# Diameter of a circle in an embedded Riemannian manifold

This question was inspired by an answer to the "Magic trick based on deep mathematics" question. I wanted to post it as a comment, but I ran out of characters! I'm sure there must be a collection of standard results related to this question, but I don't know where to start looking.

First, a quick definition. The diameter of a set $S \in \mathbb{R}^n$ is $\sup\{d(x, y) \mid x, y \in S\}$.

A sheet of paper is a good physical example of a Riemannian 2-manifold with boundary, and a table is a good physical model of (a subset of) $\mathbb{R}^2$. Embed the paper isometrically in $\mathbb{R}^2$ by laying flat on a table.

Draw the outline of a circular cup on the paper. It seems obvious that no matter how you embed the paper in $\mathbb{R}^2$, the outline of the cup will always be a metric circle, and it will always have the same diameter $D$.

Now, lift the paper into the air, embedding it isometrically in $\mathbb{R}^3$. If you let the paper flop around, the outline of the cup might not be a metric circle anymore... but will it still have diameter $D$?

Finally, cut along the outline of the cup, removing an open disk from the sheet of paper. The paper now has a second boundary component, and it's no longer simply connected. The paper has also gained a surprising property: you can bend it around in midair (that is, embed it isometrically in $\mathbb{R}^3$) so that the outline of the cup has diameter greater than $D$! What's the important property of the paper that we changed to make this possible?

I don't think you need to cut along the outline of the cup to make this work... you could probably just cut out any disk contained within the outline of the cup. So maybe simply-connectedness is the important property?

My gut tells me that if you draw two dots on the sheet of paper, the distance between the dots is maximized when the paper is flat on the table. When you bend the paper around in midair, the dots can get closer together, but they can never get farther apart. I think this is equivalent to the statement that if $\delta$ is the natural distance function on the paper, $d$ is the distance function in $\mathbb{R}^3$, and $F$ is an isometric embedding of the paper in $\mathbb{R}^3$, $d(Fx, Fy) \le \delta(x, y)$ for all points $x$ and $y$ on the paper.

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Of course, the most immediate change you made was to cut out something, so that the intrinsic diameter of your radius-$1$ circle is actually $\pi$, since that's how far someone would have to travel to get from one end to the other moving inside the paper. –  Theo Johnson-Freyd Apr 26 '10 at 16:12

For the second question:

As commented above, if you remove an open disc from the inside of the region bounded by a circle, you can bend the paper so as to increase the diameter of the circle in $\mathbb{R}^3$. This doesn't work if you remove a point rather than an open disc - but with a point removed you can instead bend the paper so as to decrease the diameter of the circle by making the removed point a cone point of a wavy cone. This gives us a good way of thinking about what's going on here: embedding $\mathbb{R}^2$ into $\mathbb{R}^3$ preserves diameter of circles, but when you ignore part of $\mathbb{R}^2$, (such as when your circle is on a small piece of paper, or when you've removed a point inside the circle) you can introduce singularities there. Perhaps what's important is probably not the fundamental group, but the type and location of singularities that would arise if the embedding was extended to all of $\mathbb{R}^2$. Any thoughts?

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The answer to the first is NO.

As mentioned by another answer you can use Do Carmo to prove a local result of the type: at points where the normal curvature is non-zero the 0-directions defines a foliation of straight lines.

With this one can prove that embedding $\mathbb{R}²$ into $\mathbb{R}^3$ preserves the diameter of circles, but one can construct a counter example in the case of a piece of paper:

Take a equilateral triangle on your table with points barely out side of the piece of paper. Then since these sides do not meet inside the piece you may fold (using very sharp foldings) the piece of paper up at the sides of the triangle such that out side of a nighborhood of the triangle the paper lies in planes which are perpendicular to the table. Note that this can not be extended to all of $\mathbb{R}^2$ because the folding lines interset! Then shrinking the circumscribed sphere slightly of the triangle to lie in the paper - this is folded up so that the diameter becomes strictly less.

For the second question I really dont see $\pi_1$ any where. What I see is the following: for me it is natural to defined the distance on a Riemmanian manifold $X$ as:

$d(x,y) = \inf_{\gamma} \textrm{len}(\gamma)$

where the infimum is over curves $\gamma$ from $x$ to $y$, and len($\gamma$) is the length of the curve. This is intrinsic and gives the same diameters for the piece of paper, but when you remove an open set it becomes different, because the curves has to avoid this set.

Since isometric embeddings preserve lengths of curves you get with this definition

$d(Fx,Fy) \leq d(x,y)$

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The embedding is not a product in general. A piece of a cone is a simples counter-example. –  Sergei Ivanov Apr 26 '10 at 11:57
Your right, I will change that. –  Thomas Kragh Apr 26 '10 at 11:59
Nice example, but I disagree with your last paragraph: the diameter is for the distance in $\mathbb{R}^3$, (otherwise your counter example would not be one). –  Benoît Kloeckner Apr 26 '10 at 12:03
Another comment: if the sheet of paper is large enough compared to the circle (more precisely, as soon as all triangles having the given circle as inscribed circle have one of their vertex in the sheet), then your construction does not work. In this case, is the answer positive? –  Benoît Kloeckner Apr 26 '10 at 12:10
@Benoît: yes, if the sheet contains a circle, say, of radius 3 times larger than the cup (with the same center), you will have to leave one of the diameters intact. –  Sergei Ivanov Apr 26 '10 at 12:21
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