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The orientation-reversed projective plane has signature minus one

Example

Assume AC, inherited from the projective-plane signature example. Let −CP2 denote CP2 with the reversed orientation. Then σ(−CP2)=−1,p1[−CP2]=−3,σ(−CP2)=p1[−CP2]3.

Facts & Assumptions

Given: AC; the closed oriented smooth 4-manifold CP2 with the complex orientation and its orientation reversal −CP2; the generator y=c1(γ∗).

[F1]

For closed oriented smooth 4k-manifolds, σ(−M)=−σ(M), where −M is M with the reversed orientation (The signature is additive under disjoint union and negates under orientation reversal).

[F2]

The fundamental class of the orientation-reversed manifold is [−M]=−[M] in H4k(M;Z): the class determined by the reversed orientation is the negative of the class determined by the original orientation (Fundamental class of a compact oriented manifold).

[F3]

Pontryagin classes are defined from the complexification of the bundle and require no orientation of the base; reversing the orientation of the manifold leaves the tangent bundle and its Pontryagin classes unchanged, and for CP2 the supplier gives p1(TCP2)=3y2 and p1[CP2]=3 (Pontryagin classes by complexification, The tangent bundle of complex projective space and its Pontryagin classes).

[F4]

The first Pontryagin number is p1[M]=⟨p1(TM),[M]⟩ and the Kronecker pairing is additive in its homology variable (Pontryagin numbers of a closed oriented manifold, Kronecker evaluation pairing).

[F5]

The middle form is QM(x,y)=⟨x⌣y,[M]⟩ and, for every closed oriented 4-manifold, σ(M)=p1[M]/3 (The middle-dimensional intersection form of a closed oriented 4k-manifold, The four-dimensional signature formula).

Verification

technique · track the two sign changes, in the fundamental class and in the orientation of the form
1.1givenF3

The projective tangent-bundle supplier gives H∗(CP2;Z)=Z[y]/(y3), ⟨y2,[CP2]⟩=1, and p1(TCP2)=3y2, so p1[CP2]=3 (The tangent bundle of complex projective space and its Pontryagin classes). Also σ(CP2)=1 by The signature and the L-genus agree on complex projective space.

1.2givenF3

The orientation reversal −CP2 has the same underlying smooth manifold and the same tangent bundle as CP2, and by [F3] the Pontryagin classes do not see the orientation of the manifold, so p1(T(−CP2))=p1(TCP2)=3y2.

1.3givenF1F2F5

By [F5] and [F2], Q−CP2(y,y)=⟨y⌣y,[−CP2]⟩=−⟨y2,[CP2]⟩=−1, so the matrix of the middle form in the basis {y} is (−1) with inertia (0,1,0) and signature −1; this is the same value as [F1] applied to σ(CP2)=1.

2.1step 1.2F2F3F4

By [F2] and [F4] applied to M=CP2 with reversed orientation, p1[−CP2]=⟨p1(T(−CP2)),[−CP2]⟩=⟨3y2,−[CP2]⟩=−3⟨y2,[CP2]⟩=−3, since ⟨y2,[CP2]⟩=1 and p1[CP2]=3 by [F3].

3.1step 2.1step 1.3F5∎

By the four-dimensional formula in [F5], σ(−CP2)=p1[−CP2]/3=−3/3=−1, in agreement with step 1.3, so all three displayed values hold.

Depends on

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