指标定理（四） (9-8)

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

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