[ zd z zdz dψ + 2i. 2 e i ψ 2 dz. (σ 2 ) 2 +(σ 3 ) 2 = (1+ z 2 ) 2

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Victoria Anderson
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1 2 S M M 4 S 2 S 2 z, w : C S 2 z = 1/w e iψ S 1 S 2 σ 1 = 1 ( ) [ zd z zdz dψ + 2i z 2, σ 2 = Re 2 e i ψ 2 dz 1 + z 2 ], σ 3 = Im [ 2 e i ψ 2 dz 1 + z 2 σ 2 σ 3 (σ 2 ) 2 (σ 3 ) 2 σ 2 σ 3 (σ 2 ) 2 +(σ 3 ) 2 = 4 dz 2 (1+ z 2 ) 2 1 S 2 1 σ 1, σ 2 σ 3 dσ 1 = σ 2 σ 3 (w, φ) = (1/z, ψ + 4 arg z) Z/2 M RP 2 M D 1 ].

2 M ds 2 = dr 2 + a 2 (σ 1 ) 2 + b 2 (σ 2 ) 2 + c 2 (σ 3 ) 2 a, b, c r [0, ) ( ) r a b c a = a2 (b c) 2 2bc, b = b2 (c a) 2 2ca, c = c2 (a b) 2 2ab, a(0) = 0 b(0) = m c(0) = m m dr σ 1 σ 2 σ 3 r r = 0 r = 0 2 Σ σ 2 σ 3 m Σ r > 0 a c b a b c (ca + ab) = 2 abc (ca)(ab), (ab + bc) = 2 abc (ab)(bc), (bc + ca) = 2 abc (bc)(ca). 2/(abc) Σ r = 0 a(0) = 0 b(0) = m = c(0) a(r) = 2r 1 2m 2 r3 + O(r 4 ), b(r) = m r 3 8m r2 + O(r 3 ), c(r) = m r + 3 8m r2 + O(r 3 ). b c q(r) = c(r) b(r) p(r) = c(r) + b(r). q(r) > 0 r 0 q(0) = 2m p(0) = 0 r > 0 (a p) > 0 p > 0 ( ) ds 2 = dr 2 + a2 zd z zdz 2 dψ + 2i z 2 + q2 + p 2 4 dz 2 [ e iψ (dz) 2 ] 4 (1 + z 2 (2 q p) Re ) 2 (1 + z 2 ) 2.

3 r = 0 a(r)/r p(r)/r q(r) r 2 a, p q a = 2(a2 q 2 ) p 2 q 2, q = 2q(p2 a 2 ) a(p 2 q 2 ), p = 2 + 2p(q2 a 2 ) a(p 2 q 2 ). a p = c + b r q = c b r a, p, q C k k N, R(X, Y, Z, W ) = Z W Y W Z Y [Z,W ] Y, X. {e i } 1 ω j i e i = ω j i e j ω j i = ωj i {ω i } ω j = ω j i ωi dω j = ω j i ωi. R j i = dωj i ωk i ω j k. R j i (X, Y ) = R(e j, e i, X, Y ) X Y M ω 0 = dr, ω 1 = a σ 1, ω 2 = b σ 2, ω 3 = c σ 3. ω 0 ω 1 ω 2 ω 3 dω 0 = 0, dω 1 = a a ω0 ω 1 + a bc ω2 ω 3,

4 dω 2 dω 3 ω0 1 = a a ω1, ω2 3 = 1 b 2 + c 2 a 2 ω 1, 2 abc ω0 2 = b b ω2, ω3 1 = 1 a 2 + c 2 b 2 ω 2, 2 abc ω0 3 = c c ω3, ω1 2 = 1 a 2 + b 2 c 2 ω 3. 2 abc 4 0 = R R3 2 = R2 0 + R1 3 = R3 0 + R2 1 (i, j, k) = (1, 2, 3) R i 0 + R k j = d(ω i 0 + ω k j ) + (ω j 0 + ωi k ) (ωk 0 + ω j i ), ω i 0 + ω k j = σ i. ω0 i + ωk j 2 ω 0 ω 1 ω 2 ω 3 Λ 2 + ω 0 ω 1 +ω 2 ω 3 ω 0 ω 2 +ω 3 ω 1 ω 0 ω 3 + ω 1 ω 2 Λ 2 + (ω 0 ω 1 + ω 2 ω 3 ) = σ 3 (ω 0 ω 2 + ω 3 ω 1 ) + σ 2 (ω 0 ω 3 + ω 1 ω 2 ), (ω 0 ω 2 + ω 3 ω 1 ) = σ 3 (ω 0 ω 1 + ω 2 ω 3 ) σ 1 (ω 0 ω 3 + ω 1 ω 2 ), (ω 0 ω 3 + ω 1 ω 2 ) = σ 2 (ω 0 ω 1 + ω 2 ω 3 ) + σ 1 (ω 0 ω 2 + ω 3 ω 1 ). SO(3) 2 Re(z) Im(e i ψ 2 + e i ψ 2 z 2 ) Re(e i ψ 2 e i ψ 2 z 2 ) 1 S = 1 + z 2 2 Im(z) Re(e i ψ 2 + e i ψ 2 z 2 ) Im(e i ψ 2 e i ψ 2 z 2 ). 1 z 2 2 Im(e i ψ 2 z) 2 Re(e i ψ 2 z) 0 σ 3 σ 2 S 1 ds = σ 3 0 σ 1, σ 2 σ {ω 0 ω 1 + ω 2 ω 3, ω 0 ω 2 + ω 3 ω 1, ω 0 ω 3 + ω 1 ω 2 }

5 2 S ω 0 ω 1 + ω 2 ω 3 = a dr σ 1 + p2 q 2 4 2i dz d z (1 + z 2 ) 2, (ω 0 ω 2 + ω 3 ω 1 ) + i(ω 0 ω 3 + ω 1 ω 2 ) = (p ei ψ 2 dz q e i ψ 2 d z) (dr ia σ 1 ) 1 + z 2 p q [ ] S 1 z 2 [ (p 2 q 2 ] ) 2i dz d z 1 + z 2 4 (1 + z 2 a dr σ1 ) 2 [ ( z dz p (dr ia σ 1 ) ) q z d z ( e iψ (dr ia σ 1 ) ) ] 2 Im (1 + z 2 ) 2 [ ] + i [ ] [ 2z (p 2 q 2 ] ) 2i dz d z 1 + z 2 4 (1 + z 2 a dr σ1 ) 2 i dz ( p (dr ia σ 1 ) ) q d z ( e iψ (dr ia σ 1 ) ) (1 + z i q z2 dz ( e iψ (dr + ia σ 1 ) ) z 2 d z ( p (dr + ia σ 1 ) ) (1 + z 2 ) 2., a(r) = 2r + r p(r) = r + r q(r) = 2m + r r = 0 2 Σ 1 z 2 [ 2im 2 ] [ dz d z 2z 2im 2 ] dz d z 1 + z 2 (1 + z 2 ) z 2 (1 + z 2 ) 2, Σ Σ [Σ] 2 [i ] S J i Σ J S 2 Σ u : S 2 M J i du = x i du J S 2 x 1, x 2 x 3 S 2 x 1 + ix 2 = 2z/(1 + z 2 ) x 3 = (1 z 2 )/(1 + z 2 ) u du J S 2 = x 1 J 1 du x 2 J 2 du x 3 J 3 du.

6 M R 1 0 R 1 0 = dω 1 0 ω 2 0 ω 1 2 ω 3 0 ω 1 3 κ(a, b, c) κ(a, b, c) R 1 0 = a a ω0 ω 1 κ(a, b, c) ω 2 ω 3, 1 2(abc) 2 [ 2a 4 a 2 (b c) 2 a 3 (b + c) + a(b c) 2 (b + c) (b + c) 2 (b c) 2]. a /a = κ(a, b, c) R 1001 = R 2301 κ(a, b, c) = κ(a, c, b) κ(a, b, c) + κ(c, a, b) + κ(b, c, a) = 0 (a, b, c) R 1001 = R 2301 = R 2332 = κ(a, b, c) = a a, R 2002 = R 3102 = R 3113 = κ(b, c, a) = b b, R 3003 = R 1203 = R 1221 = κ(c, a, b) = c c. 4 z = 0 dr 2 + a2 4 dψ2 (r e iψ, z) = (s e 2iθ, e iθ ) (s, e iθ ) R S 1 ds 2 + c 2 dθ 2 s > 0 ds 2 + b 2 dθ 2 s < 0 S 1 {Im z = 0} {Re z = 0} R 1 re iψ

7 J = ( ) + R A (R A)(V ) = R lµlν V µ h µlk h νlk V µ e ν µ,ν l l,k V = µ V µ e µ k, l µ, ν R A J Σ Σ 2, 3 0, 1 1 2m = h 022 = h 033 = h 123 = h = h 023 = h 032 = h 122 = h 133. Σ κ(a, b, c) = 3 2m 2 κ(b, c, a) = κ(c, a, b) = 3 4m 2 r = 0. R A 3 R j0j0 j=2 3 R j1j1 j=2 3 j,k=2 3 j,k=2 h 0jk h 0jk = R R 3003 (h 022 ) 2 (h 033 ) 2 = 1 m 2, h 1jk h 1jk = R R 3113 (h 123 ) 2 (h 132 ) 2 = 1 m 2, R A p Σ T p Σ z = 1 T p Σ arg z

8 z = 1 Σ Σ R A Σ C 1 ε > 0 Γ sup q Γ ( r 2 (q) + (1 + (ω 2 ω 3 )(T q Γ)) ) < ε Γ t Γ 0 = Γ Σ t r 2 ω 2 ω 3 Σ a b c r > 0 1 > r a (r) a(r) > r c (r) c(r) > r b (r) b(r) ξ r > 0 x = a c y = b c. x y x r = 0 (x(0), y(0)) = (0, 1) (x(r), y(r)) (1, 0) r x [0, 1) y [ 1, 0) r > 0 (x(r), y(r)) y < 1 + x, 0 < x < 1, 1 < y < 0. y 1 + x (x, y) = (0, 1) r = 0 r = 0 x(r) = 2 m r 1 m 2 r2 + O(r 3 ) y(r) = m r 1 2m 2 r2 + O(r 3 ). a = x2 (y 1) 2 2y, b = y2 (x 1) 2 2x, c = 1 (x y)2 2xy.

9 x(r) y(r) x = 1 c (1 x)(1 + x y) y y = 1 c (1 y)(1 + y x) x. b > 0 r > 0 c c + b b = 1 c 1 x + y y a a c c = x x = 1 c, (1 x)(1 + x y) x( y) r > 0 a r a a a = r + 1 r 3 + O(r 4 ) r = 0 2m 2 a a > r r a a a r r ( a ) (a ) 2 a < 0 r > 0 a a r r r r > 0 a = a κ(a, b, c) = 1 c 2x 4 x 2 (y 1) 2 x 3 (1 + y) + x(1 y) 2 (1 + y) (1 y) 2 (1 + y) 2 κ 0 (0, 1) (1, 0) ( x a = y + 1 x2 y 2 ) dy dx 2y 2 dx dr, dy dx 1 b r > 0 c r r Σ 2xy 2.

10 2 Σ Θ = m 2 σ 2 σ 3 = m2 bc ω2 ω 3 m 2 σ 2 σ 3 dθ = 0 Θ Σ = dvol Σ (bc) < 0 r > 0 bc m 2 r Θ Q R n λ n λ 2 λ 1 k {1,, n} { trl (Q) L R n k } k j=1 λ j L {v 1,, v k } Qv j L j {1,..., k} L Q f k p k Hess(f) p M Σ Σ dr 2 = 2r ω 0, ) Hess(r 2 ) = 2 (ω 0 ω 0 + r a a ω1 ω 1 + r b b ω2 ω 2 + r c c ω3 ω 3. r 2 r > 0 r 2 Σ Σ ε δ tr L Hess(r 2 ) δ r 2

11 p r [0, ε) L T p M N M 2 r 2 N (r 2 N ) = tr N (Hess(r 2 )) 0. r 2 N tr N Hess(r 2 ) r 2 N r 2 r Hess(r 2 ) Hess(r 2 ) r 2 2 M D D 0 M Z/2 Σ RP 2 Z/2

12 RP 2

Exercise 1 (Formula for connection 1-forms) Using the first structure equation, show that

Exercise 1 (Formula for connection 1-forms) Using the first structure equation, show that 1 Stokes s Theorem Let D R 2 be a connected compact smooth domain, so that D is a smooth embedded circle. Given a smooth function f : D R, define fdx dy fdxdy, D where the left-hand side is the integral