TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedaudited 2026-09-22

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Top Pontryagin class is the square of the Euler class

Statement

Assume AC. Let $E \to B$ be a numerable oriented real vector bundle of rank $2 n$ over a nonempty path-connected paracompact Hausdorff CW base, with Euler class $e (E) \in H^{2 n} (B; Z)$ in the given orientation. Then $p_{n} (E) = e (E)^{2} in H^{4 n} (B; Z) .$ In particular the top Pontryagin class is independent of the choice of orientation of $E$ , since in positive rank reversing the orientation negates $e (E)$ and hence leaves $e (E)^{2}$ unchanged. In rank zero the orientation is the canonical unit orientation; no reversal is asserted.

Facts & Assumptions

[A1]

The Axiom of Choice is assumed, exactly as inherited from the Chern and Euler suppliers (The Axiom of Choice).

[F1]

$p_{n} (E) = (- 1)^{n} c_{2 n} (E_{C})$ (Pontryagin classes by complexification).

[F2]

For a numerable complex rank- $m$ bundle $V$ over a path-connected CW complex one has $c_{m} (V) = e (V_{R})$ in the complex orientation (Top Chern class equals Euler class of the underlying real bundle).

[F3]

The complex orientation of $(V_{C})_{R}$ corresponds under the canonical real isomorphism $V_{C} ≅ V \oplus V$ to $(- 1)^{m}$ times the product orientation, for a real bundle $V$ of rank $2 m$ ; consequently $e ((V_{C})_{R}) = (- 1)^{m} e (V)^{2}$ , and Euler classes multiply over ordered direct sums on the general Thom bases and are natural there; reversal negates the integral Euler class only in positive rank, while $e (0_{B}) = 1$ (The complex orientation of the underlying real bundle, Naturality, orientation sign, and Whitney product for Euler classes).

[F4]

The canonical real isomorphism in clause 3 of the complex-orientation lemma is $v \otimes (a + i b) \mapsto (a v, b v)$ ; its inverse is $(x, y) \mapsto x \otimes 1 + y \otimes i$ . The formulas agree with the same real transition matrices in every chart. (The complex orientation of the underlying real bundle).

Proof

technique · direct

Given: AC and a numerable oriented real rank- $2 n$ bundle $E \to B$ .

1.1

The complexification $E_{C}$ has complex rank $2 n$ , and its underlying real bundle is $E \oplus E$ by [F4]; by [F3] the complex orientation of $(E_{C})_{R}$ differs from the product orientation by $(- 1)^{n}$ , explicitly, for a positive real frame $(v_{1}, \dots, v_{2 n})$ , changing the interleaved complex-positive frame $(v_{1}, i v_{1}, \dots, v_{2 n}, i v_{2 n})$ to the product frame $(v_{1}, \dots, v_{2 n}, i v_{1}, \dots, i v_{2 n})$ requires $2 n (2 n - 1) / 2 = n (2 n - 1)$ interchanges, whose parity is $n$ . Therefore $e ((E_{C})_{R}) = (- 1)^{n} e (E \oplus E) = (- 1)^{n} e (E)^{2}$ .

F3F4

1.2

By [F2] applied to the complex bundle $E_{C}$ of rank $2 n$ , $c_{2 n} (E_{C}) = e ((E_{C})_{R})$ .

2.1

Substituting step 1.1 into step 1.2 gives $c_{2 n} (E_{C}) = (- 1)^{n} e (E)^{2}$ , and hence by [F1] $p_{n} (E) = (- 1)^{n} c_{2 n} (E_{C}) = (- 1)^{n} (- 1)^{n} e (E)^{2} = e (E)^{2}$ .

F1step 1.1step 1.2

3.1

Orientation independence for $n > 0$ : the orientation-sign law for Euler classes multiplies $e (E)$ by $- 1$ when the orientation is reversed, while $E_{C}$ and its Chern classes do not depend on the orientation of $E$ ; hence $p_{n}$ is unchanged.

F1F3step 2.1

4.1

Boundary cases. For $n = 0$ the bundle has rank zero, $p_{0} = 1$ and $e (E) = 1$ by the rank-zero Euler convention, so the identity reads $1 = 1$ . For $n = 1$ the identity reads $p_{1} = e^{2}$ for an oriented plane bundle and the sign computation of step 1.1 has $(- 1)^{1} = - 1$ , cancelling the defining sign of $p_{1}$ . Nonorientable bundles are outside the statement, since no integral Euler class is defined; the empty base is excluded explicitly. AC is used only through [A1].

A1F1F2F3step 1.1step 2.1∎

Source notes

This is Hatcher's Proposition 3.15(b), printed pp. 94-96, with the orientation bookkeeping made explicit: the complex orientation of $E_{C}$ differs from the product orientation of $E \oplus E$ by $(- 1)^{n}$ , exactly cancelling the sign in the definition of $p_{n}$ .

Depends on

Used by

Rational cohomology of BO and BSO by Pontryagin and Euler classes Theorem

Dependency tree · two levels

28 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Hatcher, Vector Bundles & K-Theory, Proposition 3.15(b) (standard reference, not scraped)
Miller, MIT 18.906 Algebraic Topology II, Lecture 36 (standard reference, not scraped)