13
votes
Is it possible to do calculus and differential geometry the old school way, without any ortho frames or axis?
I do think you're asking a reasonable question, but many do not like your way of asking it. It would be better received if you could express it more rigorously and mathematically and showed that you ...
- 27.5k
10
votes
Accepted
Properties that only Lorentzian manifolds have
A pseudo-Riemannian manifold of signature $(p,q)$ with $p, q\geq 2$ certainly does not admit Cauchy hypersurfaces, since spacelike submanifolds have dimension at most $p$. However, you could define a ...
10
votes
Accepted
Definition of the conformal metric
Let $(M,[g])$ be a conformal manifold;
i.e. $(M,g)$ is a Riemannian manifold and $[g] = \{ u^2g \mathrel{}:\mathrel{} u \in C^\infty(M), u>0 \}$ is the set of Riemannian metrics conformal to $g$.
...
- 1,713
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 ...
- 108k
8
votes
Lorentzian analogue to Thurston geometries
Both questions are answered here:
Dumitrescu, Sorin; Zeghib, Abdelghani, Three-dimensional Lorentz geometries: Classification and completeness, Geom. Dedicata 149, 243-273 (2010). ZBL1216.53025.
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8
votes
Accepted
Diagonalization of symmetric matrices of functions
In general, this cannot be done. For example, in dimension $2$ in coordinates $(x,y)$, let
$$
G(x,y) = \left[\begin{matrix}x&y\\y&-x\end{matrix}\right].
$$
If $G$ could be diagonalized by a ...
- 108k
7
votes
Properties that only Lorentzian manifolds have
The reverse Cauchy-Schwarz inequality: for any two causal vectors (orientation in this `squared' formulation is not relevant) $v, w$ we have
$$
\vert g(v,w)\vert^2\ge g(v,v) g(w,w)
$$
is false in ...
6
votes
Accepted
Conformal equivalence degenerate metric tensors
In order to define an actual Weyl or Cotton tensor, one has to have a non-degenerate conformal structure. In the OP's case, we aren't given such a structure, so the only way to get an actual Weyl or ...
- 108k
6
votes
Accepted
$(M,g)$ is complete iff $(\tilde{M},\tilde{g})$ is complete (non-Riemannian version)
Completeness of a pseudo-Riemannian manifold (or a manifold with affine connection, more generally) means precisely the completeness of its geodesic flow on its tangent bundle. But the tangent bundle ...
- 26.3k
5
votes
Is it possible to do calculus and differential geometry the old school way, without any ortho frames or axis?
It's possible to do differential geometry in a purely intrinsic way, at least once you've gotten past the initial hurdle of defining what a manifold is. The standard definition of a manifold is a ...
- 6,001
4
votes
Accepted
Tensor Field Decomposition in Space time
The class of Lorentzian manifolds that is best suited for this kind of question is that of globally hyperbolic ones. Analyzing the space of solutions of a hyperbolic PDE that does not satisfy an ...
- 21.5k
4
votes
Accepted
Induced connection on null hypersurfaces
The simple answer is:
You can't project to a null hypersurface
This is tied to the fact that the "Lorentzian normal" vector field is in fact a tangent vector field.
In more details:
Start with ...
- 39k
4
votes
Accepted
Possible mistake in classification of marginally trapped submanifolds of $\Bbb R^{n+2}_{p+1}$
The proof of the forward implication is perhaps more complicated than necessary, but it is true.
The proof of the reverse implication is wrong. In fact here's a counterexample. Consider the space $\...
- 39k
4
votes
Accepted
Closed Semi-Riemannian manifolds with non-compact isometry group
Of course there is D’Ambra's 1988 paper "Isometry groups of Lorentz manifolds", from which the theorem you state is taken.
A later paper, taking a more general perspective, is this one by D’Ambra and ...
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4
votes
Accepted
A question on Levi-Civita connection and a fixed hyper surface
Not a full answer, but there definitely should be some integrability constraint.
Take the simplest case where $M = \mathbb{R}^3$ (the boundary is unimportant for the discussion here) and $\Sigma$ is ...
- 39k
4
votes
Accepted
A question on null geodesics in Lorentzian geometry
We argue by contradiction. Suppose there are infinitely intersections. By reversing time orientation if necessary, there exists a monotonically increasing sequence of times $s_n \in I$ such that at ...
- 39k
4
votes
Accepted
Smoothness of conformal transformations
Answers to your question are somewhat dimension and signature dependent.
For example, in dimension $2$, if $g$ is definite, then every $C^1$ conformal diffeomorphism is, in fact, real-analytic, ...
- 108k
4
votes
Usage/Application of Raychaudhuri equation in Riemann geometry or pure maths
For any 1-form $\omega$ on a pseudo-Riemannian manifold, the product rule and commutation identity say that
$$\omega^p\nabla_p\nabla_q\omega^q-\nabla^q(\omega^p\nabla_p\omega_q)=-\omega^p\omega^qR_{pq}...
- 2,247
4
votes
Accepted
Is any globally-hyperbolic manifold conformally equivalent to one with complete slices?
This is the original Nomizu-Ozeki article:
Nomizu, Katsumi; Ozeki, Hideki, The existence of complete Riemannian metrics, Proc. Am. Math. Soc. 12, 889-891 (1961). ZBL0102.16401.
Applying their proof ...
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3
votes
Accepted
Metric with a constant Chern–Pontryagin scalar
I found a solution:
\begin{equation}
g_{\mu\nu} = \left(
\begin{array}{cccc}
\frac{A}{(1-t z)^2 (x y-1)^2} & 0 & 0 & 0 \\
0 & -\frac{A}{(1-t z)^2 (x y-1)^2} & 0 & 0 \\
0 &...
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3
votes
Usage/Application of Raychaudhuri equation in Riemann geometry or pure maths
The Bochner identity is a special case of the Raychaudhuri equation, obtained when the vector field is of gradient type.
Let $v$ be a vector field and let $a^\alpha=v^\alpha{}_{;\beta} v^\beta$ be its ...
- 1,371
3
votes
Why are they called "screen" distributions?
The word "screen" refers to lightlike dimensional reductions, a worldline in $(d+1)+1$ dimensional space-time is projected onto the screen $x^{d+1}=0$ and the projected $d+1$ dimensional curve is ...
- 188k
3
votes
Accepted
Why are they called "screen" distributions?
Naturally, one should consider the quotient space $V/{\rm rad}(V)$ which consists of ${\rm rad}(V)$-rays (affine spaces parallel to ${\rm rad}(V)$). A screen space $SV$ intersects a ray in exactly one ...
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3
votes
Is any globally-hyperbolic manifold conformally equivalent to one with complete slices?
The answer by Igor is complete and very clear. Although I'd like to give a much shorter one: Yes!
Miguel Sánchez actually answers this himself in [1, Section 3.2]
[1] Sánchez https://arxiv.org/abs/...
- 61
2
votes
Limit of curvature near lightlike points
I think your conjecture is correct (assuming that "regular" means "smooth"), with one possible caveat that I'll mention below.
Since curvature is independent of parametrization, without loss of ...
- 818
2
votes
How do we calculate the gradient of this function defined using the Riemannian logarithm on a Riemannian manifold?
In order to differentiate something containing the logarithm, it's best if you can differentiate the logarithm itself.
And because you work in a neighborhood where the exponential map is a ...
- 8,086
2
votes
Accepted
Tensor component calculation
Start by writing
$$ \tag{1}\label{1} B_{nn} = \nabla^\mu \nabla^\nu C_{\mu n n \nu} - \frac{1}{2}C_{\mu n n \nu}R^{\mu\nu} . $$
Here I abuse notation to write $\nabla^\mu\nabla^\nu C_{\mu n n \nu} = n^...
- 1,713
1
vote
Accepted
Proof of equivalence between Lie triple systems and totally geodesic submanifolds
Try pages 71 and 72 of Cartan's "La géométrie des groupes de transformations" https://eudml.org/doc/235668
- 13.5k
1
vote
Accepted
A question on light cones in Lorentzian manifolds with timelike boundary
Edit: Right now I will not give a complete argument in what follows (the previous version of the answer missed the part of the question asking for $S$ to be the boundary of a spacelike hypersurface in ...
- 7,817
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