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 |
Complex, contact, Riemannian, pseudo-Riemannian and Finsler geometry, relativity, gauge theory, global analysis.
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 …
16
votes
Accepted
Hodge decomposition of smooth n-forms: is it an isomorphism of topological vector spaces?
Maybe an even more elementary argument than the one of Tobias:
The continuity of all involved operators is easy: simply all differential operators with smooth coefficients between sections of vector b …
4
votes
Accepted
Hermitian vector bundles and Hilbert $C^*$-modules
In addition to Nik Weaver's references, let me just sketch the proof which is in fact not very difficult:
A construction of Kaplansky (Rings of operators, Thm 26) shows that if $\mathcal{A}$ is a $*$ …
10
votes
How special are homogeneous spaces?
I suppose you want the action to be transitive as your title suggests. In this case, a classical theorem of Mostow (in 1950 for surfaces, in 2005 in general) says that for a compact homogeneous space …
8
votes
When (why) did we allow manifolds to be non-Hausdorff and/or non-second countable?
Non-Hausdorffness shows up in several contexts when dealing with Lie groupoids: the integration (Lie's 3rd Theorem) for Lie algebroids to Lie groupoids will typically produce a non-Hausdorff one, if i …
4
votes
Interpretation of the Schouten bracket as an integrability condition
One well-known example is the case of bivector fiels $\pi$. Then $[\pi, \pi] = 0$ is equivalent to say that $\{f, g\} = \pi(df, dg)$ is a Poisson bracket, i.e. satisfies the Jacobi identity. In this …
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 …
13
votes
Accepted
On the topology induced by a Lorentzian metric
In general, this topology is coarser than the original topology of the manifold, and, without further assumptions, strictly coarser. It coincides with the original one iff the Lorentz manifold is stro …
8
votes
2
answers
433
views
On the causal structure of spacetimes: piecewise $C^1$, $C^k$ or $C^\infty$?
This is a more technical question but it seems that there is some confusion in the literature on the choice of curves used to define the causal relations in time-oriented Lorentz manifolds: the infini …
9
votes
A vector field on the tangent bundle which is not equivalent to any second order ODE
I guess you want the topological equivalence to preserve the bundle structure of $TM \longrightarrow M$ otherwise it becomes a bit arbitrary, right?
In this case a non-zero vertical vector field will …
14
votes
Accepted
What do the differential k-forms on a product manifold look like?
Denote by $p_M: M \times N \longrightarrow M$ and $p_N: M \times N \longrightarrow N$ the canonical projections. Then you get an induced bilinear map from $\Omega^i(M) \times \Omega^j(N) \longrightarr …
2
votes
Accepted
Poisson structure on the dual Lie algebroid
Depending on you sign convention, this goes as follows. First you denote the bundle projection by $pr\colon E^* \longrightarrow X$. For a section $s \in \Gamma^\infty(E)$ you have a linear function $J …
9
votes
1
answer
2k
views
Automorphism group of a fiber bundle surjects onto diffeomorphism group?
This should surely be well-known by I have not been able to find a good reference to the following question: Given a smooth fiber bundle $\pi\colon P \longrightarrow M$ over a smooth manifold $M$ with …
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 …
6
votes
0
answers
201
views
The geometric shape of domains of flows
Consider a smooth (non-compact) manifold $M$ with a vector field $X$. Then we know that there is a open neighbourhood $U \subseteq M \times \mathbb{R}$ of $M \times \{0\}$ such that on $U$ the flow $\ …