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This is a follow-up of my previous post.

Question1: Is it always possible that to perturb the metric to a metric of positive Ricci curvature at some points (or neighborhood) and without change in other points?

Maybe the answer is obviously no. because we can do this process over and over to obtain a metric of positive Ricci curvature which seems to be impossible. Am I right?

Question2: Is there a closed manifold $M$ of positive Ricci curvature so that $M\text{#} M$ (or arbitrary number connected sum) admit no metric of positive Ricci curvature?

C.F.G
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2 Answers2

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For your first question, you can't do this for all points simultaneously otherwise you would end up with a metric with positive Ricci curvature, but these do not always exist. For example, it follows from Myers' Theorem that if a compact manifold $M$ admits a complete metric of positive Ricci curvature, then $\pi_1(M)$ is finite. However, the proposition in Jason DeVito's answer to your linked question shows you can always perturb the metric in a neighbourhood of a point $p$ so that it has positive sectional curvature (and hence positive Ricci curvature) at $p$.

As for your second question, let $M$ be a manifold with positive Ricci curvature which is not simply connected. If $n \geq 3$, then $\pi_1(M\# M) \cong \pi_1(M)\ast\pi_1(M)$ which is infinite, so by Myers' Theorem, the manifold $M\# M$ does not admit a metric of positive Ricci curvature, however it does admit a metric of positive scalar curvature by a theorem of Gromov and Lawson.

C.F.G
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Michael Albanese
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Michael's answer is fine, but I wanted to add some results in the opposite direction.

First, the following is a result of Aubin (see here)

Suppose $M$ is a complete Riemannian manifold for which $Ric\geq 0$ and for which there is some point for which $Ric_p > 0$. Then $M$ admits a metric of positive Ricci curvaure at each point.

So, as Michael points out, in general you cannot deform a metric to one with positive Ricci at some point without also messing things up elsewhere. However, if $M$ has non-negative Ricci curvature which is positive at some point, then you can. (By the way, the analogous statement for sectional curvature is FALSE in general, but is open for simply connected closed manifolds.)

Second, there are closed manifolds with positive Ricci curvature for which $M\sharp M\sharp...$ admits a metric of positive Ricci curvature for any number of connect summands (including infinitely many). One can, for example, take $M = S^2\times S^2$. This is a result of Sha and Yang: see here.

Jason DeVito - on hiatus
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    Do you know if one can prove Aubin's result using Ricci flow? – Michael Albanese Mar 17 '20 at 18:10
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    The condition $Ric \ge 0$ is not preserved by Ricci Flow in general, @MichaelAlbanese, some counter example here. But RF is used to solve a very similar question here – Arctic Char Mar 17 '20 at 22:24
  • @JasonDeVito: Due to your previous answer, and this one, one can say that "there is no any open set diffeomorphic to a ball in $\Bbb R^n$ on general manifolds of $Ric\geq 0$". Right? Very strange remark for me! – C.F.G Mar 18 '20 at 07:55
  • @C.F.G. I don't understand your comment. A manifold by definition is locally Euclidean, so has plenty of open sets diffeomorphic to balls. – Jason DeVito - on hiatus Mar 18 '20 at 13:04
  • @JasonDeVito: So, one can always deform a metric of nonnegative Ricci to positive at some points and nonnegative in other points due to your deformation proposition? If so, one can use Aubin's deformation to produce $Ric>0$ for all manifolds of $Ric\geq 0$. – C.F.G Mar 18 '20 at 13:08
  • @C.F.G: Not necessarily. For Aubin's result, you really do need to start with at least one point where $Ric > 0$. For example, the torus $T^2$ has a metric with $Ric \geq 0$, but no metric with $Ric > 0$. – Jason DeVito - on hiatus Mar 18 '20 at 17:58
  • @JasonDeVito: I am completely confused. Lets check step by step. you claim that every manifolds has open sets diffeomorphic to balls so one can equip this open ball by a metric with $Ric>0$ without losing control of Ricci in other points. By this argument now we have a manifold with $Ric\geq 0$ and $Ric>0$ inside of a ball. And now Aubin's deformation say that this manifold admit a positive Ricci metric. Which part is wrong? – C.F.G Mar 18 '20 at 20:12
  • @C.F.G.: The first part. Putting $Ric>0$ on some ball can definitely have an effect on the rest of the manifold. If, after this effect, you still have $Ric\geq 0$ everywhere, then you can deform it to $Ric > 0$ everywhere. But that "If" may or may not occur. – Jason DeVito - on hiatus Mar 18 '20 at 22:05
  • @JasonDeVito: Due to your proposition this small change on a small ball have no effect on the rest of the manifold. (I am so sorry for my comments. I just want to learn.) – C.F.G Mar 19 '20 at 05:08
  • @C.F.G.: No need to apologize for your comments! In my proposition, what "near $p$" means is that there is an even smaller open set $V$ with $p\in V\subseteq U$ for which you can adjust the metric to be whatever you want on $V$. On $V$, sectional curvatures become positive. On the complement of $U$, sectional curvatures stay whatever they originally were. But on $U\setminus V$, you lose control of the metric. – Jason DeVito - on hiatus Mar 19 '20 at 12:36
  • @JasonDeVito: Don't mention it. maybe my explanations was so bad. Thanks for useful discussion. – C.F.G Mar 19 '20 at 14:23
  • @MichaelAlbanese: In Paul Ehrlich's thesis (1976), Aubin theorem (1970) has been re-proven using deforming metric and first order derivation something like Ricci flow. – C.F.G Jul 10 '22 at 04:24