Painlevé Lax Systems and Heun Reductions

A four-pole rank-two Fuchsian system and the general Heun equation are closely related, but they are not the same object at a generic isomonodromic time. Cyclic reduction of the matrix system introduces a fifth, apparent scalar singularity. Its position moves according to Painlevé VI. A genuine four-singularity Heun equation appears only on a specified collision slice, after an integer shift of the exponent lifts and a declared scalar gauge.

In the normalization developed below, the result is particularly useful. Let $t_0$ be the fourth Heun singularity, let $\alpha_{\mathrm H}\beta_{\mathrm H}$ be the product of its two exponents at infinity, and let $\tau_{\mathrm J}(\rho;t)$ be the traceless JMU tau function for the shifted Painlevé data $\rho$. If $\theta_t=\vartheta_t-1$ is the Painlevé exponent at the collision pole, then

$$\begin{aligned} q_{\mathrm H} ={}& t_0\alpha_{\mathrm H}\beta_{\mathrm H} \\ &+ t_0(t_0-1) \left. \left[ \frac{\dd}{\dd t}\log\tau_{\mathrm J}(\rho;t) - \frac{\theta_0\theta_t}{2t} - \frac{\theta_1\theta_t}{2(t-1)} \right] \right|_{t=t_0}, \end{aligned}$$

provided the accompanying collision constraint is satisfied. That constraint may be written as a derivative condition, or as a zero of a Schlesinger-transformed tau function. Neither statement says that an arbitrary Heun equation is “equal to Painlevé VI,” and neither makes a tau zero a spectral condition without further boundary data.

First pass

Follow the route matrix system → apparent scalar pole → Painlevé VI → collision slice → Heun equation. On a first reading, use the boxed workflow near the tau constraints and postpone the rank-one Hamiltonian calculation.

The same symbols can play different roles

Several nearby quantities have similar names. This ledger fixes their use on this page.

Symbol	Role
$t$	Isomonodromic time and one true moving pole
$(q,p)$	Spectral Darboux coordinates in the traceless gauge of the preceding pages; $q$ is an apparent-pole position
$(\lambda,\mu)$	Polynomial coordinates in the rank-one residue gauge used for the collision calculation; exactly $\lambda=q$ and $\mu=P$
$K_{\mathrm{VI}}$	One polynomial Painlevé VI Hamiltonian; it differs from the bare JMU Hamiltonian by time-dependent canonical terms
$K_0$	Compact Heun accessory in the collision gauge below
$q_{\mathrm H}$	The standard Heun accessory in the house convention of Chapter 3
$\tau_{\mathrm J}$	JMU tau function of the traceless residue system, so $\partial_t\log\tau_{\mathrm J}=H_t$
$\tau_{\rho^+}$	A neighboring tau function obtained by one elementary Schlesinger transformation

In particular, the apparent coordinate $q$ is never the standard Heun accessory $q_{\mathrm H}$.

Cyclic reduction adds one apparent point

Start from the normalized traceless system

$$\frac{\partial\Phi}{\partial z} = \left( \frac{A_0}{z} + \frac{A_t}{z-t} + \frac{A_1}{z-1} \right)\Phi, \qquad A_\infty=-A_0-A_t-A_1,$$

with

$$\operatorname{spec}A_\nu = \left\{ \frac{\theta_\nu}{2}, - \frac{\theta_\nu}{2} \right\}.$$

After diagonalizing $A_\infty$, the upper-right entry has the form

$$b(z) = \frac{\chi(z-q)} {z(z-1)(z-t)}, \qquad \chi\ne0.$$

Eliminating the second component and applying the Liouville gauge gives

$$\psi''(z)+T_{\mathrm{def}}(z)\psi(z)=0.$$

Its finite-pole structure is

$$\begin{aligned} T_{\mathrm{def}}(z) ={}& \sum_{s\in\{0,t,1\}} \left[ \frac{1-\theta_s^2}{4(z-s)^2} + \frac{c_s}{z-s} \right] \\ &- \frac{3}{4(z-q)^2} + \frac{c_q}{z-q}. \end{aligned}$$

The first three points are true singularities. At $z=q$, the local exponents are $-1/2$ and $3/2$, so their difference is $2$. The coefficient

$$c_q = p + \frac12 \left( \frac1q + \frac1{q-t} + \frac1{q-1} \right)$$

and define $d_q$ by

$$T_{\mathrm{def}}(z) = -\frac{3}{4(z-q)^2} + \frac{c_q}{z-q} + d_q + O(z-q).$$

The no-log relation is

$$d_q=-c_q^2.$$

Before the Liouville gauge, the apparent exponents are $0,2$ and the local monodromy is $I$. In the displayed normal form the square-root gauge changes it to the central matrix $-I$. It remains projectively trivial, so $q$ is not a fifth point of the matrix connection.

Expansion at infinity supplies two linear constraints,

$$c_0+c_t+c_1+c_q=0,$$

and

$$\begin{aligned} \frac{1-\theta_\infty^2}{4} ={}& \sum_{s\in\{0,t,1\}} \frac{1-\theta_s^2}{4} - \frac34 \\ &+ t\,c_t+c_1+q\,c_q. \end{aligned}$$

Together with the no-log condition, these leave one scalar accessory in addition to the apparent position. This is the scalar image of the two-dimensional four-pole phase space.

Why “deformed Heun”?

The equation has four true regular singularities $0,t,1,\infty$ and one additional apparent point $q$. It is therefore often called a deformed Heun equation. At a generic time it is not the four-singularity equation solved by HeunG.

Eliminating momentum gives Painlevé VI

Use the shifted momentum from the preceding page,

$$P = p + \frac12 \left[ \frac{\theta_0}{q} + \frac{\theta_1}{q-1} + \frac{\theta_t}{q-t} \right].$$

One standard polynomial Hamiltonian is

$$\begin{aligned} t(t-1)K_{\mathrm{VI}} ={}& q(q-1)(q-t)P^2 \\ &- \Bigl[ \theta_0(q-1)(q-t) + \theta_1q(q-t) \\ &\hspace{3.2em} + (\theta_t-1)q(q-1) \Bigr]P \\ &+ \kappa(q-t), \end{aligned}$$

where

$$\kappa = \frac{ (\theta_0+\theta_1+\theta_t-1)^2 - (\theta_\infty-1)^2 }{4}.$$

The first Hamilton equation gives

$$\begin{aligned} 2q(q-1)(q-t)P ={}& t(t-1)q' \\ &+ \theta_0(q-1)(q-t) + \theta_1q(q-t) \\ &+ (\theta_t-1)q(q-1). \end{aligned}$$

Substituting this expression into the second Hamilton equation and simplifying yields

$$\begin{aligned} q'' ={}& \frac12 \left( \frac1q+\frac1{q-1}+\frac1{q-t} \right)(q')^2 \\ &- \left( \frac1t+\frac1{t-1}+\frac1{q-t} \right)q' \\ &+ \frac{q(q-1)(q-t)} {2t^2(t-1)^2} \Biggl[ (\theta_\infty-1)^2 - \frac{\theta_0^2t}{q^2} \\ &\hspace{7.0em} + \frac{\theta_1^2(t-1)}{(q-1)^2} + \frac{(1-\theta_t^2)t(t-1)} {(q-t)^2} \Biggr]. \end{aligned}$$

This is Painlevé VI in the parameter convention

$$\left( \alpha_{\mathrm{VI}}, \beta_{\mathrm{VI}}, \gamma_{\mathrm{VI}}, \delta_{\mathrm{VI}} \right) = \frac12 \left( (\theta_\infty-1)^2, - \theta_0^2, \theta_1^2, 1-\theta_t^2 \right).$$

The nonlinear function $q(t)$ is the apparent singularity of the scalar Lax equation. The fixed singular times of Painlevé VI are $0,1,\infty$. The value $q=t$ makes the displayed second-order equation look singular because eliminating momentum divided by $q-t$. In the original spectral chart $p$ generally diverges there, but the shifted polynomial momentum $P$ remains finite and resolves the collision branch. Adjacent Okamoto charts cover the complementary branch and exceptional parameter loci.

The displayed $K_{\mathrm{VI}}$ generates the chosen $(q,P)$ chart. It is not automatically the bare JMU derivative $H_t$. Their exact relation contains the canonical and gauge shifts derived on the preceding two pages.

A generic cyclic reduction has one moving apparent point $\lambda$ and is governed by Painlevé VI. On the marked slice $\lambda(t_0)=t_0$, an integer exponent shift and a scalar gauge merge that point with the true pole, leaving a four-singularity Heun equation. The standard accessory $q_{\mathrm H}$ is reconstructed from the shifted-data JMU logarithmic derivative after trace corrections; the neighboring tau zero selects the collision time.

A collision slice produces the Heun equation

The collision algebra is clearest in a second, explicitly declared gauge. Up to this point the traceless residues were denoted by $A_\nu$. In this section only, rename that connection $B(z)$ and reserve $A(z)$ for its rank-one scalar lift:

$$A(z) = B(z) + \frac12 \left( \frac{\theta_0}{z} + \frac{\theta_t}{z-t} + \frac{\theta_1}{z-1} \right)I.$$

Each finite residue now has eigenvalues $0,\theta_\nu$ rather than $\pm\theta_\nu/2$. Choose

$$A_\infty = \operatorname{diag}(\kappa_-,\kappa_+), \qquad \kappa_\pm = - \frac12 \left( \theta_0+\theta_t+\theta_1 \pm\theta_\infty \right).$$

This adds scalar multiples of the identity to the traceless residues. It preserves the projective monodromy but changes the chosen exponent lifts and multiplies tau by an elementary factor.

Write the zero of $A_{12}(z)$ as $\lambda$, and define $\mu=A_{11}(\lambda)$. The scalar lift leaves the upper-right entry unchanged and gives

$$\lambda=q, \qquad \mu = p + \frac12 \left( \frac{\theta_0}{q} + \frac{\theta_t}{q-t} + \frac{\theta_1}{q-1} \right) =P.$$

Thus the Hamiltonian $\mathcal K$ below is the same polynomial $K_{\mathrm{VI}}$ written in $(\lambda,\mu)$ coordinates, since

$$\kappa_-(1+\kappa_+)=\kappa.$$

The first component obeys

$$\begin{aligned} y'' &+ \left( \frac{1-\theta_0}{z} + \frac{1-\theta_t}{z-t} + \frac{1-\theta_1}{z-1} - \frac1{z-\lambda} \right)y' \\ &+ \Biggl[ \frac{\kappa_-(1+\kappa_+)} {z(z-1)} - \frac{t(t-1)\mathcal K} {z(z-t)(z-1)} \\ &\hspace{5.4em} + \frac{\lambda(\lambda-1)\mu} {z(z-\lambda)(z-1)} \Biggr]y =0. \end{aligned}$$

Apparency at $\lambda$ fixes the Hamiltonian

$$\begin{aligned} \mathcal K(\lambda,\mu,t) ={}& \frac{\lambda(\lambda-t)(\lambda-1)} {t(t-1)} \Biggl[ \mu^2 \\ &- \left( \frac{\theta_0}{\lambda} + \frac{\theta_1}{\lambda-1} + \frac{\theta_t-1}{\lambda-t} \right)\mu \\ &+ \frac{\kappa_-(1+\kappa_+)} {\lambda(\lambda-1)} \Biggr]. \end{aligned}$$

At an ordinary point $t_0\notin\{0,1\}$, impose

$$\lambda(t_0)=t_0, \qquad \mu(t_0)=-\frac{K_0}{\theta_t}, \qquad \theta_t\ne0.$$

Although the printed Hamiltonian appears singular at $\lambda=t$, its limit is finite:

$$\mathcal K(t_0) = - (\theta_t-1)\mu(t_0).$$

The two scalar accessory terms then combine because

$$\mu(t_0)-\mathcal K(t_0) = \theta_t\mu(t_0) = -K_0.$$

At the same time, the coefficient of $y'$ at the merged point becomes $-\theta_t/(z-t_0)$. Introduce the Heun exponent differences

$$\begin{aligned} \vartheta_0&=\theta_0, & \vartheta_t&=\theta_t+1, \\ \vartheta_1&=\theta_1, & \vartheta_\infty&=\theta_\infty-1. \end{aligned}$$

The collision equation is

$$\begin{aligned} y'' &+ \left( \frac{1-\vartheta_0}{z} + \frac{1-\vartheta_1}{z-1} + \frac{1-\vartheta_t}{z-t_0} \right)y' \\ &+ \left[ \frac{\alpha_{\mathrm H}\beta_{\mathrm H}} {z(z-1)} - \frac{t_0(t_0-1)K_0} {z(z-1)(z-t_0)} \right]y =0, \end{aligned}$$

where

$$\begin{aligned} \alpha_{\mathrm H} &= 1- \frac{ \vartheta_0+\vartheta_t+\vartheta_1-\vartheta_\infty }{2}, \\ \beta_{\mathrm H} &= 1- \frac{ \vartheta_0+\vartheta_t+\vartheta_1+\vartheta_\infty }{2}. \end{aligned}$$

Indeed,

$$\alpha_{\mathrm H}=\kappa_-, \qquad \beta_{\mathrm H}=1+\kappa_+, \qquad \alpha_{\mathrm H}-\beta_{\mathrm H} = \vartheta_\infty,$$

and

$$\alpha_{\mathrm H}\beta_{\mathrm H} = \kappa_-(1+\kappa_+),$$

and the Fuchs relation is automatic. In the house convention

$$\begin{aligned} \gamma_{\mathrm H}&=1-\vartheta_0, & \delta_{\mathrm H}&=1-\vartheta_1, \\ \epsilon_{\mathrm H}&=1-\vartheta_t, & a_{\mathrm H}&=t_0. \end{aligned}$$

Combining the two fractions in the coefficient of $y$ finally gives the standard Heun accessory

$$q_{\mathrm H} = t_0\alpha_{\mathrm H}\beta_{\mathrm H} + t_0(t_0-1)K_0.$$

The additive first term is essential. The compact coefficient $K_0$, the normal-form residue $c_{t_0}$, and the standard parameter $q_{\mathrm H}$ are affine coordinates on the same one-dimensional accessory fiber, but they are not numerically identical.

For the Chapter 3 normal form, define

$$\Delta_\nu^{\mathrm{sc}} = \frac{1-\vartheta_\nu^2}{4}, \qquad \Lambda_{\mathrm{sc}} = \Delta_\infty^{\mathrm{sc}} - \Delta_0^{\mathrm{sc}} - \Delta_1^{\mathrm{sc}} - \Delta_t^{\mathrm{sc}}.$$

Then the remaining two changes of accessory coordinate are

$$\begin{aligned} \kappa_{\mathrm H} &= q_{\mathrm H} - \frac{\gamma_{\mathrm H}}2 \left( t_0\delta_{\mathrm H} + \epsilon_{\mathrm H} \right), \\ c_{t_0}^{\mathrm{sc}} &= \frac{ t_0\Lambda_{\mathrm{sc}}-\kappa_{\mathrm H} }{ t_0(t_0-1) }. \end{aligned}$$

What happened at the collision?

The true poles did not collide with each other: $t_0$ remains distinct from $0$ and $1$. Instead, a movable apparent point met the true pole at $t_0$, and the scalar exponent lift at that pole changed by one. When $\theta_t=0$, the displayed momentum chart is singular and the collision must be taken in an adjacent gauge or by a controlled limit.

Tau constraints separate two neighboring systems

Write $\varsigma_{ij}$ for the chosen target-Heun composite-monodromy lifts and $\sigma_{ij}$ for their shifted Painlevé counterparts. The exponent and composite-monodromy lifts of the deformation system are shifted from the target Heun lifts:

$$\begin{array}{c|cccc|ccc} &\theta_0&\theta_t&\theta_1&\theta_\infty &\sigma_{0t}&\sigma_{1t}&\sigma_{01} \\ \hline \rho_{\mathrm H} &\vartheta_0&\vartheta_t&\vartheta_1&\vartheta_\infty &\varsigma_{0t}&\varsigma_{1t}&\varsigma_{01} \\ \rho &\vartheta_0&\vartheta_t-1&\vartheta_1&\vartheta_\infty+1 &\varsigma_{0t}-1&\varsigma_{1t}-1&\varsigma_{01} \end{array}$$

These integer shifts are part of an elementary Schlesinger transformation. Omitting them changes the marked scalar equation even when some projective monodromy invariants look unchanged.

Let

$$B_\nu = A_\nu-\frac{\theta_\nu}{2}I$$

be the traceless residues, and let $\tau_{\mathrm J}(\rho;t)$ be their JMU tau function. In the convention fixed on the preceding page,

$$\begin{aligned} H_t &= \frac{\operatorname{tr}(B_tB_0)}{t} + \frac{\operatorname{tr}(B_tB_1)}{t-1} \\ &= \frac{\dd}{\dd t}\log\tau_{\mathrm J}(\rho;t). \end{aligned}$$

Passing from the rank-one residues $A_\nu$ to $B_\nu$ is not invisible:

$$\operatorname{tr}(B_tB_j) = \operatorname{tr}(A_tA_j) - \frac{\theta_t\theta_j}{2}, \qquad j\in\{0,1\}.$$

Taking the smooth finite-$\mu$ collision limit gives

$$\begin{aligned} H_t(t_0) ={}& K_0 + \frac{\theta_0\theta_t}{2t_0} + \frac{\theta_1\theta_t}{2(t_0-1)}. \end{aligned}$$

Therefore the compact accessory is

$$K_0 = \left. \left[ \frac{\dd}{\dd t}\log\tau_{\mathrm J}(\rho;t) - \frac{\theta_0\theta_t}{2t} - \frac{\theta_1\theta_t}{2(t-1)} \right] \right|_{t=t_0},$$

and the standard Heun accessory is

$$\begin{aligned} q_{\mathrm H} ={}& t_0\alpha_{\mathrm H}\beta_{\mathrm H} + t_0(t_0-1) \left. \frac{\dd}{\dd t} \log\tau_{\mathrm J}(\rho;t) \right|_{t=t_0} \\ &- \frac{(t_0-1)\theta_0\theta_t}{2} - \frac{t_0\theta_1\theta_t}{2}. \end{aligned}$$

Combining this expression with the Chapter 3 normal-form dictionary causes all dependence on $\vartheta_0$, $\vartheta_1$, and $\vartheta_\infty$ outside the tau derivative to cancel:

$$c_{t_0}^{\mathrm{sc}} = - \left. \frac{\dd}{\dd t} \log\tau_{\mathrm J}(\rho;t) \right|_{t=t_0} + \frac{ (\vartheta_t-1)(2t_0-1) }{ 2t_0(t_0-1) }.$$

This compact identity belongs to the displayed cyclic component, infinity eigenline ordering, and collision branch. Another component or the other collision branch has a correspondingly shifted formula.

The affine sigma convention

Set

$$\zeta_{\mathrm J}(t) = t(t-1) \frac{\dd}{\dd t} \log\tau_{\mathrm J}(\rho;t).$$

The two collision data for this traceless JMU representative are

$$\begin{aligned} \zeta_{\mathrm J}(t_0) ={}& t_0(t_0-1)K_0 \\ &+ \frac{(t_0-1)\theta_0\theta_t}{2} + \frac{t_0\theta_1\theta_t}{2}, \\ \zeta_{\mathrm J}'(t_0) ={}& \frac{\theta_t(\theta_\infty-\theta_t)}{2}. \end{aligned}$$

For the second line, the finite-$\mu$ branch makes $B_t$ triangular with diagonal entries $-\theta_t/2,\theta_t/2$. Since $B_0+B_1=-B_\infty-B_t$, taking the trace of $B_t(B_0+B_1)$ gives the displayed derivative without choosing residue coordinates.

The scalar lift instead has the tau representative

$$\tau_A(t) = t^{\theta_0\theta_t/2} (t-1)^{\theta_t\theta_1/2} \tau_{\mathrm J}(\rho;t)$$

on a chosen logarithm branch. Its sigma variable is shifted by the exact affine function

$$\begin{aligned} \zeta_A(t) &= t(t-1)\frac{\dd}{\dd t}\log\tau_A(t) \\ &= \zeta_{\mathrm J}(t) + \frac{(t-1)\theta_0\theta_t}{2} + \frac{t\theta_t\theta_1}{2}. \end{aligned}$$

This time-dependent factor is a change of tau representative induced by the scalar gauge. It is not the time-independent JMU normalization constant. All accessory formulas on this page use $\tau_{\mathrm J}(\rho;t)$ and $\zeta_{\mathrm J}$.

The neighboring tau zero selects the collision

Let $\rho^+$ be obtained from $\rho$ by

$$(\theta_t,\theta_\infty,\sigma_{0t},\sigma_{1t}) \longmapsto (\theta_t+1,\theta_\infty-1,\sigma_{0t}+1,\sigma_{1t}+1).$$

Let $\rho^-$ denote the inverse shift, and write $\tau_{\rho^\pm}$ for the corresponding neighboring JMU tau functions. Thus $\rho^+=\rho_{\mathrm H}$ in the displayed lift table. On the nonresonant Schlesinger patch, coherent normalizations of the three neighboring tau functions obey the Toda identity

$$\begin{aligned} \zeta_{\mathrm J}'(t) - \frac{\theta_t(\theta_\infty-\theta_t)}{2} &= C\, \frac{ \tau_{\rho^+}(t)\tau_{\rho^-}(t) }{ \tau_{\mathrm J}(\rho;t)^2 }, \end{aligned}$$

where $C$ is a nonzero, $t$-independent normalization constant. At the collision the left-hand side vanishes. Provided the base tau is nonzero, one neighboring factor must vanish; choosing the $+$ factor selects $\lambda(t_0)=t_0$. Thus

$$\tau_{\rho^+}(t_0)=0.$$

This is the collision branch of the standard Toda identity, equation (3.15) of Anselmo et al. Consequently, the zero of $\tau_{\rho^+}$ locates the Heun slice, whereas the logarithmic derivative of the generally nonzero $\tau_{\mathrm J}(\rho;t)$ supplies its accessory. These are two different tau functions.

Inverse-monodromy workflow

Given a Heun position, local exponents, and target monodromy data:

1. shift the marked data from $\rho_{\mathrm H}$ to $\rho$;
2. enforce $\tau_{\rho^+}(t_0)=0$, or the equivalent $\zeta_{\mathrm J}'$ condition with the correct branch;
3. evaluate $\partial_t\log\tau_{\mathrm J}(\rho;t_0)$;
4. subtract the two trace corrections to obtain $K_0$;
5. recover $q_{\mathrm H}$ and verify it by direct monodromy or connection calculation.

If a spectral problem is intended, its boundary condition must still be translated into a constraint on $\rho_{\mathrm H}$.

An exact rational collision checkpoint

Choose the target Heun exponent differences

$$\left( \vartheta_0,\vartheta_t,\vartheta_1,\vartheta_\infty \right) = \left( \frac13,\frac12,\frac14,\frac25 \right)$$

and set $t_0=2$. The shifted Painlevé data are

$$\left( \theta_0,\theta_t,\theta_1,\theta_\infty \right) = \left( \frac13,-\frac12,\frac14,\frac75 \right).$$

Then

$$\kappa_-=\frac{79}{120}, \qquad 1+\kappa_+=\frac{31}{120},$$

so

$$\alpha_{\mathrm H}=\frac{79}{120}, \qquad \beta_{\mathrm H}=\frac{31}{120}, \qquad \alpha_{\mathrm H}\beta_{\mathrm H} = \frac{2449}{14400}.$$

Take

$$K_0=\frac13, \qquad \lambda(t_0)=t_0, \qquad \mu(t_0)=\frac23.$$

Because $\theta_t=-1/2$, the collision relation $\mu=-K_0/\theta_t$ holds. Moreover,

$$\mathcal K(t_0) = - (\theta_t-1)\mu = 1, \qquad \mu-\mathcal K=-\frac13=-K_0.$$

The resulting house-convention Heun equation is

$$\begin{aligned} y'' &+ \left( \frac{2}{3z} + \frac{3}{4(z-1)} + \frac{1}{2(z-2)} \right)y' \\ &+ \frac{ \dfrac{2449}{14400}z - \dfrac{7249}{7200} }{ z(z-1)(z-2) }y =0. \end{aligned}$$

Indeed,

$$q_{\mathrm H} = 2\left(\frac{2449}{14400}\right) + 2\left(\frac13\right) = \frac{7249}{7200},$$

and

$$\gamma_{\mathrm H}+\delta_{\mathrm H}+\epsilon_{\mathrm H} = \frac{23}{12} = \alpha_{\mathrm H}+\beta_{\mathrm H}+1.$$

The two trace corrections give

$$\begin{aligned} H_t(t_0) &= K_0 + \frac{\theta_0\theta_t}{2t_0} + \frac{\theta_1\theta_t}{2(t_0-1)} \\ &= \frac13-\frac1{24}-\frac1{16} = \frac{11}{48}. \end{aligned}$$

Hence the traceless JMU sigma data are

$$\zeta_{\mathrm J}(t_0)=\frac{11}{24}, \qquad \zeta_{\mathrm J}'(t_0)=-\frac{19}{40}.$$

Conversely, the corrected tau formula gives

$$\begin{aligned} q_{\mathrm H} ={}& \frac{2449}{7200} + \frac{11}{24} + \frac1{12} + \frac18 \\ &= \frac{7249}{7200}. \end{aligned}$$

The two Chapter 3 normal-form coordinates are

$$\kappa_{\mathrm H} = \frac{2449}{7200}, \qquad c_{t_0}^{\mathrm{sc}} = -\frac{29}{48}.$$

This benchmark checks the exponent shifts, collision sign, trace corrections, sigma derivative, standard Heun accessory, and normal-form residue with exact arithmetic.

What the reduction does and does not claim

Generic time versus Heun slice. A generic PVI Lax scalarization has an apparent fifth point. HeunG is obtained only on a collision slice or an equivalent specialization in another chart.

Local existence versus a global collision. Prescribing $\lambda(t_0)=t_0$ and $\mu(t_0)$ gives a local PVI solution away from exceptional parameters. For one fixed global monodromy point, collision times are discrete and need not occur in a chosen domain.

Gauge-specific formula versus invariant content. The printed $K_{\mathrm{VI}}=\mathcal K$, $K_0$, $c_{t_0}^{\mathrm{sc}}$, and $q_{\mathrm H}$ are different Hamiltonian or scalar coordinates. The monodromy representation is the invariant object; coefficient identities must carry their gauge dictionaries.

Tau zero versus spectral zero. The transformed tau zero enforces the apparent-pole collision. It becomes an eigenvalue or resonance condition only after separate physical boundary constraints are encoded in the monodromy data.

Common pitfalls

Calling the apparent coordinate the Heun accessory. The PVI dependent variable is $q(t)$ or $\lambda(t)$. The standard Heun accessory is $q_{\mathrm H}$ and is obtained only after specialization and an affine translation.

Dropping the integer exponent shifts. The collision changes $\theta_t$ by $+1$ and $\theta_\infty$ by $-1$ on the Heun side, together with the marked composite lifts. Reusing the unshifted tuple gives the wrong scalar equation.

Using the polynomial Hamiltonian as a tau derivative. The displayed $K_{\mathrm{VI}}$ belongs to a time-dependent Darboux chart. The JMU identity uses the bare residue Hamiltonian $H_t$.

Dropping the trace corrections. The scalar lift and the traceless JMU system have residue Hamiltonians differing by two explicit half-trace terms. They cannot be absorbed into the time-independent multiplicative normalization of tau.

Setting the wrong tau function to zero. The collision uses a Schlesinger-transformed neighbor $\tau_{\rho^+}$, while the accessory uses the logarithmic derivative of $\tau_{\mathrm J}(\rho;t)$. Conflating them produces a spurious pole.

Treating $\lambda=t$ as a true-pole collision. The true pole remains at $t_0\notin\{0,1\}$. It is the apparent point that meets it, so the event is an accessible divisor resolved by the finite-$\mu$ chart, not a degeneration of the four-punctured sphere.

Exercises

1. Recover the apparent indices

Use the rank-one scalar equation to compute the indicial roots at $z=\lambda$. Why is an additional no-log condition still required?

Solution

Put $x=z-\lambda$. The coefficient of $y'$ is $-x^{-1}+O(1)$, while the coefficient of $y$ has at most a simple pole. The indicial polynomial is therefore

$$\rho(\rho-1)-\rho = \rho(\rho-2),$$

with roots $0$ and $2$. Their integer separation permits a logarithm in the smaller-exponent solution. Expanding that solution in a Taylor series gives one resonant compatibility condition. The printed formula for $\mathcal K(\lambda,\mu,t)$ is precisely the global coefficient relation that enforces it.

2. Eliminate the Painlevé momentum

Solve the first Hamilton equation for $P$ and identify the four standard Painlevé VI parameters after eliminating it.

Solution

Differentiation of $K_{\mathrm{VI}}$ with respect to $P$ gives

$$\begin{aligned} P = \frac{1}{2q(q-1)(q-t)} \Bigl[ &t(t-1)q' \\ &+ \theta_0(q-1)(q-t) \\ &+ \theta_1q(q-t) \\ &+ (\theta_t-1)q(q-1) \Bigr]. \end{aligned}$$

Substitution into $P'=-\partial_qK_{\mathrm{VI}}$ gives the equation in the body. Comparison with the standard four-parameter form yields

$$\alpha_{\mathrm{VI}} = \frac{(\theta_\infty-1)^2}{2}, \quad \beta_{\mathrm{VI}} = -\frac{\theta_0^2}{2},$$

and

$$\gamma_{\mathrm{VI}} = \frac{\theta_1^2}{2}, \quad \delta_{\mathrm{VI}} = \frac{1-\theta_t^2}{2}.$$

3. Verify the collision cancellation

Take $\lambda\to t_0$ in $\mathcal K$ and show that the two scalar accessory terms reduce to $-t_0(t_0-1)K_0$ over the Heun denominator.

Solution

Only the term containing $(\theta_t-1)\mu/(\lambda-t)$ survives the prefactor $\lambda(\lambda-t)(\lambda-1)/[t(t-1)]$. Hence

$$\mathcal K(t_0) = -(\theta_t-1)\mu(t_0).$$

After setting $\mu(t_0)=-K_0/\theta_t$,

$$\mu(t_0)-\mathcal K(t_0) = \theta_t\mu(t_0) = -K_0.$$

The two last terms of the scalar equation therefore combine into

$$- \frac{t_0(t_0-1)K_0} {z(z-1)(z-t_0)}.$$

4. Translate the compact accessory

Combine the two fractions in the collision equation and derive the standard Heun parameter $q_{\mathrm H}$.

Solution

With $D(z)=z(z-1)(z-t_0)$,

$$\begin{aligned} \frac{\alpha_{\mathrm H}\beta_{\mathrm H}}{z(z-1)} &- \frac{t_0(t_0-1)K_0}{D(z)} \\ &= \frac{ \alpha_{\mathrm H}\beta_{\mathrm H}(z-t_0) - t_0(t_0-1)K_0 }{D(z)}. \end{aligned}$$

Comparison with $(\alpha_{\mathrm H}\beta_{\mathrm H}z-q_{\mathrm H})/D(z)$ gives

$$q_{\mathrm H} = t_0\alpha_{\mathrm H}\beta_{\mathrm H} + t_0(t_0-1)K_0.$$

5. Recover the trace corrections

Starting from the definition of $\zeta_{\mathrm J}$ and its first collision datum, derive the JMU expression for $K_0$. Then compute the affine change from $\zeta_{\mathrm J}$ to $\zeta_A$.

Solution

By definition,

$$\zeta_{\mathrm J} = t(t-1)H_t.$$

The first collision condition therefore gives

$$\begin{aligned} K_0 ={}& H_t(t_0) - \frac{\theta_0\theta_t}{2t_0} - \frac{\theta_1\theta_t}{2(t_0-1)} \\ &= \left. \left[ \frac{\dd}{\dd t}\log\tau_{\mathrm J}(\rho;t) - \frac{\theta_0\theta_t}{2t} - \frac{\theta_1\theta_t}{2(t-1)} \right] \right|_{t=t_0}. \end{aligned}$$

Multiplication by the scalar-gauge factor changes the sigma variable by

$$\zeta_A-\zeta_{\mathrm J} = \frac{(t-1)\theta_0\theta_t}{2} + \frac{t\theta_t\theta_1}{2}.$$

The second collision datum itself remains $\zeta_{\mathrm J}'(t_0) =\theta_t(\theta_\infty-\theta_t)/2$ in the declared traceless representative.

6. Audit the rational benchmark

Verify the Fuchs relation, $H_t(t_0)$, $q_{\mathrm H}$, $\zeta_{\mathrm J}'(t_0)$, and $c_{t_0}^{\mathrm{sc}}$ for the exact example.

Solution

The finite Heun residues are

$$\gamma_{\mathrm H}=\frac23, \qquad \delta_{\mathrm H}=\frac34, \qquad \epsilon_{\mathrm H}=\frac12.$$

Thus

$$\gamma_{\mathrm H}+\delta_{\mathrm H}+\epsilon_{\mathrm H} = \frac{23}{12}.$$

Also

$$\alpha_{\mathrm H}+\beta_{\mathrm H}+1 = \frac{31+79}{120}+1 = \frac{23}{12}.$$

The JMU derivative is

$$H_t(t_0) = \frac13-\frac1{24}-\frac1{16} = \frac{11}{48},$$

so $\zeta_{\mathrm J}(t_0)=2H_t(t_0)=11/24$. The standard accessory follows either from $K_0$ or from the corrected tau formula:

$$\begin{aligned} q_{\mathrm H} &= \frac{2449}{7200} + \frac{11}{24} + \frac1{12} + \frac18 \\ &= \frac{7249}{7200}. \end{aligned}$$

The derivative condition gives

$$\begin{aligned} \zeta_{\mathrm J}'(t_0) &= \frac12 \left( \frac75+\frac12 \right) \left( -\frac12 \right) = -\frac{19}{40}. \end{aligned}$$

Finally, the compact normal-form identity yields

$$c_{t_0}^{\mathrm{sc}} = -\frac{11}{48} - \frac38 = -\frac{29}{48}.$$

7. Diagnose a proposed Heun–Painlevé identity

A formula sets $\lambda=q_{\mathrm H}$ for every $t$ and evaluates an unshifted tau function at the same data. Identify at least four missing or incorrect steps.

Solution

The proposal:

1. confuses the apparent position $\lambda(t)$ with the standard Heun accessory $q_{\mathrm H}$;
2. ignores that the generic scalar equation has five singularities;
3. omits the collision condition $\lambda(t_0)=t_0$;
4. omits the integer shifts at $t$ and infinity, and the corresponding composite-lift shifts;
5. fails to distinguish the original JMU tau function from its Schlesinger-transformed neighbor;
6. misses the affine term $t_0\alpha_{\mathrm H}\beta_{\mathrm H}$ in the standard accessory.

Any spectral interpretation would additionally require boundary conditions and a spectral-parameter dictionary.

References

• M. Jimbo and T. Miwa, “Monodromy Preserving Deformation of Linear Ordinary Differential Equations with Rational Coefficients. II”, Physica D 2 (1981), 407–448. Painlevé Lax systems, elementary Schlesinger transformations, and neighboring tau functions.
• K. Okamoto, “Studies on the Painlevé Equations I. Sixth Painlevé Equation PVI”, Annali di Matematica Pura ed Applicata 146 (1987), 337–381. Polynomial Hamiltonians, affine-Weyl transformations, and regular initial-value charts.
• T. Anselmo, R. Nelson, B. Carneiro da Cunha, and D. G. Crowdy, “Accessory Parameters in Conformal Mapping: Exploiting the Isomonodromic Tau Function for Painlevé VI”, Proceedings of the Royal Society A 474 (2018), 20180080. Equations (3.3), (3.13)–(3.16), and (4.1) give the explicit deformed-Heun collision and tau constraints used here.
• B. Dubrovin and A. Kapaev, “A Riemann–Hilbert Approach to the Heun Equation”, SIGMA 14 (2018), 093. Heun equations as specified critical members of Painlevé VI isomonodromic families.
• J. Xia, S.-X. Xu, and Y.-Q. Zhao, “Isomonodromy Sets of Accessory Parameters for Heun Class Equations”, Studies in Applied Mathematics 146 (2021), 901–952. Systematic collision limits and discrete isomonodromy sets for the Heun class.
• J. Dereziński, A. Ishkhanyan, and A. Latosiński, “From Heun Class Equations to Painlevé Equations”, SIGMA 17 (2021), 056. A unified treatment of deformed Heun-class equations, apparent singularities, and their Painlevé deformations.
• K. Takemura, “Middle Convolution and Heun’s Equation”, SIGMA 5 (2009), 040. Heun equations as special loci in the four-pole Painlevé VI connection moduli and their Okamoto transformations.
• A. Litvinov, S. Lukyanov, N. Nekrasov, and A. Zamolodchikov, “Classical Conformal Blocks and Painlevé VI”, Journal of High Energy Physics 2014 (2014), 144. The Heun-monodromy problem as a Painlevé VI connection problem in the classical-block setting.
• K. Iwasaki, H. Kimura, S. Shimomura, and M. Yoshida, From Gauss to Painlevé: A Modern Theory of Special Functions, Vieweg, 1991. A systematic geometric treatment of Fuchsian systems, apparent singularities, and Painlevé VI.