Define a Riemannian metric on $\mathbb C^{n+1} - \{0\}$ in the following way: If $Z \in \mathbb C^{n+1}-\{0\}$ and $V,W \in T_Z(\mathbb C^{n+1}-\{0\})$, $$\langle V,W \rangle_Z = \frac{\text{Real}(V,W)}{(Z,Z)}.$$ Observe that hte mteric $\langle \,,\,\rangle$ restricted to $S^{2n+1} \subset \mathbb C^{n+1}-\{0\}$ coincides with the metric induced from $\mathbb R^{2n+2}$.
(b) Show that, in this metric, the sectional curvature of $\mathbb P^n(\mathbb C)$ is given by $$K(\sigma)=1+3\cos^2 \varphi,$$ where $\sigma$ is generated by the orthonormal pair $X,Y, \cos \varphi=\left\langle \overline X,i\overline Y\right\rangle$, and $\overline X,\overline Y$ are the horizontal lifts of $X$ and $Y$, respectively. In particular, $1 \le K(\sigma) \le 4$.
This is Exercise 8.12 of Riemannian Geometry by do Carmo. Yes, there is a part (a) but I did not print it here because my question here concerns only part (b).
Also, I realize that $P^n(\mathbb C)$ is the complex projective space of complex dimension $n$ and is denoted in other textbooks as $\mathbb CP^n$.
Hint for (b): Let $Z$ be the position vector describing $S^{2n+1}$. Since $(\frac d{d\theta}e^{i\theta}Z)_{\theta=0}=iZ$, $iZ \in T_Z(S^{2n+1})$ and is vertical. Let $\overline \nabla$ be the Riemannian connection of $\mathbb R^{2n+2} \approx \mathbb C^{n+1}$ and $X,Y \in \mathcal X(P^n(\mathbb C))$. Take $\alpha : (-\epsilon,\epsilon) \to S^{2n+1}$ with $\alpha(0)=Z$, $\alpha'(0)-\overline X$. Then \begin{align} (\overline \nabla_{\overline X}iZ)_Z &= \frac d{dt} iZ \circ \alpha(t))\Big\vert_{t=0} \\ &= \frac d{dt} i\alpha(t) \Big\vert_{t=0} = i\alpha'(0)=i\overline X. \end{align} Therefore, \begin{align} \left\langle [\overline X,\overline Y],iZ \right\rangle &= \left\langle \overline \nabla_{\overline X}\overline Y-\overline \nabla_{\overline Y}\overline X,iZ \right\rangle \\ &= -\left\langle i\overline X,\overline Y \right\rangle + \left\langle i\overline Y,\overline X \right\rangle = 2\cos \varphi. \end{align} Now use Exercise 10(b).
I am trying to provide the details of the last line: "Now use Exercise 10(b)". Exercise 8.10(b) merely asserts O'Neill's formula:
$$ K(\sigma)=\overline K(\overline \sigma) + \frac 34 \left|[\overline X,\overline Y]^v \right|^2 $$
So my approach is as follows: $$ K(\sigma)=\overline K(\overline \sigma) + \frac 34 \left|[\overline X,\overline Y]^v \right|^2 = 1 + \frac 34 \left|[\overline X,\overline Y]^v \right|^2 $$ and $$ 1+3\cos^2 \varphi = 1+\frac 34 |2 \cos \varphi|^2 = 1+\frac 34 \left| \left\langle [\overline X,\overline Y],iZ \right\rangle \right|^2. $$ But I am not sure how to link these together by establishing $$ \left|[\overline X,\overline Y]^v \right| = \left| \left\langle [\overline X,\overline Y],iZ \right\rangle \right|. $$ Clearly, $[\overline X,\overline Y]^v$ is a vertical field, so $\left\langle [\overline X,\overline Y],iZ \right\rangle$ has to be vertical as well?
No, $\langle [\overline{X},\overline{Y}], iZ\rangle$ is not vertical - it is not even a vector field!
The claim (which I leave to you) that you need to establish to link the two is the following:
Claim: $iZ$ is a unit vector which spans the vertical space.
Believeing this claim, the point is that $\left(\left\langle [\overline{X},\overline{Y}], iZ\right\rangle\right) i Z$ is the projection of $[\overline{X},\overline{Y}]$ to the vertical subspace. That is, $\left(\left\langle [\overline{X},\overline{Y}], iZ\right\rangle\right) i Z = [\overline{X},\overline{Y}]^v.$ Taking lengths of both sides of this equality gives the equality you need.