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18 votes
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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 ...
Jeffrey Case's user avatar
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14 votes
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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 ...
Pedro Lauridsen Ribeiro's user avatar
13 votes
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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 ...
Stefan Waldmann's user avatar
12 votes
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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 ...
Pedro Lauridsen Ribeiro's user avatar
11 votes
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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 ...
AlexArvanitakis's user avatar
10 votes
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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
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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 ...
Sigbjørn Hervik's user avatar
9 votes
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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 ...
Igor Khavkine's user avatar
9 votes
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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 ...
Fabrice Baudoin's user avatar
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 ...
Robert Bryant's user avatar
8 votes
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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 ...
Carlo Beenakker's user avatar
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.
Ian Agol's user avatar
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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: ...
Pedro Lauridsen Ribeiro's user avatar
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 ...
7 votes
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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: ...
Igor Khavkine's user avatar
7 votes
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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 ...
Igor Khavkine's user avatar
7 votes
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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 ...
Willie Wong's user avatar
6 votes
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$(M,g)$ 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 ...
Ben McKay's user avatar
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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 ...
Clemens Sämann's user avatar
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 ...
Clemens Sämann's user avatar
5 votes
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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 ...
Willie Wong's user avatar
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 ...
RBega2's user avatar
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5 votes
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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 ...
Willie Wong's user avatar
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 ...
Tim Campion's user avatar
  • 63.9k
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 $V$...
Willie Wong's user avatar
4 votes
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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 ...
Igor Khavkine's user avatar
4 votes
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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 ...
Willie Wong's user avatar
4 votes
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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 ...
Uri Bader's user avatar
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4 votes
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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 ...
Ben McKay's user avatar
  • 26.3k

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