For the final question posed:
Is it at least true that the boundary surface of a convex subset of $\mathbb{R}^3$ has non-negative curvature everywhere that its curvature is well-defined?
The answer is yes. Convexity implies that for every point $p$ on the boundary $\partial \Omega$, we can find a plane $\Pi$ through $p$ such that $\partial\Omega$ lies "on the one side" of $\Pi$. This implies that the second fundamental form is signed (either positive-semi-definite or negative-semi-definite). For surfaces in $\mathbb{R}^3$, the Gaussian curvature is the determinant of the second fundamental form, and hence is the product of the eigenvalues (the principal curvatures), and hence is always positive definite.
Now, as a general comment, convexity is more closely related to the extrinsic curvature of the boundary $\partial\Omega$ rather than the induced intrinsic curvature of it. It just happens that that for embeddings into Euclidean spaces (and especially into $\mathbb{R}^3$), we have easy relations between intrinsic and extrinsic curvatures; this drives what you observe with Gaussian curvature.
In fact, we have the following theorem:
Thm Let $\Omega\subseteq \mathbb{R}^n$ be an open domain, such that $\partial\Omega$ is an orientable smooth codimension 1 Riemannian submanifold. Then $\Omega$ is convex if and only if there exists a choice of orientation on $\partial\Omega$ such that the associated second fundamental form is positive semi-definite.
Sketch of proof:
(1) Essentially by Taylor expansion we see that "the second fundamental form is positive semi-definite everywhere" is equivalent to "locally $\partial\Omega$ is supported by a hyperplane". (The $\Leftarrow$ direction is obvious by taking local parametrization of $\partial\Omega$ as a graph over the hyperplane. As is the $\Rightarrow$ direction if the second fundamental form is positive definite. For the positive semi-definite case, when we encounter a zero eigenvalue, we need to use the fact that the form is positive semi-definite in an entire neighborhood to rule out the case of "bad" higher order Taylor coefficients.)
(2) Once we have the result of the first step, we can appeal to Tietze's theorem which establishes the equivalence between local convexity (existence of local supporting hyperplanes) and global convexity. See the link for a bit more discussion.
In the case of $n = 3$, as mentioned above the positive-semi-definiteness of the second fundamental form is directly tied to the non-negativity of the Gaussian curvature. For higher dimensions, the scalar curvature only controls the second symmetric polynomial of the eigenvalues of the second fundamental form, so is insufficient to control the local convexity of the boundary.
Almost every sentence requires a correction. To begin with:
- "A sphere can be uniquely defined as a closed two dimensional manifold with constant positive Gaussian curvature which can be determined by measuring length and angles on the surface alone without taking reference to some ambient space."
No. The issue is that in math there are several objects, all called "a sphere." In the context of your question, you can say
"A round 2-dimensional sphere is a 2-dimensional Riemannian manifold which is closed, simply-connected and of constant Gaussian curvature."
Instead of calling it "round" you can also say "homogeneous." But some adjective is needed since in RG the word "sphere" is normally reserved for a (topological or smooth) 2-dimensional manifold, without any chosen Riemannian metric.
- The next sentence is almost right:
The total surface area, $A$, and the Gaussian curvature, $K$, of the sphere are related by the Gauss-Bonnet theorem as $$ A\cdot K~=4\pi \tag{1}. $$
The adjective "local" you are using is meaningless: By the very definition, Gaussian curvature is not just local, but an infinitesimal notion, depending on the 2nd derivative of the metric tensor. Saying "local curvature" is no more meaningful than saying "local derivative" or "local Taylor series."
- The next line is almost entirely meaningless:
"This [what does "this" refer to here?] is an equivalence relation [do you know what an equivalence relation is?] that allows to define static spherically symmetric spacetime as a fiber bundle with base $\mathbb{R}^{+}$ (surface area) and fiber $\mathbb{S}^2$ (two-sphere)."
There is no equivalence relation here at all. A fiber bundle over a line is always trivial, so you are really talking about the manifold $M=S^2\times {\mathbb R}^+$. You did not even attempt to specify a semi-Riemannian metric on this manifold, so you are not getting any spacetime here. In what sense is ${\mathbb R}^+$ "surface area"? (It is like saying "${\mathbb N}$ (age).") Maybe you are talking about a 1-parameter family of round spheres $(S^2, ag_1), a\in {\mathbb R}^+$, where $g_1$ is a Riemannian metric of curvature $+1$ on $S^2$ (unique up to isometry). The parameter $a$ equals the area of the Riemannian manifold $(S^2, ag_1)$. You can then equip $M$ with the semi-Riemannian metric
$$
h(x,t)=tg_1(x) - dt^2, x\in S^2, t\in {\mathbb R}^+.
$$
One can describe such a semi-Riemannian metric abstractly, without writing an explicit formula. For instance, you can say that $(M,h)$ is a semi-Riemannian manifold admitting a surjective semi-Riemannian submersion $f: (M,h)\to ({\mathbb R}^+, -dt^2)$, whose fibers $f^{-1}(t)$ (equipped with metrics induced from $(M,h)$) are isometric to round spheres of area $t$.
I stop here.
One reference that I like is the classical textbook on semi-Riemannian geometry,
B. O'Neill, Semi-Riemannian Geometry with Applications to Relativity, Academic
Press, New York, (1983).
Best Answer
Fist of all letme recast your original formulation: \begin{equation} |K(p)| = \lim_{U \to p}\frac{\text{Area}(n(U))}{\text{Area}(U)} \end{equation} I added an absolute value since the ratio of areas is non negative. Now let $X:V \rightarrow \mathbb{R}^{3}$ be a regular parametrization of a neighborhood of $p=X(u_0,v_0)$, where V is an open subset of $\mathbb{R}^2$ . Since X(V) is an open set, $U_ε \subset X(V)$ for all ε. Since we assumed that K(p) is non null, when considering the continuity of the Gaussian curvature function $K:V \rightarrow \mathbb{R}$, we can conclude that the Gaussian curvature does not change sign for ε chosen small enough. We can define the preimage $V_{\epsilon}=X^{-1}(U_{\epsilon})$. Since X is a regular parametrization, $(u_0,v_0)$ is the unique point in all $V_{\epsilon}$ for all positive $\epsilon$.Then by using the formula of the surface are, it is possible to write: \begin{equation} Area(U_{\epsilon})=\int \int_{V_{\epsilon}} ||X_u \times X_v|| du dv \end{equation} In a similar way on the unit sphere we have \begin{equation} Area(n(U_{\epsilon}))=\int \int_{V_{\epsilon}} ||N_u \times N_v|| du dv = \int \int_{V_{\epsilon}} |K(u,v)| ||X_u \times X_v|| du dv \end{equation} By using the Mean Theorem or double integrals we can write \begin{equation} \int \int_{V_{\epsilon}}||X_u \times X_v|| du dv = ||X_u(u_{\epsilon},v_{\epsilon}) \times X_v(u_{\epsilon},v_{\epsilon})|| \end{equation} And \begin{equation} \int \int_{V_{\epsilon}}||N_u \times N_v|| du dv = |K(u^{'}_{\epsilon},v^{'}_{\epsilon})|||X_u(u^{'}_{\epsilon},v^{'}_{\epsilon}) \times X_v(u^{'}_{\epsilon},v^{'}_{\epsilon})|| \end{equation} Thus, since when $\epsilon \rightarrow 0$ both $(u_{\epsilon},v_{\epsilon})$ and $(u^{'}_{\epsilon},v^{'}_{\epsilon})$ tend to $(u_0, v_0)$ we have: \begin{equation} \lim_{U \to p} \frac{\text{Area}(n(U))}{\text{Area}(U)} = \lim_{\epsilon \to 0} \frac{\text{Area}(n(U_{epsilon}))}{\text{Area}(U_{epsilon})}= \lim_{\epsilon \to 0} \frac{|K(u_{0},v_{0})|||X_u(u_{0},v_{0}) \times X_v(u_{0},v_{0})||}{||X_u(u_{0},v_{0}) \times X_v(u_{0},v_{0})||}=K(p) \end{equation}