Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with lorentzian-geometry
Search options not deleted
user 259525
Lorentzian geometry is the geometry of Minkowski spacetime, hence essentially of a Euclidean space, but equipped not with the standard Euclidean Riemannian metric of signature $(+,+,+,…,+)$ (which yields Euclidean geometry) but with the pseudo-Riemannian metric of signature $(−,+,+,…,+).$
5
votes
1
answer
245
views
Space of spacelike embeddings as infinite-dimensional manifold
Consider a four-dimensional Lorentzian manifold $(\mathcal{M},g)$ and a $3$-dimensional compact manifold $\Sigma$, such that there exists a spacelike embedding $i:\Sigma\to\mathcal{M}$ so that $h:=i^{ …
- 1,429
2
votes
2
answers
136
views
Is any globally-hyperbolic manifold conformally equivalent to one with complete slices?
Let $(M,g)$ be a globally-hyperbolic Lorentzian manifold. By the seminal work of Geroch and Bernal-Sánchez, we know that
$$M=\mathbb{R}\times\Sigma,\,\,\,\quad g=-\beta^{2}dt^{2}+h_{t}$$
where $\Sigma …
- 1,429