Lowering indices in the Riemann tensor and Ricci tensor

[Deadstar]

Let's say I want to calculate the Ricci tensor, $$R_{bd}$$, in terms of the contractions of the Riemann tensor, $${R^a}_{bcd}$$. There are two definitions of the Riemann tensor I have, one where the $$a$$ is lowered and one where it is not, as above.

To change between the two all that I have ever seen written is 'we lower the indices' but I don't think I fully understand this. Does this mean...

$$R_{abcd} = g_{aa} {R^a}_{bcd}$$

So the answer to my original question of finding the Ricci tensor is...

$$R_{bd} = g^{ac} g_{aa} {R^a}_{bcd}$$

Following this, I also have written before me that...

$${R^b}_{bcd} = 0$$ since $$R_{abcd}$$ is symmetric on a and b. Shouldn't this be antisymmetric on a and b?

Sorry if these are basic questions but I'm finding the vagueness of 'lowering the indices' a bit confusing...

Cheers.

 

[DrGreg]

$$R_{abcd} = g_{aa} {R^a}_{bcd}$$

You can't have the same index appearing 3 times in an expression, what you mean is

$$R_{abcd} = g_{ae} {R^e}_{bcd}$$​

 

[jfy4]

The Riemann tensor

$$R^{\alpha}_{\;\beta\gamma\delta}=g^{\alpha\lambda}R_{\lambda\beta\gamma\delta}$$.

The raising and lowing of indices is done using the fundamental metric tensor $g_{\alpha\beta}$ and the inverse metric $g^{\alpha\beta}$.

The Ricci tensor

$$R^{\alpha}_{\;\beta\gamma\delta}\rightarrow R^{\alpha}_{\;\beta\alpha\delta}=R_{\beta\delta}$$

is a contraction of the first and third indices.

 

[robphy]

I would argue that $$R^{\alpha}_{\;\beta\gamma\delta}$$
is the more fundamental expression for the Riemann tensor
since this is defined from the derivative operator [without reference to a metric].
Its trace yields the Ricci tensor [as jfy4 wrote]... and this too doesn't make use of a metric.

To lower the upper index of Riemann or to get the scalar-curvature from Ricci now requires a metric.