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](https://www.jstor.org/stable/2048830?seq=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](https://link.springer.com/book/10.1007/978-1-4684-9159-3) (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).



Convex surfaces with minimal total curvature in Cartan-Hadamard 3-space
Note that curvature in sectional directions tangent to the surface, say $\Sigma$ vanish.
It seems sufficient to conclude that the usual Peterson–Codazzi equations hold for the surface.
It follows that there is a convex surface $\Sigma'$ in the Euclidian space that is isometric to $\Sigma$; plus they have identical second fundamental forms.


Now we want to use bow lemma (diffgeometric analog of arm lemma).
The bow lemma has a version for Hadamard spaces.
Applying it to plane arcs in $\Sigma'$, we get that the map $\Sigma'\to\Sigma$ is distance-noncontracting.


Applying Reshetnyak majorization theorem, to plane sections of $\Sigma'$, we get that the map $\Sigma'\to\Sigma$ is distance-preserving.


Finally, applying Kirszbraun, we get that the map $\Sigma'\to\Sigma$ extends to a distance-preserving map from the convex hull of $\Sigma'$.


