Alphabeta Math
Pipeline-generated
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

✓ 5 results · all verified · 4 also independently AI-judged
Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 1 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

The Hirzebruch Signature Theorem — Examples

1 · Prerequisites

2 · Summary

The examples test the signature theorem on the smallest closed oriented manifolds where every number can be written down. The complex projective plane is the four-dimensional normalization: its middle cohomology is one-dimensional with Q(y,y)=1, so σ(CP2)=1, while p(TCP2)=(1+y2)3 gives p1[CP2]=3, so the theorem reads 1=3/3. Reversing the orientation negates the fundamental class and hence the form and every Pontryagin number: σ(−CP2)=−1 with p1[−CP2]=−3. The product S2×S2 exhibits the hyperbolic form, with a⋅a=b⋅b=0 and a⋅b=1, and its zero signature forces p1[S2×S2]=0; the product CP2×CP2 exhibits multiplicativity, with both factors and the product of signature and L-genus equal to 1. The closing counterexample separates two invariants that are often confused: the complex projective plane and its orientation reversal have the same Euler characteristic 3 but signatures 1 and −1, so the Euler characteristic does not determine the signature, and the intersection form carries more information than the Betti-number count.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)Open item page →

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.

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)Open item page →

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.

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passOpen item page →

The signature of the product of two 2-spheres is zero: the hyperbolic intersection form

Example

Assume AC, inherited from Poincare duality, the Kuenneth suppliers and the signature definition. Equip S2 with its standard orientation and S2×S2 with the product orientation and product smooth structure. Let z∈H2(S2;Z) be the orientation class, characterized by ⟨z,[S2]⟩=1, and set a=pr1∗z, b=pr2∗z. Then H2(S2×S2;Z)=Za⊕Zb,a⋅a=0,b⋅b=0,a⋅b=1, so the middle-dimensional intersection form is the hyperbolic matrix (0110) and σ(S2×S2)=0. Consequently the first Pontryagin number vanishes: p1[S2×S2]=0.

Facts & Assumptions

Given: AC; the standard orientation of S2 with its orientation class z and fundamental class [S2]; the product orientation and product smooth structure on S2×S2; the projections pr1,pr2.

[F1]

For a commutative ring R that is a PID with either every Hq(Y;R) or every Hp(X;R) finite free over R, the external product is a graded-ring isomorphism H∗(X;R)⊗RH∗(Y;R)→H∗(X×Y;R); this uses AC (Cohomological Kunneth cross product is a ring isomorphism).

[F2]

For n≥1, H~k(Sn;G)=G for k=n and 0 otherwise, so H2(S2;Z)≅Z and H∗(S2;Z) is free (Homology of spheres). The fundamental class [S2] of Fundamental class of a compact oriented manifold restricts at every point to the local generator of the standard orientation, and for the compact connected boundaryless manifold S2 restriction to every local stalk is injective with image Z (Top homology of a connected manifold); hence under the identification H2(S2;Z)≅Z the class [S2] corresponds to ±1 and generates H2(S2;Z).

[F3]

For every space X and n≥0 the universal coefficient sequence 0→Ext⁡Z1(Hn−1(X;Z),G)→Hn(X;G)→Hom⁡Z(Hn(X;Z),G)→0 is natural with evaluation as the second map; this uses AC (Topological universal coefficient short exact sequence for cohomology). Over a field the evaluation map alone is an isomorphism (Cohomology over a field is dual to homology over that field).

[F4]

With the product orientation and product smooth structure, S2×S2 is a closed oriented smooth 4-manifold and [S2×S2]=[S2]×[S2] (Product orientations, The fundamental class of a product is the cross product of the fundamental classes, Products of smooth manifolds have a canonical product smooth structure).

[F5]

The Kronecker pairing is multiplicative under cross products: ⟨α×β,c×d⟩X×Y=⟨α,c⟩X⟨β,d⟩Y (The Kronecker pairing is multiplicative under cross products).

[F6]

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

[F7]

For every closed oriented smooth 4-manifold M, p1[M]=3 σ(M) (The four-dimensional signature formula).

Verification

technique · compute the Kuenneth basis and evaluate the four products on the product fundamental class
1.1givenF2F3

By [F2], H∗(S2;Z) is free with H0(S2;Z)=Z⟨1⟩, the class [S2] generates H2(S2;Z)≅Z, and by [F3] applied to n=2 the evaluation H2(S2;Z)→Hom⁡Z(H2(S2;Z),Z) is an isomorphism because Ext⁡Z1(H1(S2;Z),Z)=0; hence the dual generator z is the unique class in H2(S2;Z) with ⟨z,[S2]⟩=1, and it is the orientation class of the statement.

2.1step 1.1F1F2F3

By [F1] with X=Y=S2 and R=Z, the cross product is a ring isomorphism H∗(S2;Z)⊗H∗(S2;Z)→H∗(S2×S2;Z); on degree two it identifies a=pr1∗z with z⊗1 and b=pr2∗z with 1⊗z, so H2(S2×S2;Z)=Za⊕Zb, while a⌣a corresponds to (z⌣z)⊗1, which is zero because H4(S2;Z)=0 by [F3] and [F2]; b⌣b=0 likewise.

3.1step 2.1F4F5F6

By [F4] the product is a closed oriented smooth 4-manifold with fundamental class [S2]×[S2], so by [F6] and [F5], Q(a,a)=⟨a⌣a,[S2]×[S2]⟩=0, Q(b,b)=0, and Q(a,b)=⟨a⌣b,[S2]×[S2]⟩=⟨z×z,[S2]×[S2]⟩=⟨z,[S2]⟩⟨z,[S2]⟩=1.

4.1step 3.1F1F3F6

By [F1] over R and field evaluation [F3], the coefficient images of a,b form a real basis (the normalized integral class z maps to the normalized real class); thus in the basis {a,b} the matrix of Q is the hyperbolic matrix (0110); the class a+b satisfies Q(a+b,a+b)=2>0 and a−b satisfies Q(a−b,a−b)=−2<0, and Q(a+b,a−b)=0. Since a+b,a−b form a basis, this diagonalizes the form to diag⁡(2,−2), so the inertia is (1,1,0) and [F6] gives σ(S2×S2)=1−1=0.

5.1step 4.1F7∎

By [F7], p1[S2×S2]=3σ(S2×S2)=3⋅0=0, as claimed.

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)Open item page →

Multiplicativity of the signature on products of projective spaces

Example

Assume AC, inherited from the signature-product theorem. Give CP2×CP2 the product orientation and the product smooth structure. Then σ(CP2×CP2)=σ(CP2) σ(CP2)=1⋅1=1, and the L-genus of the product is also 1. Since [CP2] is the first of the polynomial generators of Ω∗SO⊗Q, this verifies multiplicativity of the signature on the square of the degree-four generator of the graded ring Ω∗SO⊗Q, whose square lies in degree eight.

Facts & Assumptions

Given: AC; the closed oriented smooth 4-manifold CP2 with the complex orientation; the product CP2×CP2 with the product orientation and product smooth structure.

[F1]

For closed oriented smooth manifolds M4a and N4b with the product orientation, σ(M×N)=σ(M) σ(N) (The signature is multiplicative under Cartesian products).

[F2]

The complex projective plane satisfies σ(CP2)=1 (The signature and the L-genus agree on complex projective space).

[F3]

For P=CP2k1×⋯×CP2kr with the product orientation, ki≥1, one has σ(P)=1=L[P] (The signature and the L-genus agree on products of complex projective spaces).

[F4]

The products CP2j1×⋯×CP2jr over partitions form a Q-basis of Ω4kSO⊗Q; equivalently Ω∗SO⊗Q is the polynomial algebra on the classes [CP2],[CP4],[CP6],… (Products of complex projective spaces span rational oriented bordism).

[F5]

Products of closed oriented smooth manifolds carry the product orientation and the canonical product smooth structure, hence are again closed oriented smooth (Product orientations, Products of smooth manifolds have a canonical product smooth structure).

Verification

technique · direct; instantiate multiplicativity on the product and compare with the L-genus value
1.2givenF1F5

By [F5], CP2×CP2 is a closed oriented smooth 8-manifold with the product orientation, so [F1] with M=N=CP2 gives σ(CP2×CP2)=σ(CP2) σ(CP2).

2.1step 1.2F2

By [F2], σ(CP2)=1, so step 1.2 gives σ(CP2×CP2)=1⋅1=1.

3.1givenF3

By [F3] with k1=k2=1, the L-genus of the product is L[CP2×CP2]=1, agreeing with the signature value of step 2.1.

4.1step 2.1step 3.1F4∎

By [F4] the class [CP2] is the first polynomial generator of Ω∗SO⊗Q, so CP2×CP2 represents its square in degree eight; steps 2.1 and 3.1 exhibit multiplicativity there, the signature of the product being the product of the factor signatures and the L-genus agreeing.

CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)Open item page →

The Euler characteristic does not determine the signature

Statement refuted

The Euler characteristic of a closed oriented four-manifold determines its signature.

Facts & Assumptions

Given: AC; the complex projective plane CP2 with the complex orientation and its orientation reversal −CP2; the Grassmannian Gr⁡1(C3) with its Schubert stratification.

[F1]

Gr⁡(1,V) is the set of 1-dimensional linear subspaces of V, and PC2 is the quotient of C3∖{0} by nonzero scalar multiplication; a point of either is exactly a complex line in C3, so CP2=Gr⁡1(C3) (The Grassmannian of r-dimensional subspaces of a finite-dimensional vector space, projective space points).

[F2]

A Schubert symbol for Gr⁡n(FN) is a strictly increasing sequence 1≤a1<⋯<an≤N, its cell satisfies e(a)≅Fd(a) with d(a)=∑i(ai−i), of real dimension d(a) for F=R and 2d(a) for F=C, and the Schubert strata form a finite CW structure on Gr⁡n(FN) (Schubert cells in real and complex Grassmannians, Schubert cells give the stable Grassmannian CW structure).

[F3]

For a finite CW complex with cn cells in dimension n, the Euler characteristic is χ(X)=∑n(−1)ncn(X) (Euler characteristic of a finite CW complex).

[F4]

σ(CP2)=1 and σ(−CP2)=−1 (The signature and the L-genus agree on complex projective space, the orientation computation in step 1.1).

[F5]

−CP2 is CP2 with the reversed orientation, so the two have the same underlying space and the same finite CW structures (Fundamental class of a compact oriented manifold).

Counterexample

technique · count the Schubert cells of the projective plane in both orientations
1.1givenalgebra

The projective tangent-bundle supplier gives H∗(CP2;Z)=Z[y]/(y3) and ⟨y2,[CP2]⟩=1 (The tangent bundle of complex projective space and its Pontryagin classes). The signature supplier computes the real middle matrix (1) and signature 1 (The signature and the L-genus agree on complex projective space). Reversing orientation negates the fundamental class (Fundamental class of a compact oriented manifold), hence the middle form (The middle-dimensional intersection form of a closed oriented 4k-manifold); its matrix is (−1), with signature −1.

1.2givenF1F2

By [F1], CP2=Gr⁡1(C3); by [F2] its Schubert symbols are the integers a1∈{1,2,3}, with d(a1)=a1−1∈{0,1,2}, so the cells have real dimensions 0,2,4 and there are no cells in odd dimensions: c0=c2=c4=1 and c1=c3=0.

1.3givenF4

By [F4], the signatures of the two closed oriented smooth four-manifolds are σ(CP2)=1 and σ(−CP2)=−1.

2.1step 1.2F3

By [F3] applied to this finite CW structure, χ(CP2)=c0−c1+c2−c3+c4=1−0+1−0+1=3.

3.1step 2.1F5

Orientation reversal changes neither the underlying space nor its cells, since −CP2 is CP2 with the reversed orientation by [F5]; hence cn(−CP2)=cn(CP2) for every n and χ(−CP2)=3 as well.

4.1step 3.1step 1.3∎

Thus CP2 and −CP2 have the same Euler characteristic 3 but different signatures, so the Euler characteristic of a closed oriented four-manifold does not determine its signature, and the intersection form carries information beyond the alternating Betti-number count.

Sources