LIE GROUPS, CH.1 p.1 Syllabus Lie Groups, General Theory by T. H. Koornwinder Korteweg-de Vries Institute, Faculty of Science, Universiteit van Amsterdam January – February 2001 Part of these notes closely follow Chapters 1, 2 and 5 in G. Segal’s lecture notes on Lie groups, which form the middle part of the book: R. Carter, G. Segal and I. Macdonald, Lectures on Lie groups and Lie algebras, London Mathematical Society Student Texts 32, Cambridge University Press, 1995. Chapter 1. Generalities about topological groups 1.1 Definition A topological group is a set G which is both a group and a topological space such that the group operations (x,y) 7→ xy:G×G → G and x 7→ x−1:G → G are continuous. AnysubgroupofatopologicalgroupGbecomesatopologicalgroupintherelativetopology of G. A map φ:G → H from a topological group G to a topological group H is called a ho- momorphism of topological groups if φ is both a group homomorphism and a continuous map. Two topological groups G and H are called isomorphic if there is an invertible map φ:G → H such that both φ and φ−1 are homomorphisms of topological groups. 1.2 Examples of topological groups (a) Any group with the discrete topology (a so-called discrete topological group). In particular, any finite group will be silently assumed to be a topological group in this way. (b) Rn as additive group with the topology of Rn. Similarly Cn as additive group with the topology of Cn. Note that Cn is isomorphic to R2n as a topological group. (c) GL(n,R), the group of n×n real invertible matrices with topology as subset of Rn2 (by considering the matrix elements T of T ∈ GL(n,R) as coordinates). ij (d) GL(n,C), the group of n×n complex invertible matrices with topology as subset of Cn2 (by considering the matrix elements T of T ∈ GL(n,C) as coordinates). ij (e) Any subgroup of GL(n,R) or of GL(n,C) in the relative topology. Among such subgroups the closed subgroups (subgroups G which are closed subsets of GL(n,R) or GL(n,C)) have much nicer properties than the non-closed subgroups, and we will mostly restrict attention to these closed subgroups. (f) Thegroups SL(n,R) := {T ∈ GL(n,R) | detT = 1} andSL(n,C) := {T ∈ GL(n,C) | detT = 1}. (g) O(n), the group of real orthogonal n×n matrices, and its subgroup SO(n) := {T ∈ O(n) | detT = 1}. (h) U(n) the group of complex unitary n×n matrices and its subgroup SU(n) := {T ∈ U(n) | detT = 1}. LIE GROUPS, CH.1 p.2 Ex. 1.3 Prove that GL(n,R) is a closed subgroup of GL(n,C) (when embedded in GL(n,C) in the obvious way). Show also that GL(n,C) is isomorphic as topological group to a certain closed subgroup of GL(2n,R). Show next that every closed subgroup of GL(n,R) is isomorphic as topological group to some closed subgroup of GL(n,C) and that every closed subgroup of GL(n,C) is isomorphic as topological group to some closed subgroup of GL(2n,R). Ex. 1.4 Show for all examples in §1.2 that these are topological groups. Show that the examples (f), (g), (h) are closed subgroups of GL(n,R) or GL(n,C). 1.5 Proposition Let G be a topological group. If {e} is a closed subset of G then G as a topological space is Hausdorff. Ex. 1.6 Prove Proposition 1.5. 1.7 Remark An important class of topological groups are the locally compact groups: these are topological groups which are locally compact Hausdorff spaces. A subclass if formed by the compact groups: these are topological groups which are compact Hausdorff spaces. Ex. 1.8 Show that the discrete topological groups are locally compact and that the compact groups among these are precisely the finite groups. Show also that GL(n,R) and GL(n,C) and all their closed subgroups are locally compact. 1.9 Proposition Let G be a topological group. Then the connected component G 0 of G containing e is a closed normal subgroup of G. The whole collection of connected components of G is precisely the collection of left cosets of G with respect to G . 0 1.10 Example The group G := O(n) is not connected, since the continuous map det:G → R maps G onto the subset {−1,1} of R, which is not connected. It can be shown that here G = SO(n). For n := 2 this is seen very easily. For general n we prove 0 below that SO(n) is connected. 1.11 Definition Let G be a group and X a set. A group action of G on X is a map (g,x) 7→ g ·x:G×X → X such that (a) (gh)·x = g ·(h·x) for all g,h ∈ G and x ∈ X; (b) e·x = x for all x ∈ X. If G is a topological group and X a topological space and if the above mapping (g,x) 7→ g · x:G×X → X is continuous then we call the group action a topological group action or a continuous group action. If a group action of G on X satisfies the property that for each x,y ∈ X there exists g ∈ G such that g ·x = y then we call the group action transitive. If G is a group with subgroup H then we denote by G/H the collection of left cosets ′ ′ gH (g ∈ G). Then G acts on G/H by the rule g · (gH) := (g g)H and this action is transitive (check this). Moreover, for g ∈ G we have: g · (eH) = eH iff g ∈ H. The set G/H with G acting on it is called a homogeneous space. Conversely, if we have a transitive group action of G on X and if we fix x ∈ X 0 then H := {g ∈ G | g · x = x } is a subgroup of G (the so-called stabilizer of x in G) 0 0 0 LIE GROUPS, CH.1 p.3 ′ and the map g · x 7→ gH:X → G/H is a well-defined bijection such that g · (g · x ) is 0 0 ′ mapped by this bijection to g ·(gH) (i.e., the actions of G on X and on G/H correspond with each other under this bijection). If G is a topological Hausdorff group and X is a topological Hausdorff space and if the transitive group action of G on X is continuous then the stabilizer subgroup H of x in G is a closed subgroup. 0 1.12 Example Some examples of transitive continuous actions of groups are: (a) The symmetric group S acting on the set {1,2,...,n}. The stabilizer of 1 in S is n n isomorphic to Sn−1. (b) The group O(n) acting on the unit sphere Sn−1 in Rn. The stabilizer subgroup of (1,0,...,0) is the group O(n−1) which is embedded as a closed subgroup of O(n) by 1 0 ... 0 0   T 7→ . :O(n−1) ֒→ O(n). . . T   0    (c) The group SO(n) acting on the unit sphere Sn−1 in Rn. The stabilizer subgroup of (1,0,...,0) is the group SO(n−1) which is embedded as a closed subgroup of SO(n) as in (b). Ex. 1.13 Prove the statements in the last two paragraphs of Definition 1.11. Prove also that the examples in §1.12 indeed give transitive continuous actions of topological groups and that the stabilizer subgroups are the ones mentioned there. 1.14 Proposition The group SO(n) is connected. Proof Clearly, SO(2) is connected. The proof will use induction with respect to n. The induction hypothesis will follow by the existence of a continuous surjective map SO(n−1)×SO(2)×SO(n−1) → SO(n). This map is obtained by observing that each T ∈ SO(n) can be written as cosθ −sinθ 0 ... 0 1 0 ... 0 1 0 ... 0 sinθ cosθ 0 ... 0   0 0     T = . 0 0 1 0 . .. A  .. .. ..  .. B   . . .   0  0    0 0 0 1    for some A,B ∈ SO(n−1) and θ ∈ [0,π). We prove this last statement. Let T ∈ SO(n) and let e ,...,e be the standard basis of Rn. Then Te = cosθe + sinθξ for some 1 n 1 1 θ ∈ [0,π) and ξ = (0,ξ ,...,ξ ) with |ξ| = 1. Then, by transitivity of the action of 2 n SO(n−1) on Sn−2, there exists A ∈ SO(n−1) such that cosθ −sinθ 0 ... 0 1 0 ... 0 sinθ cosθ 0 ... 0   0   Te = . 0 0 1 0 e . 1 .. A  .. .. ..  1   . . .  0     0 0 0 1   ′ Denote the product of the two matrices on the right-hand side of the last identity by T . Then T′ ∈ SO(n) and (T′)−1Te = e . Hence, since SO(n−1) is the stabilizer of e in 1 1 1 1 0 SO(n), there exists B ∈ SO(n−1) such that (T′)−1T = . (cid:18)0 B(cid:19) LIE GROUPS, CH.1 p.4 1.15 The group of isometriesof Rn is denoted by E(n). Then E(n) = {τ A | b ∈ Rn, A ∈ b O(n)}, where τb:v 7→ v + b:Rn → Rn denotes a translation. Note that τbA = τb′A′ iff b = b′ and A = A′, so, as sets, we can identify τ A ↔ (b,A):E(n) ↔ Rn ×O(n). Via this b bijection we can transplant the group operations (τbA)(τb′A′) = τAb′+b(AA′), unit τ0I, (τbA)−1 = τ−A−1bA−1 on E(n) to group operations (b,A)(b′,A′) = (b+Ab′,AA′), unit (0,I), (b,A)−1 = (−A−1b,A−1) on Rn ×O(n). Now Rn ×O(n), equipped with the product topology from the topologies of Rn and O(n), becomes a topological group (check this). Hence E(n) receives via the bijection a topology by which it becomes a topological group. 1.16 Definition Let H and N be groups, let Aut(N) be the group of automorphisms of N, and let α:H → Aut(N) be a group homomorphism. Then the semidirect product H ⋉N is the set N ×H made into a group with group operations (n ,h )(n ,h ) = (n α(h )n ,h h ), unit (e,e), (n,h)−1 = ((α(h−1)n)−1,h−1). 1 1 2 2 1 1 2 1 2 Now we have group embeddings n 7→ (n,e):N ֒→ H ⋉N as a normal subgroup, h 7→ (e,h):H ֒→ H ⋉N a s a subgroup. Thus we can write n instead of (n,e) and h instead of (e,h), and we can write a general element of H ⋉N as nh (n ∈ N, h ∈ H), since nh = (n,e)(e,h) = (n,h). Then we have hnh−1 = α(h)n as shorthand for (e,h)(n,e)(e,h)−1 = (α(h)n,e) (h ∈ H, n ∈ N). IfN andH aretopologicalgroupsandifthemap(n,h) 7→ α(h)n:N×H → N iscontinuous then N ×H equipped with the product topology becomes a topological group. 1.17 Definition The Heisenberg group is the group N defined as the topological space R3 embedded as a subgroup of GL(3,R) by 1 a c (a,b,c) 7→ 0 1 b. 0 0 1   The group structure on N as a subgroup of GL(3,R) then defines the following structure of topological group on R3: (a,b,c)(a′,b′,c′) = (a+a′,b+b′,c+c′+ab′), unit (0,0,0), (a,b,c)−1 = (−a,−b,−c+ab). Then the center of N equals {(0,0,c) | c ∈ R}. Let Z be the discrete subgroup {(0,0,c) | c ∈ Z} of the center. Then Z is a normal subgroup of N. In some literature the quotient group N/Z is called the Heisenberg group. Ex. 1.18 For a,b,c ∈ R define unitary operators T , M and U acting on the Hilbert a b c space L2(R) as follows: (T f)(x) := f(x−a), (M f)(x) := e2πibxf(x), (U f)(x) := e2πicf(x). a b c Show that the map (a,b,c) 7→ TaMbU−c+ab defines a group homomorphism from the Heisenberg group N (and also from N/Z) into the group of unitary operators on L2(R). LIE GROUPS, CH.2 p.1 R Chapter 2. SU(2), SO(3) and SL(2, ) 2.1 Definition The space of quaternions is a four-dimensional real vector space H with basis 1, i, j, k. We introduce an associative multiplication on H by the product rules i2 = j2 = k2 = −1 and ij = −ji = k, jk = −kj = i, ki = −ik = j. Write a general element q of H as q = t + xi + yj + zk (t,x,y,z ∈ R). Define an involution q 7→ q∗ on H by (t +xi +yj + zk)∗ := t − xi− yj − zk. Then q 7→ q∗ is linear, (q∗)∗ = q, 1∗ = 1, and ∗ is anti-multiplicative: (q q )∗ = q∗q∗. (We say that H is a real associative ∗-algebra 1 2 2 1 with identity.) We write H as the direct sum H = R +R3, where R is the real span of 1 and R3 is the real span of i,j,k. In q = t + xi +yj + zk we call t = 1(q + q∗) ∈ R the 2 real part of q and v := xi + yj + zk = 1(q − q∗) ∈ R3 the vector part of q. Then put 2 |q|2 := qq∗ = q∗q = t2 +|v|2 = t2 +x2 +y2 +z2 (the squared norm). We have |q| = |q∗| and |q q | = |q ||q |. 1 2 1 2 2.2 Proposition We have the following correspondences: a b H ←→ | a,b ∈ C (cid:26)(cid:18)−b a(cid:19) (cid:27) t+ix y +iz q := t+xi+yj +zk ←→ Q := (R-linear and bijective) (cid:18)−y +iz t−ix(cid:19) q q ←→ Q Q 1 2 1 2 ∗ ∗ q ←→ Q t = 1 trQ 2 |q|2 = detQ |q| = 1 ⇐⇒ Q ∈ SU(2) The last correspondence gives a topological group isomorphism between the group of quaternions of unit norm and the group SU(2). Since the unit sphere in R4 is connected (and simply connected), the group SU(2) is connected (and simply connected). 2.3 Proposition For v ∈ R3 ⊂ H and g ∈ H\{0} put Tv := gvg−1. Then T is g g an orthogonal transformation of R3. In particular, the map g 7→ T is a continuous g homomorphism from SU(2) (identified with the group of quaternions of unit norm) to SO(3). Proof T is an orthogonal transformation of R3 since tr(GVG−1) = trV = 0 and g det(GVG−1) = det(V). The map g 7→ T is a group homomorphism since (g g )v(g g )−1 g 1 2 1 2 = g (g vg−1)g−1. It maps SU(2) into SO(3) since the continuous image of the connected 1 2 2 1 group SU(2) must lie inside the connected component SO(3) of O(3). 2.4 Theorem The map g 7→ T is a continuous surjective group homomorphism from g SU(2) onto SO(3) with kernel consisting of 2 elements ±I. Proof Use that (t +v )(t +v ) = t t −hv ,v i + t v +t v +v ×v , 1 1 2 2 1 2 1 2 1 2 2 1 1 2 (cid:0) (cid:1) (cid:0) (cid:1) LIE GROUPS, CH.2 p.2 where v ×v is the vector product in R3 of vectors v and v in R3. Any rotation T of R3 1 2 1 2 which is not the identity can be described by a unit vector u (yielding the rotation axis) and by its angle of rotation 2θ (0 < θ < π), with rotation taken positively with respect to the direction of u. There are precisely two possible ways of describing T in such a way, namely by u,2θ and by −u,2(π −θ). On the other hand, any quaternion of unit norm not equal to ±1 can be uniquely represented as cosθ +u sinθ, with u a unit vector in R3 and 0 < θ < π. We show that, starting with this representation, the map v 7→ (cosθ + u sinθ)v(cosθ + u sinθ)−1 is a rotation about u through angle 2θ. Indeed, (cosθ+u sinθ)v(cosθ−u sinθ) = hu,vi sin2θu+cos2θv−sin2θ(u×v)×u+sin(2θ)u×v. If v = u then this equals u. If v is a unit vector orthogonal to u then this equals cos(2θ)v+ sin(2θ)u×v. Thus we get the desired rotation, because in the latter case u, v and u×v form an orthonormal system of unit vectors in the right orientation. Thus the connected and simply connected group SU(2) is the two-sheeted covering of the connected group SO(3). We call SU(2) therefore the universal covering group of SO(3). 2.5 Theorem For g ,g ∈ SU(2) (identified with the group of quaternions of unit 1 2 norm) put T v := g vg−1 (v ∈ H). Then the map (g,g ) 7→ T is a continuous g1,g2 1 2 1 2 g1,g2 surjective group homomorphism from SU(2)× SU(2) onto SO(4) with kernel consisting of 2 elements ±(I,I). Proof T is an orthogonal transformation of R4 ≃ H since det(GVG−1) = detV. g1,g2 1 2 Themap(g ,g ) 7→ T isagrouphomomorphism: (g g′)v(g g′)−1 = g (g′v(g′)−1)g−1. 1 2 g1,g2 1 1 2 2 1 1 2 2 It maps SU(2) × SU(2) into SO(4) since the continuous image of the connected group SU(2)×SU(2) must lie inside the connected component SO(4) of O(4). Next we prove that the homomorphism is surjective. Let T ∈ SO(4) and put h := T1. Then h is a quaternion of unit norm, so h−1T1 = 1 and the map v 7→ h−1Tv:H → H is in SO(4). Hence the map v 7→ h−1Tv:H → H is in SO(3) (the subgroup of SO(4) which stabilizes the first basis vector 1 of H). Hence, by Theorem 2.4 there exists g ∈ H of unit norm such that h−1Tv = gvg−1 for all v ∈ H. Hence Tv = hgvg−1. Finally we prove that the kernel of the homomorphismconsists of 2 elements. Suppose that g ,g are quaternions of unit norm such that g vg−1 = v for all v ∈ H. For v := 1 1 2 1 2 this yields g = g . Now use the result on the kernel in Theorem 2.4. 1 2 Ex. 2.6 Let R1,n be the vector space Rn+1 with elements x = (x ,x ,...,x ) and 0 1 n 1,n indefinite (Lorentzian) inner product [x,y] := x y −x y −...−x y . Let H be the set 0 0 1 1 n n + {x ∈ R1,n | [x,x] = 1, x > 0} (the upper sheet of a two-sheeted hyperboloid). Let O(1,n) 0 consist of all T ∈ GL(n+1,R) acting on R1,n such that [Tx,Tx] = [x,x] for all x ∈ R1,n. Let SO (1,n) consist of all T ∈ O(1,n) such that detT = 1 and T := [Te ,e ] > 0. 0 0,0 0 0 (SO (1,n) is the Lorentz group for 1 time dimension and n space dimensions.) Show the 0 following. (a) O(1,n) and SO (1,n) are closed subgroups of GL(n+1,R). 0 LIE GROUPS, CH.2 p.3 1,n 1,n (b) (T,v) 7→ Tv:SO (1,n) × H → H is a transitive and continuous group action 0 + + 1,n of SO (1,n) on H , and the stabilizer of e in SO (1,n) is SO(n), where SO(n) is 0 + 0 0 1 0 embedded in SO (1,n) as T 7→ . 0 (cid:18)0 T (cid:19) (c) Every T ∈ SO (1,n) can be written as 0 cosht sinht 0 ... 0 1 0 ... 0 1 0 ... 0 sinht cosht 0 ... 0   0 0     T = . 0 0 1 0 . .. A  .. .. ..  .. B   . . .   0  0    0 0 0 1    for some A,B ∈ SO(n) and some t ≥ 0. Conclude that SO (1,n) is connected. 0 Hint This is analogous to the proof of Proposition 1.14. (d) G := O(1,n) has 4 connected components and G = SO (1,n). 0 0 2.7 Proposition Every T ∈ GL(n,C) can be uniquely written as T = UH, where U ∈ U(n) and H is a positive definite hermitian matrix. This is called the polar decomposition of T. The resulting map T 7→ (U,H) is continuous. So we have homeomorphisms (U,H) 7→ UH: U(n)×{n×n pos. def. hermitian matrices} → GL(n,C) (U,H) 7→ UH: SU(n)×{n×n pos. def. hermitian matrices of det. 1} → SL(n,C) (O,S) 7→ OS: O(n)×{n×n pos. def. real symmetric matrices} → GL(n,R) (O,S) 7→ OS: SO(n)×{n×n pos. def. real symmetric matrices of det. 1} → SL(n,R) (the three last homeomorphisms by restriction). 1 Proof (sketch) First, if H is a positive definite hermitian matrix then define H2 as the unique positive definite hermitian matrix which has H as its square. The existence and 1 1 uniqueness of H2 follow because H and H2 must have the same eigenvectors while the 1 eigenvalues of H2 must be the positive square roots of the eigenvalues of H. Also note that, if T ∈ GL(n,C) then T∗T is positive definite hermitian. If T ∈ GL(n,C) can be written as T = UH with U ∈ U(n) and H a positive definite hermitian matrix, then T∗T = (UH)∗UH = H∗U∗UH = HU−1UH = H2, so ∗ 1 ∗ −1 H = (T T)2, U = T (T T) 2. Conversely, if T ∈ GL(n,C) then T = (T (T∗T)−12)(T∗T)12 = UH with U,H as above. Moreover, H is then positive definite hermitian, as we already observed, and U is unitary, since T (T∗T −21)∗ T (T∗T)−21 = (T∗T −12 T∗T (T∗T)−12 = I. (cid:0) (cid:1) (cid:0) 1 (cid:1) (cid:1) ThecontinuityofthemapH 7→ H2 fromthesetofpositivedefinitehermitianmatrices ∞ to itself can be most easily seen by using the matrix exponential exp:A 7→ Ak/k! k=0 (A a square matrix) and the inverse of this map, denoted by log. Then it caPn be shown that exp is a homeomorphism of the space of hermitian matrices onto the set of positive definite matrices, and that H21 = exp(1 log(H)) (H positive definite). We will discuss this 2 exponential map in more detail later. LIE GROUPS, CH.2 p.4 2.8 Corollary The groups GL(n,C), SL(n,C) and SL(n,R) are connected. Proof ThisusesProposition2.7andthefactthatU(n), SU(n)andSO(n)areconnected. This last thing we did not yet prove except for SO(n) and for SU(2). For general U(n) and SU(n) it can be proved by using that there is a transitive action of U(n) and SU(n) on the unit sphere in Cn with stabilizer of e given by U(n−1) and SU(n−1), respectively. 1 2.9 Proposition We have the following correspondences: R1,3 ←→ {2×2 hermitian matrices} t+x y −iz (t,x,y,z) ←→ (R-linear and bijective) (cid:18)y +iz t−x (cid:19) t+x y −iz t = 1 tr 2 (cid:18)y +iz t−x (cid:19) t+x y −iz t2 −x2 −y2 −z2 = det (cid:18)y +iz t−x (cid:19) 1,3 H ←→ {2×2 pos. def. matrices of determinant 1} + 2.10 Theorem For g ∈ SL(2,C) and A a 2×2 hermitian matrix put T(A) := gAg∗. g Under the identification of the space of 2×2 hermitian matrices with R1,3 the map T is g a linear transformation of R1,3. Then the map g 7→ T is a surjective continuous group g homomorphism from SL(2,C) onto SO(1,3) with kernel consisting of 2 elements ±I. 0 The map g 7→ T restricted to g ∈ SU(2) is essentially the double covering homomorphism g SU(2) → SO(3) discussed in Theorem 2.4. Proof If A is a 2 × 2 hermitian matrix and g ∈ SL(2,C) then gAg∗ is hermitian and ∗ det(gAg ) = det(A). HenceT ∈ O(1,3). Themapg 7→ T iseasilyseentobea continuous g g homomorphism. Since SL(2,C) is connected and SO(1,3) is the connected component of 0 I in O(1,3), we see that T ∈ SO (1,3). g 0 t+x y −iz Next we will prove the surjectivity. Put Φ:(t,x,y,z) 7→ for the (cid:18)y +iz t−x (cid:19) map from R1,3 to the 2×2 hermitian matrices. In view of Exercise 2.6(c) the surjectivity will follow if for any A of the form coshs sinhs 0 0 sinhs coshs 0 0 1 0 A =   or A = (B ∈ SO(3)) 0 0 1 0 (cid:18)0 B(cid:19)  0 0 0 1   we can find an element g ∈ SL(2,C) such that T(Φ(v)) = Φ(Av) for all v ∈ R1,3. For A g e21s 0 of the first type we can take g := 1 . For A of the second type we can take a (cid:18) 0 e−2s(cid:19) suitable h ∈ SU(2) since then h∗ = h−1 and h t+x y −iz h∗ = t 1 0 −ih ix z +iy h−1. (cid:18)y +iz t−x (cid:19) (cid:18)0 1(cid:19) (cid:18)−z +iy −ix (cid:19) LIE GROUPS, CH.2 p.5 Now apply Theorem 2.4. Finally we prove that the kernel consists of 2 elements. Suppose that g ∈ SL(2,C) ∗ ∗ such that gAg = A for all 2 ×2 hermitian matrices A. For A := I this yields gg = I. Hence g ∈ U(2)∩SL(2,C) = SU(2). Now apply again Theorem 2.4. Ex. 2.11 Consider the map g 7→ T , defined in Theorem 2.10, with g restricted to g SL(2,R). Identify R1,2 (as a subspace of R1,3) with the space of real symmetric 2 × 2 matrices (as a subspace of the space of hermitian 2×2 matrices) as follows: t+x y (t,x,y,0) ←→ . (cid:18) y t−x(cid:19) 0 −iz (a) Show that T (A) = A for all g ∈ SL(2,R) if A := and z ∈ R. Conclude g (cid:18)+iz 0 (cid:19) that the map g 7→ T restricts to a continuous homomorphism from SL(2,R) into g SO (1,2) with kernel consisting of 2 elements ±I. 0 (b) Show that the homomorphism in (a) is surjective from SL(2,R) onto SO (1,2). 0 Ex. 2.12 Construct from the surjective two-to-one homomorphisms SU(2) → SO(3) and SU(2) × SU(2) → SO(4) a surjective two-to-one homomorphism from SO(4) onto SO(3)×SO(3) and show that its kernel is ±I. a b Ex. 2.13 Let Σ := C ∪{∞} (the extended complex plane). For g = ∈ GL(2,C) (cid:18)c d(cid:19) and z ∈ Σ put az +b g.z := . cz +d Such a transformation is called a M¨obius transformation of Σ. (a) Show that the map (g,z) 7→ g.z defines a group action of G on Σ. (b) Show that the set {g ∈ GL(2,C) | g.z = z for all z ∈ Σ} equals the subgroup of nonzero multiples of I in GL(2,C). (c) Conclude from (b) that the map sending g to the transformation z 7→ g.z is a two- to-one homomorphism from SL(2,C) onto the group of M¨obius transformations of Σ and that this homomorphism has kernel ±I. Remark Put PSL(2,C) := SL(2,C)/{±I}. Then PSL(2,C) is isomorphic to the group of all M¨obius transformations of Σ. (d) Conclude from (b) that the map sending g to the transformation z 7→ g.z with g restricted to SU(2) is a two-to-one homomorphism from SU(2) into the group of M¨obius transformations of Σ and that this homomorphism has kernel ±I. Ex. 2.14 Below we give a bijection between the unit sphere S2 in R3 and the extended complex plane Σ (which thus can be considered as a sphere and is called Riemann sphere). LIE GROUPS, CH.2 p.6 By this bijection we can transplant certain orthogonal transformations of S2 to certain M¨obius transformations of Σ as follows: S2 ←→ Σ x+iy (x,y,z) ←→ 1−z eiθ x+iy +0 1−z (x cos2θ−y sin2θ,x sin2θ +y cos2θ,z) ←→ 0 x+iy +e−iθ 1−z cosθ x+iy +sinθ 1−z (x cos2θ−z sin2θ,y,x sin2θ+z cos2θ) ←→ x+iy −sinθ +cosθ 1−z Show that the above correspondences extend to an isomorphism from SO(3) onto the group of M¨obius transformations of the form g 7→ g.z with g ∈ SU(2). Ex. 2.15 The group SO (1,3) acts on R1,3. By restriction of this action SO (1,3) acts 0 0 on the forward light cone {(t,x,y,z) ∈ R1,3 | x2 +y2 +z2 = t2, t > 0}, and also on the space of light rays consisting of all half-lines {λ(t,x,y,z) | λ > 0} with (t,x,y,z) in the forward light cone. There is a bijection between the space of light rays and the unit sphere S2 in R3 given by: {λ(t,x,y,z) | λ > 0} ←→ t−1(x,y,z). Thus the action of SO (1,3) on the space of light rays can be transplanted to an action of 0 this group on S2 (which may be called the celestial sphere in this context). This defines a homomorphism T 7→ T from SO (1,3) into the group of bijective transformations (in fact 0 homeomorphisms) of S2. e (a) Show that T (x,y,z) = τ−1(ξ,η,ζ), where T(1,x,y,z) = (τ,ξ,η,ζ). (b) Show that Te = T if T ∈ SO(3), with SO(3) considered as subgroup of SO0(1,3). (c) Show that Te(x,y,z) = (cosht+z sinht)−1(x,y,sinht+z cosht) if e cosht 0 0 sinht 0 1 0 0 T =  . 0 0 1 0 sinht 0 0 cosht   (d) Transplant by the bijection S2 ↔ Σ (see Exercise 2.14) the transformation T of S2 to the transformation T of Σ. Show that T 7→ T maps SO(3) isomorphicalely onto SU(2)/{±I} and that eTe(w) = (cid:18)e012t e−012t(cid:19).w (eew ∈ Σ) if T is as in (c). Conclude e that the map T 7→ T isean isomorphism from SO (1,3) onto PSL(2,C). 0 e e Ex. 2.16 Let C1,1 be the vector space C2 with elements z = (z ,z ) and indefinite 1 2 (Lorentzian) hermitian inner product [z,w] := z w − z w . Let SU(1,1) consist of all 1 1 2 2 T ∈ SL(2,C) acting on C1,1 such that [Tz,Tz] = [z,z] for all z ∈ C1,1. Show the following.