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 riemannian-geometry
Search options not deleted
user 12482
Riemannian Geometry is a subfield of Differential Geometry, which specifically studies "Riemannian Manifolds", manifolds with "Riemannian Metrics", which means that they are equipped with continuous inner products.
4
votes
tangent and cotangent bundle
I guess what you want to prove is the following: for a given vector bundle $E$ over $M$ there is a canonical subbundle, the vertical bundle $\mathrm{Ver}(E) = \ker T\pi \subseteq TE$ of the tangent bu …
- 8,098
3
votes
Accepted
Horizontal lift of differential operator
This is a sort of standard construction you can find in several places. I don't know where this was done first though...
OK: first you can extend your horizontal lift from vector fields to all (symm …
- 8,098
3
votes
A Lie group whose Lie algebra is equal to (the Lie algebra? of )all functions with fibrewise...
The following will only deal with the Lie algebra, the question about the Lie group is far beyond my capabilities.
The symplectic structure is (I guess) the one coming from the musical isomorphism of …
- 8,098
2
votes
The imaginary exponential of a tangent field on a manifold
Let's try this. I didn't check all the estimates but the idea should be roughly as follows: since $M$ is compact, the flow $\Phi$ of $X$ is complete giving a one-parameter group action of $\mathbb{R}$ …
- 8,098
5
votes
Formal adjoint of the covariant derivative
Being a bit late for the party, here is nevertheless a small answer. In fact, there is a rather explicit way to compute adjoints of every differential operator (any order) between (smooth, compactly s …
- 8,098