No constant curvature metric on $S^2 \times S^1$

Question

I was reading the introduction to Hamilton's paper "Three-manifolds with Positive Ricci Curvature." He states that $S^2 \times S^1$ admits no metric of constant sectional curvature, and therefore represents an obstruction to improving the hypothesis of the main theorem from positive Ricci curvature to nonnegative Ricci curvature. My question is: Can someone provide a reference to the result that $S^2 \times S^1$ doesn't have a constant curvature metric or to a general result which implies it?

Answer 1

Another way to show this is the following: If $M^n$ has constant sectional curvature $\kappa$, then the universal covering of $M^n$ must be (i) $S^n$ if $\kappa>0$, (ii) $\mathbb{R}^n$ if $\kappa=0$, or (iii) $\mathbb{H}^n$ if $\kappa<0$. See Theorem 4.1 of Chapter 8 of Riemannian Geometry by Do Carmo.

But as you have already noticed, the universal covering of $S^2\times S^1$ is $S^2\times \mathbb{R}$. So it cannot have constant sectional curvature.

Answer 2

I just realized how to prove this (silly me), though I am still curious if anyone has a different solution.

It follows from the Cartan-Hadamard Theorem that if the sectional curvature is negative or $0$, the universal cover is $\mathbb{R}^n$. However, the universal cover is clearly $S^2\times \mathbb{R}$, so this can't be the case. On the other hand, it follows from the Bonnet-Myers theorem that with constant positive curvature the fundamental group must be finite, which is not so in our case where $\pi_1=\mathbb{Z}$. This proves what I wanted.