The question comes from this about metric space which is also smooth manifold. Existence of a Riemannian metric inducing a given distance.
Alexandrov proved that
Suppose that $(𝑀,𝑑)$ is a locally compact finite dimensional path-metric space with curvature locally bounded above and below and extendible geodesics. Then $𝑀$ is homeomorphic to a smooth manifold $𝑀′$ and, under this homeomorphism, the distance function
is isometric to the distance function coming from a Riemannian metric $g$ on $𝑀′$, the regularity of the metric tensor $g$ is $𝐶_{1,\alpha}$
My question is that:
Does there exist cases where a smooth manifold $M$, which is metric space, is homeomorphic and isometric to another smooth manifold $M'$, which is Riemannian metric space, and this map is not diffeomorphism? Can you write it in explicit formulas?
Best Answer
Yes! Let $f:M\to M'$ be any homeomorphism between smooth manifolds, which is not a diffeomorphism and choose any Riemannian metric $g'$ on $M'$. Let $d':M'\times M'\to\mathbb R$ be the induced metric on $M'$ and define the metric $d$ on $M$ by $d(x,y):=d'(f(x),f(y))$, which by construction is an isometry. The topology induced by $d$ is equal to the original topology on $M$, since the identity $\mathrm{Id}_M:M\to(M,d)$ is a composition of homeomorphisms
$$ \mathrm{Id}_M:M\stackrel{f}\to M'\stackrel{\mathrm{Id}_{M'}}\to (M',d')\stackrel{f^{-1}}\to (M,d) $$ For an explicit example you can set $M=M'=\mathbb R$, $f(x)=x^3$ and $g'=dx\otimes dx$ (the standard Riemannian metric on $\mathbb R$). Then $d'(x,y)=|x-y]$ and $d(x,y)=|x^3-y^3|$.