80-min Lesson Plan Math 533: Fiber bundles and Characteristic classes Spring 2016 Lesson D2: Multiplicative 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 of 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 ∈ H (B; Z) and b ∈ 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 ∗ ∞ 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 ∞ Let ι: CP → CP be the inclusion. 1 ∗ 1 ∗ 1 1 1 Then e(γ1 ) = e(ι (γ∞) = ι (e(γ∞)) = x and so e(γ∞) = x = c1(γ∞). ∞ ∗ 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 ∈ H (E,E0; Z) and u ∈ H (E,E0; Z2) be the Thom classes. Clearly, u = u (mod 2). 0 1 1 We checked by hand that u (γ1 )|RP 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 ∈ 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 ∈ A . Given a ∈ Aπ with leading term 1, define

π K(a) := K0 + K1(a1) + K2(a1, a2) + · · · ∈ A .

The sequence Ki is multiplicative if K(aa0) = K(a)K(a0) ∀A& a, a0 ∈ 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 ∈ R, we get a multiplicative sequence j Kj = λ pj ∀ j. Math 533 Lesson D2 p. 3

Given n ∈ N, let S ⊂ Z[t1, . . . , tn]. be the of symmetric polynomials. Then S' Z[σ1, . . . , σn], where σi is the i’th elementary symmetric 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 ∈ 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 ∈ 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. {Kn} 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 {Kn} 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 . Let √ z Q(z) = √ = 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 ∀k. 2k 2 2k+1 Proof. Since L is a multiplicative sequence and p(CP ) = (1 + x ) , √ !2k+1 2 2k+1 2k 2 2k+1 x  x  L(p(CP )) = L(1 + x ) = √ = . 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