18
votes
Accepted
Geometric interpretation of the Weyl tensor?
There is such an interpretation, with a few caveats. Essentially, there is a canonical connection on a certain vector bundle for which the "principal part" of the curvature is the Weyl ...
14
votes
Accepted
Is the set of Lorentzian metrics metrizable?
First of all, there is a bunch of basic things that you need to write in a slightly clearer way. If you try to topologize the set of Lorentzian metrics as you did, you need first:
Restrict to the ...
13
votes
Accepted
On the topology induced by a Lorentzian metric
In general, this topology is coarser than the original topology of the manifold, and, without further assumptions, strictly coarser. It coincides with the original one iff the Lorentz manifold is ...
12
votes
Accepted
Question on Lorentzian geometry
The signature convention $(−,+,\cdots,+)$ is more commonly used in General Relativity and Lorentzian geometry because of the desire among their practicioners to make a closer parallel to Riemannian ...
11
votes
Accepted
Fourier transform on Minkowski space
Well you seem to have worked it out but I wrote most of this before your comment happened: I claim there isn't any material difference between your "Minkowski space Fourier transform" and the usual ...
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 ...
Community wiki
10
votes
Accepted
Is the Gödel universe Wick rotatable?
$\DeclareMathOperator\SL{SL}$Clearly, as Robert Bryant indicates, it is Wick-rotatable to a different Lorentzian space. However, it is also Wick-rotatable to a Riemannian space, albeit negative ...
9
votes
Accepted
On the causal structure of spacetimes: piecewise $C^1$, $C^k$ or $C^\infty$?
The answer is that it doesn't matter, as long as the metric itself is sufficiently regular. In the following notes by Chrusciel, the basic assumption is that the metric is $C^2$:
[C] Chruściel, Piotr ...
9
votes
Accepted
Literature Request: Berger Spheres and their Construction
For a different viewpoint, the Berger Spheres or their Lorentzian analogues are well understood using the canonical variation of the metric associated with a Riemannian submersion with totally ...
9
votes
Is the Gödel universe Wick rotatable?
I may be misreading the sources that you list for the definition of Wick-rotatable, but, I believe that the following construction does fit that definition: According to the Wikipedia page that the ...
8
votes
Accepted
What's the "actual" shape of a black hole accretion disk?
The "shape" of an accretion disc is the spatial profile of the gas density $\rho$ at a given time $t$. Here is a recent calculation, arXiv:1810.0083. This image shows what a distant observer, outside ...
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.
8
votes
Topology on Minkowski space $\mathbb{R}^{4}$ and Lorentz invariant measure
Judging by your notation, I reckon you are getting the background for your questions from the Appendix to Section IX.8 of the book by M. Reed and B. Simon, Methods of Modern Mathematical Physics II: ...
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 ...
Community wiki
7
votes
Accepted
Space of spacelike embeddings as infinite-dimensional manifold
A standard reference on infinite dimensional manifolds is
Kriegl, Andreas; Michor, Peter W., The convenient setting of global analysis, Mathematical Surveys and Monographs. 53. Providence, RI: ...
7
votes
Accepted
Is every strongly causal spacetime purely electric?
These are quite orthogonal conditions. To start, one is a global condition, while the other is a local one.
Every point has a small enough neighborhood that is strongly causal (even globally ...
7
votes
Accepted
Topology and local isometry, spinning cosmic string
I think in your question, as currently formulated, the whole rotating cosmic string is a red herring.
If I interpret your notation correctly, $a$ and $\kappa$ are constants. And hence locally you can ...
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 ...
6
votes
Do (3+1)-dimensional Lorentzian manifolds admit unique smoothings?
Concerning your first three question: depending on what you exactly mean by "refinement", any $C^k$-manifold $M$ with $k\geq 1$ possesses a unique $C^\infty$-structure that is $C^k$-compatible with ...
5
votes
On the causal structure of spacetimes: piecewise $C^1$, $C^k$ or $C^\infty$?
Let me expand on Igor Khavkine's answer, especially:
"The answer is that it doesn't matter, as long as the metric itself is sufficiently regular."
In our recent paper: The future is not always ...
5
votes
Accepted
Topology on Minkowski space $\mathbb{R}^{4}$ and Lorentz invariant measure
Q1 The topology on $\mathbb{R}^4$ is the usual one. This is the general case for Lorentzian geometry: the topology is the one defined by the charts in your atlas.
Q2 Given a fixed Lorentz ...
5
votes
Synthetic differential / conformal geometry of Lorentzian manifolds?
I'm far from an expert on this area and hopefully someone more knowledgeable can answer, but it seems to me that it is impossible to give an answer unless you are more precise about what you would ...
5
votes
Accepted
Non-diffeomorphic but homeomorphic (under Lorentzian topology) Lorentzian manifolds
I would like to argue that the situation considered in the comments is "close to generic".
Let $(M,g)$ be a Lorentzian manifold that is not strongly causal; this implies that $(M,g)$ is also ...
5
votes
Can the Causal Structure recover the manifold topology for non-chronological spacetimes?
Regarding the title question: No, the topology cannot be recovered from the chronology relation for non-chronological spacetimes. For example, there are plenty of distinct Lorentzian manifolds for ...
5
votes
Lorentzian norm of the covariant derivative of a vector field is zero
Very little can be said about the vector field $V$. Below I am going to use index notation. In particular, nothing near to covariantly constant can be gained.
Complete freedom in length
If you know $V$...
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 ...
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 ...
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 ...
4
votes
Accepted
Canonical Metrics on 3- and 4-Manifolds
There is no canonical Lorentzian metric in $[g]$, because that would be a diffeomorphism invariant Lorentzian metric. The diffeomorphism group of any manifold has infinite dimension, and infinite ...
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