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The volume of a Riemannian metric on the projective plane is $2\pi$ and length of every non-contractible loop is greater than $\pi - \epsilon$ for some small, positive number $\epsilon$. Is this metric close to the canonical metric?

The question is somewhat vague on purpose. I'm mostly interested in the best constant for a bilipschitz equivalence in terms of $\epsilon$, but I also wonder whether for some sufficiently small $\epsilon$ one can conclude that the curvature is close to 1.

Stability of inequalities is a well-trodden research topic in convex geometry and I was wondering what was known about this in systolic geometry.

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    $\begingroup$ An interesting "converse" would be to ask whether there is a lower bound for the "isosystolic defect" namely the difference between the area and $\tfrac{2}{\pi}\text{sys}^2$, in the assumption that the area is normalized to $2\pi$ and the curvature in a region of area $A$ is bounded away from $1$ by $\epsilon>0$. For a related question see mathoverflow.net/questions/127599/… $\endgroup$
    – Mikhail Katz
    Commented Jul 3, 2013 at 15:36

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There is no Lipschitz or even Gromov-Hausdorff stability - just consider a round metric with long hairy tails of small area.

One can hope for stability with respect to intrinsic flat distance in the sense of Sormani-Wenger or some similar metric. This distance is basically Federer's flat distance between isometric images is $L^\infty$ (just like the Gromov-Hausdorff distance is the Hausdorff distance in $L^\infty$). The stability in this sense probably amounts to uniqueness of the equality case in the class of integral current spaces arising as limits of projective planes.

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  • $\begingroup$ Thanks Sergei. I was wondering about "sea urchins" myself and didn't know how to rule them out. Is that what the Sormani-Wenger metric can do? What about something terribly naive like a lipschitz stability for a nearly-full measure open subset? $\endgroup$
    – alvarezpaiva
    Commented Jun 18, 2013 at 12:06
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    $\begingroup$ Uniformization (along the lines of Pu's proof) gives you a conformal factor $L^2$-close to 1. This implies Lipschitz closeness on nearly full measure. $\endgroup$
    – Sergei Ivanov
    Commented Jun 18, 2013 at 16:58

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