A common intro to intrinsic curvature is to show that a standard cone has no curvature outside the vertex because it can be unrolled into a subset of the plane. Also, in polar coordinates, the cone has metric

$g_{rr} = c \qquad g_{\theta\theta} = r^2$

with the constant $c$ corresponding to the slant of the cone. But when I tried calculating the Riemann tensor components for a spherically symmetric 3-space of the form

$g_{rr} = a(r) \qquad g_{\theta\theta} = r^2 \qquad g_{\phi\phi} = r^2\sin^2\theta$

two of them came out looking funny:

${R^{\phi}}_{\theta\phi\theta} = 1 - \frac{1}{a} \qquad\qquad {R^{\theta}}_{\phi\theta\phi} = (1 - \frac{1}{a})\sin^2\theta$

They do vanish when $a(r) = 1$ which is reassuring since that's just Euclidean space. But they don't for any other constant. However, a constant other than 1 should just be the 3D version of a cone. Now it is also reassuring that the radial curvature remains zero, since the $r$ coordinate lines are the geodesics and these clearly are not deviating. But this tangential behavior is shaking my confidence in my calculation. It even gives a non-zero Ricci scalar:

$R = \frac{2}{ar}(\frac{a'}{a} + \frac{a-1}{r})$

Did I mess up or is that really how it is? If so, is there an intuitive way to understand why? And is it in any way related to the fact that all 2-spheres have the same total curvature but 3-spheres do not?

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    $\begingroup$ your calculations seem correct and a 3-cone is indeed curved. If it looks strange, maybe consider the limit case of a 3-cylinder; it is curved since a 2-sphere is curved (unlike a 1-sphere). The fact that we can locally unroll a 2-cone to the plane is just 2-dimensional $\endgroup$
    – user8268
    Sep 9, 2021 at 22:07
  • $\begingroup$ @user8268 Aha, that makes perfect sense. Thanks a lot. $\endgroup$ Sep 9, 2021 at 22:30


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