radius of tubular neighborhood Hi there,
Is there any result about the calculation of radius of tubular neighborhood of submanifold inside a Riemannian manifold?
For example, given a simple smooth curve on R^2, what's the radius of its tubular neighborhood? (One upper bound is given by the minimal curvature, but general it is not the radius)
Maybe that is what we can expect:
if the curvature of the curve is always decreasing, then the radius of the tubular neighborhood is given by the injective radius of (left) end point, it seems true, right? 
For example, they curve is given by: $xy=1, x \in [1,+\infty)$
Thanks!
 A: As has already been pointed out, while the tubular radius bounds the curvature globally from below, the curvature information is not enough to correctly estimate this radius. Many examples can be considered, often silly ones when you allow your sub-manifold to be disconnected: consider the case of two parallel line segments in the plane and note that the embedding radius depends on their distance.
Since it is not clear what exactly you are after (computing the tubular radius of the curve $xy = 1$ is not so hard) or what information you already have, I would like to mention that your problem is solvable by calculus alone! Whether this calculus is tractable or not obviously depends on your choice of Riemannian manifold $X$ and submanifold $M \subset X$.
Given $M \subset X$, the Medial Axis $A_X(M)$ of $M$ in $X$ is defined to be the collection of all $x \in X \setminus M$ such that there are multiple solutions to the following constrained optimization problem in $X$:
Minimize $\text{dist}_X(x,m)$ subject to $m \in M$.
Here is a rough sketch of what a medial axis looks like when $X = \mathbb{R}^2$ and $M$ is the Nicolaescu horseshoe. The axis itself is in blue, and the red lines are my amateurish attempts at showing equidistant $M$-points

Once you know this medial axis, the distance from $A_X(M)$ to $M$ is precisely your tubular radius. Figuring out this distance again reduces to calculus which may be intractably hard depending on the choice of $X$ and $M$.
Update Here is a simple pictorial counterexample to the claim that if the curvature is decreasing from "left to right" then the tubular radius is the injectivity radius of the "left" endpoint. A straight line going up a lampshade suffices. If you want the curvature to decrease strictly, you can wrap the initial segment of the curve around the lower edge of the lampshade and then straighten as you go up. The point is that the curvature of the ambient manifold also plays a part in restricting the tubular radius. It is likely that your conjecture applies in Euclidean space, although I don't have an immediate proof of this. 

A: The inverse of the curvature of a plane curve $C$ at a point $p$ is  the  radius of the osculator  circle to the  curve at the point $p$; see e.g. Geometry and Imagination by Hilbert and Cohn-Vossen. This suggests that the largest radius  of a tube ought to be  the inverse of the maximum of the curvature.  This  is optimal for example when the curve is a circle.   There are however  possible global obstructions, when two points, far apart along the curve, are  actually really close in $\mathbb{R}^2$; think a horseshoe.

