# Independent Components in Riemann-Christoffel Tensor

1. Sep 11, 2009

### Starproj

Help!

I am losing my mind over this problem (which is basically problem 2.6.5 in Arfken and Weber Mathematical Methods for Physicists, sixth edition). I am having difficulty using the tensor symmetric and antisymmetric relationships of the Riemann-Christoffel tensor to show that it reduces from 256 to 36 to 21 and then 20 independent components. My prof just acted like I should be able to do this in my sleep, but I am struggling. The only confirmation I can find was on Mathworld, where they offered that the number of independent components in n dimensions is given by C = (1/12)(n^2)(n^2 - 1), which is great but doesn't help me understand the steps involved.

Does anyone know of a site where this is worked out for dummies?! Or could someone perhaps help shed some light on this for me?

(I'm sorry if I put this thread under the wrong section. It was the one that made the most sense to me.)

2. Sep 11, 2009

### Amanheis

3. Sep 12, 2009

### haushofer

I'm also not very strong in this kind of things, but maybe the following helps.

So we have $R_{abcd}=-R_{bacd}=-R_{abdc}=R_{cdab}$ in n dimensions. So you can think of R as a symmetric matrix $R_{\{ab\}\{cd\}}=R_{xy}$, with each "index" compromising an antisymmetric matrix:

$$R_{xy}=+R_{yx}, \ \ x=\{ab\}, \ \ y=\{cd\}$$

Well, you know probably that an antisymmetric nxn matrix has n(n-1)/2 independent components. If not, try to fill up for example a 3x3 antisymmetric matrix A: the diagonal elements are zero because $A_{ij}=-A_{ji}\rightarrow A_{ii}=-A_{ii}$ (no sum). So in the 3x3 case you get 1+2+3 independent components, in the 4x4 case 1+2+3+4 independent components etc.

The same reasoning goes for a symmetric matrix, and there you'll find n(n+1)/2 independent components.

Our $R_{xy}$ has, viewed as a symmetrix matrx, m(m+1)/2 independent components. But a single n represents an antisymmetric matrix with m=n(n-1)/2 components. So in total we get

$$\frac{1}{2}m(m+1) = \frac{1}{2}[\frac{1}{2}n(n-1)]\frac{1}{2}[n(n-1) + 1]$$

components. However, there are still some symmetries left, namely $R_{[abcd]}=0$. These symmetries are independent of the former mentioned symmetries! A totally antisymmetrized k-tensor in dimensions has

$$n(n-1)(n-2)\ldots(n-k+1)/k!$$

independent terms, so in the very end we are left with

$$\frac{1}{2}[\frac{1}{2}n(n-1)]\frac{1}{2}[n(n-1) + 1] - \frac{1}{24}n(n-1)(n-2)(n-3)$$

components.

Hope this helps :)

4. Sep 14, 2009

### Starproj

Thank you both for taking the time to answer. It was really helpful. I'm sure you appreciate the difference between getting the answer and understandiing how you got the answer. It took a lot of plowing through it (and some hair-pulling!), but I think I figured it out.

Thanks again!