Milnor s Exotic 7-Spheres Jhan-Cyuan Syu ( June, 2017 Introduction

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1 許展銓 R 4 R 4 R n n 4 R 4

2 ξ : E π M 2n J : E E R J J(v) = v v E ξ ξ ξ R ξ : E π M M ξ : Ē π M ξ ξ R = ξ R i : E Ē i(cv) = ci(v) c C v E ξ : E π M n M c(ξ) = c i (ξ) H 2n (M, Z) i 0 c 0 (ξ) = 1 c i (ξ) H 2i (M, Z) i c i (E) = 0 i > n f : M M M M c(f ξ) = f c(ξ) c(ξ η) = c(ξ) c(η) η M c(γ) = 1 + g γ C P g H 2 (C P ) e(ξ R ) = c n (ξ) H 2n (M, Z) ξ n M

3 p(η) = i N 0 p i (η) p i (η) = ( 1) i c 2i (η C) H 4i (M, Z) η M ξ ξ C = C ξ C ξ ξ R C = C ξ ξ c i ( ξ) = ( 1) i c i (ξ) π n (X, x 0 ) = {f : n X f f(t 0 ) = x 0 }/ t 0 n [f] + [g] := [f + g] f, g : n X f(t 0 ) = g(t 0 ) = x 0 (f + g) : n X Ψ ρ ρ : n n n = n {t 0 } {t 0 } n Ψ : n n X n n n t 0 n n = n {t 0 } {t 0 } n ρ : n n n n {t 0 } {t 0 } n {t 0 } {t 0 } Ψ : n n X Ψ({ } {t 0 }) = f( ) Ψ({t 0 } { }) = g( )

4 H := {a + bi + cj + dk a, b, c, d R} {q 0,, q n } H n+1 4n+3 4n+3 H n+1 = R 4n+4 4n+3 R 4n+4 = H n+1 4n+3 n 4n+3 := {(q 0,, q n ) H n+1 (q 0,, q n ) 2 = q α 2 = 1}. (2) = 3 H 4n+3 q (q 0,, q n ) := (qq 0,, qq n ) 4n+3 H n+1, q H q = 1 (2) (2) = 3 (2) 4n+3 4n+3 / (2) 4n+3 / (2) = 4n+3 / 3 = H P n γ n H P n n := c 2 (γ n C ) = e(γ n R ) H 4 (H P n, Z). H P n H 4i (H P n, Z) = i n Z H i (H P n, Z) = 0 4 i c(γ n C ) = 1 + c 1 (γ n C ) + c 2 (γ n C ) = 1 + n p(γ n R ) c 0 (γ n C ) = 1 c 2 (γ n C ) = c i (γ n C ) = 0 i 0, 2 p(γ n R ) = p i (γ n R ) = ( 1) i c 2i (γ n R C) = ( 1) i c 2i (γ n C γ n C ) i N 0 i N 0 i N 0 π α=0 = c 0 (γ n C )c 0 (γ n C ) c 0 (γ n C )c 2 (γ n C ) c 2 (γ n C )c 0 (γ n C ) + c 2 (γ n C )c 2 (γ n C ) = 1 n n + 2 n = 1 2 n + 2 n.

5 (4) 4 4 G k π k 1 (G) k k k (4) 4 π 3 ((4)) π 3 ((4)) = π 3 ((3)) π 3 ( 3 ) = Z Z, (4) 4 (h, j) Z 2 f hj : 3 (4) f hj (v)w := v h wv j v, w H v 3 H ξ hj f hj R 4 hj ξ hj σ : 3 (4) σ : 3 (4) σ(v)w := vw σ (vw) := wv H H P 1 = 4 n = 1 γ := γ 1 σ γ σ σ σ π 3 ((4)) e(ξ hj ) = (h + j) p 1 (ξ hj ) = 2(h j) := 1 n = 1 e(γ R ) = p 1 (γ R ) = c 2 (γ C C) = c 2 (γ C γ C ) = [c 0 (γ C )c 2 ( γ C ) + c 2 (γ C )c 0 ( γ C )] = 2. e( γ R ) = p 1 ( γ R ) = 2 e(ξ hj ) = (h + j) p 1 (ξ hj ) = 2(h j)

6 k (R k+p ) k R k+p k (R k+p ) k R k+p k (R k+p ) γp k γ k (R k+p ) := {(X, v) X k (R k+p ) v X}. k (R k+p ) γ k p γ k (R k+p ) := {( X, ṽ) X k (R k+p ) ṽ X}. P(n) := {(i 1,, i r ) N r r N, i 1 i r, i i r = n} p(n) := P(n), P(n). R 2 ξ m 2n 1 (R ) m H ( 2n+1 (R ); R) = R[p 1 ( ξ 2n+1 ),, p n ( ξ 2n+1 )] H ( 2n (R ); R) = R[p 1 ( ξ 2n ),, p n 1 ( ξ 2n ), e( ξ 2n )].

7 M M M M + M M M M 1, M 2 n M 1 + ( M 2 ) = W (n + 1) W M, M [M] = [M ] n Ω n n Ω n Ω := n N 0 Ω n {G α } G α G β G α+β Ω Ω i Ω j Ω i+j ([M], [N]) [M N] [M] = [M ] Ω i [N] = [N ] Ω j M M = W N N = V M N M N = (M + W ) N M (N V ) = W N + M V = (W N M V ) [M] [N] = [M ] [N ] M, N N N f : M N A M f N A p A f 1 (N ) df(t p M) + T f(p) N = T f(p) N.

8 f A N f N A = M f {y} y N f f A N T x M dfx T f(x) N T f(x) N/T f(x) N x A f 1 (N ) N = {y} df x x A f 1 (y) f : M N y N M = m N = n f {y} f 1 (y) m n x f 1 (y) df x : T x M T yn N := df x m n M R k k L : R k R m n N T x M R k F : M N R m n x (f(x), L(x)). df x (v) = (df x (v), L(v)) df x F U x V (y, L(x)) F f 1 (y) U ({y} R m n ) V f 1 (y) m n f : M N f N {y N f {y}} N f : M N M, N C := {x M df x }

9 0 N f(c) = {y N f {y}} N f(c) N {0} {0} M R m X M M K M f : M R n f X {0} K M K (K K ) ε > 0 g : M R n g X K {0} f c K = g c K c K := M K f(x) g(x) < ε x M λ : M [0, 1] λ K 1 λ c K 0 y Rn y < ε f y g(x) = f(x) λ(x)y f c K = g c K f(x) g(x) < ε x M g K {0} x 0 g 1 (0) K 0 = g(x 0 ) = f(x 0 ) λ(x 0 )y = f(x 0 ) y x 0 f 1 (y) f y df(t x0 M) = T y R n = T 0 R n. dg(t x0 M) = d(f + λy)(t x0 M) = df(t x0 M) = T 0 R n. y 0 g K X {0} g X K {0} f X K {0} Df(x) x X K f 1 (0) Df = ( f i / x j ) ij X K

10 K (X K ) (K K ) Df K U := X K g 1 (0) y U K x U ( ) gi x j ij = ( fi y i λ ) i x j x j ij y (Df) g K g K X {0} {0} K g 1 (0) = f 1 (y) (Df) 0 f 1 (y) ξ M ξ (ξ) := (ξ)/ (ξ) (ξ) ξ 1 (ξ) ξ = 1 t 0 (ξ) (ξ) M ξ k M (ξ) (k 1) ξ M k τ : π n+k ((ξ)) Ω n (ξ) f : n+k f f 0 : n+k f 1 0 ( t 0 ) = f 1 ( t 0 ) {W 1,, W r } f 1 0 (M) f 0 (W i ) π 1 (U i ) = U i R k U i M ρ i : π 1 (U i ) R k K i W i f 1 0 (M) (K 1 K r ) f 0 W i f 1,, f r f i f 1 i ( t 0 ) f i Wi K i = f i 1 Wi K i f i K1 K r M i = 1,, r

11 π(f i (x)) M π(f 0 (x)) x f 1 0 ( t 0 ) f 0 f i f i 1 f i 1 (W i ) π 1 (U i ) = U i R k ρ i f i 1 : W i R k ρ i f i 1 (K1 K i ) W i {0}. ρ i f i 1 : W i R k ρ i f i : ρ i f i Wi K i = ρ i f i 1 Wi K i f i K1 K r {0} N := g 1 ((ξ) t 0 ) f i : W i π 1 (U i ) = U i R k π(f i (x)) ρ i f i (x) f 1, f 2,, f r g := f r g 1 (M) (K1 K r ) g K1 Kr M g M K 1 K r f 1 0 (M) n+k c (0, 1) f 0 (x) < c x / (K 1 K r ) f i f i (x) f i 1 (x) < c r x n+k. g(x) f 0 (x) < c x n+k g(x) = 0 x / (K 1 K r ) g 1 (M) (K1 K r ) g M τ([f]) = [g 1 (M)] g 1 (M) n [f 0 ] = [f 1 ] π n+k ((ξ)) f 0 M f 1 M f 1 0 (M) = f 1 1 (M) F : n+k [0, 1] (ξ) F (x, [0, 1 3 ]) = f 0(x) F (x, [ 2 3, 1]) = f 1(x). f 0 M f 1 M F n+k (0, 1 3 ] N [ 2,1) M. 3

12 F F : n+k [0, 1] (ξ) F n+k (0,1) F (x, [0, δ)) = f 0 (x), F (x, (1 δ, 1]) = f 1 (x), δ 1/3 F (M) = f 1 1 (M) f 1 0 (M) [f 1 0 (M)] = [f 1 1 (M)] Ω n τ γ k p := γ k (R k+p ) k (R k+p ) k n p n τ : π(( γ k p )) Ω n M n n M n R n+k k T N k M n R n+k T N k M n U M n R n+k T N k U = T N k γ k n γ k p ( γ k p ). g : U ( γ p k ) g M g 1 ( k (R k+p )) = M g ĝ : n+k ( γ p k ) n+k = R n+k { } n+k U t 0 [ĝ] π(( γ p k )) [M n ] = [ĝ 1 ( k (R k+p ))] C P 2i 1 C P 2ir, (i 1,, i r ) P(m) Ω 4m Ω 4m p(m)

13 X r r 1 π n (X) Q = H n (X; Q) n 2r Ω Q = Q[C P 2, C P 4, C P 6, ] Ω n π n+k (( γ p k )) π n+k (( γ p k )) Q = H n+k (( γ p k ); Q). H n+k (( γ k p ); Q) = H n+k (( γ k p ); Q) Ω n p(m) n = 4m Ω n = 0 4 n. Ω n p(m) n = 4m Ω n = p(m) n = 4m C P 2i 1 C P 2ir (i 1,, i r ) P(m) Ω 4m Q

14 F M n H n (M; F ) = F H m (M; F ) = 0 m > n. M n H k (M; R) = H n k (M; R) k = 0, 1,, n R H i (M; F ) H n i (M; F ) F. H i (M; F ) = H n i (M; F ) = H n i (M; F ) H n i (M; F ) H i (M; F ) H i (M; Z) H n i (M; Z) Z. n H n/2 (M; Z) x, y := x(y), x, y H n/2 (M; Z) x H n/2 (M; Z) x, n/2 n/2 M n M µ M H(M; Z) = Z M

15 M 4n α, β H 2n (M, R) α, β := (α β)(µ M ). H p dr (M; R) = H p (M; R). M n 4 n σ(m) n = 4k σ(m), H 2k (M; R),,, σ : Ω Z σ σ : Ω Z σ σ(m + N) = σ(m) + σ(n) σ(m N) = σ(m) σ(n) σ(m) = 0 M = W

16 M = m, N = n W = m + 1 V := M N 4 V 4 m 4 n σ(m N) = 0 σ(m) σ(n) = 0 V = 4k H 2k (V ; R) = s+t=2k H s (M; R) R H t (N; R). x, y H 2k (V ; R) xy(µ V ) := x y(µ V ) = 0 {v s i }, {w t j} H s (M; R), H t (N; R) v s i v m s j = δ ij, w t iw n t j = δ ij s m/2, t n/2 A = H m/2 (M; R) H n/2 (N; R) m, n A = 0 B := A H 2k (V ; R) B {v s i w t j s + t = 2k, s m 2, t n 2 } R H 2k+1 (W 4k+1, M 4k ) H 2k (M 4k ) H 2k (W 4k+1 ) i i H 2k (W 4k+1 ) H 2k (M 4k ) H 2k+1 (W 4k+1, M 4k ) i H 2k (M) x, y M 4k x, y W 4k+1 i (x ) = x, i (y ) = y x, y = (x y)(µ M ) = i (x y )(µ M ) = (x y )i (µ M ). A = n N 0 A n A Π := {a 0 +a 1 +a 2 + a i A i } A Π 1 := {1 + a 1 + a 2 + a i A i } A Π x A Π {K n } n N0 K n (x 1,, x n ) n K(ab) = K(a)K(b) a, b A Π 1 K(x) := 1 + K 1 (x 1 ) + K 2 (x 1, x 2 ) +. δ

17 A R[t] t 1 f(t) 1 + λ 1 t + λ 2 t 2 + R[[t]] {K n } K(1 + t) = f(t) {K n } K(1 + t) = 1 + K 1 (t) + K 2 (t, 0) + K 3 (t, 0, 0) + = 1 + λ 1 t + λ 2 t 2 + λ 3 t 3. K n (x 1,, x n ) x n 1 λ n n {K n } n N {t 1,, t n } 1 I = (i 1,, i r ) P(n) λ I := λ i1 λ ir 1,, n {t 1,, t n } { 1,, n } K n ( 1,, n ) := λ I g I ( 1,, n ) I P(n) g I g I ( 1,, n ) = t i 1 j1 t ir j r 1 j 1,, j r n g I (ab) = HJ=I g H (a)g J (b). K(ab) = K(a)K(b) {K n } f(t) {K n } m M m 4 m K(M m ) = 0 m = 4k K(M 4 k) := K k (p 1,, p k )(µ M ). g I

18 {L n } t t = t 1 45 t2 + + ( 1)n 1 2 2n B n t n +, (2n)! B k k σ(m 4k ) = L(M 4k ) σ(c P 2k ) = L(C P 2k ) k N σ(c P 2k ) = 1 L(C P 2k ) p(c P 2k ) = (1 + a 2 ) 2k+1 a := c 1 (γ 1 ) γ 1 C P 2k L(1 + a ) = a 2 a 2 L(p(C P 2k )) = ( a ) 2k+1. a a z ( z ) 2k+1 z 2 k u = z z 1 2πi L(C P 2 k) = 1 dz = du 1 u 2 dz ( z) = u 2 + u 4 + du = 1. 2k+1 2πi u 2k+1

19 f M f y f M y y f : M n R α (x 1, x 2,, x n ) U y x i (y) = 0 1 i n f(x) = f(b) x 2 1 x 2 α + x 2 α x 2 n x U \ {y}. 4 4 ξ hj 4 f hj : 3 (4) f hj (v)w := v h wv j v 3 H v h wv j H hj ξ hj h + j = 1 M 7 k := hj Mk 7 k h j = k M 7 k 7 Mk 7 f : M 7 k R y 0, y 1 M 7 k

20 f f f(y 0 ) = 0 f(y 1 ) = 1 d dt = f(). f 1 (a) a (0, 1) f 1 ([0, a]) = f 1 ([0, b]) a, b (0, 1) a (x 1,, x 7 ) f 1 ([0, a]) y 0 f() = x x 2 7. f 1 ([0, a]) = 7 f 1 ([0, 1)) = M 7 k {y 1} = 7 7 k = 7 7 k = 7 Mk 7 M 7 k M 7 k R4 3 (R 4 {0}) 3 g (R 4 {0}) 3 g : (u, v) (u, v ) ( u u, uh vu j 2 u H R 4 {0} H 3 H h + j = 1 u = u g 3 g := 3 (4) g(u)v := u h vu j. ). Mk 7 (u, v) (u, v ) f(u, v) = Re(v) = Re(u ) u := u (v ) 1 = 1 + u u 2 u u u h vu j Mk 7

21 f f (u, v ) (u, v) (0, v) f 3 (0, 1) (0, 1) k M 7 k 7 M 7 k (ξ hj ) M 7 k 7 Mk 7 = 7 Wk 8 W 8 k = (ξ hj ) H i ( 4 ) = H 4+i ((ξ hj ), hj ) = H 4+i ((ξ hj ), t 0 ). H i (Wk 8 ) = Z i = 0, 4, 8 H i (Wk 8 ) = 0 i 0, 4, 8. Wk W k 8 σ(wk 8) = 1 1 = 7p 2 p Wk 8 e(ξ h j) = p 1 (ξ hj ) = 2k := e(γ 1 R ) H 4 (H P 1, Z) γ 1 R H P 1 π : ξ hj 4 T ξ hj = π (T 4 ) π (ξ hj ) p(t 4 ) = 1 p(t ξ hj ) = π p(ξ hj ) p 1 (T ξ hj ) = π p 1 (ξ hj ) = π (2k ) = 2k = 2ke(ξ hj ). p 2 1(T Wk 8) = p2 1(T ξ hj ) = 4k 2 4k = 7p 2 0 ( 7) k 0 ( 7) k k 0 ( 7) k Mk 7 7

Math 67. Rumbos Fall Solutions to Review Problems for Final Exam. (a) Use the triangle inequality to derive the inequality

Math 67. Rumbos Fall Solutions to Review Problems for Final Exam. (a) Use the triangle inequality to derive the inequality Math 67. umbos Fall 8 Solutions to eview Problems for Final Exam. In this problem, u and v denote vectors in n. (a) Use the triangle inequality to derive the inequality Solution: Write v u v u for all