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This is the famous theorem of Gromoll and Meyer:

Theorem (Gromoll-Meyer, 1974) There is an exotic 7-sphere with nonnegative sectional curvature and positive sectional curvature at a point.

I don't understand the second part of theorem "positive sectional curvature at a point". Isn't it always possible that one cane perturb the metric such that it has positive sectional curvature at a point and we have $\sec_{\min}(M)\leq \sec_q\leq \sec_p$ for all $q$ in a small neighborhoods of $p$?

David G. Stork
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C.F.G
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    I'm not sure that I quite understand the question. Perhaps it helps to mention that (1) there is a metric on the torus $S^1 \times S^1$ with nonnegative sectional curvature (the flat metric) but (2) Gauss-Bonnet implies that any metric on the torus with positive sectional curvature at a point has negative sectional curvature at some other point. – Travis Willse Nov 01 '19 at 07:19
  • (2) is very very exotic!!! Thanks. – C.F.G Nov 01 '19 at 07:23
  • Although it is hard to illustrate, but can you explain how you perturb the (shape) of torus to make positive curvature in some points and why some points has negative sectional curvature intuitively? – C.F.G Nov 01 '19 at 07:29
  • I think your claim about perturbations is off. If you perturb the metric to get a positive sectional curvature at some point you can't guarantee that the sectional curvature increases at all points. It does increase in a small enough neighboorhood of $p$ because it is continuous but not globally. – quarague Nov 01 '19 at 08:04
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    For what it's worth, I wouldn't call (2) 'exotic'---it's just an application of G-B. I'm not sure whether gets at the kind of intuition you're looking for, but think of a torus embedded in $\Bbb R^3$ in the usual way and equipped with the induced metric. At the points on the "outside" of the torus (so, far from its axis of symmetry) the curvature is positive. At the points on the inside it is negative. Reading through a proof of G-B might help with intuition, at least for the $2$d case. – Travis Willse Nov 01 '19 at 09:35
  • @quarague Indeed, and for compact $2$-manifolds G-B again implies that for any metric you cannot increase the sectional curvature at one point without reducing it at some other point. – Travis Willse Nov 01 '19 at 09:38
  • @C.F.G If my comments here adequately get at what you're looking for, let me know, and I'll write then up a little more formally as an answer. (But if not, feel free to elaborate.) – Travis Willse Nov 01 '19 at 09:39
  • @TravisWillse, Your comments are OK. but your second comment isn't about perturbation!! – C.F.G Nov 01 '19 at 17:34
  • I suppose that's a matter of perspective: The conservation of the total Gaussian curvature implies that a perturbation of the metric (on a compact surface) that increases the curvature at one point must decrease the curvature at another. – Travis Willse Nov 01 '19 at 18:00
  • @TravisWillse: all your comments fits to my question (and a bit more) please post your answer. – C.F.G Nov 02 '19 at 06:53
  • @C.F.G I'm glad you've found my comments (and others') useful; I've posted an answer. – Travis Willse Nov 02 '19 at 06:59

2 Answers2

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The short answer is that you cannot necessarily perturb a metric to get positive sectional curvature while still maintaining non-negative/positive sectional curvature everywhere else.

My favorite example of this is due to Wilking

 Manifolds with positive sectional curvature almost everywhere, Inventiones mathematicae 148(1):117-141, 2002

He constructs a smooth Riemannian metric $g$ on $M:=\mathbb{R}P^2\times \mathbb{R}P^3$ with the following properties.

1) It is non-negatively curved everywhere.

2) If $U\subseteq M$ is the subset of all points $p$ for which every $2$-plane $\sigma \subseteq T_p M$ is positively curved, then $U$ has full measure in $M$

In other words, in the measure-theoretic sense, $M$ is positively curved almost everywhere (and non-negatively curved everywhere).

Note, however, that $M$ is non-orientable. The classical Synge theorem says that in odd dimensions, a positively curved closed Riemannian manifold must be orientable. Thus, the metric $g$ cannot be deformed to being positively curved everywhere, despite the fact that it started off with positive curvature almost everywhere.

Edit Following Travis's suggestion, I'll move part of my comment below into the answer here.

Proposition. Suppose $(M,g)$ is a Riemannian manifold and $p\in M$. Let $p\in U\subseteq M$ be an open set diffeomorphic to a ball in $\mathbb{R}^n$ Then $g$ can be deformed to a metric $g_1$ for which all sectional curvatures are positive near $p$, but $g= g_1$ outside of $U$.

Proof: Because $U$ is diffeomorphic to a ball, there is a diffeomorphism $f:U\rightarrow D^n_+\subseteq S^n$, where $D^n_+$ denotes the open northern hemisphere. Let $g_0$ denote the canonical (positively curved) metric on $S^n$.

Choose an open set $V\subseteq U$ with $\overline{V}\subseteq U$ and let $\lambda:M\rightarrow \mathbb{R}$ be a bump function with $\lambda \equiv 1$ near $p$, and $\operatorname{supp} \lambda \subseteq \overline{V}$.

The family of metrics $(1-t)g + \lambda t f^\ast g_0$ has the required properties.

C.F.G
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Jason DeVito - on hiatus
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    This is a very neat example; I hadn't seen it before. – Travis Willse Nov 01 '19 at 21:38
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    That said, it's not entirely clear to me what OP wants, but it's not clear to me that this provides a counterexample. Without knowing more about the metric, it seems conceivable that around any given point where sectional curvature vanishes the metric can be deformed so that the resulting sectional curvature is positive everywhere in some neighborhood of that point---the argument in the answer as I understand it just shows that you cannot adjust the metric to make it have positive sectional curvature at all points at once. Perhaps I am overlooking something? – Travis Willse Nov 01 '19 at 21:38
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    @TravisWillse: I agree that it may or may not be what the OP wants. Also, any metric can be deformed to positive curvature in a neighborhood of a point - you just lose control of it else where. (Sketch: wlog work on an open ball of $\mathbb{R}^n$. Let $g$ be your starting metric with whatever curvature you want. Let $g_1$ be the pullback metric from the inclusion $i:U\rightarrow S^n$ as the open northern hemisphere. Pick some smaller neighborhood $V\subseteq U$ with $\overline{V}\subseteq U$, and let $\lambda$ be a bump function supported on $\overline{V}$. Then $(1-t)g + t\lambda g_1$ – Jason DeVito - on hiatus Nov 02 '19 at 00:43
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    is the required deformation.) – Jason DeVito - on hiatus Nov 02 '19 at 00:43
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    Yes, I agree, and this is what I tried to get at in my comments under the original question. I think the content of your first comment here might be close to what OP is looking for; perhaps it would be worth adding it to your (already good) answer? – Travis Willse Nov 02 '19 at 02:08
  • Thanks @JasonDeVito. What surprised me was this part: " we lose control of it else where"!! always your answers are easy to understand for me but Wilking example is hard to imagine. – C.F.G Nov 02 '19 at 06:51
  • Excellent proposition. Why I never saw this before in books or papers? Is there any? – C.F.G Nov 04 '19 at 04:18
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    C.F.G. To my knowledge, this isn't in any book or paper. As far as papers are concerend, I image that the experts, this counts as "well known". – Jason DeVito - on hiatus Nov 04 '19 at 14:58
  • @JasonDeVito: consider product of two 2-sphere. Then Isn't it true that the product metric is positively curved almost everywhere? – C.F.G Jul 11 '22 at 18:05
  • @C.F.G: No, a product metric never has positive curvature. The "mixed" planes are always zero curvature. In more detail, given a product of Riemannian manifolds $M\times N$, if $v\in T_p M$ and $w\in T_q N$, then $(w,v)\in T_{(p,q)}(M\times N)$ and $\operatorname{sec}(v,w) = 0$. To be clear, when one says "positive curvature" at a point, the usual meaning is "all $2$-planes $\sigma\subseteq T_p M$ have positive curvature", as opposed to "there is a $2$-plane with positive curvature" or even "almost all $2$-planes have positive curvature". – Jason DeVito - on hiatus Jul 11 '22 at 18:16
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Isn't it always possible that one can perturb the metric such that it has positive sectional curvature at a point and we have "$\sec_{\min}(M)\leq \sec_q\leq \sec_p$" for all $q$ in a small neighborhoods of $p$?

This statement isn't quite precise, since $\sec_a$ is not a scalar but rather a map $$Gr(2, T_a M) \to \Bbb R .$$ In the special case that $(M, g)$ is a surface (i.e., $\dim M = 2$), however, $Gr(2, T_a M)$ is the singleton set $\{T_a M\}$, so we may identify $\operatorname{sec}_a$ with a scalar and hence view $\operatorname{sec}$ as a function $M \to \Bbb R$, namely, the Gaussian curvature, $K$, and hence interpret the inequalities in the quotation in the usual way.

A standard partition-of-unity argument shows that for any smooth surface $(M, g)$ and point $p \in M$ one can deform $g$ to $g'$ in some neighborhood of $p$ so that for all $q$ in some (a priori possibly smaller) neighborhood we have $\inf K(g) \leq K'_q \leq K'_p$, where $K'$ is the Gaussian curvature of $g'$. (See Jason DeVito's good answer for details of this argument.)

This result, however, is manifestly local, and it does not guarantee the global conclusion that $K'_a \geq \inf K(M)$ for all $a \in M$.

Example Consider the flat torus $(S^1 \times S^1, g)$, for which $K \equiv 0$; in particular $g$ has nonnegative sectional curvature. We can perturb $g$ to produce a metric $g'$ with positive Gaussian curvature $K'_p$ at a point $p$. But the Gauss-Bonnet Theorem gives that $$\int_{S^1 \times S^1} K' \,dA = 2 \pi \chi(S^1 \times S^1) = 0.$$ By continuity $K'$ is positive in some neighborhood of $p$, which thus gives a positive contribution to the integral. To satisfy the above equality, $K'$ must also be negative on some open subset of $M$.

In short, the existence of a metric on a given manifold with nonnegative sectional curvature does not imply the existence of a metric on the same manifold with nonnegative sectional curvature at a point and positive sectional curvature at some point.

Travis Willse
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  • I followed your advice and added my comment to my answer. Sorry it took so long - this weekend has been quite busy. – Jason DeVito - on hiatus Nov 03 '19 at 14:00
  • Cheers! I've updated my answer to refer to yours (rather than your comment) for the details of the partition-of-unity argument. – Travis Willse Nov 03 '19 at 19:22
  • @TravisWillse: your argument and example are so good, but I can't ignore Jason DeVito last edit (his proposition)!! what I should do in this situation for accepting answer? – C.F.G Nov 04 '19 at 03:25
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    The general rule is: Just pick whichever answer helped you more, and don't worry about the politics of which answer to accept. Everyone knows that you can only pick one, so there are generally no hurt feelings---especially among people who have been around and active as long as Jason and I have. If you really can't decide in this particular case, I encourage you to accept Jason's answer: Not only does his answer give a useful example, it also provides details of the partition-of-unity argument that establishes the positive answer to the local version of your question. – Travis Willse Nov 04 '19 at 03:51
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    @C.F.G Just to chime in, I do not care one way or another about who gets the accepted answer. My answer only contains the proposition you like at Travis's suggestion, so Travis should get all the credit ;). – Jason DeVito - on hiatus Nov 04 '19 at 15:00
  • Thanks @JasonDeVito. I just don't want anyone to be upset!! – C.F.G Nov 04 '19 at 18:25