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3.3 the L-Genus and the Todd Genus

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3.3 the L-Genus and the Todd Genus 3.3. THE L-GENUS AND THE TODD GENUS 209 3.3 The L-Genus and the Todd Genus The material in this section and the next two was first published in Hirzebruch [8]. Let B be a commutative ring with 1, and let Z, α1,...,αn,... be some independent indeterminates, all of degree 1; make new independent indeterminates qj = σj(α’s). (The σj are the symmetric functions in the α’s; for example, q1 = α1 + ···+ αn.) All computations are carried out in the ring B = B[[Z; α1,...,αn,...]]. We have the subring P = B[[Z; q1,...,qn,...,]] and in P, we have certain units (so-called one-units), namely j 1+ bjZ , where bj ∈ B. j≥1 j If Q(z) is a one-unit, 1 + j≥1 bjZ , write ∞ Q(z)= (1 + βjZ) j=1 ∞ k and call the βj’s the “roots” of Q. In the product l=1 Q(αjZ), the coefficient of Z is independent of the order of the α’s and is a formal series in the elementary symmetric functions, qj,oftheα’s. In fact, this k ··· Q coefficient has weight k and begins with bkq1 + , call the coefficients Kk (q1,q2,...,qk). We deduce that ∞ ∞ Q l 1+ Kl (q1,q2,...,ql)z = Q(αjZ). l=1 l=1 j We see that a 1-unit, Q(Z)=1+ j≥1 bjZ , yields a sequence of polynomials (in the elementary { Q}∞ symmetric functions q1,...,qk) of weights, 1, 2,...,say Kl l=1, called the multiplicative sequence of the 1-unit. { }∞ Conversely, given some sequence of polynomials, Kl l=1, it defines an operator on 1-units to 1-units, call it K. Namely, ∞ j l K(1 + qjZ )=1+ Kl(q’s)Z . j≥1 l=1 So, Q gives the operator KQ; namely, ∞ j Q l K(1 + qjZ )=1+ Kl (q’s)Z . j≥1 l=1 Claim. When Q is given, the operator KQ is multiplicative: Q j Q j Q j j K (1 + qjZ )K (1 + qj Z )=K ((1 + qjZ )(1 + qj Z )). j≥1 j≥1 j≥1 j≥1 Now, to see this, the left hand side is ∞ ∞ ∞ ∞ ∞ Q l Q m [1 + Kl (q ’s)Z ][1 + Km(q ’s)Z ]= Q(αrZ) Q(αs Z)= Q(αtZ), l=1 m=1 r=1 s=1 t=1 210 CHAPTER 3. THE HIRZEBRUCH-RIEMANN-ROCH THEOREM where we have chosen some enumeration of the α s and the α s, say α1,...,αt,...= α1,α1 ,α2,α2 ,.... But, ∞ ∞ Q n Q(αtZ)=1+ Kn (elem. symm. functions in α sandα s)Z , t=1 n=1 which is the right hand side of the assertion. If conversely, we have some endomorphism of the 1-units under multiplication, say K,lookat j Q K(1 + Z)=1+ j≥1 ajZ = Q(Z), some power series. Compute K .Wehave ∞ Q j K (1 + qjZ )= Q(αlZ), j≥1 l=1 j ∞ where 1 + j≥1 qjZ = j=1(1 + αjZ). So, as K is multiplicative, ∞ ∞ j K(1 + qjZ )=K( (1 + αjZ)) = K(1 + αjZ). j≥1 j=1 j=1 By definition of Q, the right hand side of the latter is ∞ Q j Q(αlZ)=K (1 + qjZ ) l=1 j≥1 and this proves: Proposition 3.36 The endomorphisms (under multiplication) of the 1-units are in one-to-one correspon- dence with the 1-units. The correspondence is endo K 1-unit K(1 + Z), and 1-unit Q endo KQ. We can repeat the above with new variables: X (for Z); cj (for qj); γj (for αj); and connect with the above by the relations 2 2 Z = X ; αl = γl . This means ∞ ∞ ∞ i i j r (−1) qiZ = cjX cr(−X) (∗) i=0 j=0 r=0 and if we set Q(X)=Q(X2)=Q(Z), then Q Qe Qe Kl (q1,...,ql)=K2l (c1,...,c2l)andK2l+1(c1,...,c2l+1)=0. ∗ 2 − For example, ( ) implies that q1 = c1 2c2,etc. Proposition 3.37 If B ⊇ Q, then there is one and only one power series, L(Z), so that for all k ≥ 0, the coefficient of Zk in L(Z)2k+1 is 1. In fact, √ ∞ Z 22l L(Z)= √ =1+ (−1)l−1 B Zl. (2l)! l tanh Z l=1 3.3. THE L-GENUS AND THE TODD GENUS 211 ∞ j Proof .Fork = 0, we see that L(Z) must be a 1-unit, L(Z)=1+ j=1 bjZ . Consider k = 1; then, 3 2 3 L(Z) =(1+b1Z + O(Z )) ,so 3 2 2 (1 + b1Z) + O(Z )=1+3b1Z + O(Z ), which implies b1 =1/3. Now, try for b2: We must have 1 5 1 5 1+ Z + b Z + O(Z3) = 1+ Z + b Z + O(Z3) 3 2 3 2 5 4 1 1 3 = 1+ Z +5 1+ Z b2Z + O(Z ) 3 3 10 = junk + +5b Z2 + O(Z3). 9 2 Thus, 10 1 5b =1− = − , 2 9 9 i.e., b2 = −1/45. It is clear that we can continue by induction and obtain the existence and uniqueness of the power series. Now, let √ Z M(Z)= √ . tanh Z Then, M(Z)2k+1 is a power series and the coefficient of Zk is (by Cauchy) 1 M(Z)2k+1 k+1 dZ. 2πi |Z|= Z √ Let t = tanh Z. Then, √ 1 dt = sech2 Z √ dZ, 2 Z so √ √ M(Z)2k+1 Z2 Zdt 2dt dZ = √ = √ . zk+1 t2k+1Zsech2 Z t2k+1sech2 Z However, sech2Z =1− tanh2 Z =1− t2,so M(Z)2k+1 2dt dZ = . zk+1 t2k+1(1 − t2) When t goes once around the circle |t| = small(), Z goes around twice around, so 1 2dt = twice what we want 2k+1 − 2 2πi |t|=small() t (1 t ) and our answer is 1 dt 1 t2kdt = + other zero terms = 1, 2k+1 − 2 2k+1 − 2 2πi |t|=small() t (1 t ) 2πi |t|=small() t (1 t ) as required. 212 CHAPTER 3. THE HIRZEBRUCH-RIEMANN-ROCH THEOREM Recall that 1 1 L(Z)=1+ Z − Z2 + O(Z3). 3 45 Let us find L1(q1)andL2(q1,q2). We have 2 ··· 1+L1(q1)Z + L2(q1,q2)Z + = L(α1Z)L(α2Z) 1 − 1 2 2 ··· 1 − 1 2 2 ··· = 1+ α1Z α1Z + 1+ α2Z α2Z + 3 45 3 45 1 1 1 =1+(α + α )Z + − (α2 + α2)+ α α Z2 + O(Z3). 3 1 2 45 1 2 9 1 2 We deduce that 1 L (q )= q 1 1 3 1 2 2 2 − 2 − and since α1 + α2 =(α1 + α2) 2α1α2 = q1 2q2,weget 1 1 L (q ,q )=− (7q − q2)=− (7q − q2). 2 1 2 45 2 1 32 · 5 2 1 Here are some more L-polynomials: 1 L = (62q − 13q q +2q3) 3 33 · 5 · 7 3 1 2 1 1 L = (381q − 71q q − 19q2 +22q q2 − 3q4) 4 34 · 52 · 7 4 3 1 2 2 1 1 1 L = (5110q − 919q q − 336q q + 237q q2 + 127q2q − 83q q3 +10q5). 5 35 · 52 · 7 · 11 5 4 1 3 2 3 1 2 1 2 1 1 Geometric application:LetX be an oriented manifold and let TX be its tangent bundle. Take a multi- plicative sequence, {Kl}, in the Pontrjagin classes of TX : p1,p2,.... Definition 3.3 The K-genus (or K-Pontrjagin genus)ofX is 0ifdimRX ≡ 0 (mod 4), Kn(p1,...,pn)[X]ifdimRX =4n. (a 4n rational cohomology class applied to a 4n integral homology class gives a rational number). When Kl = Ll (our unique power series, L(Z)), we get the L-genus of X, denoted L[X]. 2n Look at PC , of course, we mean its tangent bundle, to compute characteristic classes. Write temporarily T 2n Θ= PC aU(2n)-bundle. We make ζ(Θ) (remember, ζ :U(2n) → O(4n)), then we know i j k pi(ζ(Θ))(−Z) = cj(Θ)X ck(Θ)(−X) , i j k 2 2n with Z = X . Now, for projective space, PC , 2n 2n+1 2n+1 1+c1(Θ)t + ···+ c2n(Θ)t + t =(1+t) . Therefore, 2n 2 i 4n+1 4n+2 2n+1 2n+1 2 2n+1 pi(ζ(Θ))(−X ) + terms in X ,X =(1+X) (1 − X) =(1− X ) . i=0 3.3. THE L-GENUS AND THE TODD GENUS 213 Hence, we get 2n +1 p (ζ(Θ)) = H2i, 1 ≤ i ≤ n. i i Let KL be the multiplicative homomorphism coming from the 1-unit, L. Then L 2 i 2 l K (1 + pi(−X ) )= Ll(p1,...,pl)(−X ) i j = KL((1 − X2)2n+1) = KL(1 − X2)2n+1 = L(−X2)2n+1 = L(−Z)2n+1. n n n The coefficient of Z in the latter is (−1) and by the first equation, it is (−1) Ln(p1,...,pn). Therefore, we have Ln(p1,...,pn)=1, for every n ≥ 1. Thus, we’ve proved 2n Proposition 3.38 On the sequence of real 4n-manifolds: PC , n =1, 2,..., the L-genus of each, namely Ln(p1 ,...,p n),is1.TheL-genus is the unique genus having this property. Alternate form: If we substitute 2n+1 pj = j in the L-polynomials, we get 2n +1 2n +1 L ,..., =1. n 1 n Now, for the Todd genus. Proposition 3.39 If B ⊇ Q, then there is one and only one power series, T (X), having the property: For all k ≥ 0, the coefficient of Xk in T (X)k+1 is 1.
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