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A Cartan-Hadamard 3-space $M$ is a complete simply connected 3-dimensional Riemannian manifold with nonpositive sectional curvature. A (smooth) convex surface $\Gamma\subset M$ is an embedded topological sphere with nonnegative second fundamental form $\mathrm{I\!I}$. The total (Gauss-Kronecker) curvature of $\Gamma$ is defined as $$ \mathcal{G}(\Gamma):=\int_\Gamma\det(\mathrm{I\!I}). $$ It follows quickly from Gauss' equation and Gauss-Bonnet theorem that $\mathcal{G}(\Gamma)\geq 4\pi$. Suppose that $\mathcal{G}(\Gamma)= 4\pi$. Does it follow that the compact region of $M$ bounded by $\Gamma$ is Euclidean, i.e., all its sectional curvatures are zero?

Note 1: Schroeder and Strake showed in this paper (see Theorem 2) that the answer is yes, provided that $\Gamma$ is strictly convex, i.e., the second fundamental form is positive definite. Strict convexity appears to be an essential feature of the proof.

Note 2: On page 66 of Lectures on Manifolds on Nonpositive Curvature (see Exercise (b)), Gromov poses a more general question for total absolute curvature of closed surfaces in Cartan-Hadamard $3$-spaces (the term "absolute" is not explicitly mentioned). The convex hull trick in a paper of Kleiner reduces the problem to the convex case.

A Cartan-Hadamard 3-space $M$ is a complete simply connected 3-dimensional Riemannian manifold with nonpositive sectional curvature. A (smooth) convex surface $\Gamma\subset M$ is an embedded topological sphere with nonnegative second fundamental form $\mathrm{I\!I}$. The total (Gauss-Kronecker) curvature of $\Gamma$ is defined as $$ \mathcal{G}(\Gamma):=\int_\Gamma\det(\mathrm{I\!I}). $$ It follows quickly from Gauss' equation and Gauss-Bonnet theorem that $\mathcal{G}(\Gamma)\geq 4\pi$. Suppose that $\mathcal{G}(\Gamma)= 4\pi$. Does it follow that the compact region of $M$ bounded by $\Gamma$ is Euclidean, i.e., all its sectional curvatures are zero?

Note 1: Schroeder and Strake showed in this paper (see Theorem 2) that the answer is yes, provided that $\Gamma$ is strictly convex, i.e., the second fundamental form is positive definite. Strict convexity appears to be an essential feature of the proof.

Note 2: On page 66 of Lectures on Manifolds on Nonpositive Curvature (see Exercise (b)), Gromov poses a more general question for total absolute curvature of closed surfaces in Cartan-Hadamard $3$-spaces (the term "absolute" is not explicitly mentioned).

A Cartan-Hadamard 3-space $M$ is a complete simply connected 3-dimensional Riemannian manifold with nonpositive sectional curvature. A (smooth) convex surface $\Gamma\subset M$ is an embedded topological sphere with nonnegative second fundamental form $\mathrm{I\!I}$. The total (Gauss-Kronecker) curvature of $\Gamma$ is defined as $$ \mathcal{G}(\Gamma):=\int_\Gamma\det(\mathrm{I\!I}). $$ It follows quickly from Gauss' equation and Gauss-Bonnet theorem that $\mathcal{G}(\Gamma)\geq 4\pi$. Suppose that $\mathcal{G}(\Gamma)= 4\pi$. Does it follow that the compact region of $M$ bounded by $\Gamma$ is Euclidean, i.e., all its sectional curvatures are zero?

Note 1: Schroeder and Strake showed in this paper (see Theorem 2) that the answer is yes, provided that $\Gamma$ is strictly convex, i.e., the second fundamental form is positive definite. Strict convexity appears to be an essential feature of the proof.

Note 2: On page 66 of Lectures on Manifolds on Nonpositive Curvature (see Exercise (b)), Gromov poses a more general question for total absolute curvature of closed surfaces in Cartan-Hadamard $3$-spaces (the term "absolute" is not explicitly mentioned).

A Cartan-Hadamard 3-space $M$ is a complete simply connected 3-dimensional Riemannian manifold with nonpositive sectional curvature. A (smooth) convex surface $\Gamma\subset M$ is an embedded topological sphere with nonnegative second fundamental form $\mathrm{I\!I}$. The total (Gauss-Kronecker) curvature of $\Gamma$ is defined as $$ \mathcal{G}(\Gamma):=\int_\Gamma\det(\mathrm{I\!I}). $$ It follows quickly from Gauss' equation and Gauss-Bonnet theorem that $\mathcal{G}(\Gamma)\geq 4\pi$. Suppose that $\mathcal{G}(\Gamma)= 4\pi$. Does it follow that the compact region of $M$ bounded by $\Gamma$ is Euclidean, i.e., all its sectional curvatures are zero?

Note 1: Schroeder and Strake showed in this paper (see Theorem 2) that the answer is yes, provided that $\Gamma$ is strictly convex, i.e., the second fundamental form is positive definite. Strict convexity appears to be an essential feature of the proof.

Note 2: On page 66 of Lectures on Manifolds on Nonpositive Curvature (see Exercise (b)), Gromov poses a more general question for total absolute curvature of closed surfaces in Cartan-Hadamard $3$-spaces (the term "absolute" is not explicitly mentioned). The convex hull trick in a paper of Kleiner reduces the problem to the convex case.

A Cartan-Hadamard 3-space $M$ is a complete simply connected 3-dimensional Riemannian manifold with nonpositive sectional curvature. A (smooth) convex surface $\Gamma\subset M$ is an embedded topological sphere with nonnegative second fundamental form $\mathrm{I\!I}$. The total (Gauss-Kronecker) curvature of $\Gamma$ is defined as $$ \mathcal{G}(\Gamma):=\int_\Gamma\det(\mathrm{I\!I}). $$ It follows quickly from Gauss' equation and Gauss-Bonnet theorem that $\mathcal{G}(\Gamma)\geq 4\pi$. Suppose that $\mathcal{G}(\Gamma)= 4\pi$. Does it follow that the compact region of $M$ bounded by $\Gamma$ is Euclidean, i.e., all its sectional curvatures are zero?

Schroeder and Strake showed in this paper (see Theorem 2) that the answer is yes, provided that $\Gamma$ is strictly convex, i.e., the second fundamental form is positive definite. Strict convexity appears to be an essential feature of the proof.

A Cartan-Hadamard 3-space $M$ is a complete simply connected 3-dimensional Riemannian manifold with nonpositive sectional curvature. A (smooth) convex surface $\Gamma\subset M$ is an embedded topological sphere with nonnegative second fundamental form $\mathrm{I\!I}$. The total (Gauss-Kronecker) curvature of $\Gamma$ is defined as $$ \mathcal{G}(\Gamma):=\int_\Gamma\det(\mathrm{I\!I}). $$ It follows quickly from Gauss' equation and Gauss-Bonnet theorem that $\mathcal{G}(\Gamma)\geq 4\pi$. Suppose that $\mathcal{G}(\Gamma)= 4\pi$. Does it follow that the compact region of $M$ bounded by $\Gamma$ is Euclidean, i.e., all its sectional curvatures are zero?

Note 1: Schroeder and Strake showed in this paper (see Theorem 2) that the answer is yes, provided that $\Gamma$ is strictly convex, i.e., the second fundamental form is positive definite. Strict convexity appears to be an essential feature of the proof.

Note 2: On page 66 of Lectures on Manifolds on Nonpositive Curvature (see Exercise (b)), Gromov poses a more general question for total absolute curvature of closed surfaces in Cartan-Hadamard $3$-spaces (the term "absolute" is not explicitly mentioned).