Lie Groups by Hall - Unofficial Solutions For Chapter 1 [PDF]

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Math 360 Assignment 1 Zachary Eman 8 September 2017 . P P Pn Pn+k Pn+k : We see [x, y]n,k := ni=1 xi yi − n+k i=n+1 xi yi = i=1 xi (yi )+ i=n+1 xi (−yi ) = i=1 xi (gy)i = hx, gyi. We can therefore restate the denition of O(n, k) as the set of matrices A such that hAx, g(Ay)i = hx, gyi for all vectors x, y ∈ Rn+k . Now, hAu, vi = hu, AT vi for any real matrix and vectors, so in particular the condition becomes hx, gyi = hAx, gAyi = hx, AT gAyi. This can hold for all x, y only if AT gA = g . This is equivalent to g −1 AT g = A−1 . Since g is self-inverse, the criterion becomes gAT g = A−1 as desired.

1.1

2n : We see ω(x, y) = ni=i xi yn+i + 2n i=n+1 xi (−yi−n ) = i=1 xi (Ωy)i = hx, Ωyi. The arguments of the previous exercise now show that Sp(n; R) is the set of matrices A such that Ω−1 AT Ω = A−1 . In this case, Ω−1 = −Ω, yielding the desired criterion −ΩAT Ω = A−1

P

1.2

P

P

  a b 1.3: By the previous problem, SpR (1) = {A ∈ GLR (2)| − ΩA Ω = A }. Writing A = , the   c d      d −b d −b 0 −1 a c 0 1 1 . The left-hand product equals , condition becomes = ad−bc 1 0 b d −1 0 −c a −c a so clearly the equation holds just if ad−bc = det(A) = 1. This exactly the condition for A to be in SLR (2), so we conclude SpR (1) = SLR (2) as subsets of GLR (2). The exact same reasoning also shows SpC (1) = SLC (2). Finally, Sp(1) := SpC (1) ∩ U (2) = SLC (2) ∩ U(2) =: SU(2). T

−1

1.4: Assume A ∈ O(2). The column vectors c1 , c2 of A are orthonormal, so in particular hc1 , c1 i = 1. We can thus write c1 in polar form (cos θ, sin θ)T for some angle θ. The set of vectors orthogonal to a nonzero vector 2 T in R2 forms a 1-dimensional subspace  of R . (−  sin θ,cos θ) is orthogonal to c1 , so c2 = λ(− sin θ, cos θ)

for some λ. But 1 = hc2 , c2 i = hλ  cos θ sin θ

  − sin θ cos θ or cos θ sin θ

− sin θ − sin θ ,λ i = λ2 , so λ = ±1. We conclude A is of the form cos θ cos θ

 sin θ . Any matrix of this form has orthonormal columns, so the condition is − cos θ

necessary and sucient. Now, let A be a matrix in SO(2). Since SO(2) ⊂ O(2), A is of one of the two above forms. If it were of 2 2 the second form has  determinant  − cos θ − sin θ = −1 and so would not be in SO(2). Therefore A must be expressible as

cos θ sin θ

− sin θ . Any matrix of this form has orthonormal columns and determinant 1, so cos θ

the condition is necessary and sucient.

: Let U have an orthonormal basis of eigenvectors u1 , ..., un , v1 , ..., vn satisfying

1.6

Jui = vi ∀i ∃θi ∈ R : U ui = eiθi ui , U vi = e−iθi vi

and ω(ui , vj ) = 0, ω(ui , uj ) = ω(vi , vj ) = δij

We prove that U is unitary by showing that it preserves the inner product of basis eigenvectors. We compute hU ui , U uj i = heiθi ui , eiθj uj i = ei(θj −θi ) δij = δij

1

hU ui , U vj i = heiθi ui , e−iθi vj i = 0 hU vi , U vj i = he−iθi vi , e−iθj vj i = ei(θi −θi j δij = δij

By conjugate linearity of the inner product, U preserves the inner product of any two vectors and is in U(2n). To prove that U ∈ Sp(n) we perform analogous computations for ω : ω(U ui , U uj ) = ω(eiθi ui , eiθj uj ) = ei(θi +θj ) ω(ui , uj ) = 0 = ω(ui , uj ) ω(U vi , U vj ) = ω(e−iθi vi , e−iθj vj ) = e−i(θi +θj ) ω(vi , vj ) = 0 = ω(vi , vj ) ω(U ui , U vj ) = ω(eiθi ui , e−iθj vj ) = ei(θi −θj ) ω(ui , vj ) = δij = ω(ui , vj ) ω(U vi , U uj ) = −ω(U uj , U vi ) = −ω(uj , vi ) = ω(vi , uj )

where the skew-symmetry of ω is exploited in the nal case. Since ω is bilinear, U preserves its operation on any two vectors and is in Sp(n). : We rst note that since a is irrational, |Ea | = ∞: suppose for the sake of contradiction otherwise. Ea is generated by g = e2πia and so is cyclic. If it were nite, there would be some m with g m = e2πima = 1. But then ma would be an integer l, and a = l/m would be rational. We note additionally that for any θ, φ ∈ [0, 2π), |eiθ − eiφ | ≤ |θ − φ|. Now, let p ∈ S 1 and take  > 0. Select N ∈ N such that 2π/N <  and divide the circle into intervals Ik = {eiθ |N k ≤ θ < N (k + 1)} for k = 0, 1, .., N − 1. The Ik are disjoint and cover S 1 , so by the innite pigeonhole principle there is some index m such that innitely many elements of Ea lie in Im . Take a1 = e2πian1 , a2 = e2πian2 Ea ∩ Ik two such elements. We may represent them in the form a1 = eiθ1 , a2 = eiθ2 with θ1 , θ2 ∈ Ik . Consequently |a1 − a2 | ≤ |θ1 − θ2 | < 2π/N < . We have also ˜ −1 −1 −1 |1−a−1 ˜. For each n ∈ N, write a ˜n = eiθn 1 a2 | = |a1 (a1 −a2 )| = |a1 ||a1 −a2 | = |a1−a2 | < . Write a1 a2 = a with θ˜n ∈ [0, 2π). Let n˜ be minimal such that θ˜n˜ < θ˜n˜ −1 . Then the intervals I˜n = {eiθ |θ˜n ≤ θ < θ˜n+1 } cover S 1 and have width less than . Therefore, if p ∈ I˜k , |p − a ˜k | <  as needed.     an bn a b 1.10: Let An = be a sequence of matrices in M2 (R). An → A = exactly if an → a, bn → b cn dn c d and so on for c and d as sequences of real numbers. In the case where An ∈ G, we have  bn = cn = 0 a b itn itn a for all n and a sequence of real numbers tn such that an = e , dn = e . If An → , we must c d have b = lim bn = 0, c = lim cn = 0. Since the absolute value is a continuous function, we must also have iθ iφ |a| = lim |an | = 1, |d| = lim|dn | = 1. This means there exist real numbers θ, φ such  iθthat a= e , d = e , eiθ 0 e 0 ¯ ⊆ T . Now, let and A is in the set T = { |θ, φ ∈ R}. This shows G ∈ T . Let k 0 eiφ 0 eiφ be a sequence of positive reals tending to zero. For each natural number k, select an integer nk such that |eia(θ+2πnk ) − eiφ |= |eia2πnk − ei(φ−aθ) |