0
$\begingroup$

Take the set in $R^3$ such that for any point $(x,y,z)$ in it, $\sqrt{x^2+y^2} = z$.

It is well known, that the smooth flat metric (induced from $R^3$) on the complement of (0,0,0) in this cone doesn't extend to the origin. Which means that there is a pair of smooth vector fields $v_1, v_2$ such that $d(v_1,v_2)$ is not smooth. What is a simple example of such a pair?

I tried to find one by writing this metric in a smooth chart around the origin and got bogged down in calculations. I produced something like $dx^2+dy^2 -\frac{x^2+y^2}{xy}$dxdx, but not sure if that's correct (here (x,y) are the same coordinates as on R^3, i.e. I just projected the cone).

I have some intuition about why this metric is not smooth in terms of lengths of curves: the length changes "sharply" around the "sharp" point. Also, I know of other ways to see that this metric is not extendable smoothly: the holonomy is not trivial even though the curvature is. But I want to get a feeling for this singularity in terms of vector fields.

$\endgroup$
1
$\begingroup$

The cone is a topological manifold (homeomorphic to $\mathbb R^2$), but it's not a smooth submanifold of $\mathbb R^3$. Thus it has a smooth structure, but not an obvious one that's adapted to the present situation.

The question whether the induced metric on the cone is smooth doesn't make sense until you decide on a smooth structure for the cone. One such smooth structure is obtained by choosing the global chart obtained by restricting the $x$ and $y$ coordinates to the cone, and declaring that to be a smooth chart. In that chart, the metric has the formula $$ dx^2 + dy^2 + \frac{(x\,dx + y\, dy)^2}{x^2+y^2} $$ and, for example, the inner product of $\partial/\partial x$ with itself is $1 + x^2/(x^2+y^2)$, which is not smooth at the origin.

But there are many other smooth structures you could choose, and they might give different results.

The real point here, though, is that there is no smooth structure on the cone under which this metric extends smoothly to the origin. Probably the easiest way to see this is to note that if $g$ is any smooth metric on a $2$-manifold $M$ and $p\in M$, then the circle $C_r$ of radius $r$ around $p$ (that is, the set of points whose Riemannian distance from $p$ is exactly $r$) has length satisfying $$ \frac{L(C_r)}{2\pi r} \to 1 \text{ as } r\to 0. $$ This can be proved by doing the computation in normal coordinates centered at $p$.

However, in the cone, the circle at a Riemannian distance $r$ from the origin is the set $\{(x,y,z): z>0,\ x^2 + y^2 = z^2 = r^2\}$, which has length $\pi r\sqrt{2}$, so the above limit is $1/\sqrt{2}$ instead of $1$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Can your proof be made to work in $C^1$ or it requires $C^2$? $\endgroup$ – mathquest Feb 26 '17 at 23:57
  • $\begingroup$ Yes, you are right about the possible existence of other charts (compatible with the annulus where the metric is smooth) such that this metric is $C^0$ or $C^1$. But is there? $\endgroup$ – mathquest Feb 27 '17 at 0:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.