Resolving Singularities and Monodromy Reduction of Fuchsian Connections

Abstract

We study monodromy reduction of Fuchsian connections from a sheave theoretic viewpoint, focusing on the case when a singularity of a special connection with four singularities has been resolved. The main tool of study is based on a bundle modification technique due to Drinfeld and Oblezin. This approach via invariant spaces and eigenvalue problems allows us not only to explain Erdélyi’s classical infinite hypergeometric expansions of solutions to Heun equations, but also to obtain new expansions not found in his papers. As a consequence, a geometric proof of Takemura’s eigenvalues inclusion theorem is obtained. Finally, we observe a precise matching between the monodromy reduction criteria giving those special solutions of Heun equations and that giving classical solutions of the Painlevé VI equation.

Appendices

Appendix A. Fuchsian Relations

We point out that there is a difference between the Fuchsian relation of a Fuchsian connection and a Fuchsian differential equation [2] that is derived from the Fuchsian connection. Since we cannot find a reference for this fact, a proof is provided here.

Let a connection $\nabla$ relative to the canonical basis have the matrix representation

$$\begin{aligned} -\sum _{j=1}^{n} \frac{A_j}{x-a_j}:=-A(x) =\left( \begin{matrix} A_{11} &{} A_{12}\\ A_{21} &{} A_{22} \end{matrix} \right) \end{aligned}$$

(A.1)

such that $a_{n+1}=\infty$ and that the residue matrices $A_j\ (j=1,\cdots , n+1)$ satisfy the Fuchsian relation:

$$\begin{aligned} \sum _{j=1}^{n+1} {{\,\mathrm{Tr}\,}}(A_j)=0. \end{aligned}$$

(A.2)

Lemma A.1

Let ${{\,\mathrm{Tr}\,}}(A_j)=\alpha _{j1}+\alpha _{j2}$ be the trace of the residue matrix $A_j\ (j=1,\cdots n+1)$. Let y be the first component of a horizontal section of the connection defined above. Let the Riemann scheme of the differential equation

$$\begin{aligned} y^{\prime \prime }-\big ({{\,\mathrm{Tr}\,}}\big (A(x)\big )+\frac{A^\prime _{12}(x)}{A_{12}(x)}\big )y^\prime +\Big (\det A(x)-A_{11}(x)\log ^\prime \big (A_{12}(x)/A_{11}(x)\big )\Big )y=0 \end{aligned}$$

(A.3)

derived from the connection above and satisfied by y be of the form

$$\begin{aligned} P \left( \begin{matrix} a_1 &{} \cdots &{}a_j &{}\cdots &{} a_n &{} \infty &{}b_1 &{} \cdots &{}b_{n-1} \\ \beta _{11} &{} \cdots &{}\beta _{j1} &{} \cdots &{} \beta _{n1} &{}\beta _{n+1,\, 1} &{} 0 &{} \cdots &{} 0\\ \beta _{22} &{} \cdots &{}\beta _{j2} &{} \cdots &{} \beta _{n2} &{}\beta _{n+1,\, 2} &{} 2 &{} \cdots &{} 2 \end{matrix} ;\ x \right) , \end{aligned}$$

where $b_j\ (j=1,\cdots , n-1)$ are the apparent singularities. If $b_j=a_j\ (j=1,\cdots , n-1)$, then we have

$$\begin{aligned} \begin{aligned}&\alpha _{j1}+\alpha _{j2} =\beta _{j1}+\beta _{j2}-1\quad (j=1,\cdots , n-1)\\&\alpha _{n1}+\alpha _{n2} =\beta _{n1}+\beta _{n2},\\&\alpha _{n+1,\, 1}+\alpha _{n+1,\, 2} =\beta _{n+1,\, 1} +\beta _{n+1,\, 2}. \end{aligned} \end{aligned}$$

(A.4)

In particular,

$$\begin{aligned} \sum _{j=1}^{n+1} (\beta _{j1}+\beta _{j2})=(n-1)+\sum _{j=1}^{n+1} {{\,\mathrm{Tr}\,}}(A_j). \end{aligned}$$

Proof

It is sufficient to compute the coefficient of $y^\prime$. Let

$$\begin{aligned} A_{12}(x)=:\mathrm {(const.)} \frac{\prod _{j=1}^{n-1}(x-b_j)}{\prod _{j=1}^n(x-a_j)}. \end{aligned}$$

Hence

$$\begin{aligned} -{{\,\mathrm{Tr}\,}}\big (A(x)\big )-\frac{A^\prime _{12}(x)}{A_{12}(x)} =\sum _{j=1}^n \frac{1-\alpha _{j1}-\alpha _{j2}}{x-a_j} -\sum _{j=1}^{n-1}\frac{1}{x-b_j} \end{aligned}$$

(A.5)

where the $b_j\ (j=1,\cdots , n-1)$ are apparent singularities which are inherited from $A^\prime _{12}/A_{12}$ and they do not contribute to the monodromy of the equation (A.3). Let us now assume $b_j=a_j\ (j=1,\cdots , n)$. Hence,

$$\begin{aligned} \begin{aligned} -{{\,\mathrm{Tr}\,}}\big (A(x)\big )-\frac{A^\prime _{12}(x)}{A_{12}(x)}&=\sum _{j=1}^{n-1} \frac{-\alpha _{j1}-\alpha _{j2}}{x-a_j}+\frac{1-\alpha _{n1}-\alpha _{n2}}{x-a_n} \\&=\sum _{j=1}^{n-1} \frac{1-\beta _{j1}-\beta _{j2}}{x-a_j}+\frac{1-\beta _{n1}-\beta _{n2}}{x-a_n}, \end{aligned} \end{aligned}$$

where we have identified $\alpha _{j1}+\alpha _{j2}=\beta _{j1}+\beta _{j2}-1\ (j=1,\cdots , n-1)$, $\alpha _{n1}+\alpha _{n2}=\beta _{n1}+\beta _{n2}$ and $\alpha _{n+1,\, 1}+\alpha _{n+1,\, 2} =\beta _{n+1,\, 1} +\beta _{n+1,\, 2}$. Thus, the sum of the indicial roots of the Riemann scheme P above yields

$$\begin{aligned} \begin{aligned} \sum _{j=1}^{n+1} (\beta _{j1}+\beta _{j2})&=\sum _{j=1}^{n-1} (\beta _{j1}+\beta _{j2}) +(\beta _{n1}+\beta _{n2}) +(\beta _{n+1,\, 1}+\beta _{n+1,\, 2})\\&= n-1+\sum _{j=1}^{n-1} (\alpha _{j1}+\alpha _{j2}) +(\alpha _{n1}+\alpha _{n2}) +(\alpha _{n+1,\, 1}+\alpha _{n+1,\, 2})\\&=n-1+\sum _{j=1}^{n+1} {{\,\mathrm{Tr}\,}}(A_j). \end{aligned} \end{aligned}$$

$\square$

Remark A.2

It is clear that one can also include $\infty$ as one of the apparent singular points in the above theorem. We show below an example in terms of a hypergeometric equation where the apparent singularity is placed at infinity.

Example A.3

Consider $Y(x)=\begin{bmatrix}y_1(x)\\ y_2(x)\end{bmatrix}$ satisfying the following $2\times 2$ Fuchsian system

$$\begin{aligned} \frac{dY}{dx}-\left[ \dfrac{A_0}{x}+\dfrac{A_1}{x-1}\right] Y=0, \end{aligned}$$

(A.6)

corresponding to the connection of hypergeometric type ($A_0,A_1$), where $A_0=\begin{bmatrix}u_0+\gamma &{}1\\ -u_0(u_0+\gamma )&{}-u_0\end{bmatrix}$, $A_1=\begin{bmatrix}u_1+\delta &{}-1\\ u_1(u_1+\delta )&{}-u_1\end{bmatrix}$, $u_0=\frac{\alpha (\alpha +\gamma )}{\alpha -\beta }$, $u_1=\frac{\alpha (\alpha +\delta )}{\alpha -\beta }$ such that

$$\begin{aligned}A_\infty =-A_0-A_1 =\begin{bmatrix}\beta &{}0\\ 0&{}\alpha \end{bmatrix}, \text{ where } \alpha +\beta =\gamma +\delta ,\end{aligned}$$

then $y_1(x)$ satisfies

$$\begin{aligned} \frac{d^2y}{dx^2}+ \Big (\frac{1-\gamma }{x}+\frac{1-\delta }{x-1}\Big )\frac{dy}{dx}+ \frac{\beta (\alpha +1)}{x(x-1)}y=0. \end{aligned}$$

(A.7)

and $y_2(x)$ satisfies

$$\begin{aligned} \frac{d^2y}{dx^2}+ \Big (\frac{1-\gamma }{x}+\frac{1-\delta }{x-1}\Big )\frac{dy}{dx}+ \frac{\alpha (\beta +1)}{x(x-1)}y=0. \end{aligned}$$

(A.8)

Two linearly independent local solutions are identified as: $\begin{bmatrix}y_1&y_2\end{bmatrix}^T$ and $\begin{bmatrix}\tilde{y_1}&\tilde{y_2}\end{bmatrix}^T$, where

and

Appendix B. Convergence of Erdélyi’s Expansions

The Heun function can be expanded as

(B.1)

whose coefficients $X_m$ satisfy a three-term recursion relation

$$\begin{aligned} {\left\{ \begin{array}{ll} L_0X_0+M_0X_1=0\\ K_mX_{m-1}+L_mX_m+M_mX_{m+1}&{}=0,\ m=1,2,\cdots \end{array}\right. }, \end{aligned}$$

(B.2)

where $K_m,L_m,M_m$ are given in [14, (5.3)]

$$\begin{aligned} \begin{aligned} K_{m+1}&:=a\frac{(\alpha +m)(\beta +m)(\varepsilon +m)(\alpha +\beta -\delta +m)}{(\alpha +\beta -\delta +2m)(\alpha +\beta -\delta +2m+1)}\\ L_m\&:=am(\gamma +m-1) \Big \{\frac{(\alpha +m)(\alpha -\delta +m+1)+(\beta +m)(\beta -\delta +m+1)}{(\alpha +\beta -\delta +2m-1)(\alpha +\beta -\delta +2m+1)}\\&\quad - \frac{1}{\alpha +\beta -\delta +2m-1}\Big \}- m(\alpha +\beta -\delta +m)-\alpha \beta q\\&\quad +a\frac{\alpha \beta (\gamma +2m)-\varepsilon m (\delta -m-1)}{\alpha +\beta -\delta +2m+1)}\\ M_{m-1}&:= \frac{am(\alpha -\delta +m)(\beta -\delta +m)(\gamma +m-1)}{(\alpha +\beta -\delta +2m-1)(\alpha +\beta -\delta +2m)}. \end{aligned} \end{aligned}$$

The convergence of the above series is given by

Theorem B.1

([14]) Suppose that the series (B.1) is non-terminating, and the branch of the square root is chosen such that its real part is nonnegative. Let $k=\bigg |\frac{1-\sqrt{1-a}}{1+\sqrt{1-a}}\bigg |\ne 1$. Then, it converges uniformly compacta on

$$\begin{aligned} \Omega _0=\Big \{x\in {\mathbb {C}}:\bigg |\frac{1-\sqrt{1-x}}{1+\sqrt{1-x}}\bigg |<\min (k,k^{-1})\Big \}, \end{aligned}$$

(B.3)

where $\Omega _0$ denotes a neighbourhood of 0 but excluding $x=1$ (see the remark below). Moreover, if the accessory parameter q in (1.4) satisfies the infinite continued fraction

$$\begin{aligned} L_0/M_0-\frac{K_1/M_1}{L_1/M_1-}\frac{K_2/M_2}{L_2/M_2-}\frac{K_3/M_3}{L_3/M_3-}\cdots =0, \end{aligned}$$

(B.4)

which contains q implicitly, then the series (B.1) converges in a larger region

$$\begin{aligned} \Omega _1=\Big \{x\in {\mathbb {C}}:\bigg |\frac{1-\sqrt{1-x}}{1+\sqrt{1-x}}\bigg |<\max (k,k^{-1})\Big \} \end{aligned}$$

(B.5)

except possibly on the branch cut $[1, +\infty ).$

We include a brief proof since the argument may not be easily found in modern literature.

Proof

We apply Poincaré’s theorem and Perron’s theorem to the three-term recurrence relation (B.2), see, for example, [17, Theorems 1.1, 2.1-2.2] to yield,

$$\begin{aligned} \lim _{m\rightarrow \infty }\left| \frac{X_{m+1}}{X_{m}}\right| = {\left\{ \begin{array}{ll} \min (k,k^{-1}) &{}\text{ if } (B.4) \text{ holds }\\ \max (k,k^{-1}) &{}\text{ otherwise } \end{array}\right. }.\end{aligned}$$

After applying Watson’s asymptotic representation (see [41, §9]), we derive

$$\begin{aligned} \frac{\varphi _{m+1}^1(x)}{\varphi _{m}^1(x)}\sim \frac{1-\sqrt{1-x}}{1+\sqrt{1-x}}\ \text{ as } m\rightarrow \infty \end{aligned}$$

(B.6)

(see [14, (4.7)]). Thus, the result follows from the ratio test applied to the cases $\Omega _0$ and $\Omega _1$ separately. $\square$

Remark B.2

(Description of $\Omega _0$ and $\Omega _1$). Note that for $m>0$, $\big |\frac{1-y}{1+y}\big |=m$ is equivalent to the circle equation

$$\begin{aligned}|y-y_0|=r, \text{ where } y_0=\frac{1+m^2}{1-m^2} \text{ and } r=\frac{2m}{|1-m^2|}. \end{aligned}$$

Let $m_0:=\min (k,k^{-1})<1$ and $m_1:=\max (k,k^{-1})>1$. Then,

$$\begin{aligned} \Big \{y\in {\mathbb {C}}:\bigg |\frac{1-y}{1+y}\bigg |<m_0\Big \} \end{aligned}$$

is the open disk $D_0$ centred at $y_0=\frac{1+m_0^2}{1-m_0^2}>0$ with radius $r=\frac{2m_0}{1-m_0^2}$ containing $y=1$. In particular, $D_0$ is contained in the half-plane $\{\text{ Re } y>0\}$. Since $\text{ Re }\sqrt{1-x}$ is always taken to be nonnegative, $\Omega _0$ is a neighbourhood of 0, not containing $x=1,\infty$. On the other hand,

$$\begin{aligned} \Big \{y\in {\mathbb {C}}:\bigg |\frac{1-y}{1+y}\bigg |<m_1\Big \} \end{aligned}$$

is the complement of the closed disk $D_1$ centred at $y_0=-\frac{m_1^2+1}{m_1^2-1}<0$ with radius $r=\frac{2m_1}{m_1^2-1}$. In particular, the complement contains the half-plane $\{\text{ Re } y\ge 0\}$. Since $\text{ Re }\sqrt{1-x}$ is always taken to be nonnegative, $\Omega _1={\mathbb {C}}\backslash [1,\, \infty )$.

Remark B.3

(A second linearly independent series solution). Erdélyi also considered the series solution $\displaystyle \sum \nolimits _{m=0}^\infty X_m\varphi _m$ other than (B.1) by replacing $\varphi _m^1$ with another linearly independent solution $\varphi _m$, which can be any linear combination of $\varphi _m^2,\cdots ,\varphi _m^6$ defined in [14, Eqn(4.2)]. In this case, we have the following asymptotic representation (see [14, Eqn(4.8)]) instead of (B.6)

$$\begin{aligned} \frac{\varphi _{m+1}(x)}{\varphi _{m}(x)}\sim \frac{1+\sqrt{1-x}}{1-\sqrt{1-x}}\ \text{ as } m\rightarrow \infty . \end{aligned}$$

In order to study the domain of convergence, we consider

$$\begin{aligned} \Omega _0^-=\Big \{x\in {\mathbb {C}}:\bigg |\frac{1+\sqrt{1-x}}{1-\sqrt{1-x}}\bigg |<m_0\Big \} \text{ and } \Omega _1^-=\Big \{x\in {\mathbb {C}}:\bigg |\frac{1+\sqrt{1-x}}{1-\sqrt{1-x}}\bigg |<m_1\Big \}. \end{aligned}$$

Note that

$$\begin{aligned} \Big \{y\in {\mathbb {C}}:\bigg |\frac{1+y}{1-y}\bigg |<m_0\Big \} \end{aligned}$$

is the open disk $D_0^-$ centred at $y_0=-\frac{1+m_0^2}{1-m_0^2}<0$ with radius $r=\frac{2m_0}{1-m_0^2}$ containing $y=-1$, and

$$\begin{aligned} \Big \{y\in {\mathbb {C}}:\bigg |\frac{1+y}{1-y}\bigg |<m_1\Big \} \end{aligned}$$

is the complement of the closed disk $D_1^-$ centred at $y_0=\frac{m_1^2+1}{m_1^2-1}>0$ with radius $r=\frac{2m_1}{m_1^2-1}$. In particular, $D_0^-$ is contained in the half-plane $\{\text{ Re } y<0\}$, and $D_1^-\subseteq \{\text{ Re } y>0\}$ contains $y=1$, but not $y=0,\, \infty$. Since $\text{ Re }\sqrt{1-x}$ is always taken to be nonnegative, $\Omega _0^-=\emptyset$ and $\Omega _1^-$ is the domain containing $x=1,\infty$, but not $x=0$. As the coefficients $X_m$ satisfy the same three-term recurrence relation (B.2), by the similar argument in the proof of Theorem B.1, the series

$$\begin{aligned} {\left\{ \begin{array}{ll} \text{ converges } \text{ on } \Omega _1^- &{}\text{ if } (B.4) \text{ holds }\\ \text{ diverges } &{}\text{ otherwise }. \end{array}\right. } \end{aligned}$$

We conclude that the two series $\sum _{m=0}^\infty X_m\varphi _m^1(x)$ and $\sum _{m=0}^\infty X_m\varphi _m$ both converge in $\Omega _1^-$ when (B.4) holds. $\square$