bio | website | maths.swan.ac.uk/staff/ejb |
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location | Swansea, Wales | |
age | ||
visits | member for | 2 years, 5 months |
seen | May 22 at 12:39 | |
stats | profile views | 154 |
May 8 |
answered | Example of a covariant derivative on a non-projective bundle |
May 8 |
revised |
Example of a covariant derivative on a non-projective bundle
for clarity |
May 8 |
comment |
Example of a covariant derivative on a non-projective bundle
Thanks - there is no chance for a locally trivial bundle to do this. But is there any sheaf-like structure (expressed as a module with connection over the algebra of functions) for which it can be done? |
May 8 |
asked | Example of a covariant derivative on a non-projective bundle |
May 8 |
awarded | Benefactor |
May 6 |
accepted | Identifying a Hopf algebra cohomology theory |
May 2 |
comment |
Identifying a Hopf algebra cohomology theory
OK, I have added more explanation - apologies for that. There is a way to reformulate the complex with no explicit invariants and no algebra structure. |
May 2 |
revised |
Identifying a Hopf algebra cohomology theory
better explanation of problem |
May 2 |
comment |
Identifying a Hopf algebra cohomology theory
The required action for the van Est spectral sequence is the full tensor product coaction (this is needed to get a non-trivial dependence on the coaction). I see now that my question was ambiguous on this point. If I remember correctly, the sequence without taking the invariants is acyclic - will check. |
May 1 |
comment |
Identifying a Hopf algebra cohomology theory
The product appears implicitly in the coaction on the tensor product. I am checking the reference - thanks! |
May 1 |
awarded | Promoter |
Apr 18 |
awarded | Yearling |
Apr 18 |
revised |
Identifying a Hopf algebra cohomology theory
spelling! |
Apr 17 |
asked | Identifying a Hopf algebra cohomology theory |
Mar 5 |
comment |
How rigid can a rigid object be in GR?
That is a good way of thinking about it - and may well give a method for finding the answer. |
Mar 4 |
comment |
How rigid can a rigid object be in GR?
The problem is, what if we cannot start in flat space? What is the best that we can do with the coordinate system? The black hole is just an example of a localised curvature - the details are not important. |
Mar 4 |
comment |
How rigid can a rigid object be in GR?
There is a coordinate system which is formed by taking a 3D space like submanifold, and declaring it to be time zero, and then taking the geodesic motion in 4D perpendicular to the submanifold, with time coordinate proper time. This gives a nicely behaved coordinate system, at least locally. If we can start in flat space, and then move into more complicated geometry, we can set an initial metric on the 3D slice to be the usual Euclidian metric. |
Mar 4 |
comment |
How rigid can a rigid object be in GR?
The question is a bit ambiguous, but I was hoping that there was an existing body of literature, and I did not want to make it too restrictive. |
Mar 3 |
asked | How rigid can a rigid object be in GR? |
Nov 3 |
asked | model structure of noncommutative non-negatively graded DGAs |