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Let $M$ be a compact $3$-dimensional manifold diffeomorphic to a ball. Suppose that $M$ has nonpositive (sectional) curvature and its boundary $\partial M$ is convex, or even that $M$ is a Riemannian ball, i.e., $\partial M$ is equidistant from a point in $M$. Is it possible to make the curvature negative everywhere by a $C^2$ perturbation of the metric?

Note: This does not seem possible in dimension $2$. Consider for instance the Monkey saddle, which has an isolated point of zero curvature. This point happens to be a branch point of the Gauss map. Hence it cannot be removed by a perturbation of the surface. But I am not sure if all perturbations of the metric here correspond to a perturbation of the surface in $R^3$, or even admit an isometric embedding in $R^3$.

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  • $\begingroup$ What do you mean by convexity if $M$ in general? Do you mean convex boundary? Also, do you mean that $M$ is connected with nonempty boundary? $\endgroup$
    – Moishe Kohan
    Commented May 20, 2022 at 22:51
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    $\begingroup$ @Kohan: yes, convex boundary. Yes, connected with nonempty boundary. $\endgroup$
    – Mohammad Ghomi
    Commented May 20, 2022 at 23:33
  • $\begingroup$ You’re allowing the metric be perturbed at the boundary too? $\endgroup$
    – Deane Yang
    Commented May 21, 2022 at 4:11
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    $\begingroup$ @Deane Yang: yes, the metric may be perturbed at the boundary. $\endgroup$
    – Mohammad Ghomi
    Commented May 21, 2022 at 4:15
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    $\begingroup$ Regarding your last comment, it is not known whether any perturbation of the saddle surface metric can be isometrically embedded into $\mathbb{R}^3$. It’s a hyperbolic Monge-Ampère PDE, so it’s possible that it is. $\endgroup$
    – Deane Yang
    Commented May 22, 2022 at 0:58

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Choose a strongly convex function $f\colon M\to \mathbb{R}$; you can take $f=\mathrm{dist}_p^2$. Consider the product $M\times (-\mathbb R)$ (this is a space-time manifold). Observe that, for small $\varepsilon>0$, the graph of $\varepsilon\cdot f$ with the induced metric from $M\times (-\mathbb R)$ has strictly negative curvature.

Convexity of boundary should hold for small $\varepsilon$, assuming that the boundary was strongly convex; the latter holds for balls.

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  • $\begingroup$ This should work in dimension 2 also? (So Ghomi’s comment on the monkey saddle doesn’t seem correct) $\endgroup$
    – Ian Agol
    Commented Nov 17 at 18:25
  • $\begingroup$ @IanAgol, yes, meaning no --- the monkey saddle cannot be perterbed as a surface in the space, but can be perterbed as an abstract surface. $\endgroup$
    – Anton Petrunin
    Commented Nov 19 at 3:19

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