Riemann curvature coefficients using Cartan structure equation

[snypehype46]

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.

 

[romsofia]

This should help I believe: http://sites.science.oregonstate.edu/physics/coursewikis/GDF/book/gdf/components.html

 

[romsofia]

Also, to add onto this, your notation (if it's from a book, id recommend reviewing the above wiki before going back) is probably leading you to so many issues. You have to PICK which a, b you are using in order to calculate it. But the other side of the equation is ijkl? So let's say you let a = 1, and b = 3. What's the ijkl in your notation? If you wrote it yourself, stop. You can't just randomly throw different names for indices and hope it sticks. If you have $\Omega^a_b$, then you better have an a in an upper index somewhere, and and b in a lower index somewhere. This will make life so much easier for you.

Here are the steps:
1) Identify basis one forms
2) Use first structure equation to calculate connection coefficients
3) Use second structure equation to get your curvature (your $\Omega^a_b$)

Once you have these calculated, THEN you set it equal to your Riemannian junk, and whalla, you have something you calculated = Riemannian junk times one forms. You can identify what you $R^a_{bcd}$ is easily, but you first have to calculate what your $\Omega^a_b$ is, which comes from taking exterior derivatives of you connection coefficients, which involves taking exterior derivatives of your basis one forms. And in another thread you talked about why do some terms "not matter". It isn't that they "don't matter", it could be that when you calculate those connection coefficients, it's zero.

EDIT: I responded to your other thread, they weren't zero in this case!