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 connections
Search options answers only
not deleted
user 40297
Ehresmann connections; covariant derivatives; connections on vector bundles, principal bundles, ∞-bundles, submersions, bundle gerbes; holonomy and higher holonomy; parallel transport; torsion; curvature. See also the tags [principal-bundles], [vector-bundles], [gerbes], [curvature], [geodesics], [characteristic-classes], [torsion].
29
votes
Manifolds admitting flat connections
For question 1 : if a vector bundle admits a flat connection, it comes from a representation of the fundamental group. Thus if $TM$ admits a flat connection and $M$ is simply-connected, $TM$ is trivi …
- 38k
10
votes
Accepted
Question about a proof in Berthelot's crystalline book
Suppose you have a commutative ring $R$, a square zero ideal $I\subset R$, a $R$-module $M$ and an endomorphism $u$ of $M$ which is the identity modulo $IM$.
Then $v:= 1_M-u$ maps $M$ to $IM$, hence $ …
- 38k