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Mar
18
accepted Sheaf cohomology of projective spaces
Feb
9
awarded  Critic
Jan
27
asked Can one exchange fibre and base space in a fibre bundle?
Jan
23
awarded  Tumbleweed
Jan
23
revised Connection in fibre bundle from discontinuous group action
added 37 characters in body
Jan
23
comment All almost complex structures on a manifold
@JimBelk $\mathcal O(k)$ is the complex line bundle (real rank 2 bundle) of Chern class $k$ over $\mathbb P^1\cong S^2$ (see the notation on Wikipedia). I loosely expect the space to be finite-dimensional, because the space of complex structures on $\mathcal O(k)$, $k$ neg., is finite-dimensional (this one has to calculate). Even if it isn't finite-dimensional, though, one should be able to construct at least a one-parameter family of almost complex structures for the theorem to be of any practical value.
Jan
22
reviewed Reviewed How is statistical uncertainty calculated for the modulus function?
Jan
22
reviewed Reviewed How do I prove this statement is tautology without using truth tables?
Jan
22
reviewed Reviewed How to derive the Descartes equation of a line in a coordinate system? $ax+by+c=0$
Jan
22
awarded  Custodian
Jan
22
reviewed Reviewed Where is the flaw in this argument of a proof that 1=2? (Derivative of repeated addition)
Jan
22
comment Intuitively what is it if making a modification of a torus?
As far as I can tell, your modification is not an equivalence relation.
Jan
22
awarded  Disciplined
Jan
22
revised All almost complex structures on a manifold
edited tags
Jan
22
revised All almost complex structures on a manifold
edited tags; edited title
Jan
22
asked All almost complex structures on a manifold
Jan
20
asked Connection in fibre bundle from discontinuous group action
Jan
16
asked Sheaf cohomology of projective spaces
Dec
13
comment Derived functors and coboundary operator
@ZhenLin I'm not exactly sure what your point is, but as far as I understand, you can calculate group cohomology either from an injective resolution (which is difficult to find) or from a projective resolution with the Hom functor applied to the whole resolution.
Dec
13
comment Derived functors and coboundary operator
@ZhenLin In principle, yes. I added more details in the question body.