Isometry embedding

147 Views Asked by At

Problem: Let $(M,g)$ be a compact Riemannian manifold. Then clearly $(M,d_R)$ is a metric space, where $$ d_R(x,y)=\|x-y\| \quad \forall x,y\in M $$ Now let's see the following:

1. The Kuratowski embedding allows every metric space to be seen as a subset of some Banach space. If we fix $x_0\in M$, then the map $$\Phi:(M,d_R)\hookrightarrow({C^0_b},d_{\infty})$$ defined by $$\Phi(x)={f_x}, \qquad \forall x \in M, $$ is an isometry embedding (but not canonical because it depends on $x_0$), where $${f_x}(y)=d_R(x,y)-d_R(x_0,y), \qquad \forall y\in M,$$ and $${C^0_b}(M) = \{ f:M \to \mathbb R \colon f \text{ is continuous and } \|f\|_\infty = \sup_{x \in M}|f(x)|\leqslant c \text{ for some } c > 0 \}.$$

2. Optimal transport: Let $\mu,\nu \in \mathcal P(M)$, then the optimal distance between $\mu$ and $\nu$ is defined by $$\mathcal {C_g}_{R} (\mu,\nu)= \inf_{\pi\in \Pi} \left\{ \int d_R(x,y)d\pi(x,y) \right\}=W_1(\mu,\nu)$$ where $W_1(\mu,\nu)$ is the Wasserstein distance between $\mu $ and $\nu$, and the coupling or transport plan is given by $$\Pi(\mu,\nu)=\{\pi\in \mathcal P(M\times M): \ (\text{proj}_1)_{\#} \pi =\mu, \quad (\text{proj}_2)_{\#} \pi = \nu \}$$ Further, as a conjecture is given: The map $$\Psi \colon (M,d_R)\hookrightarrow \mathcal (\mathcal P(M),\mathcal {C_g}_{R})$$ defined by $$\Psi(x)= \delta_x, \qquad \forall x\in M$$ is also an isometric embedding, where $\delta_x$ is a Dirac measure at the point $x$.

My question is: What is the relation between these two isometry embeddings?