# Ricci tensor from Ricci 1-form

1. Dec 31, 2011

### graupner1000

Hi all,

once again I'm stuck on something I am quite certain is silly, but here it goes. My confusion pertains to the equation

$Ric=R^{a}\otimes e_{a}$

where $Ric$ is the Ricci tensor, $R^{a}$ is the Ricci 1-form and $e_{a}$ are the elements of an orthonormal basis.

Now, lets say for arguments sake that $a=0,1,2$ and I have a Ricci 1-form that looks something like this (What I'm actually trying to work out is a lot larger but follows a similar pattern)

$R^{a}=\left[ \begin{array}{c} Ae_{0} + Be_{1} \\ Be_{0} - Ae_{1} \\ e_{2} \end{array} \right]$

where $A$ and $B$ are constants. The next step would be to take the tensor product of $R^{a}$ and $e_{a}$ and this is where the problem lies. My instinct would be to treat this as an outer product so you end up with something like

$R^{a}\otimes e_{a}=\left[ \begin{array}{ccc} (Ae_{0} + Be_{1})e_{0} & (Ae_{0} + Be_{1})e_{1} & (Ae_{0} + Be_{1})e_{2} \\ (Ae_{0} - Be_{1})e_{0} & (Ae_{0} - Be_{1})e_{1} & (Ae_{0} - Be_{1})e_{2} \\ e_{2}e_{0} & e_{2}e_{1} & e_{2}e_{2} \end{array} \right]$

But that seems to be ignoring the sum over $a$ (or is this the operation it implies?) and more importantly, I really doubt there should be multiplication between the elements, i.e does

$(Ae_{0} + Be_{1})e_{0}$
imply
$(Ae_{0} + Be_{1})\otimes e_{0}$
or
$(Ae_{0} + Be_{1})\wedge e_{0}$

As said, this is a really silly thing to be stuck with and probably means that I've missed(read not paid attention to) something really basic so any help would be very much appreciated.

2. Dec 31, 2011

### torquil

I think it should be (using the Einstein summation convention for repeated indices):

$$Ric=R^{a}\otimes e_{a} := R^0 \otimes e_0 + R^1 \otimes e_1 + R^2 \otimes e_2 = (Ae_{0} + Be_{1}) \otimes e_0 + (Be_{0} - Ae_{1}) \otimes e_1 + e_2 \otimes e_2 = \dots$$

This gives you a 2-tensor, as you are supposed to get.

3. Jan 6, 2012

### graupner1000

Back again. Thanks for your answer, that was one thing I was thinking about. But is there any way to write that in a "traditional" matrix form?

4. Jan 7, 2012

### torquil

Just use the correspondence between the coefficients and basis-expansion of a 2-tensor.

$$A = A_{\mu\nu} \omega^\mu \otimes \omega^\nu$$

to indentify the matrix components as the coefficients in this expansion.

Btw, I'm not sure what you meant by "$R^a$ is the Ricci 1-form"? Do you have three "Ricci 1-forms", one for each value of a, or are these the components of one "Ricci 1-form". If the latter is the case, your 3-component expression for $R^a$ doesn't make sense, since you have put basis elements in the components.

Sorry, this terminology is a bit unusual for me, I'm used to the curvature forms like what is done here:

http://www.uio.no/studier/emner/mat...dervisningsmateriale/Kursmateriell/fys307.pdf

I haven't heard of a Ricci 1-form before.

5. Jan 10, 2012

### graupner1000

This is the terminology I have been taught, but it might have other names elsewhere. The Ricci 1-form is the contraction of the curvature form (or Ricci 2-form):

$R_{a}=i_{b}R^{b}_{ a}$

(Using R twice might not be the best convention) where $R^{b}_{ a}$ is given by Cartan's second structure equation.

My example has three components just because I needed an example. What I am actually trying to work out is considerably larger and I couldn't be asked to write out the entire thing.