# Contracting my 2-form?

1. Jun 3, 2006

### Oxymoron

I have

$$R^{ij} = c(e^i \wedge e^j)$$

as my curvature 2-form, (and c represents "constant" curvature). I would like to contract to form the Ricci 1-forms, $P_a$, which in turn, would allow me to write out the Ricci tensor as

$$\mbox{Ric} = P_a \otimes e^a$$

and eventually, the Einstein tensor.

My question is, how would I go about contracting my 2-form? I think it is fairly easy and Im just missing something. I mean there is a big difference between saying that $R^i{}_i$ is the Ricci 1-form because I contracted the 2-form, $R_{ij}$ and writing down exactly what just happened.

2. Jun 3, 2006

### Oxymoron

Perhaps all I have to do is take the interior product of the 2-form.

So could I write

$$P_a = i(X_i)R^i{}_i = ce^i \wedge e^j \wedge e^i = ce^j$$

So could I say that the contracted 2-form, $R_{ij}$, equals

$$P_a = R^i{}_i = ce^j[/itex] therefore [tex]\mbox{Ric} = ce^j \wedge e^a$$

Then, the Einstein tensor is

$$G_{ij} = R_{ij} - \frac{1}{2}\mbox{Ric}g_{ij}$$
$$= ce^i \wedge e^j - \frac{c}{2}e^i \wedge e^j g_{ij}$$

Last edited: Jun 3, 2006
3. Jun 3, 2006

### Perturbation

Just contract one index with something or other.

Your notation seems a bit sloppy to me. For one thing $R^{ij}$ are the components of the Ricci tensor, which are not equal to $c(e^i \wedge e^j)$, but $\mathbf{R}(\vec{e}_i, \vec{e}_j)$, to write the components as a wedge product doesn't make any sense, as components are just numbers. Furthermore the wedge product is anti-symmetric, the Ricci tensor, at least in Riemannian geometry (as opposed to Cartan-Riemann), is symmetric. The wedge product of two basis forms is not the basis for the dual tangent space ($\tilde{e}^i\wedge\tilde{e}^j$), $\tilde{e}^i$ is. $\tilde{e}^i\wedge\tilde{e}^j$ is a notation used for differential forms, it being shorthand for $\tilde{e}^i\otimes\tilde{e}^j-\tilde{e}^j\otimes\tilde{e}^i$, so that all the components of the form are anti-symmetric.

For example, I can write the exterior derivative of a one-form $\tilde{\omega}$ as

$$d\tilde{\omega}=\partial_i\omega_j\tilde{e}^i\wedge\tilde{e}^j$$

But its components are

$$\omega_{ij}=\partial_i\omega_j-\partial_j\omega_i$$

Not $\partial_i\omega_j$.

Unless you're using abstract indices or something here, in which case I don't see why you've bothered expanding the tensor into sums over components and basis forms, as the whole point of abstract indices is to remove the inference that tensors are dependant on coordinate frames made by writing $V_j$ etc.

Your outer product also doesn't make any sense. If $P_a$ are the components of your one forms and $\tilde{e}^a$ your dual vector basis, the form is simply $P_a\tilde{e}^a$. The outer product doesn't come into it; the outer product is a rule for making higher rank tensors from lower rank tensors, such as one-forms and vectors.

If you want to make a one-form from the Ricci tensor $\mathbf{R}$ supply it with a vector argument, that way you're left with one more vector "slot" to fill, making it a one-form: $\mathbf{R}(\vec{e}_i,-)$.

Furthermore $R^i_i$ is a scalar, not a one-form. It's equivalent to $g_{ij}R^{ij}$, the double sum of the components of the Ricci tensor with those of the metric.

If your Ricci scalar is a constant and your Ricci tensor has all the same components, then the Einstein tensor is just $c(1-\frac{1}{2}g_{ij})$. Don't worry about there being no indices on the one, $g_{ij}$ is just a number as well, and when it comes to actually explicitly computing the ij-th component of the Einstein tensor you won't have indices in there any who. If it bothers you that much, make up your own notation for the matrix with ones in every entry. Though the Ricci tensor components aren't likely to be constant in every coordinate system, so the previous expression isn't the best way of writing it.

I may well be getting confused because of your notation here.

Last edited: Jun 4, 2006
4. Jun 4, 2006

### Oxymoron

Wow. Thanks for your reply Perturbation. It sounds like I have a lot of things wrong.

Basically what happened is this: I was given a metric

$$g=-\mbox{d}t\otimes\mbox{d}t + f^2\hat{g}$$

where $\hat{g}$ is the metric on some 3-space of constant curvature. and f is a function of t only. an orthonormal coframe:

$$e^0 = \mbox{d}t \quad\quad e^i = f\hat{e}_i$$

then I differentiated to get

$$\mbox{d}e^i = f^{\prime}e^0 \wedge e^i + f\mbox{d}\hat{e}_i \quad\quad \mbox{d}e^0 = 0$$

where i = 1,2,3

then I calculated the connection 1-forms:

$$\omega_{ab} = \hat{\omega}_{ab} + \mbox{some other stuff}$$

$$\omega_{0b} = -\omega_{b0} = -\frac{f^{\prime}}{f}\hat{e}^k$$

where k = 1,2,3. Then I got my curvature 2-forms:

$$R_{ab} = \mbox{d}\omega_{ab} + \omega_{ak}\wedge\omega^k{}_a + \omega_{a0}\wedge\omega^0{}_a$$

So I differentiated $\omega_{ab}$ and substituted everything and got:

$$R_{ab} = \left(\frac{f^{\prime}}{f}\right)^2 e_a \wedge e_b$$

and are you saying that this last equation does not make sense?

Last edited: Jun 4, 2006
5. Jun 4, 2006

### George Jones

Staff Emeritus
I have not checked to see whether I agree with it, but it does make sense. It also sucks notationally :tongue: , and this is causing confusion.

In your equation, the indices label which second rank tensor. In the more standard interpretation of the notation for $R_{ab}$, the indices label which component (with respect to a basis) of *one* particular second rank tensor. This is why, e.g., the curvature 2-forms are denoted by $\theta$ in Frankel, by $\rho$ in Szekeres, and by $\mathcal{R}$ in Misner, Thorne, and Wheeler.

I also have notational problems with $e^i = f\hat{e}_i$ and

$$\omega_{0b} = -\omega_{b0} = \frac{f^{\prime}}{f}\hat{e}^k.$$

Last edited: Jun 4, 2006