In this post we inspect how the Fubini-Study metric works and compute an example. Professor Mihai Paun for helpful discussions. Recall that from projective space $\P^n$ there are natural maps
\[[x_0:x_1:\cdots:x_n]\tov{\vp_i}\left(\frac{x_0}{x_i},\dots,\widehat{\frac{x_i}{x_i}},\dots,\frac{x_n}{x_i}\right)
\]
for $i=0,\dots,n$. The maps land in $\C^n$ with coordinates $(z_1,z_2,\dots,z_n)$. We use $\vp_0$ as the main map, and conflate notation for objects in $\P^n$ and in $\C^n$ under $\vp_0$. Most of this post deals with the $n=2$ case.
The metric
The metric used on $\P^n$ is the Fubini-Study metric. Directly from Section 3.1 of Huybrechts, for $n=2$ the associated differential 2-form and its image in $\C^2$ are
\begin{align*}
\omega & = \frac i{2\pi}\partial \bar\partial \log\left(1+\left|\frac{x_1}{x_0}\right|^2+\left|\frac{x_2}{x_0}\right|^2\right), \\
\vp_0(\omega) & = \frac i{2\pi}\partial \bar\partial \log\left(1+\left|z_1\right|^2+\left|z_2\right|^2\right) \\
& = \underbrace{\frac{i}{2\pi (1+|z_1|^2+|z_2|^2)^2}}_{\lambda_2}\sum_{k,\ell=1}^2\underbrace{(1+|z_1|^2+|z_2|^2)\delta_{k\ell} -\overline{z_k}z_\ell}_{\chi_{k\ell}}dz_k\wedge d\overline{z_\ell}. \hspace{1cm} (1)
\end{align*}
A Hermitian metric on a complex manifold $X$ may be described as a 2-tensor $h=g-i\omega$, where $g$ is a Riemannian metric (also a 2-tensor) on the underlying real manifold and $\omega$ is a Kahler form, a 2-form. As in Lemma 3.3 of Voisin, the relationship between $g$ and $\omega$ is given by
\[
g(u,v)=\omega(u,Iv)=\omega(Iu,v), \hspace{1cm} (2)
\]
where $I:T_xX\to T_xX$ is a tangent space endomorphism defined by
\[
\begin{array}{r c l}
I|_{T^{1,0}_xX} & = & i\cdot \id, \\
\frac{\dy}{\dy z_i} & \mapsto & i\frac{\dy}{\dy z_i},
\end{array}
\hspace{1cm}
\begin{array}{r c l}
I|_{T^{0,1}_xX} & = & -i\cdot \id, \\
\frac{\dy}{\dy \overline{z_i}} & \mapsto & -i\frac{\dy}{\dy \overline{z_i}},
\end{array}
\]
as in Proposition 1.3.1 of Huybrechts.
An application
Let $\gamma:[0,1]\to \C^2$ be a path, described as $\gamma(t)=(\gamma_1(t),\gamma_2(t))$. The derivative of $\gamma$ with respect to $t$, in the basis $\frac{\dy}{\dy z_1}$, $\frac{\dy}{\dy \overline{z_1}}$, $\frac{\dy}{\dy z_2}$, $\frac{\dy}{\dy \overline{z_2}}$ is given by
\[
\frac{d\gamma_1}{dt} = \frac{du_1}{dt}\frac\dy{\dy x_1} + i\frac{dv_1}{dt}\frac\dy{\dy y_1} = \frac{du_1}{dt}\left(\frac\dy{\dy \overline{z_1}}+\frac\dy{\dy z_1}\right) + i\frac{dv_1}{dt}\left(\frac\dy{\dy \overline{z_1}} -\frac{\dy}{\dy z_1}\right) = \underbrace{\left(\frac{du_1}{dt} + i\frac{dv_1}{dt}\right)}_{\gamma_1'}\frac\dy{\dy \overline {z_1}} + \underbrace{\left(\frac{du_1}{dt}-i\frac{dv_1}{dt}\right)}_{\overline \gamma_1'}\frac\dy{\dy z_1},\]
and analogously for $\gamma_2$. Hence
\[
\frac{d\gamma}{dt} =
\overline \gamma_1'\frac\dy{\dy z_1} + \gamma_1'\frac\dy{\dy \overline{z_1}} + \overline \gamma_2' \frac\dy{\dy z_2} + \gamma_2' \frac{\dy}{\dy \overline{z_2}}. \hspace{1cm} (3)
\]
The length of $\gamma$ is
\[
\int_0^1\sqrt{g\left(\frac{d\gamma}{dt},\frac{d\gamma}{dt}\right)}\ dt = \int_0^1\sqrt{\omega\left(\frac{d\gamma}{dt},I\frac{d\gamma}{dt}\right)}\ dt,
\]
using equation (2). Recall that the pairing of vectors with covectors is given by\[
\left(d\alpha_1\wedge \cdots \wedge d\alpha_n\right)\left(\frac\dy{\dy \beta_1},\dots,\frac\dy{\dy \beta_n}\right) = \det\begin{bmatrix}
d\alpha_1\frac\dy{\dy \beta_1} & d\alpha_1\frac{\dy}{\dy \beta_2} & \cdots & d\alpha_1\frac\dy{\dy \beta_n} \\
d\alpha_2\frac\dy{\dy \beta_1} & d\alpha_2\frac{\dy}{\dy \beta_2} & \cdots & d\alpha_2\frac\dy{\dy \beta_n} \\
\vdots & \vdots & \ddots & \vdots \\
d\alpha_n\frac\dy{\dy \beta_1} & d\alpha_n\frac{\dy}{\dy \beta_2} & \cdots & d\alpha_n\frac\dy{\dy \beta_n}
\end{bmatrix}
\ \ = \ \
\det\left(d\alpha_i\frac\dy{\dy \beta_j}\right),
\]
for $\alpha_i,\beta_j$ a basis of the underlying real manifold (as in the previous post "Vector fields," 2016-10-10). The components of the vector (3) may be viewed as given in directions $z_1,\overline{z_1}, z_2,\overline{z_2}$, respectively, which also indicates how the coefficient functions $\chi_{k\ell}$ act on (3). Apply the definition of $\omega$ from equation (1), and note that we are always at the tangent space to the point $\gamma(t)=(\gamma_1(t),\gamma_2(t))$, to get that
\begin{align*}
& \omega\left(\frac{d \gamma}{dt},I\frac{d\gamma}{dt}\right) \\
& = \lambda_2(\gamma(t)) \sum_{k,\ell=1}^2 \chi_{k\ell}(\gamma(t)) dz_k\wedge d\overline{z_\ell}\left(\overline \gamma_1'\frac\dy{\dy z_1} + \gamma_1'\frac\dy{\dy \overline{z_1}} + \overline \gamma_2' \frac\dy{\dy z_2} + \gamma_2' \frac{\dy}{\dy \overline{z_2}}, i\overline \gamma_1'\frac\dy{\dy z_1} - i\gamma_1'\frac\dy{\dy \overline{z_1}} + i\overline \gamma_2' \frac\dy{\dy z_2} -i\gamma_2' \frac{\dy}{\dy \overline{z_2}}\right) \\
& = \lambda_2(\gamma(t)) \sum_{k,\ell=1}^2 \chi_{k\ell}(\gamma(t))\det
\begin{bmatrix}
\overline \gamma_k'(t) & i\overline \gamma_k'(t) \\[5pt] \gamma_\ell'(t) & -i\gamma_\ell'(t)
\end{bmatrix} \\
& = \frac{(1+|\gamma_2(t)|^2)|\gamma_1'(t)|^2 - \overline\gamma_1(t)\gamma_2(t)\overline\gamma_1'(t)\gamma_2'(t) - \overline\gamma_2(t)\gamma_1(t)\overline \gamma_2'(t)\gamma_1'(t) + (1+|\gamma_1(t)|^2) |\gamma_2'(t)|^2}{\pi\left(1+\left|\gamma_1(t)\right|^2+\left|\gamma_2(t)\right|^2\right)^2}.\end{align*}
Unfortunately this expression does not simplify too much. In $\P^n$, with $\gamma = (\gamma_1,\dots,\gamma_n):[0,1]\to \C^n$, we have that
\[
g\left(\frac{d \gamma}{dt},\frac{d\gamma}{dt}\right) = \lambda_n(\gamma(t)) \sum_{k,\ell=1}^n \chi_{k\ell}(\gamma(t))\det
\begin{bmatrix}
\overline \gamma_k'(t) & i\overline \gamma_k'(t) \\[5pt] \gamma_\ell'(t) & -i\gamma_\ell'(t)
\end{bmatrix}.
\]
\left(d\alpha_1\wedge \cdots \wedge d\alpha_n\right)\left(\frac\dy{\dy \beta_1},\dots,\frac\dy{\dy \beta_n}\right) = \det\begin{bmatrix}
d\alpha_1\frac\dy{\dy \beta_1} & d\alpha_1\frac{\dy}{\dy \beta_2} & \cdots & d\alpha_1\frac\dy{\dy \beta_n} \\
d\alpha_2\frac\dy{\dy \beta_1} & d\alpha_2\frac{\dy}{\dy \beta_2} & \cdots & d\alpha_2\frac\dy{\dy \beta_n} \\
\vdots & \vdots & \ddots & \vdots \\
d\alpha_n\frac\dy{\dy \beta_1} & d\alpha_n\frac{\dy}{\dy \beta_2} & \cdots & d\alpha_n\frac\dy{\dy \beta_n}
\end{bmatrix}
\ \ = \ \
\det\left(d\alpha_i\frac\dy{\dy \beta_j}\right),
\]
for $\alpha_i,\beta_j$ a basis of the underlying real manifold (as in the previous post "Vector fields," 2016-10-10). The components of the vector (3) may be viewed as given in directions $z_1,\overline{z_1}, z_2,\overline{z_2}$, respectively, which also indicates how the coefficient functions $\chi_{k\ell}$ act on (3). Apply the definition of $\omega$ from equation (1), and note that we are always at the tangent space to the point $\gamma(t)=(\gamma_1(t),\gamma_2(t))$, to get that
\begin{align*}
& \omega\left(\frac{d \gamma}{dt},I\frac{d\gamma}{dt}\right) \\
& = \lambda_2(\gamma(t)) \sum_{k,\ell=1}^2 \chi_{k\ell}(\gamma(t)) dz_k\wedge d\overline{z_\ell}\left(\overline \gamma_1'\frac\dy{\dy z_1} + \gamma_1'\frac\dy{\dy \overline{z_1}} + \overline \gamma_2' \frac\dy{\dy z_2} + \gamma_2' \frac{\dy}{\dy \overline{z_2}}, i\overline \gamma_1'\frac\dy{\dy z_1} - i\gamma_1'\frac\dy{\dy \overline{z_1}} + i\overline \gamma_2' \frac\dy{\dy z_2} -i\gamma_2' \frac{\dy}{\dy \overline{z_2}}\right) \\
& = \lambda_2(\gamma(t)) \sum_{k,\ell=1}^2 \chi_{k\ell}(\gamma(t))\det
\begin{bmatrix}
\overline \gamma_k'(t) & i\overline \gamma_k'(t) \\[5pt] \gamma_\ell'(t) & -i\gamma_\ell'(t)
\end{bmatrix} \\
& = \frac{(1+|\gamma_2(t)|^2)|\gamma_1'(t)|^2 - \overline\gamma_1(t)\gamma_2(t)\overline\gamma_1'(t)\gamma_2'(t) - \overline\gamma_2(t)\gamma_1(t)\overline \gamma_2'(t)\gamma_1'(t) + (1+|\gamma_1(t)|^2) |\gamma_2'(t)|^2}{\pi\left(1+\left|\gamma_1(t)\right|^2+\left|\gamma_2(t)\right|^2\right)^2}.\end{align*}
Unfortunately this expression does not simplify too much. In $\P^n$, with $\gamma = (\gamma_1,\dots,\gamma_n):[0,1]\to \C^n$, we have that
\[
g\left(\frac{d \gamma}{dt},\frac{d\gamma}{dt}\right) = \lambda_n(\gamma(t)) \sum_{k,\ell=1}^n \chi_{k\ell}(\gamma(t))\det
\begin{bmatrix}
\overline \gamma_k'(t) & i\overline \gamma_k'(t) \\[5pt] \gamma_\ell'(t) & -i\gamma_\ell'(t)
\end{bmatrix}.
\]
An example
Here we compute the distance between two points in $\P^2$. Let $\gamma$ be the straight line segment connecting $p=[p_0:p_1:p_2]$ and $q=[q_0:q_1:q_2]$. The word "straight" is used loosely, and means the segment may be parametrized as
\[
\gamma(t) = [(1-t)p_0+tq_0:(1-t)p_1+tq_1:(1-t)p_2+tq_2],
\]
so $\gamma(0)=p$ and $\gamma(1)=q$. The image of $\gamma$ under $\vp_0$ and its derivative are given by
\[
\vp_0(\gamma(t)) = \left(\frac{(1-t)p_1+tq_1}{(1-t)p_0+tq_0}, \frac{(1-t)p_2+tq_2}{(1-t)p_0+tq_0}\right) = (\gamma_1,\gamma_2),
\hspace{2cm}
\gamma_i' = \frac{q_ip_0-q_0p_i}{((1-t)p_0+tq_0)^2}.
\]
If, for example, $p=[1:1:0]$ and $q=[1:0:1]$, then
\[
\text{length}(\gamma) = \frac{3}{4\pi}\int_0^1\frac1{(t^2-t+1)^2}\ dt = \frac{9+2\pi\sqrt 3}{18\pi}.
\]
A further goal is to consider the path $\gamma$ as lying on a projective variety, beginning with a complete intersection. This would allow some of the $dz_i$ to be expressed in terms of other $dz_j$.
References: Huybrechts (Complex geometry, Section 3.1), Voisin (Hodge theory and complex algebraic geometry 1, Chapter 3.1), Wells (Differential analysis on complex manifolds, Chapter V.4)
No comments:
Post a Comment