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Signature and first Pontryagin number of the complex projective plane

Example

Assume AC, inherited from the signature and characteristic-class suppliers. Give CP2 its complex orientation, let γ→CP2 be the tautological complex line, and put y=c1(γ∗). Then H∗(CP2;Z)=Z[y]/(y3) with ⟨y2,[CP2]⟩=1, the middle-dimensional intersection form of CP2 is the 1×1 matrix (1) in the basis {y}, so σ(CP2)=1; and p(TCP2)=(1+y2)3, so p1(TCP2)=3y2, the first Pontryagin number is p1[CP2]=3, and 1=σ(CP2)=p1[CP2]3.

Facts & Assumptions

Given: AC, the tautological complex line γ→CP2, the class y=c1(γ∗), and the complex orientation of CP2.

[F1]

The in-run supplier gives H∗(CPn;Z)=Z[x]/(xn+1) with x=c1(γ∗), ⟨xn,[CPn]⟩=1, and p(TCPn)=(1+x2)n+1 in the truncated ring (The tangent bundle of complex projective space and its Pontryagin classes).

[F2]

The middle-dimensional intersection form is QM(x,y)=⟨x⌣y,[M]⟩ on H2k(M;R), and the signature is the difference p−q of the positive and negative inertia indices of the nondegenerate symmetric form QM (The middle-dimensional intersection form of a closed oriented 4k-manifold, The signature of a closed oriented manifold of dimension divisible by four).

[F3]

The first Pontryagin number is p1[M]=⟨p1(TM),[M]⟩∈Z, the Kronecker evaluation of the Pontryagin class on the fundamental class (Pontryagin numbers of a closed oriented manifold, Kronecker evaluation pairing).

[F4]

For every closed oriented 4-manifold M, σ(M)=p1[M]/3 (The four-dimensional signature formula), and on projective spaces the signature and the L-genus agree: σ(CP2k)=1=L[CP2k] for every k≥0 (The signature and the L-genus agree on complex projective space).

Verification

technique · direct; the middle cohomology is one-dimensional and the Pontryagin class is read off the splitting
1.1givenF1F2

By [F1] with n=2, H∗(CP2;Z)=Z[y]/(y3) and ⟨y2,[CP2]⟩=1; hence H2(CP2;R)=Ry and, by [F2], the matrix of QCP2 in the basis {y} is the 1×1 matrix with entry Q(y,y)=⟨y⌣y,[CP2]⟩=1.

1.2givenF1

By [F1], p(TCP2)=(1+y2)3=1+3y2+3y4+y6, and the terms 3y4 and y6 vanish because y3=0 in H∗(CP2;Z)=Z[y]/(y3); hence p1(TCP2)=3y2.

2.1step 1.1F2F4

The 1×1 matrix (1) has inertia (1,0,0), so [F2] gives σ(CP2)=1−0=1, in agreement with [F4].

2.2step 1.2F1F3

By [F3], p1[CP2]=⟨3y2,[CP2]⟩=3⟨y2,[CP2]⟩=3⋅1=3.

3.1step 2.1step 2.2F4∎

By the four-dimensional formula in [F4], σ(CP2)=p1[CP2]/3=3/3=1, so 1=σ(CP2)=p1[CP2]/3, as claimed.

Depends on

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