指标定理（五） (9-1)

注意：指标定理（5/5）

21

is elliptic. The adjoint of d + d*：ΓE₊ → TE₋ is d + d*：ΓE₋ → ΓE₊ . The analytical index is

indα(d+d*：ΓE₊ → ΓE₋)

＝dimℂker(d＋d*：ΓE₊ → ΓE₋) – dimℂker(d＋ d*：ΓE₋ → ΓE₊)

＝dimℂ(𝓡 ₊²ᵏ) – dimℂ(𝓡 ₋²ᵏ).

＝σ(X)，

where the next-to-last equality is because the components of harmonic forms away from the middle-dimensional cohomology have the same dimension and so cancel. Let me explain this in detail.

We have a decomposition ∧(T* ⨂ ℂ)＝E₊⨁E₋. Note that τ：∧ᵖ(T* ⨂ ℂ) → ∧⁴ᵏ⁻ᵖ(T* ⨂ ℂ). Elements of Γ(⨁²ᵏ⁻¹ ₛ₌₀ (∧ˢ(T* ⨂ ℂ) ⨁∧⁴ᵏ⁻ˢ(T* ⨂ ℂ))∩E₊) are of the form ∑²ᵏ⁻¹ₛ₌₀(αₛ＋ταₛ) where αₛ ∈ Γ(∧ˢ(T* ⨂ ℂ)). For arbitrary α ∈ Γ(∧ˢ(T* ⨂ ℂ)⨁∧⁴ᵏ⁻ˢ(T* ⨂ ℂ))，0 ≤ s＜2k，we have α＝1

──

2

((α＋τα)＋(α – τα))，where τ(α＋τα)＝(α＋τα) and τ(α–τα)＝ –(α–τα).

Thus for all 0 ≤ s＜2k，we have a decomposition

∧ˢ(T* ⨂ ℂ) ⨁ ∧⁴ᵏ⁻ˢ(T* ⨂ ℂ)＝Eˢ₊ ⨁ Eˢ₋

into eigenbundles. If α ∈ Γ(∧ˢ(T* ⨂ ℂ))，0 ≤ s s＜2k satisfies α＋τα ∈ ker(d＋d*)＝ker(Δ)， since Δ preserves degree，we must have α ∈ ker(Δ) and τα ∈ ker(Δ). Thus α – τα ∈ ker(d＋d*)＝ker(Δ). The map α ＋ τα ↦ α – τα and its inverse α – τα ↦ α ＋ τα give us the desired bijective correspondence between ker(d＋d*)∩Γ(Eˢ₊) and ker(d＋d*)∩Γ(Eˢ₋).

Next，we continue to calculate the topological index. Again by splitting we have

ch(E₊) – ch(E₋)＝∏²ᵏⱼ₌₁(eˣʲ – e⁻ˣʲ)，where we split TX ⨂ ℂ＝⨁²ᵏⱼ₌₁(lⱼ ⨁ ˉlⱼ)，and xⱼ＝c₁(lⱼ). Thus TX ≅ ⨁lᵢ implies

₂ₖ

indₜ(d+d*)＝(∏(eˣʲ – e⁻ˣʲ)

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