指标定理（三） (10-9)

indₜ(M × N，V ⨂ W)＝indₜ(M，V) · indₜ(N，W). We construct a cobordism ring Ω(B∪) as follows. We say that (M，V) ~ (N，W)， if there is a manifold K such that ∂K＝M ⊔ N， and there’s a complex vector bundle P restricting to V and W on M and N respectively. iv) [M，V]＝0 in Ω(B∪) implies indₜ(M，V)＝0. Ω(B∪) ⨂ ℚ is a polynomial ring generated by [ℂP²ⁿ，1] and [S²ⁿ，αₙ] where αₙ ∈ ˉK(S²ⁿ)＝ℤ is a generator. We find that

indₜ[ℂP²ⁿ，1]＝1，indₜ[S²ⁿ，αₙ]＝2ⁿ.

Now intₜ is completely determined. In order to finish the proof，it suffices to show that the same properties hold for intα.

Here comes a problem. In order to define intα on K(T*M，T*M₀)， we need to show that every class can be represented by a differential operator. But this is not true in general. However, this is true for pseudodifferential operators.

Note that this is a generalization of the proof of the Hirzebruch index theorem，which we will talk about later.

3 Examples:de Rham Complex and Dolbeault Com-plex

3.1 de Rham complex

Let X be a compact，differentiable manifold of dimension k and T its tangent bun-dle. Let Eᵢ＝∧ⁱ (T* ⨂ ℂ). The exterior derivative d yields an elliptic complex，d＝(d：ΓEᵢ → ΓEᵢ₊₁). Locally ∂f

d：fdxⁱ¹∧ · · · dxⁱᵏ ↦df∧dxⁱ¹∧ · · · dxⁱᵏ

∂f

＝── dxʲ∧dxⁱ¹∧ · · · dxⁱᵏ， ∂xʲ

where the Einstein convention is used. Thus it is of order one，and

σ(x，υ)＝iυⱼdxʲ∧dxⁱ¹ ∧ · · · dxⁱᵏ＝iυ ∧ dxⁱ¹ ∧ · · · dxⁱᵏ.

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