Context:
Let $\Psi: \mathbb{R}^d \rightarrow \mathscr{H}$ be a $C^k$-embedding of $\mathbb{R}^d$ into a Hilbert space $\mathscr{H}$.
We may view $\mathscr{M}:=Im(\Psi)$ as a submanifold of $\mathscr{H}$.
Define the metrics: \begin{align} d_1: \mathscr{M} \times \mathscr{M} & \rightarrow [0,\infty)\\ d_1(m,n) & \mapsto \|\Psi^{-1}(m)-\Psi^{-1}(n)\|_{\mathbb{R}} \end{align} and the pullback metric $d_2:=\Psi^{\star}(d_{\mathscr{H}})$ of the metric induced by the inner-product in $\mathscr{H}$.
Question:
My question is are the two following metrics equivalent? That is, does there exist a constants $K,k>0$ satisfying: \begin{equation} (\forall m,n \in \mathscr{M})\, d_1(m,n) \leq K d_2(m,n) \leq kd_1(m,n). \end{equation}
Reasoning:
I was thinking we could use the diffeomorphism $\Psi$ (onto $\mathscr{M}$) to define a norm on $\mathbb{R}$ and then use the fact that all norm typologies are equivalent on finite dimensional vector spaces. However I'm not quite sure how to do this/ or if this would actually work...
Intuition: This seems to intuitively make sense when $\mathscr{H}=\mathbb{R}^D$ for some $D>>d$ with the usual euclidean inner product, but my intuition is wavering when the inner-product and space are more foreign...
However, both $\mathbb{R}^d$ and $\mathscr{H}$ have flat metrics... so this again makes me feel like this should be true...
Thanks for your help, ahead of time.
They are not equivalent. Indeed your $d_1$ has no relation to how $\mathscr{M}$ sits inside $\mathscr{H}$.
Counterexample: Let $\Psi : \mathbb R \to \mathbb R^2$, $\Psi(t) = (t, t^2)$. Let $a_n = (n, n^2)$. Then $d_1(a_n, a_m) = |n-m|$, while
$$\begin{split} d_2(a_n, a_m) &= \sqrt{(n-m)^2 + (n^2-m^2)^2}\\ & = \sqrt{1 + (m+ n)^2}|n-m| \\ &= \sqrt{1 + (m+ n)^2} d_1(a_n, a_m). \end{split}$$
Thus there does not exist $M>0$ so that
\begin{equation} d_2(x, y) \le Md_1(x, y),\ \ \ \forall x, y\in \mathscr M. \end{equation}