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Math-2231 · mathematics · Geometry

Scalar Curvature Question [?12]: Q-Non-Essentiality of Manifolds with Sc > 0

Status Partially solved · unsolvedmath.comLLM status Open unverified

Statement

Definitions and mathematical statement

Shared notation.
$\mathbb N$, $\mathbb Z$, $\mathbb Q$, $\mathbb R$, and $\mathbb C$ denote the natural numbers, integers, rationals, reals, and complex numbers, respectively. All additional parameters and hypotheses are specified in the question.
Statement recorded in the supplied source.
Let $X$ be a closed oriented Riemannian manifold of strictly positive scalar curvature, and let $\Gamma$ be a discrete group. For every continuous map $f:X\to B\Gamma$, must $f_*[X]=0$ in $H_{\dim X}(B\Gamma;\mathbb Q)$? The classifying space $B\Gamma$ has fundamental group $\Gamma$ and contractible universal cover. The assertion concerns rational homology and includes both spin and nonspin manifolds.

Formulation references: [S1], [S2].

Short English statement

Let $X$ be a closed oriented Riemannian manifold of strictly positive scalar curvature, and let $\Gamma$ be a discrete group. For every continuous map $f:X\to B\Gamma$, must $f_*[X]=0$ in $H_{\dim X}(B\Gamma;\mathbb Q)$? The classifying space $B\Gamma$ has fundamental group $\Gamma$ and contractible universal cover. The assertion concerns rational homology and includes both spin and nonspin manifolds.

Sources

Catalogue entry

Statement licence: CC-BY-4.0 · Source file

Contributions 1

Math-2231-A1 · Literature report · LLM-generated · unknown ·

OpenAI claim summary for Math-2231

Positive scalar curvature and inessentiality: Any map from a closed oriented manifold to a group classifying space factors, up to homotopy, through its own fundamental-group classifying map. The theorem makes that rational fundamental class zero, hence every specified realization is zero; no spin hypothesis is imposed.

With the hypotheses and conventions above, the claimed result addresses the whole mathematical assertion on this page.

Scalar Curvature Question [?12]: Q-Non-Essentiality of Manifolds with Sc > 0

OpenAI's claim

Positive scalar curvature and inessentiality: OpenAI claims that positive scalar curvature forces the rational fundamental class to vanish under the classifying map, without spin or group restrictions. Consequences include the aspherical obstruction, flatness rigidity at nonnegative scalar curvature, and zero real simplicial volume in the nonnegative case.

Mathematical statement of the claimed result

Mathematical claim 1.
For a closed connected oriented smooth $n$-manifold, $n\ge2$, admitting a metric with $\operatorname{Scal}>0$, the classifying map $c_M:M\to B\pi_1(M)$ is claimed to satisfy \[(c_M)_*[M]=0\in H_n(B\pi_1(M);\mathbb Q).\] Its explicit aspherical corollary rules out positive scalar curvature on every closed smooth aspherical manifold, including nonorientable ones by passage to an orientable cover.

Outline of the argument

Positive scalar curvature and inessentiality: An essential rational class is detected on an auxiliary cover at a finite word-metric scale. Compactly supported cohomology and relative duality produce a finite-dimensional comparison with a nonzero local degree. Transverse realization and scalar-curvature reduction contradict that degree; the nonnegative aspherical rigidity statement then uses splitting after the positive-curvature obstruction is established.

Formal verification

Positive scalar curvature and inessentiality: The pinned catalogue lists no Lean formalization for this result.

This site has not run the Lean development. These links describe OpenAI’s selected formal statements and should be read with their stated scope.

Status caveat

OpenAI reports that an internal model produced the manuscripts. Their arguments have not been independently verified as proofs by this site. Any portion outside the documented formal scope may contain mistakes, and the catalogue label reports the announced mathematical claim.

Sources

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