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✓ 21 results · all verified · 20 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

1 · Prerequisites

2 · Summary

This page develops the Hirzebruch signature theorem: on every closed oriented smooth $4 k$ -manifold the signature, defined as the inertia difference of the middle-dimensional intersection form, equals the evaluation of the $L$ -class on the fundamental class. The route isolates the two inputs. The geometric side defines the middle form $Q_{M} (x, y) = ⟨ x ⌣ y, [M] ⟩$ and proves it symmetric and nondegenerate, so that Sylvester's law gives the signature; the additivity, product and bordism-invariance properties are then proved from the form, and the duality argument with the half-dimensional isotropic subspace of a boundary supplies the vanishing on null-cobordisms. The algebraic side builds the formal series $x / \tanh x$ , the completed fourfold-graded cohomology ring, the Hirzebruch $L$ -polynomials and the total $L$ -class, and proves the multiplicative-sequence axioms by universal polynomial identities and computes the resulting class on complex projective space. The two sides meet in the two agreements proved on this page: the signature and the $L$ -genus agree on projective spaces and on products of projective spaces, and since those products form a rational basis of oriented bordism, the theorem follows. The closing corollaries record the four- and eight-dimensional formulas and their divisibility consequences, and the final remark keeps the zero extension in other dimensions distinct from the geometric definition. Every use of Poincare duality, of the characteristic numbers supplied by the preceding pair, and of the Pontryagin classes is a direct supplier use recorded in the proof contracts; the axiom of choice is inherited from the duality and characteristic-class suppliers and is declared at each consumer. The companion page verifies the theorem on explicit small manifolds.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6.1-sol)Open item page →

The middle-dimensional intersection form of a closed oriented 4k-manifold

Definition

Let $M$ be a closed oriented smooth manifold of dimension $4 k$ , $k \geq 0$ , with orientation $o$ and fundamental class $[M] \in H_{4 k} (M; Z)$ determined by $o$ (Fundamental class of a compact oriented manifold, Top homology of a connected manifold). The middle-dimensional intersection form of $M$ is $Q_{M} : H^{2 k} (M; R) \times H^{2 k} (M; R) ⟶ R, Q_{M} (x, y) : = ⟨ x ⌣ y, [M] ⟩,$ where $⌣$ is the singular cup product of Singular cohomology ring and $⟨ -, - ⟩$ is the Kronecker evaluation pairing of Kronecker evaluation pairing, read on the fundamental class. The pairing is well defined: the cup product is defined on cohomology classes and the evaluation on classes is independent of cocycle and cycle representatives by The kronecker pairing is independent of cocycle and cycle representatives. Since $\dim M = 4 k$ and both variables have degree $2 k$ , the product $x ⌣ y$ has degree $4 k$ and pairs with the top-degree class $[M]$ ; the coefficient field is $R$ , with $Z \subset R$ as the coefficient image.

Integral form. On images of integral classes the formula restricts to the integral pairing $⟨ x ⌣ y, [M] ⟩$ on $H^{2 k} (M; Z)$ , valued in $Z$ . The torsion subgroup of $H^{2 k} (M; Z)$ lies in the kernel of that integral pairing; this is proved in the symmetry and nondegeneracy lemma following on this page, not assumed here, and it is what lets the real form be controlled by the free quotient.

Geometric identification. For closed oriented embedded submanifolds $A, B \subset M$ of complementary dimension $2 k$ , with Poincaré duals $P D [A], P D [B] \in H^{2 k} (M; Z)$ of their fundamental classes, one has $Q_{M} (P D [A], P D [B]) = I (A, B) = ⟨ A, B ⟩_{M},$ the geometric intersection number, with the cohomology-first, front-evaluation cap and cup conventions and the geometric factor order fixed by The geometric intersection number is the Poincare-dual cup pairing; no third sign convention is introduced (The cap-product order is fixed by the AT convention, not minted here).

Disconnected and empty manifolds. If $M = M_{1} ⊔ \dots ⊔ M_{r}$ is a disjoint union of closed oriented components, the orientation restricts to each component (orientability is componentwise, Every manifold is F2-orientable and orientability is componentwise) and $Q_{M}$ is defined componentwise, on the summands of $H^{2 k} (M; R) = ⨁_{j} H^{2 k} (M_{j}; R)$ . For the empty manifold one sets $Q_{\emptyset} = 0$ . The form is determined by this displayed formula alone; its symmetry, nondegeneracy and additivity properties are not part of the definition and are proved in the lemma following on this page. The pairing formula itself uses no choice principle. The geometric identification assumes AC (The Axiom of Choice), inherited from its Poincare-duality supplier.

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

The middle-dimensional intersection form is symmetric and nondegenerate

Statement

Assume AC (The Axiom of Choice), inherited from Poincare duality. Let $M$ be a closed oriented smooth $4 k$ -manifold with middle form $Q_{M}$ (The middle-dimensional intersection form of a closed oriented 4k-manifold). Then: (1) $Q_{M} (x, y) = Q_{M} (y, x)$ for all $x, y \in H^{2 k} (M; R)$ ; (2) the adjoint maps $x \mapsto Q_{M} (x, -)$ and $y \mapsto Q_{M} (-, y)$ are isomorphisms onto the full $R$ -linear dual, so $Q_{M}$ is nondegenerate and $H^{2 k} (M; R)$ is finite dimensional; (3) on the free quotient $H^{2 k} (M; Z) / Tor$ the integral pairing is unimodular, and the torsion subgroup lies in its kernel; (4) for $M = M_{1} ⊔ \dots ⊔ M_{r}$ with induced orientations, $Q_{M}$ is the orthogonal direct sum $⨁_{j} Q_{M_{j}}$ under $H^{2 k} (M; R) ≅ ⨁_{j} H^{2 k} (M_{j}; R)$ . All statements hold verbatim with $Q$ in place of $R$ .

Facts & Assumptions

Given: AC; a closed oriented smooth $4 k$ -manifold $M$ with fundamental class $[M]$ and middle form $Q_{M}$ ; the field $F$ is $R$ or $Q$ .

[F1]

$Q_{M} (x, y) = ⟨ x ⌣ y, [M] ⟩$ on $H^{2 k} (M; R) \times H^{2 k} (M; R)$ , restricting on integral classes to the integral pairing, and $Q_{\emptyset} = 0$ (The middle-dimensional intersection form of a closed oriented 4k-manifold).

[F2]

Cup product is graded commutative: $u ⌣ v = (- 1)^{p q} v ⌣ u$ for $u \in H^{p}$ , $v \in H^{q}$ (Singular cohomology is graded commutative).

[F3]

Assume AC. For a closed $F$ -oriented $n$ -manifold with $F$ a field, the pairing $H^{p} \times H^{n - p} \to F$ , $(a, b) \mapsto ⟨ a ⌣ b, [M] ⟩$ , is perfect with both adjoints onto the full dual, and the groups are finite-dimensional; for $R = Z$ and an integral orientation the same formula induces a unimodular pairing on the free quotients $H^{p} (M; Z) / Tor$ and $H^{n - p} (M; Z) / Tor$ , which are finite free abelian groups with both adjoints to the integer duals isomorphisms (Poincaré duality gives a nonsingular cup pairing).

[F4]

Cap with the compatible compact orientation classes gives the duality isomorphisms $D_{M} : H_{c}^{p} (M; R) \to H_{n - p} (M; R)$ , and for compact $M$ one has $D_{M} (a) = a \cap [M]$ (Poincaré duality for oriented topological manifolds).

[F5]

For a closed $R$ -oriented $n$ -manifold with $R$ a commutative PID, every $H_{q} (M; R)$ and $H^{p} (M; R)$ is finitely generated and vanishes outside degrees $0, \dots, n$ (Finite generation from cap with a finite fundamental cycle).

[F6]

The Kronecker pairing is additive in both variables, independent of representatives, and natural: $⟨ f^{*} α, z ⟩ = ⟨ α, f_{*} z ⟩$ (Kronecker evaluation pairing, The kronecker pairing is independent of cocycle and cycle representatives).

[F7]

The fundamental class of a disjoint union corresponds to the finite tuple of component fundamental classes: for $M = ⨆_{j = 1}^{r} M_{j}$ with inclusions $i_{j}$ , the class $[M]$ is $\sum_{j} (i_{j})_{*} [M_{j}]$ , and the orientation of $M$ restricts to the given orientation on each component (Fundamental class of a compact oriented manifold).

[F8]

For every field $k$ and space $X$ , evaluation is an isomorphism $H^{n} (X; k) \to {Hom}_{k} (H_{n} (X; k), k)$ (Cohomology over a field is dual to homology over that field).

[F9]

Singular homology of a disjoint union splits: $H_{n} (⨆_{α} X_{α}; G) ≅ ⨁_{α} H_{n} (X_{α}; G)$ (The singular homology of a disjoint union is the direct sum).

Proof

technique · direct; symmetry from graded commutativity, nondegeneracy from Poincare duality, and the componentwise clause from naturality

1.1givenF1F2F6

Symmetry: for $x, y \in H^{2 k} (M; F)$ the cup product has $p = q = 2 k$ in [F2], so $x ⌣ y = (- 1)^{4 k^{2}} y ⌣ x = y ⌣ x$ because $4 k^{2}$ is even; since the Kronecker evaluation is additive in the first variable by [F6], $Q_{M} (x, y) = Q_{M} (y, x)$ .

1.2givenF1F3F4F5

Nondegeneracy and finite-dimensionality: with $p = n - p = 2 k$ the perfectness clause of [F3] says that $H^{2 k} (M; F)$ is finite dimensional and that the adjoint maps $x \mapsto ⟨ x ⌣ -, [M] ⟩$ and $y \mapsto ⟨ - ⌣ y, [M] ⟩$ into the full $F$ -linear dual are isomorphisms; these maps are exactly $x \mapsto Q_{M} (x, -)$ and $y \mapsto Q_{M} (-, y)$ by [F1], the duality isomorphisms underlying the perfectness being the cap isomorphisms of [F4]; finite generation for $F = R, Q$ also follows from [F5] with $R = F$ .

1.3givenF1F3F5F6algebra

Integral clause: the second clause of [F3] gives that the integral pairing descends to a unimodular pairing on $H^{2 k} (M; Z) / Tor$ with both adjoints to the integer duals isomorphisms, with these groups finite free abelian. The torsion subgroup lies in the kernel: if $n x = 0$ in $H^{2 k} (M; Z)$ , then $n (x ⌣ y) = (n x) ⌣ y = 0$ by bilinearity, so $x ⌣ y$ is torsion in $H^{4 k} (M; Z)$ , and any group homomorphism from a torsion group to the torsion-free group $Z$ is zero, whence $⟨ x ⌣ y, [M] ⟩ = 0$ for every $y$ by [F1].

1.4givenF1F6F7F8F9

Componentwise clause: write $M = ⨆_{j = 1}^{r} M_{j}$ with inclusions $i_{j}$ . By [F7] $[M] = \sum_{j} (i_{j})_{*} [M_{j}]$ , and by [F6] evaluated on this sum, $Q_{M} (x, y) = \sum_{j} ⟨ x ⌣ y, (i_{j})_{*} [M_{j}] ⟩ = \sum_{j} ⟨ i_{j}^{*} (x ⌣ y), [M_{j}] ⟩ = \sum_{j} ⟨ i_{j}^{*} x ⌣ i_{j}^{*} y, [M_{j}] ⟩ = \sum_{j} Q_{M_{j}} (i_{j}^{*} x, i_{j}^{*} y)$ , using naturality of the cup product. The restriction maps $(i_{j}^{*})_{j}$ identify $H^{2 k} (M; F)$ with $⨁_{j} H^{2 k} (M_{j}; F)$ : by [F8] and [F9] the group is ${Hom}_{F} (⨁_{j} H_{2 k} (M_{j}; F), F) ≅ \prod_{j} H^{2 k} (M_{j}; F)$ , a finite product, hence the direct sum, and naturality of the duality isomorphism in the inclusions identifies the factors with the restrictions. Under this identification the displayed identity says precisely that $Q_{M}$ is the orthogonal direct sum of the forms $Q_{M_{j}}$ : classes from distinct components pair to zero and each summand carries its own form.

2.1step 1.1step 1.2step 1.3step 1.4given∎

Steps 1.1-1.4 prove clauses (1)-(4); replacing $F$ by $Q$ throughout uses the field clauses of [F3] and [F5] verbatim, while clause (3) remains the assertion about integral coefficients. For $k = 0$ the pairing is $Q_{M} (x, y) = \sum_{j} ε_{j} x_{j} y_{j}$ on the component-wise constant classes, the signed count; for $M = \emptyset$ all groups vanish and $Q_{\emptyset} = 0$ , and both assertions hold trivially.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6.1-sol)Open item page →

The signature of a closed oriented manifold of dimension divisible by four

Definition

Assume AC (The Axiom of Choice), inherited from the nondegeneracy of the middle form. Let $M$ be a closed oriented smooth manifold of dimension $4 k$ , $k \geq 0$ , and let $Q_{M}$ be its middle-dimensional intersection form on $H^{2 k} (M; R)$ (The middle-dimensional intersection form of a closed oriented 4k-manifold, The middle-dimensional intersection form is symmetric and nondegenerate). Let $(p, q, z)$ be the inertia data of the symmetric form $Q_{M}$ (Positive and negative definiteness, the inertia $(p, q, r)$ , rank $p + q$ , and signature $p - q$ of a real symmetric bilinear or quadratic form): $p$ the maximal dimension of a positive-definite subspace, $q$ the maximal dimension of a negative-definite subspace, and $z$ the dimension of the radical. By Sylvester's law of inertia (Sylvester's law of inertia: every real symmetric form is congruent to $diag (I_{p}, - I_{q}, 0_{r})$ , and $(p, q, r)$ is unique) these integers are intrinsic to $Q_{M}$ and $p + q + z = \dim_{R} H^{2 k} (M; R)$ . The signature of $M$ is $σ (M) : = p - q \in Z .$

The middle form is nondegenerate by The middle-dimensional intersection form is symmetric and nondegenerate, so $z = 0$ and $p + q = \dim_{R} H^{2 k} (M; R)$ ; the signature can equivalently be read as the difference of the numbers of positive and negative diagonal entries in any basis diagonalizing $Q_{M}$ . For $k = 0$ the group $H^{0} (M; R)$ is free on the components of $M$ and the pairing is the signed count of components, so $σ (M^{0})$ is the number of positively oriented components minus the number of negatively oriented components, and $σ (\emptyset) = 0$ .

The signature is defined by this item only for dimensions divisible by four; no value is assigned in other dimensions (see the closing bookkeeping remark on this page). The value is independent of the chosen diagonalizing basis and of reading the form over $Q$ or over $R$ by The signature is independent of the diagonalizing basis and unchanged by scalar extension from the rationals to the reals ↗, the lemma named in justified_by.

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

The signature is independent of the diagonalizing basis and unchanged by scalar extension from the rationals to the reals

Statement

Assume AC, inherited from finite generation and the coefficient-duality suppliers. Let $M$ be a closed oriented smooth $4 k$ -manifold. (1) Any basis of $H^{2 k} (M; R)$ diagonalizing $Q_{M}$ has the same number $p$ of positive entries and the same number $q$ of negative entries, and $σ (M) = p - q$ equals the intrinsic inertia difference of $Q_{M}$ ; in particular the definition The signature of a closed oriented manifold of dimension divisible by four is well posed. (2) With $V_{Q} = H^{2 k} (M; Q)$ and $V_{R} = V_{Q} \otimes_{Q} R$ , the inertia data of $Q_{M}^{Q}$ and of $Q_{M} = Q_{M}^{Q} \otimes R$ agree; more generally, scalar extension of a finite-dimensional symmetric bilinear form along an ordered-field extension preserves inertia, so the signatures computed over $Q$ and over $R$ coincide.

Facts & Assumptions

Given: AC; a closed oriented smooth $4 k$ -manifold $M$ with middle form $Q_{M}$ (over $R$ ) and its rational counterpart $Q_{M}^{Q}$ on $H^{2 k} (M; Q)$ .

[F1]

$Q_{M} (x, y) = ⟨ x ⌣ y, [M] ⟩$ on $H^{2 k} (M; R)$ , and the inertia of a symmetric bilinear form is the triple $(p, q, r)$ of counts of positive, negative and zero diagonal entries in a diagonalizing basis, with signature $p - q$ (The middle-dimensional intersection form of a closed oriented 4k-manifold, Positive and negative definiteness, the inertia $(p, q, r)$ , rank $p + q$ , and signature $p - q$ of a real symmetric bilinear or quadratic form).

[F2]

Every symmetric bilinear form on a finite-dimensional real vector space is congruent to exactly one normal form $diag (I_{p}, - I_{q}, 0_{r})$ , and the counts are independent of the diagonalizing basis (Sylvester's law of inertia: every real symmetric form is congruent to $diag (I_{p}, - I_{q}, 0_{r})$ , and $(p, q, r)$ is unique).

[F3]

Two symmetric forms of the same finite dimension are congruent exactly when they have the same inertia (Two real symmetric bilinear forms are congruent if and only if they have the same inertia); every symmetric bilinear form over a field of characteristic not two has an orthogonal basis (Every symmetric bilinear form on a finite-dimensional space over a field of characteristic not $2$ has an orthogonal basis).

[F4]

If $F$ is a subfield of the ordered field $K$ with the order induced from $K$ (Ordered field, Subfield: a subring of a field closed under inverses of its nonzero elements, and therefore a field with the restricted operations), then for $a \in F$ the sign of $a$ is the same in $F$ and in $K$ , since the positive cone of $F$ is the restriction of that of $K$ ; and if $e_{1}, \dots, e_{n}$ is an $F$ -basis of $V$ , then $e_{i} \otimes 1$ is a $K$ -basis of $V \otimes_{F} K$ , with the same diagonal entries for a diagonal form (The elementary tensors of two bases form the product basis of the tensor product).

[F5]

For a closed oriented $4 k$ -manifold, $H_{j} (M; Z)$ and $H^{j} (M; Z)$ are finitely generated, and for a divisible coefficient group $F$ evaluation gives a natural isomorphism $H^{2 k} (M; F) ≅ {Hom}_{Z} (H_{2 k} (M; Z), F)$ (Finite generation from cap with a finite fundamental cycle, Cohomology with a divisible abelian coefficient group is Hom of homology).

[F6]

Every finitely generated abelian group is $Z^{b} \oplus T$ with $T$ finite, and for torsion-free $F$ one has ${Hom}_{Z} (T, F) = 0$ and ${Hom}_{Z} (Z^{b}, F) ≅ F^{b}$ (The fundamental theorem of finitely generated abelian groups from PID modules).

[F7]

The Kronecker pairing evaluates representatives and is natural in the cohomology variable under coefficient homomorphisms $u : G \to G^{'}$ : $⟨ u_{*} α, z ⟩ = u ⟨ α, z ⟩$ ; the cup product is the coefficientwise front/back cochain formula, so it commutes with coefficient change (The kronecker pairing is independent of cocycle and cycle representatives, Kronecker evaluation pairing, Singular cup product on cochains).

[F8]

The fundamental class is the unique class restricting to the prescribed local generator at every point; for the real orientation obtained from the integral one by coefficient change, the image of $[M]_{Z}$ is $[M]_{R}$ , because coefficient change carries the integral local generator to the real one and preserves local restrictions (Fundamental class of a compact oriented manifold).

Proof

technique · direct; scalar extension of a diagonal form, then the coefficient-change bridge via divisible-coefficient duality

1.1givenF3F4algebra

Let $B$ be a symmetric form on a finite-dimensional space over an ordered field $F$ , and let $K \supseteq F$ carry an extending order. By [F3] choose an orthogonal $F$ -basis with diagonal entries $b_{i}$ . Write $V = V_{+} \oplus V_{-} \oplus V_{0}$ for its positive, negative and zero coordinate subspaces, of dimensions $p, q, r$ . The radical is $V_{0}$ . A positive-definite subspace projects injectively to $V_{+}$ : a vector with zero positive coordinates has $B (v, v) \leq 0$ , so cannot be a nonzero vector in such a subspace. Rank-nullity and the subspace dimension bound (Rank-nullity: $\dim_{F} V = nullity T + rank T$ , If $\dim_{F} V = n$ and $U$ is a linear subspace of $V$ , then $U$ is finite-dimensional, $\dim_{F} U \leq n$ , and $\dim_{F} U = n$ if and only if $U = V$ ) give dimension at most $p$ , attained by $V_{+}$ . Similarly the maximal negative dimension is $q$ . Thus $(p, q, r)$ is intrinsic over any ordered field. By [F4], the extended basis has the same diagonal entries and signs over $K$ , so these intrinsic dimensions, and hence the signature, are preserved.

1.2givenF1F2F3

Basis independence and well-posedness: by [F2] the diagonal entries' sign counts $(p, q, r)$ of any diagonalizing basis of $Q_{M}$ are intrinsic to $Q_{M}$ , and by [F3] congruent forms have equal inertia, so $p$ , $q$ are the same for every diagonalizing basis; since $Q_{M}$ is symmetric and nondegenerate by The middle-dimensional intersection form is symmetric and nondegenerate, $r = 0$ and $σ (M) = p - q$ is the intrinsic inertia difference of $Q_{M}$ as claimed in The signature of a closed oriented manifold of dimension divisible by four.

1.3givenF5F6

Coefficient bridge: by [F5] applied with the divisible groups $F = Q$ and $F = R$ , evaluation gives natural isomorphisms $H^{2 k} (M; Q) ≅ {Hom}_{Z} (H_{2 k} (M; Z), Q)$ and $H^{2 k} (M; R) ≅ {Hom}_{Z} (H_{2 k} (M; Z), R)$ . Writing $H_{2 k} (M; Z) = Z^{b} \oplus T$ with $T$ finite by [F6], both groups are $Q^{b}$ and $R^{b}$ , and the map induced by the coefficient inclusion $Q ↪ R$ is the natural inclusion $Q^{b} ↪ R^{b}$ . Hence the natural map $H^{2 k} (M; Q) \otimes_{Q} R \to H^{2 k} (M; R)$ is an isomorphism $Q^{b} \otimes_{Q} R ≅ R^{b}$ .

2.1givenF1F7F8

Form compatibility: for $x, y \in H^{2 k} (M; Q)$ , with images $x_{R}, y_{R}$ under coefficient change, $Q_{M} (x_{R}, y_{R}) = Q_{M}^{Q} (x, y)$ . Indeed the cup product formula is coefficientwise by [F7], so $x_{R} ⌣ y_{R}$ is the coefficient-change image of $x ⌣ y$ ; the real fundamental class is the coefficient-change image of the integral one by [F8], and the Kronecker pairing is natural in coefficients by [F7]; evaluating the rational cup class on the integral fundamental class gives the rational number $Q_{M}^{Q} (x, y)$ , whose image in $R$ is the real evaluation. Since the $x_{R}$ span $H^{2 k} (M; R)$ over $R$ by step 1.3, the real form is the scalar extension $Q_{M}^{Q} \otimes_{Q} R$ of the rational one.

3.1step 1.1step 1.3step 2.1

Inertia agreement: by step 2.1 the real form $Q_{M}$ is the scalar extension of the rational form $Q_{M}^{Q}$ along $Q \subseteq R$ , so step 1.1 gives that their inertia triples agree; in particular the signatures over $Q$ and over $R$ coincide.

4.1step 1.1step 1.2step 1.3step 2.1step 3.1∎

Steps 1.1 and 2.1 prove the basis-independence and well-posedness clause, and steps 1.1, 1.3 and 2.1 prove the scalar-extension clause for $M$ ; the general statement for arbitrary finite-dimensional symmetric forms over an ordered field is exactly step 1.1. If the vector space of the form is zero, its diagonal data are empty and its signature is $0$ . For a zero-manifold the middle group need not vanish; both coefficient fields give its signed point count.

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

A nondegenerate symmetric form with a half-dimensional isotropic subspace has zero signature

Statement

Let $V$ be a finite-dimensional real vector space with a symmetric bilinear form $B$ (Bilinear forms, and symmetric, skew-symmetric, and alternating bilinear forms), and let $W \subseteq V$ be totally isotropic: $B (w, w^{'}) = 0$ for all $w, w^{'} \in W$ . If $B$ is nondegenerate (The matrix, left and right radicals, rank, and nondegeneracy of a bilinear form on a finite-dimensional space) and $2 \dim W = \dim V$ , then $sign (B) = b_{+} - b_{-} = 0$ , with $b_{+} = b_{-} = \frac{1}{2} \dim V$ ; equivalently, in a suitable basis $B$ is an orthogonal direct sum of hyperbolic planes.

Facts & Assumptions

Given: A finite-dimensional real vector space $V$ with a symmetric bilinear form $B$ , and a totally isotropic subspace $W \subseteq V$ with $2 \dim W = \dim V$ and $B$ nondegenerate.

[F1]

If a basis diagonalizes $B$ with $p$ positive, $q$ negative and $r$ zero diagonal entries, the inertia is $(p, q, r)$ , the rank is $p + q$ , the signature is $p - q$ , and Sylvester's law makes the triple independent of the diagonalizing basis (Positive and negative definiteness, the inertia $(p, q, r)$ , rank $p + q$ , and signature $p - q$ of a real symmetric bilinear or quadratic form).

[F2]

Every symmetric bilinear form on a finite-dimensional real vector space is congruent to exactly one normal form $diag (I_{p}, - I_{q}, 0_{r})$ with $p + q + r = \dim V$ (Sylvester's law of inertia: every real symmetric form is congruent to $diag (I_{p}, - I_{q}, 0_{r})$ , and $(p, q, r)$ is unique).

[F3]

In a basis $B$ with coordinate columns $x = [u]_{B}$ and $y = [v]_{B}$ one has $B (u, v) = x^{T} [B]_{B} y$ ; the left and right radicals are the sets of vectors pairing to zero with everything, and $B$ is nondegenerate exactly when both radicals vanish (The matrix, left and right radicals, rank, and nondegeneracy of a bilinear form on a finite-dimensional space).

[F4]

A symmetric bilinear form satisfies $B (u, v) = B (v, u)$ for all $u, v$ (Bilinear forms, and symmetric, skew-symmetric, and alternating bilinear forms).

[F5]

On $R^{N}$ the Euclidean inner product is positive definite: $⟨ x, x ⟩ = \sum_{k < N} x_{k}^{2} \geq 0$ , and $⟨ x, x ⟩ = 0$ holds only for $x = 0$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[F6]

For a linear map $T : U \to U^{'}$ with $U$ finite-dimensional, $\dim U = \dim \ker T + \dim im T$ (Rank-nullity: $\dim_{F} V = nullity T + rank T$ ).

[F7]

A subspace of a finite-dimensional space is finite-dimensional and has dimension at most that of the ambient space (If $\dim_{F} V = n$ and $U$ is a linear subspace of $V$ , then $U$ is finite-dimensional, $\dim_{F} U \leq n$ , and $\dim_{F} U = n$ if and only if $U = V$ ).

Proof

Proof technique: direct; diagonal normal form and the positivity of sums of squares.

1.1givenF1

If $\dim V = 0$ then $V = 0$ and $W = 0$ , the inertia is $(0, 0, 0)$ , so $sign (B) = 0$ and $b_{+} = b_{-} = 0 = \frac{1}{2} \dim V$ .

1.2givenF2F3

Suppose $\dim V = n$ . By [F2], fix a basis $e_{1}, \dots, e_{n}$ of $V$ in which $B$ has matrix $diag (I_{p}, - I_{q}, 0_{r})$ with $p + q + r = n$ ; thus $B (e_{i}, e_{j}) = ε_{i} δ_{i j}$ with $ε_{i} = 1$ for $i \leq p$ , $ε_{i} = - 1$ for $p < i \leq p + q$ , and $ε_{i} = 0$ otherwise. No $ε_{i}$ is zero: if $ε_{i} = 0$ , then $B (e_{i}, e_{j}) = 0$ for every $j$ , so $B (e_{i}, -)$ vanishes on the basis and hence on $V$ , putting the nonzero vector $e_{i}$ in the left radical, contrary to nondegeneracy. Hence $r = 0$ and $p + q = n$ .

1.3givenF3F4algebra

Let $w = \sum_{i = 1}^{n} a_{i} e_{i} \in W$ . Since $W$ is totally isotropic, $0 = B (w, w) = \sum_{i \leq p} a_{i}^{2} - \sum_{i > p} a_{i}^{2}$ by the diagonal matrix formula [F3]; equivalently $\sum_{i \leq p} a_{i}^{2} = \sum_{i > p} a_{i}^{2}$ .

2.1step 1.3F3F5

The coordinate projection $π_{+} : W \to V_{+} : = span (e_{1}, \dots, e_{p})$ , $π_{+} (w) = \sum_{i \leq p} a_{i} e_{i}$ , is injective: if $π_{+} (w) = 0$ , then $\sum_{i \leq p} a_{i}^{2} = 0$ , so by step 1.3 also $\sum_{i > p} a_{i}^{2} = 0$ , and by the positive definiteness [F5] applied to the coordinate vector of $π_{-} (w) = \sum_{i > p} a_{i} e_{i}$ we get $π_{-} (w) = 0$ ; hence $w = 0$ .

3.1step 2.1F6F7given

By [F6] applied to the injective linear map $π_{+}$ , whose image lies in the $p$ -dimensional space $V_{+}$ with basis $e_{1}, \dots, e_{p}$ , we get $\dim W = \dim im π_{+} \leq \dim V_{+} = p$ by [F7].

4.1step 1.3step 2.1step 3.1F5F6F7given

The projection $π_{-} : W \to span (e_{p + 1}, \dots, e_{p + q})$ onto the negative coordinates is injective by the same argument: if $π_{-} (w) = 0$ , then $\sum_{i > p} a_{i}^{2} = 0$ , so by step 1.3 also $\sum_{i \leq p} a_{i}^{2} = 0$ , so $π_{+} (w) = 0$ and step 2.1 gives $w = 0$ . Applying [F6] and [F7] as in step 3.1 to the subspace $span (e_{p + 1}, \dots, e_{p + q})$ of dimension $q$ gives $\dim W \leq q$ .

5.1step 3.1step 4.1F1given

Since $2 \dim W = n = p + q$ and $\dim W \leq \min (p, q)$ , both $p$ and $q$ equal $\dim W = \frac{1}{2} \dim V$ . By [F1] the inertia is $(m, m, 0)$ with $m = \dim W$ , so $b_{+} = b_{-} = m = \frac{1}{2} \dim V$ and $sign (B) = p - q = 0$ .

6.1step 1.2step 5.1F3F4algebra∎

Equivalently, the basis can be chosen hyperbolic: with $m = \dim W$ and $u_{i} : = e_{i} + e_{m + i}$ , $g_{i} : = \frac{1}{2} (e_{i} - e_{m + i})$ for $1 \leq i \leq m$ , bilinearity and [F4] give $B (u_{i}, u_{i}) = 1 - 1 = 0$ , $B (g_{i}, g_{i}) = \frac{1}{4} (1 - 1) = 0$ and $B (u_{i}, g_{i}) = \frac{1}{2} (1 + 1) = 1$ , while orthogonality of the diagonal basis makes the planes $span (u_{i}, g_{i})$ pairwise orthogonal with $V = ⨁_{i = 1}^{m} span (u_{i}, g_{i})$ . Thus $B$ is an orthogonal direct sum of $m$ hyperbolic planes.

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The restriction image on a cobordism boundary is Lagrangian

Statement

Assume AC. Let $W$ be a compact oriented smooth manifold of dimension $4 k + 1$ , $k \geq 0$ , with boundary $M = \partial W$ carrying the induced orientation and inclusion $i : M ↪ W$ . Let $K = im (i^{*} : H^{2 k} (W; R) \to H^{2 k} (M; R))$ and let $Q_{M}$ be the middle-dimensional intersection form of The middle-dimensional intersection form of a closed oriented 4k-manifold. Then $K = K^{⊥_{Q_{M}}}$ . In particular $K$ is a totally isotropic subspace of dimension one half of $\dim H^{2 k} (M; R)$ , hence is Lagrangian.

Facts & Assumptions

Given: AC; a compact oriented smooth $(4 k + 1)$ -manifold $W$ with boundary $M = \partial W$ , inclusion $i$ , the induced boundary orientation, and the image $K = im i^{*}$ in $V : = H^{2 k} (M; R)$ .

[F1]

$Q_{M} (x, y) = ⟨ x ⌣ y, [M] ⟩$ for $x, y \in H^{2 k} (M; R)$ , and $K^{⊥} = {x \in V : Q_{M} (k, x) = 0 for all k \in K}$ (The middle-dimensional intersection form of a closed oriented 4k-manifold).

[F2]

The long exact cohomology sequence of the pair is $\dots \to H^{2 k} (W, M; R) \overset{j}{\to} H^{2 k} (W; R) \overset{i^{*}}{\to} H^{2 k} (M; R) \overset{δ}{\to} H^{2 k + 1} (W, M; R) \to \dots$ , exact at every term; the connector sends a cocycle class $[x]$ to $[δ \tilde{x}]$ for any cochain extension $\tilde{x}$ of a representative (Long exact sequence of a pair in singular cohomology).

[F3]

The cup product on cochains satisfies the Leibniz identity $δ (φ ⌣ ψ) = δ φ ⌣ ψ + (- 1)^{∣ φ ∣} φ ⌣ δ ψ$ and restricts naturally to subspaces (Cup product Leibniz identity, Singular cup product on cochains).

[F4]

The singular coboundary is defined on chains by $(δ φ) (σ) = φ (\partial σ)$ , so $(δ φ) (z) = φ (\partial z)$ for every finite chain $z$ ; the Kronecker pairings on $W$ and on $M$ evaluate a cocycle class on a cycle class by evaluating representatives and are well defined and independent of the chosen representatives (Singular cohomology with coefficients, Kronecker evaluation pairing, The kronecker pairing is independent of cocycle and cycle representatives).

[F5]

The relative fundamental class satisfies $\partial [W, M] = [M] \in H_{4 k} (M; R)$ (Relative fundamental class and boundary orientation).

[F6]

Relative cap and evaluation: for a relative $p$ -cocycle $α \in Z^{p} (W, M; R)$ , a relative $n$ -cycle $z$ with $\partial z \in C_{n - 1} (M)$ , and an absolute $(n - p)$ -cochain $ξ$ , the front-evaluation/back-face formulas of the cohomology-first convention give the cochain identity $(α ⌣ ξ) (z) = ξ (α \cap z)$ (Relative cap products with quotient domains displayed, Singular cup product on cochains). By the boundary identity $\partial (α \cap z) = (- 1)^{p} (α \cap \partial z - δ α \cap z)$ (Cap product boundary identity) the chain $α \cap z$ is a cycle: the first term vanishes because $α$ vanishes on chains in $M$ , the second because $δ α = 0$ as a relative cocycle. Its class is the relative cap product $α \cap [W, M]$ , which is the Poincaré–Lefschetz map $T_{p} (α)$ of Poincaré–Lefschetz duality.

[F7]

Poincare-Lefschetz duality gives isomorphisms $T_{p} : H^{p} (W, M; R) \to H_{n - p} (W; R)$ , $a \mapsto a \cap [W, M]$ , for every $p$ , in particular an isomorphism $T_{2 k + 1}$ out of $H^{2 k + 1} (W, M; R)$ ; their representative independence and naturality are as stated there (Poincaré–Lefschetz duality).

[F8]

$Q_{M}$ is symmetric, nondegenerate and finite-dimensional, with both adjoints $x \mapsto Q_{M} (x, -)$ , $y \mapsto Q_{M} (-, y)$ isomorphisms onto the full dual; the Kronecker pairing over the field $R$ satisfies $H^{2 k} (W; R) ≅ {Hom}_{R} (H_{2 k} (W; R), R)$ , so a class $z \in H_{2 k} (W; R)$ with $⟨ a, z ⟩ = 0$ for all $a$ is zero (The middle-dimensional intersection form is symmetric and nondegenerate, Cohomology over a field is dual to homology over that field, The kronecker pairing is independent of cocycle and cycle representatives).

[F9]

For a finite-dimensional vector space and a subspace $U$ , the annihilator $U^{\circ} = {f : f ∣_{U} = 0}$ has $\dim U^{\circ} = \dim V - \dim U$ ; a projection onto a finite-dimensional subspace exists without choice, and rank-nullity computes the dimension of a kernel (Assuming choice, $^{\circ} (U^{\circ}) = U$ ; in finite dimension, $\dim U^{\circ} = \dim V - \dim U$ , The annihilator $U^{\circ} \leq V^{*}$ of $U \leq V$ and the preannihilator $^{\circ} S \leq V$ of $S \leq V^{*}$ , Finite-dimensional subspaces admit projections without Choice, Rank-nullity: $\dim_{F} V = nullity T + rank T$ ).

[F10]

Cup product is graded commutative, so $u ⌣ v = v ⌣ u$ whenever the degrees are even (Singular cohomology is graded commutative).

Proof

technique · direct; identify the orthogonal complement with the kernel of the cohomology connector by a cochain evaluation computation, then count dimensions

1.1givenF1F2

Exactness at $H^{2 k} (M; R)$ in [F2] reads $im i^{*} = \ker δ$ , so $K = \ker δ$ ; in particular $K \leq V : = H^{2 k} (M; R)$ is a linear subspace and $x \in K$ holds exactly when $δ x = 0$ .

1.2givenF1F2F3F4F5F6F10

Pairing identity: for $x \in H^{2 k} (M; R)$ and $a \in H^{2 k} (W; R)$ , $Q_{M} (i^{*} a, x) = ⟨ a, T_{2 k + 1} (δ x) ⟩$ . Indeed choose cocycle representatives $X \in Z^{2 k} (M; R)$ of $x$ and $A \in Z^{2 k} (W; R)$ of $a$ , an extension $\tilde{X} \in C^{2 k} (W; R)$ of $X$ , and a relative cycle $z$ with $[z] = [W, M]$ and $[\partial z] = [M]$ , possible by [F5]. Since $X$ is a cocycle, $δ \tilde{X}$ vanishes on chains in $M$ , and $δ x$ is the class of the relative cocycle $δ \tilde{X}$ by [F2]; hence by [F6] the chain $δ \tilde{X} \cap z$ is a cycle representing $T_{2 k + 1} (δ x) = δ x \cap [W, M]$ . Then $⟨ a, T_{2 k + 1} (δ x) ⟩ = A (δ \tilde{X} \cap z) = (δ \tilde{X} ⌣ A) (z) = (δ (\tilde{X} ⌣ A)) (z) = (\tilde{X} ⌣ A) (\partial z) = (X ⌣ i^{*} A) (\partial z)$ , where the equalities use, in order, the absolute Kronecker pairing on $W$ [F4], the cochain identity of [F6], the Leibniz rule [F3] with $δ A = 0$ , the definition of the coboundary [F4], and the fact that $\tilde{X}$ restricts to $X$ on $M$ while $A$ restricts to $i^{*} A$ ; since $\partial z$ represents $[M]$ and $X ⌣ i^{*} A$ is a cocycle representing $x ⌣ i^{*} a$ on $M$ , the last value is $⟨ x ⌣ i^{*} a, [M] ⟩$ by [F4]. Graded commutativity [F10] (degrees $2 k, 2 k$ ) and [F1] give $⟨ x ⌣ i^{*} a, [M] ⟩ = ⟨ i^{*} a ⌣ x, [M] ⟩ = Q_{M} (i^{*} a, x)$ , as asserted.

1.3givenF7F8

Tools for the two inclusions: by [F8], $V$ is finite-dimensional and the adjoint map $Φ : V \to V^{*}$ , $Φ (x) = Q_{M} (x, -)$ , is an isomorphism; by [F8], a homology class $z \in H_{2 k} (W; R)$ that pairs to zero with every cohomology class is zero; and by [F7] the map $T_{2 k + 1}$ is injective.

2.1step 1.1step 1.2step 1.3F8

Orthogonal complement equals the kernel: for $x \in V$ , $x \in K^{⊥}$ holds exactly when $Q_{M} (i^{*} a, x) = 0$ for all $a \in H^{2 k} (W; R)$ ; by step 1.2 this is equivalent to $⟨ a, T_{2 k + 1} (δ x) ⟩ = 0$ for all such $a$ , hence by step 1.3 to $T_{2 k + 1} (δ x) = 0$ , hence to $δ x = 0$ by the injectivity in step 1.3, and hence to $x \in K$ by step 1.1. Therefore $K^{⊥} = K$ .

3.1step 2.1F1F9

Dimension and isotropy: by step 2.1, $Φ (K^{⊥}) = K^{\circ}$ is the annihilator of $K$ in $V^{*}$ , so $\dim K = \dim K^{⊥} = \dim K^{\circ} = \dim V - \dim K$ by the annihilator dimension formula [F9]; hence $2 \dim K = \dim V = \dim H^{2 k} (M; R)$ . Since $K = K^{⊥}$ , $Q_{M} (x, y) = 0$ for all $x, y \in K$ , so $K$ is totally isotropic; a totally isotropic subspace of half the dimension of a nondegenerate finite-dimensional form is Lagrangian.

4.1step 2.1step 3.1given∎

If $M = \emptyset$ (in particular if $W$ is empty), $V = H^{2 k} (M; R) = 0$ , so both $K$ and $K^{⊥}$ are zero, so the identity and the dimension count hold trivially; for $k = 0$ the formula counts the oriented boundary points and gives $K = K^{⊥}$ of dimension $\frac{1}{2} \dim H^{0} (M; R)$ . Thus steps 2.1 and 3.1 prove $K = K^{⊥_{Q_{M}}}$ and the Lagrangian property in all cases, AC being used only through the inherited duality and field-dual suppliers.

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The signature is additive under disjoint union and negates under orientation reversal

Statement

Assume AC. Let $M, N$ be closed oriented smooth $4 k$ -manifolds. Then $σ (M ⊔ N) = σ (M) + σ (N)$ and $σ (- M) = - σ (M)$ , where $- M$ is $M$ with the reversed orientation. More generally $σ$ is additive over disjoint unions with arbitrary orientation signs.

Facts & Assumptions

Given: AC; closed oriented smooth $4 k$ -manifolds $M, N$ with middle forms $Q_{M}, Q_{N}$ and signatures $σ$ .

[F1]

$σ (M) = p - q$ is the inertia difference of the nondegenerate symmetric form $Q_{M}$ on $H^{2 k} (M; R)$ (The signature of a closed oriented manifold of dimension divisible by four).

[F2]

$Q_{M} (x, y) = ⟨ x ⌣ y, [M] ⟩$ (The middle-dimensional intersection form of a closed oriented 4k-manifold).

[F3]

For $M = M_{1} ⊔ \dots ⊔ M_{r}$ the fundamental class is $\sum_{j} (i_{j})_{*} [M_{j}]$ and the middle form is orthogonal direct sum: $Q_{M} (x, y) = \sum_{j} Q_{M_{j}} (i_{j}^{*} x, i_{j}^{*} y)$ under $H^{2 k} (M; R) ≅ ⨁_{j} H^{2 k} (M_{j}; R)$ (The middle-dimensional intersection form is symmetric and nondegenerate, Fundamental class of a compact oriented manifold, The singular homology of a disjoint union is the direct sum).

[F4]

If a nondegenerate symmetric bilinear form is an orthogonal direct sum of forms $B_{1}, B_{2}$ , then the inertia triples add: $p = p_{1} + p_{2}$ , $q = q_{1} + q_{2}$ , $r = r_{1} + r_{2}$ ; this follows because the union of diagonalizing bases diagonalizes the sum, and by Sylvester's law the inertia is intrinsic (Two real symmetric bilinear forms are congruent if and only if they have the same inertia).

[F5]

Reversing the orientation negates the fundamental class, $[- M] = - [M]$ , while the underlying smooth manifold and its tangent data are unchanged; orientability is componentwise (Fundamental class of a compact oriented manifold, Every manifold is F2-orientable and orientability is componentwise).

Proof

technique · direct; the form splits orthogonally over components and changes sign under orientation reversal

1.1givenF1F3F4

Disjoint union: for $M ⊔ N$ the form is the orthogonal direct sum $Q_{M} \oplus Q_{N}$ under $H^{2 k} (M ⊔ N; R) ≅ H^{2 k} (M; R) \oplus H^{2 k} (N; R)$ by [F3], and both summands are nondegenerate. By [F4] the inertia data add, so $p (M ⊔ N) = p (M) + p (N)$ and $q (M ⊔ N) = q (M) + q (N)$ ; hence $σ (M ⊔ N) = p (M) + p (N) - q (M) - q (N) = σ (M) + σ (N)$ by [F1].

1.2givenF1F2F5

Orientation reversal: by [F5], $[- M] = - [M]$ , so by [F2] $Q_{- M} (x, y) = ⟨ x ⌣ y, [- M] ⟩ = - ⟨ x ⌣ y, [M] ⟩ = - Q_{M} (x, y)$ . Multiplication by $- 1$ is an isomorphism of $H^{2 k} (M; R)$ carrying positive-definite subspaces of $Q_{M}$ to negative-definite subspaces of $- Q_{M}$ and conversely, so $p (- M) = q (M)$ and $q (- M) = p (M)$ ; hence $σ (- M) = q (M) - p (M) = - σ (M)$ by [F1].

2.1step 1.1step 1.2given∎

More generally, for a finite disjoint union with signs $⨆_{j} ε_{j} M_{j}$ , where $ε_{j} M_{j}$ means $M_{j}$ with its given orientation when $ε_{j} = + 1$ and the reversed orientation when $ε_{j} = - 1$ , steps 1.1 and 1.2 applied successively give $σ (⨆_{j} ε_{j} M_{j}) = \sum_{j} ε_{j} σ (M_{j})$ ; the empty union has signature $0$ and the zero-dimensional case $k = 0$ is the signed count of components, consistent with both steps.

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The signature of an oriented boundary vanishes, so the signature is an oriented cobordism invariant

Statement

Assume AC, inherited from Poincare-Lefschetz duality. Let $W$ be a compact oriented smooth $(4 k + 1)$ -manifold with boundary $M = \partial W$ carrying the induced orientation and inclusion $i : M ↪ W$ . Then $σ (M) = 0$ . Consequently, if $M_{0}, M_{1}$ are closed oriented $4 k$ -manifolds that are oriented cobordant (Oriented smooth cobordism), then $σ (M_{0}) = σ (M_{1})$ : by Oriented smooth cobordism, $\partial V = (- M_{0}) ⊔ M_{1}$ ; reversing the orientation of $V$ gives $\partial (- V) = M_{0} ⊔ (- M_{1})$ as an oriented boundary, so $0 = σ (M_{0} ⊔ (- M_{1})) = σ (M_{0}) - σ (M_{1})$ by The signature is additive under disjoint union and negates under orientation reversal. Hence $σ$ descends to a well-defined additive map on oriented bordism classes $Ω_{4 k}^{S O} \to Z$ and vanishes on null-cobordant classes (Null-cobordant closed manifolds).

Facts & Assumptions

Given: AC; a compact oriented smooth $(4 k + 1)$ -manifold $W$ with boundary $M = \partial W$ and inclusion $i$ , with the induced boundary orientation.

[F1]

For a closed oriented $4 k$ -manifold the signature is the inertia difference $σ (M) = p - q$ of the nondegenerate middle form $Q_{M}$ (The signature of a closed oriented manifold of dimension divisible by four, The middle-dimensional intersection form of a closed oriented 4k-manifold).

[F2]

With $K = im (i^{*} : H^{2 k} (W; R) \to H^{2 k} (M; R))$ , one has $K = K^{⊥_{Q_{M}}}$ , and $K$ is a totally isotropic subspace of half the dimension of $H^{2 k} (M; R)$ (The restriction image on a cobordism boundary is Lagrangian).

[F3]

A nondegenerate symmetric form with a totally isotropic subspace of exactly half the dimension has zero signature (A nondegenerate symmetric form with a half-dimensional isotropic subspace has zero signature).

[F4]

The signature is additive over disjoint unions and negates under orientation reversal: $σ (X ⊔ Y) = σ (X) + σ (Y)$ and $σ (- X) = - σ (X)$ (The signature is additive under disjoint union and negates under orientation reversal).

[F5]

Oriented cobordism is the equivalence relation generated by oriented bordisms: for a bordism $V$ from $M_{0}$ to $M_{1}$ , $\partial V = (- M_{0}) ⊔ M_{1}$ , and a null-cobordant closed manifold bounds a compact oriented manifold (Oriented smooth cobordism, Null-cobordant closed manifolds, Unoriented and oriented bordism groups).

Proof

technique · direct; the boundary restriction image is a Lagrangian subspace, so the isotropic lemma kills the signature

1.1givenF1F2F3

Boundary vanishing: by [F2] the subspace $K$ of $H^{2 k} (M; R)$ is totally isotropic for the nondegenerate form $Q_{M}$ and has dimension one half of $\dim H^{2 k} (M; R)$ ; hence $σ (M) = 0$ by [F1] and [F3].

2.1step 1.1F4F5

Cobordism invariance: let $V$ be an oriented cobordism from $M_{0}$ to $M_{1}$ , so that $\partial V = (- M_{0}) ⊔ M_{1}$ by [F5]. Reversing the orientation of $V$ makes its boundary $M_{0} ⊔ (- M_{1})$ , so step 1.1 gives $0 = σ (M_{0} ⊔ (- M_{1})) = σ (M_{0}) + σ (- M_{1}) = σ (M_{0}) - σ (M_{1})$ by [F4]; hence $σ (M_{0}) = σ (M_{1})$ .

3.1step 1.1step 2.1F4F5∎

Descent: steps 1.1 and 2.1 show that $σ$ is constant on oriented cobordism classes and vanishes on null-cobordant manifolds (which bound by [F5]); with additivity [F4] it descends to a well-defined additive map $Ω_{4 k}^{S O} \to Z$ . The empty manifold has $σ (\emptyset) = 0$ and for $k = 0$ the boundary of an oriented 1-manifold has signed count zero, consistent with step 1.1.

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

The tensor product of nondegenerate real symmetric forms has multiplicative signature

Statement

Let $V, W$ be finite-dimensional real vector spaces with nondegenerate symmetric bilinear forms $B$ and $C$ (Bilinear forms, and symmetric, skew-symmetric, and alternating bilinear forms) of inertia data $(p, q)$ and $(p^{'}, q^{'})$ . The symmetric form $B \otimes C$ on $V \otimes_{R} W$ defined on simple tensors by $(B \otimes C) (v \otimes w, v^{'} \otimes w^{'}) = B (v, v^{'}) C (w, w^{'})$ is nondegenerate with inertia data $(p p^{'} + q q^{'}, p q^{'} + p^{'} q)$ , so $sign (B \otimes C) = sign (B) sign (C) .$

Facts & Assumptions

Given: Finite-dimensional real vector spaces $V, W$ , nondegenerate symmetric bilinear forms $B$ on $V$ and $C$ on $W$ , with inertia data $(p, q)$ and $(p^{'}, q^{'})$ .

[F1]

If a basis diagonalizes a symmetric bilinear form with $p$ positive, $q$ negative and $r$ zero diagonal entries, then its inertia is $(p, q, r)$ and its signature is $p - q$ (Positive and negative definiteness, the inertia $(p, q, r)$ , rank $p + q$ , and signature $p - q$ of a real symmetric bilinear or quadratic form).

[F2]

Every symmetric bilinear form on a finite-dimensional real vector space is congruent to exactly one normal form $diag (I_{p}, - I_{q}, 0_{r})$ (Sylvester's law of inertia: every real symmetric form is congruent to $diag (I_{p}, - I_{q}, 0_{r})$ , and $(p, q, r)$ is unique).

[F3]

Every symmetric bilinear form on a finite-dimensional vector space over a field of characteristic not $2$ has a basis whose distinct vectors are pairwise orthogonal (Every symmetric bilinear form on a finite-dimensional space over a field of characteristic not $2$ has an orthogonal basis).

[F4]

A symmetric bilinear form satisfies $B (u, v) = B (v, u)$ for all $u, v$ (Bilinear forms, and symmetric, skew-symmetric, and alternating bilinear forms).

[F5]

In a basis with coordinate columns $x, y$ one has $B (u, v) = x^{T} [B] y$ ; the form is nondegenerate exactly when its radical vanishes, equivalently when its matrix in any basis is invertible (The matrix, left and right radicals, rank, and nondegeneracy of a bilinear form on a finite-dimensional space).

[F6]

If $(v_{i})$ and $(w_{j})$ are bases of $V$ and $W$ over the commutative ring $R$ , then $(v_{i} \otimes w_{j})$ is a basis of $V \otimes_{R} W$ (The elementary tensors of two bases form the product basis of the tensor product).

[F7]

The elementary tensors generate $V \otimes_{R} W$ and are additive and $R$ -balanced in each variable (The tensor product $M \otimes_{R} N$ from the additive group underlying the free $Z$ -module on $M \times N$ , elementary tensors, and finite tensor sums, Universal property of the tensor product for balanced maps into abelian groups).

[F8]

Two real symmetric forms of the same finite dimension are congruent exactly when they have the same inertia (Two real symmetric bilinear forms are congruent if and only if they have the same inertia).

Proof

technique · direct; diagonalize both factors and read off the inertia of the tensor product in the product basis

1.1givenF3F5

By [F3] there are bases $v_{1}, \dots, v_{n}$ of $V$ and $w_{1}, \dots, w_{m}$ of $W$ diagonalizing $B$ and $C$ : $B (v_{i}, v_{i^{'}}) = b_{i} δ_{i i^{'}}$ and $C (w_{j}, w_{j^{'}}) = c_{j} δ_{j j^{'}}$ for real $b_{i}, c_{j}$ . No $b_{i}$ is zero: a zero entry would make $v_{i}$ pair to zero with every basis vector, hence lie in the radical by [F5], contradicting nondegeneracy; the same holds for the $c_{j}$ . After reordering, $b_{i} > 0$ for $i \leq p$ , $b_{i} < 0$ for $p < i \leq n$ , and $c_{j} > 0$ for $j \leq p^{'}$ , $c_{j} < 0$ for $p^{'} < j \leq m$ .

1.2givenF4F5F6

By [F6] the products $v_{i} \otimes w_{j}$ form a basis of $V \otimes_{R} W$ . Define a bilinear form $D$ on $V \otimes_{R} W$ by prescribing its diagonal matrix in this basis, $D (v_{i} \otimes w_{j}, v_{i^{'}} \otimes w_{j^{'}}) : = b_{i} c_{j} δ_{i i^{'}} δ_{j j^{'}}$ , and extending by the matrix formula of [F5]: this is a well-defined bilinear form, and it is symmetric because the diagonal matrix is symmetric.

2.1step 1.2F5F7algebra

For $v = \sum_{i} a_{i} v_{i}$ , $w = \sum_{j} d_{j} w_{j}$ , $v^{'} = \sum_{i^{'}} a_{i^{'}}^{'} v_{i^{'}}$ , $w^{'} = \sum_{j^{'}} d_{j^{'}}^{'} w_{j^{'}}$ in $V$ and $W$ , bilinearity and the diagonal prescription give $D (v \otimes w, v^{'} \otimes w^{'}) = \sum_{i, i^{'}, j, j^{'}} a_{i} d_{j} a_{i^{'}}^{'} d_{j^{'}}^{'} b_{i} c_{j} δ_{i i^{'}} δ_{j j^{'}} = (\sum_{i} a_{i} a_{i}^{'} b_{i}) (\sum_{j} d_{j} d_{j}^{'} c_{j}) = B (v, v^{'}) C (w, w^{'})$ . Hence $D$ is exactly the form $B \otimes C$ of the statement on simple tensors, and by [F7] it is the unique bilinear form with these values on elementary tensors.

2.2step 1.1step 1.2F1algebra

Inertia: the basis $(v_{i} \otimes w_{j})$ is diagonal for $D$ with entries $b_{i} c_{j}$ , none zero. The entry $b_{i} c_{j}$ is positive exactly when $b_{i}$ and $c_{j}$ have the same sign, which happens for $p p^{'}$ pairs of positive entries and $q q^{'}$ pairs of negative entries; it is negative for the $p q^{'}$ pairs with $b_{i} > 0 > c_{j}$ and the $p^{'} q$ pairs with $c_{j} > 0 > b_{i}$ . By [F1] the inertia of $D$ is therefore $(p p^{'} + q q^{'}, p q^{'} + p^{'} q, 0)$ .

3.1step 1.1step 1.2step 2.1F4F5

Nondegeneracy: let $u = \sum_{i, j} x_{i j} v_{i} \otimes w_{j}$ satisfy $D (u, u^{'}) = 0$ for all $u^{'}$ . Evaluating on $u^{'} = v_{i} \otimes w_{j}$ and using the diagonal prescription gives $x_{i j} b_{i} c_{j} = 0$ , and $b_{i} c_{j} \neq 0$ , so every $x_{i j} = 0$ and $u = 0$ . By symmetry the same computation with the second variable shows the right radical vanishes, so $D$ is nondegenerate.

4.1step 3.1step 2.2F1F2F8∎

Consequently $sign (B \otimes C) = (p p^{'} + q q^{'}) - (p q^{'} + p^{'} q) = (p - q) (p^{'} - q^{'}) = sign (B) sign (C)$ ; if $V = 0$ then $p = q = 0$ , and if $W = 0$ then $p^{'} = q^{'} = 0$ ; in either case the tensor-product basis is empty and both sides vanish. By [F8] the inertia computation pins down the congruence class of $D$ , and [F2] gives its normal form.

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The signature is multiplicative under Cartesian products

Statement

Assume AC. For closed oriented smooth manifolds $M^{4 a}$ and $N^{4 b}$ , with $M \times N$ carrying the product orientation (Product orientations, Products of smooth manifolds have a canonical product smooth structure), $σ (M \times N) = σ (M) σ (N) .$

Facts & Assumptions

Given: AC; closed oriented smooth $M^{4 a}$ , $N^{4 b}$ , $k = a + b$ , and the product orientation on $M \times N$ .

[F1]

$σ$ is the inertia difference of the nondegenerate middle form $Q_{M}$ (The signature of a closed oriented manifold of dimension divisible by four, The middle-dimensional intersection form is symmetric and nondegenerate), and $Q_{P} (u, v) = ⟨ u ⌣ v, [P] ⟩$ (The middle-dimensional intersection form of a closed oriented 4k-manifold).

[F2]

Since $M, N$ are closed, their (co)homology is finitely generated, in particular finite free over the field $R$ ; the cross product is a ring isomorphism $H^{*} (M; R) \hat{\otimes} H^{*} (N; R) \to H^{*} (M \times N; R)$ for the graded tensor multiplication $(α \otimes β) (α^{'} \otimes β^{'}) = (- 1)^{∣ β ∣ ∣ α^{'} ∣} (α ⌣ α^{'}) \otimes (β ⌣ β^{'})$ (Cohomological Kunneth cross product is a ring isomorphism, Finite generation from cap with a finite fundamental cycle).

[F3]

The Kronecker pairing is multiplicative under cross products: $⟨ α \times β, c \times d ⟩ = ⟨ α, c ⟩ ⟨ β, d ⟩$ , and the fundamental class of a product is $[M \times N] = [M] \times [N]$ (The Kronecker pairing is multiplicative under cross products, The fundamental class of a product is the cross product of the fundamental classes).

[F4]

Over the field $R$ , Poincare duality gives $\dim_{R} H^{i} (M; R) = \dim_{R} H^{4 a - i} (M; R)$ and similarly for $N$ , and the pairings $H^{i} (M) \times H^{4 a - i} (M) \to R$ and $H^{j} (N) \times H^{4 b - j} (N) \to R$ are perfect for every $i, j$ (Poincaré duality gives a nonsingular cup pairing).

[F5]

A nondegenerate symmetric form with a totally isotropic subspace of exactly half the dimension has signature zero (A nondegenerate symmetric form with a half-dimensional isotropic subspace has zero signature), and the tensor product of nondegenerate symmetric forms has multiplicative inertia (The tensor product of nondegenerate real symmetric forms has multiplicative signature).

[F6]

The tangent bundle of a product splits canonically as $T (M \times N) ≅ T M ⊞ T N$ for the product smooth structure (Canonical tangent and cotangent splittings for products, Products of smooth manifolds have a canonical product smooth structure).

Proof

technique · direct; split the middle cohomology of the product by Kunneth and kill the off-middle blocks by the isotropic lemma

1.1givenF2

Kunneth decomposition: by [F2], $H^{2 k} (M \times N; R) = ⨁_{i = 0}^{2 k} V_{i}$ with $V_{i} = H^{i} (M; R) \otimes H^{2 k - i} (N; R)$ , and the ring structure is the graded tensor product.

1.2givenF1F2F3

For $x \otimes y \in V_{i}$ and $x' \otimes y' \in V_{j}$ with $i + j = 4 a$ , the ring formula [F2] and the matching-degree evaluation [F3] give $Q_{M \times N} (x \otimes y, x' \otimes y') = (- 1)^{(2 k - i) j} ⟨ x ⌣ x', [M] ⟩ ⟨ y ⌣ y', [N] ⟩$ . If $i + j \neq 4 a$ , one factor cup class has degree greater than the dimension of its manifold, so vanishes by [F2]. Thus $V_{i}$ pairs only with $V_{4 a - i}$ ; the factors in the displayed formula are complementary-degree cup pairings, rather than middle forms unless $i = j = 2 a$ .

2.1step 1.2F1

Middle block: taking $i = j = 2 a$ in step 1.2, the sign is $(- 1)^{∣ y ∣ ∣ x^{'} ∣} = (- 1)^{2 b \cdot 2 a} = + 1$ , so $Q_{M \times N}$ restricted to $V_{2 a} = H^{2 a} (M) \otimes H^{2 b} (N)$ is exactly $Q_{M} \otimes Q_{N}$ .

2.2givenF2F4step 1.2

Off-middle part is nondegenerate: by [F4] $\dim V_{i} = \dim V_{4 a - i}$ for every $i$ , and the pairing of $V_{i}$ with $V_{4 a - i}$ induced by $Q_{M \times N}$ is the tensor product of the perfect pairings $H^{i} (M) \times H^{4 a - i} (M) \to R$ and $H^{2 k - i} (N) \times H^{4 b - 2 k + i} (N) \to R$ , hence perfect: in dual bases the tensor pairing matrix is a nonzero scalar times an identity matrix, the scalar being the fixed Koszul sign. Set $V_{i} = 0$ when $i$ is outside $0, \dots, 2 k$ ; if a block has no complementary index in that range it is zero by the dimension bounds in [F2]. Therefore $Z = ⨁_{i \neq 2 a} V_{i}$ is nondegenerate: a class in $Z$ pairing to zero with all of $Z$ must have every block component zero, testing against the complementary block.

2.3step 1.2F4

Half-dimensional isotropic subspace: $W = ⨁_{i < 2 a} V_{i} \subseteq Z$ is totally isotropic by step 1.2, since $i, j < 2 a$ give $i + j < 4 a$ and hence vanishing pairing; and $2 \dim W = \dim Z$ because $Z$ is the direct sum of the pairs $V_{i} \oplus V_{4 a - i}$ with $i < 2 a$ and $\dim V_{i} = \dim V_{4 a - i}$ by [F4].

3.1step 1.2step 2.1step 2.2step 2.3F1F5

The off-middle blocks contribute nothing: by steps 2.2 and 2.3, $Z$ carries a nondegenerate symmetric form with the half-dimensional totally isotropic subspace $W$ , so $sign (Q_{M \times N} ∣_{Z}) = 0$ by [F5]. Moreover $Z ⊥ V_{2 a}$ by step 1.2, and $V_{2 a}$ is nondegenerate because $Q_{M \times N}$ is nondegenerate and $Z$ is nondegenerate with $H^{2 k} (M \times N; R) = Z \oplus V_{2 a}$ ; hence the inertia data of $Q_{M \times N}$ are the sums of those of $Q ∣_{Z}$ and $Q ∣_{V_{2 a}}$ .

4.1step 2.1step 3.1F5F6∎

Therefore $σ (M \times N) = sign (Q ∣_{Z}) + sign (Q ∣_{V_{2 a}}) = 0 + sign (Q_{M} \otimes Q_{N}) = σ (M) σ (N)$ , using step 3.1, step 2.1 and the multiplicativity of inertia under tensor products [F5]; the product smooth structure and orientation used are those of [F6] and the statement.

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The formal hyperbolic tangent series and the even series $x / \tanh x$

Definition

Work in $Q ⟦ x ⟧$ with the formal exponential and logarithm of Formal exponential, logarithm, and binomial powers over a commutative $Q$ -algebra and the identities of Formal $\exp$ and $\log$ are inverse homomorphisms and formal binomial powers obey the expected addition laws. Define the formal hyperbolic tangent and the formal area hyperbolic tangent by $T (x) : = \frac{\exp (x) - \exp (- x)}{\exp (x) + \exp (- x)} \in Q ⟦ x ⟧, A (z) : = \sum_{j \geq 0} \frac{z^{2 j + 1}}{2 j + 1} \in Q ⟦ z ⟧ .$

The denominator $\exp (x) + \exp (- x)$ has constant term $2$ , a unit of $Q$ , so the quotient is a well-defined power series by A formal power series is a unit exactly when its constant coefficient is a unit; $T (0) = 0$ and the linear coefficient of $T$ is $\frac{1}{2} (1 - (- 1)) / \frac{1}{2} (1 + 1) = 1$ . Substituting $- x$ into the defining quotient replaces the numerator by its negative and fixes the denominator, so $T (- x) = - T (x)$ : $T$ is odd. Hence $S (x) : = \frac{T (x)}{x} \in Q ⟦ x ⟧$ is even with constant term $1$ , and $Q (x) : = \frac{x}{T (x)} = \frac{1}{S (x)} \in Q ⟦ x^{2} ⟧, Q (x) = 1 + \frac{x^{2}}{3} - \frac{x^{4}}{45} + O (x^{6}) .$ The displayed coefficients of $Q$ are the normalisation recorded here; they are verified in the inverse-series lemma following on this page, which is the result named in justified_by.

The series $A$ is summable degreewise (Summable families of formal series are locally finite in every coefficient range), $A (0) = 0$ , and its linear coefficient is $1$ ; its coefficients are the evaluation of a family whose $j$ -th term has order $2 j + 1$ , so no convergence question arises. The symbol $x / \tanh x$ used by the sources denotes exactly the element $Q (x)$ ; no analytic convergence, contour, or branch is involved. The residue calculus of Formal Laurent series $K ((x))$ , their order, derivative, and residue applies to $T$ because $T = x S$ has order $1$ . No choice principle is used.

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The formal hyperbolic tangent and artanh series are inverse, with the artanh derivative

Statement

Let $T$ and $A$ be the formal series of The formal hyperbolic tangent series and the even series $x / \tanh x$ . Then in $Q ⟦ \cdot ⟧$ , $T \circ A = z, A \circ T = x,$ so $T$ and $A$ are mutually inverse formal power series; moreover $A^{'} (z) = \frac{1}{1 - z^{2}} = \sum_{j \geq 0} z^{2 j}, T^{'} (x) = 1 - T (x)^{2},$ and $Q (x) = x / T (x)$ has the expansion $Q (x) = 1 + x^{2} / 3 - x^{4} / 45 + O (x^{6})$ .

Facts & Assumptions

Given: The series $T (x) = (\exp x - \exp (- x)) / (\exp x + \exp (- x))$ and $A (z) = \sum_{j \geq 0} z^{2 j + 1} / (2 j + 1)$ of The formal hyperbolic tangent series and the even series $x / \tanh x$ , with $S = T / x$ , $Q = x / T = 1 / S$ .

[F1]

$A (z) = \frac{1}{2} (\log (1 + z) - \log (1 - z))$ holds coefficientwise: the logarithmic series has $\frac{1}{2} ((- 1)^{n - 1} + 1) z^{n} / n$ , which vanishes for even $n$ and equals $z^{2 j + 1} / (2 j + 1)$ for $n = 2 j + 1$ . This is the recorded relation between the two definitions (The formal hyperbolic tangent series and the even series $x / \tanh x$ , Formal $\exp$ and $\log$ are inverse homomorphisms and formal binomial powers obey the expected addition laws).

[F2]

In a commutative $Q$ -algebra, $\exp (u + v) = \exp (u) \exp (v)$ , $\exp$ and $\log$ are mutually inverse, and $(1 + u)^{c} = \exp (c \log (1 + u))$ (Formal $\exp$ and $\log$ are inverse homomorphisms and formal binomial powers obey the expected addition laws).

[F3]

A series $f \in x R ⟦ x ⟧$ has a two-sided compositional inverse if and only if its linear coefficient is a unit, and then that inverse is unique (A zero-constant formal series has a compositional inverse exactly when its linear coefficient is a unit).

[F4]

Composition is associative when the inner series have zero constant coefficient: $(f \circ g) \circ h = f \circ (g \circ h)$ , and $f \circ x = f$ , $x \circ f = f$ (Substitution by a zero-constant series is a ring homomorphism, and composition is associative when both inner series have zero constant coefficient, Composition $f \circ g$ of formal series when the outer series is a polynomial or the inner series has zero constant term).

[F5]

The formal derivative is additive and satisfies the product, power, quotient and chain rules $D (f g) = (D f) g + f D g$ , $D (f^{m}) = m f^{m - 1} D f$ for $m \geq 1$ , $D (f / g) = ((D f) g - f D g) / g^{2}$ for a unit $g$ , and $D (f \circ g) = (D f \circ g) D g$ (Formal differentiation is linear and satisfies product, power, quotient, chain, and coefficient-recovery laws, The formal derivative $D (\sum a_{n} x^{n}) = \sum_{n \geq 1} n a_{n} x^{n - 1}$ ).

[F6]

A series is a unit exactly when its constant coefficient is a unit, and $\sum_{j \geq 0} (- u)^{j}$ is the inverse of $1 + u$ (A formal power series is a unit exactly when its constant coefficient is a unit).

[F7]

Summable families may be regrouped and reindexed, and infinite products of factors $1 + u_{k}$ with ${ord}_{x} (u_{k}) \to + \infty$ may be regrouped (Summable formal families may be regrouped and rearranged, distribute over multiplication, and have well-defined locally finite products).

Proof

technique · direct; exponentiate $A$, compare the defining quotient of $T$, then differentiate and expand

1.1givenF1

The identity $A (z) = \frac{1}{2} (\log (1 + z) - \log (1 - z))$ holds by [F1].

1.2givenF6F7algebra

Expansion of $Q$ : from $\exp (\pm x)$ with coefficients $(\pm 1)^{n} / n!$ (Formal exponential, logarithm, and binomial powers over a commutative $Q$ -algebra), the even and odd parts are $\exp x + \exp (- x) = 2 + x^{2} + x^{4} / 12 + O (x^{6})$ and $\exp x - \exp (- x) = 2 x + x^{3} / 3 + x^{5} / 60 + O (x^{7})$ . Hence $S = T / x = (2 + x^{2} / 3 + x^{4} / 60 + O (x^{6})) / (2 + x^{2} + x^{4} / 12 + O (x^{6})) = (1 + x^{2} / 6 + x^{4} / 120 + O (x^{6})) (1 - x^{2} / 2 + (1 / 4 - 1 / 24) x^{4} + O (x^{6})) = 1 - x^{2} / 3 + 2 x^{4} / 15 + O (x^{6})$ , and inverting this unit by [F6] gives $Q = 1 / S = 1 + x^{2} / 3 + (1 / 9 - 2 / 15) x^{4} + O (x^{6}) = 1 + x^{2} / 3 - x^{4} / 45 + O (x^{6})$ .

2.1step 1.1F2

Exponentiating: by [F2], $\exp (A) = \exp (\frac{1}{2} \log (1 + z)) \exp (- \frac{1}{2} \log (1 - z)) = (1 + z)^{1 / 2} (1 - z)^{- 1 / 2}$ and similarly $\exp (- A) = (1 + z)^{- 1 / 2} (1 - z)^{1 / 2}$ , where the binomial powers are the series of [F2].

2.2step 1.1F5F6F7algebra

Derivative of $A$ : differentiating $A = \frac{1}{2} (\log (1 + z) - \log (1 - z))$ coefficientwise, $D \log (1 + z) = \sum_{n \geq 1} (- 1)^{n - 1} z^{n - 1} = \sum_{j \geq 0} (- z)^{j} = (1 + z)^{- 1}$ and $D \log (1 - z) = - (1 - z)^{- 1}$ by [F5] and [F6], so $A^{'} = \frac{1}{2} ((1 + z)^{- 1} + (1 - z)^{- 1}) = \frac{1}{2} ((1 - z) + (1 + z)) / (1 - z^{2}) = 1 / (1 - z^{2})$ ; and $(1 - z^{2}) \sum_{j \geq 0} z^{2 j} = 1$ coefficientwise, so $1 / (1 - z^{2}) = \sum_{j \geq 0} z^{2 j}$ by uniqueness of inverses in [F6].

3.1step 2.1F2F6

Both $\exp (A)$ and $\exp (- A)$ have constant term $1$ , so the denominator $\exp (A) + \exp (- A)$ has constant term $2$ , a unit of $Q$ ; the quotient defining $T (A)$ is therefore well defined by [F6]. Multiplying numerator and denominator by the unit $(1 + z)^{1 / 2} (1 - z)^{1 / 2}$ turns it into $((1 + z) - (1 - z)) / ((1 + z) + (1 - z)) = 2 z / 2 = z$ , so $T \circ A = z$ .

4.1step 3.1F3F4

The series $T$ has linear coefficient $1$ , a unit of $Q$ , so by [F3] it has a unique two-sided compositional inverse $g$ , with $g \circ T = x = T \circ g$ . Associativity [F4] applied to the inner series $T, A, g$ (all with zero constant term) gives $A = (g \circ T) \circ A = g \circ (T \circ A) = g \circ z = g$ , so $A = g$ and $A \circ T = x$ .

4.2step 3.1F5algebra

Derivative of $T$ : the termwise derivative of $\exp (\pm x)$ is $\pm \exp (\pm x)$ because $D (\sum_{n} (\pm 1)^{n} x^{n} / n!) = \sum_{n \geq 1} (\pm 1)^{n} x^{n - 1} / (n - 1)!$ ; with $N = \exp x - \exp (- x)$ and $D_{0} = \exp x + \exp (- x)$ this gives $N^{'} = D_{0}$ and $D_{0}^{'} = N$ . The quotient rule [F5] applied to $T = N / D_{0}$ (with $D_{0}$ a unit) gives $T^{'} = (N^{'} D_{0} - N D_{0}^{'}) / D_{0}^{2} = (D_{0}^{2} - N^{2}) / D_{0}^{2} = 1 - T^{2}$ .

5.1step 3.1step 1.2step 4.1step 4.2step 2.2∎

Steps 3.1 and 4.1 give $T \circ A = z$ and $A \circ T = x$ ; steps 2.2 and 4.2 give the two derivative formulas; and step 1.2 gives the recorded expansion of $Q = x / T$ . All identities are coefficientwise identities between formal series; the zero series, the case of a single variable, and the degenerate cases $z = 0$ are included as the constant coefficients of the same computations, and no analytic convergence or choice principle is involved.

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The coefficient identity $[z^{2 k}] (z / \tanh z)^{2 k + 1} = 1$ for every $k$

Statement

For every $m \geq 0$ , in the formal series $Q (z) = z / T (z) = z / \tanh z$ of The formal hyperbolic tangent series and the even series $x / \tanh x$ , $[z^{m}] (\frac{z}{\tanh z})^{m + 1} = {\begin{cases} 1, & m even, \\ 0, & m odd. \end{cases}$ Equivalently, for every $k \geq 0$ , $[z^{2 k}] (\frac{z}{\tanh z})^{2 k + 1} = 1, [z^{2 n + 1}] (\frac{z}{\tanh z})^{2 n + 2} = 0 (n \geq 0),$ and the formal residue ${res}_{z} (\tanh z)^{- (2 k + 1)} d z$ equals $1$ .

Facts & Assumptions

Given: The series $T$ and $A$ of The formal hyperbolic tangent series and the even series $x / \tanh x$ , the integer $m \geq 0$ , and the Bernoulli-free coefficient computation below.

[F1]

$T = x S$ with $S$ an even power series of constant term $1$ , so $S$ is a unit and $Q = x / T = 1 / S$ ; in particular $T$ has order $1$ and linear coefficient $1$ (The formal hyperbolic tangent series and the even series $x / \tanh x$ ).

[F2]

For a field $K$ , $K ((x))$ consists of the Laurent series with support bounded below, with finite convolution in each degree, derivative $D (x^{n}) = n x^{n - 1}$ , and residue ${res}_{x} (f) = [x^{- 1}] f$ (Formal Laurent series $K ((x))$ , their order, derivative, and residue).

[F3]

The inverse-series lemma gives $T \circ A = z$ , $A \circ T = x$ , and $A^{'} (z) = 1 / (1 - z^{2}) = \sum_{j \geq 0} z^{2 j}$ (The formal hyperbolic tangent and artanh series are inverse, with the artanh derivative).

[F4]

Over a field $K$ containing $Q$ , for $g \in x K ⟦ x ⟧$ with nonzero linear coefficient and $F \in K ((x))$ for which $F \circ g$ is formed by Laurent substitution, ${res}_{x} ((F \circ g) D g) = {res}_{x} (F)$ (Formal residues satisfy integration by parts, logarithmic differentiation, and change of variables).

[F5]

A power series is a unit exactly when its constant coefficient is a unit, and inverses are unique; $(1 - z^{2}) \sum_{j \geq 0} z^{2 j} = 1$ (A formal power series is a unit exactly when its constant coefficient is a unit).

[F6]

Summable families may be regrouped and reindexed, and coefficient extraction is the functional $[z^{n}]$ of Formal power series over a commutative ring and the coefficient-extraction functional $[x^{n}]$ (Summable formal families may be regrouped and rearranged, distribute over multiplication, and have well-defined locally finite products).

Proof

technique · direct; reduce the coefficient to a residue and change variables by the inverse series

1.1givenF1F2F5F6algebra

Since $T = u S$ with $S$ a unit of $Q ⟦ u ⟧$ by [F1], the series $F (u) : = T (u)^{- (m + 1)} = u^{- (m + 1)} S (u)^{- (m + 1)}$ is a well-defined element of $Q ((u))$ of order $- (m + 1)$ . As $(u / T (u))^{m + 1} = u^{m + 1} T (u)^{- (m + 1)}$ in Laurent series, $F (z) = z^{- (m + 1)} (z / T (z))^{m + 1}$ and therefore ${res}_{z} F (z) = [z^{- 1}] z^{- (m + 1)} (z / T (z))^{m + 1} = [z^{m}] (z / T (z))^{m + 1}$ .

2.1step 1.1F1F2F3F5

The composition $F \circ A$ is formed by Laurent substitution: $A = z U$ with $U$ a unit, so $A^{- (m + 1)} = z^{- (m + 1)} U^{- (m + 1)}$ has finitely many terms in each degree and $S^{- (m + 1)} \circ A$ is a unit power series, and every coefficient of $F \circ A$ is a finite sum. By [F3], $T \circ A = z$ , hence $F \circ A = (T \circ A)^{- (m + 1)} = z^{- (m + 1)}$ ; and $D A = A^{'} = 1 / (1 - z^{2}) = \sum_{j \geq 0} z^{2 j}$ , the last identity by [F3] and [F5].

3.1step 1.1step 2.1F3F4F5F6

The change-of-variables identity [F4] applies over $K = Q$ with $g = A$ , whose linear coefficient is $1 \neq 0$ : ${res}_{z} ((F \circ A) A^{'}) = {res}_{z} F$ . By step 2.1 the left side is ${res}_{z} (z^{- (m + 1)} (1 - z^{2})^{- 1}) = [z^{- 1}] z^{- (m + 1)} \sum_{j \geq 0} z^{2 j} = [z^{m}] \sum_{j \geq 0} z^{2 j}$ , which is $1$ when $m$ is even and $0$ when $m$ is odd, since $z^{2 j}$ has degree $2 j$ . Combining with step 1.1 gives $[z^{m}] (z / T (z))^{m + 1} = 1$ for even $m$ and $0$ for odd $m$ .

4.1step 1.1step 3.1given∎

Taking $m = 2 k$ gives $[z^{2 k}] (z / \tanh z)^{2 k + 1} = 1$ for every $k \geq 0$ ; taking $m = 2 n + 1$ gives $[z^{2 n + 1}] (z / \tanh z)^{2 n + 2} = 0$ ; and step 1.1 with $m = 2 k$ identifies ${res}_{z} T (z)^{- (2 k + 1)} = [z^{2 k}] (z / T (z))^{2 k + 1} = 1$ , the residue form asserted. The value $k = 0$ reads $[z^{0}] Q = 1$ , the constant term of the unit series $Q$ . All computations are coefficientwise identities between formal Laurent series over $Q$ ; no analytic contour and no choice principle is involved.

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The completed cohomology ring in degrees divisible by four

Definition

Let $B$ be a space (a topological space; no finite-dimensionality or CW assumption is made). The completed fourfold-graded cohomology group of $B$ with rational coefficients is the product ${\hat{H}}^{4 *} (B; Q) : = \prod_{j \geq 0} H^{4 j} (B; Q),$ the set of all sequences $a = (a_{0}, a_{1}, a_{2}, \dots)$ with $a_{j} \in H^{4 j} (B; Q)$ (Singular cohomology ring). Addition is componentwise, $(a + b)_{j} : = a_{j} + b_{j},$ and multiplication is the convolution of the graded cup product, $(a \cdot b)_{n} : = \sum_{i + j = n} a_{i} ⌣ b_{j} .$

The product is well defined: for fixed $n$ the sum has exactly $n + 1$ terms, so no infinite sum occurs in any component, and each $a_{i} ⌣ b_{j}$ lies in $H^{4 n} (B; Q)$ because cup product adds degrees. The unit is $1 : = (1, 0, 0, \dots),$ with $1 \in H^{0} (B; Q)$ the unit of Singular cohomology ring. For a continuous map $f : B \to C$ the pullback is $f^{*} : {\hat{H}}^{4 *} (C; Q) \to {\hat{H}}^{4 *} (B; Q), f^{*} (a)_{j} : = f^{*} (a_{j}),$ the componentwise pullback of the ordinary cohomology rings.

The ring laws and the naturality assertions are not part of this definition: they are proved in The completed fourfold-graded cohomology ring is natural and satisfies the ring laws ↗, the lemma named in justified_by. The construction is a product of abelian groups and a prescribed formula; nothing is selected, so no choice principle is used.

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The completed fourfold-graded cohomology ring is natural and satisfies the ring laws

Statement

For every space $B$ , the operations of The completed cohomology ring in degrees divisible by four make ${\hat{H}}^{4 *} (B; Q)$ a commutative unital ring. For every continuous map $f : B \to C$ , componentwise pullback is a unital ring homomorphism $f^{*} : {\hat{H}}^{4 *} (C; Q) \to {\hat{H}}^{4 *} (B; Q)$ . The assertion is valid for arbitrary, possibly infinite-dimensional spaces.

Facts & Assumptions

Given: Spaces $B$ and $C$ , a continuous map $f : B \to C$ , and elements $a = (a_{j})$ , $b = (b_{j})$ , $c = (c_{j})$ of ${\hat{H}}^{4 *} (C; Q)$ .

[F1]

The completed group is ${\hat{H}}^{4 *} (B; Q) = \prod_{j \geq 0} H^{4 j} (B; Q)$ with componentwise addition, product $(a \cdot b)_{n} = \sum_{i + j = n} a_{i} ⌣ b_{j}$ , and unit $(1, 0, 0, \dots)$ ; for fixed $n$ the sum has $n + 1$ terms (The completed cohomology ring in degrees divisible by four).

[F2]

The singular cohomology ring has multiplication induced by the cochain cup product, with unit the class of the constant cochain $1$ , and its multiplication is associative and distributive over addition (Singular cohomology ring).

[F3]

Cup product is natural and unital: $f^{*} (u ⌣ v) = f^{*} u ⌣ f^{*} v$ and $f^{*} 1 = 1$ (Cup product is natural, unital and associative).

[F4]

Cup product is graded commutative: $u ⌣ v = (- 1)^{p q} v ⌣ u$ for $u \in H^{p}$ , $v \in H^{q}$ (Singular cohomology is graded commutative).

Proof

technique · direct; compare finite sums in each degree

1.1givenF1F2

Addition and multiplication are well defined: both are given by finite operations in each degree $4 n$ , since $a_{i} ⌣ b_{n - i} \in H^{4 n} (B; Q)$ for $0 \leq i \leq n$ and these are the only contributing pairs, so no infinite sum occurs. Addition is associative and commutative and has the zero sequence as neutral element because these hold in each group $H^{4 n} (B; Q)$ .

1.2givenF1F2

Associativity: for every $n$ , $((a \cdot b) \cdot c)_{n} = \sum_{i + j + k = n} (a_{i} ⌣ b_{j}) ⌣ c_{k}$ and $(a \cdot (b \cdot c))_{n} = \sum_{i + j + k = n} a_{i} ⌣ (b_{j} ⌣ c_{k})$ , two finite sums over the same triples that agree termwise by associativity of the cup product.

1.3givenF1F2

Unit: $(1 \cdot a)_{n} = 1 ⌣ a_{n} = a_{n}$ and $(a \cdot 1)_{n} = a_{n} ⌣ 1 = a_{n}$ for every $n$ , since $1 \in H^{0} (B; Q)$ is the unit of the graded cohomology ring.

1.4givenF1F2

Distributivity: $(a \cdot (b + c))_{n} = \sum_{i + j = n} a_{i} ⌣ (b_{j} + c_{j}) = \sum_{i + j = n} a_{i} ⌣ b_{j} + \sum_{i + j = n} a_{i} ⌣ c_{j}$ by bilinearity of the cup product and additivity of finite sums.

2.1step 1.1F4algebra

Commutativity: $(a \cdot b)_{n} = \sum_{i + j = n} a_{i} ⌣ b_{j} = \sum_{i + j = n} (- 1)^{16 i j} b_{j} ⌣ a_{i} = (b \cdot a)_{n}$ , because $4 i \cdot 4 j = 16 i j$ is even and so every sign is $+ 1$ .

2.2step 1.1F1F3

Naturality: for each $n$ , $f^{*} ((a \cdot b)_{n}) = \sum_{i + j = n} f^{*} (a_{i} ⌣ b_{j}) = \sum_{i + j = n} f^{*} a_{i} ⌣ f^{*} b_{j} = (f^{*} a \cdot f^{*} b)_{n}$ , and $f^{*} 1 = 1$ componentwise; hence $f^{*}$ is a unital ring homomorphism.

3.1step 1.2step 1.3step 1.4step 2.1step 2.2given∎

Steps 1.1-1.4 and 2.1 give the commutative unital ring laws, and step 2.2 gives naturality, in every degree and hence componentwise; nothing was assumed about the dimension or CW type of $B$ , and the statements include the empty space and the zero ring, where all groups vanish and the same finite computations apply with zero elements. Only the prescribed formulas are used, so no choice principle is invoked.

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The Hirzebruch L-polynomials and the total L-class of a real vector bundle

Definition

Work over $Q$ with the series $Q (x) = x / \tanh x = \sum_{j \geq 0} q_{2 j} x^{2 j}$ of The formal hyperbolic tangent series and the even series $x / \tanh x$ , so that $q_{0} = 1$ and, as recorded and justified in that definition, $q_{2} = 1 / 3$ and $q_{4} = - 1 / 45$ . Give the $i$ -th polynomial variable weight $4 i$ .

The L-polynomials. For each $k \geq 0$ let $L_{k} \in Q [p_{1}, p_{2}, p_{3}, \dots]$ denote the polynomial, homogeneous of weight $4 k$ , characterized by the following condition: for every $N \geq 1$ and all indeterminates $x_{1}, \dots, x_{N}$ , with $e_{j}$ the elementary symmetric polynomials (The elementary symmetric polynomials $e_{0}, e_{1}, \dots, e_{n}$ , Symmetric polynomials as the invariants of variable permutations), $L_{k} (e_{1} (x_{1}^{2}, \dots, x_{N}^{2}), \dots, e_{k} (x_{1}^{2}, \dots, x_{N}^{2})) = [weight 4 k] \prod_{i = 1}^{N} Q (x_{i}) .$ Here $[weight 4 k]$ selects the homogeneous component of weight $4 k$ .

Existence and uniqueness. Give each $x_{i}$ weight $2$ and put $u_{i} = x_{i}^{2}$ , of weight $4$ . For $N \geq \max (1, k)$ , the weight- $4 k$ component of $\prod_{i} Q (x_{i})$ is a symmetric polynomial of ordinary degree $k$ in the $u_{i}$ . The fundamental theorem of symmetric polynomials (Fundamental theorem of symmetric polynomials: unique expression as a polynomial in $e_{1}, \dots, e_{n}$ ) expresses it uniquely in $e_{1} (u), \dots, e_{N} (u)$ . Since $e_{j}$ has ordinary degree $j$ , uniqueness of homogeneous components makes this expression homogeneous of weighted degree $k$ and excludes every $e_{j}$ with $j > k$ . Define $L_{k}$ using $N = \max (1, k)$ . Setting additional roots to zero leaves the product unchanged because $Q (0) = 1$ ; injectivity of substitution in at least $k$ variables shows that the same polynomial works for every larger $N$ . Setting roots to zero then gives the identity for $N < k$ too, where $e_{j} = 0$ for $j > N$ ; injectivity is asserted only when $N \geq k$ . In particular $L_{0} = 1$ . For two variables the weight- $4$ part of $Q (x_{1}) Q (x_{2})$ is $q_{2} (x_{1}^{2} + x_{2}^{2}) = q_{2} e_{1}$ , and its weight- $8$ part is $q_{4} (x_{1}^{4} + x_{2}^{4}) + q_{2}^{2} x_{1}^{2} x_{2}^{2}$ , which the identities $x_{1}^{4} + x_{2}^{4} = e_{1}^{2} - 2 e_{2}$ and $x_{1}^{2} x_{2}^{2} = e_{2}$ rewrite as $q_{4} (e_{1}^{2} - 2 e_{2}) + q_{2}^{2} e_{2}$ ; substituting $q_{2} = 1 / 3$ , $q_{4} = - 1 / 45$ and the corresponding powers of $p_{1}, p_{2}$ gives $L_{1} = \frac{p_{1}}{3}, L_{2} = \frac{7 p_{2} - p_{1}^{2}}{45} .$

The total L-class. Assume AC (The Axiom of Choice), inherited from the Pontryagin and Chern constructions, including their bundle and Thom suppliers; no claim is made that DC alone supplies those constructions. For a real vector bundle $E \to B$ of finite rank over a CW-type base with Pontryagin classes $p_{i} (E) \in H^{4 i} (B; Z)$ of Pontryagin classes by complexification, regarded in rational cohomology, the total L-class is the element $L (E) : = (L_{k} (p_{1} (E), \dots, p_{k} (E)))_{k \geq 0} \in {\hat{H}}^{4 *} (B; Q)$ of the completed ring of The completed cohomology ring in degrees divisible by four; its degree- $4 k$ component is written $L_{k} (E) : = L_{k} (p_{1} (E), \dots, p_{k} (E))$ . The sequence is well defined because each $L_{k} (p_{1} (E), \dots, p_{k} (E))$ is a class in $H^{4 k} (B; Q)$ computed from the given Pontryagin classes, and $p_{i} (E) = 0$ whenever $2 i > rank E$ (Naturality, stability, and mod-two reduction of Pontryagin classes), and if $B$ has a finite-dimensional CW model, its cohomology vanishes above that dimension, so only finitely many components are nonzero. The coefficients $q_{2 j}$ belong to $Q$ , so no integrality of $L (E)$ is asserted; see Formal power series over a commutative ring and the coefficient-extraction functional $[x^{n}]$ for the coefficient notation. Naturality, stability and multiplicativity of $L$ are not part of this definition; they are proved in the lemma named in justified_by.

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The L-polynomials are well defined and form a multiplicative, natural and stable sequence

Statement

Assume AC. Let $L_{k}$ and the completed total class $L (E)$ be as in The Hirzebruch L-polynomials and the total L-class of a real vector bundle, and let $E, F \to B$ be numerable real vector bundles over a path-connected paracompact Hausdorff CW base $B$ . The same assertions hold for paracompact Hausdorff CGWH bases of CW type, and componentwise for their disjoint unions. Pullbacks are between bases in this scope. Then:

Each $L_{k}$ is a well-defined homogeneous polynomial of weight $4 k$ in $p_{1}, \dots, p_{k}$ , independent of the number of formal roots, with $L_{0} = 1$ ; $L (E)$ is natural under pullback, stable ( $L (E \oplus ε^{r}) = L (E)$ ) and equal to $1$ for trivial $E$ .
Multiplicativity. $L (E \oplus F) = L (E) L (F)$ in ${\hat{H}}^{4 *} (B; Q)$ , equivalently $L_{k} (p (E \oplus F)) = \sum_{i + j = k} L_{i} (p (E)) L_{j} (p (F))$ .
Rank two and complex lines. If $E$ is an oriented real rank-two bundle with Euler class $e$ , then $p_{1} (E) = e^{2}$ and $L (E) = \sum_{k \geq 0} q_{2 k} e^{2 k} = Q (e) = e / \tanh e$ ; a Whitney sum of oriented rank-two bundles has $L = \prod_{i} Q (e_{i})$ . In particular a complex line bundle $ℓ$ with $c_{1} (ℓ) = t$ has underlying real L-class $Q (t) = t / \tanh t$ .

Facts & Assumptions

Given: AC; the L-polynomials $L_{k} \in Q [p_{1}, p_{2}, \dots]$ and the total L-class of The Hirzebruch L-polynomials and the total L-class of a real vector bundle, built from the series $Q (x) = \sum_{j} q_{2 j} x^{2 j} = x / \tanh x$ ; numerable real bundles over the stated base.

[F1]

$L_{k}$ is determined by the identity $L_{k} (e_{1} (x_{1}^{2}, \dots, x_{N}^{2}), \dots, e_{k} (x_{1}^{2}, \dots, x_{N}^{2})) = [weight 4 k] \prod_{i = 1}^{N} Q (x_{i})$ for every $N \geq 1$ , and $L_{k} \in Q [p_{1}, \dots, p_{k}]$ is homogeneous of weight $4 k$ ; for a bundle $E$ , $L_{k} (E) = L_{k} (p_{1} (E), \dots, p_{k} (E))$ and $L (E) = (L_{k} (E))_{k \geq 0}$ in the completed ring (The Hirzebruch L-polynomials and the total L-class of a real vector bundle).

[F2]

The substitution $T_{k} \mapsto e_{k}$ is an $Q$ -algebra isomorphism from $Q [T_{1}, \dots, T_{N}]$ onto the symmetric polynomials in $x_{1}, \dots, x_{N}$ (Fundamental theorem of symmetric polynomials: unique expression as a polynomial in $e_{1}, \dots, e_{n}$ ).

[F3]

The elementary symmetric polynomials satisfy $e_{0} = 1$ , $e_{j} = 0$ for $j > N$ , and $e_{r} (z_{1}, \dots, z_{M}) = \sum_{a + b = r} e_{a} (z_{1}, \dots, z_{m}) e_{b} (z_{m + 1}, \dots, z_{M})$ for $M = m + n$ ; the last identity is the coefficient comparison in $\prod_{i \leq m} (1 + t z_{i}) \prod_{j} (1 + t z_{j})$ (The elementary symmetric polynomials $e_{0}, e_{1}, \dots, e_{n}$ ).

[F4]

For numerable real bundles over a CW base: $p_{i} (f^{*} E) = f^{*} p_{i} (E)$ for continuous $f$ , $p_{i} (E \oplus ε^{r}) = p_{i} (E)$ , and $p_{i} (E) = 0$ whenever $2 i > rank E$ (Naturality, stability, and mod-two reduction of Pontryagin classes).

[F5]

On a CW base, over a coefficient ring in which $2$ is invertible, in particular over $Q$ , the total Pontryagin class is multiplicative: $p (E \oplus F) = p (E) p (F)$ , i.e. $p_{r} (E \oplus F) = \sum_{a + b = r} p_{a} (E) p_{b} (F)$ (Pontryagin Whitney product away from two).

[F6]

${\hat{H}}^{4 *} (B; Q)$ is a commutative unital ring under convolution product, and pullback is a unital ring homomorphism (The completed fourfold-graded cohomology ring is natural and satisfies the ring laws, The completed cohomology ring in degrees divisible by four).

[F7]

For a numerable oriented real bundle of rank $2 n$ with Euler class $e$ , the top Pontryagin class is $p_{n} = e^{2}$ (Top Pontryagin class is the square of the Euler class).

[F8]

For a complex rank- $n$ bundle $E$ over a path-connected CW complex, with the complex orientation of the underlying real bundle, $c_{n} (E) = e (E_{R})$ (Top Chern class equals Euler class of the underlying real bundle).

[F9]

Chern and Euler classes of numerable bundles are natural, and their Whitney sums multiply; the Chern classes are obtained from the projective-bundle relation and the Euler class from the zero-section pullback of the Thom class (Naturality, normalization, and Whitney sum for Chern classes, Naturality, orientation sign, and Whitney product for Euler classes, Chern classes from the projective-bundle relation, Euler class by zero-section pullback of the Thom class).

Proof

technique · direct; a universal polynomial identity from the defining product, then substitution of Pontryagin classes

1.1givenF1F2F3algebra

For fixed $k$ , take $m, n \geq \max (1, k)$ and let $x_{1}, \dots, x_{m}, y_{1}, \dots, y_{n}$ be indeterminates; put $A = \prod_{i \leq m} Q (x_{i})$ and $B = \prod_{j \leq n} Q (y_{j})$ in $Q ⟦ x, y ⟧$ . Then $(\prod_{i} Q (x_{i})) (\prod_{j} Q (y_{j})) = A B$ , so its homogeneous component of weight $4 k$ is the finite sum $\sum_{i + j = k} ([weight 4 i] A) ([weight 4 j] B)$ ; the elementary symmetric function of the combined squared variables is $e_{r} (x_{1}^{2}, \dots, x_{m}^{2}, y_{1}^{2}, \dots, y_{n}^{2}) = \sum_{a + b = r} e_{a} (x^{2}) e_{b} (y^{2})$ by [F3]. Applying the defining identity of [F1] to the combined family of roots and to each block separately, we obtain the universal identity $L_{k} (\sum_{a + b = 1} e_{a} (x^{2}) e_{b} (y^{2}), \dots, \sum_{a + b = k} e_{a} (x^{2}) e_{b} (y^{2})) = \sum_{i + j = k} L_{i} (e (x^{2})) L_{j} (e (y^{2}))$ in the symmetric polynomial ring of the two blocks.

1.2F1F2F4F6

Well-definedness and trivial values: $L_{k}$ is a well-defined homogeneous weight- $4 k$ polynomial in $p_{1}, \dots, p_{k}$ , independent of the number $N$ of formal roots, by the existence and uniqueness argument of the definition of [F1] together with the injectivity in [F2]; $L_{0} = 1$ is the weight- $0$ component. Evaluating the defining identity at $x_{1} = 0$ (so all $e_{j} (0) = 0$ for $j \geq 1$ ) gives $L_{k} (0, \dots, 0) = [weight 4 k] Q (0) = 0$ for $k \geq 1$ ; hence for a trivial bundle, whose positive Pontryagin classes vanish by [F4], the total class is $L (ε^{r}) = (1, 0, 0, \dots)$ , the unit of the completed ring.

1.3givenF1F4F7algebra

Rank two: let $E \to B$ be a numerable oriented real bundle of rank two over a path-connected paracompact Hausdorff CW base, with Euler class $e$ . By [F7] $p_{1} (E) = e^{2}$ , and $p_{i} (E) = 0$ for $i \geq 2$ since $2 i > rank E$ by [F4]. Substituting into the defining identity of [F1] with $N = 1$ gives $L_{k} (E) = L_{k} (e^{2}, 0, \dots, 0) = [weight 4 k] Q (x) = q_{2 k} e^{2 k}$ after putting $x^{2} = e^{2}$ , so $L (E) = \sum_{k \geq 0} q_{2 k} e^{2 k} = Q (e) = e / \tanh e$ .

1.4F4F5F6F7F8F9algebra

To extend the bundle identities to a path-connected paracompact Hausdorff CGWH base $B$ of CW type, use homotopy inverse maps $h : K \to B$ , $g : B \to K$ from a CW model. The CW-type transport in Pontryagin numbers of a closed oriented manifold gives $p_{i} (E) = g^{*} p_{i} (h^{*} E)$ , naturality between these bases, and stability. The Whitney identity for $h^{*} E, h^{*} F$ on $K$ therefore pulls back to $B$ , and all polynomial L-identities do too. Euler and Chern naturality in [F9] give $e (E) = g^{*} e (h^{*} E)$ and the analogous Chern identity, so the rank-two and complex-line formulas also transport. For a disjoint union, each singular simplex lies in one component; cochains, their differentials and cup products are componentwise products, hence every class and identity assembles componentwise, without selecting representatives for a family of classes.

2.1step 1.1F2F5F6

Multiplicativity: since the two blocks $e_{a} (x^{2})$ and $e_{b} (y^{2})$ are jointly algebraically independent by two applications of [F2], the identity of step 1.1 is equivalent, under the inverse substitution $P_{a} \mapsto e_{a} (x^{2})$ , $P_{b}^{'} \mapsto e_{b} (y^{2})$ , to the polynomial identity $L_{k} (P_{1}^{''}, \dots, P_{k}^{''}) = \sum_{i + j = k} L_{i} (P) L_{j} (P^{'})$ in $Q [P_{1}, \dots, P_{k}, P_{1}^{'}, \dots, P_{k}^{'}]$ , where $P_{r}^{''} : = \sum_{a + b = r} P_{a} P_{b}^{'}$ . For bundles $E, F$ over $B$ , [F5] gives $p_{r} (E \oplus F) = \sum_{a + b = r} p_{a} (E) p_{b} (F)$ in $H^{4 r} (B; Q)$ , so substituting $P_{a} \mapsto p_{a} (E)$ , $P_{b}^{'} \mapsto p_{b} (F)$ into that polynomial identity yields $L_{k} (E \oplus F) = \sum_{i + j = k} L_{i} (E) L_{j} (F)$ for every $k$ ; collecting degrees via the convolution product of [F6] gives $L (E \oplus F) = L (E) L (F)$ in ${\hat{H}}^{4 *} (B; Q)$ .

2.2step 1.2F4F6

Naturality: for a continuous $f : B^{'} \to B$ and a bundle $E \to B$ , [F4] gives $p_{i} (f^{*} E) = f^{*} p_{i} (E)$ ; since $f^{*}$ is a unital ring homomorphism on completed cohomology by [F6] and $L_{k}$ is a polynomial, $L_{k} (f^{*} E) = L_{k} (f^{*} p_{1} (E), \dots) = f^{*} L_{k} (E)$ , and componentwise $L (f^{*} E) = f^{*} L (E)$ .

2.3step 1.2F4

Stability: [F4] gives $p_{i} (E \oplus ε^{r}) = p_{i} (E)$ for every $i$ , so $L_{k} (E \oplus ε^{r}) = L_{k} (E)$ for every $k$ and $L (E \oplus ε^{r}) = L (E)$ ; the trivial-bundle case is step 1.2.

2.4step 1.3F8F9

Complex line: let $ℓ \to B$ be a complex line bundle with $c_{1} (ℓ) = t$ , regarded as an oriented real rank-two bundle through the complex orientation. By [F8] its Euler class is $e (ℓ_{R}) = c_{1} (ℓ) = t$ , so step 1.3 gives $L (ℓ_{R}) = Q (t) = t / \tanh t$ .

3.1step 1.3step 2.1

Whitney sums of rank-two bundles: iterating step 2.1 and using step 1.3, an oriented Whitney sum $E_{1} \oplus \dots \oplus E_{r}$ of numerable oriented rank-two bundles has $L (E_{1} \oplus \dots \oplus E_{r}) = \prod_{i = 1}^{r} L (E_{i}) = \prod_{i = 1}^{r} Q (e_{i})$ , with $e_{i}$ the Euler class of $E_{i}$ .

4.1step 1.2step 1.3step 2.1step 2.2step 2.3step 2.4step 3.1step 1.4∎

Steps 1.2, 2.2 and 2.3 give well-definedness, naturality, stability and the trivial-bundle value; step 2.1 gives multiplicativity; steps 1.3, 2.4 and 3.1 give the rank-two, complex-line and Whitney-sum evaluations. All statements are identities between prescribed polynomials in characteristic classes over $Q$ ; the empty Whitney sum is the trivial bundle, and the rank-zero and rank-two boundary cases are included by $L_{0} = 1$ and by $p_{i} = 0$ for $2 i > rank$ , and AC is used only as declared through the Pontryagin-class construction and its multiplicativity supplier.

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The total L-class and the L-genus of a smooth manifold

Definition

Assume AC, inherited from the Pontryagin-class construction (Pontryagin classes by complexification) and used only there and in the admissibility supplied by Smooth manifolds have CW homotopy type.

Let $M$ be a smooth manifold: finite-dimensional, Hausdorff and second countable, possibly with boundary, possibly disconnected, possibly empty. By Smooth manifolds have CW homotopy type such an $M$ is a CW-type base, its tangent bundle $T M$ is a numerable finite-rank real bundle, and the Pontryagin classes $p_{i} (T M) \in H^{4 i} (M; Z)$ are defined for all $i \geq 0$ , with $p_{0} = 1$ and $p_{i} (T M) = 0$ whenever $2 i > \dim M$ (Pontryagin classes by complexification, Naturality, stability, and mod-two reduction of Pontryagin classes). The total L-class of $M$ is $L (M) : = L (T M) \in {\hat{H}}^{4 *} (M; Q),$ the total L-class of the tangent bundle (The Hirzebruch L-polynomials and the total L-class of a real vector bundle); its component of degree $4 j$ is $L_{j} (T M) = L_{j} (p_{1} (T M), \dots, p_{j} (T M))$ , and for disconnected $M$ the class is taken componentwise. Naturality and stability on CW-type bases are supplied by the transport argument of Pontryagin numbers of a closed oriented manifold, and the L-identities extend to these bases by The L-polynomials are well defined and form a multiplicative, natural and stable sequence. The construction is well defined because each component is a polynomial in the Pontryagin classes and the completed ring is the product of the groups $H^{4 j} (M; Q)$ (The completed cohomology ring in degrees divisible by four, The completed fourfold-graded cohomology ring is natural and satisfies the ring laws).

The L-genus. Let $M$ be a closed oriented smooth manifold of dimension $4 k$ , $k \geq 0$ , with fundamental class $[M] \in H_{4 k} (M; Z)$ . The L-genus of $M$ is the rational characteristic number $L [M] : = ⟨ L_{k} (T M), [M] ⟩ \in Q,$ the degree- $4 k$ evaluation of the total L-class under the Kronecker pairing (Kronecker evaluation pairing, Fundamental class of a compact oriented manifold). Expanding the weight- $4 k$ polynomial $L_{k} = \sum_{∣ J ∣ = k} c_{J} p_{J}$ with $c_{J} \in Q$ and $p_{J} = p_{j_{1}} \dots p_{j_{r}}$ (The Hirzebruch L-polynomials and the total L-class of a real vector bundle) gives $L [M] = \sum_{∣ J ∣ = k} c_{J} p_{J} [M],$ so the L-genus is a rational linear combination of the Pontryagin numbers $p_{J} [M]$ of Pontryagin numbers of a closed oriented manifold.

Zero extension. Since $L (M) = \sum_{j \geq 0} L_{j} (T M)$ is concentrated in degrees divisible by four, a closed oriented manifold whose dimension is not divisible by four has no degree- $4 j$ component in its dimension and no L-genus of the above form; one declares $L [M] : = 0$ in that case as a bookkeeping extension only (see the closing bookkeeping remark on this page). Naturality and multiplicativity of $L$ are properties proved in The L-polynomials are well defined and form a multiplicative, natural and stable sequence, not part of this definition.

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The L-genus is an oriented rational bordism ring homomorphism

Statement

Assume AC. The L-genus The total L-class and the L-genus of a smooth manifold satisfies: (1) $L [M ⊔ N] = L [M] + L [N]$ and $L [- M] = - L [M]$ for closed oriented manifolds of dimension divisible by four; (2) $L [M] = 0$ whenever $M = \partial W$ for a compact oriented smooth manifold $W$ ; (3) $L [M \times N] = L [M] L [N]$ for closed oriented $M^{4 a}, N^{4 b}$ . Consequently the L-genus is well defined on oriented bordism classes, is additive, and extends to a unital $Q$ -algebra homomorphism $L : Ω_{*}^{S O} \otimes Q ⟶ Q,$ assigning to each closed oriented $4 k$ -manifold its L-genus and the value $0$ in dimensions not divisible by four.

Facts & Assumptions

Given: AC; closed oriented smooth manifolds in the dimensions named; the L-genus of The total L-class and the L-genus of a smooth manifold.

[F1]

$L [M] = ⟨ L_{k} (T M), [M] ⟩$ in dimension $4 k$ , and $L_{k}$ is a homogeneous weight- $4 k$ polynomial $L_{k} = \sum_{∣ J ∣ = k} c_{J} p_{J}$ with $c_{J} \in Q$ , so $L [M] = \sum c_{J} p_{J} [M]$ ; for dimensions not divisible by four the value is declared $0$ (The total L-class and the L-genus of a smooth manifold, Pontryagin numbers of a closed oriented manifold).

[F2]

The bundle-level total L-class is stable, multiplicative and natural, and equals $1$ for trivial bundles (The L-polynomials are well defined and form a multiplicative, natural and stable sequence).

[F3]

Every Pontryagin number of a closed oriented boundary vanishes: $p_{J} [\partial W] = 0$ (Oriented boundaries have zero Pontryagin numbers, All characteristic numbers vanish on null-cobordant manifolds).

[F4]

The fundamental class of a disjoint union is componentwise, $[M ⊔ N] = [M] + [N]$ , and reversing the orientation negates it, $[- M] = - [M]$ ; the class of each component of a disjoint union is computed on that component (Fundamental class of a compact oriented manifold, The singular homology of a disjoint union is the direct sum).

[F5]

The tangent bundle of a product splits canonically: $T (M \times N) ≅ T M ⊞ T N$ for the product smooth structure (Canonical tangent and cotangent splittings for products).

[F6]

$[M \times N] = [M] \times [N]$ for closed oriented manifolds with the product orientation (The fundamental class of a product is the cross product of the fundamental classes).

[F7]

The Kronecker pairing is multiplicative under cross products: $⟨ α \times β, [M] \times [N] ⟩ = ⟨ α, [M] ⟩ ⟨ β, [N] ⟩$ (The Kronecker pairing is multiplicative under cross products).

[F8]

Characteristic numbers of products expand over the Kunneth splitting, so pairing the degree- $4 k$ part of $L (T M) L (T N)$ with $[M \times N]$ gives the product of the factor evaluations (Characteristic numbers of products satisfy the Whitney-sum and Kunneth product formulas).

[F10]

Closed oriented manifolds have rational cohomology zero above their dimension (Finite generation from cap with a finite fundamental cycle).

[F9]

Oriented cobordism classes form the graded ring $Ω_{*}^{S O}$ under disjoint union and Cartesian product, with unit the class of a positively oriented point, and $Ω_{*}^{S O} \otimes Q$ is its rationalization; a null-cobordant manifold is the boundary of a compact oriented manifold (Unoriented and oriented bordism groups, Cartesian product makes bordism a graded ring, Product boundary formula for oriented manifolds, Null-cobordant closed manifolds).

Proof

technique · direct; reduce each of the three bordism axioms to the bundle-level multiplicativity and the product formulas

1.1givenF1F2F4algebra

Additivity and orientation: let $M, N$ be closed oriented $4 k$ -manifolds. The restriction of the tangent bundle of $M ⊔ N$ to $M$ is $T M$ and to $N$ is $T N$ , so the polynomial $L_{k} (T (M ⊔ N))$ restricts to $L_{k} (T M)$ and $L_{k} (T N)$ ; with $[M ⊔ N] = [M] + [N]$ by [F4] this gives $L [M ⊔ N] = L [M] + L [N]$ . For the opposite orientation, $T (- M) = T M$ (the tangent bundle does not see the orientation), while $[- M] = - [M]$ by [F4], so $L [- M] = - L [M]$ .

1.2givenF1F3

Boundary vanishing: if $M = \partial W$ with $W$ compact oriented and $\dim M = 4 k$ , then, expanding by [F1], $L [M] = \sum_{∣ J ∣ = k} c_{J} p_{J} [M] = \sum_{∣ J ∣ = k} c_{J} \cdot 0 = 0$ because every Pontryagin number of a boundary vanishes by [F3].

1.3givenF2F5F6F7F8F10

Let $M^{4 a}, N^{4 b}$ be closed oriented with the product orientation, and write $π_{M}, π_{N}$ for the projections. The tangent splitting [F5] and naturality and multiplicativity [F2] give $L (T (M \times N)) = π_{M}^{*} L (T M) π_{N}^{*} L (T N)$ . Its degree- $4 (a + b)$ part is $\sum_{i + j = a + b} L_{i} (T M) \times L_{j} (T N)$ . By [F10] only $i = a, j = b$ can survive. Evaluating this term on $[M] \times [N]$ using [F6] and the matching-degree identity [F7] gives $L [M \times N] = L [M] L [N]$ . This uses the class identity from [F8], without its separate monomial-number expansion.

1.4givenF1F2F5F6F7F10algebra

Multiplicativity with the zero convention: if $4 ∤ (m + n)$ , the product and at least one factor have value zero by [F1]. Otherwise suppose $m = \dim M$ is not divisible by four, write $m = 4 p + r$ with $1 \leq r \leq 3$ , and let $N$ have dimension $n = 4 k - m$ so that $M \times N$ has dimension $4 k$ . Then $L [M] = 0$ by [F1]. In the degree- $4 k$ part of $L (T M) L (T N)$ only the terms $L_{i} (T M) L_{j} (T N)$ with $i + j = k$ occur, and $L_{i} (T M) \in H^{4 i} (M; Q) = 0$ whenever $4 i > m$ , so $i \leq p$ ; similarly $L_{j} (T N) = 0$ whenever $4 j > n = 4 (k - p) - r$ , so $j \leq k - p - 1$ because $r \geq 1$ ; thus $i + j \leq k - 1 < k$ and no term of total degree $4 k$ survives. Hence $L [M \times N] = 0 = L [M] L [N]$ , and the same argument with the roles of $M$ and $N$ interchanged covers $4 ∤ \dim N$ .

2.1step 1.1step 1.2F1F2F9

Descent: the value $L [M]$ is additive under disjoint union and vanishes on oriented boundaries by steps 1.1 and 1.2, so it is constant on oriented cobordism classes: for a cobordism $V$ from $M_{0}$ to $M_{1}$ with $\partial V = - M_{0} ⊔ M_{1}$ by [F9], step 1.2 gives $0 = L [\partial V] = L [- M_{0} ⊔ M_{1}] = - L [M_{0}] + L [M_{1}]$ by step 1.1. Hence $L$ induces a well-defined additive map $Ω_{n}^{S O} \to Q$ for each $n$ and, extended $Q$ -linearly, a functional on $Ω_{*}^{S O} \otimes Q$ that assigns the value $0$ in degrees not divisible by four. It is unital: $L [p t] = 1$ because the tangent bundle of a point is trivial and $L$ of a trivial bundle is $1$ by [F2].

3.1step 1.3step 1.4step 2.1F9∎

By step 2.1 the functional is defined and additive on the rationalized oriented bordism groups, and by steps 1.3 and 1.4 it is multiplicative on products of closed oriented manifolds, with unit value $1$ ; since products of such classes generate $Ω_{*}^{S O} \otimes Q$ biadditively, this makes $L$ a unital $Q$ -algebra homomorphism $Ω_{*}^{S O} \otimes Q \to Q$ , assigning each closed oriented $4 k$ -manifold its L-genus and the value $0$ in dimensions not divisible by four. The empty disjoint union and the zero class satisfy both sides trivially.

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The total L-class of complex projective space is a power of $x / \tanh x$

Statement

Assume AC. Let $γ \to {C P}^{n}$ be the tautological complex line and $y = c_{1} (γ^{*}) \in H^{2} ({C P}^{n}; Z)$ the standard generator with $H^{*} ({C P}^{n}; Z) = Z [y] / (y^{n + 1})$ and $⟨ y^{n}, [{C P}^{n}] ⟩ = 1$ (The tangent bundle of complex projective space and its Pontryagin classes), and give ${C P}^{n}$ its complex orientation. Then in $H^{4 *} ({C P}^{n}; Q)$ , $L (T {C P}^{n}) = Q (y)^{n + 1} = (\frac{y}{\tanh y})^{n + 1} .$ Its degree- $4 j$ component is $c_{j} y^{2 j}$ with $c_{j} = \sum_{i_{1} + \dots + i_{n + 1} = j} q_{2 i_{1}} \dots q_{2 i_{n + 1}} = [y^{2 j}] (y / \tanh y)^{n + 1}$ , the multinomial coefficient sum; for $j = 1$ this is $(n + 1) q_{2}$ .

Facts & Assumptions

Given: AC, the tautological line $γ \to {C P}^{n}$ , the generator $y = c_{1} (γ^{*})$ , and the complex orientation.

[F1]

The total L-class of a smooth manifold is $L (M) = L (T M) \in {\hat{H}}^{4 *} (M; Q)$ , with degree- $4 j$ component $L_{j} (T M)$ , and $L$ of a bundle is a polynomial in its Pontryagin classes (The total L-class and the L-genus of a smooth manifold, The Hirzebruch L-polynomials and the total L-class of a real vector bundle).

[F2]

$L$ is stable, multiplicative and natural, and for a complex line bundle $ℓ$ with $c_{1} (ℓ) = t$ the underlying real L-class is $L (ℓ_{R}) = Q (t) = t / \tanh t$ (The L-polynomials are well defined and form a multiplicative, natural and stable sequence).

[F3]

The in-run supplier establishes the complex bundle isomorphism $\underline{C} \oplus T {C P}^{n} ≅ (γ^{*})^{\oplus (n + 1)}$ and $p (T {C P}^{n}) = (1 + y^{2})^{n + 1}$ (The tangent bundle of complex projective space and its Pontryagin classes).

[F4]

The projective tangent-bundle supplier gives $H^{*} ({C P}^{n}; Z) = Z [y] / (y^{n + 1})$ for $y = c_{1} (γ^{*})$ and $⟨ y^{n}, [{C P}^{n}] ⟩ = 1$ (The tangent bundle of complex projective space and its Pontryagin classes). Rational coefficient change gives the same truncated ring over $Q$ : the integral groups are finite free, and the universal coefficient sequence identifies both coefficient groups with duals of the integral homology free quotients. Thus $H^{4 j} ({C P}^{n}; Q) = Q y^{2 j}$ for $2 j \leq n$ and is zero otherwise (Topological universal coefficient short exact sequence for cohomology).

[F5]

For a complex line bundle, the Euler class of the underlying oriented real bundle is the first Chern class (Top Chern class equals Euler class of the underlying real bundle, Euler class by zero-section pullback of the Thom class, Naturality, orientation sign, and Whitney product for Euler classes).

[F6]

$Q (u) = \sum_{j \geq 0} q_{2 j} u^{2 j}$ with $q_{0} = 1$ (The Hirzebruch L-polynomials and the total L-class of a real vector bundle).

Proof

technique · direct; split the stable tangent bundle and apply the rank-two evaluation

1.1givenF1F2F3

Step [F3] gives a complex bundle isomorphism $\underline{C} \oplus T {C P}^{n} ≅ (γ^{*})^{\oplus (n + 1)}$ . Since $L$ is computed from Pontryagin classes and is therefore unchanged under bundle isomorphism, and since $L$ is stable, $L (T {C P}^{n}) = L (T {C P}^{n} \oplus \underline{C}) = L ((γ^{*})^{\oplus (n + 1)})$ in the completed ring of ${C P}^{n}$ .

1.2givenF2F4F5

The dual tautological bundle $γ^{*}$ is a complex line bundle with $c_{1} (γ^{*}) = y$ by [F4], so by [F5] its underlying oriented real bundle has Euler class $y$ and the complex-line clause of [F2] gives $L (γ_{R}^{*}) = Q (y) = y / \tanh y$ .

2.1step 1.1step 1.2F2

By multiplicativity in [F2], applied $(n + 1)$ times to the Whitney sum of copies of $γ_{R}^{*}$ , $L ((γ^{*})^{\oplus (n + 1)}) = L (γ^{*})^{n + 1} = Q (y)^{n + 1} = (y / \tanh y)^{n + 1}$ .

3.1step 2.1F4F6algebra∎

Degree components: writing $Q (y)^{n + 1} = (\sum_{i \geq 0} q_{2 i} y^{2 i})^{n + 1}$ and extracting coefficients first in the formal indeterminate $y$ and then reducing modulo $y^{n + 1}$ by [F4], the degree- $4 j$ component is the multinomial sum $c_{j} y^{2 j}$ with $c_{j} = \sum_{i_{1} + \dots + i_{n + 1} = j} q_{2 i_{1}} \dots q_{2 i_{n + 1}} = [y^{2 j}] Q (y)^{n + 1}$ , because a product of $n + 1$ factors of weights $4 i_{a}$ has weight $4 j$ exactly when $i_{1} + \dots + i_{n + 1} = j$ ; the case $j = 1$ gives $c_{1} = (n + 1) q_{2}$ , and components with $2 j > n$ vanish by [F4] and the convention on Pontryagin classes. Hence $L (T {C P}^{n}) = Q (y)^{n + 1}$ with the displayed components.

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The L-genus of complex projective space of even complex dimension is one

Statement

Assume AC, inherited from the L-class and projective-space characteristic-class suppliers. For every $k \geq 0$ , with the complex orientation of ${C P}^{2 k}$ , $L [{C P}^{2 k}] = ⟨ L_{k} (T {C P}^{2 k}), [{C P}^{2 k}] ⟩ = 1.$

Facts & Assumptions

Given: AC, the integer $k \geq 0$ , and the complex orientation of ${C P}^{2 k}$ .

[F1]

$L (T {C P}^{2 k}) = Q (y)^{2 k + 1} = (y / \tanh y)^{2 k + 1}$ in $H^{4 *} ({C P}^{2 k}; Q)$ , where $y = c_{1} (γ^{*})$ is the standard generator (The total L-class of complex projective space is a power of $x / \tanh x$ ).

[F2]

The in-run supplier gives $H^{*} ({C P}^{n}; Z) = Z [y] / (y^{n + 1})$ , $⟨ y^{n}, [{C P}^{n}] ⟩ = 1$ , and $p (T {C P}^{n}) = (1 + y^{2})^{n + 1}$ (The tangent bundle of complex projective space and its Pontryagin classes).

[F3]

For every $m \geq 0$ , $[z^{m}] (z / \tanh z)^{m + 1} = 1$ when $m$ is even and $0$ when $m$ is odd (The coefficient identity $[z^{2 k}] (z / \tanh z)^{2 k + 1} = 1$ for every $k$ ).

[F4]

The L-genus in dimension $4 k$ is $L [M] = ⟨ L_{k} (T M), [M] ⟩$ , the degree- $4 k$ evaluation of the total L-class under the Kronecker pairing, and $L (T {C P}^{2 k})$ has degree- $4 k$ component lying in $H^{4 k} ({C P}^{2 k}; Q) = Q y^{2 k}$ (The total L-class and the L-genus of a smooth manifold, Kronecker evaluation pairing, The Hirzebruch L-polynomials and the total L-class of a real vector bundle).

Proof

technique · direct; reduce the genus to a single power-series coefficient

1.1givenF1F2F3

By [F1], $L (T {C P}^{2 k}) = Q (y)^{2 k + 1} = (y / \tanh y)^{2 k + 1}$ , whose degree- $4 k$ component is $c y^{2 k}$ with $c = [y^{2 k}] (y / \tanh y)^{2 k + 1}$ , since $H^{4 k} ({C P}^{2 k}; Q)$ is one-dimensional spanned by $y^{2 k}$ by [F2]. Taking $m = 2 k$ in [F3] gives $c = 1$ .

2.1step 1.1F2F4

Evaluating: $L [{C P}^{2 k}] = ⟨ c y^{2 k}, [{C P}^{2 k}] ⟩ = c ⟨ y^{2 k}, [{C P}^{2 k}] ⟩ = c = 1$ , using $⟨ y^{2 k}, [{C P}^{2 k}] ⟩ = 1$ from [F2] and the $Q$ -linearity of the Kronecker pairing [F4]. For $k = 0$ the manifold is a point with $y \in H^{2} ({C P}^{0}) = 0$ , the total class is $Q (0) = 1$ , and the evaluation on $[p t]$ is $1$ , the same computation with an empty product.

3.1step 1.1step 2.1given∎

Steps 1.1 and 2.1 compute $L [{C P}^{2 k}] = ⟨ L_{k} (T {C P}^{2 k}), [{C P}^{2 k}] ⟩ = 1$ for every $k \geq 0$ , as asserted; AC is used only through the inherited L-class and projective-space suppliers.

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

The signature and the L-genus agree on complex projective space

Statement

Assume AC, inherited from the signature form and L-genus suppliers. For every $k \geq 0$ , with the complex orientation of ${C P}^{2 k}$ , $σ ({C P}^{2 k}) = 1 = L [{C P}^{2 k}] .$

Facts & Assumptions

Given: AC, the integer $k \geq 0$ , the complex orientation of ${C P}^{2 k}$ , and the generator $y = c_{1} (γ^{*})$ .

[F1]

$σ (M) = p - q$ is the difference of the positive and negative inertia indices of the nondegenerate symmetric form $Q_{M}$ on the middle cohomology (The signature of a closed oriented manifold of dimension divisible by four, The middle-dimensional intersection form is symmetric and nondegenerate).

[F2]

$Q_{M} (x, y) = ⟨ x ⌣ y, [M] ⟩$ (The middle-dimensional intersection form of a closed oriented 4k-manifold, Kronecker evaluation pairing).

[F3]

The in-run supplier gives $H^{*} ({C P}^{n}; Z) = Z [y] / (y^{n + 1})$ , and the coefficient bridge of The signature is independent of the diagonalizing basis and unchanged by scalar extension from the rationals to the reals gives $H^{2 k} ({C P}^{2 k}; R) = R y^{k}$ , and $⟨ y^{2 k}, [{C P}^{2 k}] ⟩ = 1$ (The tangent bundle of complex projective space and its Pontryagin classes).

[F4]

The L-genus satisfies $L [{C P}^{2 k}] = 1$ for every $k \geq 0$ (The L-genus of complex projective space of even complex dimension is one).

Proof

technique · direct; the middle cohomology is one-dimensional with value one

1.1givenF1F2F3

By [F3], $H^{2 k} ({C P}^{2 k}; R)$ is the one-dimensional space spanned by $y^{k}$ , and $Q_{{C P}^{2 k}} (y^{k}, y^{k}) = ⟨ y^{2 k}, [{C P}^{2 k}] ⟩ = 1$ by [F2]. Hence the matrix of $Q$ in the basis ${y^{k}}$ is the $1 \times 1$ matrix $(1)$ , its inertia is $(1, 0, 0)$ , and [F1] gives $σ ({C P}^{2 k}) = 1 - 0 = 1$ .

1.2givenF4

The L-genus value is $L [{C P}^{2 k}] = 1$ by [F4].

2.1step 1.1step 1.2∎

Steps 1.1 and 1.2 give $σ ({C P}^{2 k}) = 1 = L [{C P}^{2 k}]$ for every $k \geq 0$ ; for $k = 0$ the manifold is a point, the middle group is $H^{0} = R$ with $Q (1, 1) = 1$ , and both values are $1$ .

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

The signature and the L-genus agree on products of complex projective spaces

Statement

Assume AC, inherited from the signature-product and L-genus suppliers. Let $r \geq 1$ and $k_{1}, \dots, k_{r} \geq 1$ with $k_{1} + \dots + k_{r} = k$ and let $P = {C P}^{2 k_{1}} \times \dots \times {C P}^{2 k_{r}}$ carry the product orientation. Then $σ (P) = 1 = L [P] .$

Facts & Assumptions

Given: AC, integers $k_{1}, \dots, k_{r} \geq 1$ with sum $k$ , and the product $P = {C P}^{2 k_{1}} \times \dots \times {C P}^{2 k_{r}}$ with the product orientation and product smooth structure.

[F1]

The signature is multiplicative under Cartesian products: $σ (X \times Y) = σ (X) σ (Y)$ for closed oriented manifolds of dimensions divisible by four (The signature is multiplicative under Cartesian products).

[F2]

The L-genus is a unital $Q$ -algebra homomorphism on rational oriented bordism, in particular multiplicative: $L [X \times Y] = L [X] L [Y]$ , with the product orientation (The L-genus is an oriented rational bordism ring homomorphism).

[F3]

$σ ({C P}^{2 k_{i}}) = 1 = L [{C P}^{2 k_{i}}]$ for each factor (The signature and the L-genus agree on complex projective space, The L-genus of complex projective space of even complex dimension is one).

[F4]

The product orientation and product smooth structure are those of Product orientations and Products of smooth manifolds have a canonical product smooth structure.

Proof

technique · direct; iterate multiplicativity over the factors

1.1givenF1F3F4

Signature: by [F1], applied inductively to the product and using [F4], $σ (P) = \prod_{i = 1}^{r} σ ({C P}^{2 k_{i}}) = \prod_{i = 1}^{r} 1 = 1$ by [F3].

1.2givenF2F3F4

L-genus: by [F2], $L [P] = \prod_{i = 1}^{r} L [{C P}^{2 k_{i}}] = \prod_{i = 1}^{r} 1 = 1$ by [F3].

2.1step 1.1step 1.2∎

Steps 1.1 and 1.2 give $σ (P) = 1 = L [P]$ , for every $r \geq 1$ , every partition $k_{1} + \dots + k_{r} = k$ and every $k \geq 1$ ; the products are closed oriented smooth of dimension $4 k$ and the case of a single factor is [F3].

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The Hirzebruch signature theorem

Statement

Assume AC. For every closed oriented smooth manifold $M$ of dimension $4 k$ , $k \geq 0$ , $σ (M) = ⟨ L_{k} (T M), [M] ⟩ = L [M],$ the Hirzebruch signature theorem. With the signature extended by zero in dimensions not divisible by four, the signature and the L-genus define the same unital $Q$ -algebra homomorphism $Ω_{*}^{S O} \otimes Q \to Q$ . In particular the signature of a closed oriented $4 k$ -manifold is a rational polynomial in its Pontryagin numbers and depends only on the oriented bordism class.

Facts & Assumptions

Given: AC; a closed oriented smooth $4 k$ -manifold $M$ ; the signature $σ$ and the L-genus $L$ of the pair.

[F1]

The signature vanishes on oriented boundaries and is additive and orientation-reversing, hence descends to a well-defined additive map on rationalized oriented bordism classes in each degree $4 k$ (The signature of an oriented boundary vanishes, so the signature is an oriented cobordism invariant, The signature is additive under disjoint union and negates under orientation reversal, The signature of a closed oriented manifold of dimension divisible by four).

[F2]

The L-genus is a unital $Q$ -algebra homomorphism $Ω_{*}^{S O} \otimes Q \to Q$ , additive, multiplicative and vanishing on boundaries (The L-genus is an oriented rational bordism ring homomorphism, The total L-class and the L-genus of a smooth manifold).

[F3]

For $k = 0$ the basis is the positively oriented point. For every $k \geq 1$ the products ${C P}^{2 k_{1}} \times \dots \times {C P}^{2 k_{r}}$ over partitions $k_{1} + \dots + k_{r} = k$ , $k_{i} \geq 1$ , form a $Q$ -basis of $Ω_{4 k}^{S O} \otimes Q$ ; equivalently $Ω_{*}^{S O} \otimes Q$ is the polynomial algebra on the classes $[{C P}^{2 k}]$ , $k \geq 1$ (Products of complex projective spaces span rational oriented bordism, Unoriented and oriented bordism groups, Cartesian product makes bordism a graded ring).

[F4]

On each basis product $P_{J} = {C P}^{2 k_{1}} \times \dots \times {C P}^{2 k_{r}}$ of [F3], with $r \geq 1$ , $k_{i} \geq 1$ and the product of the complex orientations, $σ$ and $L$ both take the value $1$ : $σ (P_{J}) = 1 = L [P_{J}]$ (The signature and the L-genus agree on products of complex projective spaces, The signature and the L-genus agree on complex projective space).

[F5]

The L-genus of a closed oriented $4 k$ -manifold is $L [M] = ⟨ L_{k} (T M), [M] ⟩$ , a rational linear combination of its Pontryagin numbers (The total L-class and the L-genus of a smooth manifold, The Hirzebruch L-polynomials and the total L-class of a real vector bundle).

Proof

technique · direct; compare the two additive functionals on the spanning family of projective-space products

1.1givenF1F2

Both $σ$ and $L$ are well-defined $Q$ -linear functionals on $Ω_{4 k}^{S O} \otimes Q$ : for $σ$ this is the descent statement [F1], extended $Q$ -linearly; for $L$ it is the homomorphism property [F2].

1.2givenF3F4

For $k \geq 1$ , every partition $J$ of $k$ has a nonempty projective-space product, and [F4] gives $σ (P_{J}) = 1 = L [P_{J}]$ . For $k = 0$ the basis is the positively oriented point, and both values are $1$ by the $k = 0$ case of The signature and the L-genus agree on complex projective space.

2.1step 1.1step 1.2F2F3

Equality of functionals: by [F3] the classes $[P_{J}]$ form a $Q$ -basis of $Ω_{4 k}^{S O} \otimes Q$ , and by steps 1.1 and 1.2 the two $Q$ -linear functionals $σ$ and $L$ agree on every basis element; hence $σ = L$ on $Ω_{4 k}^{S O} \otimes Q$ . Since this holds for every $k$ and both functionals are declared zero in dimensions not divisible by four, they agree on all of $Ω_{*}^{S O} \otimes Q$ ; by [F2] and [F3] the common functional is the unital $Q$ -algebra homomorphism recorded in the statement, since it is multiplicative on the polynomial generators according to [F4] and [F2].

3.1step 2.1F1F5∎

Restating: for the closed oriented $4 k$ -manifold $M$ , its class in $Ω_{4 k}^{S O} \otimes Q$ maps to $σ (M)$ under the signature functional and to $L [M] = ⟨ L_{k} (T M), [M] ⟩$ under the L-genus by [F5], and step 2.1 shows these values are equal; the value depends only on the oriented bordism class and is a rational polynomial in the Pontryagin numbers of $M$ by [F5]. The empty manifold and $k = 0$ give the value $1$ on a positively oriented point and $0$ on the empty manifold, consistent with both sides.

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The four-dimensional signature formula

Statement

Assume AC, inherited from the Hirzebruch signature theorem. For every closed oriented smooth $4$ -manifold $M$ , $σ (M) = \frac{1}{3} ⟨ p_{1} (T M), [M] ⟩ = \frac{p_{1} [M]}{3} .$ Consequently $3$ divides the Pontryagin number $p_{1} [M] \in Z$ .

Facts & Assumptions

Given: AC; a closed oriented smooth $4$ -manifold $M$ .

[F1]

$σ (M) = L [M] = ⟨ L_{1} (T M), [M] ⟩$ (The Hirzebruch signature theorem).

[F2]

$L_{1} = p_{1} / 3$ (The Hirzebruch L-polynomials and the total L-class of a real vector bundle).

[F3]

$p_{1} [M] = ⟨ p_{1} (T M), [M] ⟩ \in Z$ is the first Pontryagin number, computed by the Kronecker pairing, which is linear (Pontryagin numbers of a closed oriented manifold, Kronecker evaluation pairing).

[F4]

The signature $σ (M) = p - q$ is by definition the difference of two nonnegative integers, hence an integer (The signature of a closed oriented manifold of dimension divisible by four).

Proof

technique · direct; specialise the signature theorem to $k=1$

1.1givenF1F2F3

By [F1] and [F2], $σ (M) = ⟨ L_{1} (T M), [M] ⟩ = ⟨ p_{1} (T M) / 3, [M] ⟩ = \frac{1}{3} ⟨ p_{1} (T M), [M] ⟩ = p_{1} [M] / 3$ , using the $Q$ -linearity of the Kronecker pairing [F3].

2.1step 1.1F3F4

Divisibility: $σ (M) \in Z$ and $p_{1} [M] \in Z$ with $3 σ (M) = p_{1} [M]$ , so $3$ divides $p_{1} [M]$ .

3.1step 1.1step 2.1∎

Steps 1.1 and 2.1 prove the displayed formula and the divisibility statement for every closed oriented smooth $4$ -manifold.

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

The eight-dimensional signature formula

Statement

Assume AC, inherited from the Hirzebruch signature theorem. For every closed oriented smooth $8$ -manifold $M$ , $σ (M) = \frac{1}{45} (7 p_{2} [M] - p_{(1, 1)} [M]) .$ Here $p_{(1, 1)} [M] : = ⟨ p_{1} (T M) ⌣ p_{1} (T M), [M] ⟩$ ; it is not the square of a degree-four evaluation on $[M]$ . Consequently $45$ divides $7 p_{2} [M] - p_{(1, 1)} [M]$ in $Z$ .

Facts & Assumptions

Given: AC; a closed oriented smooth $8$ -manifold $M$ , and the Pontryagin classes $p_{1}, p_{2}$ of $T M$ .

[F1]

$σ (M) = L [M] = ⟨ L_{2} (T M), [M] ⟩$ (The Hirzebruch signature theorem).

[F2]

$L_{2} = (7 p_{2} - p_{1}^{2}) / 45$ (The Hirzebruch L-polynomials and the total L-class of a real vector bundle).

[F3]

The Pontryagin numbers $p_{(1, 1)} [M] = ⟨ p_{1} ⌣ p_{1}, [M] ⟩$ and $p_{2} [M] = ⟨ p_{2}, [M] ⟩$ are integers. The Kronecker pairing is $Q$ -linear in the cohomology variable; no multiplicativity of evaluation on a single fundamental class is asserted (Pontryagin numbers of a closed oriented manifold, Kronecker evaluation pairing).

[F4]

$σ (M)$ is an integer by The signature of a closed oriented manifold of dimension divisible by four.

Proof

technique · direct; specialise the signature theorem to $k=2$

1.1givenF1F2F3

By [F1], [F2] and the linearity of the Kronecker pairing [F3], $σ (M) = ⟨ (7 p_{2} - p_{1}^{2}) / 45, [M] ⟩ = \frac{1}{45} (7 ⟨ p_{2}, [M] ⟩ - ⟨ p_{1} ⌣ p_{1}, [M] ⟩) = \frac{1}{45} (7 p_{2} [M] - p_{(1, 1)} [M])$ .

2.1step 1.1F3F4

Divisibility: the left side is an integer by [F4] and $7 p_{2} [M] - p_{(1, 1)} [M]$ is an integer by [F3], so $45$ divides $7 p_{2} [M] - p_{(1, 1)} [M]$ .

3.1step 1.1step 2.1∎

Steps 1.1 and 2.1 prove the displayed formula and the divisibility statement.

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

The signature theorem imposes divisibility constraints on Pontryagin numbers

Statement

Assume AC, inherited from the Hirzebruch signature theorem. For every closed oriented smooth manifold $M$ of dimension $4 k$ the L-genus value $σ (M) = L [M] = ⟨ L_{k} (T M), [M] ⟩$ is an integer and is the rational polynomial in the Pontryagin numbers of $M$ determined by $L_{k}$ . In low dimensions: (1) every closed oriented $4$ -manifold satisfies $3 ∣ p_{1} [M]$ ; (2) every closed oriented $8$ -manifold satisfies $45 ∣ (7 p_{2} [M] - p_{(1, 1)} [M])$ , where $p_{(1, 1)} [M] = ⟨ p_{1} (T M)^{2}, [M] ⟩$ ; (3) more generally $L_{k} (T M)$ is a rational polynomial in the Pontryagin classes and the integrality of its evaluation on $[M]$ is the arithmetic constraint recorded here, for every $k \geq 0$ . No integrality or divisibility is asserted for the analogous expressions for arbitrary bundles.

Facts & Assumptions

Given: AC; a closed oriented smooth manifold $M$ of dimension $4 k$ ; the tangent Pontryagin classes $p_{i} = p_{i} (T M)$ and the L-polynomial $L_{k}$ .

[F1]

The signature theorem gives $σ (M) = L [M] = ⟨ L_{k} (T M), [M] ⟩$ for every closed oriented smooth $4 k$ -manifold (The Hirzebruch signature theorem).

[F2]

The total L-class is $L (M) = \sum_{j \geq 0} L_{j} (T M)$ , a polynomial in the Pontryagin classes with rational coefficients; explicitly $L_{1} = p_{1} / 3$ and $L_{2} = (7 p_{2} - p_{1}^{2}) / 45$ (The total L-class and the L-genus of a smooth manifold, The Hirzebruch L-polynomials and the total L-class of a real vector bundle).

[F3]

The Pontryagin numbers $p_{I} [M] = ⟨ \prod p_{i_{j}}, [M] ⟩$ are integers and the Kronecker pairing is linear over $Q$ in its cohomology variable (Pontryagin numbers of a closed oriented manifold, Kronecker evaluation pairing).

[F4]

The signature $σ (M) = p - q$ is the difference of two nonnegative integers, hence an integer (The signature of a closed oriented manifold of dimension divisible by four).

[F5]

The low-dimensional cases are already established: $3 ∣ p_{1} [M]$ for closed oriented $4$ -manifolds and $45 ∣ (7 p_{2} [M] - p_{(1, 1)} [M])$ for closed oriented $8$ -manifolds (The four-dimensional signature formula, The eight-dimensional signature formula).

Proof

technique · direct; specialise the L-polynomial to each degree and read off the integrality constraint

1.1givenF1F2F3F4

By [F1] and [F2], $σ (M) = ⟨ L_{k} (T M), [M] ⟩$ is the evaluation of the rational polynomial $L_{k}$ in the Pontryagin classes on the fundamental class; by [F3] this evaluation is the corresponding rational linear combination of the Pontryagin numbers $p_{I} [M]$ , and by [F4] its value is an integer.

1.2F2F3F4F5

In dimension four, [F2] gives $L_{1} = p_{1} / 3$ , so $3 σ (M) = p_{1} [M]$ ; both sides are integers by [F4] and [F3], hence $3 ∣ p_{1} [M]$ , in agreement with the first clause of [F5].

1.3F2F3F4F5

In dimension eight, [F2] gives $L_{2} = (7 p_{2} - p_{1}^{2}) / 45$ , so $45 σ (M) = 7 p_{2} [M] - p_{(1, 1)} [M]$ ; both sides are integers by [F4] and [F3] applied to the products $p_{1} ⌣ p_{1}$ and $p_{2}$ , hence $45 ∣ (7 p_{2} [M] - p_{(1, 1)} [M])$ , in agreement with the second clause of [F5].

2.1step 1.1F1F2F3F4

In general degree $4 k$ , [F2] makes $L_{k} (T M)$ a rational polynomial in the Pontryagin classes, [F3] turns its evaluation on $[M]$ into the corresponding rational combination of Pontryagin numbers, and [F1] and [F4] identify that combination with the integer $σ (M)$ ; thus for every $k \geq 0$ the integrality of $⟨ L_{k} (T M), [M] ⟩$ is exactly the arithmetic constraint stated in clause (3), and no further arithmetic conclusion is drawn.

3.1givenstep 2.1∎

Scope of the assertion: the identification of $L [M]$ with an integer uses the tangent bundle and the fundamental class of a closed oriented manifold, so nothing here asserts integrality or divisibility for $L$ of an arbitrary real vector bundle; the final sentence of the statement is this scope boundary, not a vanishing claim.

RemarkRemark: Literature-sourcedProof: Not applicableOpen item page →

The zero extension of the signature is bookkeeping, not a geometric definition

Remark

Assume AC, inherited from the signature definition and theorem. The signature $σ (M)$ is intrinsically defined only for closed oriented smooth manifolds of dimension divisible by four, through the middle-dimensional form $Q_{M}$ (The signature of a closed oriented manifold of dimension divisible by four). Declaring $σ (M) = 0$ when $4 ∤ \dim M$ is a bookkeeping extension used by the sources to make the induced map $Ω_{*}^{S O} \to Z$ additive in all degrees (Unoriented and oriented bordism groups, The signature is additive under disjoint union and negates under orientation reversal); it does not extend the middle-form definition of The signature of a closed oriented manifold of dimension divisible by four beyond dimensions $4 k$ . The geometric identity in The Hirzebruch signature theorem applies in dimensions $4 k$ , while its algebraic formulation explicitly includes this zero extension: on rational oriented bordism, the extended signature equals the L-genus and is a unital $Q$ -algebra homomorphism. Thus the zero extension is multiplicative in all degrees. It supplies no middle-form signature or top $L_{k}$ evaluation outside dimensions divisible by four: in dimensions $4 k + 2$ the middle cup pairing is alternating rather than symmetric, so it has no inertia signature in the sense used here, and the zero value assigned to those dimensions is bookkeeping for the ring structure, not a geometric pairing.

5 · Examples, counterexamples and false statements

None yet.

The Hirzebruch Signature Theorem

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

The middle-dimensional intersection form of a closed oriented 4k-manifold

Definition

The middle-dimensional intersection form is symmetric and nondegenerate

Statement

Facts & Assumptions

Proof

The signature of a closed oriented manifold of dimension divisible by four

Definition

The signature is independent of the diagonalizing basis and unchanged by scalar extension from the rationals to the reals

Statement

Facts & Assumptions

Proof

A nondegenerate symmetric form with a half-dimensional isotropic subspace has zero signature

Statement

Facts & Assumptions

Proof

The restriction image on a cobordism boundary is Lagrangian

Statement

Facts & Assumptions

Proof

The signature is additive under disjoint union and negates under orientation reversal

Statement

Facts & Assumptions

Proof

The signature of an oriented boundary vanishes, so the signature is an oriented cobordism invariant

Statement

Facts & Assumptions

Proof

The tensor product of nondegenerate real symmetric forms has multiplicative signature

Statement

Facts & Assumptions

Proof

The signature is multiplicative under Cartesian products

Statement

Facts & Assumptions

Proof

The formal hyperbolic tangent series and the even series x/tanh⁡xx/tanhx

Definition

The formal hyperbolic tangent and artanh series are inverse, with the artanh derivative

Statement

Facts & Assumptions

Proof

The coefficient identity [z2k](z/tanh⁡z)2k+1=1[z2k](z/tanhz)2k+1=1 for every kk

Statement

Facts & Assumptions

Proof

The completed cohomology ring in degrees divisible by four

Definition

The completed fourfold-graded cohomology ring is natural and satisfies the ring laws

Statement

Facts & Assumptions

Proof

The Hirzebruch L-polynomials and the total L-class of a real vector bundle

Definition

The L-polynomials are well defined and form a multiplicative, natural and stable sequence

Statement

Facts & Assumptions

Proof

The total L-class and the L-genus of a smooth manifold

Definition

The L-genus is an oriented rational bordism ring homomorphism

Statement

Facts & Assumptions

Proof

The total L-class of complex projective space is a power of x/tanh⁡xx/tanhx

Statement

Facts & Assumptions

Proof

The L-genus of complex projective space of even complex dimension is one

Statement

Facts & Assumptions

Proof

The signature and the L-genus agree on complex projective space

Statement

Facts & Assumptions

The formal hyperbolic tangent series and the even series $x / \tanh x$

The coefficient identity $[z^{2 k}] (z / \tanh z)^{2 k + 1} = 1$ for every $k$

The total L-class of complex projective space is a power of $x / \tanh x$