Liouville Normal Form and SL(2) Opers

Removing the first derivative does more than simplify a scalar ODE. It exposes a constant Wronskian, a traceless companion system, and a projective connection whose coordinate law contains the Schwarzian derivative. These are the local ingredients of an $SL(2)$ oper.

The construction is always valid on a simply connected regular coordinate patch. Compatible local normal-form coefficients obeying the Schwarzian overlap law define a global projective connection, equivalently $PSL(2,\mathbb C)$-oper data. A single-valued trace-zero gauge of a given $GL(2)$ system is controlled by its determinant local system, while lifting the underlying projective local system to $SL(2)$ is a separate lifting problem. A scalar realization on $K^{-1/2}$ requires a theta characteristic; for a projective oper, that theta choice supplies a compatible $SL(2)$-oper lift, and changing the theta characteristic gives the corresponding $\mathbb Z_2$ twist.

Removing the first derivative

Start with

$$y''+p(z)y'+q(z)y=0.$$

Set

$$y(z)=f(z)\psi(z), \qquad \frac{f'}f=-\frac12p.$$

A local choice is

$$f(z) =\exp\left( -\frac12\int^z p(\zeta)\,\dd\zeta \right).$$

Direct differentiation gives

$$\begin{aligned} y' &=f\left(\psi'-\frac12p\psi\right),\\ y'' &=f\left[ \psi''-p\psi' +\left( \frac14p^2-\frac12p' \right)\psi \right]. \end{aligned}$$

The $\psi'$ terms cancel, leaving

$$\psi''+T(z)\psi=0, \qquad T=q-\frac12p'-\frac14p^2.$$

This is the book’s scalar normal form. Its companion system is

$$\Psi' = \begin{pmatrix} 0 & 1\\ -T & 0 \end{pmatrix} \Psi, \qquad \Psi= \begin{pmatrix} \psi\\ \psi' \end{pmatrix}.$$

The matrix is traceless. If $Y=(y,y')^{\mathsf T}$, the system gauge relating the two state vectors is

$$Y =f \begin{pmatrix} 1 & 0\\ -p/2 & 1 \end{pmatrix} \Psi.$$

Its determinant is $f^2$, which is precisely the local determinant-line factor removed by the transformation. More explicitly, if $G$ denotes this system gauge, then

$$\det G=f^2, \qquad (\log\det G)'=-p =\operatorname{tr}A_{\mathrm{comp}},$$

so the gauge locally trivializes the induced determinant connection.

Constant Wronskians and the oper matrix

For two normal-form solutions,

$$\Wr[\psi_1,\psi_2]'=0.$$

After a constant rescaling of one basis vector, a local fundamental matrix can therefore be normalized to

$$\Phi= \begin{pmatrix} \psi_1 & \psi_2\\ \psi_1' & \psi_2' \end{pmatrix}, \qquad \det\Phi=1.$$

On any domain where the chosen normal-form system is itself single-valued, tracelessness forces its monodromy to lie in $SL(2,\mathbb C)$. This concerns the normal-form local system; it does not assert a single-valued gauge equivalence with the original $GL(2)$ system. Forgetting the lift gives its $PSL(2,\mathbb C)$ projectivization. Distinct global lifts, when they exist, differ by a $\mathbb Z_2$ local system; the constant basis change $\Phi\mapsto-\Phi$ does not change the lift.

In a nonsingular patch, an $SL(2)$ oper can be described as:

• a rank-two bundle $E$ with a trivialized determinant;

• a flat connection $\nabla$ preserving that determinant trivialization;

• a line subbundle $L\subset E$ such that the induced map

  $$L\longrightarrow(E/L)\otimes K$$

  is an isomorphism.

Here $K$ is the canonical bundle. The last condition says that $L$ is a cyclic line: differentiating a local section of $L$ supplies the missing direction. In an oper-adapted frame, the connection takes the companion form above. Singular or parabolic opers require additional data at the punctures and are introduced only when needed.

The determinant condition and oper transversality imply

$$\begin{aligned} E/L&\simeq L^{-1},\\ L&\simeq L^{-1}\otimes K,\\ L^{\otimes2}&\simeq K. \end{aligned}$$

Thus the oper line is a theta characteristic, and the scalar operator acts on its dual $L^{-1}\simeq K^{-1/2}$.

Coordinate covariance and half-densities

Let $z=z(w)$ be locally biholomorphic and write

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

Simply composing $\psi$ with $z(w)$ introduces a first derivative. The normal-form dependent variable is instead

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

with a local branch of $s^{1/2}$. A calculation gives

$$\widetilde\psi''(w) +\widetilde T(w)\widetilde\psi(w)=0,$$

where

$$\begin{aligned} \widetilde T(w) &=s(w)^2T(z(w))\\ &\quad+\frac12\{z,w\}, \end{aligned}$$

and

$$\{z,w\} =\frac{z'''}{z'} -\frac32\left(\frac{z''}{z'}\right)^2$$

is the Schwarzian derivative.

The rule

$$\widetilde\psi =\left(\frac{\dd z}{\dd w}\right)^{-1/2}\psi$$

is the transformation law for a local section of $K^{-1/2}$. Accordingly, the normal-form operator is globally of the type

$$\partial^2+T: K^{-1/2}\longrightarrow K^{3/2},$$

once a square root $K^{1/2}$ has been chosen. Such a theta characteristic is not unique. Without choosing it, the projective connection still makes sense, but a particular scalar half-density equation is not globally fixed.

Why the Schwarzian is inevitable

The Schwarzian obeys the chain rule

$$\{z,u\} =\left(\frac{\dd w}{\dd u}\right)^2\{z,w\} +\{w,u\}.$$

This is exactly the cocycle required for successive coordinate changes to preserve the normal-form equation. A Möbius transformation has zero Schwarzian, so projective coordinate changes act without an inhomogeneous term.

If $T_1$ and $T_2$ are two projective connections, their Schwarzian terms cancel:

$$\widetilde T_1-\widetilde T_2 =\left(\frac{\dd z}{\dd w}\right)^2 (T_1-T_2).$$

Thus the difference of two projective connections is an ordinary quadratic differential. Projective connections form an affine space modeled on quadratic differentials; there is no coordinate-independent choice of “zero” projective connection on a general Riemann surface.

An exponential-coordinate check

Take the free equation

$$\psi''(z)=0$$

and set $z=\ee^w$. Since

$$\{\ee^w,w\}=-\frac12,$$

the transformed coefficient is

$$\widetilde T(w)=-\frac14.$$

The original solutions $1$ and $z$ become

$$\widetilde\psi_-(w)=\ee^{-w/2}, \qquad \widetilde\psi_+(w)=\ee^{w/2},$$

and both satisfy

$$\widetilde\psi''-\frac14\widetilde\psi=0.$$

This elementary example detects both the half-density power and the sign of the Schwarzian term.

Ratios of solutions and projective coordinates

Let $\psi_1,\psi_2$ be independent normal-form solutions and define

$$F(z)=\frac{\psi_1(z)}{\psi_2(z)}.$$

Where $\psi_2\neq0$, the Schwarzian of the ratio is

$$\{F,z\}=2T(z).$$

To verify it, use the constant Wronskian to write

$$F'=-\frac{\Wr[\psi_1,\psi_2]}{\psi_2^2}.$$

Then

$$\frac{F''}{F'} =-2\frac{\psi_2'}{\psi_2},$$

and substitution into

$$\{F,z\} =\left(\frac{F''}{F'}\right)' -\frac12\left(\frac{F''}{F'}\right)^2$$

gives $-2\psi_2''/\psi_2=2T$.

A constant change of solution basis sends $F$ to a Möbius transform. If

$$\begin{pmatrix} \widetilde\psi_1 & \widetilde\psi_2 \end{pmatrix} = \begin{pmatrix} \psi_1 & \psi_2 \end{pmatrix} \begin{pmatrix} a & b\\ c & d \end{pmatrix},$$

then

$$\widetilde F =\frac{aF+c}{bF+d}.$$

Because the Schwarzian is Möbius-invariant, $T$ does not depend on the chosen solution basis. The multivalued map $F$ is a local developing map for the projective structure, and monodromy acts on it projectively.

Regular singularities in oper form

Suppose

$$T(z) =\frac{t_{-2}}{x^2} +\frac{t_{-1}}x+O(1), \qquad x=z-z_0.$$

The normal-form exponents solve

$$\sigma(\sigma-1)+t_{-2}=0.$$

It is convenient to write

$$t_{-2}=\frac{1-\theta^2}{4},$$

so that

$$\sigma_\pm=\frac{1\pm\theta}{2}.$$

For the displayed scalar half-density lift, the local eigenvalues are

$$\mu_\pm =\exp(2\pi\ii\sigma_\pm) =-\exp(\pm\pi\ii\theta).$$

The projective conjugacy class retains their ratio $\mu_+/\mu_-=\ee^{2\pi\ii\theta}$; forgetting the scalar or $SL(2)$ lift loses their common sign. At resonance, $\theta$ still does not determine whether local monodromy is scalar or has a nontrivial Jordan part. The coefficient $t_{-1}$ is not fixed by the local conjugacy class alone; in a global Fuchsian oper it contributes to accessory data.

Schrödinger form and the quantum Schwarzian

For

$$\left[ -\hbar^2\partial_z^2+V(z) \right]\psi=E\psi,$$

the projective coefficient is

$$T(z)=\frac{E-V(z)}{\hbar^2}.$$

After $z=z(w)$,

$$\begin{aligned} \widetilde T(w) &=\left(\frac{\dd z}{\dd w}\right)^2 \frac{E-V(z(w))}{\hbar^2}\\ &\quad+\frac12\{z,w\}. \end{aligned}$$

Equivalently, if the transformed equation is written $[-\hbar^2\partial_w^2+\widetilde V]\widetilde\psi =\widetilde E\,\widetilde\psi$, then

$$\begin{aligned} \widetilde V(w)-\widetilde E &=\left(\frac{\dd z}{\dd w}\right)^2 \bigl(V(z(w))-E\bigr)\\ &\quad-\frac{\hbar^2}{2}\{z,w\}. \end{aligned}$$

The last term is sometimes called the quantum Schwarzian correction. Dropping it for an $\hbar$-independent coordinate change with an $O(1)$ Schwarzian is a leading-order $\hbar\to0$ approximation, not an exact coordinate transformation at finite $\hbar$.

Only the difference $\widetilde V-\widetilde E$ is fixed by this calculation. If $s(w)=\dd z/\dd w$ is nonconstant and $E$ varies as a spectral parameter, taking $\widetilde E$ constant generally makes $\widetilde V$ depend on $E$ through $(1-s^2)E$. A nonlinear coordinate change therefore does not usually preserve a parameter-independent potential–energy split.

Sturm–Liouville to Schrödinger form

For a real Sturm–Liouville expression, assume for the classical derivation that $P,R\in C^2$, $Q\in C$, and $P,R>0$ on the interval:

$$-\bigl(P(x)y'(x)\bigr)' +Q(x)y(x) =\lambda R(x)y(x), \qquad P>0,\quad R>0,$$

define the Liouville coordinate and scale factor by

$$\frac{\dd t}{\dd x} =\sqrt{\frac{R(x)}{P(x)}}, \qquad r(x)=\bigl(P(x)R(x)\bigr)^{1/4}.$$

With

$$u(t)=r(x)y(x),$$

one obtains

$$-\frac{\dd^2u}{\dd t^2} +\widehat Q(t)u =\lambda u,$$

where

$$\widehat Q(t) =\frac{Q(x(t))}{R(x(t))} +\frac1{r(x(t))} \frac{\dd^2r(x(t))}{\dd t^2}.$$

This transformation also changes the interval, measure, and boundary behavior. When $t$ is one-to-one,

$$\int_{t(a)}^{t(b)} |u(t)|^2\,\dd t = \int_a^b |y(x)|^2R(x)\,\dd x.$$

Thus $y\mapsto u$ is unitary from $L^2((a,b),R\,\dd x)$ to $L^2((t(a),t(b)),\dd t)$. Operator equivalence still requires transporting the domains and endpoint or boundary conditions. In a complex problem the square root, coordinate image, and zeros or poles of $P$ and $R$ must also be declared.

A normalization ledger for oper form

Before using an oper or Schrödinger potential, record:

Item	Required choice
Local scalar gauge	Primitive of $p$ and branch of its exponential
Global trace-zero gauge of a given $GL(2)$ system	Horizontal determinant trivialization; a scalar gauge additionally needs a single-valued square root
Projective coefficient	Sign convention in $\psi''+T\psi=0$
$SL(2)$ local-system lift	Lift of the projective holonomy; when lifts exist, their choices differ by a $\mathbb Z_2$ local system
Scalar oper	Theta characteristic $L\simeq K^{1/2}$; changing it gives the corresponding $\mathbb Z_2$ twist of the compatible $SL(2)$-oper lift
Projective coordinate	Ordered solution basis, defined up to Möbius action
Singular data	Exponent difference, resonance, and accessory coefficients
Spectral form	Which quantities are called $V$, $E$, and $\hbar$
Operator statement	Measure, interval or contour, domain, and boundary conditions

Common pitfalls

Treating the gauge exponential as single-valued. A local primitive removes $y'$, but its continuation can acquire a scalar factor. The projective connection globalizes more readily than a chosen scalar normal-form basis.

Transforming $T$ as a quadratic differential. A projective connection has the additional Schwarzian term. Only the difference of two projective connections transforms quadratically.

Calling every traceless system an oper. An oper also contains a cyclic line satisfying a transversality condition. A flat $SL(2)$ bundle without that line is not yet an oper.

Ignoring the function space in a Liouville transformation. The transformed differential expression may be simple while endpoint classification or the operator domain changes. Spectral equivalence needs the corresponding map of Hilbert spaces and domains.

Exercises

1. Derive normal form. Substitute $y=f\psi$ into $y''+py'+qy=0$ and determine $f$ by cancelling $\psi'$. Verify the displayed $T$.

Solution

Substitution gives

$$\psi'' +\left(2\frac{f'}f+p\right)\psi' +\left( \frac{f''}f+p\frac{f'}f+q \right)\psi=0.$$

The first derivative vanishes when $f'/f=-p/2$. Since

$$\frac{f''}f =-\frac12p'+\frac14p^2,$$

the remaining coefficient is

$$T =-\frac12p'+\frac14p^2 -\frac12p^2+q =q-\frac12p'-\frac14p^2.$$

2. Check the Schwarzian sign. Starting from $\psi''+T\psi=0$, set $\widetilde\psi=s^{-1/2}\psi(z(w))$ with $s=z'$. Derive $\widetilde T=s^2T+\{z,w\}/2$.

Solution

Writing $\psi(z(w))=s^{1/2}\widetilde\psi$ and differentiating twice, the coefficient of $\widetilde\psi'$ cancels. The remaining equation is

$$\widetilde\psi'' +\left[ s^2T +\frac12\frac{s''}{s} -\frac34\left(\frac{s'}s\right)^2 \right]\widetilde\psi=0.$$

Because

$$\{z,w\} =\frac{s''}{s} -\frac32\left(\frac{s'}s\right)^2,$$

the bracket is $s^2T+\{z,w\}/2$.

3. Recover the equation from a developing map. Suppose $F$ is locally univalent and set $T=\{F,z\}/2$. Show that

$$\psi_2=(F')^{-1/2}, \qquad \psi_1=F(F')^{-1/2}$$

solve $\psi''+T\psi=0$, up to a consistent square-root branch.

Solution

Let $h=F''/F'$. For $\psi_2=(F')^{-1/2}$,

$$\frac{\psi_2''}{\psi_2} =-\frac12h'+\frac14h^2 =-\frac12\{F,z\} =-T.$$

Thus $\psi_2$ solves the equation. Since $\Wr[F\psi_2,\psi_2]=-1$, the function $\psi_1=F\psi_2$ is independent of $\psi_2$. Its Wronskian is constant, so

$$\begin{aligned} 0 &=\Wr[\psi_1,\psi_2]'\\ &=\psi_1\psi_2''-\psi_1''\psi_2\\ &=-\psi_2(\psi_1''+T\psi_1). \end{aligned}$$

Hence $\psi_1$ solves the equation wherever $\psi_2\neq0$, and therefore throughout the local patch by analytic continuation.

4. Derive the Sturm–Liouville transform. Use $\dd t/\dd x=\sqrt{R/P}$ and $u=ry$ to obtain the displayed $\widehat Q$.

Solution

Since $r^2=\sqrt{PR}$,

$$P\frac{\dd y}{\dd x} =r\frac{\dd u}{\dd t} -u\frac{\dd r}{\dd t}.$$

Differentiating with respect to $x$ gives

$$\frac{\dd}{\dd x} \left( P\frac{\dd y}{\dd x} \right) =\sqrt{\frac RP} \left( r\frac{\dd^2u}{\dd t^2} -u\frac{\dd^2r}{\dd t^2} \right).$$

Substitution into the Sturm–Liouville equation and multiplication by $r/R$ produce

$$-u_{tt} +\left( \frac QR+\frac{r_{tt}}r \right)u =\lambda u.$$

References

• NIST DLMF, Change of variables and Liouville transformations, for first-derivative removal, the Schwarzian coordinate law, and Sturm–Liouville normal form.
• A. Beilinson and V. Drinfeld, Opers, for the bundle, connection, and filtration definition of opers.
• V. Ovsienko and S. Tabachnikov, Projective Differential Geometry Old and New, for projective structures, Schwarzian derivatives, and differential operators on densities.
• G. Teschl, Mathematical Methods in Quantum Mechanics, for one-dimensional Schrödinger and Sturm–Liouville operators.