Logo
ScienceStack
Table of Contents
Fetching ...
Sign In

Leafwise positive scalar curvature and the Rosenberg index

Guangxiang Su, Zelin Yi

Abstract

Let $M$ be a closed spin manifold, in this paper, we show that if there is a foliation $(M,F)$ and a Riemannian metric on $M$ that has leafwise positive scalar curvature then the Rosenberg index of $M$ is zero.

Leafwise positive scalar curvature and the Rosenberg index

Abstract

Let be a closed spin manifold, in this paper, we show that if there is a foliation and a Riemannian metric on that has leafwise positive scalar curvature then the Rosenberg index of is zero.

Paper Structure

This paper contains 12 sections, 25 theorems, 158 equations.

Table of Contents

  1. Introduction
  2. Main Theorem
  3. Outline of the proof
  4. Organization of the paper
  5. Sobolev space and elliptic regularity
  6. Kasparov product
  7. Bismut-Lebeau analytic localization
  8. $A_{T,1}$
  9. $A_{T,2}$ and $A_{T,3}$
  10. $A_{T,4}$
  11. Equivalent Kasparov module
  12. Invertibility

Key Result

Theorem 1.1

If $M$ admits positive scalar curvature, then $\langle \widehat{A}(M) \operatorname{ch}(E), [M] \rangle = 0$ for all unitary flat vector bundles $E$.

Theorems & Definitions (48)

  • Theorem 1.1
  • Definition 1.2
  • Theorem 1.3
  • Remark 1.4
  • Theorem 1.5
  • Theorem 1.6
  • Proposition 2.1
  • Proposition 2.2
  • Remark 3.1
  • Proposition 3.2
  • ...and 38 more