Is there any surface (or general manifold) on which the value of $\pi$ is rational?

Question

Let $X(u,v)=(x(u,v),y(u,v),z(u,v))$ be some map $X:U\to S$

Let $\pi(u,v,d):=C(u,v,d)/d$, with the following definitions:

A "circle" is the set of all points $p\in S$ of distance $d/2$ to some fixed point $q=X(u,v)$ (where distance is measured using the metric $G=dX^TdX$),

And $C(u,v,d)$ is the arclength of the curve (parameterized in local coordinates by $\gamma(u,v,d)$) defined by this set.

(Feel free to suggest other more general definitions if you think they might be helpful)

First off:

1. Is there any surface other than the plane for which $\pi(u,v,d)=const$?
2. Is there any other surface on which $\pi(u,v,d)=\pi(u,v)$ (i.e. its constant per point)

I'm essentially asking if the fact that $\pi$ is irrational is somehow "built-in" to reality, or are there some more "rational" realities where, say, $\pi==22/7$.

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Appendix:

I'm reminded vaguely of the mean value theorem for harmonic functions (and solutions of the heat equation), which states that the mean value of any harmonic function $f$ over any set of concentric circles is constant. Assuming this even applies in the case of a harmonic function over some general metric space (does it?), can we somehow interpret $C/d$ as the (reciprocal?) mean value over circles of a harmonic function, and thereby at least prove (2)?

The mean around some point $(u,v)$ would be calculated as:

$\mu(u,v,d)=\frac{\int_\gamma f(\gamma(u,v,d)) ds}{C(u,v,d)}$

Perhaps the above question can then be rephrased as follows:

Is there any surface on which we can define some harmonic function for which $\mu(u,v,d)=d/C(u,v,d)$, in which case we get by the mean value theorem that $\pi$ is constant per point, and further is there any such surface+harmonic function combination for which $\mu(u,v,d)=d/C=const.$ ?

Answer 1

Moise Cohen and Andrew D. Hwang's answers together tell the story.

First, Andrew's: it's possible to have $\pi$, at some point, be rational. But when you look at the point in question, it's not a manifold point (but see below).

Next, Moishe's remark: He's observing that as $d$ goes to zero, your circle and its circumference become more and more like a circle/circumference in the tangent plane, where $\pi$ turns out to be ... well, $\pi$.

So if you want $\pi$ to be rational (and constant) everywhere, it's not going to happen, by Moishe's remark; if you want it to be rational and constant at some point, that's possible (by Andrew's example), but that point will be a non-manifold point.

Regarding "non-manifold points": the cone can be thought of as being in continuous 1-1 correspondence with a disk (project along the axial direction), and the disk is certainly a smooth manifold. Can't we just pull back this smoothness structure to make the cone a smooth manifold? The answer is "Sure...but then the inclusion of the cone into 3-space is not an embedding." So if you're going to consider the topology and geometry of your surface as being inherited from the ambient space, the arguments above all hold.

By the way: great question!