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Positive curvature forces positive Euler characteristic

Statement

Assume the axiom of choice. Let $(M, g)$ be a nonempty closed oriented Riemannian surface whose Gaussian curvature is strictly positive everywhere, $K > 0$ on $M$ . Then $χ (M) > 0$ , where $χ$ is the Euler characteristic of Topological well-definedness of the surface Euler characteristic. No genus formula and no classification conclusion is asserted here.

Facts & Assumptions

Given: A nonempty closed oriented Riemannian surface $(M, g)$ with $K > 0$ at every point.

[A1]

full AC is assumed; it is inherited exactly through the global Gauss-Bonnet theorem; the compact-support positivity proposition is choice-free, and no additional choice is used (The Axiom of Choice).

[F1]

For every closed oriented Riemannian surface, $\int_{M} K d A = 2 π χ (M)$ (Global Gauss-Bonnet for closed oriented surfaces).

[F2]

If a compactly supported top form on an oriented smooth manifold is nonnegative on the positive determinant ray and is not the zero form, then its integral is strictly positive (Positivity of the oriented integral).

[F3]

On an oriented Riemannian surface the area form $d A$ is the positive unit top form of the orientation: in every positively oriented chart it has a strictly positive coordinate coefficient, and it is nonzero at every point (Riemannian volume form on an oriented manifold, The riemannian volume form is the unique positive unit top form).

Proof

technique · exhibit the strictly positive curvature form and apply positivity of the oriented integral

1.1F3given

Since $M$ is closed, the smooth top form $ω : = K d A$ is compactly supported. Because $M \neq \emptyset$ there is a point $p$ , and at $p$ both factors are nonzero: $K (p) > 0$ and $d A_{p} \neq 0$ by [F3]; hence $ω \neq 0$ . In every positively oriented chart $d A$ has strictly positive coordinate coefficient by [F3], so $K d A$ is nonnegative on the positive determinant ray, indeed strictly positive there.

2.1F2step 1.1

By [F2] applied to the nonzero nonnegative compactly supported top form $ω = K d A$ of step 1.1, $\int_{M} K d A > 0$ .

3.1F1step 2.1algebra

By [F1], $2 π χ (M) = \int_{M} K d A > 0$ , and since $2 π > 0$ it follows that $χ (M) > 0$ . This conclusion concerns only the integer $χ (M)$ : no genus, normal form or classification statement is derived.

4.1A1step 3.1∎

No new choice is made: the assumption full AC entered only through the global Gauss-Bonnet theorem; [F2] uses the choice-free finite-chart compact-support integral.

Source locator

Lee, Riemannian Manifolds: An Introduction to Curvature, Chapter 9, Theorem 9.7, printed pp. 167-172, gives $\int_{M} K d A = 2 π χ (M)$ ; Datar, Lectures on Riemannian Geometry, Lecture 2, Theorem 2.2.4, printed pp. 14-15, states the same identity. The strict positivity of the integral for strictly positive curvature is the positivity of the oriented integral of Positivity of the oriented integral, applied to $ω = K d A$ with the area form of Riemannian volume form on an oriented manifold. The item deliberately stops at the sign of $χ$ and asserts no classification consequence.

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Sources

John M. Lee, Riemannian Manifolds: An Introduction to Curvature (1997) (standard reference, not scraped)
Ved Datar, Lectures on Riemannian Geometry (standard reference, not scraped)