Find # of Independent Components of Riemann Curvature in D Dimensions

dextercioby
Science Advisor
Insights Author
Messages
13,388
Reaction score
4,035
How do i find the number of independent components of the Riemann curvature tensor in D space-time dimensions.

One is given that the Riemann tensor is an (2,2) irreducible rep of GL(4, \mathbb{R}) and obeys Bianchi I

R_{[\mu\nu|\rho]\lambda}=0

Been trying this problem for 3 days and couldn't come up with a reasonable answer. :frown:


Daniel.
 
Physics news on Phys.org
(1/12)*n^2*(n^2-1) where n=D.
 
Tx, but i found it solved in Nakahara pages 231-232.

Daniel.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top