I think what you are looking for is a Poincare Lemma for the co-differential, if I get this right. So the problem is not so much about using vector fields or forms, when you have a metric available (as you say) to switch between them.
Essentially the divergence \mathrm{div}(A) of a vector...
I think I figured out a proof of this: Assume M is non-Hausdorff and P admits a connection. Let x,y be two points in M which cannot be separated by disjoint open sets in M. Then there are two curves c_1 and c_2 in M, c_1 closed at x and c_2 connecting x and y, which coincide everywhere but at x...
Well, differentiable structures can be defined on non-Hausdorff manifolds in the same way as on Hausdorff ones. As it is for example done in the book by N.J. Hicks: Note on Differential Geometry (Van Nostrand Reynhold, 1965). The only thing one does not have at hand in this case is...
Suppose (P,M,\pi,G) is a G-principal bundle. With this I mean a locally trivial fibration (G acts freely on P) over M=P/G with total space P and typical fibre G, as well as a differentiable surjective submersion \pi\colon P\to M. In this case M is nearly a manifold, but may be non-Hausdorff...
Sorry, I can't find anything in there, which supports this claim of vanishing curvature at p. Note that curvature is a tensor, whereas Christoffel symbols are not tensors. If the components of a tensor vanish at a point in one coordinate system, they vanish in all coordinate systems, hence the...
Now that seems rather strange to me. As I always understood that curvature is exactly the quantity, which we can NOT gauge to zero at some point p by choosing normal coordinates about p. We get coordinate lines which are geodesics and have orthonormal tangent vectors at p, as well as vanishing...
Aha, yes there appear derivatives of the structure functions c^i_{jk}. But the Riemannian normal coordinates about p can always be chosen such that these derivatives vanish at p, right?
Hi there,
I was doing some calculations with tensors and ran into a result which seems a bit odd to me. I hope someone can validate this or tell me where my mistake is.
So I have a normal orthonormal frame field \{E_i\} in the neighbourhood of a point p in a Riemannian manifold (M,g), i.e...
I see. Indeed this holds only for spherical triangles consisting of arcs. But in this case there is indeed more detailed information on your problem necessary...
As far as I know, you have a spherical triangle if it lives on the surface of a sphere. It does not matter if the sides are (parts of) great circles or not. The formula for the sum of angles is valid for all spherical triangles, but it depends on the area.
There is an explicit formula for the SUM of angles in a spherical triangle. The sum is of course larger than Pi (=180 degrees). If I remember correctly there is an additional summand in the formula depending on the area A of the triangle and the curvature radius R of the sphere. Something like...
I am not sure what you would like to know. Your mapping F(x,t) is such that F(x,0)=x, hence F(.,0)=id on R^n - {0} and F(x,1)=x/|x| is in the (n-1)-sphere. This is the reason the t's and (t-1)'s are put there as they are.
Yes, I think Hehl and Obukhov meant a global product structure M=\mathbb{R}\times N in the book, such that the 3-dim hypersurfaces of constant t\in\mathbb{R} are spacelike everywhere. Maybe they believed that this was always possible. I wasn't aware of the articles that followed and questioned...
As the manifold in question should be a Lorentzian manifold, especially a spacetime, things are a bit more complicated than just asking if there exists any 3-dim foliation. This foliation should also be transverse to the time direction. Meaning the leaves should be spacelike slices.
So if...