Can a Riemannian Manifold Allow the Existence of a Square Circle?

[andrewkirk]

In certain philosophy discussions the concept of a square circle sometimes comes up as an example of something that can be proven not to exist.

It occurred to me that the impossibility of its existence depends on:
1. the definitions one uses for square and circle; and
2. the geometry in which one is working.

It is well known that, given any widely accepted definition of the two terms, a square circle cannot exist in Euclidean 2-space. However I wondered whether it might be possible for one to exist on a suitably defined Riemannian manifold.

Say we take the following definitions, which match my intuitive concept of a circle and a square:

- A circle of radius r centred on a point P in the plane is the set of all points whose distance from P (measured along a geodesic) is r.

- A square of side L is a set of points comprising the union of four geodesic segments, each of length L, with each end point of each segment being common with the end point of one other segment, no intersections of segments other than on the ends, and the tangent vectors to any two intersecting line segments are orthogonal at the point of intersection.

Is it possible to define a Riemannian metric for a two-dimensional manifold that would allow one or more sets to satisfy both definitions?
Would we have to lose differentiability at some points?

 

[lavinia]

In certain philosophy discussions the concept of a square circle sometimes comes up as an example of something that can be proven not to exist.

It occurred to me that the impossibility of its existence depends on:
1. the definitions one uses for square and circle; and
2. the geometry in which one is working.

It is well known that, given any widely accepted definition of the two terms, a square circle cannot exist in Euclidean 2-space. However I wondered whether it might be possible for one to exist on a suitably defined Riemannian manifold.

Say we take the following definitions, which match my intuitive concept of a circle and a square:

- A circle of radius r centred on a point P in the plane is the set of all points whose distance from P (measured along a geodesic) is r.

- A square of side L is a set of points comprising the union of four geodesic segments, each of length L, with each end point of each segment being common with the end point of one other segment, no intersections of segments other than on the ends, and the tangent vectors to any two intersecting line segments are orthogonal at the point of intersection.

Is it possible to define a Riemannian metric for a two-dimensional manifold that would allow one or more sets to satisfy both definitions?
Would we have to lose differentiability at some points?

You can define a metric - not Riemannian - on Euclidean space whose unit circle is a square

On a Riemannian manifold the set of point equidistant from a central point - if the distance is small enough - is a smooth closed curve. It is a circle by definition on a manifold but may not be a circle when the manifold is embedded in Euclidean space.

your definition of a square can not be a smooth curve.

 

[andrewkirk]

your definition of a square can not be a smooth curve.

Thank you for your reply.

I agree the square will not be a smooth curve, because the tangent is discontinuous at the vertices.

However smoothness is not part of the above definition of a circle. Is it possible to prove that smoothness is a consequence of the definition, bearing in mind that for smoothness to make a square circle impossible, we would need to show that all circles must be smooth, not just some (eg small ones)?

 

[lavinia]

Thank you for your reply.

I agree the square will not be a smooth curve, because the tangent is discontinuous at the vertices.

However smoothness is not part of the above definition of a circle. Is it possible to prove that smoothness is a consequence of the definition, bearing in mind that for smoothness to make a square circle impossible, we would need to show that all circles must be smooth, not just some (eg small ones)?

In a non-Riemannian metric the set of point equidistant from a point does not have to be a circle.

however on a Riemannian manifold it must be a smooth curve.