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Lecture 5. Lie Groups

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Lecture 5. Lie Groups Lecture 5. Lie Groups In this lecture we will make a digression from the development of geometry of manifolds to discuss an very important special case. 5.1 Examples Recall that a Lie Group is a group with the structure of a smooth manifold such that the composition from M ×M → M and the inversion from M → M are smooth maps. Example 5.1.1 The general linear, special linear, and orthogonal groups. The general linear group GL(n) (or GL(n, R)) is the set of non-singular n × n matrices with real components. The is a Lie group under the usual multipli- cation of matrices. We showed in Example 2.1.1 that the multiplication on GL(n)isasmooth map from GL(n)×GL(n) → GL(n). To show that GL(n) is a Lie group, we nned to show that the inversion is also a smooth map. 2 GL(n)isanopen subset of M n Rn ,soiscovered by a single chart. With respect to this chart, the inversion is given by the formula j 1 − M −1 = sgnσ sgnτMτ(1)M τ(2) ...Mτ(n 1)δi δτ(n) i n! det M σ(1) σ(2) σ(n−1) σ(n) j σ,τ j where δi =1ifi = j and 0 otherwise, and the sum is over pairs of permuta- tions σ and τ of the set {1,...,n}. Each component of M −1 is therefore given by sums of productss of components of M, divided by the non-zero smooth function det M: 1 − det M = sgnσ sgnτMτ(1)M τ(2) ...Mτ(n 1)M τ(n) n! σ(1) σ(2) σ(n−1) σ(n) σ,τ −1 j and so M i is a smooth function of the components of M, and GL(n)is a Lie group. From Example 3.2.1 we know that the multiplications on SL(n) and O(n) are also smooth; these are both submanifolds of GL(n), so it follows that the 36 Lecture 5. Lie Groups restriction of the inversion to each of these is smooth, so SL(n) and O(n) are Lie groups. Example 5.1.2 The group O(2). Recall that the orthogonal group O(n) (or O(n, R)isthe group of n × n matrices with real components satisfying T × M M =In.Inparticular O(2) is the group of orthogonal 2 2 matrices. ab M = is in O(2) if the rows (or columns) of M form an orthonor- cd mal basis: a2 + b2 =1; c2 + d2 =1; ac + bd =0. In particular M is determined by its first row [ab] and its determinant δ = det M = ±1: ab M = . −δb δa This gives a natural map ϕ : O(2) → S1 ×{−1, 1} given by ab ϕ =((a, b),ad− bc) ∈ S1 ×{−1, 1}. cd Then ϕ is smooth and has the smooth inverse − ab ϕ 1 ((a, b),δ)= . −δb δa This shows that O(2) is diffeomorphic to S1 ×{−1, 1}.Ingeometric terms, this map takes an orientation-preserving orthogonal transformation (i.e. a ro- tation) to its angle of rotation, or an orientation-reversing orthogonal trans- formation (i.e. a reflection) to twice the angle between the line of reflection and the positive x-axis. Example 5.1.3 The torus Tn The Torus Tn is the product of n copies of S1 = R/ ∼ where a ∼ b ⇔ a − b ∈ Z, with the group structure given by ([a1],...,[an])+([b1],...,[bn]) = ([a1 + b1],...,[an + bn]). This can also be naturally embedded as a subgroup of the group (C\{0})n (with the usual complex multiplication on each factor) by the map 2πix1 2πixn F ([x1],...,[xn]) = e ,...,e . Remark. The previous example is one of an important class of examples of groups (and more generally of manifolds) which arise in the following way: Suppose M is a smooth manifold, and G is a group which acts on M – that 5.1 Examples 37 is, there is a map ρ : G × M → M, such that the map ρg : M → M given by ρg(x)=ρ(g, x)isadiffeomorphism, and ρg ◦ ρh = ρhg. Equivalently, ρ is a group homomorphism from G to the group of diffeomorphisms of M. The action of G on M is called totally discontinuous if for any x ∈ M there is a neighbourhood U of x in M such that ρg(U) ∩ U = ∅ for g = e.For example, the action of the group Zn on Rn given by ρ((z1,...,zn), (x1,...,xn)) = (x1 + z1,...,xn + zn) is totally discontinuous, but the action of Z on S1 given by ρ(k, z)=eikαz is not totally discontinuous if α/∈ Q. If G acts totally discontinuously on M, then the quotient space M/ ∼ with x ∼ y ⇔ y = ρg(x) for some g ∈ G can be made a differentiable manifold with the same dimension as M,inaunique way such that the projection from M onto M/ ∼ is a local diffeomorphism: Let A be an atlas for M. Construct an atlas B for M/ ∼ as follows: For each [x] ∈ M/ ∼ choose a representative x ∈ M, and choose a chart ϕ : W → Z in A about x. Choose U such that ρg(U) ∩ U = ∅ for all g = e, and takeϕ ˜([y]) = ϕ(x) for each y ∈ (U ∩ W )/ ∼. ϕ˜ is well-defined since y ∈ U ⇒ [y] ∩ U = {y}, and for the same reasonϕ ˜ is 1:1onW ∩U. The transition maps between these charts are just restrictions of transition maps of charts in A, and so are smooth. Note that the Klein bottle and the M¨obius strip are given by the quotient of the plane by totally discontinuous actions, and the real projective space RP n is the quotient of the sphere Sn by the totally discontinuous action of Z2 generated by the antipodal map. Example 5.1.4 The group SO(3). SO(3) is the group of orientation-preserving orthogonal transformations of three-dimensional space. Note that an orthog- onal transformation is a real unitary transformation, and so is conjugate to a diagonal unitary transformation with spectrum closed under conjugation. In the three-dimensional case this means that the spectrum consists of a pair of conjugate points on the unit circle, together with 1 or −1. In the case of a special orthogonal transformation, the spectrum must be {1,eiα,e−iα} for some α ∈ [0,π]. In particular such a transformation fixes some vector in R3, and is a rotation by angle α about this axis. Thus an element in SO(3) is determined by a choice of unit vector v and an angle α. Example 5.1.5 S3 as a Lie group. The three-dimensional sphere S3 can be made into a Lie group as follows: We can think of S3 as contained in R4, which we identify with the quaterions H = {x0 + x1i + x2j + x3k} where i2 = j2 = k2 = −1 and ij = k, jk = i, and ki = j.Wethink of x in H as having a real part x0 and a vector part x1i + x2j + x3k corresponding to a vector in R3. The multiplication on H can be rewritten in these terms as 38 Lecture 5. Lie Groups (x, x) · (y, y)=(xy − x · y,xy + yx + x × y) . (5.1) Given x ∈ H, the conjugate x¯ of x is given by (x, x)=(x, −x). Then the identity (5.1) gives (x, x) · (x, x)=x2 + x2 which is the same as the norm of x as an element of R4.Wealso have (x, x) · (y, y)=(xy − x · y, −xy − yx − x × y) =(y, −y) · (x, −x) = (y, y) · (x, x). Therefore |x · y|2 = x · y · x · y =¯y · x¯ · x · y = |x|2|y|2. In particular, for x and y in S3 we have xy ∈ S3.Itfollows that the restriction of the multiplication in H to S3 makes S3 into a Lie group. This example will reappear later in a more familiar guise. Remark. We have seen two examples of spheres which are also Lie groups: S1 is a Lie group, and so is S3 (one could also say that S0 = {1, −1} is a 0- dimensional Lie group). This raises the question: Are all spheres Lie groups? If not which ones are? The following proposition, which also introduces some important concepts in the theory of Lie groups, will be used to show that the two-dimensional sphere S2 can’t be made into a Lie group: Proposition 5.1.1 Let G beaLie group. Then there exist n smooth vec- tor fields E1,...,En on G each of which is everywhere non-zero, such that E1,...,En form a basis for TxM for every x ∈ M. Proof. Consider the left and right translation maps from G to itself given by lg(g )=gg and rg(g )=g g for all g ∈ G. These are smooth maps, and −1 −1 (lg) = lg−1 and (rg) = rg−1 are smooth, so lg and rg are diffeomorphisms from G to G. The derivative of a diffeomorphism is a linear isomorphism, and in particular this gives us isomorphisms Delg,Derg : TeG → TgG. Choose a basis {e1,...,en} for TeG. Then define Ei(g)=Delg(ei) for each i and each g ∈ G. The vector fields E1,...,En have the properties claimed in the proposition. We say that a vector field V on a Lie group G is left-invariant if it satisfies Delg(Ve)=Vg for all g ∈ G.Itisstraightforward to show that the set of left-invariant vector fields form a vector space with dimension equal 5.1 Examples 39 to the dimension of G (in fact {E1,...,En} are a basis).
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