Problems: Cycles, Residues, and Coordinate Covariance

This page is the chapter’s stress test. Each problem forces several parts of the formal WKB ledger to interact: the normalized cover, its punctures, a declared chain, a meromorphic differential, and any endpoint or coordinate data needed to define its integral. The solutions are complete enough for self-study, but each heading first states a task that can be attempted independently.

Problems 5–9 address a subtle point directly. A normal-form equation is not carried between coordinates by treating its coefficient as a scalar. The wavefunction is an inverse half-density, the full coefficient acquires a Schwarzian term, and only the branch-antisymmetric Riccati one-form is tensorial. Open regularized integrals require one more layer: their endpoint conventions must be transported as well.

Page question

What must be transported with a WKB period so that a change of cycle, parameter, or coordinate preserves the mathematical object rather than only the appearance of its formula?

Status ledger

Statement used below	Status
Genus, puncture ranks, residues, and intersection matrices	Exact algebraic-topological statements
Riccati recursion, Schwarzian law, and coefficientwise covariance	Exact formal identities on charts where the coordinate map is locally biholomorphic
Picard–Fuchs reduction	Exact away from the stated discriminant, modulo exact forms
Open finite parts	Defined only after local parameters, logarithm branches, and subtraction scales are fixed
Borel sums, Stokes automorphisms, and exact spectra	Not inferred here; these begin in Chapter 9

Working conventions and route through the problems

Unless a problem says otherwise, use

$$\begin{aligned} \hbar^2\psi_{zz} &=R(z,\hbar)\psi, \qquad R=R_0+\hbar^2R_2+\cdots, \\ P^2+\hbar P_z &=R, \qquad y^2=R_0. \end{aligned}$$

The branch-antisymmetric momentum and formal WKB form are

$$P_{\mathrm{even}} = \frac{P^{(+)}-P^{(-)}}2, \qquad \Omega = P_{\mathrm{even}}\,\dd z.$$

Orient small loops positively in their local complex coordinate. On a compact oriented cover, write the intersection pairing as $\gamma\mathbin{\cdot}\delta$. For a positive parameter loop and a positive Dehn twist, retain the convention fixed on Page 4:

$$T_\delta(\gamma) = \gamma - (\gamma\mathbin{\cdot}\delta)\delta.$$

The problems form four passes through the chapter.

Pass	Problems	What is being audited
Geometry	1–3	Compactification, punctures, relative classes, residues, and integral bases
Period operators	4	Exact forms, Picard–Fuchs reduction, and endpoint terms
Coordinate covariance	5–9	Half-densities, Schwarzian terms, quantum residues, and finite-part data
Logical scope	10	Which conclusions are formal, topological, regularized, or analytic

Two routes through the page

For a first pass, do Problem 1, the rational warm-up in Problem 2, Problems 5–6, the concrete specialization in Problem 8, and Problems 9–10. The Weber extension, lattice calculation, operator reduction, and quantum-residue problem form the full expert route.

Problem 1 · A handle, two punctures, and a radical

Let

$$\Sigma_k: \qquad y^2=(z^2-1)(z^2-k^2), \qquad k\notin\{0,1,-1\},$$

and let $\lambda_0=y\,\dd z$.

1. Compactify and normalize the curve. Determine its genus and the number of points above $z=\infty$.
2. With $u=1/z$, expand $\lambda_0$ at both points over infinity. Determine the pole orders and residues.
3. Remove those two points to form $\Sigma_k^\circ$. Determine $\operatorname{rank}H_1(\Sigma_k^\circ;\mathbb Z)$ and identify the radical of its intersection form.
4. Let $\tau(z,y)=(z,-y)$ be the deck involution. Determine its action on a positive small loop $\ell_+$ around one point at infinity.
5. Explain why a nonzero puncture class can nevertheless have zero $\lambda_0$-period.

Solution

The four finite zeros $\pm1,\pm k$ are distinct and simple. A two-sheeted cover of the sphere branched at four points has

$$2g-2 = 2(-2)+4,$$

so $g=1$. Because the polynomial has even degree, infinity is not a branch point. There are two points $\infty_+$ and $\infty_-$, distinguished by the sign of $y/z^2$.

Choose the sheet label $\sigma=+1$ or $-1$ and put $u=1/z$. Then

$$\begin{aligned} y &= \sigma u^{-2} \sqrt{(1-u^2)(1-k^2u^2)} \\ &= \sigma u^{-2} \left[ 1-\frac{1+k^2}{2}u^2+O(u^4) \right]. \end{aligned}$$

Since $\dd z=-u^{-2}\dd u$,

$$\lambda_0 = -\sigma \left[ u^{-4} - \frac{1+k^2}{2}u^{-2} +O(1) \right]\dd u.$$

Thus $\lambda_0$ has a fourth-order pole and zero residue at each point over infinity. In particular, the absence of a $u^{-1}\dd u$ term is a local computation; it does not follow merely from the global residue theorem.

A genus-$g$ surface with $n>0$ punctures has first-homology rank $2g+n-1$. Therefore

$$\operatorname{rank} H_1(\Sigma_k^\circ;\mathbb Z) = 2+2-1 = 3.$$

If $A\mathbin{\cdot}B=1$ is a handle basis and $\ell_\pm$ are the puncture loops, then

$$\ell_++\ell_-=0.$$

The handle block is nondegenerate, while the puncture direction $\mathbb Z\ell_+$ is the radical of the intersection form. The deck map exchanges the two points and preserves complex orientation, so

$$\tau_*\ell_+ = \ell_- = -\ell_+.$$

Finally,

$$\oint_{\ell_+}\lambda_0 = 2\pi\ii \operatorname{Res}_{\infty_+}\lambda_0 =0.$$

Homology records possible integration contours; it does not promise that one chosen differential detects every class. Here $\ell_+$ is a nonzero anti-invariant homology class on the punctured cover, but $\lambda_0$ has zero residue on it.

Problem 2 · Closed zero does not mean relative zero

A rational warm-up

On

$$X=\mathbb P^1\setminus\{0,-1,\infty\},$$

consider

$$\omega_c = \left( \frac ct+ \frac{2}{(t+1)^2} \right)\dd t,$$

and let $\beta$ run along the positive real axis from $0$ to $\infty$.

1. Compute the three residues.
2. Define and evaluate a finite part by subtracting only the two logarithmic endpoint divergences.
3. Replace $\beta$ by $\beta+n\ell_0$, where $\ell_0$ is a positive loop around $0$.
4. Set $c=0$ and explain why exactness does not force the relative integral to vanish.

Solution

Write

$$\omega_c = c\,\dd\log t + \dd F, \qquad F(t)=\frac{t-1}{t+1}.$$

The residues at $0$, $-1$, and $\infty$ are respectively

$$c, \qquad 0, \qquad -c.$$

With the positive-real logarithm, define

$$\begin{aligned} \operatorname{FP} \int_\beta\omega_c :={}& \lim_{\substack{\epsilon\to0^+\\R\to+\infty}} \left[ \int_\epsilon^R\omega_c -c\log R +c\log\epsilon \right]. \end{aligned}$$

The logarithms cancel and $F(R)-F(\epsilon)\to1-(-1)$, hence

$$\operatorname{FP} \int_\beta\omega_c =2.$$

Adding the puncture loop changes the homology class and gives

$$\operatorname{FP} \int_{\beta+n\ell_0}\omega_c = 2+2\pi\ii n c.$$

For $c=0$, the differential is the globally exact rational form $\dd F$. Its closed periods vanish, but

$$\int_\beta\dd F = F(\infty)-F(0) =2.$$

Exactness annihilates an absolute closed cycle. A relative chain remembers the values of the primitive at its boundary.

Expert extension: the same distinction in the Weber model

Rationalize $y^2=z^2-a^2$ by

$$z=\frac a2(t+t^{-1}), \qquad y=\frac a2(t-t^{-1}).$$

Verify that $t=\pm1$ are the turning points, while $t=0,\infty$ are the two points above $z=\infty$. Show that

$$\lambda_0 = \frac{a^2(t^2-1)^2}{4t^3}\,\dd t$$

has residues $-a^2/2$ at $t=0$ and $a^2/2$ at $t=\infty$. Then prove

$$\lambda_2 = -\frac{ t(3t^4+14t^2+3) }{ 2a^2(t-1)^4(t+1)^4 }\,\dd t = \dd H_2,$$

where

$$H_2(t) = \frac{9t^4+12t^2-1}{ 12a^2(t-1)^3(t+1)^3 }.$$

What are the closed $\lambda_2$-periods and the endpoint difference from $t=0$ to $t=\infty$?

Solution

Substitution gives the displayed $\lambda_0$. Reading the coefficient of $t^{-1}\dd t$ gives the residue at zero; transforming with $u=1/t$ gives the opposite residue at infinity. Therefore a positive loop around the $t=\infty$ puncture has classical period

$$2\pi\ii\operatorname{Res}_{t=\infty}\lambda_0 = \ii\pi a^2.$$

Differentiating $H_2$ produces the stated rational form. Its integral around every closed cycle on this punctured sphere is zero, whereas a path from $0$ to $\infty$ that avoids $t=\pm1$ gives

$$H_2(\infty)-H_2(0) = -\frac{1}{12a^2}.$$

The value is an endpoint constant of a rational primitive. It is the first coefficient in the pole-to-pole Weber correction computed on Page 7, while the corresponding higher closed period vanishes.

Problem 3 · Mathieu’s physical cycles are not a basis

For $-2\Lambda^2<E<2\Lambda^2$, put

$$m = \frac{E+2\Lambda^2}{4\Lambda^2}, \qquad 0<m<1.$$

The oriented physical cycles from Page 7 satisfy

$$\delta_-\mathbin{\cdot}\delta_+=2, \qquad A=\delta_-, \qquad \delta_+=A+2B,$$

with $A\mathbin{\cdot}B=1$.

1. Lattice index. Show from the change-of-generators matrix that $\delta_-,\delta_+$ span an index-two sublattice.

2. Twist formula. Let $J=\left(\begin{smallmatrix}0&1\\-1&0\end{smallmatrix}\right)$ and represent $d=aA+bB$ by the column $(a,b)^{\mathsf T}$. Derive the matrix of $T_d$ from the stated Picard–Lefschetz convention.

3. Two monodromies. Compute the twists about $\delta_-$ and $\delta_+$, and check that each fixes its vanishing cycle and preserves $J$.

4. Period Wronskians. Use the classical periods

$$\begin{aligned} J_- &= 8\Lambda \left[ \mathbf E(m)-(1-m)\mathbf K(m) \right], \\ J_+ &= 8\Lambda \left[ \mathbf E(1-m)-m\mathbf K(1-m) \right] \end{aligned}$$

and

$$\partial_EJ_- = \frac{\mathbf K(m)}{\Lambda}, \qquad \partial_EJ_+ = -\frac{\mathbf K(1-m)}{\Lambda}$$

to compute the ordered energy Wronskians $W_E(f,g)=f\,\partial_Eg-(\partial_Ef)g$ of the physical and primitive period pairs. You may use Legendre’s relation

$$\begin{aligned} &\mathbf K(m)\mathbf E(1-m) + \mathbf E(m)\mathbf K(1-m) \\ &\qquad - \mathbf K(m)\mathbf K(1-m) = \frac\pi2. \end{aligned}$$

Solution

The columns of the physical generators in the $(A,B)$ basis form

$$C = \begin{pmatrix} 1&1\\ 0&2 \end{pmatrix}.$$

Since $|\det C|=2$, their span has index two in $H_1(\Sigma;\mathbb Z)$. This is why solving $B=(\delta_+-\delta_-)/2$ is legitimate but treating an arbitrary half-cycle as integral is not.

For coordinate columns $v$ and $d$, the intersection is $v^{\mathsf T}Jd$. Hence

$$T_d(v) = v-(v^{\mathsf T}Jd)d = \left(I+dd^{\mathsf T}J\right)v.$$

The two vanishing-cycle columns are

$$d_-= \begin{pmatrix}1\\0\end{pmatrix}, \qquad d_+= \begin{pmatrix}1\\2\end{pmatrix}.$$

Therefore

$$T_- = \begin{pmatrix} 1&1\\ 0&1 \end{pmatrix}, \qquad T_+ = \begin{pmatrix} -1&1\\ -4&3 \end{pmatrix}.$$

Direct multiplication gives

$$T_-d_-=d_-, \qquad T_+d_+=d_+,$$

and, for either sign,

$$\det T_\pm=1, \qquad T_\pm^{\mathsf T}JT_\pm=J.$$

Now set $\Pi_-=\ii J_-$ and $\Pi_+=J_+$. Legendre’s relation,

$$\begin{aligned} &\mathbf K(m)\mathbf E(1-m) + \mathbf E(m)\mathbf K(1-m) \\ &\qquad - \mathbf K(m)\mathbf K(1-m) = \frac\pi2, \end{aligned}$$

gives

$$W_E(\Pi_-,\Pi_+) := \Pi_-\Pi_+'-\Pi_-'\Pi_+ = -4\pi\ii.$$

Because

$$\Pi_A=\Pi_-, \qquad \Pi_B=\frac{\Pi_+-\Pi_-}{2},$$

the primitive Wronskian is

$$W_E(\Pi_A,\Pi_B) = -2\pi\ii.$$

The same factor of two occurs in the intersection pairing. As a further orientation check,

$$\frac{\partial_E\Pi_B}{\partial_E\Pi_A} = -\frac12 + \frac{ \ii\mathbf K(1-m) }{ 2\mathbf K(m) },$$

whose imaginary part is positive in the real chamber. Reversing the parameter loop replaces either Picard–Lefschetz matrix by its inverse; it does not alter the intersection lattice.

Problem 4 · Reduce the operator, but retain the primitive

For the Mathieu curve

$$y^2 = 2\Lambda^2\cos(2x)-E, \qquad \lambda_0=y\,\dd x,$$

define

$$\begin{aligned} \Delta &=E^2-4\Lambda^4, \\ G &=\frac{\Lambda^2\sin(2x)}{2y}, \\ \mathcal L &=\Delta\partial_E^2+\frac14. \end{aligned}$$

1. Picard–Fuchs certificate. Verify $\mathcal L\lambda_0=\dd_xG$.

2. Left-ideal comparison. Express the differences between the following three operators as left multiples of $\mathcal L$, retaining the order of operator composition:

$$\begin{aligned} \mathcal D_{\mathrm{hi}} &= -\frac14\partial_E -3E\partial_E^2 -\frac53\Delta\partial_E^3, \\ \mathcal D_{\mathrm{poly}} &= \frac16\partial_E +\frac E3\partial_E^2, \\ \mathcal D_{\mathrm{red}} &= \frac16\partial_E -\frac{E}{12\Delta}. \end{aligned}$$

3. Exact representatives. Given the pointwise identity $\lambda_2=\mathcal D_{\mathrm{hi}}\lambda_0$, find explicit primitives that relate the other two representatives to $\lambda_2$.

4. Chain dependence. State separately what follows on a flat closed cycle and on a relative path.

5. Parameter-dependent coordinates. Let $x=x(w)$ be independent of $E$ and transport the Picard–Fuchs certificate. Then identify the extra term when $x=x(w,E)$.

Solution

Since $\partial_E^2\lambda_0=-\dd x/(4y^3)$, the left side of the certificate is

$$\left[ -\frac{\Delta}{4y^3} +\frac y4 \right]\dd x.$$

Putting it over the denominator $y^3$ and using $y^2=2\Lambda^2\cos(2x)-E$ reproduces $\dd_xG$.

Operator composition is noncommutative because $\Delta$ depends on $E$. The exact left-ideal identities are

$$\mathcal D_{\mathrm{hi}} - \mathcal D_{\mathrm{poly}} = -\frac53\, \partial_E\circ\mathcal L$$

and

$$\mathcal D_{\mathrm{poly}} - \mathcal D_{\mathrm{red}} = \frac{E}{3\Delta}\mathcal L.$$

Consequently,

$$\lambda_2 = \mathcal D_{\mathrm{poly}}\lambda_0 + \dd_xF_{\mathrm{poly}},$$

where

$$F_{\mathrm{poly}} = -\frac{5\Lambda^2\sin(2x)}{12y^3} = -\frac53\partial_EG.$$

Off the discriminant $\Delta=0$, one may instead write

$$\lambda_2 = \mathcal D_{\mathrm{red}}\lambda_0 + \dd_xF_{\mathrm{red}},$$

with

$$F_{\mathrm{red}} = F_{\mathrm{poly}} + \frac{E}{3\Delta}G.$$

If $\gamma(E)$ is Gauss–Manin flat and closed, the exact terms vanish:

$$\Pi_{\gamma,2} = \mathcal D_{\mathrm{poly}}\Pi_{\gamma,0} = \mathcal D_{\mathrm{red}}\Pi_{\gamma,0}.$$

The second equality is unavailable at $\Delta=0$. On a fixed relative path $\beta:p\to q$ with finite endpoint values,

$$\begin{aligned} \int_\beta\lambda_2 ={}& \mathcal D_{\mathrm{poly}} \int_\beta\lambda_0 \\ &+ F_{\mathrm{poly}}(q) -F_{\mathrm{poly}}(p). \end{aligned}$$

At singular endpoints, the last line means the finite endpoint constants in the declared local subtraction scheme. Dropping it would silently turn a relative identity into a closed-period identity.

For an $E$-independent coordinate $x=x(w)$, pullback commutes with $\partial_E$, so

$$\mathcal L\widetilde\lambda_0 = \dd_w\bigl(G\circ x\bigr).$$

For an $E$-dependent map, let a general form be $\alpha=f(x,E)\,\dd x$. At fixed $w$,

$$\left. \partial_E\widetilde\alpha \right|_w = x^* \left( \left.\partial_E\alpha\right|_x \right) + \dd_w(fx_E).$$

Thus closed periods remain compatible when the cycle and coordinate family are transported consistently, but open paths acquire boundary terms. For example, take $f=x$, $x=Ew$, and $0\leq w\leq1$. Then

$$\int_0^1E^2w\,\dd w = \frac{E^2}{2},$$

whose derivative $E$ is entirely the endpoint term $[fx_E]_{0}^{1}$; the fixed-$x$ derivative of $x\,\dd x$ is zero.

Problem 5 · Derive the projective transformation law

Let $z=z(w)$ and

$$s(w)=\frac{\dd z}{\dd w}\neq0$$

on the chart under consideration. Begin with a solution of $\hbar^2\psi_{zz}=R\psi$.

1. Pull $\psi$ back as a scalar, $\chi(w)=\psi(z(w))$, and show why the resulting equation is not in normal form.
2. Find the power of $s$ that removes the first derivative. Derive the transformed coefficient, including its sign.
3. Transform both Riccati branches. Which combination is a one-form, and which is an affine connection?
4. Prove that the ordered Wronskian is preserved.
5. Prove the Schwarzian chain rule and use it to check two successive coordinate changes.

Solution

The scalar pullback obeys

$$\hbar^2\chi'' - \hbar^2\frac{s'}s\chi' = s^2R\chi.$$

The first-derivative term disappears for the inverse-half-density field

$$\widetilde\psi(w) = s(w)^{-1/2}\psi(z(w)),$$

where a local branch of $s^{1/2}$ has been chosen. Direct differentiation gives

$$\hbar^2\widetilde\psi'' = \widetilde R\widetilde\psi,$$

with

$$\begin{aligned} \widetilde R(w,\hbar) &= s^2R(z(w),\hbar) - \frac{\hbar^2}{2}\{z,w\}, \\ \{z,w\} &= \frac{s''}{s} - \frac32 \left( \frac{s'}s \right)^2. \end{aligned}$$

The minus sign is forced by cancellation of the first derivative; it is not a convention that can be changed independently.

For $P=\hbar\,\partial_z\log\psi$, the two branches transform by

$$\widetilde P^{(\pm)} = sP^{(\pm)} - \frac{\hbar}{2}\frac{s'}s.$$

The common affine term cancels in the half-difference but remains in the half-sum:

$$\begin{aligned} \widetilde P_{\mathrm{even}}\,\dd w &= P_{\mathrm{even}}\,\dd z, \\ \widetilde P_{\mathrm{amp}}\,\dd w &= P_{\mathrm{amp}}\,\dd z - \frac{\hbar}{2}\dd\log s. \end{aligned}$$

Thus $\Omega=P_{\mathrm{even}}\,\dd z$ is the phase one-form. The amplitude term is a connection in the chosen half-density trivialization.

For the book’s convention $\Wr[f,g]=fg'-f'g$,

$$\Wr_w \left[ s^{-1/2}f(z(w)), s^{-1/2}g(z(w)) \right] = \Wr_z[f,g].$$

The derivatives of $s^{-1/2}$ cancel between the two terms, and the remaining factor $s$ cancels the two half-density factors.

Now let $z=f(w)$ and $w=g(u)$. Expansion of three derivatives proves

$$\{f\circ g,u\} = g'(u)^2 \{f,w\}\big|_{w=g(u)} + \{g,u\}.$$

Substituting this identity into the potential law shows that transforming first from $z$ to $w$ and then from $w$ to $u$ gives the same coefficient as the direct transformation. Locally compatible square roots give the same conclusion for the half-density:

$$(f'(g)g')^{-1/2} = g'^{-1/2}f'(g)^{-1/2}.$$

The geometric ledger is therefore:

Object	Transformation type
$\psi$	Inverse half-density
$R_0\,\dd z^2$	Quadratic differential
Full $R$	Projective coefficient; not a scalar or quadratic differential
$\mathcal T=-2R/\hbar^2$	Projective connection, $\widetilde{\mathcal T}=s^2\mathcal T+\{z,w\}$
$\Omega$	One-form
$P_{\mathrm{amp}}\,\dd z$	Affine connection one-form

Only $R_0$ transforms tensorially, because the Schwarzian first enters at order $\hbar^2$. Also note that the transformed object is $R=V-E$. If one insists on writing $\widetilde R=\widetilde V-E$, then

$$\widetilde V_E = E+s^2(V-E) - \frac{\hbar^2}{2}\{z,w\},$$

which generally depends on $E$. A nonlinear coordinate change does not preserve the special decomposition “potential minus a constant energy.”

Problem 6 · Audit a non-Möbius pullback through order ℏ²

The first even correction can be written in either form

$$\begin{aligned} p_2 ={}& \frac{R_2}{2y} + \frac{y_{zz}}{4y^2} - \frac{3y_z^2}{8y^3} \\ ={}& \frac{R_2}{2y} + \frac{R_{0,zz}}{8y^3} - \frac{5R_{0,z}^2}{32y^5}. \end{aligned}$$

1. Coefficientwise covariance. Under $z=z(w)$, substitute

$$\widetilde R_0=s^2R_0, \qquad \widetilde R_2=s^2R_2-\frac12\{z,w\}, \qquad \widetilde y=sy$$

and prove $\widetilde p_2=sp_2$.

2. Missing projective term. Identify the error if the Schwarzian term is omitted.

3. Quadratic laboratory. Test the result on the constant equation

$$\hbar^2\psi_{zz}=\kappa^2\psi$$

under the non-Möbius map $z=w^2$. Carry out the normal-form calculation on a simply connected sector with $w\neq0$, where both the map and the chosen square root are single-valued. Problem 7 then continues the resulting data around the punctured annulus.

Solution

When derivatives of $\widetilde y=sy(z(w))$ are expanded, all mixed terms containing $s'y_z$ cancel. The purely coordinate-dependent part left by the two derivative terms is

$$\frac{s''}{4s^2y} - \frac{3(s')^2}{8s^3y} = \frac{\{z,w\}}{4sy}.$$

The Schwarzian contribution to $\widetilde R_2/(2\widetilde y)$ is its negative. Therefore

$$\widetilde p_2=sp_2, \qquad \widetilde\lambda_2 = \widetilde p_2\,\dd w = p_2\,\dd z.$$

If the Schwarzian is omitted, the uncancelled error is

$$\widetilde p_2^{\mathrm{naive}} - sp_2 = \frac{\{z,w\}}{4sy}.$$

Now take $z=w^2$, so

$$s=2w, \qquad \{w^2,w\} = -\frac{3}{2w^2}.$$

The correctly transformed equation is

$$\hbar^2\widetilde\psi'' = \left( 4\kappa^2w^2 + \frac{3\hbar^2}{4w^2} \right) \widetilde\psi,$$

and its two exact pullback solutions are

$$\widetilde\psi_\pm(w) = (2w)^{-1/2} \exp\!\left( \pm\frac{\kappa w^2}{\hbar} \right).$$

Their exact Riccati momenta are

$$\widetilde P^{(\pm)} = \pm2\kappa w - \frac{\hbar}{2w}.$$

Thus

$$\widetilde P_{\mathrm{even}} = 2\kappa w, \qquad \widetilde\Omega = 2\kappa w\,\dd w = \kappa\,\dd z.$$

At order $\hbar^2$, the derivative terms built from $\widetilde y=2\kappa w$ contribute

$$-\frac{3}{16\kappa w^3},$$

whereas $\widetilde R_2=3/(4w^2)$ contributes $3/(16\kappa w^3)$. They cancel, as they must because the original constant problem has $p_2=0$. Omitting the Schwarzian manufactures the spurious exact form

$$\widetilde\lambda_2^{\mathrm{naive}} = -\frac{3}{16\kappa w^3}\,\dd w = \dd\!\left( \frac{3}{32\kappa w^2} \right).$$

Its closed integral happens to vanish. That accident does not repair the pointwise formal equation; the next problem gives a case in which the period itself is wrong.

Problem 7 · Quantum residues survive a coordinate change

On the punctured $z$-plane, take

$$R(z)=\frac{\kappa^2}{z^2}, \qquad \kappa\neq0,$$

and choose the formal square root that tends to $\kappa$ as $\hbar\to0$.

1. Solve the Riccati equation exactly and compute the formal period around a positive loop $\gamma$ about $z=0$.
2. Pull the equation back by $z=w^2$. Compare the period of a full positive $w$-loop with the original one and explain the factor of two.
3. Repeat the transformed calculation without the Schwarzian term.
4. Relate the sign of the inverse half-density around $w=0$ to the amplitude connection.

Solution

The exact Riccati branches are

$$P^{(\pm)}(z,\hbar) = \frac{ \hbar/2\pm\nu(\hbar) }{z}, \qquad \nu(\hbar) = \sqrt{ \kappa^2+\frac{\hbar^2}{4} }.$$

Hence

$$\Omega = \nu(\hbar)\frac{\dd z}{z}$$

and

$$\begin{aligned} \Pi_\gamma(\hbar) &= 2\pi\ii\nu(\hbar) \\ &= 2\pi\ii \left[ \kappa + \frac{\hbar^2}{8\kappa} - \frac{\hbar^4}{128\kappa^3} + \frac{\hbar^6}{1024\kappa^5} +\cdots \right]. \end{aligned}$$

Under $z=w^2$, the correct coefficient and even momentum are

$$\widetilde R = \frac{ 4\kappa^2+3\hbar^2/4 }{w^2}, \qquad \widetilde P_{\mathrm{even}} = \frac{2\nu(\hbar)}{w}.$$

A full $w$-loop projects to a loop that winds twice around $z=0$. Accordingly,

$$\oint_{|w|=\epsilon} \widetilde\Omega = 4\pi\ii\nu(\hbar) = 2\Pi_\gamma(\hbar).$$

This is covariance with the chain transported correctly. Comparing a full loop in each coordinate would compare different chains.

If the Schwarzian is omitted, the transformed coefficient is $4\kappa^2/w^2$. Its branch-antisymmetric exact momentum is

$$\widetilde P_{\mathrm{even}}^{\mathrm{naive}} = \frac1w \sqrt{ 4\kappa^2+\frac{\hbar^2}{4} }.$$

This differs from $2\nu/w$ beginning at order $\hbar^2$, so the loop period loses part of every quantum residue correction. Unlike the spurious exact form in Problem 6, this error is detected by a closed period.

Finally, the amplitude identity

$$P_{\mathrm{amp}}\,\dd z = -\frac{\hbar}{2} \dd\log P_{\mathrm{even}}$$

and the argument principle imply

$$\oint_\gamma P_{\mathrm{amp}}\,\dd z = -\pi\ii\hbar \left( N_{\mathrm{zero}}-N_{\mathrm{pole}} \right).$$

For $\widetilde P_{\mathrm{even}}=2\nu/w$, the loop contains one pole and no zero, giving $+\pi\ii\hbar$. The WKB prefactor $\widetilde P_{\mathrm{even}}^{-1/2}\propto w^{1/2}$ therefore changes sign. Equivalently, the coordinate half-density $(2w)^{-1/2}$ changes sign around the excluded ramification point. The phase one-form is single-valued on the punctured spectral cover; the local half-density trivialization can still have sign holonomy.

Problem 8 · A finite part remembers the endpoint chart

At an initial endpoint $p$, suppose a meromorphic one-form has the local expansion

$$\lambda = \left( \frac{a_{-2}}{t^2} + \frac rt +O(1) \right)\dd t.$$

Use the singular primitive with no constant term,

$$F_t^{\mathrm{sing}} = -\frac{a_{-2}}t + r\log\frac{t}{\mu_t},$$

and compatible logarithm branches. Change endpoint coordinate by

$$t = cu+du^2+O(u^3), \qquad c\neq0.$$

1. Derive the difference between the independently normalized $t$- and $u$-finite parts.
2. State what changes at a terminal endpoint.
3. Specialize to $\lambda=\dd z/z^2$ and $z=w+cw^2$.
4. Explain which data must be transported for an invariant open regularized integral.

Solution

Express the $t$-primitive in the $u$-coordinate:

$$\begin{aligned} -\frac{a_{-2}}t + r\log\frac{t}{\mu_t} ={}& -\frac{a_{-2}}{cu} + r\log\frac{u}{\mu_u} \\ &+ \frac{a_{-2}d}{c^2} + r\log\frac{c\mu_u}{\mu_t} +O(u). \end{aligned}$$

The $u$-prescription “singular part with no constant” discards the constant on the second line. Therefore, at an initial endpoint,

$$\operatorname{FP}_t \int_p^q\lambda - \operatorname{FP}_u \int_p^q\lambda = \frac{a_{-2}d}{c^2} + r\log\frac{c\mu_u}{\mu_t}.$$

At a terminal endpoint the sign reverses because the endpoint primitive enters with the opposite sign. The residue $r$ itself is coordinate invariant, but the finite part depends on a logarithm branch and scale. A pole of order two also remembers the second jet $d$, even when $r=0$.

For the concrete form

$$\lambda = \frac{\dd z}{z^2} = \dd\!\left(-\frac1z\right),$$

integrated from $z=0$ to a regular point $z=b$,

$$\operatorname{FP}_z \int_0^b\lambda = \lim_{\epsilon\to0} \left[ \int_\epsilon^b\frac{\dd z}{z^2} - \frac1\epsilon \right] = -\frac1b.$$

If $z=w+cw^2$ and one independently subtracts only $1/\epsilon$ in the $w$-coordinate, then

$$\operatorname{FP}_w \int_0^b\lambda = -\frac1b-c.$$

Indeed,

$$-\frac{1}{w+cw^2} = -\frac1w+c+O(w).$$

Transporting the original convention means subtracting the full endpoint divergence $1/\epsilon-c+O(\epsilon)$, which restores $-1/b$. In general one must transport the local parameter, all finite constants induced by the required principal part, the logarithm branch, and the scale $\mu$. The differential alone is insufficient data for an open finite part.

Problem 9 · Complete the Mathieu coordinate audit

Return to

$$\hbar^2\psi_{xx} = \left[ 2\Lambda^2\cos(2x)-E \right]\psi$$

and set $w=\exp(2\ii x)$.

1. Scalar pullback. Substitute directly while treating $\psi$ as a scalar and exhibit the first-derivative term.

2. Normal-form restoration. Put $x=(2\ii)^{-1}\log w$, restore normal form with the inverse half-density, and compute the Schwarzian term.

3. Classical differential. With

$$Y^2 = w(\Lambda^2w^2-Ew+\Lambda^2),$$

recover the classical form used on Page 7.

4. Controlled failure. Omit the Schwarzian deliberately. Compute the resulting error in $\lambda_2$ and in every transported closed period.

Solution

Since $\partial_x=2\ii w\partial_w$, direct scalar substitution gives

$$-4\hbar^2 \left( w^2\psi_{ww}+w\psi_w \right) = \left[ \Lambda^2(w+w^{-1})-E \right]\psi.$$

The $w\psi_w$ term shows that this is not yet a normal-form equation. For

$$x(w)=\frac{\log w}{2\ii}, \qquad s=\frac{\dd x}{\dd w}=\frac{1}{2\ii w},$$

one finds

$$\{x,w\} = \frac{1}{2w^2}.$$

With $\widetilde\psi=s^{-1/2}\psi$, the normal-form coefficient is

$$\widetilde R = -\frac{ \Lambda^2(w+w^{-1})-E }{4w^2} - \frac{\hbar^2}{4w^2}.$$

The leading square root is

$$\widetilde y = s y = \frac{Y}{2\ii w^2},$$

after a compatible sheet choice. Therefore

$$\widetilde\lambda_0 = \widetilde y\,\dd w = \frac{Y}{2\ii w^2}\,\dd w,$$

exactly as on Page 7.

The correct transformed coefficient has $\widetilde R_2=-1/(4w^2)$. If the Schwarzian is omitted, the difference in the first quantum coefficient comes only from the source term:

$$\begin{aligned} \lambda_2^{\mathrm{naive}} - \lambda_2 &= \frac{ \widetilde R_2^{\mathrm{naive}} - \widetilde R_2 }{2\widetilde y}\,\dd w \\ &= \frac{\ii}{4Y}\,\dd w. \end{aligned}$$

But differentiating the classical form at fixed $w$ gives

$$\partial_E\widetilde\lambda_0 = \frac{\ii}{4Y}\,\dd w.$$

Thus, for every Gauss–Manin-flat closed cycle away from the discriminant,

$$\Pi_{\gamma,2}^{\mathrm{naive}} - \Pi_{\gamma,2} = \partial_E\Pi_{\gamma,0}.$$

This is a useful diagnostic because the mistake survives integration. It is not a change of basis, an exact-form ambiguity, or a choice of regularization. It is the omitted projective term in the differential equation.

Problem 10 · Sort the claims before using them

Classify each statement as true or false. If false, replace it by a correct statement with the missing hypothesis or data.

1. The compact genus counts all useful WKB periods.
2. A residue-free meromorphic one-form on a compact Riemann surface is exact.
3. An exact one-form integrates to zero on every path.
4. The full normal-form coefficient $R$ transforms as a scalar.
5. Closed formal WKB periods are coordinate covariant.
6. A finite-part open integral is determined by the meromorphic one-form and relative homology class alone.
7. Reduction modulo a Picard–Fuchs operator is valid on closed and relative paths without modification.
8. Formal coordinate covariance proves covariance of Borel sums and exact quantization conditions.

Solution

1. False. Puncture loops and relative endpoint directions can add period data beyond the $2g$ compact handle cycles. The declared differential may still vanish on some of those classes.
2. False. Residue-free only excludes simple-pole contributions. On positive-genus curves, holomorphic and second-kind differentials can have nonzero handle periods. On the sphere, a residue-free meromorphic one-form is rationally exact.
3. False. A globally exact form $\dd F$ vanishes on every closed cycle but gives $F(q)-F(p)$ on an open path. A locally exact form with a multivalued primitive, such as $\dd z/z$, is not globally exact and need not have zero closed periods.
4. False. The scalar pullback introduces a first derivative. Normal form is restored by an inverse half-density, and the full coefficient acquires a Schwarzian term.
5. True with hypotheses. Use a local biholomorphism, the compatible inverse-half-density transformation, the Schwarzian-corrected equation, and the transported closed cycle. The claim is coefficientwise formal.
6. False. Singular endpoints also require local parameters, subtraction constants, scales, and logarithm branches. Those data must be transported under a coordinate change.
7. False. Exact representatives vanish on suitable closed cycles. Relative paths retain endpoint values of their primitives, and singular endpoints retain finite endpoint constants.
8. False. A formal identity does not choose a Borel direction, prove summability, cross a Stokes wall, impose a boundary condition, or establish an exact spectrum. Those are analytic questions for the next chapter.

Reproducible capstone audit

The cycles, residues, and coordinate-covariance check verifies:

• the quartic expansions and puncture residues;
• the rational and Weber exact-form identities and endpoint constants;
• the Mathieu lattice index, intersection form, Picard–Lefschetz matrices, period Wronskians, and Picard–Fuchs reductions;
• the half-density law, Schwarzian cocycle, and order-$\hbar^2$ covariance identity;
• the quadratic-map exact solutions, the inverse-square quantum residue, and the coordinate winding factor;
• the nonlinear finite-part shift and the Mathieu Schwarzian error.

Run

python3 public/code/advanced-ode/cycles-residues-coordinate-covariance-check.py

The script uses symbolic identities and high-precision special-function checks. It is not a proof of the topological classifications, and it does not perform Borel summation or spectral computation.

Exit checklist

Before calling a formal WKB period “the same in another coordinate,” you should now be able to record all of the following:

• the normalized cover, punctures, and integral cycle or relative chain;
• sheet, orientation, and Gauss–Manin transport conventions;
• the inverse-half-density branch and Schwarzian-corrected normal form;
• whether the object is $R_0\,\dd z^2$, the projective coefficient, the phase one-form, or the amplitude connection;
• every endpoint local parameter, subtraction scale, logarithm branch, and finite constant;
• whether an operator identity holds pointwise, modulo an exact form, only after closed-cycle integration, or only away from a discriminant.

If any item is missing, the safest conclusion is not “invariance” but “the comparison has not yet been defined.”

References

• NIST Digital Library of Mathematical Functions, §1.13(iv), Change of Variables, especially equations 1.13.18–1.13.21 for the Liouville transformation, Schwarzian derivative, and chain rule.
• E. Frenkel, “Affine Kac–Moody Algebras, Integrable Systems and Their Deformations,” §2, for the coordinate-free description of a second-order oper between half-density bundles and its projective-connection coefficient.
• K. Iwaki and T. Nakanishi, “Exact WKB Analysis and Cluster Algebras”, Journal of Physics A 47 (2014) 474009, §2.1, Proposition 2.7, Proposition 2.8, and §3.1 for coordinate changes, the odd Riccati form, regularized endpoints, and cycle/path pairings.
• K. Iwaki, T. Koike, and Y.-M. Takei, “Voros Coefficients for the Hypergeometric Differential Equations and Eynard–Orantin’s Topological Recursion, Part I: For the Weber Equation”, Annales Henri Poincaré 24 (2023), §4, especially Theorem 4.10, for the Weber rational cover and its complete pole-to-pole Voros coefficient.
• T. Bridgeland and I. Smith, “Quadratic Differentials as Stability Conditions”, Publications Mathématiques de l’IHÉS 121 (2015), §§2.3–2.4, especially Lemmas 2.2–2.3 for spectral-cover homology, residue classes, and the radical of the intersection form.
• A. Hatcher, Algebraic Topology, §2.1, for relative homology and the long exact sequence of a pair.
• O. Forster, Lectures on Riemann Surfaces, §§9–10 and §18 for meromorphic differentials, residues, and the global residue theorem.
• M. Bertola, “Two-Matrix Model with Semiclassical Potentials and Extended Whitham Hierarchy”, Journal of Physics A 39 (2006), §2 and Appendix D for regularized Abelian integrals with declared local parameters and logarithmic normalizations.
• J. Carlson, S. Müller-Stach, and C. Peters, Period Mappings and Period Domains, 2nd ed., Chapter 1 §1.1, equation (1.13), for the Picard–Lefschetz sign convention used here.
• NIST Digital Library of Mathematical Functions, §19.4(i), for the complete elliptic-integral derivatives and Legendre relation used in Problem 3; its modulus $k$ satisfies $k^2=m$ here.
• K. Imaizumi, “Exact WKB Analysis and TBA Equations for the Mathieu Equation”, Physics Letters B 806 (2020), §§2–3, for the two Mathieu cycle sectors, their same-sign crossings, and the first quantum-period operator; Page 7 gives the complete normalization translation.
• P. A. Griffiths, “On the Periods of Certain Rational Integrals: I”, Annals of Mathematics 90 (1969), 460–495, and Part II, 496–541, for reduction of rational differentials modulo exact forms.
• F. Fischbach, A. Klemm, and C. Nega, “WKB Method and Quantum Periods beyond Genus One”, Journal of Physics A 52 (2019), §2.2, for cohomological Picard–Fuchs reduction and quantum-period differential operators.

Chapter 9 begins with the extra analytic data deliberately absent from these problems: Gevrey bounds, Borel transforms, summation directions, and lateral sums.