I am working on a physical paper about solutions of Einstein field equations in case of static spacetime with perfect fluid spheres, and wanted to invent a new definition of spherical symmetry there.
My question is about mathematical correctness of the below definition.
"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. The total surface area, $A$, and the local Gaussian curvature, $K$, of the sphere are related by Gauss-Bonnet theorem as $$ A\cdot K~=4\pi \tag{1}. $$ This is an equivalence relation 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). That fiber bundle consists of all spheres of different areas and curvatures that fulfill the condition (1).
Due to this equation the sphere with zero surface area has an infinite Gaussian curvature. However, it does not inevitable imply singularity because the expression $0\cdot \infty$ is well defined being equal $4\pi$. While the sphere is not contractible to a point, the sphere with vanishing surface area remains a sphere despite of its infinite curvature. It is just a fiber assigned to the minimal sphere. It marks the central location similar to the point in a spherically symmetric foliation of $\mathbb{R}^3$ but opposite to it, where the sphere degenerates to a non-dimensional point, it retains its topology. The principal difference between foliation and fibration is that in foliation the equivalence classes, the leaves, can have a different topology, whereas in a fiber bundle the corresponding equivalence classes, the fibers, always have the same topology.
Between different fibers there is no pointwise identification, whereas each sphere has assigned area $A$ via canonical projection. The relation between different fibers is given by the metric metric. The infinitesimal length element in such metric can be written in the form $$ ds^2=e^{2\nu}c^2dt^2-e^{2\lambda}dr^2-r^2d\Omega^2 \tag{2},$$ with curvature radius, $r$, and the infinitesimal surface element, $d\Omega$, on the curvature unit sphere. The metric functions $e^{\nu}$ and $e^{\lambda}$ describe the local sphere (fiber) time and length scales respectively. Einstein's field equations describe they relation to each other."
Any critical comment or remark regarding possible mathematical incorrectness of that idea would be welcome.
Best Answer
Almost every sentence requires a correction. To begin with:
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 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."
"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).