This post contains calculations that continue on the ideas from the previous post "Fubini--Study metric," 2017-03-05. First we suppose that $\gamma$ lies on a curve $C\subset \P^2$, with the curve defined as the zero locus of a polynomial $P$. Taking the derivative of $P$ on $\C^2$ gives $P_{z_1}dz_1 + P_{z_2}dz_2=0$, which can be manipulated to give
\begin{align*}
dz_2 & = \frac{-P_{z_1}}{P_{z_2}}dz_1, & \frac\dy{\dy z_2} & = \frac{-P_{z_2}}{P_{z_1}} \frac\dy{\dy z_1},\\
d\overline{z_2} & = \frac{-\overline{P_{z_1}}}{\overline{P_{z_2}}}d\overline{z_1}, & \frac\dy{\dy \overline{z_2}} & = \frac{-\overline{P_{z_2}}}{\overline{P_{z_1}}} \frac\dy{\dy \overline{z_1}}.
\end{align*}
Using the above and an equation from the mentioned post, for $e = \frac\dy{\dy z_1} + \frac\dy{\dy \overline {z_1}} + \frac\dy{\dy z_2} + \frac\dy{\dy \overline{z_2}}$, we get
\begin{align*}
\frac{d \gamma}{dt} & = \left(\overline\gamma_1'-\frac{P_{z_2}}{P_{z_1}}\overline \gamma_2'\right)\frac\dy{\dy z_1} + \left(\gamma_1' - \frac{\overline{P_{z_2}}}{\overline{P_{z_1}}}\gamma_2'\right)\frac\dy{\dy \overline{z_1}} \\
\left(\sum_{k,\ell=1}^2\chi_{k\ell}(\gamma)dz_k\wedge d\overline{z_\ell}\right)(e,e) & = 1+|\gamma_2|^2 + \frac{\overline{P_{z_1}}}{\overline{P_{z_2}}} \overline \gamma_1\gamma_2 + \frac{P_{z_1}}{P_{z_2}}\gamma_1\overline\gamma_2 + \left|\frac{P_{z_1}}{P_{z_2}}\right|^2 \left(1+|\gamma_1|^2\right) = 1 +\left|\frac{P_{z_1}}{P_{z_2}}\right|^2 + \left|\frac{P_{z_1}}{P_{z_2}}\gamma_1+\gamma_2\right|^2, \\
(dz_1\wedge d\overline{z_1})\left(\frac{d\gamma}{dt},I\frac{d\gamma}{dt}\right) & = \det
\begin{bmatrix}
\overline\gamma_1'-\frac{P_{z_2}}{P_{z_1}}\overline \gamma_2' & i\left(\overline\gamma_1'-\frac{P_{z_2}}{P_{z_1}}\overline \gamma_2'\right) \\[5pt]
\gamma_1' - \frac{\overline{P_{z_2}}}{\overline{P_{z_1}}}\gamma_2' & -i\left(\gamma_1' - \frac{\overline{P_{z_2}}}{\overline{P_{z_1}}}\gamma_2'\right)
\end{bmatrix} = -2i \left|\gamma_1' - \frac{\overline{P_{z_2}}}{\overline{P_{z_1}}}\gamma_2'\right|^2.
\end{align*}
Hence
\[
g\left(\frac{d\gamma}{dt},\frac{d\gamma}{dt}\right) = \frac{\left(1 +\left|\frac{P_{z_1}}{P_{z_2}}\right|^2 + \left|\frac{P_{z_1}}{P_{z_2}}\gamma_1+\gamma_2\right|^2\right)\left|\gamma_1' - \frac{\overline{P_{z_2}}}{\overline{P_{z_1}}}\gamma_2'\right|^2}{\pi\left(1+|\gamma_1|^2+|\gamma_2|^2\right)^2}.
\]
Now we move to $\P^n$, and consider $X\subset \P^n$ a complete intersection of codimension $r$, or the zero set of polynomials $P_1=0,\dots,P_r=0$. Expressing some covectors in terms of others reduces the number of determinants we calculated above from $2n$ to $2(n-r)$. Then
\begin{align*}
P_{1,z_1}dz_1 + \cdots + P_{1,z_n}dz_n & = 0, & dz_n & = c_{n,1}dz_1 + \cdots + c_{n,n-r}dz_{n-r}, \\
& \ \ \vdots & & \ \ \vdots \\
P_{r,z_1}dz_1 + \cdots + P_{r,z_n}dz_n & = 0, & dz_{n-r+1} & = c_{n-r+1,1}dz_1 + \cdots + c_{n-r+1,n-r}dz_{n-r},
\end{align*}
for the $c_{i,j}$ some combinations of the $P_{k,z_\ell}$. By orthonormality of the basis vectors, and assuming that the $c_{i,j}$ are all non-zero, we find
\[
\frac\dy{\dy z_i} = \sum_{j=1}^{n-r} \frac1{(n-r)c_{i,j}}\frac\dy{\dy z_j},\hspace{2cm}
\frac\dy{\dy \overline{z_i}} = \sum_{j=1}^{n-r} \frac1{(n-r)\overline{c_{i,j}}}\frac\dy{\dy \overline{z_j}},
\]
for all integers $n-r<i\leqslant n$. This allows us to rewrite the path derivative as
\begin{align*}
\frac{d\gamma}{dt} & = \sum_{i=1}^n \overline \gamma_i'\frac\dy{\dy z_i} +\gamma_i'\frac\dy{\dy \overline{z_i}} \\
& = \sum_{i=1}^{n-r} \left(\overline \gamma_i'\frac\dy{\dy z_i} +\gamma_i'\frac\dy{\dy \overline{z_i}}\right) +\sum_{i=n-r+1}^n \left(\sum_{j=1}^{n-r} \frac{\overline \gamma_i'}{(n-r)c_{i,j}}\frac\dy{\dy z_j} + \sum_{j=1}^{n-r} \frac{\gamma_i'}{(n-r)\overline{c_{i,j}}}\frac\dy{\dy \overline{z_j}}\right) \\
& = \sum_{i=1}^{n-r}\left(\overline\gamma_i' + \sum_{j=n-r+1}^n \frac{\overline\gamma_j'}{(n-r)c_{j,i}}\right)\frac\dy{\dy z_i} + \left(\gamma_i'+\sum_{j=n-r+1}^n \frac{\gamma_j'}{(n-r)\overline{c_{j,i}}}\right)\frac\dy{\dy \overline{z_i}}.
\end{align*}
In the case of a curve in $\P^n$, when $r=n-1$, let $c_{1,1}=1$ and $e = \frac\dy{\dy z_1} + \frac\dy{\dy \overline {z_1}} + \cdots + \frac\dy{\dy z_n} + \frac\dy{\dy \overline{z_n}}$ to get
\begin{align*}
\frac{d\gamma}{dt} & = \left(\sum_{j=1}^n \frac{\overline\gamma_j'}{c_{j1}}\right)\frac\dy{\dy z_1} + \left(\sum_{j=1}^n \frac{\gamma_j'}{\overline{c_{j1}}}\right)\frac\dy{\dy \overline{z_1}},\\
\left(\sum_{k,\ell=1}^n\chi_{k\ell}(\gamma)dz_k\wedge d\overline{z_\ell}\right)(e,e) & = \sum_{k,\ell=1}^n \left(1+\sum_{i=1}^n |\gamma_i|^2\right)\delta_{k\ell} - \overline{\gamma_kc_{\ell1}}\gamma_\ell c_{k1}, \\
(dz_1\wedge d\overline{z_1})\left(\frac{d\gamma}{dt},I\frac{d\gamma}{dt}\right) & = \det
\begin{bmatrix}
\sum_{j=1}^n \frac{\overline\gamma_j'}{c_{j1}} & i \sum_{j=1}^n \frac{\overline\gamma_j'}{c_{j1}} \\[5pt]
\sum_{j=1}^n \frac{\gamma_j'}{\overline{c_{j1}}} & -i\sum_{j=1}^n \frac{\gamma_j'}{\overline{c_{j1}}}
\end{bmatrix} = -2i \left|\sum_{j=1}^n \frac{\gamma_j'}{\overline{c_{j1}}}\right|^2.
\end{align*}
Hence
\[
g\left(\frac{d\gamma}{dt},\frac{d\gamma}{dt}\right) = \frac{\left(\sum_{k,\ell=1}^n \left(1+\sum_{i=1}^n |\gamma_i|^2\right)\delta_{k\ell} - \overline{\gamma_kc_{\ell1}}\gamma_\ell c_{k1}\right)\left|\sum_{j=1}^n \frac{\gamma_j'}{\overline{c_{j1}}}\right|^2}{\pi \left(1+\sum_{i=1}^n |\gamma_i|^2\right)^2}.
\]
The terms $\overline{\gamma_kc_{\ell1}}\gamma_\ell c_{k1}$ may be rearranged into terms $|\gamma_kc_{\ell1}-\gamma_\ell c_{k1}|^2$, but it does not provide any enlightening results, similarly to the rest of this post.
\begin{align*}
dz_2 & = \frac{-P_{z_1}}{P_{z_2}}dz_1, & \frac\dy{\dy z_2} & = \frac{-P_{z_2}}{P_{z_1}} \frac\dy{\dy z_1},\\
d\overline{z_2} & = \frac{-\overline{P_{z_1}}}{\overline{P_{z_2}}}d\overline{z_1}, & \frac\dy{\dy \overline{z_2}} & = \frac{-\overline{P_{z_2}}}{\overline{P_{z_1}}} \frac\dy{\dy \overline{z_1}}.
\end{align*}
Using the above and an equation from the mentioned post, for $e = \frac\dy{\dy z_1} + \frac\dy{\dy \overline {z_1}} + \frac\dy{\dy z_2} + \frac\dy{\dy \overline{z_2}}$, we get
\begin{align*}
\frac{d \gamma}{dt} & = \left(\overline\gamma_1'-\frac{P_{z_2}}{P_{z_1}}\overline \gamma_2'\right)\frac\dy{\dy z_1} + \left(\gamma_1' - \frac{\overline{P_{z_2}}}{\overline{P_{z_1}}}\gamma_2'\right)\frac\dy{\dy \overline{z_1}} \\
\left(\sum_{k,\ell=1}^2\chi_{k\ell}(\gamma)dz_k\wedge d\overline{z_\ell}\right)(e,e) & = 1+|\gamma_2|^2 + \frac{\overline{P_{z_1}}}{\overline{P_{z_2}}} \overline \gamma_1\gamma_2 + \frac{P_{z_1}}{P_{z_2}}\gamma_1\overline\gamma_2 + \left|\frac{P_{z_1}}{P_{z_2}}\right|^2 \left(1+|\gamma_1|^2\right) = 1 +\left|\frac{P_{z_1}}{P_{z_2}}\right|^2 + \left|\frac{P_{z_1}}{P_{z_2}}\gamma_1+\gamma_2\right|^2, \\
(dz_1\wedge d\overline{z_1})\left(\frac{d\gamma}{dt},I\frac{d\gamma}{dt}\right) & = \det
\begin{bmatrix}
\overline\gamma_1'-\frac{P_{z_2}}{P_{z_1}}\overline \gamma_2' & i\left(\overline\gamma_1'-\frac{P_{z_2}}{P_{z_1}}\overline \gamma_2'\right) \\[5pt]
\gamma_1' - \frac{\overline{P_{z_2}}}{\overline{P_{z_1}}}\gamma_2' & -i\left(\gamma_1' - \frac{\overline{P_{z_2}}}{\overline{P_{z_1}}}\gamma_2'\right)
\end{bmatrix} = -2i \left|\gamma_1' - \frac{\overline{P_{z_2}}}{\overline{P_{z_1}}}\gamma_2'\right|^2.
\end{align*}
Hence
\[
g\left(\frac{d\gamma}{dt},\frac{d\gamma}{dt}\right) = \frac{\left(1 +\left|\frac{P_{z_1}}{P_{z_2}}\right|^2 + \left|\frac{P_{z_1}}{P_{z_2}}\gamma_1+\gamma_2\right|^2\right)\left|\gamma_1' - \frac{\overline{P_{z_2}}}{\overline{P_{z_1}}}\gamma_2'\right|^2}{\pi\left(1+|\gamma_1|^2+|\gamma_2|^2\right)^2}.
\]
Now we move to $\P^n$, and consider $X\subset \P^n$ a complete intersection of codimension $r$, or the zero set of polynomials $P_1=0,\dots,P_r=0$. Expressing some covectors in terms of others reduces the number of determinants we calculated above from $2n$ to $2(n-r)$. Then
\begin{align*}
P_{1,z_1}dz_1 + \cdots + P_{1,z_n}dz_n & = 0, & dz_n & = c_{n,1}dz_1 + \cdots + c_{n,n-r}dz_{n-r}, \\
& \ \ \vdots & & \ \ \vdots \\
P_{r,z_1}dz_1 + \cdots + P_{r,z_n}dz_n & = 0, & dz_{n-r+1} & = c_{n-r+1,1}dz_1 + \cdots + c_{n-r+1,n-r}dz_{n-r},
\end{align*}
for the $c_{i,j}$ some combinations of the $P_{k,z_\ell}$. By orthonormality of the basis vectors, and assuming that the $c_{i,j}$ are all non-zero, we find
\[
\frac\dy{\dy z_i} = \sum_{j=1}^{n-r} \frac1{(n-r)c_{i,j}}\frac\dy{\dy z_j},\hspace{2cm}
\frac\dy{\dy \overline{z_i}} = \sum_{j=1}^{n-r} \frac1{(n-r)\overline{c_{i,j}}}\frac\dy{\dy \overline{z_j}},
\]
for all integers $n-r<i\leqslant n$. This allows us to rewrite the path derivative as
\begin{align*}
\frac{d\gamma}{dt} & = \sum_{i=1}^n \overline \gamma_i'\frac\dy{\dy z_i} +\gamma_i'\frac\dy{\dy \overline{z_i}} \\
& = \sum_{i=1}^{n-r} \left(\overline \gamma_i'\frac\dy{\dy z_i} +\gamma_i'\frac\dy{\dy \overline{z_i}}\right) +\sum_{i=n-r+1}^n \left(\sum_{j=1}^{n-r} \frac{\overline \gamma_i'}{(n-r)c_{i,j}}\frac\dy{\dy z_j} + \sum_{j=1}^{n-r} \frac{\gamma_i'}{(n-r)\overline{c_{i,j}}}\frac\dy{\dy \overline{z_j}}\right) \\
& = \sum_{i=1}^{n-r}\left(\overline\gamma_i' + \sum_{j=n-r+1}^n \frac{\overline\gamma_j'}{(n-r)c_{j,i}}\right)\frac\dy{\dy z_i} + \left(\gamma_i'+\sum_{j=n-r+1}^n \frac{\gamma_j'}{(n-r)\overline{c_{j,i}}}\right)\frac\dy{\dy \overline{z_i}}.
\end{align*}
In the case of a curve in $\P^n$, when $r=n-1$, let $c_{1,1}=1$ and $e = \frac\dy{\dy z_1} + \frac\dy{\dy \overline {z_1}} + \cdots + \frac\dy{\dy z_n} + \frac\dy{\dy \overline{z_n}}$ to get
\begin{align*}
\frac{d\gamma}{dt} & = \left(\sum_{j=1}^n \frac{\overline\gamma_j'}{c_{j1}}\right)\frac\dy{\dy z_1} + \left(\sum_{j=1}^n \frac{\gamma_j'}{\overline{c_{j1}}}\right)\frac\dy{\dy \overline{z_1}},\\
\left(\sum_{k,\ell=1}^n\chi_{k\ell}(\gamma)dz_k\wedge d\overline{z_\ell}\right)(e,e) & = \sum_{k,\ell=1}^n \left(1+\sum_{i=1}^n |\gamma_i|^2\right)\delta_{k\ell} - \overline{\gamma_kc_{\ell1}}\gamma_\ell c_{k1}, \\
(dz_1\wedge d\overline{z_1})\left(\frac{d\gamma}{dt},I\frac{d\gamma}{dt}\right) & = \det
\begin{bmatrix}
\sum_{j=1}^n \frac{\overline\gamma_j'}{c_{j1}} & i \sum_{j=1}^n \frac{\overline\gamma_j'}{c_{j1}} \\[5pt]
\sum_{j=1}^n \frac{\gamma_j'}{\overline{c_{j1}}} & -i\sum_{j=1}^n \frac{\gamma_j'}{\overline{c_{j1}}}
\end{bmatrix} = -2i \left|\sum_{j=1}^n \frac{\gamma_j'}{\overline{c_{j1}}}\right|^2.
\end{align*}
Hence
\[
g\left(\frac{d\gamma}{dt},\frac{d\gamma}{dt}\right) = \frac{\left(\sum_{k,\ell=1}^n \left(1+\sum_{i=1}^n |\gamma_i|^2\right)\delta_{k\ell} - \overline{\gamma_kc_{\ell1}}\gamma_\ell c_{k1}\right)\left|\sum_{j=1}^n \frac{\gamma_j'}{\overline{c_{j1}}}\right|^2}{\pi \left(1+\sum_{i=1}^n |\gamma_i|^2\right)^2}.
\]
The terms $\overline{\gamma_kc_{\ell1}}\gamma_\ell c_{k1}$ may be rearranged into terms $|\gamma_kc_{\ell1}-\gamma_\ell c_{k1}|^2$, but it does not provide any enlightening results, similarly to the rest of this post.