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EDIT: Let $g$ be a smooth Riemannian metric on $S^2$ which is invariant under the antipodal involution $x\mapsto -x$. Assume it has Gauss curvature at least $-1$ and diameter $D$.

Does there exist a constant $c(D)>0$ (but independent of the metric) such that $\sup_x dist(x,-x)\geq c(D)$?

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  • $\begingroup$ @alesia : Thanks, corrected. It remained from an earlier version of the question I wanted to ask. $\endgroup$
    – asv
    Commented Aug 6, 2019 at 18:07
  • $\begingroup$ MKO: since you accepted Anton's answer I assume you understand it. Could you explain why the diameter of $\Delta$ is at least $D$? $\endgroup$
    – Igor Belegradek
    Commented Aug 7, 2019 at 18:25
  • $\begingroup$ @IgorBelegradek: I gave more details in my answer below. Actually, as pointed out below, the esimate of diameter of $\Delta$ is $D/2$. But may be after Anton corrected his answer my remarks are not necessary. $\endgroup$
    – asv
    Commented Aug 8, 2019 at 13:46

2 Answers 2

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The answer is "yes".

Let $g$ be the induced Riemannian metric on $\mathbb R\mathrm P^2=\mathbb S^2/\mathbb Z_2$. Denote by $\ell$ the systole of $(\mathbb R\mathrm P^2,g)$ and let $\gamma$ be the corresponding closed geodesic. Let us cut $(\mathbb R\mathrm P^2,g)$ along $\gamma$. We obtain a disc $\Delta$ with diameter is at least $D/2$ and its boundary has length $2\cdot\ell$.

Note that we can assume that $\ell<D/100$. Therefore there is a point $x$ in $\Delta$ that lies on distance at least $D/10$ from the boundary. It remains to note that the distance from $x$ to its antipodal point is more than $D/10$.

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    $\begingroup$ Why is the diameter of $\Delta$ at least $D$? $\endgroup$
    – Igor Belegradek
    Commented Aug 7, 2019 at 1:45
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    $\begingroup$ it should be at least $D/2$ for sure $\endgroup$
    – alesia
    Commented Aug 7, 2019 at 3:13
  • $\begingroup$ @IgorBelegradek fixed, thank you. $\endgroup$
    – Anton Petrunin
    Commented Aug 8, 2019 at 12:14
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This is not an answer. Following the request of Igor Belegradek I would like to elaborate on Anton Petrunin's answer and prove that $diam (\Delta) \geq D/2$.

First of all $diam (\mathbb{R}\mathbb{P}^2)\geq D/2$. This is a special case of the more general estimate proved in the final answer to this post: Diameter of m-fold cover

Let $f\colon S^2\to \mathbb{R}\mathbb{P}^2$ be the canonical map. Let $a\colon S^2\to S^2$ be the antipodal involution. Then $\tilde\gamma:=f^{-1}(\gamma)$ is a closed connected (!) geodesic on $S^2$, it is $a$-invariant. By the Jordan theorem its complement consists of two disks whose closures will be denoted by $\Delta $ and $\Delta'$. (Say, the first disk is the one Anton constructed, I guess.) Then $$a(\Delta)=\Delta'.(1)$$ Indeed otherwise $a(\Delta)=\Delta$. In that case the group $\mathbb{Z}_2$ would act freely on the disk $\Delta$ which is imposible since in that case the quotient $\Delta/\mathbb{Z}_2$ would be a compact manifold (possibly with boundary) of Euler characteristic 1/2 which is absurd.

Let $x,y\in \mathbb{R}\mathbb{P}^2$ be such that $dist(x,y)\geq D/2$. Let $\tilde x,\tilde y \in \Delta$ be their lifts (they can be chosen to belong to $\Delta$ due to (1)). Then one has $$diam(\Delta)\geq dist(\tilde x,\tilde y)\geq \min\{dist(\tilde x,\tilde y),dist (\tilde x,a(\tilde y))\}=dist(x,y)\geq D/2.$$ QED

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