Excersise [14.30]

Pibe · 07 Jun 2009, 09:07

Regards to all you!

I´m not sure if the formula $\frac{1}{12}n^2(n^2-1)$ about the quantity of independents components of Riemann tensor is right. I think this value must be $\frac{1}{12}n^2(n-1)^2$

I say this because:

$R_{abcd}=-R_{bacd}\Rightarrow R_{aacd}=-R_{aacd}\Rightarrow R_{aacd}=0$ (1)
Moreover a=1,..,n

and

and

then, there are n^3

components whose value is 0.

The second antisymmetry tells us that $R_{abcd}=-R_{abdc}\Rightarrow R_{abcc}=-R_{abcc}\Rightarrow R_{abcc}=0$
So, there are n^3

components whose vale is 0, but the components $R_{aacc}=0$ were already counted in (1); and a=1,..,n

and

. Then there are n^3-n^2

more null components.

So, R has n^4-(2n^3-n^2)=n^2(n^2-2n+1)=n^2(n-1)(n-1)=n^2(n-1)^2

not-null components (that is, independent components)

But the first antisymmetry $R_{abcd}=-R_{bacd}$ tells us that half the components of R are dependent of another half, so the independent componets are $\frac{1}{2}n^2(n-1)^2$

And the second antisymmetry $R_{abcd}=-R_{abdc}$ tells us the same, so remain $\frac{1}{4}n^2(n-1)^2$ elements. The interchange symmetry $R_{abcd}=-R_{cdab}$ tells the same, so remain $\frac{1}{8}n^2(n-1)^2$ elements.

The Bianchi symmetry $R_{abcd}+R_{bcad}+R_{cabd}=0$ implies that each component depends of another two elements, so only 2/3 of remaining indepependent components are really independent.

Then, finally, there are $\frac{2}{3}\frac{1}{8}n^2(n-1)^2=\frac{1}{12}n^2(n-1)^2$ independent components in Riemann tensor R.

This is my reasoning. Is there misprint in the book? Am I right or I'm wrong? Please I wish someone answer me.
See you soon!!