How is the Ricci Tensor Derived from the Ricci 1-Form?

[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, let's 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.

 

[torquil]

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, let's 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.

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.

 

[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?

 

[torquil]

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

<br /> A = A_{\mu\nu} \omega^\mu \otimes \omega^\nu<br />

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.

 

[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.