Why is $\delta^b_e\nabla_c {R_{cad}}^e$ equal to $\nabla^b {R_{cadb}}$ - ie how to contract the Bianchi identity properly

Question

My lecture notes, when discussing the Riemann curvature tensor, says:

\begin{aligned} 0 &=\delta_{e}^{b}\left(\nabla_{a} R_{b c d}{ }^{e}+\nabla_{c} R_{a b d}{ }^{e}+\nabla_{b} R_{c a d}{ }^{e}\right) \\ &=\nabla_{a} R_{c d}-\nabla_{c} R_{a d}+\nabla^{b} R_{c a d b} \end{aligned}

I don't get how we transformed the least term:

$$\bbox[5px,border:3px solid red]{\delta_{e}^{b}\nabla_{b}R_{c a d}{ }^{e}=\nabla^{b} R_{c a d b}}$$

I would like to know how this transformation happened. (One of my main concerns is, how did the $b$ ended up being a superscript.) I know that there might be relabelling going on but I don't know what identity we used to make the rewriteing in the red box possible.

I have tried this:

\begin{aligned} & \delta_e^b \nabla_{b} {R_{c a d}}^{e} \\ =& \delta_{e}^{b} \nabla_{b} g^{e i} R_{c a d i} \\ =& \nabla_{b} g^{b i} R_{c a d i} \\ =& \nabla^{i} R_{c a d i} \\ =& \nabla^{b} R_{c a d b} \end{aligned}

where in the last row I have relabelled $i$ to $b$. Is this correct?

Answer 1

It becomes more apparent when thinking about $\delta_e^b$ as

$$ \delta_e^b = g_{e\alpha}g^{\alpha b} $$

Then you just lower one and raise another

$$ \delta^b_e\nabla_b {R_{cad}}^e = g_{e\alpha}g^{\alpha b}\nabla_b {R_{cad}}^e $$ $$ = g_{e\alpha}\nabla^\alpha {R_{cad}}^e = \nabla^\alpha R_{cad\alpha}$$