Isometries between leaf spaces (Q2256260)

Let \((M,F)\) be a foliated manifold and \(M/F\) the leaf space. Question: Given an isometry \(\varphi:M/F\to M'/F'\) between the quotients of two Riemannian manifolds, is \(\varphi\) smooth? The aim of the work is to prove a sufficient condition for an isometry \(\varphi: M/F\to M'/F'\) to be smooth. Theorem 1. ``Let \((M_i,F_i)\), \(i= 1,2\), be singular Riemannian foliations with closed leaves, and \(\varphi: M_1/F_1\to M_2/F_2\) be an isometry. If \(M_1/F_1\) has no boundary, then \(\varphi\) is smooth''. The general case, Theorem 2: ``Let \(M_1\) and \(M_2\) be complete Riemannian manifolds and \((M_1,F_1)\), \((M_2,F_2)\) be singular Riemannian foliations with closed leaves. Assume that there exists an isometry \(\varphi: M_1/F_1\to M_2/F_2\) that preserves the codimension of the leaves. Then \(\varphi\) is a smooth map''. The authors reduce the problem to the case of foliations \((\mathbb{R}^n,F_i)\), \(i= 1,2\) with an isometry \(\varphi: \mathbb{R}^n/F_1\to \mathbb{R}^n/F_2\) preserving the dimension of leaves. Finally, one checks that for any smooth basic function \(f\in C^\infty(\mathbb{R}^n,F_2)\), the pullback \(\varphi^* f\in C^0(\mathbb{R}^n,F_1)\) satisfies the elliptic equation \(\Delta\varphi^* f= \varphi^*\Delta f\).