Tubular neighbourhood of a smooth curve - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T08:35:13Z http://mathoverflow.net/feeds/question/50319 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/50319/tubular-neighbourhood-of-a-smooth-curve Tubular neighbourhood of a smooth curve Pratik 2010-12-25T04:08:13Z 2010-12-25T18:11:00Z <p>In plane for a smooth non self intersecting curve $C$ the tubular neighbourhood can be constructed with non intersecting starigh line segments normal to the curve from points of the curve. If we use the fact that the $\varepsilon$-tubular neighbourhood is constructed diffeomorphically by extending the normals we could arrive at a necessary condition that $\varepsilon &lt; 1/K$, where $K=\sup_{p\in C} k(p)$ and $k(p)$ is the curvature of curve at $p$. </p> <p>Suppose the curve $C$ is in a surface embedded in $R^3$. The tubular neighbourhood can be constructed similarly along the normal geodesics to the curve. Is it true that for $K=\sup k_g(p)$ where $k_g(p)$ is geodesic curvature of curve at $p$, the necessary condition is $\varepsilon &lt; 1/K$?</p> http://mathoverflow.net/questions/50319/tubular-neighbourhood-of-a-smooth-curve/50326#50326 Answer by Dick Palais for Tubular neighbourhood of a smooth curve Dick Palais 2010-12-25T05:34:47Z 2010-12-25T07:30:13Z <p>No. </p> <p>Consider the example of a right circular cylinder as the surface in question and let $C$ be a "generator" (i.e., a line on the cylinder parallel to the axis). Then the geodesic curvature is everywhere zero, but if epsilon is greater than half the circumference there is no epsilon-tubular neighborhood.</p> http://mathoverflow.net/questions/50319/tubular-neighbourhood-of-a-smooth-curve/50361#50361 Answer by Pietro Majer for Tubular neighbourhood of a smooth curve Pietro Majer 2010-12-25T18:11:00Z 2010-12-25T18:11:00Z <p>As shown in Richard Palais' answer, the radius $\epsilon$ of the larger tubular neighborhood can be much smaller than $1/K$.</p> <p>On the other hand, at least for smooth embeddings of $\mathbb{S}^1$, the best uniform radius $\epsilon$ is also never larger that $1/K$. So in this sense yours is a necessary condition (with weak inequality though). </p> <p>Consider a simple closed curve in, say, a Hilbert space $H$, with a $C^2$ arc-length parametrization $\gamma:\mathbb{S}^1\to H$. Thus, $1/K=\inf_s |\ddot \gamma(s)|$. </p> <p>Let $r > 1 / K$. So there is $s _0$ such that $r > |\ddot\gamma (s_0) |$. Consider the point $x _ 0 \in H$ at abscissa $r$ on the normal at $\gamma (s _0 )$:</p> <p>$$x_0 : = \gamma(s_ 0) + r \frac {\ddot \gamma(s _0) } {| \ddot \gamma(s_0)|}\ .$$</p> <p>Computing the second derivative of $|\gamma(s)-x_0|^2$ at $s = s_0$ we find that the distance from $x _ 0$ to $\gamma(s)$ has a strict local maximum at $s _ 0$. Thus, in particular $s_0$ is different form the global minimum, which is also a point whose normal meets $x_0$. Therefore $x_0$ belongs to at least two normal disks of radius $r$, and in conclusion $r$ is larger than the radius of any tubular nbd. </p> <p>(All that with minor changes also holds for Riemann or Hilbert manifolds; and for non-closed curves too, with some care at the end-points).</p>