20070604, 21:22  #1 
May 2005
Argentina
BA_{16} Posts 
Covariant Derivation
Can someone explain me in simple terms what is the covariant derivation?
For example if we take as a Manifold the unit circle in R^2, what would be its covariant derivation? 
20070604, 23:56  #2  
Feb 2007
2^{4}·3^{3} Posts 
Quote:
e.g. for R_ab^cd DR_ab^cd = dR_ab^cd+R_eb^cd phi_a^e +R_ae^cd phi_a^e +R_ab^ed phi_e^c+R_ab^ce phi_e^d sorry I may have not 100% mainstream conventions , also that expression might simplify (to zero of course, but suppose R was sth else than d phi + phi phi), i dont remember well : I did that in an earlier life... Last fiddled with by m_f_h on 20070604 at 23:57 

20070605, 18:20  #3 
May 2005
Argentina
2×3×31 Posts 
I don't get it

20070605, 20:45  #4 
∂^{2}ω=0
Sep 2002
República de California
2·3·1,931 Posts 
The proper term is "covariant differentiation."
Any decent text or webpage on differential geometry should have an adequate description. 
20070606, 15:41  #5 
May 2005
Argentina
2·3·31 Posts 
I know, I wanted a less formal definition, maybe with some numerical example, to make it less abstract and more easy to understand it.

20070607, 17:04  #6  
Feb 2007
2^{4}×3^{3} Posts 
Quote:
PS: it seems there is some explicit calculation on http://en.wikipedia.org/wiki/Connection_(mathematics) PPS: well, not much... I think you have to plug in those into the formulae on the page "covariant derivative" In fact, there are different notions of covariant derivatives. In general, "covariant" is w.r.t. some local ("gauge") transformation. In general relativity, there are 2 such transformations to be considered : local Lorentz transformations (SU(2) or SO(3,1) acting on "Lorenz" indices), and local coordinate transformations (diffeomorphisms ; acting on "Einstein indices"). "of course", both are linked... Last fiddled with by m_f_h on 20070607 at 17:19 
