I am currently working through the proof of the following equation for the curvature of a Riemannian submersion:
\begin{align} \langle \overline R(\overline X,\overline Y)\overline Z,\overline W \rangle &= \langle R(X,Y)Z,W \rangle - \frac 14 \left\langle [\overline X,\overline Z]^v,[\overline Y,\overline W]^v \right\rangle \\ &\qquad + \frac 14 \left\langle [\overline Y,\overline Z]^v,[\overline X,\overline W]^v \right\rangle - \frac 12 \left\langle [\overline Z,\overline W]^v,[\overline X,\overline Y]^v \right\rangle. \end{align}
The full context of this question, including do Carmo's notation reproduced here, comes from Problems 8.8-8.10 of do Carmo's Riemannian Geometry (PDF file of those problems). But my question focuses only on Problem 8.10(a).
I am trying to fill in the details of do Carmo's hint leading up to the formula. I am reproducing the hint below but I am also attempting to fill in the details of the hint. My questions below center around filling in such details.
Hint: Observe that $\overline X \langle \overline \nabla_{\overline Y}\overline Z,\overline W \rangle = X\langle \nabla_Y Z,W\rangle$. Therefore, \begin{align} \tag{1} \langle \overline \nabla_{\overline X} \overline \nabla_{\overline Y} \overline Z,\overline W\rangle &= \overline X\langle \overline \nabla_{\overline Y}\overline Z,\overline W\rangle - \langle \overline \nabla_{\overline Y}\overline Z,\overline \nabla_{\overline X}\overline W\rangle \\ &= \langle \nabla_X \nabla_Y Z,W\rangle - \frac 14 \langle [\overline Y,\overline Z]^v,[\overline X,\overline W]^v \rangle. \end{align} On the other hand, if $T \in \mathcal X(\overline M)$ is vertical, \begin{align} \tag{2} \langle \overline \nabla_T \overline X,\overline Y \rangle = \langle \overline \nabla_\overline X T,\overline Y \rangle + \langle [T,\overline X],\overline Y\rangle = -\langle T,\overline \nabla_{\overline X} \overline Y \rangle. \end{align} Therefore, \begin{align} \tag{3} \langle \overline \nabla_{[\overline X,\overline Y]} \overline Z,\overline W\rangle &= \langle \overline \nabla_{[\overline X,\overline Y]^h} \overline Z,\overline W \rangle + \langle \overline \nabla_{[\overline X,\overline Y]^v} \overline Z,\overline W \rangle \\ &= \langle \overline \nabla_{[X,Y]} Z,W \rangle - \frac 12 \langle [\overline X,\overline Y]^v,[\overline Z,\overline W]^v \rangle. \end{align} Putting the above together, we obtain the desired equation.
My thoughts so far:
Once the above three identities in the hint are established, I would certainly know how to apply them to the definition of the curvture of $\overline R$ to establish the desired equation.
But I cannot thoroughly understand how those above identities are established.
In $(1)$, I worked from a term on the RHS of $(1)$---and tried to apply the formula from Exercise 8.9(b) (can be viewed in PDF file linked above)---as follows: \begin{align} &\frac 14 \langle [\overline Y,\overline Z]^v,[\overline X,\overline W]^v \rangle \\ &\qquad = \frac 14 \langle 2(\overline \nabla_{\overline Y}\overline Z-\overline{(\nabla_Y Z)}),2(\overline \nabla_{\overline X}\overline W-\overline{(\nabla_X W)})\rangle \\ &\qquad = \langle \overline \nabla_{\overline Y}\overline Z-\overline{(\nabla_Y Z)},\overline \nabla_{\overline X}\overline W-\overline{(\nabla_X W)}\rangle \\ &\qquad = \langle \overline \nabla_{\overline Y}\overline Z,\overline \nabla_{\overline X}\overline W \rangle - \langle \nabla_{\overline Y}\overline Z, \overline{(\nabla_X W)} \rangle - \langle \overline{(\nabla_Y Z)} , \overline \nabla_{\overline X} \overline W \rangle + \langle \overline{(\nabla_Y Z)}, \overline{(\nabla_X W)} \rangle \end{align} but I'm not sure how to proceed (though I see a term on the LHS of $(1)$, which is a good sign).
In $(2)$, the first equality is simple. But it is the second equality of $(2)$ that is giving me problems. I tried working from the RHS again (like I did in $(1)$): \begin{align} -\langle T, \overline \nabla_{\overline X}\overline Y \rangle = -\frac 12 \langle T,[\overline X,\overline Y] \rangle = -\frac 12 \langle T,\overline \nabla_{\overline X}\overline Y \rangle + \frac 12 \langle T,\overline \nabla_{\overline Y} \overline X \rangle, \end{align} which would imply algebraically that $-\langle T, \overline \nabla_{\overline X} \overline Y \rangle = \langle T, \overline \nabla_{\overline Y} \overline X \rangle$, or $\overline \nabla_{\overline Y}\overline X = -\overline \nabla_{\overline X}\overline Y$ if it holds for all $T$. I'm not sure how this would help me get to establishing \begin{align} -\langle T,\overline \nabla_{\overline X} \overline Y \rangle=\langle \overline \nabla_\overline X T,\overline Y \rangle + \langle [T,\overline X],\overline Y\rangle, \end{align} though.
In $(3)$, the first equality is easy because one need only recognize that $[\overline X,\overline Y] = [\overline X,\overline Y]^h+[\overline X,\overline Y]^v$. However, the second equality of $(3)$ is not easy for me to figure out. But I am guessing that I use something similar to the work I did in $(1)$.
This is a long post, so my work can be at times sloppy (though I hope not). Please ask me if there is something that you read that you need me to clarify better.
A direct computation shows for $(1)$: \begin{align*} &\left\langle\overline\nabla_{\overline{X}}\overline\nabla_{\overline{Y}}\overline{Z},\overline{W}\right\rangle \\ &\qquad = \overline{X}\left\langle\overline{\nabla}_{\overline{Y}}\overline{Z} ,\overline{W}\right\rangle - \left\langle\overline\nabla_{\overline{Y}}\overline{Z},\overline\nabla_{\overline{X}}\overline{W}\right\rangle \quad \text{metric compatibility} \\ &\qquad = X\left\langle\nabla_{{Y}}{Z} ,{W}\right\rangle - \left\langle\overline\nabla_{\overline{Y}}\overline{Z},\overline\nabla_{\overline{X}}\overline{W}\right\rangle \quad \text{8.9(c)} \\ &\qquad = X\left\langle\nabla_{{Y}}{Z} ,{W}\right\rangle - \left\langle\overline{\nabla_YZ} + \frac{1}{2}[\overline{Y},\overline{Z}]^v,\overline{\nabla_XW} + \frac{1}{2}[\overline{X},\overline{W}]^v\right\rangle \quad \text{8.9(b)}\\ &\qquad = X\left\langle\nabla_{{Y}}{Z} ,{W}\right\rangle - \left\langle\overline{\nabla_YZ},\overline{\nabla_XW}\right\rangle - \left\langle\overline{\nabla_YZ}, \frac{1}{2}[\overline{X},\overline{W}]^v\right\rangle \\ &\qquad -\left\langle\frac{1}{2}[\overline{Y},\overline{Z}]^v,\overline{\nabla_XW} \right\rangle - \frac{1}{4}\left\langle[\overline{Y},\overline{Z}]^v,[\overline{X},\overline{W}]^v\right\rangle \quad \text{bilinearity of $\langle\cdot,\cdot\rangle$} \\ &\qquad = X\left\langle\nabla_{{Y}}{Z} ,{W}\right\rangle - \left\langle\overline{\nabla_YZ},\overline{\nabla_XW}\right\rangle - \frac{1}{4}\left\langle[\overline{Y},\overline{Z}]^v,[\overline{X},\overline{W}]^v\right\rangle \quad \text{$\langle\overline{X},T\rangle=0$ in 8.9(c)} \\ &\qquad = X\left\langle\nabla_{{Y}}{Z} ,{W}\right\rangle - \left\langle\overline{\nabla_YZ},\overline{\nabla_XW}\right\rangle - \frac{1}{4}\left\langle[\overline{Y},\overline{Z}]^v,[\overline{X},\overline{W}]^v\right\rangle \\ &\qquad = X\left\langle\nabla_{{Y}}{Z} ,{W}\right\rangle - \left\langle{\nabla_YZ},{\nabla_XW}\right\rangle - \frac{1}{4}\left\langle[\overline{Y},\overline{Z}]^v,[\overline{X},\overline{W}]^v\right\rangle \quad \text{definition of horizontal lift}\\ &\qquad = \langle{\nabla_X\nabla_YZ,W}\rangle - \frac{1}{4}\left\langle[\overline{Y},\overline{Z}]^v,[\overline{X},\overline{W}]^v\right\rangle \quad \text{metric compatibility} \end{align*}
The only confusing thing in the above is that there are two different Riemannian metrics in play, and it would have been more clear to explitictly refer to them as $g$ and $\overline{g}.$
Now for $(2)$: \begin{align*} \left\langle\overline{\nabla}_T\overline{X},\overline{Y}\right\rangle& =\left\langle\overline{\nabla}_{\overline{X}}T,\overline{Y}\right\rangle + \left\langle[T,\overline{X}],\overline{Y}\right\rangle \quad \text{torsion free}\\ & = -\left\langle T,\overline{\nabla}_{\overline{X}}\overline{Y} \right\rangle + \overline{X}\left\langle T,\overline{Y} \right\rangle + \left\langle[T,\overline{X}],\overline{Y}\right\rangle \quad \text{metric compatibility}\\ & = -\left\langle T,\overline{\nabla}_{\overline{X}}\overline{Y} \right\rangle + \left\langle[T,\overline{X}],\overline{Y}\right\rangle \\ & = -\left\langle T,\overline{\nabla}_{\overline{X}}\overline{Y} \right\rangle \quad \text{8.9(c)} \end{align*}
Finally, for $(3)$: \begin{align*} \left\langle{\overline\nabla_{[\overline{X},\overline{Y}]}\overline{Z},\overline{W}}\right\rangle & = \left\langle\overline\nabla_{[\overline{X},\overline{Y}]^h}\overline{Z},\overline{W}\right\rangle + \left\langle\overline\nabla_{[\overline{X},\overline{Y}]^v}\overline{Z},\overline{W}\right\rangle \\ & = \left\langle\overline\nabla_{[\overline{X},\overline{Y}]^h}\overline{Z},\overline{W}\right\rangle - \left\langle[\overline{X},\overline{Y}]^v,\overline{\nabla}_{\overline{Z}}\overline{W}\right\rangle \quad \text{by (2)}\\ & = \left\langle\overline\nabla_{[\overline{X},\overline{Y}]^h}\overline{Z},\overline{W}\right\rangle - \left\langle[\overline{X},\overline{Y}]^v,\overline{\nabla_ZW}+\frac{1}{2}[\overline{Z},\overline{W}]^v\right\rangle \\ & = \left\langle\overline\nabla_{[\overline{X},\overline{Y}]^h}\overline{Z},\overline{W}\right\rangle - \frac{1}{2}\left\langle[\overline{X},\overline{Y}]^v,[\overline{Z},\overline{W}]^v\right\rangle \end{align*} Now note that $[\overline{X},\overline{Y}]-\overline{[X,Y]}$ is vertical, and so $([\overline{X},\overline{Y}]-\overline{[X,Y]})^h=0$. In particular, we see that $[\overline{X},\overline{Y}]^h=\overline{[X,Y]}$. We finish the computation for $(3)$: \begin{align*} \left\langle\overline\nabla_{[\overline{X},\overline{Y}]^h}\overline{Z},\overline{W}\right\rangle - \frac{1}{2}\left\langle[\overline{X},\overline{Y}]^v,[\overline{Z},\overline{W}]^v\right\rangle & = \left\langle\overline\nabla_{\overline{[X,Y]}}\overline{Z},\overline{W}\right\rangle - \frac{1}{2}\left\langle[\overline{X},\overline{Y}]^v,[\overline{Z},\overline{W}]^v\right\rangle \\ & = \left\langle\nabla_{{[X,Y]}}{Z},{W}\right\rangle - \frac{1}{2}\left\langle[\overline{X},\overline{Y}]^v,[\overline{Z},\overline{W}]^v\right\rangle. \quad \text{8.9(c)} \end{align*} Note that this final formula in $(3)$ has $\nabla$ instead of $\overline\nabla$, but I think that is how it should be since you can now derive the desired curvature relationship.