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1 GEOMETRY HW 8 CLAY SHONKWILER Consider the Heisenberg group x z 0 y which is a Lie group diffeomorphic to R 3 a: Find the left invariant vector fields X, Y, Z on R 3 whose value at the identity is the standard basis in R 3 Answer: Note that, if g a c 0 b then, if x, y, z R 3, L g x, y, z a c 0 b x z 0 y, x + a z + ay + c 0 y + b which corresponds to x + a, y + b, z + ay + c R 3 That is, so L g x, y, z x + a, y + b, a + ay + c, dl g a Hence, the left-invariant vector field corresponding to e, 0, 0 in the tangent space of the identity is simply Xg dl g I e , 0, 0 0 a 0 x Similarly, Y g dl g I e a 0 0 0,, a y + a z,

2 2 CLAY SHONKWILER and Zg dl g I e a 0 0 0, 0, z b: Compute the Lie brackets [X, Y ], [X, Z] and [Y, Z] Answer: If H represents the Heisenberg group and A : ɛ, ɛ H is a path through the origin, then [X, Y ] [X, Z] da dt t0 0 a a a 2 for some a, a 2, a 3, since the diagonal entries are constant Hence, elements of the Lie algebra of H, h, are matrices of this term Associating with R 3, we see that and Hence, Xg Y g Zg a XY Y X a Similarly, XZ ZX 0 0, 0 0 a

3 and [Y, Z] Y Z ZY 0 0 a GEOMETRY HW a c: Find the same Lie brackets in a simpler fashion, using material we learned in class Answer: Now, consider h as a Lie subalgebra of gl3, R M3, 3, R R 9 Then the standard basis on the copy of R 3, e, e 2, e 3, considered above corresponds, when we embed R 3 into R 9 appropriately, to the matrices 0 0,, respectively In class, we saw that, for A, B Mn, n, R, [A, B] AB BA Thus, we don t even need to explicitly compute X, Y, Z, as, [X, Y ] [X, Z] [Y, Z] ,,, which agrees with the results computed in b note that this is valid only because X, Y, Z are left-invariant and thus the Lie brackets are left-invariant as well,

4 4 CLAY SHONKWILER 2 a: Show that the Lie algebra of the Lie group S 3 is isomorphic to the one for SO3 Proof On the last homework, we proved that φ : q [v qvq ] is a surjective Lie group homomorphism from S 3 to SO3 Thus, φ induces a Lie algebra homomorphism dφ : s 3 so3 Furthermore, we showed that SO3 is diffeomorphic to RP 3, so S 3 and SO3 have the same dimension Therefore, by Sard s theorem, the regular values of φ are dense in SO3 That is to say, dφ is surjective on a dense subset of the tangent spaces Now, since dφ is linear, this implies that, in fact, dφ is surjective on all of the tangent spaces of SO3, specifically dφ : s 3 so3 is surjective Since dφ is a Lie algebra homomorphism and the dimensions of s 3 and so3 are equal, this implies that dφ is a vector space isomorphism Now, if V, W s 3, then dφ[v, W ] [dφv, dφw ] and, if dφv Y, dφw Z, then dφ [Y, Z] dφ [dφv, dφw ] dφ dφ[v, W ] [V, W ] [dφ Y, dφ Z], so dφ preserves the multiplicative structure of the Lie algebras and is, therefore, a Lie algebra isomorphism b: Show that the Lie algebra of SO3 and SU2 are isomorphic, and both are isomorphic to R 3,, where is the cross product a + bi c + di Proof Suppose A SU2 Then e + fi g + hi g + hi c di A e fi a + bi A a bi e fi c di g hi So g a, h b, e c, d f, or a + bi c + di A c + di a bi 0 i a + b + c 0 0 i 0, 0 i + d i 0

5 GEOMETRY HW 8 5 Now, if we call these matrices 0 0 i 0 i 0 i 0 j 0 0 i k i 0 a + bi c + di then it s clear that all elements SU2 can c + di a bi be written in the form a + bi + cj + dk Furthermore, i, j, k are all of order 2 and i 0 i ij k 0 i 0 i 0 i 0 0 i 0 ik j 0 i i i i 0 jk i, 0 i 0 0 i so we see that the map a + bi + cj + dk a + bi + cj + dk is a well-defined map from SU2 to the quaternions Furthermore, a 2 + b 2 + c 2 + d 2 a + bi c + di c + di a bi so this is a map into the unit quaternions, which correspond to the sphere S 3 This map is clearly bijective and a group homomorphism, so we see that SU2 and S 3 are isomorphic as groups Furthermore, both the map and its inverse are smooth since the coordinate functions are smooth, so this is, in fact, a Lie group isomorphism Therefore, since SU2 and S 3 are isomorphic as Lie groups, they have the same Lie algebra; we showed in a that S 3 and SO3 have isomorphic Lie algebras, so we conclude that the Lie algebras of SU2 and SO3 are isomorphic Now, as demonstrated in Spivak pg 378, the Lie algebra of O3 and, therefore, the connected component containing the identity, namely SO3 consists of all skew-symmetric matrices Therefore, define the map f : so3 r 3 by 0 a b a 0 c a, b, c b c 0

6 6 CLAY SHONKWILER f[a, B] This map is certainly bijective and, thus, a vector space isomorphism Furthermore, if A a 0 c, B d 0 f 0 a b 0 d e b c 0 e f 0 so3, then fab BA 0 a b f a 0 c b c 0 f f 0 d e d 0 f e f 0 ad be bf af ce ad cf ae cd bd be cf 0 ce bf af cd bf ce 0 bd ae cd af ae bd 0 bf ce, dc af, ae bd a, b, c d, e, f 0 d e d 0 f e f 0 Hence, f is a Lie algebra isomorphism Therefore, we conclude that r 3 is isomorphic to so3 and, since so3 is isomorphic to su2, to su2 as well Show that a: S SUn and Un have the same Lie algebra Proof Define the map f : S SUn Un by 3 λ, A λa Since λa λa, we see that λaλa λa λa λ λaa Id, so this map really does map into UN Furthermore, if λ, A, γ, B S SUn, fλ, Aγ, B fλγ, AB λγab λaγb fλ, Afγ, B, so this is a group homomorphism Furthermore, this is clearly a smooth map, so it is a Lie group homomorphism In addition, if B Un, and we let x /n denote the principal branch ie x /n n x and x /n is the first such value that we encounter when traversing S counter-clockwise starting at, then det det B /n B n det B /n det B det B det B 0 a b a 0 c b c 0 ad be ce dc bf ad cf bd af ae be cf

7 and, furthermore, det B /n B det B /n B GEOMETRY HW 8 7 t det B /n det B /n BBt Id where is the complex conjugate of ; hence, det B /n det B /n det B /n, det B /n B B, so f is surjective Now, if α, A ker f, then αa Id, so A is a diagonal matrix In fact, A must be a scalar matrix, and even at that α, A ker f only if α is an nth root of unity ζ i n and A ζn n i Id Since detζn n i Id ζn n i n, we note that all such pairs are in the kernel of f Hence, ker f {ζ i n, ζ i nid ζ n a primitive nth root of unity} so f is n-to-one Since f is a surjective Lie group homomorphism with discrete kernel, we know that Un and S SUn/ker f are isomorphic as Lie groups Furthermore, the argument given in 2a above demonstrates that S SUn and S SUn/ker f have the same Lie algebra, so we see that the Lie algebras of S SUn and Un are isomorphic b: Show that Un is diffeomorphic to S SUn but not isomorphic as Lie groups Proof To see that these two manifolds are not isomorphic as Lie groups, we consider the centers of each The ZUn consists simply of the scalar matrices On the other hand, since S is abelian, ZS SUn is given by S C where C is the set of scalar matrices in SUn Now, of the scalar matrices in Un, the only ones of order n are those with primitive nth roots of unity on their diagonals; since there are φn such where φ is the Euler phi-function, we see that there are φn elements of order n in ZUn On the other hand, ζ i n, ζ n ZS SUn for ζ n a primitive nth root of unity and i n, and each such element is of order n since the order of ζ i n divides n, so we see that there are at least nφn elements of order n in ZS SUn Hence, we conclude that ZUn is not isomorphic to ZS SUn and, therefore, that Un and S SUn are not isomorphic as groups and, hence, they are not isomorphic as Lie groups

8 8 CLAY SHONKWILER Now, to show that these two Lie groups are diffeomorphic, define the map f : Un S SUn by A e2πiα, e 2πiα A, where e 2πiα det A Note that e 2πiα 0 0 det A 0 0 e 2πiα 0 0 det 0 0 e 2πiα e 2πiα det A Also, since taking the conjugate transpose reverses the order, it is clear that the image of A under φ is self-adjoint, so we see that this really is a map into S SUn Since the determinant map is continuous, this is a smooth map Now, if we define ψ : S SUn Un by e 2πiα, A e 2πiα A then ψ is also clearly a smooth map Furthermore, ψ φa e 2πiα 0 0 ψ e2πiα, A 0 0 e 2πiα 0 0 e 2πiα IdA A A

9 GEOMETRY HW 8 9 and e 2πiα 0 0 φ ψe 2πiα, A φ A 0 0 e 2πiα 0 0 e 2πiα 0 0 e2πiα, A e 2πiα, A so φ ψ Id and ψ φ Id; since both are smooth, we see that ψ φ and that φ is a diffeomorphism, so S SUn is diffeomorphic to Un Show that every Lie group is orientable 4 Proof Let G be a Lie group and let {e,, e n } denote a basis for T e G Then let V,, V n denote the left invariant vector fields on G such that V i e e i for all i,, n Then, since the X i are left-invariant and L g gives a vector space isomorphism on the tangent spaces, we see that the X i must be linearly independent everywhere Hence, for all a, v T G, a, v n i a ix i a, so we can define the map f : T G G R n, where n a i X i a a, a,, a n i The second coordinate is clearly linear and this map is certainly smooth with smooth inverse, so we see that T G is trivial and, thus, that G is necessarily orientable 5 Show that F : S 3 S 3 SO4 defined by F q, r {v H qvr H} is a homomorphism which is onto with kernel {±, } Conclude that SO4 is diffeomorphic to S 3 S 3 /{x, y x, y} Proof First, note that, if q, r, p, s S 3 S 3, then F q, rp, s F qp, rs [v qpvrs ] [v qpvs r ] [v qvr ] [v pvs ] F q, r F p, s,

10 0 CLAY SHONKWILER so, assuming we can show it is well-define, F is a group homomorphism Now, if q a + bi + cj + dk and r e + fi + gj + hk, then r r, so: qr a + bi + cj + dke fi gj hk ae + bf + cg + dh + ibe af ch + dg +jce ag df + bh + kde ah bg + cf qir a + bi + cj + dkie fi gj hk a + bi + cj + dkf + ei + hj gk af be ch + dg + iae + bf cg dh +jah + cf + de + bg + kdf ag + bh ce qjr a + bi + cj + dkje fi gj hk a + bi + cj + dkg hi + ej + fk ag + bh ce df + ibf ah + cf de +jae + cg dh bf + kaf + dg + be + ch qkr a + bi + cj + dkke fi gj hk a + bi + cj + dkh + gi fj + ek ah bg + cf de + iag + bh + ce + df +jch af be + dg + kae + dh bf cg Hence, F q, r is given by the matrix ae + bf + cg + dh af be ch + dg ag + bh ce df ah bg + cf de be af ch + dg ae + bf cg dh bf ah + cf de ag + bh + ce + df ce ag df + bh ah + cf + de + bg ae + cg dh bf ch af be + dg de ah bg + cf df ag + bh ce af + dg + be + ch ae + dh bf cg Now, we can extend this to a map from S 3 S 3 to R 6 ; the coordinate functions of this map are clearly smooth, so restricting the range doesn t affect smoothness, and, hence, we see that F is smooth If q, r as above, then q, r q, r q, r, so, substituting appropriately in the above matrix, we see that F q, r F q, r t, so the image of F really is contained in O4 Furthermore, F, Id 4 and thus, since F is continuous and S 3 S 3 is connected, the image of F must be contained in the connected component of O4 containing the identity, namely SO4 Therefore, F : S 3 S 3 SO4 is a Lie group homomorphism Now, if q, r as above in terms of a, b, and q, r ker F, then, looking at the diagonal entries and recalling that q q r r, we see that ae + bf + cg + dh ae + bf cg dh ae + cg dh bf ae + dh bf cg a 2 + b 2 + c 2 + d 2 e 2 + f 2 + g 2 + h 2 Thus, it must be the case that ae and b c d f g h 0, so we see that q, r ±, ;

11 GEOMETRY HW 8 that is, ker F {±, } Therefore, if only we can show that F is surjective, we will be able to conclude that SO4 S 3 S 3 /ker F S 3 S 3 /{x, y x, y} To do so, it suffices to show that, for an arbitrary oriented, orthonormal basis of R 4 there exists [v qvr ] Image F such that [v qvr ] maps the arbitrary basis to the standard basis, {, i, j, k} To that end, let {w, x, y, z} be an oriented, orthonormal basis for R 4 Let q rw Then F q, rw, x, y, z, rw xr, rw yr, rw zr Now, since w is of unit length and linearly independent of x, y, z, {, w x, w y, w z} is also an oriented basis for R 4 Thus, we need only find r such that the right side of equation above is simply, i, j, k However, this is precisely what we needed to show on the last homework to see that our map from S 3 to SO3 was surjective; having reduced to this problem, then, we see that, in fact, F is surjective Therefore, we conclude that, indeed, SO4 is diffeomorphic to S 3 S 3 /{x, y x, y} DRL 3E3A, University of Pennsylvania address: shonkwil@mathupennedu

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