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The real projective plane is not unoriented null-cobordant

Statement refuted

It is false that the real projective plane is null-cobordant: there is no compact smooth $3$ -manifold whose boundary is ${R P}^{2}$ , so not every closed surface bounds. The counterexample computes the Stiefel-Whitney number $w_{1}^{2} [{R P}^{2}] = ⟨ w_{1} (T {R P}^{2})^{2}, [{R P}^{2}] ⟩ = 1$ and concludes that the class of ${R P}^{2}$ is a nonzero element of $Ω_{2}^{O}$ (Unoriented and oriented bordism groups).

Facts & Assumptions

Given: The real projective plane ${R P}^{2}$ with its smooth structure and tangent bundle, the trivial real rank-three bundle over a point, and AC (The Axiom of Choice) for the Stiefel-Whitney class construction.

[F1]

${R P}^{2}$ is the projectivisation $P (E)$ of the trivial rank-three real bundle $E$ over a point, with tautological degree-one class $x = x_{E} \in H^{1} ({R P}^{2}; F_{2})$ ; the mod-two projective bundle theorem makes $H^{*} ({R P}^{2}; F_{2})$ a free module over $H^{*} (p t; F_{2}) = F_{2}$ with basis $1, x, x^{2}$ and unique monic relation $x^{3} + c_{1} x^{2} + c_{2} x + c_{3} = 0$ with $c_{i} \in H^{i} (p t) = 0$ for $i > 0$ ; hence $x^{3} = 0$ , $x \neq 0$ and $x^{2} \neq 0$ (Real projective bundle and tautological line, Tautological degree-one class on a real projective bundle, Mod-two real projective bundle theorem).

[F2]

A numerable real bundle $E$ has $w_{1} (E) = 0$ exactly when $E$ is orientable, and for a closed $n$ -manifold with $n \geq 2$ an orientation of the tangent determinant lines is equivalent to an atlas with positive transition Jacobians; ${R P}^{2}$ is not orientable (The first Stiefel–Whitney class classifies orientability, Positive oriented atlases characterize orientations except for one-manifolds with boundary, Positive-dimensional real projective space is orientable exactly in odd dimension).

[F3]

${R P}^{2}$ is a connected closed smooth surface (any two lines are joined by the projectivization of a path in the sphere), hence an admissible base whose tangent bundle is numerable, and it carries the canonical mod-two fundamental class $[M] \in H_{2} ({R P}^{2}; F_{2})$ of its canonical mod-two orientation (Smooth manifolds have CW homotopy type, Every manifold is F2-orientable and orientability is componentwise, Fundamental class of a compact oriented manifold).

[F4]

For a connected closed $2$ -manifold the pairing $H^{2} (M; F_{2}) \times H^{0} (M; F_{2}) \to F_{2}$ , $⟨ a ⌣ b, [M] ⟩$ , is perfect, $H^{0} (M; F_{2}) = F_{2} \cdot 1$ , and the Kronecker evaluation is $F_{2}$ -bilinear (Poincaré duality gives a nonsingular cup pairing, Kronecker evaluation pairing).

[F5]

A Stiefel-Whitney number of a closed smooth $n$ -manifold is $w^{I} [M] = ⟨ w^{I} (T M), [M] ⟩$ for a degree- $n$ monomial, and a closed manifold with at least one nonzero Stiefel-Whitney number is not null-cobordant (Stiefel-Whitney numbers of a closed manifold, Boundaries have zero Stiefel-Whitney numbers, Null-cobordant closed manifolds).

Counterexample

1.1F1

(The mod-two cohomology of ${R P}^{2}$ .) Model ${R P}^{2}$ as the projectivisation of the trivial rank-three bundle over a point. By [F1], $H^{*} ({R P}^{2}; F_{2})$ is free over $F_{2}$ with basis $1, x, x^{2}$ , where $x$ is the tautological degree-one class, and the only relation is $x^{3} = 0$ ; in particular $x \neq 0$ and $x^{2} \neq 0$ , and $H^{1} ({R P}^{2}; F_{2}) = F_{2} \cdot x$ has exactly two elements.

2.1F2F3step 1.1

( $w_{1} (T {R P}^{2}) = x \neq 0$ .) Suppose $w_{1} (T {R P}^{2}) = 0$ . The tangent bundle of a closed smooth manifold is numerable over an admissible base by [F3], so by [F2] the vanishing of $w_{1}$ would make $T {R P}^{2}$ orientable, and for the closed surface ${R P}^{2}$ the orientation of the tangent determinant lines would give an atlas with positive transition Jacobians, making ${R P}^{2}$ orientable. This contradicts [F2], since ${R P}^{2}$ is not orientable. Hence $w_{1} (T {R P}^{2}) \neq 0$ ; by step 1.1 it is the unique nonzero element, $w_{1} (T {R P}^{2}) = x$ .

3.1F3F4step 1.1step 2.1

( $w_{1}^{2} [{R P}^{2}] = 1$ .) By step 2.1, $w_{1} (T {R P}^{2})^{2} = x^{2}$ , and $x^{2} \neq 0$ by step 1.1. Apply [F4] with $M = {R P}^{2}$ : the pairing $H^{2} \times H^{0} \to F_{2}$ is perfect and $H^{0} = F_{2} \cdot 1$ , so the adjoint map $a \mapsto ⟨ a \cdot 1, [{R P}^{2}] ⟩ = ⟨ a, [{R P}^{2}] ⟩$ is an isomorphism $H^{2} ({R P}^{2}; F_{2}) \to F_{2}$ ; a nonzero class therefore has evaluation $1$ . Hence $w_{1}^{2} [{R P}^{2}] = ⟨ x^{2}, [{R P}^{2}] ⟩ = 1 \neq 0$ .

4.1F5step 3.1∎

(Conclusion: ${R P}^{2}$ is not null-cobordant.) The Stiefel-Whitney number $w_{1}^{2} [{R P}^{2}] = 1$ is nonzero, so by [F5] the closed surface ${R P}^{2}$ is not null-cobordant: it is not the boundary of any compact smooth $3$ -manifold, and in particular not every closed surface bounds. Therefore the cobordism class of ${R P}^{2}$ is a nonzero element of $Ω_{2}^{O}$ , and the claim that ${R P}^{2}$ is null-cobordant is refuted.

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Sources

John Milnor and James Stasheff, Characteristic Classes (original pagination) (standard reference, not scraped)
Daniel S. Freed, Bordism: Old and New (lecture notes, UT Austin, Fall 2012) (standard reference, not scraped)