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

A classical question in quantitative topology is to bound the mapping degree $\mathrm{deg}(f)$ in terms of its Lipchitz constant $\mathrm{Lip}(f)$. For a closed, oriented manifold $M$, the flexible exponent $\alpha (M)$ is the infimum of $\alpha \ge 0$ such that $\mathrm{\mid }\mathrm{deg}f\mathrm{\mid }\le C(\mathrm{Lip}f{)}^{\alpha }$ 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 $\alpha (M)$: $$\alpha (M)=\{\begin{array}{ll}3& M\text{ modeled on }{\mathbb{S}}^3,{\mathbb{E}}^3,{\mathbb{S}}^2\times {\mathbb{E}}^1,\\ \frac{8}{3}& M\text{ modeled on Nil},\\ 2& M\text{ modeled on Sol},\\ 1& M\text{ modeled on }{\mathbb{H}}^2\times {\mathbb{E}}^1,\\ 0& M\text{ modeled on }{\mathbb{H}}^3,\widetilde{\mathrm{S}{\mathrm{L}}_2}\mathrm{.}\end{array}$$ To prove $\alpha (M)=8\mathrm{/}3$ 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.