80-min Lesson Plan Math 533: Fiber bundles and Characteristic classes Spring 2016 Lesson D2: Multiplicative sequences Unit D: Pontryagin classes Objective. At the end of this lesson, students will be able to: • Calculate the genus associated to a formal power series Assignments: • Read { MS section 19. • HW due Tuesday Materials and Equipment needed: • none Introduction. Today, we will associate a multiplicative sequence of polynomials to every formal power series. By evaluating these polynomials on Pontryagin classes and then inte- grating, we associate a rational number to every closed oriented manifold. Outline. introduction: 5 min lecture. Read introduction. Give reading. Relations between characteristic classes: 10 min lecture. ∗ ∗ ∗ ∗ Note: There is a natural map β : H (B; Z) ! H (B; Z2). Given a 2 H (B; Z) and b 2 H (B; Z2), we say that a = b (mod 2) , b = β(a). n Proposition. Given C ! E ! B, e(ER) = cn(E): ∗ 1 2 ∗ 1 Proof. Let H (CP ) = Z[x]=x and H (CP ) = Z[x] Since R e(T P 1) = P dim Hi( P 1) = 2, CP 1 C i C 1 1 we have c1(T CP ) = 2x = e(T CP ). 1 1 2 2 Define f : CP ! CP by f([z0; z1]) = [z0; z1]. ∗ 1 1 ∗ 1 1 ∗ 1 1 Since f (c1(γ1 )) = c1(T CP ), f (γ1 ) ' T CP ; hence f (e(γ1 )) = e(T CP ) = 2x. 1 1 Therefore, e(γ1 ) = x = c1(γ1 ). 1 1 Let ι: CP ! CP be the inclusion. 1 ∗ 1 ∗ 1 1 1 Then e(γ1 ) = e(ι (γ1) = ι (e(γ1)) = x and so e(γ1) = x = c1(γ1). 1 ∗ 1 Given any line bundle C ! L ! B, there exists g : B ! CP so L ' g (γ ). Therefore, ∗ 1 ∗ 1 ∗ 1 ∗ 1 e(L) = e(g (γ )) = g (e(γ )) = g (c1(γ )) = c1(g (γ ) = c1(L): n Let E = ⊕i=1Li. Then Y Y e(E) = e(Li) = c1(Li) = cn(E): i i n Proposition. Given oriented vector bundle R ! E ! B, e(E) = wn(E) (mod 2): Math 533 Lesson D2 p. 2 n 0 n 0 Proof. Let u 2 H (E; E0; Z) and u 2 H (E; E0; Z2) be the Thom classes. Clearly, u = u (mod 2). 0 1 1 We checked by hand that u (γ1 )jRP 1 = w1(γ1 ). The claim now follows as in the argument above. n Proposition. Given C ! E ! B, w(ER) = c(E) (mod 2): Proof. Fix C ! L ! B. Since LR is oriented, w1(LR) = 0. By the claims above, c1(L) = e(LR) = w2(LR) (mod 2). Therefore, w(LR) = c(L) (mod 2). Q Q Q Let E = ⊕iLi. Then c(E) = c(⊕Li) = c(Li) = w((Li)R) (mod 2) and w((Li)R) = w(⊕(Li)R) = w(ER). n Proposition. Given C ! E ! B, 2 k pk(ER) = ck(E) − 2ck−1(E)ck+1(E) + ··· + (−1) 2c2k(E): k Recall: Given bundle R ! E ! C,(E ⊗ C)R = E ⊕ E. If E is oriented, the orientations agree iff k(k − 1)=2 is even. Multiplicative sequences: 30 min lecture. Fix a sequence K0;K1;::: of polynomials Ki 2 Q[p1; p2; : : : ; pi] of degree i, where pj has degree j for all j. i even Let A = ⊕i≥0A be a graded, (honestly) commutative Q-algebra, e.g., A = H (M; Q): π i Let A be formal sums a = a0 + a1 + ::: , where ai 2 A . Given a 2 Aπ with leading term 1, define π K(a) := K0 + K1(a1) + K2(a1; a2) + · · · 2 A : The sequence Ki is multiplicative if K(aa0) = K(a)K(a0) 8A& a; a0 2 A with leading term 1: 2 For example, suppose K(a) = 1 + c1a1 + c2a2 + c1;1a1 + ::: . Then 0 0 0 0 aa = 1 + (a1 + a1) + (a2 + +a2 + a1a1) + ::: and so 0 0 0 0 2 0 2 0 K(aa ) = 1 + c1(a1 + a1) + c2(a2 + a2 + a1a1) + c1;1(a1 + (a1) + 2a1a1) + ::: Similarly, 0 2 0 0 0 2 K(a)K(a ) = (1 + c1a1 + c2a2 + c1;1a1 + ::: )(1 + c1a1 + c2a2 + c1;1(a )1 + ::: ) 0 0 2 0 2 2 0 = 1 + c1(a1 + a1) + c2(a2 + a2) + c1;1(a1 + (a1) ) + c1a1a1 + :::: These agree up to degree 2 iff 2 c2 + 2c1;1 = c1: Example. Given λ 2 R, we get a multiplicative sequence j Kj = λ pj 8 j: Math 533 Lesson D2 p. 3 Given n 2 N, let S ⊂ Z[t1; : : : ; tn]: be the ring of symmetric polynomials. Then S' Z[σ1; : : : ; σn]; where σi is the i'th elementary symmetric polynomial in t1; : : : ; tn, i.e. the unique polynomial of degree i s.t. X σi = (1 + t1) ··· (1 + tn): i 2 Fix a formal power series Q 2 Q[[z]] s.t. Q = 1 + λ1z + λ2z + ::: Qm Fix k ≥ 0. Given m > k, degree k part of i=1 Q(ti) is a symmetric polynomial in the ti. Therefore, there is a unique polynomial Kk 2 Q[p1; p2; : : : ; pk] such that Kk(σ1; : : : ; σk) is the de- Qm gree k part of i=1 Q(ti). (This does not depend on m.) Proposition. fKng is a multiplicative sequence. 0 00 Proof. Let σi, σi, and σi be the elementary symmetric polynomials in t1; : : : ; tn, tn+1; : : : ; tn+n0 , and t1; : : : ; tn+n0 respectively. Then 00 X 0 σi = σjσk: j+k=i Moreover, n+n0 n n+n0 00 Y Y Y 0 K(1 + σ1 + ::: ) = Q(ti) = Q(ti) Q(ti) = K(1 + σ1 + ::: )K(1 + σ1 + ::: ): i=1 i=1 i=n+1 0 Since σi and σi are independant, this completes the proof. The characteristic power series of a a multiplicative sequence fKng is K(1 + z). Claim: These maps are inverses. Given a multiplicative sequence K ande a closed oriented manifold M n, let ( R K (p(TM)) if n = 4` K[M] = M ` 0 otherwise L-genus of projective space: 20 min small group. Let p z Q(z) = p = x = 1 + z=3 − z2=45 + :::: tanh( z) Let L(p) be the multiplicative sequence associated to Q. To do: • Find L1 and L2 2 4 • Find L[CP ] and L[CP ]. 2 2 • Bonus: Find L[CP × CP ];::: . L-genus of projective space: 5 min lecture. Lemma. 2k L[CP ] = 1 8k: 2k 2 2k+1 Proof. Since L is a multiplicative sequence and p(CP ) = (1 + x ) , p !2k+1 2 2k+1 2k 2 2k+1 x x L(p(CP )) = L(1 + x ) = p = : tanh( x2) tanh x Math 533 Lesson D2 p. 4 2k We need the coefficient of x in this power series. To find it, extend x to C, divide the series by 2πix2k+1, and integrate around the origen. Let u = tanh x; then dx = du=(1 − u2) = (1 + u2 + u4 + ::: )du. Hence, 1 I dx 1 I 1 + u2 + u4 + ::: = du = 1: 2πi (tanh x)2k+1 2πi u2k+1 .