Flexible exponent of geometric 3-manifolds and Legendrian maps of Seifert spaces

TL;DR

This paper determines the flexible exponent for various geometric 3-manifolds, revealing how their self-maps' degrees relate to Lipschitz constants, and introduces Legendrian maps for Nil manifolds.

Contribution

It provides a complete classification of the flexible exponent for all Thurston geometric 3-manifolds and constructs Legendrian maps to establish the exponent for Nil manifolds.

Findings

Flexible exponents are explicitly calculated for all Thurston geometric 3-manifolds.

Legendrian maps are constructed for Nil manifolds, showing the exponent is 8/3.

Any Legendrian map cannot be a diffeomorphism.

Abstract

A classical question in quantitative topology is to bound the mapping degree $\operatorname{deg}(f)$ in terms of its Lipchitz constant $\operatorname{Lip}(f)$. For a closed, oriented manifold $M$, the flexible exponent $\alpha(M)$ is the infimum of $\alpha\geq 0$ such that $|\operatorname{deg} f|\leq C(\operatorname{Lip} f)^\alpha$ holds for all differentiable map $f:M\to M$. The flexible exponent measures how effectively a manifold can wrap itself through self-maps. For geometric 3-manifolds $M$ in the sense of Thurston, we give the complete result for $\alpha(M)$: \[ \alpha(M)= \begin{cases} 3 & M \text{ modeled on } \mathbb S^3,\mathbb E^3,\mathbb S^2\times\mathbb E^1,\\ \frac83 & M \text{ modeled on Nil},\\ 2 & M \text{ modeled on Sol},\\ 1 & M \text{ modeled on }\mathbb H^2\times\mathbb E^1,\\ 0 & M \text{ modeled on } \mathbb H^3,\widetilde{\rm SL_2}. \end{cases} \] To prove $\alpha(M)=8/3$ for Nil 3-manifold $M$, we construct the so-called Legendrian map: a smooth self-map $f: M\to M$ such that $f$ is homotopic to the identity and $f$ maps all $S^1$-fibers into the orthogonal contact plane field simultaneously. Moreover, we prove that any Legendrian map must not be a diffeomorphism.