Hi, cosmicstring, you asked:
cosmicstring1 said:
I have Christoffel symbols for a metric and I want to find the connection 1-forms.
I suggested starting from a frame field, a la Cartan, as explained in MTW. I didn't answer the question you asked, I told you where to find the answer to the question you
should have asked. You replied:
cosmicstring1 said:
Yes, I read that section from MTW but it does not give the answer.
Sure it does--- look at the sections on computing curvature. This contains what you need to answer the question you asked, as well as showing you a much better way to do all this stuff.
cosmicstring1 said:
But Christoffel symbols do not share the symmetries of the conn. 1-forms. Do you know any way to make this possible?
I don't think that even makes sense!
But there is one small point concerning the symmetry properties of the matrix of connection one-forms which MTW slur over. Flanders doesn't cover it either because he only discusses Riemannian manifolds. Namely, the rule for adjoint is a bit different for Lorentzian vice Riemannian. You need to know this when you use the symmetries of the connection one-forms in order to read off the connection one-forms from the exterior derivatives of the basis one-forms, as suggested by Cartan.
By definition, if A is some linear operator and (\cdot , \cdot) is some inner product, the adjoint A^\ast is the linear operator satisfying
(Ax, y) = (x, A^{\ast} y)
for all x,y. If you write this out in matrix notation, with the inner product written (x,y) = x^T L y, you obtain
x^T \, A^T \, L \, y = x^T \, L \, A^{\ast} \, y
This holds for all x,y, so
A^\ast = L^{-1} \, A^T \, L
Now apply this when L = \operatorname{diag} (-1,1,1,1) to obtain the
Lorentzian transpose. Now apply this to your matrix of one-forms when reading off the connection one-forms as suggested by Cartan.
In the past I have gone through all this stuff in great detail--- you can Google for old posts by myself in sci.physics.*