3 HW Unitary group. 3.2 Symplectic Group. ] GL (n, C) to be unitary:
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1 3 HW3 3.1 Unitary group GL (n, C) is a group. U (n) inherits associativity from GL (n, C); evidently 1l 1l = 1l so 1l U (n). We need to check closure & invertibility: SG0) Let U, V U (n). Now (UV) (UV) = V U UV = V V = 1l, where unitarity was used. Hence UV U (n). SG1) Furthermore, is an involution, so, letting U U (n), ( U ) U = UU = 1l since left and right matrix inverses coincide and U is unitary. Hence U U (n). Hence U (n) is a subgroup. To show that it is a proper subgroup, observe that { 1l, 1l} U (n), so U (n) is nontrivial, and 1l GL (n, C) (det 1l = n = 0) but (1l) (1l) = 41l, so 1l / U (n), so U (n) GL (n, C). Hence U (n) is a proper subgroup. [please don t write anything about U (n), use dimensions if you want at the end (if you re sure you re correct), but U (n) = ℵ 1 ] To find out the dimension, let us write down the condition for u = [ u ij ] GL (n, C) to be unitary: u ik u ij = δ kj. For the n diagonal elements, writing u ik in polar form u ik = r ik e iϕ ik (and restoring explicit sums) i r ik e iϕ ikr ik e iϕ ik = r ik r ik = 1 k. This places n independent real constraints. For the n (n 1) off-diagonal elements, both the real and imaginary parts must cancel separately, and we obtain ( n n ) real constraints. However, the (strictly) upper and lower diagonal parts of the product yield equivalent equations, and hence the number of independent constraints is halved, giving 1 ( n n ) = n n real constraints. Since the real dimension of GL (n, C) is n, we obtain dim R U (n) = n n ( n n ) = n. Note: Connection to QM/QFT 3. Symplectic Group Proceeding as in #1, we show that Sp Sp Ω (n, R) = { M GL (n, R) M T ΩM = Ω }. G0) Let M, N Sp. Then (MN) T Ω (MN) = N T M T ΩMN = N T ΩN = Ω. Hence MN Sp. G1) Follows from associativity (used already above) G) Observe that 1l T Ω1l = Ω. Hence 1l Sp. G3) Let M Sp. Then M T ΩM = Ω. Multiply from left by ( M 1) T and from right by M 1. We get ( M 1 ) T M T ΩMM 1 = ( MM 1) T ΩMM 1 = 1l T Ω1l = Ω. Hence M 1 Sp. Hence Sp is a group. To find the dimension, write M Sp as a block matrix and write down the symplectic condition: AT B T C T D T 1l 1l A C B = AT C C T A D B T C D T C A T D C T B = B T D D T B 1l 1l. 7
2 Note that off-diagonal blocks are equivalent, and the diagonal ones require A T C,B T D to be symmetric (not the matrices A, B, C, D! Reality check: that s way too many conditions!). One way to see the constraints more clearly is to define X = A + ic, Y = B + id. Then the 11-block is equivalent to ImX X = 0 and the -block to ImY Y = 0. Each yields 1 ( n n ) real conditions (note that these are complex matrices), so together we get 1 ( n n ) constraints. The off-diagonal condition becomes ImX Y = 1l, which gives 1 n real conditions. Put together, dim R Sp = 4n ( n n ) n = n + n. Note: Connection to Hamiltonian mechanics, Kähler manifolds Conjugacy Classes [solution done in more detail since the problem is quite lengthy] Problem: Consider the Lie group SL (, R) = {A GL (, R) det A = 1}. (1) a) Find a parametrisation of this group manifold satisfying a constraint c 0 x0 + c 3x3 + x 1 + x = c, where c 0, c n, c { 1, 1}. b) Find its conjugacy classes and identify them as -dimensional pseudospheres. Solution: a) Let A SL (, R) and write A = m 11 m 1 m 1 m = x 0 x 1 x + x 3 x x 3 x 0 + x 1, () where the representation is made possible by the well-known fact that f : R : (x, y) (x y, x + y) is a bijection. Since A SL (, R). det A = (x 0 x 1 ) (x 0 + x 1 ) (x x 3 ) (x + x 3 ) = x0 + x 3 x 1 x = 1. (3) This is of the required form, as multiplying it by 1 we obtain x0 x 3 + x 1 + x = 1, (4) 8
3 that is, (c 0, c n, c) = ( 1, 1, 1): SL (, R) is the 3-dimensional anti-de Sitter space AdS 3, and the above parametrisation is of the requested form. b Setup) Let A SL(, R), and write it in the parametrisation. The conjugacy class of A is given by Cl (A) = { } B SL (, R) S SL (, R) : B = S 1 AS. (5) If B Cl (A) we often write A B. 1 To preface the computation we recall the elementary fact that the trace of a matrix is invariant under similarity transformations, which follows from the cyclic property: A B = Tr (A) = Tr (B) However, it is vital to note that the implication does not hold in the other direction. To show this, note that Cl (±1l) = {±1l}, (6) which is a general fact: The conjugacy classes of scalar multiples of the identity are singletons. On the other hand, there are certainly other elements of SL (, R) with a trace of ±. We will have to keep this mind when computing the conjugacy classes. Pseudospheres) Regardless, the trace of A, Tr (A) = x 0 χ, will be an important tool in what follows. This ties rather nicely into the classification of the conjugacy classes as two-dimensional pseudospheres, as there are three different geometrical objects one can associate with a conjugacy classes depending on the value of χ. We will begin with this classification before proceeding into the computation of the explicit conjugacy classes: i) Suppose χ >, x0 > 1. Then 4 yields x 3 + x 1 + x = x 0 1 > 0 = x 3 + x0 1 x + x0 1 7 describes the de Sitter space ds, which is topologically a cylinder R 1 S 1. ii) Suppose χ =, x0 = 1. Then 4 yields x 1 x0 1 = 1. (7) x0 x 3 + x 1 + x = 1 = x 3 + x + x 1 = 0. (8) 1 Note that the matrix S generating the similarity transformation must be in SL (, R) and that the relation is indeed an equivalence relation (conjugacy classes partition a group into equivalence classes; this is an easy exercise to show if you haven t done it already). 9
4 8 is not a pseudosphere, but it describes the lightcone in 1 + -dimensional Minkowski space. iii) Suppose χ <, x0 < 1. Then 4 yields x3 + x 1 + x = x 0 1 < 0 = 9 describes the hyperbolic space H. In 7,9 the quantities x x0 x 1 x 0 + x 1 1 x0 = 1. (9) x 0 1, 1 x 0 R + describe, in a certain sense, the size of the associated manifold. I encourage you to look up pictures of these spaces. Eigenvalues) We follow by examining the eigenvalues of each of the conjugacy class classes i)-iii). In general, the eigenvalues our two-by-two matrix A, λ +, λ C, can be written as λ + = 1 ( ) χ χ 4, λ = 1 ( ) χ + χ 4. There are, however, additional requirements on these numbers: A SL (, C) = det A = λ + λ = 1 = 0 = λ + = 1 λ, arg (λ +) = arg (λ ). (10) It follows from 10 that we can parametrise λ + = re iϕ, λ = r 1 e iϕ ; r R +, ϕ ( π, π]. Furthermore, χ = λ + + λ R. We can now consider the different conjugacy class classes: i) Suppose χ >. Now either r > 1 or r 1 > 1 we can choose the former wlog. Now r = r 1 and χ = re iϕ + r 1 e iϕ R = ϕ = 0: A has two distinct real eigenvalues. (11) ii) Suppose χ =. Now r = r 1 = 1, and χ = e iϕ + e iϕ = ± = ϕ = 0 or ϕ = π: The eigenvalue has multiplicity two: A has two identical real eigenvalues of magnitude 1. (1) However, we would expect there to be several equivalence classes for both cases ±1. iii) Suppose χ <. Again r = r 1 = 1, but we can set χ = e iϕ + e iϕ = cos ϕ < with arbitrary ϕ ( π, π)\ {0}: 1 We will show that A dg (r, 1/r), and it always holds that dg (r, 1/r) dg (1/r, r) by conjugation under SL (, R). 1 10
5 A has a complex conjugate pair of eigenvalues. (13) Conjugacy Classes) We can finally begin searching for the actual conjugacy classes. In general, it is useful to note that given two linearly independent column vectors u, v and a matrix Q = (u, v) formed by them, det Q = 0, and we may in fact scale the vectors so that det Q = 1 3. Furthermore, Q 1 is given by ( Q 1) T = (u, ) s.t. u span T (v) and span T (u);, v = u, u = 1: Q 1 Q = u ( ) u v = u, u u, v = 1., u, v We will keep denoting χ TrA 4. i) Suppose χ >. By 11, we can find two linearly independent eigenvectors u, v: Au = λu, Av = 1 λ v and form Q, Q 1 SL (, R) as above. Hence = u A Q 1 AQ = u A ( u ) v (λu, 1λ ) v = λ u 1, u λ u (, v = dg λ, 1 ). λ 1, u λ, v λ Since the order of the eigenvalues was arbitrary (see fn. ), we can conclude that any matrix A SL (, R) ( ) with a fixed trace χ : χ > is in the conjugacy class of dg λ, λ 1, and due to the invariance of the trace there can be no other matrices in this conjugacy class. That is, the conjugacy classes of the ds -type are [ ( )] dg λ, λ 1 λ>1. ii) Suppose χ =. By 1, the eigenvalue is unique and λ { 1, 1}. Let u be an associated eigenvector Au = λu and let v be a vector v / span (u), scaled so that Q = (u, v) SL (, R). Even though v is no longer an eigenvector, Q still follows the constructions from earlier and we may conjugate to observe that A Q 1 AQ = u (λu, Av) = λ u, u u, Av. (14) λ, u, Av In 14 we observe that the 1-component vanishes regardless (since by construction span T (u)) and u, u = 1. Computing the determinant condition det A = 1 = det Q 1 AQ = det λ u, Av = λ, Av = 1 =, Av = 1 0, Av λ = λ. 3 We will keep calling them u, v. 4 Suppose χ > and similar statements are implicitly taken to mean Let A SL (, R) and suppose TrA > ; so that we can examine properties of generic elements. 11
6 However, the 1-element u, Av R in 14 is still arbitrary; let us call it σ. We may now conjugate Q 1 AQ with an arbitrary element P SL (, R) in the standard parametrisation 5, Q 1 AQ P 1 QAQP = d b λ σ a b = λ + cdσ c a 0 λ c d c σ d σ (15) λ cdσ We make the following observations: We can change the diagonal elements to be arbitrarily large or small (so long as they are subject to χ = ±) by choosing appropriate c, d by conjugation, and we can likewise scale the magnitude of the off-diagonal elements. However, the parameters appear in the offdiagonals squared, so that their signs are determined by σ and cannot be changed by SL (, R)-conjugation. In addition, we cannot set both c = d = 0 (as the matrix P would no longer be in SL (, R)), which is seen to be consistent with 6. And, naturally, due to transitivity 15 implies corresponding conjugacy relations for A. So, the conjugacy classes for both values of λ = ±1consist of three classes, determined by the sign (and non-zeroness) of σ, but don t split further, as varying c, d in 15 allows one to vary the magnitude of the elements, and we already knew that A Q 1 AQ = λ σ, where σ R is the only free parameter. That 0 λ is, the six conjugacy classes are 1 0, 1 1, 1 1, 1 0, 1 1, Note that the associated manifold is the light cone. 8 makes it intuitively reasonable that in this case there wouldn t be infinitely many classes, as the size parameter is missing when compared to ds and H. ii) Suppose finally χ <. By 13, the eigenvalues are determined by an angle, λ { e iϕ, e iϕ} ; but they are complex. We can still find an associated complex eigenvector: z C, so that Az = e iϕ z. Since A is real, taking the conjugate of the eigenvalue equation shows that zis the other eigenvector, A z = e iϕ z, and of course z, zare linearly independent. However, for the construction of a matrix Q as in the previous parts we need real-valued vectors. This can be achieved by taking the real and imaginary parts of the two eigenvalue equations: ARez = ReAz = Ree iϕ z = cos ϕrez sin ϕimz AImz = ImAz = Ime iϕ z = sin ϕrez + cos ϕimz. (16) If we now denote the real vectors Rez = u, Imz = v, form a matrix Q = (u, v), we get as earlier, using 16, A Q 1 AQ = u (Au, Av) = u (cos ϕu sin ϕv, cos ϕv + sin ϕu) = cos ϕ u, u sin ϕ u, v cos ϕ u, v + sin ϕ u, u = cos ϕ cos ϕ, u sin ϕ, v cos ϕ, v + sin ϕ, u sin ϕ sin ϕ R ϕ. (17) cos ϕ 5 Note that ad bc = 1 when computing the inverse and the conjugation. 1
7 ? So, A R ϕ. By the invariance of trace, the only question that remains is whether or not R ϕ Rϕ (since χ = cos ϕ = cos ( ϕ)). We can again conjugate by an arbitrary SL (, R)-matrix P Q 1 AQ P 1 QAQP = d b cos ϕ sin ϕ a b c a sin ϕ cos ϕ c d ( cos ϕ + (ab + cd) sin ϕ = ( a + c ) sin ϕ b + d ) sin ϕ, cos ϕ (ab + cd) sin ϕ which answers our question: The same thing happens as for χ =, that is, the signs of the off-diagonal elements cannot be changed as the parametres appear only squared, so that R ϕ R ϕ. By transitivity we see that the H -equivalence classes are given by cos ϕ sin ϕ sin ϕ cos ϕ ϕ ( π,0) (0,π) and there can be no other classes since 17 shows that each H -type matrix is in the same conjugacy class as a rotation matrix. Note that the exclusion of { π, 0, π} is necessary to avoid overlap with χ =., 13
8 Summary) We found out that an arbitrary matrix A SL (, R) can be parametrised with four real numbers {x i } 3 i=0 s.t. A = x 0 x 1 x + x 3 x x 3 x 0 + x 1, x0 x 3 + x 1 + x = 1, (18) allowing us to identify the manifold structure of SL (, R) with AdS 3. Furthremore, the conjugacy classes of SL (, R) and the associated spaces are given by {[ ( dg λ, 1 )]}, ds ; λ λ>1 1 0, 1 1, 1 1, 1 0, 1 1, 1 1, 1 + d light cone; cos ϕ sin ϕ 3.4 Mobius transformations sin ϕ cos ϕ ϕ ( π,0) (0,π), H. (Note: singularities irrelevant. Usually formulated in compactified C-plane Ċ. ) a) G0) Let f A, f B Mob, associated SL matrices the map f B f A a c b, a b. With direct computation, we obtain d c d ( f B f A ) (z) =... = (a a + b c) z + a b + b d (c a + cd ) z + c d + dd = f BA (z), where BA is just the regular matrix product. Since SL is a group, BA SL, hence we see f BA Mob; closure. G1) Associativity is trivila (eg by G0) & assc. of matrix multiplication). G) Observe that f 1l = id C : z z (not z 1, as some people claimed). Eg. by G0), f A f 1l = f A1l = f A = f 1l f A. Since SL is a group, 1l SL and f 1l Mob; f 1l is the unit element of Mob. G3) Since SL is a group, A 1 SL and by G0) f A f A 1 = f AA 1 = f 1l = f A 1 f A. Hence f A 1 = f 1 A. b) Follows immediately from G0). f (AB) = f AB = f A f B = f (A) f (B). c) ker f = {A SL (, C) f A = id C }. By resolving the condition f A (z) = z z from the general form, we see az + b = cz + dz z. From lin ind. c = b = 0, a = d. Since det A = 1, we get d = a = ±1. Hence ker f = {1l, 1l} = Z. By the 1st isomorphism theorem and b), Mob = SL (, C) /Z PSL (, C). If one defines Mob w/ GL-matrices, the quotient relation is slightly more nontrivial. Note: Special orthochronal Lorentz group, other stuff. 14
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