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Closed odd-dimensional manifolds have zero Euler characteristic

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let M be a closed smooth n-manifold with n odd. Then χ(M)=0 (Euler characteristic of a compact manifold).

Facts & Assumptions

Given: A closed smooth n-manifold M with n odd.

[F1]

There is an excellent Morse function f on M, and for any Riemannian metric g the field X:=grad⁡gf vanishes exactly at the (finitely many, nondegenerate) critical points of f (Every compact smooth manifold admits an excellent Morse function, Every smooth manifold admits a riemannian metric, The Riemannian gradient is the metric dual of the differential, The Riemannian gradient vanishes exactly at the critical points, Morse functions and excellent Morse functions).

[F2]

The field −X has the same zeros as X, and ind⁡p(−X)=(−1)nind⁡pX; for odd n this is −ind⁡pX (Negation scales the local index by (−1)n, Isolated zero and local index of a vector field).

[F3]

Poincare-Hopf: for a smooth field with only isolated zeros on a closed manifold, the index sum equals χ(M) (Poincare-Hopf for closed manifolds).

Proof

1.1F1F3algebra

Choose an excellent Morse function f and a Riemannian metric g; the field X=grad⁡gf has only isolated (indeed nondegenerate) zeros, so [F3] gives χ(M)=∑pind⁡pX.

2.1F2F3step 1.1algebra∎

The field −X has the same zero set, and since n is odd [F2] gives ind⁡p(−X)=−ind⁡pX; applying [F3] to −X gives χ(M)=∑pind⁡p(−X)=−∑pind⁡pX=−χ(M) by step 1.1. Hence 2χ(M)=0 in Z, so χ(M)=0.

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