# Intrinsic diameter

The length of a circle is $L$. What is its diameter? The answer could be $L/\pi$ if one uses the metric induced by the standard embedding into the plane. But a geometer is more likely to answer $L/2$, referring to the intrinsic distance: the infimal length of paths joining two points. From this point of view, all simple closed curves of length $L$ are isometric, and in particular all of them have length $L/2$.

In this post, the words distance and diameter will always refer to the intrinsic version.

Here’s a less smooth but still very reasonable (Lipschitz) curve: boundary of a rectangle. Once again, the diameter is half of the perimeter. Such is the distance between two diagonally opposite vertices, marked red and green below.

Let’s move one dimension up: multiply a curve of length $L$ by a line segment of length $h$ to obtain a cylinder. The diameter is $\sqrt{L^2/4+h^2}$, as one sees by cutting the cylinder and flattening it.

The 2-dimensional analog of the rectangulary boundary is the surface of a rectangular box, denoted $S$. Let $a\le b\le c$ be its dimensions. Any two points on $S$ lie in a cylinder with height $h\in\lbrace a,b,c\rbrace$ and circumference $2(a+b+c-h)$. The diameter of such a cylinder is greatest when $h=a$, which yields $\mathrm{diam}\,S\le \sqrt{a^2+(b+c)^2}$. On the other hand, the endpoints of any spatial diagonal of the box are at distance $\sqrt{(a+b)^2+c^2}$ from each other, as one can see by cutting a few edges and flattening the box. Thus, $\displaystyle (*) \qquad \sqrt{(a+b)^2+c^2}\le \mathrm{diam}\,S\le \sqrt{a+(b+c)^2}$

In particular, the cube $a=b=c$ has diameter $a\sqrt{5}$, attained by any pair of centrally symmetric vertices. This can be seen directly from the sketch below (by symmetry, one of two points may be assumed to be within the orange triangle).

But for boxes other than cubes the estimate (*) has some room in it. Intuition may suggests that centrally symmetric vertices is where the diameter is to be found, i.e., that $\sqrt{a+(b+c)^2}$ is the true value. However, this is disproved by french fry $1\times \epsilon\times \epsilon$. The distance between opposite vertices is $\sqrt{1+4\epsilon^2}=1+O(\epsilon^2)$ while the distance between the centers of square faces is $1+\epsilon$.

For more general boxes one can think of other candidates for diameters, such as this pair of black points, which are at distance $\sqrt{(a+c)^2+(b-a)^2}$. Unfortunately I don’t have the right degree to handle this stuff.