Lengths of paths on projective varieties

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.

Optimal sampling and arrangement on an n-sphere

The goal of this post is to create a "good" algorithm for sampling and arranging points on the $n$-sphere. We find the $\epsilon$-covering number of the $n$-sphere and arrange the points in a Hamiltonian path of small pairwise consecutive distance. This post relates to several previous posts:

• 2017-02-12: Generalizing planar detection to $k$-plane detection
• 2016-12-22: Sampling points uniformly on parametrized manifolds
• 2016-05-26: Reconstructing a manifold from sample data, with noise

Thanks to Professor Cheng Ouyang for a helpful discussion.

Although rejection sampling is a standard method to sample points uniformly on the $n$-sphere (sample points uniformly on the $(n+1)$-cube, check if the norm is less than or equal to 1, if it is, normalize the point to the $n$-sphere), this is not best for our scenario (the arranging part). A better suited approach is to take a parametrization $f$ from an $n$-cube into $\R^{n+1}$ of the unit $n$-sphere. We use
\[
\begin{array}{r c l}
f\ :\ [0,2\pi]^{n-1}\times[0,\pi) & \to & \R^{n+1}, \\
(\alpha_1,\dots,\alpha_n) & \mapsto & \big(\cos(\alpha_1), \\ && \sin(\alpha_1)\cos(\alpha_2),\\ && \vdots\\ && \sin(\alpha_1)\cdots\sin(\alpha_{n-1})\cos(\alpha_n), \\ && \sin(\alpha_1)\cdots\sin(\alpha_{n-1})\sin(\alpha_n)\big).
\end{array}
\]
Adapting the main Proposition from the "Sampling points" post, we have following proposition.

Proposition: The probability density function $g_n:[0,2\pi]^{n-1}\times[0,\pi] \to \R_{\geqslant0}$, defined as
\[
g_n(\alpha_1,\dots,\alpha_n)=\frac{\prod_{k=1}^{n-1}|\sin^{n-k}(\alpha_k)|}{2^{n-1}\pi\prod_{k=1}^{n-1}\int_0^\pi \sin^{n-k}(\alpha_k)\ d\alpha_k},
\]
is uniform on the natural embedding of the unit $n$-sphere $S^n$ in $\R^{n+1}$.

The denominator of $g_n$ does not seem to have closed form, though the ratios between consecutive terms are given by the denominators of $\Gamma(\frac{\ell+3}2)/\Gamma(\frac{\ell+2}2)$ and $\ell!!/(\ell+1)!!$, with appropriate powers of $\pi$. The first few terms of this sequence are
\[
4\pi,4\pi^2,\frac{32}3\pi^2,8\pi^3,\frac{256}{15}\pi^3,\frac{32}3\pi^4,\frac{2048}{105}\pi^4,\dots.
\]
Next, recall the $n$-surface of an $n$-sphere and $k$-volume of a $k$-ball are
\[
\text{surf}(n,r) = \frac{2\pi^{(n+1)/2}r^n}{\Gamma((n+1)/2)},\hspace{2cm}
\text{vol}(k,r) = \frac{\pi^{k/2}r^k}{\Gamma((k+2)/2)}.
\]
Adapting Proposition 3.2 of Niyogi, Smale and Weinberger, similarly to the "Reconstructing a manifold" post, we have the following proposition.

Proposition: A collection of $N$ points sampled uniformly from $S^n$ is $\epsilon$-dense in $S^n$ with certainty $1-\delta$, given
\[
N \geqslant \frac{\text{surf}(n,1)}{(1-\frac{\epsilon^2}{16})^{n/2}\text{vol}(n,\frac\epsilon2)}\log\left(\frac{\text{surf}(n,1)}{\delta(1-\frac{\epsilon^2}{64})^{n/2}\text{vol}(n,\frac\epsilon4)}\right).
\] Bauer and Polthier sample points "evenly" on the 2-hemisphere and then connect them with a winding path, which winds around the hemisphere 6 times. Generalizing this approach, suppose we wanted to have a path that wind around the $n$-sphere $\ell$ times and has a small distance between consecutive vertices of the path. The following algorithm describes one way of doing this.

---

Algorithm: SpherePathFinder
Input: Positive integers $n,\ell$ and real numbers $\epsilon,\delta\in (0,1)$
Output: A path on $S^n$ that winds around $\ell$ times, whose vertices are $\epsilon$-dense on $S^n$ with certainty $1-\delta$

Sample $\lceil N\rceil$ points on $[0,2\pi]^{n-1}\times[0,\pi]$ according to $g_n$ in a set $X$
Initiate an empty path $P=()$
for $k_n\in\{1,\dots,\ell\}$:
    for $k_{n-1}\in\{1,\dots,2\ell\}$:
       $\vdots$
           for $k_2\in\{1,\dots,2\ell\}$:
               Set $L=\{\alpha\in X\ :\ \alpha_n\in[(k_n-1)\frac\pi\ell,k_n\frac\pi\ell], \alpha_{n-t}\in[(k_{n-t}-1)\frac{2\pi}{2\ell},k_{n-t}\frac{2\pi}{2\ell}],1<t<n-1\}$
               Order $L$ by increasing values of $\alpha_1$
               Append $L$ to the end of $P$ and set $X=X\setminus L$
Return $P$

---

Since the sample space is $[0,2\pi]^{n-1}\times[0,\pi]$, finding the appropriate points in the nested for loop is very easy. We conclude with an experimental example with $n=2$, $\ell=12$, $\epsilon=.1$, and $\delta=.01$. We must sample at least 87 points, and we do so below.

Example: To demonstrate the results of the SpherePathFinder algorithm, we sample 100, 300, and 600 points on the 2-sphere. Only the paths are shown, which wind around 12 times. The range of distances $d$ between consecutive ordered points is also given, with an average $\widetilde d$.

As $N$ increases and the winding number stays the same, the path gets more and more jagged. To make the path smoother, we would need to increase the number of times the path winds around the sphere.

References: Bauer and Polthier (Detection of Planar Regions in Volume Data for Topology Optimization), Niyogi, Smale, and Weinberger (Finding the homology of submanifolds with high confidence from random samples), Sloane (OEIS A036069, A004731), Wikipedia (article "N-sphere")

The Fubini-Study metric and length in projective space

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
\[
g\left(\frac{d \gamma}{dt},\frac{d\gamma}{dt}\right) = \lambda_n(\gamma(t)) \sum_{k,\ell=1}^n \chi_{k\ell}(\gamma(t))\det
\begin{bmatrix}
\overline \gamma_k'(t) & i\overline \gamma_k'(t) \\[5pt] \gamma_\ell'(t) & -i\gamma_\ell'(t)
\end{bmatrix}.
\]

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)

Differential 1-forms are closed if and only if they are exact

 Preliminary exam prep

The title refers to 1-forms in Euclidean $n$-space $\R^n$, for $n\geqslant 2$. This theorem is instructive to do in the case $n=2$, but we present it in general. We will use several facts, most importantly that the integral of a function $f:X\to Y$ over a curve $\gamma:[a,b]\to X$ is given by
\[
\int_\gamma f\ dx_1\wedge \cdots \wedge dx_k = \int_a^b (f\circ \gamma)\ d(x_1\circ \gamma)\wedge \cdots \wedge d(x_n\circ \gamma),
\]
where $x_1,\dots,x_n$ is some local frame on $X$. We will also use the fundamental theorem of calculus and one of its consequences, namely
\[
\int_a^b \frac{\dy f}{\dy t}(t)\ dt = f(b)-f(a).
\]

Theorem: A 1-form on $\R^n$ is closed if and only if it is exact, for $n\geqslant 2$.

Proof: Let $\omega = a_1dx_1+\cdots a_ndx_n\in \Omega^1_{\R^n}$ be a 1-form on $\R^n$. If there exists $\eta\in \Omega^0_{\R^n}$ such that $d\eta = \omega$, then $d\omega = d^2\eta = 0$, so the reverse direction is clear. For the forward direction, since $\omega$ is closed, we have
\[
0 = d\omega = \sum_{i=1}^n \frac{\dy a_1}{\dy x_i}dx_i \wedge dx_1 + \cdots + \sum_{i=1}^n \frac{\dy a_n}{\dy x_n} dx_i\wedge dx_n
\ \ \ \implies\ \ \
\frac{\dy a_i}{\dy x_j} = \frac{\dy a_j}{\dy x_i}\ \forall\ i\neq j.
\]
Now fix some $(\textbf{x}_1,\dots,\textbf{x}_n)\in \R^n$, and define $f\in \Omega^0_{\R^n}$ by
\[
f(\textbf x_1,\dots,\textbf x_n) = \int_{\gamma(\textbf x_1,\dots,\textbf x_n)}\omega,
\]
for $\gamma$ the composition of the paths
\[
\begin{array}{r c l}
\gamma_1\ :\ [0,\textbf x_1] & \to & \R^n, \\
t & \mapsto & (t,0,\dots,0),
\end{array}
\hspace{5pt}
\begin{array}{r c l}
\gamma_2\ :\ [0,\textbf x_2] & \to & \R^n, \\
t & \mapsto & (\textbf x_1,t,0,\dots,0),
\end{array}
\hspace{5pt}\cdots\hspace{5pt}
\begin{array}{r c l}
\gamma_n\ :\ [0,\textbf x_n] & \to & \R^n, \\
t & \mapsto & (\textbf x_1,\dots,\textbf x_{n-1},t).
\end{array}
\]
By applying the definition of a pullback and the change of variables formula (use $s=\gamma_i(t)$ for every $i$),
\begin{align*}
\int_{\gamma(\textbf x_1,\dots,\textbf x_n)}\omega  & = \sum_{i=1}^n \int_{\gamma_i} a_1 dx_1 + \cdots + \sum_{i=1}^n \int_{\gamma_i}a_n dx_n \\
& = \sum_{i=1}^n \int_{\gamma_i} a_1(x_1,\dots,x_n)\ dx_1 + \cdots + \sum_{i=1}^n \int_{\gamma_i}a_n(x_1,\dots,x_n)\ dx_n \\
& = \sum_{i=1}^n \int_0^{\textbf x_i} a_1(\gamma_i(t))\ d(x_1\circ \gamma_i)(t) + \cdots + \sum_{i=1}^n \int_0^{\textbf x_i}a_n(\gamma_i(t))\ d(x_n\circ \gamma_i)(t) \\
& = \int_0^{\textbf x_1} a_1(\gamma_1(t))\gamma'_1(t)\ dt + \cdots + \int_0^{\textbf x_n}a_n(\gamma_n(t))\gamma_n'(t)\ dt \\
& = \int_{(0,\dots,0)}^{(\textbf x_1,0,\dots,0)} a_1(s)\ ds + \cdots + \int_{(\textbf x_1,\dots,\textbf x_{n-1},0)}^{(\textbf x_1,\dots,\textbf x_n)}a_n(s)\ ds \\
& = \int_0^{\textbf x_1} a_1(s,0,\dots,0)\ ds + \cdots + \int_0^{\textbf x_n}a_n(\textbf x_1,\dots,\textbf x_{n-1}, s)\ ds.
\end{align*}
To take the derivative of this, we consider the partial derivatives first. In the last variable, we have
\[
\frac{\dy f}{\dy \textbf x_n} = \frac\dy{\dy \textbf x_n}\int_0^{\textbf x_n}a_n(\textbf x_1,\dots,\textbf x_{n-1}, s)\ ds = a_n(\textbf x_1,\dots,\textbf x_n) = a_n.
\]
In the second-last variable, applying one of the identities from $\omega$ being closed, we have
\begin{align*}
\frac{\dy f}{\dy \textbf x_{n-1}}  & = \frac\dy{\dy \textbf x_{n-1}}\int_0^{\textbf x_{n-1}}a_{n-1}(\textbf x_1,\dots,\textbf x_{n-2}, s,0)\ ds  +  \frac\dy{\dy \textbf  x_{n-1}}\int_0^{\textbf x_n}a_n(\textbf x_1,\dots,\textbf x_{n-1}, s)\ ds \\
& = a_{n-1}(\textbf x_1,\dots,\textbf x_{n-1},0) + \int_0^{\textbf x_n}\frac{\dy a_n}{\dy \textbf x_{n-1}}(\textbf x_1,\dots,\textbf x_{n-1}, s)\ ds \\
& = a_{n-1}(\textbf x_1,\dots,\textbf x_{n-1},0) + \int_0^{\textbf x_n}\frac{\dy a_{n-1}}{\dy s}(\textbf x_1,\dots,\textbf x_{n-1}, s)\ ds \\
& = a_{n-1}(\textbf x_1,\dots,\textbf x_{n-1},0) + a_{n-1}(\textbf x_1,\dots,\textbf x_n) - a_{n-1}(\textbf x_1,\dots,\textbf x_{n-1}, 0) \\
& = a_{n-1}(\textbf x_1,\dots, \textbf x_n) \\
& = a_{n-1}.
\end{align*}
This pattern continues. For the other variables we have telescoping sums, and we compute the partial derivative in the first variable as an example:
\begin{align*}
\frac{\dy f}{\dy \textbf x_1}  & = \frac{\dy}{\dy \textbf x_1}\int_0^{\textbf x_1}a_1(s,0,\dots,0)\ ds + \sum_{i=2}^n\frac\dy{\dy \textbf x_1}\int_0^{\textbf x_i}a_i(\textbf x_1,\dots,\textbf x_{i-1}, s,0,\dots,0)\ ds \\
& = a_1(\textbf x_1,0,\dots,0) + \sum_{i=2}^n \int_0^{\textbf x_i} \frac {\dy a_i}{\dy \textbf x_1} (\textbf x_1,\dots,\textbf x_{i-1}, s,0,\dots,0)\ ds \\
& = a_1(\textbf x_1,0,\dots,0) + \sum_{i=2}^n \int_0^{\textbf x_i} \frac {\dy a_1}{\dy s} (\textbf x_1,\dots,\textbf x_{i-1}, s,0,\dots,0)\ ds \\
& = a_1(\textbf x_1,0,\dots,0) + \sum_{i=2}^n \left(a_1(\textbf x_1,\dots,\textbf x_i, 0,\dots,0) - a_1(\textbf x_1,\dots,\textbf x_{i-1},0,\dots,0)\right) \\
& = a_1(\textbf x_1,\dots,\textbf x_n) \\
& = a_1.
\end{align*}
Hence we get that
\[
df = \frac{\dy f}{\dy x_1} dx_1 + \cdots + \frac{\dy f}{\dy x_n}dx_n = a_1dx_1 + \cdots + a_ndx_n = \omega,
\]
so $\omega$ is exact. $\square$

References: Lee (Introduction to smooth manifolds, Chapter 11)