Sp(n), THE SYMPLECTIC GROUP CONNIE FAN 1. Introduction t ¯ 1.1. Definition. Sp(n)= (n, H)= A Mn(H) A A = I is the symplectic group. O { 2 | } 1.2. Example. Sp(1) = z M1(H) z =1 { 2 || | } = z = a + ib + jc + kd a2 + b2 + c2 + d2 =1 { | } 3 ⇠= S 2. The Lie Algebra sp(n) t ¯ 2.1. Definition. AmatrixA Mn(H)isskew-symplecticif A + A =0. 2 t ¯ 2.2. Definition. sp(n)= A Mn(H) A + A =0 is the Lie algebra of Sp(n) with commutator bracket [A,B]{ 2 = AB -| BA. } Proof. This was proven in class. ⇤ 2.3. Fact. sp(n) is a real vector space. Proof. Let A, B sp(n)anda, b R 2 2 (aA + bB)+taA + bB = a(A + tA¯)+b(B + tB¯)=0 ⇤ 1 2.4. Fact. The dimension of sp(n)is(2n +1)n. Proof. Let A sp(n). 2 a11 + ib11 + jc11 + kd11 a12 + ib12 + jc12 + kd12 ... a1n + ib1n + jc1n + kd1n . A = . .. 0 . 1 an1 + ibn1 + jcn1 + kdn1 an2 + ibn2 + jcn2 + kdn2 ... ann + ibnn + jcnn + kdnn @ A Then A + tA¯ = 2a11 (a12 + a21)+i(b12 b21)+j(c12 c21)+k(d12 d21) ... (a1n + an1)+i(b1n bn1)+j(c1n cn1)+k(d1n dn1) − − − − − − (a21 + a12)+i(b21 b12)+j(c21 c12)+k(d21 d12)2a22 ... (a2n + an2)+i(b2n bn2)+j(c2n cn2)+k(d2n dn2) − − − − − − 0 . 1 . .. B . C B C B(a1n + an1)+i(b1n bn1)+j(c1n cn1)+k(d1n dn1) ... 2ann C @ − − − A t 2 A + A¯ =0,so: axx =0, x 0degreesoffreedom 8 ! n(n 1) − axy = ayx,x= y 2 degrees of freedom − 6 !n(n 1) − bxy = byx,x= y 2 degrees of freedom 6 ! n(n 1) − cxy = cyx,x= y 2 degrees of freedom 6 ! n(n 1) − dxy = dyx,x= y 2 degrees of freedom b ,c ,d unrestricted,6 ! x 3n degrees of freedom xx xx xx 8 ! In total, dim(sp(n)) = 2n2 + n = n(2n +1) ⇤ 2.5. Proposition. If A sp(n), then eA Sp(n), ie. exp : sp(n) Sp(n) 2 2 ! t t Proof. I = e0 = eA+ A¯ = eAe A¯ = eA teA · ⇤ 2.6. Proposition. If G = Sp(n), then TG = sp(n) Proof. The proof was done in class. T skew matrices and skew matrices T ,soT = skew matrices . G ⇢{ } { }⇢ G G { } ⇤ 2.7. Corollary. dim(Sp(n)) = (2n +1)n Proof. dim(Sp(n)) = dim(TSp)=dim(sp(n)) = (2n +1)n ⇤ 3. Interesting Isomorphisms 3.1. Invariants. In order for a group to be isomorphic to Sp(n), it must have the same invariants to preserve structure. Rank and dimension are numerical invariants. The center is a subgroup invariant. When comparing Sp(n) to other matrix groups with the same rank for some rank 4, dim U < dim SU < dim SO(even) < dim SO(odd) = dim Sp [1] ≥ Thus, for odd dimensions, SO(n)= A O(n) det A= 1 might be isomorphic. However, SO(n) has center I , while Sp(n) has{ center2 1,| so that means} for rank 4, Sp(n) is not isomorphic to U(n),{ } SU(n), and SO(n). ± ≥ What about for rank 1, 2, and 3? For rank 1: Sp(1), SU(2), and SO(3) all have dimension 3. Sp(1) and SO(3) are not isomor- phic because they have di↵erent centers, but as proved in the homework [PS 2], SO(3) = Sp(1)/ I . ⇠ {± } 3.2. Proposition. Sp(1) = SU(2) = A U(2) det A= 1 ⇠ { 2 | } Proof. Let z Sp(1), z = a + ib + jc + kd. 2 In our homework [PS 1] we showed the map φ : H M2(C)givenby a + ib c + id ! φ(a + ib + jc + kd)= c + id a ib is an injective algebra✓ homomorphism.− − ◆ To show that the map is also an isomorphism φ : Sp(1) SU(2), I show 1) z,z Sp(1), φ(z) SU(2) ! Let8z = 2a + ib + jc + k2d.Thena2 + b2 + c2 + d2 =1. 3 a + ib c + id a ib c id φ(z)tφ(z)= c + id a ib c − id− a +−ib ✓− − ◆✓ − ◆ a2 + b2 + c2 + d2 0 = 0 a2 + b2 + c2 + d2 ✓ ◆ 10 = 01 ✓ ◆ det(φ(z)) = 1. Thus, φ(z) SU(2). 2 2) Next I show φ is surjective, ie. for every A SU(2), there is some z Sp(1) such that A = φ(z). 2 2 ↵ Let A = SU(2),↵,δ,β,γ C γδ2 2 det(A)=1,✓ so ↵◆ βγ =1. − Also, A SU(2) U(2) = (2, C), so the rows of A form an orthonormal basis for C2. Thus, δ =¯2 ↵ and γ⇢= β¯. O Let ↵ = a + ib and β −= c + id,thensetz = a + ib + jc + kd. z Sp(1) since a2 + b2 + c2 + d2 = 1. Also, A = φ(z). 2 ⇤ For rank 2, Sp(2) and SO(5) both have dimension 10. For rank 3, Sp(3) and SO(7) both have dimension 21. Again, however, the centers are di↵erent so they are not isomorphic. There are some isometries to groups that we did not go over in class. Of particular interest are isometries to the Spin(n) group. k 3.3. Definition. The real algebra Ck (called the Cli↵ord algebra)ofdimension2 is generated by e ,e ,...,e such that e2 = 1ande e = e e if i = j. 1 2 k i − j i − i j 6 3.4. Example. C0 = R For C1,letthebasisbe 1,e . Let 1 be the multiplicative{ identity.} e2 = 1. Multiplication is (a + be)(c + de)=(ac bd)+(ad + bc)e − − So C1 ⇠= C k 1 3.5. Proposition. If Ck⇤ denotes the group of units in Ck,thenS − Ck⇤. k 1 ⇢ (S − is the unit sphere in R) Proof. See [1] p. 135. ⇤ k 1 3.6. Definition. Pin(k) is the subgroup of Ck⇤ generated by S − . 4 3.7. Definition. ↵(e )= e is an automorphism of C i − i k 3.8. Definition. For u P in(k)andx Rk, 2 2 ⇢(u)(x)=↵(u)xu⇤ *isconjugationinCk k 1 3.9. Proposition. If u S − Pin(k) and u = 1, 2 ⇢ 6 ± then ⇢(u)isreflectioninRk in the hyperplane perpendicular to u. Proof. See [1] p. 136. ⇤ 1 3.10. Definition. Spin(k)= ⇢− (SO(k)) From a topological perspective, Spin(n) is the double cover of SO(n). 3.11. Example. C1 = C C1⇤ = C 0 0 −{ } S = e1, e1 P in(1){ = −e ,e} 2 = 1,e3 = e ,e4 =1 { 1 1 − 1 − 1 1 } ⇢(e1)=⇢( e1)isthereflection Spin(1) =−1, 1 { − } 2 2 2 2 Spin(3) = a + be1e2 + ce1e3 + de2e3 a + b + c + d =1 The assignment{ | } e e i 1 2 7! e1e3 j e e 7! k 2 3 7! gives an isomorphism Spin(3) ⇠= Sp(1) Also, Spin(4) = Sp(1) Sp(1) ⇠ ⇥ Spin(5) ⇠= Sp(2) n(n 1) − Like SO(n), Spin(n) has dimension 2 . Also, the center of Spin(n) for odd n is 1 , so Spin(2n+1) is a good candidate for isomorphism with Sp(n). However, for n 3thisis{± } ≥ not the case. If Spin(2n+1) Sp(n), then Spin(2n+1) Sp(n) .Thenormalizerofamaximal ⇠= center ⇠= center Sp(n) Spin(2n+1) torus in center splits for n=1,2, while the normalizer in a maximal torus for center splits for all n Z+. 2 Proof. See chapter 11 of [1]. ⇤ 3.12. Summary. Here is a summary of isomorphisms of Sp(n) that were mentioned: 3 Sp(1) ⇠= S , SU(2), Spin(3) Sp(2) ⇠= Spin(5) SO(3) = Sp(1)/ I ⇠ {± } Spin(4) ⇠= Sp(1) Sp(1) Basically, the isomorphisms⇥ occur in low dimensions and are “accidental”. 5.