Is there any example of a manifold with a positive isotropic curvature but it possibly obtains a negative Ricci curvature at some point and the direction? If we see the definition of the positive isotropic curvature condition, it doesn't imply the positivity of the Ricci curvature (it is true if $(M\times \mathbb{R},g+dr^2)$ has a positive isotropic curvature, but it is stronger condition than the original one). But I've not heard about the example of this case.

Here are more details. The reference I used is [Simon Brendle - Ricci flow and the sphere theorem] but you can also find it from https://arxiv.org/abs/0705.0766.

  • $R$ is said to have a positive isotropic curvature if it satisfies $R_{ikki}+R_{illi}+R_{jkkj}+R_{jllj}-2R_{ijkl}>0$ for all orthonormal 4-frame $\{e_i,e_j,e_k,e_l\}$. If we take a sum for all $i, j, k, l$, we can see that it implies the positivity of scalar curvature on $M$.

  • Also, we can observe that $(M\times \mathbb{R},g+dr^2)$ has a positive isotropic curvature if and only if the curvature $R$ of $g$ satisfies $R_{ikki}+\lambda^2R_{illi}+R_{jkkj}+\lambda^2R_{jllj}-2\lambda R_{ijkl}>0$ for all orthonormal 4-frame $\{e_i,e_j,e_k,e_l\}$ on $M$ and for all $\lambda \in [0,1]$. After taking $\lambda=0$, we can see it implies the positivity of Ricci curvature on $M$.

  • As it is mentioned in Richard Hamilton’s paper https://www.intlpress.com/site/pub/files/_fulltext/journals/cag/1997/0005/0001/CAG-1997-0005-0001-a001.pdf, $S^4, S^3\times \mathbb{R}$ have a positive isotropic curvature and $S^2\times S^2, S^2\times \mathbb{R}^2, \mathbb{C}P^2$ are the example of manifolds with nonnegative isotropic curvature.

  • There is a topological restriction of having a positive isotropic curvature metric. For example, Micallef and Moore proved that a compact, simply connected manifold with positive isotropic curvature is homeomorphic to $S^n$.

  • In 2007, Simon Brendle and Richard Schoen proved the differentiable sphere theorem using this curvature condition and the Ricci flow(as proved by Simon Brendle, these conditions are preserved under the Ricci flow).

  • $\begingroup$ Although I once knew what positive isotropic curvature is, I no longer do. Perhaps you could include a definition and maybe a reference, too? $\endgroup$
    – Deane Yang
    Mar 13, 2019 at 3:27
  • 1
    $\begingroup$ @DeaneYang Sorry about that. I just added the definition and the reference. $\endgroup$
    – Jae Ho Cho
    Mar 13, 2019 at 4:01
  • $\begingroup$ Many thanks! I remember this from when Micallef and Moore did their work. $\endgroup$
    – Deane Yang
    Mar 13, 2019 at 4:26

1 Answer 1


If you have a look into https://arxiv.org/pdf/1711.05167.pdf , page 4, it's written there:

"Conversely, it follows from work of Micallef and Wang [29] that every manifold which is diffeomorphic to a connected sum of quotients of $S^n$ and $S^{n−1} × \mathbb R$ admits a metric with positive isotropic curvature."

Such manifolds can have fundamental group that is not virtually abeliean (i.e. $\pi_1$ doesn't have an abelian subgroup of finite index). For example $\mathbb Z_2*\mathbb Z_3$ is the fundamental group of $\mathbb RP^{2n+1}\#(S^{2n+1}/\mathbb Z_3)$ and it is not virtually abelian. At the same time any compact manifold that admits a metric of non negative Ricci curvature has virtually abelian $\pi_1$ (Cheeger-Gromoll splitting https://projecteuclid.org/euclid.jdg/1214430220). So any metric on $\mathbb RP^{2n+1}\#(S^{2n+1}/\mathbb Z_3)$ has negative Ricci curvature at some point in some direction.


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