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 dg.differential-geometry
Search options not deleted
user 39404
Complex, contact, Riemannian, pseudo-Riemannian and Finsler geometry, relativity, gauge theory, global analysis.
1
vote
Why do we use the less simple convention for the definition of a vector bundle connection?
The "usual" definition tantamounts to saying that $[\nabla, \omega ]=d\omega$, where $[,]$ is the (graded) commutator of (graded) endomorphisms.
- 153
4
votes
2
answers
541
views
Existence of connections making a bundle endomorphism parallel
Let $M$ be a manifold, $TM$ its tangent bundle, and $N:TM\to TM$ a vector bundle morphism. It is possible to find a torsionless linear connection $\nabla$ on $TM$ such that $\nabla N=0$?
- 153
0
votes
0
answers
69
views
Cohomology operators inducing local basis of $1-$forms
Suppose that $\partial$ is a non-trivial ($\partial \neq 0$) cohomology operator on an $m-$dimensional manifold $M$ (that is: $\partial:\Omega(M)\to\Omega(M)$ is a degree $1$ derivation such that $\pa …
- 153