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I'm trying to find the “round” metric for an $S^{2}\times S^{1}$ space. One can think of this as the 3-sphere punctured by a 3d type catenoid. I was thinking I could start with the standard (round) metric of a three-sphere of radius R which is:

$$ds^{2}=R^{2}\left(d\psi^{2}+sin^{2}\left(\psi\right)\left[d\theta^{2}+sin^{2}(\theta)d\phi^{2}\right]\right)$$

And modify it by making the smallest possible two-sphere embedded around the origin to have radius a (which corresponds to minimal radius of the “throat” of the catenoid piercing the three-sphere.

$$ds^{2}=\left(R^{2}d\psi^{2}+\left\{ R^{2}sin^{2}\left(\psi\right)+a^{2}\right\} \left[d\theta^{2}+sin^{2}(\theta)d\phi^{2}\right]\right)$$

And making sure that two points $P(\psi)=P(\psi+2\pi)$ are the same point. This gives us:

$$ds^{2}=\left(R^{2}d\psi^{2}+\left\{ R^{2}sin^{2}\left(\psi/2\right)+a^{2}\right\} \left[d\theta^{2}+sin^{2}(\theta)d\phi^{2}\right]\right)$$ Something seems off here however. I'm sure there's a simple way to do this, I'm just not seeing it

Arctic Char
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R. Rankin
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1 Answers1

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There isn't a round metric on $\mathbb S^2 \times \mathbb S^1$. More generally, there is no metric on $\mathbb S^2 \times \mathbb S^1$ with positive Ricci curvature: If $g$ is such a metric on $\mathbb S^2 \times \mathbb S^1$, then the pullback metric $\pi^* g$ on the universal cover of $\mathbb S^2 \times \mathbb S^1$ is complete and has strictly positive Ricci curvature. By Bonnet Myers theorem, it has bounded diameter. But this is impossible since $\pi_1 (\mathbb S^2 \times \mathbb S^1)$ is infinite.

Arctic Char
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  • Thank you. Though by round metric I simply meant a modified version of that on the three-sphere (hence the quotes). Since all oriented three-manifolds are parallelizable there should be a rather simple expression for the metric and line element in terms of everywhere orthonormal basis (with the same parameters R and a I introduced). Perhaps a better way to view this is as a 0 surgery on the three sphere (an integer Dehn surgery) – R. Rankin Nov 21 '21 at 19:45
  • Such a metric should have positive Ricci curvature everywhere away from the "throat" which pierces the three sphere. – R. Rankin Nov 21 '21 at 20:53
  • Can you make this precise? @R.Rankin As of now I don't know what you are asking. – Arctic Char Nov 22 '21 at 12:33
  • I'm trying to find the metric in terms of three angular parameters as on the three-sphere and two scale parameters (One corresponding to the radius of the largest allowed 2-sphere on the manifold and one corresponding to the smallest ) – R. Rankin Nov 22 '21 at 19:15