A cone with a $90^\circ$ vertex angle can be parameterized by

$$(x, y, z) = (z \cos \theta, z \sin \theta, z).$$

The metric on the cone can then be found to be

$$dx^2 + dy^2 + dz^2 = 2 dz^2 + z^2 d \theta.$$

The only non-zero Christoffel symbols are

$$\Gamma^z_{\space \space \theta \theta} = -\frac{1}{z} \hspace{1 cm}\Gamma^\theta_{\space \space \theta z} = \frac{z}{2}$$

and the only non-zero independent component of the Riemann curvature tensor is

$$R_{z \theta z \theta} = 2 z^{-2} + 1.$$

It's possible I made a mistake in the above computations, but I believe they are correct.

This shows that a cone is not locally flat.

However, a cone in real life can be formed by curving a piece of paper. For example, to compute the surface area of a cone, you can consider the surface area of a pac-man shape laying flat on a table. When the top and bottom of the mouth of the pac-man are brought together, the paper is in the shape of a cone.

The Christoffel symbols and Riemann tensor contain no information on the embedding of a manifold in ambient space. Therefore, it would seem to me that the Riemann tensor would have to be $0$, as it would be for flat piece of paper. How can this be?

  • $\begingroup$ A circular cone, just as a circular cylinder has Gaussian curvature zero. Gaussian curvature is intrinsic. $\endgroup$ – Lord Shark the Unknown Feb 18 '18 at 21:00
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    $\begingroup$ Your calculations are incorrect. For starters, it looks like your Christoffel symbols are incorrect; I think they should be $\Gamma^z_{\theta\theta}-\frac12z$ and $\Gamma^\theta_{\theta z}=\frac1z$ instead. In the calculation of $R$, the terms coming from $\partial \Gamma$ should cancel out those coming from $\Gamma\Gamma.$ $\endgroup$ – Anthony Carapetis Feb 18 '18 at 22:27

As Anthony Carapetis correctly points out, there was indeed a mistake in my calculation and the Riemann tensor of a cone is indeed $0$. I will say a few more words for those who stumble across this answer and want an extra lesson about cones and curvature.

I mistakenly thought that a cone cannot have $0$ Riemann curvature because if you take a vector tangent to the cone and parallel transport it $360^\circ$ around the cone, it will not come back in the same configuration as when you started. (Think about the cone in terms of the pac-man to convince yourself of this.) Because the cone has non-trivial parallel transport properties, I assumed it could not be flat.

Let's introduce a slope parameter $a$ to the cone. When $a = 1$, we have the cone from before. When $a = 0$ we have a flat plane.

$$(x, y, z) = (r \cos \theta, r \sin \theta, ar).$$

$$dx^2 + dy^2 + dz^2 = (1 + a^2) dr^2 + r^2 d \theta$$

$$\Gamma^r_{\space \space \theta \theta} = \frac{-r}{1 + a^2} \hspace{1 cm} \Gamma^\theta_{\space \space \theta r} = \frac{1}{r}$$

$$R_{r \theta r \theta} = 0$$

Let us now consider parallel transporting a vector $u^\mu = (u^r, u^\theta)$ around the cone. Our path will be

$$(r, \theta) = (A, t)$$

and the velocity vector around this path will therefore be

$$(v^r, v^\theta) = (0, 1).$$

From the parallel transport equation

$$\frac{d u^\nu}{dt} = - \Gamma^\nu_{\space \space \mu \sigma} u^\mu v^\sigma$$

we can easily work out the following two equations governing the evolution of $u^\mu$:

$$\frac{d^2 u^r}{dt^2} = -\frac{1}{1 + a^2} u^r$$ $$\frac{d^2 u^\theta}{dt^2} = -\frac{1}{1 + a^2} u^\theta$$

When $t = 2 \pi$ we will have completed a full loop around the cone.

We can see from the two equations above that when $a = 0$, the vector $u^\mu$ is returned to its original state. Otherwise, the vector will in general be different. (The solution for each equation is a linear combination of $\cos(t / \sqrt{1 + a^2} )$ and $\sin(t / \sqrt{1 + a^2} )$.)

I mistakenly thought that if the curvature was $0$ everywhere this cannot happen. Now I realize that this is only true for a simply connected smooth region. (The tip of the cone is the problem point.) Otherwise, the proof that zero curvature $\implies$ flat space does not work. My failure to realize this is why I was not more skeptical of my incorrect work. I hope this is interesting to others.

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    $\begingroup$ Morally the Gaussian curvature of a surface at a smooth point should be thought as the limit of the angular displacement of the vector as it parallel transports along a loop around that point (maybe divided by the area the loop encloses?), as the radius of the loop goes to zero. In this case the trouble is the vertex is not a smooth point. If one smooths it out, however, these two notions of curvature should give the same value. $\endgroup$ – Balarka Sen Feb 19 '18 at 9:34

I was preparing an answer to this question on the same line of the answer of user1379857. This one is perfect, so I've nothing to add, but since I builded a little model that illustrate how the vertex is a special point and that a vector parallel transported around this special point it comes back to the start position with a residual angle, I add here the images that can be useful.

enter image description here

enter image description here

The first image illustrate how a vector is moved on a plane witout an angle (the pac-man in the answer of user1379857 ), the second image illustrate how this is equivalent to path on a cone when the two sides of the angle are glued.

Note that this is true only if the closed path contain the vertex, so that the region is not smooth.


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