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Riemannian Geometry is a subfield of Differential Geometry, which specifically studies "Riemannian Manifolds", manifolds with "Riemannian Metrics", which means that they are equipped with continuous inner products.
3
votes
Reference request: $\operatorname{Sym}^2_0(T^*M) \simeq \Lambda_- \otimes \Lambda_+$
There is also a quick abstract proof via representation theory: $S^2_0(\mathbb{R}^4)$ is a 9-dimensional representation of $\mathrm{SO}(4)/\{\pm I_4\}\simeq \mathrm{SO}(3)\times\mathrm{SO}(3)$ and, h …
15
votes
Accepted
Example of non homogenous manifold with a finitely generated algebra of natural functions
No. Here's a counterexample: Let $f(r)$ satisfy the equation $f'' + 2 f^3 = 0$ with the initial conditions $f(0)=1$ and $f'(0)=0$. Then $f$ satisfies $(f')^2+f^4=1$ and is periodic with period
$$
L …
5
votes
Accepted
Asymptotic parametrization for negatively curved surfaces
As asked, the answer to the question is 'no'. The simply-connected cover $f:\mathbb{R}^2\to S$ of Sherck's first surface $S$ (which is defined in $\mathbb{R}^3$ by the equation $\mathrm{e}^{z} \cos x …
3
votes
Accepted
How badly does the geodesic exponential map fail to be $C^2$ on Finsler manifolds
Note: I am making the standard assumptions in Finsler geometry, i.e., that $F:TM\to\mathbb{R}$ is smooth away from the zero section and $F$ is 'strictly convex'.
Although $\partial_i\partial_j\exp^k_ …
4
votes
Smooth isometric immersions of the a hemisphere in $\mathbb R^3$
Actually, it's not enough to specify the isometric immersion along a curve. You have to specify more data than that. For example, you can reflect the unit sphere across its equator, and that will gi …
17
votes
When are these base spaces isomorphic?
Here's a perhaps more serious example: There is a compact $4$-manifold that fibers over both $\mathbb{RP}^3$ and $(S^1\times S^2)/\mathbb{Z}_2$ where, in each case, the fibers are circles. (The $\ma …
13
votes
Isometry group of a left-invariant Riemannian metric on $\mathrm{SU}(2)$
The $\sigma_i$ as defined above satisfy $\mathrm{d}\sigma_i = -2\,\sigma_j\wedge\sigma_k$ when $(i,j,k)$ is an even permutation of $(1,2,3)$. For later use, let $E_i$ be the dual (left-invariant) fram …
10
votes
Accepted
Understanding exterior differential systems
Here's an expansion of my comment that the natural formulation of this problem as an EDS is on the coframe bundle $\pi: P\to M$, which, I hope, will be helpful. Also, because it will match my usual n …
6
votes
Accepted
Systems of (hyperbolic) 2nd order PDEs with lower order constraints
Yes, there is a standard procedure to analyze such systems, essentially, it is Cartan's method of prolongation combined with his theory of involutive systems. There are other approaches as well, but …
5
votes
Frobenius theorem and the size of integral manifold
Your equations are equivalent to the $1$-form equations
$$
\mathrm{d}f = X_0(f,g)\,\mathrm{d}s + Y_0(f,g)\,\mathrm{d}t
\quad \text{and}\quad
\mathrm{d}g = X_1(f,g)\,\mathrm{d}s + Y_1(f,g)\,\mathrm{ …
14
votes
Accepted
How to learn intrinsic torsion
General $G$-structures are not usually treated in textbooks, other than the basic definitions, so it's not surprising that there would be few if any textbooks that treat the intrinsic torsion of $G$-s …
5
votes
Is there an absolute geometry that underlies spherical, Euclidean and hyperbolic geometry?
I'm not sure exactly what you mean by 'absolute geometry' that unifies spherical, Euclidean, and hyperbolic geometry, but in the sense of Klein's identification of geometries with group actions, so th …
1
vote
Accepted
Exterior differential systems on compact three-manifolds and Cartan-Kähler theory
In their comments to my first answer, the OP has clarified that they did not mean to regard the metric $h$ as a given, but, rather, an output of the problem of prescribing coframings by specifying the …
3
votes
Exterior differential systems on compact three-manifolds and Cartan-Kähler theory
Let me phrase the problem as I understand the given data and then describe how the 'theory of exterior differential systems' would be applied.
One starts with a compact Riemannian $3$-manifold $(M,h)$ …
2
votes
Accepted
Parametrization of $k$-ruled submanifold: can we choose the base to be orthogonal to the rul...
The answer is already 'no' in the first nontrivial case: A 3-manifold in $\mathbb{R}^d$ (where $d>3$) that is ruled by lines (i.e., $k=1$).
One can see this as follows: As the OP notes, one can writ …