注意:指标定理(4/5)
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Now we compute its topological index. We need to compute the Chern character of an induced bundle,and we consider this in a general setting.
Let ρ:∪ₙ → ∪ₘ be a homomorphism,and E be a complex vector bundle of rank n. Denote the induced bundle by E × ᵨ ℂᵐ. We want to find how the Chern classes of the two bundles are related.
Without loss of generality,we assume that E=l₁⨁ · · · ⨁lₙ,that is to say,the structure group has been reduced to the torus Tⁿ. ρ defines a representation of the torus and the representation decomposes into one-dimensional ones,since Tⁿ is compact and abelian.
We first consider α:Tⁿ= S¹ × · · · × S¹ → S¹. Suppose it is of the form (z₁,. . .,zₙ) ↦
∏zᵢᵏⁱ,kᵢ ∈ ℤ. The S¹-bundle ET × α S¹ → BT has first chern class c(α) ∈ H²(BT;ℤ),which is the image of the generator in H²(BS¹;ℤ). H*(BT;ℤ)=ℤ[t₁,. . .,tₙ] is the tensor product of copies of H*(BS¹;ℤ).
Let ф:Tⁿ → Uₙ be the embedding. The bundle ET ×α ℂ → BT has transition functions ∏ fᵢᵏⁱ when ET × ф ℂⁿ=l₁⨁ · · · ⨁lₙ → BT has transition functions (f₁,. . .,fₙ),according to the definition of α. Thus ET × α ℂ → BT is isomorphic to l₁ᵏⁱ ⨂ · · · ⨂lₙᵏⁿ .Thus c(α)=∑kᵢtᵢ.
In general,if α=(α¹,. . .,αᵐ):Tⁿ → Tᵐ,
ch(ET × α ℂᵐ)=∑eᶜ⁽αʲ⁾=∑eΣᵢᵏʲᵢᵗᵢ .
ⱼ ⱼ
The total Chern class is
c(ET × α ℂᵐ)=∏(1+c(αʲ))=∏(1+∑kʲᵢtᵢ).
ʲ ⱼ
By splitting principle,this is true for general rank n complex vector bundles. Here are some examples:i) ρ:∪ₙ → ∪ₙ is conjugation (of complex numbers). E × ᵨ
ℂⁿ=E*=ˉE. kⁱⱼ=–δⁱⱼ.
ₙ ₙ
c(E*)=∏(1 – xᵢ)=∑(–1)ⁱcᵢ(E).
ᵢ₌₁ ᵢ₌₀
ii) For ∧ᵏ E,u ∈ Tⁿ acts as eᵢ₁ ∧ · · · ∧ eᵢₖ ↦ ue,∧ · · · ∧ ueᵢₖ.
c(αᵢ₁,. . .,ᵢₖ)=xᵢ₁+· · ·+xᵢₖ and c(∧ᵏE)
=∏ (1+(xᵢ₁+· · ·+xᵢₖ)).
1≤i₁<· · ·<iₖ≤n
iii) For ∧ᵏE*,
c(∧ᵏE*)=∏ (1–(xᵢ₁+· · ·+xᵢₖ)).
1≤i₁<· · ·<iₖ≤n
ch(∧ᵏE*)=∑eˉ(xᵢ₁+· · ·+xᵢₖ)).
1≤i₁<· · ·<iₖ≤n
iv)
ⁿ ⁿ
∑ ch(∧ᵏE*) · tᵏ=∏(1+te⁻ˣⁱ).
ₖ₌₀ ᵢ₌₁
Now let’s consider an oriented real bundle E of rank 2n. By splitting principle, we may assume that it splits into a direct sum of oriented plane bundles (or complex line
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bundles) and talk about its Chern roots. We want to calculate ∑ⁿₖ₌₀ ch(∧ᵏE* ⨂ ℂ) · tᵏ Let ρ:Tⁿ ⊂ SO₂ₙ → ∪₂ₙ be the inclusion,where Tⁿ denotes the standard maximal torus. This Tⁿ ⊂ ∪₂ₙ is a conjugate subgroup of a torus in the standard maximal torus in ∪₂ₙ, and the conjugation restricts to
(cos(2πtᵣ) –sin(2πtᵣ)
) ↦(ᵉⁱ²πᵗʳ 0)
(sin(2πtᵣ) cos(2πtᵣ) ( 0 ₑ⁻ⁱ²πᵗʳ)
on each S¹. Thus the weights of ρ:Tⁿ ⊂ SO₂ₙ → ∪₂ₙ are (x₁,–x₁,. . .,xₙ,–xₙ). where the xᵢ’s are the Chern roots. By the same argument as above,
ⁿ ₙ
∑ch(∧ᵏE* ⨂ ℂ) · tᵏ=∏(1+te⁻ˣⁱ)(1+teˣⁱ). ᵢ₌₁
ₖ₌₀
Settingt=–1,we have
ₙ ₙ
∑ch(∧ᵏE* ⨂ ℂ) · (–1)ᵏ=∏(1 – e⁻ˣⁱ)(1 – eˣⁱ). ᵢ₌₁
ₖ₌₀
Thus for the de Rham complex defined for an oriented compact 2n-dimensional manifold,
ₘ
indₜ(d)=〈((∑(–1)ⁱ · ch(∧ⁱT* ⨂ ℂ) )
ᵢ₌₀
ₙ xⱼ 1
∏(─── · ───).[X]〉
ⱼ₌₁ 1 – e⁻ˣʲ 1– eˣʲ
ₙ
=〈(∏(1 – e⁻ˣⁱ)(1 – eˣⁱ)
ᵢ₌₁
ₙ xⱼ 1
∏ (─── · ───)),[X]〉
ⱼ₌₁ 1 – e⁻ˣʲ 1 – eˣʲ
=e(X)
=indα(d).
3.2 Dolbeault complex
Let X be a complex n-dimensional manifold,Tℝ its tangent bundle as a real manifold, and T=T¹,⁰ the holomorphic tangent bundle. Recall that
__
T*ℝ ⨂ ℂ ≅ T* ⨁ T*,
__
∧ⁱ(T*ℝ ⨂ ℂ) ≅ ∧ⁱ(T* ⨁ T*) ≅ ⨁ₚ₊q₌ᵢ (∧ᴾT*
__
⨂ ∧qT*).
__
Let ∧ᴾ,q=∧ᴾT* ⨂ ∧qT*,Aᴾ,q:=Γ
__
(∧ᴾT* ⨂ AqT*).The exterior derivative d: Aᴾ,q → Aᴾ⁺¹,q ⨁ Aᴾ,q⁺¹ splits into d=∂+ˉ∂,with ∂:Aᴾ,q → Aᴾ⁺¹,q and ˉ∂: Aᴾ,q → Aᴾ,q⁺¹.
One easily check that as before,ˉ∂ defines an elliptic complex of differential operators for each fixed p. This is called the Dolbeault complex. Let Hᴾ,q be the q-th cohomology group of this complex,and hᴾ,q be its dimension.
Definition 5 For fired p,χᴾ;=∑ⁿq₌₀(–1)q h ᴾ,q is defined to be the αnαlyticαl index of ˉ∂. χ⁰ is αlso cαlled the αrithmetic genus.
Now we would like to find what the topological index is by the index theorem. First set p=0.
__
∧qT*=AqT. By the calculations in the former subsection we have
ₙ ₙ
∑ch(∧ᵏT) · tᵏ=∏(1+teˣⁱ).
ₖ₌₀ ᵢ₌₁
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Thus
ₘ
χ⁰=〈((∑(–1)ⁱ · ch(∧ᵏT)
ᵢ₌₀
ₙ xⱼ 1
∏(─── · ───)),[X]〉
ⱼ₌₁ 1 – e⁻ˣʲ 1 – e⁻ˣʲ
ₙ xⱼ
=〈∏(───)),[X]〉
ⱼ₌₁ 1 – e⁻ˣʲ
=Td(T)[X]=Td(X).
For general p,define χy=∑ⁿₚ₌₀ χᴾ·yᴾ to be a formal linear combination of χᴾ. Formally χy is the analytical index of the elliptic complex (Cq,ˉ∂),whose direct summands consist of yᴾ copies of the p-th complex for each p, i.e. Cq=⨁ₚyᴾ · ∧ᴾ,q.
Thus
∑(–1)q ch(Cq)=∑(–1)q yᴾ ch(∧ᴾT*)ch(∧qT)
p,q
= (∑(–1)q ch(∧q T))(∑ yᴾch(∧ᴾT*))
q p
=∏(1 – eˣʲ)(1+ye⁻ˣʲ),
ⱼ
and consequently
ₙ xⱼ
χy=∏((1+y · e⁻ˣʲ) ───) [X].
ⱼ₌₁ 1 – e⁻ˣʲ
That is,
ₙ xⱼ
χy=∫ₓ∏((1+y · e⁻ˣʲ) ───).
ⱼ₌₁ 1 – e⁻ˣʲ
This finishes the calculation.
For y=0。the above formula tells us that
χ₀=χ(X,𝓞 )=∫ₓ Td(X).
Indeed this is a special case of Hirzebruch-Riemann-Roch theorem,as we will see later.
For y=–1 and X an n-dimensional compact Kähler manifold,this yields the Gauss-Bonnet formula
ₙ
χ–1=∑(–1)ᴾ⁺q hᴾ,q=e(X)=∫ₓ∏ xⱼ=∫ₓ cₙ(x). ⱼ₌₁
For y=1 and X a 2n-dimensional compact Kähler manifold,the χy-genus is
ₙ xⱼ
χ₁=χ(X,⨁Ωᴾₓ)=∫ₓ∏((1+eˉˣʲ)───)
ⱼ₌₁ 1 – e⁻ˣʲ
=∫ₓL(X),
where L(X)=L(p(X)) is the L-class explained later. This is the Hirzebruch signature theorem.
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4 Hirzebruch Signature Theorem
4.1 Multiplicative sequence
Let A be a fixed commutative ring with unit, and A* a graded A-algebra. Write AΠ for the ring of formal power series α₀+α₁+. . ., where αᵢ ∈ Aⁱ,and AΠ₀ for its subset containing elements with leading term 1. AΠ₀ form a group under multiplication.
Now consider a sequence of polynomials
K₁(x₁),K₂(x₁,x₂),K₃(x₁,x₂,x₃),. . .
with coefficients in ∧ such that,if the variable xᵢ is assigned degree i,then each Kₙ (x₁,. . .,xₙ)is homogeneous of degree n. Given an element α ∈ AΠ with leading term 1, define a new element K(α) ∈ AΠ by the formula
K(α)=1+K₁(α₁)+K₂(α₁,α₂)+ . . .
Definition 6 These Kₙ form α multiplicαtiυe sequence if K(αb)=K(α)K(b) for αll grαded ∧-αlgebrαs A* αnd αll α,b ∈ AΠ₀(i.e.K:AΠ₀ → AΠ₀ is α group homomorphism).
Theorem 11 Giυen α formαl pοωer series f(t)=1+λ₁t+ . . . ωith coefficients in ∧,there ezists α unigue multiplicαtiυe sequence such thαt K(1+t)=f(t) for αll 1+t ∈ AΠ₀.
We omit the proof. Note that if α=(1+t₁) . . . (1+tₙ),then K(α)=f(t₁) . . . f(tₙ). tᵢ
Consequently,the Todd class ∏ᵢ ───
1–e⁻ᵗⁱ
∈ HΠ(B∪ₙ;ℚ) is indeed given by the mul-tiplicative sequence associated to the series
x
td(x)=───,
1–e⁻ˣ
where we take ∧=ℚ,AΠ=HΠ(B∪ₙ;ℚ). We may write
Td(c)=∏ td(tᵢ) ∈ HΠ(B∪ₙ;ℚ).
where c is the total Chern class and tᵢ’ s are the Chern roots.
Here you may notice somet hing strange.tᵢ is indeed in H²(B∪ₙ;ℚ) but not H¹(B∪ₙ;ℚ). To avoid this,we may replace H*(B∪ₙ;ℚ) by H²*(B∪ₙ;ℚ):= ⨁ₖ H²ᵏ(B∪ₙ;ℚ) with grad-ing given by k. We will take this for granted when considering Chern classes,and use H⁴* when considering Pontrjagin classes.
4.2 Hirzebruch signature theorem
We define the L-genus of a 4n-dimensional smooth compact oriented manifold to be
L[M]=〈L(p(M)),[M]〉=〈Lₙ(p₁(M), . . . ,pₙ(M)),[M]〉
where L is the multiplicative sequence associated to
√x 2²ᵏB₂ₖ x x²
──── =∑ ────xᵏ=1+─ – ─
tanh(√x) k≥0 (2k)! 3 45
+ . . . .
For a 4n-dimensional smooth compact oriented manifold,we define its signature σ(M) to be the signature of the symmetric bilinear form defined on H²ⁿ(M⁴ⁿ;ℚ),
(α,b) ↦〈α∪b,[M]〉.
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Theorem 12 For α 4n-dimensionαl smooth compαct oriented mαnifold.
L[M]=σ(M).
Both sides are algebra homomorphisms from Ω*.⨂ to ℚ (where Ω* is the oriented cobordism ring),so we only need to check the result on the generators ℂP²ᵏ of Ω* ⨂ ℚ.
H²ⁿ(ℂP²ⁿ;ℚ) is generated by a single element and we easily see that its signature is one.
For a complex vector bundle ξ viewed as a real bundle,
1 – p₁+p₂ – · · · =(1 – c₁+c₂ – . . . )(1+C,c₁+c₂+ . . . ).
In particular,for the tangent bundle of ℂPⁿ,
1 – p₁+p₂ – · · · =(1 – α)ⁿ⁺¹(1+α)ⁿ⁺¹=(1 – α²)ⁿ⁺¹,
1+p₁+p₂+ · · · =(1+α²)ⁿ⁺¹. Consequently for ℂP²ᵏ,L(p)=
α
(──)²ᵏ⁺¹,
tαnh(α)
where α is a generator of the cohomology ring.
To calculate〈L(p(ℂP²ᵏ)),[ℂP²ᵏ]〉,it suffices to calculate the coefficient of α²ᵏ
α
in(───)²ᵏ⁺¹.
tαnh(α)
The result follows from direct calculation using contour integration in complex analysis, and this completes the proof. ▢
In the proof above we are using the fact that the total Chern class of ℂPⁿ is of the form (1+α)ⁿ⁺¹ where α is a generator. We sketch a proof.Let L be the tautological line bundle(which is the dual of the hyperplane bundle H).Let ε be the trivial line bundle,and ω the complementary rank n bundle of L ⊂ εⁿ⁺¹. Thus TℂPⁿ=Hom(L,ω) and
TℂPⁿ ⨁ ε=Hom(L,ω ⨁ L)=Hom(L,εⁿ⁺¹)=(n+1)H.
Taking total Chern class of both sides we have c(TℂPⁿ)=(1+c₁(H))ⁿ⁺¹,as desired.
Now we show that this is indeed a special case of the Atiyah-Singer index theorem. Recall Aⁱ:=Γ(∧ⁱ(T* ⨂ ℂ)),〈α,b〉= ∫ₓ α∧ *ˉb.
Define τ:=(–1)ⁱ⁽ⁱ⁻¹⁾/²⁺ᵏ* and d*:= – * d*=–τdτ,where dim(X)=4k,and * is the Hodge-* operator. Since τ² is the identity, ∧(T* ⨂ ℂ)=E₊ ⨁ E₋ splits into eigenbundles with eigenvalues +1 and –1 with respect to τ. Since
T(d+d*)=τd – ττdτ=τd – dτ= –(dτ – τd)
= – (dτ+d*τ)= – (d+d*)τ,
one can consider d+d*:ΓE₊ → ΓE₋.
Recall H²ᵏ(X;ℂ) is isomorphic to the vector space of harmonic forms 𝓡 ²ᵏ(X)=ker(d+d*) ⊂ Γ(∧²ᵏ(T* ⨂ ℂ)). One verifies that 𝓡 ²ᵏ(X) splits into eigenspaces 𝓡 ₊²ᵏ(X)⨁
𝓡 ₋²ᵏ(X) with respect to τ:Γ(∧²ᵏ(T* ⨂ ℂ)) → Γ(∧²ᵏ(T* ⨂ ℂ)). Restricting to the space of real harmonic forms H²ᵏ(X;ℝ) ⊂ H²ᵏ(X;ℂ), this gives exactly the decomposition into positive and negative parts with respect to the intersection form. To see this,for example, for a real harmonic 2k-form α in 𝓡 ₊²ᵏ(X),we have
0<〈α,α〉=∫ₓ α ∧ *α=∫ₓ α ∧ τα=∫ₓ α ∧ α.
Hence σ(X)=dimℂ(𝓡 ₊²ᵏ)–dimℂ(𝓡 ₋²ᵏ). We have an elliptic operator d+d*:ΓE₊ → ΓE₋ since its square is the Laplace operator Δ:Γ(∧(T* ⨂ ℂ)) → Γ(∧(T* ⨂ ℂ)) which
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