But yes, you can, since μ and ν are not Lorentz indices, and the first theorem (that v^a = ∂_b s^{ab}) only makes a statement about Lorentz indices.
U is antisymmetric in the first and second index, and separately in the third and fourth, but not in the first and third. So only in that case...
True, but that's not a point on the manifold (sure you meant φ=2π, no?). My φ coordinate is in the range [0, 2π), so it includes 0 but excludes 2π. So everything ok there.
(if you want strictest mathematical rigor: φ ∊ ℝ/{2πℤ})
With mathematical theorems you must be careful in intepretation, they assure only and exactly only what they state. For each manifold of dimension n (so in my toy example n=2) there is an atlas containing n+1 charts, so you don't need more. Of course you can take more if you wish, and there are...
Yes, with your notation T^{ab} comes out zero. With the notation I proposed also, I see it now. But note that in the original problem the derivatives should act on the first and third index of U, so that T doesn't vanish. So you should take W^c_{μν} = ∂_d U^c_μ^d_ν, sorry for the confusion before.
So now you know that there are intrinsically curved manifolds where you only need one chart. The topology doesn't fully specify the metric; there are infinitely many possibilities to bestow a given topological manifold with a metric.
We didn't understand each other saying "trivial" :)
There...
Yes, I see. I'm sorry, don't know more.
From n = 3/2 +- 1/2 \sqrt{2/3 ( m^2 - 5/2 )}, you see that 2/3 ( m^2 - 5/2 ) must be a perfect odd square for n to be integer, so you can equally solve 2/3 ( m^2 - 5/2 ) = (2k+1)^2, with k = 0,1,2,... This can be reexpressed as m = +- \sqrt{6 k^2 + 6 k...
The original theorem was that for any v^a with d_a v^a = 0, there is an antisymmetric tensor s^{ab} such that v^a = d_b s^{ab} = - d_b s^{ba}. So you should have T^a_μ = ∂_c W^{ac}_μ, which is minus the result you gave. Of course you can always redefine W to be -W, but let's stick with mine for...
Ok, integer values and Pell's equation, I refer you to https://www.physicsforums.com/showthread.php?t=78855 and http://en.wikipedia.org/wiki/Chakravala_method
For which variable should you solve? I assume for n, then from 3/2 (2n-3)^2 + 5/2 = m^2 I get (2n-3)^2 = 2/3 ( m^2 - 5/2 ) and then 2n-3 = +- \sqrt{2/3 ( m^2 - 5/2 )}, so n = 3/2 +- 1/2 \sqrt{2/3 ( m^2 - 5/2 )}. If you should solve for m, m = +- \sqrt{3/2 (2n-3)^2 + 5/2}.
But since that's...
Could you show your calculation in a bit more detail, please? Otherwise it is difficult seeing where you went wrong, as I haven't done the calculation.
I don't want to offend you in any way, if you felt I have done this. But it seems to me that you don't understand some (subtle) points, and I want to explain them to you.
I may have misinterpreted your statement
you gave in your post before the last, but to me you were stating that H is a...
The parameter H is intrinsically defined in de Sitter space: it's R/12 in the four-dimensional case, where R is the curvature scalar. I don't need any embedding for it, period.
If you want to embed de Sitter space, then H is the radius as seen from the embedding space, but H is not a...
True; technically this is a coordinate singularity is you use polar coordinates for the sphere. However, you can for example obtain its area by integrating dA = sin θ dθ dφ over the whole sphere, and so it's an integrable singularity. Those typically don't cause any problems, so let's include...
(To avoid confusion: space in the following is short for spacetime.)
Global coordinates for me are coordinates that cover the whole manifold, p.ex. cartesian coordinates for Minkowski space. Every point of Minkowski space is described by one coordinate tuple (x,y,z,t), and to every coordinate...