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Suppose that a closed geodesic $\gamma$ is the fixed-point set of an isometric involution on $(\mathbb{S}^3,g)$. Assume that sectional curvature of $g$ is at least $1$. Is it true that $$\mathrm{length}\,\gamma\leqslant2\cdot\pi\ ?$$

Comments.

  • The same question can be asked about polyhedral metrics on $\mathbb{S}^3$ with curvature at least 1 in the sense of Alexandrov.

  • If instead of an isometric involution we have an $\mathbb{S}^1$-action, then the answer is yes. It follows since the quotient space $(\mathbb{S}^3,g)/\mathbb{S}^1$ is a convex disc with curvature at least 1 in the sense of Alexandrov, and $\gamma$ projects isometrically to its boundary.

  • If we go one dimesion up, that is consider involution of $(\mathbb{S}^4,g)$ with fixed 2-sphere, then the fixed sphere has sectional curvature at least 1; therefore, its area cannot exceed the area of $\mathbb{S}^2$.

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    $\begingroup$ Why cannot we suspend the involution? Its fixed point set is then the suspension on the given circle. Doesn't an area bound on the 2d suspension gives a bound on the length of the equator? $\endgroup$
    – Igor Belegradek
    Commented May 13 at 18:26
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    $\begingroup$ This is naive, but can you take the spherical warped product and apply the 4D case? $\endgroup$
    – Ian Agol
    Commented May 13 at 18:28
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    $\begingroup$ @IgorBelegradek and IanAgol: the spherical suspension, say $\Sigma^4$ in an Alexandrov space with curvature $\ge 1$ and it is homeomorphic to $\mathbb{S}^4$. It is expected that the fixed point set of codimension 2 in $\Sigma^4$ is an Alexandrov space with curvature $\ge 1$, but it is unknown. If the codimension $\ge 3$ then there are counterexamples. (By the way codimesion 1 case is equivalent to the conjecture that boundary of Alexandrov space is an Alexandrov space.) $\endgroup$
    – Anton Petrunin
    Commented May 13 at 18:36

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