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][1] (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][2] (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][3] reduces the problem to the convex case.


  [1]: https://www.jstor.org/stable/2048830?seq=1
  [2]: https://link.springer.com/book/10.1007/978-1-4684-9159-3
  [3]: https://link.springer.com/content/pdf/10.1007/BF02100598.pdf