bio | website | |
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location | ||
age | ||
visits | member for | 4 years, 5 months |
seen | Jun 19 '13 at 21:33 | |
stats | profile views | 485 |
Jun 13 |
answered | Seifert Fibrations and their associated Spectral Sequence |
Jun 13 |
asked | Krull dimension in equivariant cohomology |
Oct 15 |
asked | H-spaces without rational homology |
Oct 16 |
awarded | Yearling |
Jun 27 |
accepted | Cohomology of finite quotients of Lie groups |
Jun 22 |
comment |
Cohomology of finite quotients of Lie groups
It sounds very interesting. Unfortunately, I do not know what gerbes are have to believe you without understanding the reason. Your answer shows that the problem is probably very difficult in general. |
Jun 22 |
comment |
Cohomology of finite quotients of Lie groups
@Konrad: This sounds intersting and would help me to understand the situation at least a little bit. Is it difficult to see? Or is there a reference for this claim? |
Jun 22 |
comment |
Cohomology of finite quotients of Lie groups
In fact, I would not know how to use that $\Gamma$ is in the center. For istance, I do not know if the map on $H^3$ is a surjection, if $G=SU(3)$ and $\Gamma$ the center of $G$. |
Jun 22 |
comment |
Cohomology of finite quotients of Lie groups
No, I would not like to assume that $\Gamma$ is in the center. If my Lie group is $Spin$, as in the question, and the order of the group not divisible by $2$, it cannot be in the center anyway. |
Jun 22 |
asked | Cohomology of finite quotients of Lie groups |
Dec 15 |
awarded | Scholar |
Dec 15 |
awarded | Supporter |
Dec 15 |
accepted | Regular simplex in projective space |
Dec 15 |
comment |
Regular simplex in projective space
Many thanks for the surprising answer. |
Dec 15 |
comment |
Regular simplex in projective space
No, I really meant the projective space and am very surprised by the negative answer to my question. |
Dec 14 |
asked | Regular simplex in projective space |
Dec 9 |
awarded | Student |
Dec 9 |
comment |
Covering a subset by submanifolds
Yes, one considers $K$ with the subspace topology. The subset $K$ satisfies the assumption and conclusion with $l=2$ but not with $l=1$ (at any point of $[0,1] \times 0$ |
Dec 9 |
asked | Covering a subset by submanifolds |
Oct 15 |
awarded | Teacher |