$\mathbb{RP}^n\times \mathbb{S}^3$ cannot have everywhere positive sectional curvature

curvaturedifferential-geometryriemannian-geometry

I'm triying to solve the following exercise:

Show that $\mathbb{RP}^n\times \mathbb{S}^3$ does not admit a metric with everywhere positive sectional curvature.

I know how to prove that $\mathbb{RP}^n\times \mathbb{S}^3$ cannot have a metric $g$ with $K\leq 0$ at every point.

If there exists such $g$, taking the universal cover $\pi:\mathbb{S}^n\times \mathbb{S}^3\to \mathbb{RP}^n\times \mathbb{S}^3$ and considering the pullback metric $\widetilde{g}:=\pi^*(g)$ we have that $(\mathbb{S}^n\times \mathbb{S}^3,\widetilde{g})$ is complete, simply connected with $\widetilde{K}\leq 0$. By Hadamard's theorem, $\mathbb{S}^n\times \mathbb{S}^3$ is diffeomorphic to $\mathbb{R}^{n+3}$ (absurd).

I also know that any product metric on $\mathbb{RP}^n\times \mathbb{S}^3$ is such that for all $p$ we have $K(\sigma)=0$ for some $2$-dimensional tangent subspace at $p$, so the product metric is excluded.

That's all I could think of.

Best Answer

This follows immediately from Synge's Theorem.

Synge's Theorem: Let $M$ be a compact Riemannian manifold with positive sectional curvature. Then:

  • if $M$ is even-dimensional and orientable, then $M$ is simply connected.
  • if $M$ is odd-dimensional, then it is orientable.

When $n$ is odd, $M = \mathbb{RP}^n\times S^3$ is even-dimensional, so by Synge's Theorem, it cannot admit a metric of positive sectional curvature as $M$ is not simply connected.

When $n$ is even, $M$ is non-orientable, so by Synge's Theorem, it cannot admit a metric of positive sectional curvature.

It is worth noting that $\mathbb{RP}^n\times S^3$ does admit a metric of non-negative sectional curvature, for example, the product of the constant sectional curvature metrics.

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