- #1

snypehype46

- 12

- 1

- Homework Statement
- Computing the Riemann curvature tensor using Cartan's structure equations

- Relevant Equations
- $$\frac{1}{2} \Omega_{ab} (\theta^a \wedge \theta^b) = -\frac{1}{4} R_{ijkl} (dx^i \wedge dx^j)(dx^k \wedge dx^l)$$

To calculate the Riemann coefficient for a metric ##g##, one can employ the second Cartan's structure equation:

$$\frac{1}{2} \Omega_{ab} (\theta^a \wedge \theta^b) = -\frac{1}{4} R_{ijkl} (dx^i \wedge dx^j)(dx^k \wedge dx^l)$$

and using the tetrad formalism to compute the coefficients of the curvature tensor.

Now I'm trying to properly understand this method, I was doing this exercise for which I obtained:

$$\frac{1}{2} \Omega_{ab} (\theta^a \wedge \theta^b) = -\frac{1}{4} A (dx \wedge du)^2 -\frac{1}{2}B(dx \wedge du) (dy \wedge du) - \frac{1}{4} (dy \wedge du)^2$$

However, from here I'm not quite how I would read the coefficient for the Riemann tensor. From here it seems for example that we have:

$$-\frac{1}{4} A (dx \wedge du)^2 = -\frac{1}{4} R_{xuxu}(dx \wedge du)^2$$

$$-\frac{1}{2}B(dx \wedge du) (dy \wedge du) = -\frac{1}{4}R_{xuyu}(dx \wedge du)(dy \wedge du)$$

The answer is supposed to be ##R_{xuxu} = \frac{1}{2}A## and ##R_{xuyu}= \frac{1}{2} B##, however I don't quite understand how this would be obtained from the equation above. I assume it has something to do with the symmetries of the Riemann tensor but not quite sure.

$$\frac{1}{2} \Omega_{ab} (\theta^a \wedge \theta^b) = -\frac{1}{4} R_{ijkl} (dx^i \wedge dx^j)(dx^k \wedge dx^l)$$

and using the tetrad formalism to compute the coefficients of the curvature tensor.

Now I'm trying to properly understand this method, I was doing this exercise for which I obtained:

$$\frac{1}{2} \Omega_{ab} (\theta^a \wedge \theta^b) = -\frac{1}{4} A (dx \wedge du)^2 -\frac{1}{2}B(dx \wedge du) (dy \wedge du) - \frac{1}{4} (dy \wedge du)^2$$

However, from here I'm not quite how I would read the coefficient for the Riemann tensor. From here it seems for example that we have:

$$-\frac{1}{4} A (dx \wedge du)^2 = -\frac{1}{4} R_{xuxu}(dx \wedge du)^2$$

$$-\frac{1}{2}B(dx \wedge du) (dy \wedge du) = -\frac{1}{4}R_{xuyu}(dx \wedge du)(dy \wedge du)$$

The answer is supposed to be ##R_{xuxu} = \frac{1}{2}A## and ##R_{xuyu}= \frac{1}{2} B##, however I don't quite understand how this would be obtained from the equation above. I assume it has something to do with the symmetries of the Riemann tensor but not quite sure.