General Relativity asymmetry identity

[binbagsss]

Homework Statement

I have $R^{u}_{o b u } F^{o}_a - R^{u}_{oau}F^{o}_{b}$

and I want to show that this equal to $2R_{o[aF^{o}_b]}$

where $[ ]$ denotes antisymmetrization , and $F_{uv}$ is a anitymstric tensor

Homework Equations

Since $F_{uv}$is antisymetric the antisymetrization $2R_{o[aF^{o}_b}]$ reduces to $3(R^{u}_{o b u } F^{o}_a - R^{u}_{oau}F^{o}_{b} + R_{oo}F^a_b)$

The Attempt at a Solution

[/B]
Contracting to get the Ricci tensor and using that $F_{uv}$ is antisymetric

$R^{u}_{o b u } F^{o}_a - R^{u}_{oau}F^{o}_{b}=R_{o b } F^{o}_a + R_{oa}F^{o}_{b}$
$=R{o b } F^{a}_o + R_{oa}F^{o}_{b}$

I can see that there needs to be a $R_{oo}F^a_b$ term added somewhere, which I can't see where this is going to come from unless this is zero ? which I can't see that it is, and even then Id get a factor of $3$ rather than the $2$ needed.

Many thanks in advance.

 

[TSny]

Go to this link and then scroll up into the previous section about 10 lines.
https://en.wikipedia.org/wiki/Ricci_calculus#Differentiation

It says,
As with symmetrization, indices are not antisymmetrized when they are not on the same level, for example;
$$ A_{[\alpha }B^{\beta }{}_{\gamma ]}={\dfrac {1}{2!}}\left(A_{\alpha }B^{\beta }{}_{\gamma }-A_{\gamma }B^{\beta }{}_{\alpha }\right)$$
I don't know if you are meant to adopt this convention.

 

[binbagsss]

Go to this link and then scroll up into the previous section about 10 lines.
https://en.wikipedia.org/wiki/Ricci_calculus#Differentiation

It says,
As with symmetrization, indices are not antisymmetrized when they are not on the same level, for example;
$$ A_{[\alpha }B^{\beta }{}_{\gamma ]}={\dfrac {1}{2!}}\left(A_{\alpha }B^{\beta }{}_{\gamma }-A_{\gamma }B^{\beta }{}_{\alpha }\right)$$
I don't know if you are meant to adopt this convention.

ahh right thank you,
and you can surely get this from the antisymmetrization of indices on the same level by raising an index?

 

[TSny]

you can surely get this from the antisymmetrization of indices on the same level by raising an index?

I don't think so unless I'm overlooking something.

Suppose we have a tensor $C_{\alpha}{} ^{\mu}{} _{\beta}$. Then we can antisymmetrize over $\alpha$ and $\beta$ to produce another tensor $C_{[\alpha}{} ^{\mu}{} _{\beta]}$.

But it would not generally be true that $C_{[\alpha}{} ^{\mu}{} _{\beta]} = g^{\mu \tau} C_{[\alpha \tau \beta]}$.

Instead, you would have to write $C_{[\alpha}{} ^{\mu}{} _{\beta]} = g^{\mu \tau} C_{[\alpha |\tau |\beta]}$ where we use another convention that indices located between vertical bars are to be ignored in the antisymmetrization.

That's how I see it, anyway.