Recent content by nughret

Does anyone know of any sources which provide a proof, or outline of, the Coleman-Mandula theorem and the Haag-Lopuszanski-Sohnius Theorem?

What about the manifold (RP2)X[0,1)? However there isn't a compact manifold of which it is a boundary. For more information about this sort of stuff cobordism theory is what you should read into

I am working through a book on Kahler manifolds and for one of the proofs it states that the maximum exterior power of TM is m (where M has complex dimension). Could you explain why this is the case rather than the maximum exterior power being 2m.

To find the isometry group of AdS(n,m) consider its embedding in M(n,m+1) (Minkowski spacewith one extra time dimension). We can consider AdS as the set of points in M such that x2 = -R2. Now consider the isometries of M that leave AdS invarient; it is not hard to show that this is O(n,m+1)...

For vector fields V = Viei, W [V,W] = (VjWi,j - Vi,jWi + VjWkCijk)ei Where Cijk = ([ej,ek]d we can find these in some neighbourhood by expanding the non-commuting basis in terms of a coordinate basis

Yes I see your point Status, and as you say the connection vanishing in some basis implies only zero curvature not zero torsion. In regards to the formula for the Lie bracket using Z = [X,Y] if Z(f) = X(Y(f)) - Y(X(f)), gives all required terms. Using a non commutative basis we then find the...

Of course. I was referring to the problem of being able to find smooth local orthonormal bases through any point. The proof is used in the more general setting of vector bundles and as i said i will try to recall it, we have to consider the orthogonal subgroup of the general linear group and you...

In the case of a Riemannian manifold I am aware of a beautiful proof which gives a positive answer to the case of finding on orthonormal base. My memory of the details are a bit hazy and we use the principle frame bundle (or some prolongation of this) and then consider whether we can find a...

My interpretation of the question was this: given any point x in our manifold, given that we can find some chart around x in which the connection vanishes show that the torsion tensor is zero everywhere. The above posts provide an outline of the proof. If you do not believe such a proof is...

I guess though you are considering local considersations only so you will be able to find some domain where this is possible

These equations are not gaurenteed to give you a coordinate change as the 1-form dT may not be exact in order to actually find the quantity T = T(t,r) , thenecessary co-ordinate change. I will come up with an example if necessary, it is similar to thecase of holonomic and non-holonomic...

This is incorrect; write out the equation in local co-ords(in which connection vanishes) then (DXY)i = Yi,jXj

Re reading i see that we must also make the identification (l,1,n) ~' (l,1,n') But then we will just get ; Cone(L(X,x)xN(X,x))/~'

Consider decomposing a path,p, into a loop,l, and a non looping part,n, and a time,t, at the point where we connect these paths: p = l (+t) n where for two paths a,b such that a(1)=b(0) we have (for t not equal 1,0) (a (+t) b) (s) = a(s/t) , 0<s<t b((s-t)/(1-t)) ...

Yes your example is clear and i to was running into similar problems when trying to consider where to map (c(x),t) for different values of t on a general space. The task was set by someone who is knowledgeable about the subject so maybe if instead of the cone we consider (Cone(L(X,x)))/~...