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1 Albrai Topoloy: Solution St #5 1) Lt G,, and x X. Baus X is Hausdor and x φ (x), thr ar opn nihborhoods U x o x and U φ(x) o φ (x) whih ar disjoint. Thn V,x : U x φ 1 (U φ(x)), is also an opn nihborhood o x. Obsrv that V x φ (V x ) is mpty baus i y V x, thn y U x whil y φ (V x ) implis that y φ (φ 1 (U φ(x))) U φ(x) and U x U φ(x). Lt G {, 1,..., n }. Thn W x n i1 V i,x is th dsird opn nihborhood o x. 2) For i 1, 2, pik x i p 1 (x i ). Baus X is path-onntd, thr is a ontinuous map α : I X with α() x 1 and α(1) x 2. Din α p α. For any ỹ p 1 (x 1 ), lt αỹ : I X b th uniqu lit o α with αỹ() ỹ. Baus αỹ(1) p 1 (x 2 ), w an din Φ : p 1 (x 1 ) p 1 (x 2 ) by Φ(ỹ) αỹ(1). W prov that Φ is a bijtion by dinin a map Ψ : p 1 (x 2 ) p 1 (x 1 ) as ollows. For ỹ p 1 (x 2 ), lt βỹ : I X b th uniqu lit o ᾱ with βỹ () ỹ. Din Ψ : p 1 (x 2 ) p 1 (x 1 ) by Ψ(ỹ ) βỹ (1). I now laim that Φ and Ψ ar invrss. Obsrv that or any ỹ p 1 (x 1 ), both α 1 ỹ and β Φ(ỹ) ar lits o α 1 with initial point Φ(ỹ). Hn thy ar qual. Thus, Ψ(Φ(ỹ)) β Φ(ỹ) (1) α 1 ỹ (1) α ỹ() ỹ. Thus, Ψ Φ is th idntity on p 1 (x 1 ). Th arumnt that Φ Ψ is th idntity on p 1 (x 2 ) is symmtri (usin α 1 or α and (α 1 ) 1 α. Hn Φ is a bijtion. 3) Lt p : R S 1 C b th univrsal ovrin map, p(t) 2πit. By th litin ritrion and baus R is simply-onntd, a map : (X, x ) (S 1, 1) admits a lit : X R i and only i Im(π 1 ()) is trivial. I π 1 (X) is init, thn any homomorphism π 1 () : π 1 (X, x ) π 1 (S 1, 1) Z must b trivial. (I π 1 ()([γ]), thn or I N π 1 (X, x ), thn or any [γ] π 1 (X, x ), [γ] N. Hn π 1 ()([γ] N ) Nπ 1 ()([γ]), so π 1 ()([γ]).) Thus, any : X S 1, admits a lit : X R. Lt F : X [, 1] R b th homotopy F (x, t) t (x). Thn F : X [, 1] dind by F p F will iv th dsird homotopy. 4a) Obsrv tha a 2 + b 2 aā + b b 1, thn (ā ) a b b aā + b b ab + ab b a ā b + ā b b b + āa so Hn, w an omput: a b or all SU(2), 1 a b 1 1 (ā ) b b a (.1) 1 (ā ) a b ix y a b a b ix y b ȳ ix ȳ ix b a a b ixā + y b ixb + ay ix b āȳ ixa + bȳ i( a 2 b 2 )x + a by ābȳ a 2 y + b 2 ȳ 2iabx ā 2 ȳ b 2 y 2iā bx i( a 2 b 2 )x a by + ābȳ 1

2 2 Th prdin ormula implis that or all A SU(2) and X su(2), Ad(A)X su(2). W now hk that (A, X) Ad(A)X dins an ation o SU(2) on su(2). Equation (.1) implis that Ad(I 2 )X X. For A, B SU(2) and X su(2), Ad(AB)X (AB)X(AB) 1 ABXB 1 A 1 Ad(A) (Ad(B)X) (Ad(A)Ad(B))X, ompltin th proo that Ad dins a roup ation. An asir approah to showin Ad(A)X su(2) is ivn as ollows. For N M n (C), din N N t to b th matrix whos ntris ar ivn by th takin th omplx onjuats o th ntris o N t, th transpos o N. Not that (A ) A and (AB) B A (sin (AB) t B t A t ). Thus, Thn, or A SU(2) and X su(2), so Ad(A)X su(2). A SU(2) A A 1 and X su(2) X X. (Ad(A)X) (AXA ) AX A AXA Ad(A)X, 4b) Baus Ad is a homomorphism, Ad(A 1 ) Ad(A) 1. Thror, to prov that Ad(A) SO(3), w must show that Ad(A 1 ) Ad(A) t. Considr th basis or su(2) ivn by i 1 i σ 1, σ i 2, σ 1 3. i Thn quation (.1) implis that or a b A Ad(A)σ 1 i( a 2 b 2 ) 2iab 2iā b i( a 2 b 2 ) ( a 2 b 2 )σ Im(ab)σ 2 2R(ab)σ 3 a b āb a Ad(A)σ b 2 ā 2 b 2 a b + āb 2 Im(a b)σ 1 + R(a 2 + b 2 )σ 2 + Im(a 2 + b 2 )σ 3 ia b + iāb ia Ad(A)σ 3 2 ib 2 iā 2 ȳ i b 2 ia b iāb 2 R(a b)σ 1 Im(a 2 b 2 )σ 2 + R(a 2 b 2 )σ 3. Thus, with rspt to th basis {σ 1, σ 2, σ 3 } o su(2), ( a 2 b 2 ) 2 Im(a b) 2 R(a b) (.2) Ad(A) 2 Im(ab) R(a 2 + b 2 ) Im(a 2 b 2 ) 2 R(ab) Im(a 2 + b 2 ) R(a 2 b 2 ) Applyin this ormula to a b A, and A 1 (ā ) b b a

3 3 w t, usin Im(ab) Im(ā b), R(āb) R(a b) ( a 2 b 2 ) 2 Im(ā b) 2 R(ā b) Ad(A 1 ) 2 Im(āb) R(a 2 + b 2 ) Im(ā 2 b 2 ) 2 R(āb) Im(ā 2 + b 2 ) R(ā 2 b 2 ) ( a 2 b 2 ) 2 Im(a b) 2 R(a b) 2 Im(ab) R(a 2 + b 2 ) Im(a 2 b 2 ) 2 R(ab) Im(a 2 + b 2 ) R(ā 2 b 2 ) Ad(A) t, as rquird. Nxt, w hk that dt(ad(a)) 1 or all A SU(2). Baus dt(y t ) dt(y ), th quality Y 1 Y t satisid or all Y Ad(A) with A SU(2) implis that dt(ad(a)) 2 1 so dt(ad(a)) ±1. Howvr, SU(2) is path-onntd, dt(ad(i 2 )) 1, and th map A dt(ad(a)) is ontinuous, so th ima o SU(2) undr th map dt(ad( )) must b onntd and hn qual to 1. Hn, Ad dins a homomorphism rom SU(2) to SO(3). 4) I Ad(A) I 3, thn rom (.1) w s that a 2 b 2 1 whil rom th dinition o SU(2), w hav a 2 + b 2 1. Thus, a 1 and b so b. Thn R(a 2 ) 1 whil Im(a 2 ) so a 2 1 whih implis a ±1. Hn, Kr(Ad) {±I 2 }. Th surjtivity o Ad is hardr. Lt Y SO(3) b ivn by v 1 w 1 u 1 v 1 w 1 Y v 2 w 2 u 2 v v 2, w w 2, u v 3 w 3 u 3 v 3 w 3 t u 1 u 2, u 3 so Y 1 v, Y 2 w, and Y 3 u whr { 1, 2, 3 } is th standard basis. Not that or, th standard innr produt on R 3, th quality Y t Y 1 implis that or all v, w R 3, Y v, Y w v, Y t Y w v, w, so Y is an isomtry. Hn th vtors v, w, u ar an orintd orthonormal basis. Morovr, th quality 1 dt(y t ) u ( v w) implis that u v w and so SO(3) is in bijtiv orrspondn with th st o orthonormal pairs, v, w. From (.2), w s tha Ad(A) Y, w must hav a 2 + b 2 1 and a 2 b 2 v 1. Hn (.3) a 2 (1 + v 1 )/2, b 2 (1 v 1 )/2. W writ a a iθa and b b iθ b. Th qualitis v 2 2 Im(ab) and v 3 2 R(ab) imply that (.4) v 3 + iv 2 2ab 2 1 v 1 2 i(θa+θb). 2 Baus v 2 1, w hav v2 2 + v3 2 1 v1. 2 Thus, i w writ v 3 + iv 2 v2 2 + v2 3 iθ b, w an solv (.4) by sttin θ v θ a + θ b. Thus, ivn Y SO(3), w an ind A SU(2) with Ad(A) 1 v Y 1. Thn i w is a unit lnth vtor, orthoonal to v, w s that Ad(A 1 ) w is a unit lnth vtor, orthoonal to Ad(A 1 ) v 1. Hn, thr is θ with Ad(A 1 // os(θ) ) w. sin(θ)

4 4 Thn, w omput that Ad( )) 2 Hn, Not that so whih implis that // os(θ) Ad(A 1 ) w. sin(θ) w Ad(A ) 2. Ad( ) 1 1, Ad(A ) 1 v, Ad(A ) 2 w, ompltin th proo o th surjtivity o Ad. Th vtor w must satisy, Ad(A ) Y, w 1, w 2, w 3. 4d) By dinition, SU(2) an b idntiid with S 3 {(a, b) C 2 : a 2 + b 2 1}, and is thus simply-onntd. Th map Ad : SU(2) SO(3) idntiis SO(3) with SU(2)/{±I 2 } whih is a quotint by a init roup ation and thus a ovrin spa by problm (#1). Hn, SU(2) SO(3) is th univrsal ovr o SO(3) with Aut(SU(2) SO(3)) Z 2. Thus, π 1 (SO(3), ) Aut(SU(2) SO(3)) Z 2. 5) W iv th proo undr th assumption that n 2, i 1, and j 2. Th nral as ollows asily rom this. W brak th homotopy into two stps, illustratd by th ollowin diarams: (.5)

5 5 (.6) (.7) I w din th homomorphism φ a,b : [a, b] : [a, b] [, 1] by φ a,b (s) (s a)/(b a), and γ : [, 1] X is a path, thn th path γ φ a,b runs throuh th path γ durin th intrval [a, b]. Lttin t b th homotopy paramtr, th homotopy btwn th maps ivn by diarams (.5) and (.6) is dind by lttin run throuh th loops durin th intrval [, 1 t/2] and lttin run throuh th loops durin th intrval [t/2, 1]: t (2, j 1 t/2 ) i 1/2 and 1 t/2 x i 1/2 and 1 t/2 1 F 12 (,, t) (2 1, tj t/2 1 t/2 ) i 1/2 1 and 1 t/2 1 x i 1/2 and 1 t/2 Lttin t b th homotopy paramtr, th homotopy btwn th maps ivn by diarams (.6) and (.7) is ivn by lttin run throuh th loops durin th intrval [, (1 + t)/2] and lttin run throuh th loops durin th intrval [(1 t)/2, 1]. W thus din th homotopy btwn th diarams (.6) and (.7) by ( 2ti 1+t, 2) i (1 + t)/2 and 1/2 x i (1 + t)/2 1 and 1/2 F 23 (,, t) ( 2ti 1+t 1+t, 2 1) i (1 t)/2 1 and 1/2 1 x i (1 t)/2 and 1/2 1

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