Relationship between radii in Riemannian manifolds S.T. Yau raised an (somewhat) interesting question in a lecture today, which I thought I would ask for thoughts on from the good people here.
The context is as follows: Consider a 3-manifold $M$ with boundary and $\sigma$ be a loop that bounds a disk in $M$. Then we define the radius $\text{Rad}(M)$ of $M$ as:
$$ \text{Rad}(M):=\sup_{\sigma}\inf\big\{\epsilon>0 \ \big| \ [\sigma]=0\in\pi_1(T_{\epsilon}(\sigma)) \big\}, $$
where $T_{\epsilon}(\sigma)$ is the $\epsilon$-tubular neighborhood around $\sigma$. It turns out that this is useful for estimating the first eigenvalue of an operator important in GR:

Thm: (See Prop. 1, Schoen-Yau (1983)) Suppose $M$ is a Riemannian 3-manifold and $\Omega\subset M$ is a bounded domain. Then the first eigenvalue of $-\nabla+\tfrac{R}{2}$ satisfies:
$$ \lambda_1\leq \frac{2}{3\pi\cdot\text{Rad}^2(\Omega)}. $$

At any rate, Yau remarked that one could, of course, generalize this definition by replacing loops with higher-dimensional spheres:

We define the $i$-radius $\text{Rad}_i(M)$ of $M$ as:
$$ \text{Rad}_i(M):=\sup_{\sigma}\inf\big\{\epsilon>0 \ \big| \ [\sigma]=0\in\pi_i(T_{\epsilon}(\sigma)) \big\}, $$

Then the question is:
$\fbox{$\textbf{Q: What relationship, if any, is there between the various } \text{Rad}_i(M)?$}$
Thoughts? Anything jump out as even likely to be true?

Note: Of course one could also increase the dimension of $M$ - the only reason for focusing on 3-manifolds here is because, well, they tend to be more relevant for people doing GR, and many of the results Yau was discussing during this lecture depended on certain maps being conformal, which was only the case (generally speaking) in this context.
Note 2: It seems to me that if we focus on a map $\sigma:S^{i+1}\rightarrow M$ that is an isometry, and we consider the collection of $\sigma':S^i\rightarrow M$ (again focusing on isometries) sitting inside of (the image of) $\sigma$, then the $\sup$ over such $\sigma'$ will be obtained by the $\sigma'$ that are great circles. In that case, the $\inf\{\epsilon\}$ will be the same for both $\sigma$ and $\sigma'$… and so it seems like $\text{Rad}_{i+1}(M)=\text{Rad}_i(M)$ (and an inductive argument would yield that all the $i$-radii agree)… but for general $\sigma:S^{k+1}\rightarrow M$ (that might be "lumpier") it isn't so clear… hmm…

 A: As a special case, there is no obvious relationship between $\mathrm{Rad}_1(M)$ and $\mathrm{Rad}_2(M)$.  For example:


*

*Let $M$ be an $\epsilon$-neighborhood of the $xy$-plane in $\mathbb{R}^3$.  Then clearly $\mathrm{Rad}_1(M)$ is infinite, but $\mathrm{Rad}_2(M)$ is on the order of $\epsilon$, since any image of a sphere in a plane is contractible.

*For the second example, consider the union of the spherical cylinder
$$
x_1^2 + x_2^2 + x_3^2 \,=\, \epsilon^2,\qquad x_4 \geq 0
$$
in $\mathbb{R}^4$ with the ball
$$
x_1^2 + x_2^2 + x_3^2 \,\leq\, \epsilon^2,\qquad x_4 = 0.
$$
Think of this as an infinite cylinder with a cap, except in one dimension higher, so the cross-sections of the cylinder are spheres, and the cap is a ball.
This isn't a smooth manifold, since there's a spherical "seam" at the place where the cylinder is glued to the ball.  However, metrically it does have the property that $\mathrm{Rad}_2(M)$ is infinite (since you can use spherical cross-sections with arbitrarily large $x_4$ value) while $\mathbb{Rad}_1(M)$ is on the order of $\epsilon$ (since each spherical cross-section of radius $\epsilon$ is simply connected).
It's not hard to find a $C^\infty$ manifold in $\mathbb{R}^4$ with the same basic shape, e.g. the manifold
$$
x_4 \;=\; \frac{x_1^2 + x_2^2 + x_3^2}{\epsilon^2-x_1^2 - x_2^2 - x_3^2} \qquad\text{for}\qquad x_1^2+x_2^2+x_3^2 < \epsilon^2.
$$
