Skip to main content
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
Search options not deleted user 135778

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 $(−,+,+,…,+).$

2 votes
1 answer
309 views

$(M,g)$ is complete iff $(\tilde{M},\tilde{g})$ is complete (non-Riemannian version)

I'm not sure if this question is too low level for Math Overflow (so feel free to move this to SE if you think it is). Inspired by this and this question I'm wondering if the following statement is a …
user450093's user avatar