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Let $(M^3,g)$ be a compact Riemannian $3$-manifold with boundary. In a paper by L. Ambrozio, he considers the set $\mathcal{F}_M$ of all immersed disks in $M$ whose boundaries are curves in $\partial M$ that are homotopically non-trivial. Two quantities are defined:

$$\mathcal{A}(M,g) = \inf_{D \in \mathcal{F}_M} \operatorname{Area}(D) \quad \text{ and } \quad \mathcal{L}(M,g) = \inf_{D \in \mathcal{F}_M} \operatorname{Length}(\partial D).$$

He then proves the following theorem:

If $\mathcal{F}_M \neq \emptyset$ and $\partial M$ is mean-convex (positive mean curvature), then $$\frac{1}{2} \mathcal{A}(M,g) \inf_M R_M + \mathcal{L}(M,g) \inf_{\partial M} H^{\partial M} \leq 2 \pi,$$ where $R_M$ is the scalar curvature of $M$ and $H^{\partial M}$ is the mean curvature of $\partial M$. Moreover, if equality holds, then the universal cover of $(M,g)$ is isometric to a cylinder $(\Sigma_0 \times \mathbb{R}, g_0 + dt^2)$, where $(\Sigma_0, g_0)$ is a disk with constant Gaussian curvature $\inf_M R_M/ 2$ and $\partial \Sigma_0$ has constant geodesic curvature $\inf_{\partial M} H^{\partial M}$ in $\Sigma_0$.

My question is whether the converse of the last statement in this theorem holds. Namely: if the universal cover of $(M, g)$ is isometric to a cylinder $(\Sigma_0 \times \mathbb{R}, g_0 + dt^2)$, where $(\Sigma_0, g_0)$ is a disk with constant Gaussian curvature $\inf_M R_M/ 2$ and $\partial \Sigma_0$ has constant geodesic curvature $\inf_{\partial M} H^{\partial M}$ in $\Sigma_0$, does equality hold?

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    $\begingroup$ Have a look at the Gauss-Bonnet Theorem. en.wikipedia.org/wiki/Gauss%E2%80%93Bonnet_theorem $\endgroup$
    – Ian Agol
    Sep 22, 2020 at 4:13
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    $\begingroup$ @IanAgol is $\mathcal{A}(M,g)$ realized by a slice? $\endgroup$ Sep 22, 2020 at 4:23
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    $\begingroup$ Yes. The equality of the formula is equivalent to Gauss-Bonnet for a slice. $\endgroup$
    – Ian Agol
    Sep 22, 2020 at 4:54
  • $\begingroup$ It is intuitive that $\mathcal{A}$ is realized by a slice, but how to prove it? $\endgroup$ Sep 22, 2020 at 15:43
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    $\begingroup$ Apply the gauss-bonnet $\endgroup$
    – Ian Agol
    Sep 22, 2020 at 15:52

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