Since N(σ(x)) = N(x) for σ Aut(O) and x O, G 2 acts on S 6. This action is actually transitive. Take x S 6 and then choose u S 6 such that u is orthogonal to x. Thenpick v S 6 such that (x, v) =(u, v) =(ux, v) = 0, where (, ) denotes the inner product. It follows that (x, uv) = 0, and so x is orthogonal to the subalgebra H O generated by {1,u,v,uv}. As H H there exists an algebra isomorphism, τ : H H. Furthermore, O = H H, so we can apply Proposition 1.3 and extend τ to σ Aut(O) =G 2 such that σ() =x. Thus,G 2 acts transitively on S 6. Let G be the stabilizer of O and W be the orthogonal complement of in V. We have that (w, 1) = 0 for all w W,soW V. Furthermore, (w, ) = 0 for all w W, so W W. As W and W have the same dimension, W = W. Take the basis of W to be {i, j, k}. IfweviewC = R R O, thenw is a C-vector space. Define the Hermition inner product on W,by<v,w>=(v, w)+(v, w), for v, w W. 3

4 Now take σ G.Sinceσ() = and automorphisms preserve the inner product, we have <σ(v),σ(w) > =(σ(v),σ(w)) + (σ(v),σ(w)) =(σ(v),σ(w)) + (σ(v),σ(w)) =(v, w)+(v, w) =< v,w>. Thus, G U(W ) U(3). Now take σ G (W ) There exist eigenvalues u, v, w W that are orthonormal with respect to the Hermitian inner product. Since u, v, and uv are orthonormal with respect to the Hermitian form, we can take w = uv. Automorphisms preserve the norm, so the eigenvalues of these eigenvectors must have norm 1. Recall that W is an R R = C-vector space, so the eigenvalues of u, v are e θ and e φ,respectively(θ, φ R). Therefore, σ(uv) =σ(u)σ(v) = e θ ue φ v = (cos θu + sinθu)(cos φv + sinφv) = cos θ cos φ(uv)+sinθ sin φ(u)(v) + cos θ sin φu(v)+sinθ cos φ(u)v = cos θ cos φ(uv)+sinθ sin φ( vū) + cos θ sin φ(ūv)+sinθ cos φ(vu) = cos θ cos φ(uv) sin θ sin φ(uv) cos θ sin φ(uv) sin θ cos φ(uv) = (cos(θ + φ) sin(θ + φ))uv = e (θ+φ) uv With respect to the basis {u, v, uv}, σ is a diagonal matrix with the diagonal entries being the respective eigenvalues. Therefore, det σ = e θ e φ e (θ+φ) = e 0 = 1, and so G SU(W ) SU(3). Actually, if σ Aut(O) such that σ(i) =e α,σ(j) =e β,σ(k) =e γ,whereα+β+γ = 0, and σ() =, thenσ is an element of the maximal torus of T of G, which turns out to also be a maximal torus of G 2. Next we show that the dimension of G is 8, and so SU(3) G G 2. From above, we know that automorphisms that fix and k and send i to e α i and j to e α j, are in G (see figure below). This family of automorphisms rotates the points on the two lines of the multiplication triangle that point to k and do not contain. It is a one-parameter subgroup corresponding to a line in g. 4

5 We have five other family of automorphisms that fix by looking at the respective lines and rotations for the other five basis vectors of O that are not equal to. There are three families of automorphisms that rotate points on the lines pointing to, while fixing (see figure below). Only two of these three rotations are linearly independent. In fact, g is a direct sum of the eight lines corresponding to these families of automorphisms. 5

6 We can now use the dimension of G to find the dimension of G 2. Since G 2 acts transitively on S 6 and S 6, S 6 is homeomorphic to G 2 /G.Thus,g 2 = g T (S 6 ) and dim G 2 =dimg +dims 6 = = 14. In fact, G 2 has the smallest dimension of the exceptional Lie groups. 4 SO 4 Let G H be the stabilizer of H O. For σ G H, σ(h) =H, so we have the restriction map r : G H O(H) definedbyr(σ) =σ H.Ifσ acts trivially on H then σ fixes and u for all u H. Thus, u = σ(u) =σ(u)σ() =σ(u) and we have σ(u) =u for all u H. Therefore, the restriction map is injective and G H O(H) O 4. Now look at the restriction map r : G H Aut(H) =SO 3 defined by r (σ) =σ H. Take τ Aut(H), then by Proposition 1.3, τ extends to σ Aut(O) such that σ() = and σ(h) =τ(h) =H. Thus,σ G H with r (σ) =τ, sor is surjective. Let K =kerr and take σ K. As σ fixes all elements of H, it is completely determined by σ(). As σ preserves the inner product and is orthogonal to H, wehaveσ() H. So σ() =u σ, some u σ H. As σ preserves the norm, 1 = N() =N(σ()) = N(u σ )= N(u σ )N() =N(u σ ), and so u σ S 3 H. 6

7 Now take τ also in K. We will show that u τ σ = u τ u σ : u τ σ = τ(σ()) = τ(u σ )=τ(u σ )τ() =u σ (u τ )=u σ (ū τ )=(baru σ ū τ )=(u τ u σ )=(u τ u σ ). Furthermore, σ() = for σ K if and only if σ = id. Thus, φ : K S 3 defined by φ(σ) =u σ is an injective homomorphism. To show that φ is also surjective, we take u S 3 and use Proposition 1.3 again. Since is orthogonal to H, u is orthogonal to H, so we can extend the identity map on H to σ Aut(O) such that σ() =u and σ H = id so σ K. Thus, we have the short exact sequence, 1 S 3 G H SO 3 1. Since S 3 and SO 3 are connected, G H is as well. Thus, G H O 4 must actually be contained in the connected component of O 4, SO 4. Furthermore, dim G H = dims 3 + dim SO 3 =3+3=6=dimSO 4,soSO 4 G H G 2. 5 Root System The root system for SU(3) and SO 4 can be seen in the root system of G 2. 7