TITLE: diameter of Riemannian manifolds
QUESTION [6 upvotes]: Let $M$ be a compact Riemannian manifold of dimension $n$. If $M\subset N$ is an embedding in another Riemannian manifold, we can define the diameter $d(M,N$) of $M$ in $N$ as the longest distance in $N$ of two points in $M$. For example if $S^2$ is a sphere of radius $R$ in ${\mathbb{R}}^3$, we have
$$d(S^2,S^2)=\pi R , \qquad d(S^2,{\mathbb{R}}^3)=2R.$$
If $d>0$ is an integer we can then define
$$e(M,d)=\inf\big(d(M,N);\quad M\subset N \quad \text{and} \quad \dim(N)=\dim(M)+d\big)$$
If there a way (a formula) to compute $e(M,d)$ ? Is it true that
$$\lim_{d\to\infty}e(M,d) \ = \ 0 \ ?$$

REPLY [2 votes]: This is just an expanded version of Einar Rødland's comment. Note that it suffices to show that $e(M,1) = 0$. Let $N = M \times \mathbb R$ and $M\subseteq N$ is identified with $M \times \{0\}$. For each $\epsilon >0$, let $g = g_\epsilon$ be a metric on $N$ given by 
$$g_\epsilon = f^2_\epsilon (t) ( g_M + dt^2),$$
where $ f_\epsilon$ is a positive function so that 
$$f_\epsilon (t) = \begin{cases} 1 & \text{if } t\in (-\epsilon/8,\epsilon/8), \\ \epsilon/2d(M) &\text{if } t= \epsilon/4\\ \le 1 &\text{otherwise.}\end{cases}$$
Note that $M \subset N$ is an isometric embedding. Let $x,y\in M$ and $\gamma$ be a minimal geodesic in $M$ joining $x, y$. Then the (piecewise smooth) curve 
$$ (x,0) \to (x,\epsilon/4) \overset{(\gamma(\cdot), \epsilon/4)}{\to } (y,\epsilon/4) \to (y,0)$$ 
has length less than $\epsilon/4 + (\epsilon/2d(M)) \times d_M(x,y) + \epsilon /4 \le \epsilon$.
Since $x, y$ are arbitrary, $e(M,1) \le \epsilon$. Thus $e(M,1) = 0$ as $\epsilon $ is arbitrary.