Affine spheres and finite gap solutions of Tzitzèica equation

We start with description of affine spheres in equiaffine differential geometry. For more details we refer to the textbook [18] by Nomizu and Sasaki.

Let ${{\bf{R}}}^{3}$ be a Cartesian 3-space. We denote by $\det :{{\bf{R}}}^{3}\times {{\bf{R}}}^{3}\times {{\bf{R}}}^{3}\longrightarrow {\bf{R}}$ the determinant function, which defines a volume element of ${{\bf{R}}}^{3}$. Then the triplet ${{\bf{A}}}^{3}=({{\bf{R}}}^{3},D,\det )$ is an equiaffine 3-space, i.e., it satisfies $D\ \det \,=\,0$, where D is the canonical covariant differentiation for ${{\bf{R}}}^{3}$. An immersion $\psi :M\longrightarrow {{\bf{A}}}^{3}$ of an oriented 2-manifold is said to be an affine immersion if $(M,\psi )$ admits a complementary subbundle ${ \mathcal N }$ of the tangent bundle TM of M in the pull-back bundle ${\psi }^{* }(T{{\bf{A}}}^{3})$, that is, if we have ${\psi }_{* }(T{{\bf{A}}}^{3})={TM}\oplus { \mathcal N }$. A local section ξ of ${ \mathcal N }$ is called a transversal vector field of $(M,\psi )$. In the following, we assume that there is a global transversal vector field ξ to $(M,\psi )$ so that ${ \mathcal N }={\cup }_{p\in M}{\bf{R}}{\xi }_{p}$. We then call a triplet $(M,\psi ,\xi )$ an affine immersion.

For any vector fields X and Y on M we have the Gauss formula

$$\begin{eqnarray*}&&{D}_{X}\ {\psi }_{* }Y={\psi }_{* }({{\rm{\nabla }}}_{X}Y)+h(X,Y)\xi .\end{eqnarray*}$$

We easily verify that ${\rm{\nabla }}$ defines a torsion free linear connection on M and h is a symmetric tensor field on M. The symmetric tensor field h is called the affine fundamental form derived from ξ. We also have the Weingarten formula

$$\begin{eqnarray*}&&{D}_{X}\xi =-{\psi }_{* }({SX})+{ \mathcal T }(X)\xi .\end{eqnarray*}$$

The endomorphism field S of TM is called the affine shape operator with respect to ξ. The 1-form ${ \mathcal T }$ is called the transversal connection form. For example, an immersion $\psi :M\longrightarrow {{\bf{A}}}^{3}\setminus \{0\}$ with the property that the position vector field ψ is transversal to M is an affine immersion and in particular it is called centro-affine immersion. For an affine immersion $(M,\psi ,\xi )$, we may define a volume element ϑ on M by $\vartheta (X,Y)=\det (X,Y,\xi )$ for any vector fields X and Y on M. We address here some known facts about affine immersions.

• For an affine immersion, the rank of the affine fundamental form is independent of the choice of the transversal vector field.
• The triplet $(M,{\rm{\nabla }},\vartheta )$ is equiaffine, i.e., ${\rm{\nabla }}\vartheta =0$ if and only if ${ \mathcal T }=0$.
• Let $\psi :M\longrightarrow {{\bf{A}}}^{3}$ be an affine immersion with non-degenerate affine fundamental form h. Then there is a unique (up to sign) transversal vector field ξ such that ${ \mathcal T }=0$ and the volume element of the semi-Riemannian surface $(M,h)$ coincides with ϑ. This ξ is called the Blaschke normal vector filed and the pair $(\psi ,\xi )$ is called a Blaschke immersion. The affine fundamental form h with respect to the Blaschke normal vector field is traditionally called the Blaschke metric of $(M,\psi )$. For a Blaschke immersion $(M,\psi ,\xi )$ we have the following fundamental equations.

$$\begin{eqnarray*}\begin{array}{rcl}R(X,Y)Z & = & h(Y,Z){SX}-h(Z,X){SY},\qquad \qquad (\mathrm{Gauss}\ \mathrm{equation})\\ ({{\rm{\nabla }}}_{X}h)(Y,Z) & = & ({{\rm{\nabla }}}_{Y}h)(X,Z),\qquad \qquad \qquad \qquad (\mathrm{Codazzi}\ \mathrm{equation})\\ h({SX},Y) & = & h(X,{SY}).\qquad \qquad \qquad \qquad \qquad (\mathrm{Ricci}\ \mathrm{equation})\end{array}\end{eqnarray*}$$

Here R is the curvature tensor field of the connection ${\rm{\nabla }}$. The Codazzi equation implies that $C:= {\rm{\nabla }}h$ is a symmetric covariant tensor field. We call the C cubic form of $(M,\psi ,\xi )$.

Now let $(M,\psi ,\xi )$ be a Blaschke immersion with indefinite Blaschke metric h. Then we can take an asymptotic coordinate system $(x,t)$ on a simply connected domain ${\bf{D}}\subset M$ with respect to the Lorentz conformal structure determined by h. Namely, the Blaschke metric h is represented as $h={e}^{u}\ {dx}\ {dt}$ with respect to $(x,t)$. It then follows from Fact 2 and Fact 3 that the cubic form C is represented as $C={{Adx}}^{3}+{{Bdt}}^{3}$.

Definition 2.1. A Blaschke immersion $(M,\psi ,\xi )$ is said to be an affine sphere if $S=k\ I$ for some constant k. Moreover, a affine sphere $(M,\psi ,\xi )$ is said to be proper if $k\ne 0$ and improper if $k=0$.

• Let $(M,\psi ,\xi )$ be a proper affine sphere. Then all the Blaschke normals meet in one point.

In the following, we consider proper affine spheres with indefinite Blaschke metric h. In particular, for such proper affine spheres we have $\xi =-H\psi$, where $H=\tfrac{1}{2}\mathrm{trace}\ S$ is the affine mean curvature and it is a negative constant. By scaling the affine metric, without loss of generality, we may assume that $H=-1$, hence we may take $\xi =\psi$. Note that $(M,\psi ,\xi )$ is centro-affine.

Set ${\partial }_{x}=\tfrac{\partial }{\partial x}$, ${\partial }_{t}=\tfrac{\partial }{\partial t}$, then the induced connection ${\rm{\nabla }}$ is given by

$$\begin{eqnarray*}&&{{\rm{\nabla }}}_{{\partial }_{x}}{\partial }_{x}={u}_{x}{\partial }_{x}+{{Ae}}^{-u}\ {\partial }_{t},\quad {{\rm{\nabla }}}_{{\partial }_{x}}{\partial }_{t}=0={{\rm{\nabla }}}_{{\partial }_{t}}{\partial }_{x},\quad {{\rm{\nabla }}}_{{\partial }_{t}}\ {\partial }_{t}={{Be}}^{-u}{\partial }_{x}+{u}_{t}\ {\partial }_{t}.\end{eqnarray*}$$

Set ${\psi }_{t}={\psi }_{* }({\partial }_{t})$ and ${\psi }_{x}={\psi }_{* }({\partial }_{x})$ then the Gauss-Weingarten equations are given by

$$\begin{eqnarray}&&{\psi }_{{xx}}={u}_{x}\ {\psi }_{x}+{{Ae}}^{-u}\ {\psi }_{t},\ {\psi }_{{tx}}={e}^{u}\psi ={\psi }_{{xt}},\ {\psi }_{{tt}}={{Be}}^{-u}\ {\psi }_{x}+{u}_{t}\ {\psi }_{t}.\end{eqnarray} \tag{ 2.1 }$$

Here we introduce a matrix valued function (called a framing) ${ \mathcal F }=({\psi }_{t}\ {\psi }_{x}\ \psi )$. Since ψ is a Blaschke immersion, we see that $\det { \mathcal F }={e}^{u}\ne 0$. We rewrite (2.1) in the following way [19, 20]:

$$\begin{eqnarray}{{ \mathcal F }}^{-1}\displaystyle \frac{\partial { \mathcal F }}{\partial t}=\left(\begin{array}{ccc}{u}_{t} & 0 & 1\\ {{Be}}^{-u} & 0 & 0\\ 0 & {e}^{u} & 0\end{array}\right)=: { \mathcal U },\quad {{ \mathcal F }}^{-1}\displaystyle \frac{\partial { \mathcal F }}{\partial x}=\left(\begin{array}{ccc}0 & {{Ae}}^{-u} & 0\\ 0 & {u}_{x} & 1\\ {e}^{u} & 0 & 0\end{array}\right)=: { \mathcal V }\end{eqnarray} \tag{ 2.2 }$$

The compatibility condition of the system (2.2) is given by

$$\begin{eqnarray}&&{u}_{{xt}}={e}^{u}-{AB}\ {e}^{-2u},\quad {A}_{t}=0,\quad {B}_{x}=0.\end{eqnarray} \tag{ 2.3 }$$

The first equation is the Gauss equation and the last two equations are Coddazi equations. Throughout this paper we assume that indefinite proper affine spheres are weakly regular, that is, ${AB}\ne 0$ [20]. Then the Coddazi equation means that $A=A(x)$ and $B=B(t)$. Therefore, choosing the coordinate system $(x,t)$ we may assume that the cubic form is represented as $C={{dx}}^{3}+{{dt}}^{3}$ from the first step. In this case, we have $A=B=1$ and (2.3) is called the Tzitzèica equation [6].

The compatibility condition (2.3) is invariant under the replacements

$$\begin{eqnarray}{ \mathcal U }(\nu )=\left(\begin{array}{ccc}{u}_{t} & 0 & 1\\ \nu {e}^{-u} & 0 & 0\\ 0 & {e}^{u} & 0\end{array}\right),\quad { \mathcal V }(\nu )=\left(\begin{array}{ccc}0 & {\nu }^{-1}{e}^{-u} & 0\\ 0 & {u}_{x} & 1\\ {e}^{u} & 0 & 0\end{array}\right),\quad \nu \in {{\bf{R}}}^{* },\end{eqnarray} \tag{ 2.4 }$$

respectively. We denote by ${ \mathcal F }(\nu )$ the solution to the system

$$\begin{eqnarray*}&&{\partial }_{t}\ { \mathcal F }(\nu )={ \mathcal F }(\nu )\ { \mathcal U }(\nu ),\ {\partial }_{x}\ { \mathcal F }(\nu )={ \mathcal F }(\nu )\ { \mathcal V }(\nu ).\end{eqnarray*}$$

Using λ with ${\lambda }^{3}=\nu$ and setting

$$\begin{eqnarray*}F(\lambda ):= { \mathcal F }(\nu )\left(\begin{array}{ccc}{\lambda }^{-1}{e}^{-\tfrac{u}{2}} & 0 & 0\\ 0 & \lambda {e}^{-\tfrac{u}{2}} & 0\\ 0 & 0 & 1\end{array}\right),\end{eqnarray*}$$

we obtain a new framing $F(\lambda )$ with

$$\begin{eqnarray*}&&{\partial }_{t}F(\lambda )=F(\lambda )\ U(\lambda ),\quad {\partial }_{x}F(\lambda )=F(\lambda )\ V(\lambda ),\qquad (\nu ={\lambda }^{3})\end{eqnarray*}$$

where

$$\begin{eqnarray}U(\lambda )=\left(\begin{array}{ccc}\displaystyle \frac{1}{2}{u}_{t} & 0 & \lambda {e}^{\tfrac{u}{2}}\\ \lambda {e}^{-u} & -\displaystyle \frac{1}{2}{u}_{t} & 0\\ 0 & \lambda {e}^{\tfrac{u}{2}} & 0\end{array}\right),\quad V(\lambda )=\left(\begin{array}{ccc}-\displaystyle \frac{1}{2}{u}_{x} & {\lambda }^{-1}{e}^{-u} & 0\\ 0 & \displaystyle \frac{1}{2}{u}_{x} & {\lambda }^{-1}{e}^{\tfrac{u}{2}}\\ {\lambda }^{-1}{e}^{\tfrac{u}{2}} & 0 & 0\end{array}\right).\end{eqnarray} \tag{ 2.5 }$$

Thus for any indefinite proper afine sphere $\psi :M\to {{\bf{A}}}^{3}$, there exists a map $F(\lambda ):{\bf{D}}\times {{\bf{R}}}^{* }\to {SL}(3,{\bf{R}})$ satisfying (2.5). We call $F(\lambda )$ a coordinate extended framing of an indefinite proper affine sphere $(M,\psi )$. Coordinate extended framings are examples of 'extended framings' which will be introduced in later section (definition 3.2).

Before closing this section, we state the assumption on u throughout the paper. We assume that $u(x,t)=u(t,x)$ with respect to the coordinate system $(x,t)$ so that the cubic form is given by $C={{dx}}^{3}+{{dt}}^{3}$. Of course, this assumption is independent of introducing the non-zero parameter ν.

Let $G={SL}(3,{\bf{R}})$ be a real special linear group of degree 3 and ${\mathfrak{g}}={\mathfrak{sl}}(3,{\bf{R}})$ its Lie algebra. We denote by ${{\mathfrak{g}}}^{{\bf{C}}}={\mathfrak{sl}}(3,{\bf{C}})$ the complexification of ${\mathfrak{g}}$. Define two automorphisms σ and γ of ${{\mathfrak{g}}}^{{\bf{C}}}$ by

$$\begin{eqnarray*}\sigma (\xi )=-\mathrm{Ad}\ \left(\begin{array}{ccc}0 & 1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 1\end{array}\right)\cdot {}^{t}\xi ,\quad \gamma (\xi )=\mathrm{Ad}\ \left(\begin{array}{ccc}1 & 0 & 0\\ 0 & {\omega }^{2} & 0\\ 0 & 0 & \omega \end{array}\right)\cdot \xi ,\qquad \quad \xi \in {{\mathfrak{g}}}^{{\bf{C}}},\end{eqnarray*}$$

where $\omega =\exp (2\pi \sqrt{-1}/3)$. We then see that $\tau =\sigma \gamma =\gamma \sigma$ is an automorphism of order 6 [20]. Let $\epsilon =-{\omega }^{2}$ be the sixth root of unity. Denote by ${{\mathfrak{g}}}_{j}^{{\bf{C}}}$ be the ${\epsilon }^{j}$-eigenspace of τ, where $j=0$, 1, 2, 3, 4, 5. Set ${{\mathfrak{g}}}_{j}={{\mathfrak{g}}}_{j}^{{\bf{C}}}\cap {\mathfrak{g}}$. For example, we have ${{\mathfrak{g}}}_{0}$ consists of the marices of the form

$$\begin{eqnarray*}\left(\begin{array}{ccc}a & 0 & 0\\ 0 & -a & 0\\ 0 & 0 & 0\end{array}\right),\ a\in {\bf{R}},\end{eqnarray*}$$

which is a Lie subalgebra of ${\mathfrak{g}}$ and is isomorphic to ${\mathfrak{o}}(1,1)$ in ${\mathfrak{sl}}(3,{\bf{R}})$. Let K be a subgroup of G, which consists of the matrices of the form

$$\begin{eqnarray*}\left(\begin{array}{ccc}s & 0 & 0\\ 0 & {s}^{-1} & 0\\ 0 & 0 & 1\end{array}\right),\ s\in {{\bf{R}}}^{* }\end{eqnarray*}$$

and is isomorphic to ${SO}(1,1)\subset {SL}(3,{\bf{R}})$. We denote by ${{SO}}^{0}(1,1)$ the identity component of ${SO}(1,1)$. Then ${{SO}}^{0}(1,1)$ is isomorphic to $\exp ({{\mathfrak{g}}}_{0})=: {K}_{+}$. We denote by ${SO}(2,1)$ the Lorentzian group with Lie algebra ${{\mathfrak{g}}}_{0}\oplus {{\mathfrak{g}}}_{2}\oplus {{\mathfrak{g}}}_{4}$ and its identity component by ${{SO}}^{0}(2,1)$. We then easily see the following fact.

Proposition 3.1. The automorphism $\tau$ of order 6 gives a semi-Riemannian 6-symmetric space structure on ${SL}(3,{\bf{R}})/{{SO}}^{0}(1,1)$. The involution $\sigma$ gives ${SL}(3,{\bf{R}})/{{SO}}^{0}(2,1)$ the outer semi-Riemannian symmetric space structure. Moreoevr, we have the homogeneous projection $\pi :{SL}(3,{\bf{R}})/{{SO}}^{0}(1,1)\longrightarrow {SL}(3,{\bf{R}})/{{SO}}^{0}(2,1)$.

Let $\hat{\sigma }$ be an involution on ${{\mathfrak{g}}}^{{\bf{C}}}$ defined by

$$\begin{eqnarray*}\hat{\sigma }=\mathrm{Ad}\ \left(\begin{array}{ccc}0 & 1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 1\end{array}\right).\end{eqnarray*}$$

We then have $\hat{\sigma }({{\mathfrak{g}}}_{j})={{\mathfrak{g}}}_{-j}$. We consider a twisted loop algebra

$$\begin{eqnarray*}&&{\rm{\Lambda }}{{\mathfrak{g}}}_{\tau }^{{\bf{C}}}=\{\xi :{S}^{1}\longrightarrow {{\mathfrak{g}}}^{{\bf{C}}}\ | \ \xi (\epsilon \lambda )=\tau \xi (\lambda )\ {\rm{for}}\,{\rm{all}}\ \lambda \in {S}^{1}\ {\rm{and}}\ | | \xi | | \,\lt \,\infty \},\end{eqnarray*}$$

where $| | \xi | | ={\sum }_{n\in {\bf{Z}}}| | {\xi }_{n}| |$ for $\xi (\lambda )={\sum }_{n\in {\bf{Z}}}{\xi }_{n}{\lambda }^{n}$ and $| | {\xi }_{n}| | ={\max }_{j\in \{\mathrm{1,2,3}\}}{\sum }_{i=1}^{3}| {({\xi }_{n})}_{{ij}}|$. Then, ${\rm{\Lambda }}{{\mathfrak{g}}}_{\tau }^{{\bf{C}}}$ is a Banach Lie algebra (see [21, 22]). Set

$$\begin{eqnarray*}&&{\rm{\Lambda }}{{\mathfrak{g}}}_{\tau }=\left\{\xi \in {\rm{\Lambda }}{{\mathfrak{g}}}_{\tau }^{{\bf{C}}}| \xi =\displaystyle \sum _{j\in {\bf{Z}}}{\xi }_{j}{\lambda }^{j}\ {\rm{with}}\ {\xi }_{j}\in {{\mathfrak{g}}}_{[j]}\right\},\end{eqnarray*}$$

where $[j]\in {\bf{Z}}/6{\bf{Z}}$.

We decompose each element $\xi \in {\rm{\Lambda }}{{\mathfrak{g}}}_{\tau }$ as $\xi ={\xi }_{+}+{\xi }_{0}+{\xi }_{-}$, where ${\xi }_{+}(\lambda )={\sum }_{j\gt 0}{\xi }_{j}{\lambda }^{j},{\xi }_{-}(\lambda )={\sum }_{j\lt 0}{\xi }_{j}{\lambda }^{j}$. Next, we extend the involution $\hat{\sigma }$ to ${\rm{\Lambda }}{{\mathfrak{g}}}_{\tau }$ by

$$\begin{eqnarray*}&&(\hat{\sigma }\xi )(\lambda )=\hat{\sigma }(\xi ({\lambda }^{-1})).\end{eqnarray*}$$

We may write ξ as

$$\begin{eqnarray}&&\xi =({\xi }_{+}+\hat{\sigma }({\xi }_{+}))+({\xi }_{-}-\hat{\sigma }({\xi }_{+})+{\xi }_{0}).\end{eqnarray} \tag{ 3.1 }$$

We now define the following Banach Lie subalgebras of ${\rm{\Lambda }}{{\mathfrak{g}}}_{\tau }$ :

$$\begin{eqnarray}&&{{\rm{\Lambda }}}^{\hat{\sigma }}{{\mathfrak{g}}}_{\tau }=\{\xi \in {\rm{\Lambda }}{{\mathfrak{g}}}_{\tau }\ | \ \hat{\sigma }\xi =\xi \},\ {{\rm{\Lambda }}}_{* 0}^{-}{{\mathfrak{g}}}_{\tau }=\left\{\xi \in {\rm{\Lambda }}{{\mathfrak{g}}}_{\tau }| \xi =\displaystyle \sum _{j\leqq 0}{\xi }_{j}{\lambda }^{j}\right\}.\end{eqnarray} \tag{ 3.2 }$$

It then follows from (3.1) that

$$\begin{eqnarray}&&{\rm{\Lambda }}{{\mathfrak{g}}}_{\tau }={{\rm{\Lambda }}}^{\hat{\sigma }}{{\mathfrak{g}}}_{\tau }\oplus {{\rm{\Lambda }}}_{* 0}^{-}{{\mathfrak{g}}}_{\tau },\end{eqnarray} \tag{ 3.3 }$$

which is a vector space decomposition into Banach Lie subalgebras.

For a positive integer $d\equiv 1(\mathrm{mod}\ 6)$, we define the vector subspace ${{\rm{\Lambda }}}_{d}^{\hat{\sigma }}$ of ${{\rm{\Lambda }}}^{\hat{\sigma }}{{\mathfrak{g}}}_{\tau }$ by

$$\begin{eqnarray}&&{{\rm{\Lambda }}}_{d}^{\hat{\sigma }}=\left\{\xi \in {{\rm{\Lambda }}}^{\hat{\sigma }}{{\mathfrak{g}}}_{\tau }| \xi =\displaystyle \sum _{j=-d}^{d}{\xi }_{j}{\lambda }^{j}\right\}.\end{eqnarray} \tag{ 3.4 }$$

Let ${\rm{\Lambda }}{G}_{\tau }$ be a twisted loop group defined by

$$\begin{eqnarray*}&&{\rm{\Lambda }}{G}_{\tau }=\{g:{S}^{1}\longrightarrow {G}^{{\bf{C}}}\ | \ g(\epsilon \lambda )=\tau g(\lambda ),\ \overline{g(\overline{\lambda })}=g(\lambda )\ {\rm{for}}\,{\rm{all}}\ \lambda \in {S}^{1}\ {\rm{and}}\ | | g| | \,\lt \,\infty \}\end{eqnarray*}$$

whose Lie algebra is ${\rm{\Lambda }}{{\mathfrak{g}}}_{\tau }$. Here we denote the Lie group automorphism corresponding to τ by the same letter. Define subgroup ${{\rm{\Lambda }}}^{\hat{\sigma }}{G}_{\tau }$ of ${\rm{\Lambda }}{G}_{\tau }$ by

$$\begin{eqnarray*}&&{{\rm{\Lambda }}}^{\hat{\sigma }}{G}_{\tau }=\{g\in {\rm{\Lambda }}{G}_{\tau }\ | \ \hat{\sigma }g=g\}.\end{eqnarray*}$$

Consider a map $g:{\bf{D}}\subset {{\bf{R}}}^{2}\ni (x,t)\,\longmapsto \,g(x,t)\in {\rm{\Lambda }}{G}_{\tau }$. We extend the involution $\hat{\sigma }$ to the above map by the rule

$$\begin{eqnarray}&&({\hat{\sigma }}^{* }g)(x,t)=\hat{\sigma }(g(t,x)).\end{eqnarray} \tag{ 3.5 }$$

Define the mapping spaces ${\rm{\Lambda }}{G}_{\tau }({\bf{D}})$ and ${{\rm{\Lambda }}}^{{\hat{\sigma }}^{* }}{G}_{\tau }({\bf{D}})$ by

$$\begin{eqnarray}&&{\rm{\Lambda }}{G}_{\tau }({\bf{D}})=\{g:{\bf{D}}\longrightarrow {\rm{\Lambda }}{G}_{\tau }\},\quad {{\rm{\Lambda }}}^{{\hat{\sigma }}^{* }}{G}_{\tau }({\bf{D}})=\{a\in {\rm{\Lambda }}{G}_{\tau }({\bf{D}})| {\hat{\sigma }}^{* }a=a\},\end{eqnarray} \tag{ 3.6 }$$

$$\begin{eqnarray}&&{\rm{\Lambda }}{{\mathfrak{g}}}_{\tau }({\bf{D}})=\{\xi :{\bf{D}}\longrightarrow {\rm{\Lambda }}{{\mathfrak{g}}}_{\tau }\},\quad \quad \ {{\rm{\Lambda }}}^{{\hat{\sigma }}^{* }}{{\mathfrak{g}}}_{\tau }({\bf{D}})=\{\xi \in {\rm{\Lambda }}{{\mathfrak{g}}}_{\tau }({\bf{D}})\ | \ {\hat{\sigma }}^{* }\xi =\xi \}.\end{eqnarray} \tag{ 3.7 }$$

Now we return to indefinite proper affine spheres. Let $\psi :{\bf{D}}\subset {{\bf{R}}}^{2}\to {{\bf{A}}}^{3}$ be an indefinite proper affine sphere parametrized by global asymptotic coordinates $(x,t)$ as before. The coordinate extended framing $F(\lambda )$ can be extended analytically on ${{\bf{C}}}^{* }$. The (extended) map $F(\lambda )$ is uniquely determined by the values on ${S}^{1}\subset {{\bf{C}}}^{* }$ (cf. [20], p. 234). Hence $F(\lambda )$ is regarded as a map into ${\rm{\Lambda }}{G}_{\tau }$. In addition, since we assumed that $u(x,t)=u(t,x)$, it follows from (2.5) that ${\hat{\sigma }}^{* }(U(\lambda ))=V(\lambda )$. Therefore, the coordinate extended framing $F(\lambda )$ can be considered as a map $F\in {{\rm{\Lambda }}}^{{\hat{\sigma }}^{* }}{G}_{\tau }({\bf{D}})$. Moreover, since

$$\begin{eqnarray*}&&\psi =F\cdot \left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)={{ \mathcal F }(\nu )| }_{\nu =1}\cdot \left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)={F(\lambda )| }_{\lambda =1}\cdot \left(\begin{array}{c}0\\ 0\\ 1\end{array}\right),\end{eqnarray*}$$

we see that the 3rd column vector of the coordinate extended framing $F(\lambda ){| }_{\lambda =1}$ gives a Blaschke immersion $\psi :{\bf{D}}\longrightarrow {{\bf{A}}}^{3}$. Here we give the following definition (compare with [20, 23]).

Definition 3.2. An element $F\in {{\rm{\Lambda }}}^{{\hat{\sigma }}^{* }}{G}_{\tau }({\bf{D}})$ which satisfies ${F}^{-1}{dF}=\lambda {\alpha }_{1}{dt}+{\alpha }_{0}+{\lambda }^{-1}{\alpha }_{-1}{dx}$ with ${\alpha }_{-1}={\hat{\sigma }}^{* }({\alpha }_{1})$ is said to be an extended framing for indefinite proper affine sphere.

Remark 3.3. One can check that every extended framing $F\in {{\rm{\Lambda }}}^{{\hat{\sigma }}^{* }}{G}_{\tau }(D)$ induces a harmonic map (nonlinear σ-model) $F\cdot {K}_{+}:{\bf{D}}\to G/{K}_{+}$ (see proposition 3.1, [24]).

Next we introduce the following notion for affine spheres.

Definition 3.4. An indefinite proper affine sphere is said to be of finite type if its extended framing a is obtained from the following differential equation under certain initial condition:

$$\begin{eqnarray*}&&\left\{\begin{array}{l}d\xi =[\xi ,{\alpha }_{\xi }],\quad \xi \in {{\rm{\Lambda }}}_{d}^{{\hat{\sigma }}^{* }}({\bf{D}})=\left\{\xi \in {{\rm{\Lambda }}}^{{\hat{\sigma }}^{* }}{{\mathfrak{g}}}_{\tau }({\bf{D}})| \xi =\displaystyle \sum _{j=-d}^{d}{\xi }_{j}{\lambda }^{j}\right\},\\ {\alpha }_{\xi }=\left(\lambda {\xi }_{d}+\displaystyle \frac{1}{2}{\xi }_{d-1}\right)\ {dt}+\left({\lambda }^{-1}{\hat{\sigma }}^{* }({\xi }_{d})+\displaystyle \frac{1}{2}{\hat{\sigma }}^{* }({\xi }_{d-1})\right)\ {dx},\end{array}\right.\end{eqnarray*}$$

where $d\equiv 1\ ({\rm{mod}}\ 6)$.

$a\in {{\rm{\Lambda }}}^{{\hat{\sigma }}^{* }}{G}_{\tau }({\bf{D}})$

${a}^{-1}{da}=\alpha$

Example 3.5 We present a trivial solution $u=0$ of Tzitzèica equation and corresponding affine sphere. Take a matrix $A=\left(\begin{array}{ccc}0 & 0 & 1\\ 1 & 0 & 0\\ 0 & 1 & 0\end{array}\right)$. Note that $\hat{\sigma }(A)={A}^{-1}$. Then we easily see that $a(x,t)=\exp (t\lambda A+x{\lambda }^{-1}{A}^{-1})$ is an extended framing of an indefinite proper affine sphere ${\psi }_{0}$ of finite type. After some elementary calculation we obtain

$$\begin{eqnarray*}&&{\psi }_{0}=a{| }_{\lambda =1}\cdot \left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)=\left(\begin{array}{ccc}-\displaystyle \frac{1}{3} & -\displaystyle \frac{\sqrt{3}}{3} & \displaystyle \frac{1}{3}\\ -\displaystyle \frac{1}{3} & \displaystyle \frac{\sqrt{3}}{3} & \displaystyle \frac{1}{3}\\ \displaystyle \frac{2}{3} & 0 & \displaystyle \frac{1}{3}\end{array}\right)\cdot \left(\begin{array}{c}\cos \left(\displaystyle \frac{\sqrt{3}}{2}(x-t)\right){e}^{-\tfrac{t+x}{2}}\\ \sin \left(\displaystyle \frac{\sqrt{3}}{2}(x-t)\right){e}^{-\tfrac{t+x}{2}}\\ {e}^{t+x}\end{array}\right).\end{eqnarray*}$$

Hence the resulting surface ${\psi }_{0}$ is affine congruent to the hexenhut $Z({X}^{2}+{Y}^{2})=1$ in ${{\bf{A}}}^{3}(X,Y,Z)$ (see Figure 1). The hexenhut ${\psi }_{0}$ corresponds to the trivial solution $u=0$ to the Tzitzèica equation [25].

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Remark 3.6. One can establish the Symes method (also called AKS-scheme) for constructing indefinite proper affine sphere of finite type. This scheme was established in a separate publication [26].

In this section, we give a solution of Tzitzèica equation (1.1) in terms of elliptic functions and represent affine spheres in terms of elliptic functions. Although these formulas have been known in some literature, our purpose here is to represent them in terms of Riemann theta function. This achievement gives a nice story for the description of the solutions in terms of the spectral curves of higher genus and Prym-theta functions.

Introducing new coordinates $\hat{x}$, $\hat{y}$ by

$$\begin{eqnarray*}&&\hat{x}=\displaystyle \frac{1}{2}(x+t),\quad \hat{y}=\displaystyle \frac{1}{2}(x-t),\end{eqnarray*}$$

we have $4\tfrac{\partial }{\partial x}\tfrac{\partial }{\partial t}=\tfrac{{\partial }^{2}}{\partial {\hat{x}}^{2}}-\tfrac{{\partial }^{2}}{\partial {\hat{y}}^{2}}$.

We now assume that a solution u depends only on the parameter $\hat{y}$ and write $u=u(\hat{y})$. We suppose that an initial condition ${e}^{u(0)}=\tfrac{\ \alpha \ }{2},{u}_{\hat{y}}(0)=0$, where α is a real number with $\alpha \gt 2$. If we set $Y(\hat{y})={e}^{u(\hat{y})}$, then the equation (1.1) may rewritten as

$$\begin{eqnarray*}&&{\left(\displaystyle \frac{{dY}}{d\hat{y}}\right)}^{2}=-8\left({Y}^{3}-{{aY}}^{2}+\displaystyle \frac{1}{2}\right)=-8(Y-{\zeta }_{1})(Y-{\zeta }_{2})(Y-{\zeta }_{3}),\end{eqnarray*}$$

where

$$\begin{eqnarray*}&&a=\displaystyle \frac{{\alpha }^{3}+4}{2{\alpha }^{2}},\ {\zeta }_{1}=\displaystyle \frac{1-\sqrt{{\alpha }^{3}+1}}{{\alpha }^{2}},\ {\zeta }_{2}=\displaystyle \frac{\alpha }{2},\ {\zeta }_{3}=\displaystyle \frac{1+\sqrt{{\alpha }^{3}+1}}{{\alpha }^{2}}.\end{eqnarray*}$$

If we set $Y={\zeta }_{2}-({\zeta }_{2}-{\zeta }_{3}){X}^{2}$ and ${p}^{2}=\tfrac{{\zeta }_{3}-{\zeta }_{2}}{{\zeta }_{1}-{\zeta }_{2}}$ we obtain

$$\begin{eqnarray*}&&{\left(\displaystyle \frac{{dX}}{d\hat{y}}\right)}^{2}=2({\zeta }_{2}-{\zeta }_{1})(1-{p}^{2}{X}^{2})(1-{X}^{2}),\end{eqnarray*}$$

hence, setting $X(\hat{y})=\widetilde{X}(\sqrt{2}\sqrt{{\zeta }_{2}-{\zeta }_{1}}\ \hat{y})$ and $\hat{z}=\sqrt{2}\sqrt{{\zeta }_{2}-{\zeta }_{1}}\ \hat{y}$, we have

$$\begin{eqnarray*}&&{\left(\displaystyle \frac{d\widetilde{X}}{d\hat{z}}\right)}^{2}=(1-{p}^{2}{\widetilde{X}}^{2})(1-{\widetilde{X}}^{2}).\end{eqnarray*}$$

Then, we see that $\widetilde{X}(\hat{z})=\mathrm{sn}\ (\hat{z},p)$, where $\mathrm{sn}\ (\hat{z},p)$ is the Jacobi $\mathrm{sn}$-function with modulus p. Therefore, we obtain a solution of (1.1) as follows.

$$\begin{eqnarray}&&\exp (u(\hat{y}))=Y(\hat{y})={\zeta }_{2}-({\zeta }_{2}-{\zeta }_{3})\ {\mathrm{sn}}^{2}\ (\sqrt{2}\sqrt{{\zeta }_{2}-{\zeta }_{1}}\ \hat{y},p).\end{eqnarray} \tag{ 4.1 }$$

The function $Y(\hat{y})$ can be extended as a function of complex variables. Extending $\hat{y}$ as a complex valued function, we see that $Y(\hat{y})$ is a doubly-periodic function with the periods $2{\omega }^{0},2{\omega }^{{\prime} }{}^{0}$, where

$$\begin{eqnarray}&&{\omega }^{0}=\displaystyle \frac{K(p)}{\sqrt{2}\sqrt{{\zeta }_{2}-{\zeta }_{1}}},\quad {\omega }^{{\prime} }{}^{0}=\displaystyle \frac{\sqrt{-1}{K}^{{\prime} }(p)}{\sqrt{2}\sqrt{{\zeta }_{2}-{\zeta }_{1}}},\end{eqnarray} \tag{ 4.2 }$$

and $K(p)$ is the complete elliptic integral of 1st kind, ${K}^{{\prime} }(p)=K({p}^{{\prime} })$ and ${p}^{{\prime} }=\sqrt{1-{p}^{2}}$. The Y can be essentially represented by Weierstrass $\wp$-function. In fact, if we set $P(z)=P(\sqrt{2}\sqrt{-1}\hat{y})=Y(\hat{y})-\tfrac{\ a\ }{3}$ then we have

$$\begin{eqnarray*}&&{\left(\displaystyle \frac{{dP}}{{dz}}\right)}^{2}=4(P-{\eta }_{1})(P-{\eta }_{2})(P-{\eta }_{3}),\end{eqnarray*}$$

where ${\eta }_{j}={\zeta }_{j}-\tfrac{\ a\ }{3}$, $j=1$, 2, 3 with the property ${\eta }_{1}+{\eta }_{2}+{\eta }_{3}=0$. The solution of this equation is given by Weierstrass $\wp$-function $\wp (z)$, which is related to the Jacobi $\mathrm{sn}$-function by

$$\begin{eqnarray*}&&\wp (z)={\eta }_{1}+\displaystyle \frac{{\eta }_{2}-{\eta }_{1}}{{\mathrm{sn}}^{2}(\sqrt{{\eta }_{2}-{\eta }_{1}}\ z,k)},\end{eqnarray*}$$

where ${k}^{2}=\tfrac{{\eta }_{1}-{\eta }_{3}}{{\eta }_{1}-{\eta }_{2}}=\tfrac{{\zeta }_{1}-{\zeta }_{3}}{{\zeta }_{1}-{\zeta }_{2}}=1-{p}^{2}={({p}^{{\prime} })}^{2}$, hence we have $k={p}^{{\prime} }$ and ${k}^{{\prime} }=p$. As it is well-known, $\wp$-function is doubly-periodic with periods $2\omega (k),2{\omega }^{{\prime} }(k)$, where

$$\begin{eqnarray*}&&\omega (k)=\displaystyle \frac{K(k)}{\sqrt{{\eta }_{2}-{\eta }_{1}}},\quad {\omega }^{{\prime} }(k)=\displaystyle \frac{\sqrt{-1}{K}^{{\prime} }(k)}{\sqrt{{\eta }_{2}-{\eta }_{1}}}.\end{eqnarray*}$$

Since we know that $k={p}^{{\prime} }$ and ${k}^{{\prime} }=p$, the above equation yields the following.

$$\begin{eqnarray}&&\omega (k)=-\sqrt{2}\sqrt{-1}{\omega }^{{\prime} }{}^{0},\quad {\omega }^{{\prime} }(k)=\sqrt{2}\sqrt{-1}{\omega }^{0}.\end{eqnarray} \tag{ 4.3 }$$

We see that

$$\begin{eqnarray}&&\wp (\sqrt{2}\sqrt{-1}\hat{y})=Y(\hat{y}+{\omega }^{{\prime} }{}^{0})-\displaystyle \frac{\ a\ }{3},\end{eqnarray} \tag{ 4.4 }$$

where we used the well-known relations

$$\begin{eqnarray*}&&\mathrm{sn}\ (\sqrt{-1}v,{p}^{{\prime} })=\sqrt{-1}\displaystyle \frac{\mathrm{sn}\ (v,p)}{\mathrm{cn}\ (v,p)},\quad \mathrm{sn}\ (v+\sqrt{-1}{K}^{{\prime} }(p),p)=\displaystyle \frac{1}{p\ \mathrm{sn}\ (v,p)}.\end{eqnarray*}$$

Remark 4.1.  $Y(\hat{y})={\zeta }_{2}+\displaystyle \frac{{\zeta }_{1}-{\zeta }_{2}}{{\mathrm{sn}}^{2}\ (\sqrt{2}\sqrt{{\zeta }_{2}-{\zeta }_{1}}\ \hat{y},p)}$ is also a solution of (1.1). The Weierstrass $\wp (z)$-function is defined by

$$\begin{eqnarray*}&&\wp (z)=\displaystyle \frac{1}{{z}^{2}}+\displaystyle \sum _{\omega \in L,\omega \ne 0}\left(\displaystyle \frac{1}{{(z-\omega )}^{2}}-\displaystyle \frac{1}{{\omega }^{2}}\right),\quad L={\bf{Z}}\omega (k)+{\bf{Z}}{\omega }^{{\prime} }(k).\end{eqnarray*}$$

The sum of the right hand side is absolutely and uniformly convergent on a compact subset of ${\bf{C}}$. It is an elliptic function.

On the other hand, rewriting (2.1) by the new coordinates $\hat{x}$, $\hat{y}$ and differentiating the rewritten equation one more time with respect to $\hat{x}$ we have ${\psi }_{\hat{x}\hat{x}\hat{x}}=2a{\psi }_{\hat{x}}+2\psi$. Let ${\mu }_{1}$, ${\mu }_{2}$ and ${\mu }_{3}$ be the solutions of the characteristic equation ${\mu }^{3}-2a\mu -2=0$, that is,

$$\begin{eqnarray}&&{\mu }_{1}=\displaystyle \frac{\ 1+\sqrt{{\alpha }^{3}+1}\ }{\alpha },\quad {\mu }_{2}=-\displaystyle \frac{\ 2\ }{\alpha },\quad {\mu }_{3}=\displaystyle \frac{\ 1-\sqrt{{\alpha }^{3}+1}\ }{\alpha }.\end{eqnarray} \tag{ 4.5 }$$

We then may write $\psi (\hat{x},\hat{y})=({e}^{{\mu }_{1}\hat{x}}{C}_{1}(\hat{y}),\ {e}^{{\mu }_{2}\hat{x}}{C}_{2}(\hat{y}),\ {e}^{{\mu }_{3}\hat{x}}{C}_{3}(\hat{y}))$. Substituting this expression into the rewritten equation of (2.1), in particular, into ${\psi }_{\hat{x}\hat{y}}=\tfrac{1}{2}{u}_{\hat{y}}{\psi }_{\hat{y}}-{e}^{u}{\psi }_{\hat{y}}$, we obtain $(2{e}^{u}+{\mu }_{j}^{2}-2a){C}_{j}^{{\prime} }(\hat{y})={u}_{\hat{y}}{e}^{u}{C}_{j}(\hat{y})$. Using the relations

$$\begin{eqnarray*}\begin{array}{rcl}{u}_{\hat{y}}{e}^{u} & = & -2({\zeta }_{2}-{\zeta }_{3})\mathrm{sn}\ (v)\displaystyle \frac{d}{d\hat{y}}\mathrm{sn}\ (v)=2({\zeta }_{2}-{\zeta }_{3})\mathrm{cn}\ (v)\displaystyle \frac{d}{d\hat{y}}\mathrm{cn}\ (v)\\ & = & 2({\zeta }_{2}-{\zeta }_{1})\mathrm{dn}\ (v)\displaystyle \frac{d}{d\hat{y}}\mathrm{dn}\ (v)\quad {\rm{with}}\ v=\sqrt{2}\sqrt{{\zeta }_{2}-{\zeta }_{1}}\ \hat{y},\end{array}\end{eqnarray*}$$

which, together with $\det { \mathcal F }={e}^{u}$, yield the explicit parametrization

$$\begin{eqnarray}&&\psi (\hat{x},\hat{y})=\left(\begin{array}{c}{c}_{1}\exp ({\mu }_{1}\hat{x})\ \mathrm{dn}\ (\sqrt{2}\sqrt{{\zeta }_{2}-{\zeta }_{1}}\ \hat{y},p)\\ {c}_{2}\exp ({\mu }_{2}\hat{x})\ \mathrm{sn}\ (\sqrt{2}\sqrt{{\zeta }_{2}-{\zeta }_{1}}\ \hat{y},p)\\ {c}_{3}\exp ({\mu }_{3}\hat{x})\ \mathrm{cn}\ (\sqrt{2}\sqrt{{\zeta }_{2}-{\zeta }_{1}}\ \hat{y},p)\end{array}\right),\end{eqnarray} \tag{ 4.6 }$$

where c₁, c₂ and c₃ are non-zero real numbers with

$$\begin{eqnarray*}&&{c}_{1}{c}_{2}{c}_{3}={(\sqrt{2}\sqrt{{\zeta }_{2}-{\zeta }_{1}}({\zeta }_{3}-{\zeta }_{1}))}^{-1}.\end{eqnarray*}$$

This is a rotational surface as an orbit of the group ${SO}(1,1)$ of hyperbolic rotations (Lorentz boosts) up to affine transformations. When α tends to 2, the solution converges to a trivial solution $u=0$.

Remark 4.2. Analogously, we may get a solution of (1.1) which depends only on the variable $\hat{x}$ as follows.

$$\begin{eqnarray*}&&{e}^{u(\hat{x})}={\zeta }_{1}+({\zeta }_{3}-{\zeta }_{1}){\mathrm{sn}}^{2}\ (\sqrt{2}\sqrt{{\zeta }_{2}-{\zeta }_{1}}\hat{x},q)\quad {\rm{with}}\quad {q}^{2}=\displaystyle \frac{{\zeta }_{1}-{\zeta }_{3}}{{\zeta }_{1}-{\zeta }_{2}}.\ \end{eqnarray*}$$

The corresponding affine sphere is obtained as follows.

$$\begin{eqnarray*}&&\psi (\hat{x},\hat{y})=\left(\begin{array}{c}{c}_{1}\sqrt{{e}^{u}-a}\ \exp \left(-\displaystyle \int \displaystyle \frac{d\hat{x}}{{e}^{u}-a}\right)\\ {c}_{2}\cos (\sqrt{2a}\hat{y})\ \exp \left(\displaystyle \frac{1}{2}u-\displaystyle \int {e}^{-u}d\hat{x}\right)\\ {c}_{3}\sin (\sqrt{2a}\hat{y})\ \exp \left(\displaystyle \frac{1}{2}u-\displaystyle \int {e}^{-u}d\hat{x}\right)\end{array}\right).\end{eqnarray*}$$

Therefore, this is a rotational surface as a orbit of the rotational group SO(2) of elliptic type up to affine transformations. When α tends to 2, the above solution converges to the 1-soliton solution

$$\begin{eqnarray*}&&{e}^{u(\hat{x})}=1-\displaystyle \frac{3}{2{\cosh }^{2}(\sqrt{3}\hat{x})}.\end{eqnarray*}$$

of Tzitzèica equation. The rotational surface corresponding to the 1-soliton is called the Jonas Kelch (Figure 2, see also [25]).

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In the rest of the section 4, we represent the formulas (4.1) and (4.6) in terms of the Riemann theta function.

We consider an elliptic curve ${ \mathcal C }$ defined by ${B}^{2}=4(\eta -{\eta }_{1})(\eta -{\eta }_{2})(\eta -{\eta }_{3})$. The curve is parametrized by $(B,\eta )=({\wp }^{{\prime} }(z),\wp (z))$. We may choose a cycle $\{{a}_{1},{b}_{1}\}$ of ${ \mathcal C }$ so that the following holds :

$$\begin{eqnarray}&&\left\{\begin{array}{l}{\displaystyle \int }_{{a}_{1}}\displaystyle \frac{d\eta }{B}={\displaystyle \int }_{{a}_{1}}\displaystyle \frac{d\wp (z)}{{\wp }^{{\prime} }(z)}={\displaystyle \int }_{{a}_{1}}{dz}=2\omega (k),\\ {\displaystyle \int }_{{b}_{1}}\displaystyle \frac{d\eta }{B}={\displaystyle \int }_{{b}_{1}}\displaystyle \frac{d\wp (z)}{{\wp }^{{\prime} }(z)}={\displaystyle \int }_{{b}_{1}}{dz}=2{\omega }^{{\prime} }(k).\end{array}\right.\end{eqnarray} \tag{ 4.7 }$$

If we take ${w}_{1}=\displaystyle \frac{\pi \sqrt{-1}}{\omega (k)}\displaystyle \frac{d\eta }{B}$, then we easily see that

$$\begin{eqnarray*}&&{\int }_{{a}_{1}}{w}_{1}=2\pi \sqrt{-1},\quad {\int }_{{b}_{1}}{w}_{1}=2\pi \sqrt{-1}\displaystyle \frac{{\omega }^{{\prime} }(k)}{\omega (k)}=: {\rm{\Pi }}(k).\end{eqnarray*}$$

Then, the elliptic curve has a Riemann period matrix $(2\pi \sqrt{-1}\ {\rm{\Pi }}(k))$. Setting $\widetilde{B}=\sqrt{2}\sqrt{-1}B,\eta =\zeta -\displaystyle \frac{\ a\ }{3}$, we obtain ${\widetilde{B}}^{2}=-8(\zeta -{\zeta }_{1})(\zeta -{\zeta }_{2})(\zeta -{\zeta }_{3})$, which is denoted by ${{ \mathcal C }}^{0}$. We may choose a cycle $\{{a}_{1}^{0},{b}_{1}^{0}\}$ of ${{ \mathcal C }}^{0}$ by ${a}_{1}^{0}={b}_{1}$ and ${b}_{1}^{0}=-{a}_{1}$, so that the following relations hold:

$$\begin{eqnarray}&&\left\{\begin{array}{l}{\displaystyle \int }_{{a}_{1}^{0}}\displaystyle \frac{d\zeta }{\widetilde{B}}={\displaystyle \int }_{{b}_{1}}\displaystyle \frac{d\eta }{\sqrt{2}\sqrt{-1}B}=\displaystyle \frac{2{\omega }^{{\prime} }(k)}{\sqrt{2}\sqrt{-1}}=2{\omega }^{0},\\ {\displaystyle \int }_{{b}_{1}^{0}}\displaystyle \frac{d\zeta }{\widetilde{B}}=-{\displaystyle \int }_{{a}_{1}}\displaystyle \frac{d\eta }{\sqrt{2}\sqrt{-1}B}=\displaystyle \frac{-2\omega (k)}{\sqrt{2}\sqrt{-1}}=2{\omega }^{{\prime} }{}^{0}.\end{array}\right.\end{eqnarray} \tag{ 4.8 }$$

If we take ${w}_{1}^{0}=\tfrac{\pi \sqrt{-1}}{{\omega }^{0}}\tfrac{d\zeta }{\widetilde{B}}$, then we see that

$$\begin{eqnarray}&&{\int }_{{a}_{1}^{0}}{w}_{1}^{0}=2\pi \sqrt{-1},\quad {\int }_{{b}_{1}^{0}}{w}_{1}^{0}=2\pi \sqrt{-1}\displaystyle \frac{{\omega }^{{\prime} }{}^{0}}{{\omega }^{0}}=-2\pi \displaystyle \frac{{K}^{{\prime} }(p)}{K(p)}=: {\rm{\Pi }}(p).\end{eqnarray} \tag{ 4.9 }$$

Thus, the curve ${{ \mathcal C }}^{0}$ has a Riemann period matrix $(2\pi \sqrt{-1}\ {\rm{\Pi }}(p))$. Let ${\zeta }_{w}$ be the Weierstrass ζ-function defined by

$$\begin{eqnarray*}&&{\zeta }_{w}(z)=\displaystyle \frac{1}{z}+\displaystyle \sum _{\omega \in L,\ \omega \ne 0}\left(\displaystyle \frac{1}{z-\omega }+\displaystyle \frac{1}{\omega }+\displaystyle \frac{z}{{\omega }^{2}}\right).\end{eqnarray*}$$

It is a meromorphic function on ${\bf{C}}$ and a odd function, but not an elliptic function. In fact, we have the following properties :

$$\begin{eqnarray}&&\left\{\begin{array}{c}{\zeta }_{w}^{{\prime} }(z)=-\wp (z),\\ {\zeta }_{w}(z+2\omega )={\zeta }_{w}(z)+2{\zeta }_{w}(\omega ),\\ {\zeta }_{w}(z+2{\omega }^{{\prime} })={\zeta }_{w}(z)+2{\zeta }_{w}({\omega }^{{\prime} }),\\ {\zeta }_{w}(\omega +{\omega }^{{\prime} })={\zeta }_{w}(\omega )+{\zeta }_{w}({\omega }^{{\prime} }),\\ {\zeta }_{w}(\omega ){\omega }^{{\prime} }-{\zeta }_{w}({\omega }^{{\prime} })\ \omega =\displaystyle \frac{\pi \sqrt{-1}}{2},\qquad \quad ({\rm{Legendre\mbox{'}s}}\,{\rm{relation}})\end{array}\right.\end{eqnarray} \tag{ 4.10 }$$

where $\omega =\omega (k)$ and ${\omega }^{{\prime} }={\omega }^{{\prime} }(k)$. In the sequel, we keep on using this notation when there is no confusion. The last equation in (4.10) follows from the 2nd and 3rd equations in (4.10) and from the fact ${\int }_{{\rm{\Gamma }}}{\zeta }_{w}(z)\ {dz}=2\pi \sqrt{-1}$, where Γ is a fundamental parallelogram which contains a pole of ${\zeta }_{w}(z)$ inside. Let ${{\rm{\Omega }}}_{\infty }^{0}$ be an Abelian differential of 2nd kind defined by

$$\begin{eqnarray}&&{{\rm{\Omega }}}_{\infty }^{0}=-\left(\wp (z)+\displaystyle \frac{{\zeta }_{w}({\omega }^{{\prime} })}{{\omega }^{{\prime} }}\right)\displaystyle \frac{\ d\zeta \ }{\widetilde{B}}.\end{eqnarray} \tag{ 4.11 }$$

Set ${{\bf{U}}}^{0}={\int }_{{b}_{1}^{0}}{{\rm{\Omega }}}_{\infty }^{0}$. It follows from (4.10) and ${b}_{1}^{0}=-{a}_{1}$ that

$$\begin{eqnarray*}\begin{array}{rcl}{\displaystyle \int }_{{b}_{1}^{0}}{{\rm{\Omega }}}_{\infty }^{0} & = & -{\displaystyle \int }_{z}^{z+2\omega }\left(-\wp (z)-\displaystyle \frac{{\zeta }_{w}({\omega }^{{\prime} })}{{\omega }^{{\prime} }}\right)\displaystyle \frac{{dz}}{\sqrt{2}\sqrt{-1}}\\ & = & -\displaystyle \frac{1}{\sqrt{2}\sqrt{-1}}{\left[{\zeta }_{w}(z)-\displaystyle \frac{{\zeta }_{w}({\omega }^{{\prime} })}{{\omega }^{{\prime} }}z\right]}_{z}^{z+2\omega }\\ & = & -\displaystyle \frac{1}{\sqrt{2}\sqrt{-1}}\left(2{\zeta }_{w}(\omega )-2{\zeta }_{w}({\omega }^{{\prime} })\displaystyle \frac{\omega }{{\omega }^{{\prime} }}\right)\\ & = & -\displaystyle \frac{\pi \sqrt{-1}}{\sqrt{2}\sqrt{-1}{\omega }^{{\prime} }}=\displaystyle \frac{\pi \sqrt{-1}}{2{\omega }^{0}},\end{array}\end{eqnarray*}$$

hence we obtain

$$\begin{eqnarray}&&{{\bf{U}}}^{0}=\displaystyle \frac{\pi \sqrt{-1}}{2{\omega }^{0}}.\end{eqnarray} \tag{ 4.12 }$$

Similarly, we see that ${\int }_{{a}_{1}^{0}}{{\rm{\Omega }}}_{\infty }^{0}=0$. Therefore, ${{\rm{\Omega }}}_{\infty }^{0}$ is the normalized Abelian differential of 2nd kind.

Remark 4.3. Alternatively, since we have ${{\rm{\Omega }}}_{\infty }^{0}=\left(-\tfrac{\ {z}^{-2}\ }{\sqrt{2}\sqrt{-1}}+O(1)\right)\ {dz}$, it follows from the reciprocity laws for differentials of 1st and 2nd kind that

$$\begin{eqnarray*}&&{{\bf{U}}}^{0}=-\displaystyle \frac{1}{\sqrt{2}\sqrt{-1}}{\left.\displaystyle \frac{d}{{dz}}\right|}_{z=0}{\int }_{{P}_{0}}^{P}{w}_{1}^{0}=-\displaystyle \frac{1}{\sqrt{2}\sqrt{-1}}\displaystyle \frac{\pi \sqrt{-1}}{\sqrt{2}\sqrt{-1}{\omega }^{0}}=\displaystyle \frac{\pi \sqrt{-1}}{2{\omega }^{0}}.\end{eqnarray*}$$

For ${\rm{\Pi }}={\rm{\Pi }}(p)$, $z\in {\bf{C}}$, set $\tau =\tfrac{{\rm{\Pi }}}{2\pi \sqrt{-1}},v=\tfrac{z}{2\pi \sqrt{-1}}$. The Riemann theta function $\theta (z;{\rm{\Pi }})$ for elliptic curve with the period matrix $(2\pi \sqrt{-1}\ {\rm{\Pi }})$ is defined by

$$\begin{eqnarray*}&&\theta (z;{\rm{\Pi }})=\displaystyle \sum _{m\in {\bf{Z}}}\exp \left(\displaystyle \frac{1}{2}{\rm{\Pi }}{m}^{2}+{mz}\right),\end{eqnarray*}$$

which absolutely and uniformly converges on a compact subset of ${\bf{C}}$. The Jacobi theta functions are described in terms of the Riemann theta function as follows (see [27] and [28]).

$$\begin{eqnarray}&&\left\{\begin{array}{c}{\theta }_{0}(v| \tau )=\theta (z+\pi \sqrt{-1};{\rm{\Pi }}),\\ {\theta }_{1}(v| \tau )=-\exp \left(\displaystyle \frac{1}{8}{\rm{\Pi }}+\displaystyle \frac{1}{2}(z+\pi \sqrt{-1})\right)\ \theta \phantom{\rule{}{0.25ex}}(z+\pi \sqrt{-1}+\displaystyle \frac{1}{2}{\rm{\Pi }};{\rm{\Pi }}\phantom{\rule{}{0.25ex}}),\\ {\theta }_{2}(v| \tau )=\exp \left(\displaystyle \frac{1}{8}{\rm{\Pi }}+\displaystyle \frac{1}{2}z\right)\ \theta \left(z+\displaystyle \frac{1}{2}{\rm{\Pi }};{\rm{\Pi }}\right),\\ {\theta }_{3}(v| \tau )=\theta (z;{\rm{\Pi }}).\end{array}\right.\end{eqnarray} \tag{ 4.13 }$$

We then see that ${\theta }_{3}(v+\tfrac{1}{2})={\theta }_{0}(v),{\theta }_{2}(v+\tfrac{1}{2})=-{\theta }_{1}(v)$ (see [28]). Although the Jacobi theta functions are not elliptic functions, there are some relations between the Jacobi theta functions and the Jacobi elliptic functions as follows.

$$\begin{eqnarray}\left\{\begin{array}{cl}\mathrm{sn}(u,p)=\displaystyle \frac{{\theta }_{3}(0){\theta }_{1}(v)}{{\theta }_{2}(0){\theta }_{0}(v)}, & \mathrm{cn}(u,p)=\displaystyle \frac{{\theta }_{0}(0){\theta }_{2}(v)}{{\theta }_{2}(0){\theta }_{0}(v)},\\ \mathrm{dn}(u,p)=\displaystyle \frac{{\theta }_{0}(0){\theta }_{3}(v)}{{\theta }_{3}(0){\theta }_{0}(v)}, & {\rm{where}}\ u=2K(p)v.\end{array}\right.\end{eqnarray} \tag{ 4.14 }$$

Moreover, the following beautiful formula is known

$$\begin{eqnarray}&&\displaystyle \frac{{d}^{2}}{{{du}}^{2}}\mathrm{log}{\theta }_{0}\left(\left.\displaystyle \frac{u}{2K(p)}\right|\tau \right)={\mathrm{dn}}^{2}(u,p)-\displaystyle \frac{\ E(p)\ }{K(p)},\end{eqnarray} \tag{ 4.15 }$$

where $E(p)={\int }_{0}^{K(p)}{\mathrm{dn}}^{2}(u,p)\ {du}$ is the complete elliptic integral of 2nd kind. We now calculate

$$\begin{eqnarray}\begin{array}{rcl}-2\displaystyle \frac{\partial }{\partial x}\displaystyle \frac{\partial }{\partial t}\mathrm{log}{\theta }_{0}\left(\left.\displaystyle \frac{1}{2{\omega }^{0}}\hat{y}\right|\tau \right) & = & \displaystyle \frac{1}{2}\left(\displaystyle \frac{{\partial }^{2}}{\partial {\hat{y}}^{2}}-\displaystyle \frac{{\partial }^{2}}{\partial {\hat{x}}^{2}}\right)\mathrm{log}{\theta }_{0}\left(\left.\displaystyle \frac{1}{2{\omega }^{0}}\hat{y}\right|\tau \right)\\ & = & \displaystyle \frac{1}{2}\displaystyle \frac{{\partial }^{2}}{\partial {\hat{y}}^{2}}\mathrm{log}{\theta }_{0}\left(\displaystyle \frac{1}{2{\omega }^{0}}\hat{y}| \tau \right)\\ & = & \displaystyle \frac{1}{2}{\left(\displaystyle \frac{1}{2{\omega }^{0}}\right)}^{2}{(2K(p))}^{2}\left({\mathrm{dn}}^{2}\left(\displaystyle \frac{K(p)}{{\omega }^{0}}\hat{y},p\right)-\displaystyle \frac{\ E(p)\ }{K(p)}\right)\\ & = & ({\zeta }_{2}-{\zeta }_{1})\left(1-{p}^{2}{\mathrm{sn}}^{2}(\sqrt{2}\sqrt{{\zeta }_{2}-{\zeta }_{1}}\ \hat{y},p)-\displaystyle \frac{\ E(p)\ }{K(p)}\right)\\ & = & ({\zeta }_{2}-{\zeta }_{1})\left(1-\displaystyle \frac{\ E(p)\ }{K(p)}\right)-({\zeta }_{2}-{\zeta }_{3}){\mathrm{sn}}^{2}\ (\sqrt{2}\sqrt{{\zeta }_{2}-{\zeta }_{1}}\ \hat{y},p),\end{array}\end{eqnarray} \tag{ 4.16 }$$

where we have used (4.2) and ${p}^{2}=\tfrac{{\zeta }_{3}-{\zeta }_{2}}{{\zeta }_{1}-{\zeta }_{2}}$. If we set $C={\zeta }_{1}+\tfrac{E(p)}{K(p)}({\zeta }_{2}-{\zeta }_{1})$ then we have

$$\begin{eqnarray*}\begin{array}{rcl}C-2\displaystyle \frac{\partial }{\partial x}\displaystyle \frac{\partial }{\partial t}\mathrm{log}{\theta }_{0}\left(\left.\displaystyle \frac{1}{2{\omega }^{0}}\hat{y}\right|\tau \right) & = & {\zeta }_{2}-({\zeta }_{2}-{\zeta }_{3}){\mathrm{sn}}^{2}(\sqrt{2}\sqrt{{\zeta }_{2}-{\zeta }_{1}}\ \hat{y},p)\\ & = & Y(\hat{y})=\exp (u(\hat{y})),\end{array}\end{eqnarray*}$$

which is a solution of (1.1) (see (4.1)). On the other hand, for ${\bf{e}}=\pi \sqrt{-1}$ we see that

$$\begin{eqnarray*}&&\theta ({{\bf{U}}}^{0}(x-t)+{\bf{e}};\,{\rm{\Pi }}(p))=\theta \left(\displaystyle \frac{\pi \sqrt{-1}}{{\omega }^{0}}\hat{y}+\pi \sqrt{-1};{\rm{\Pi }}(p)\right)={\theta }_{0}\left(\displaystyle \frac{1}{2{\omega }^{0}}\hat{y}| \tau \right).\end{eqnarray*}$$

Thus, we obtain the following formula.

$$\begin{eqnarray}&&\exp (\ u(\hat{y}))=C-2\displaystyle \frac{\partial }{\partial x}\displaystyle \frac{\partial }{\partial t}\mathrm{log}\theta ({{\bf{U}}}^{0}(x-t)+{\bf{e}};\,{\rm{\Pi }}(p)),\end{eqnarray} \tag{ 4.17 }$$

where $C={\zeta }_{1}+\tfrac{\ E(p)\ }{K(p)}({\zeta }_{2}-{\zeta }_{1})$ and ${\bf{e}}=\pi \sqrt{-1}$.

Lemma 4.4. The $C$ in (4.17) is given by

$$\begin{eqnarray}&&C=\displaystyle \frac{\ a\ }{3}-\displaystyle \frac{\ {\zeta }_{w}({\omega }^{{\prime} })\ }{{\omega }^{{\prime} }}.\end{eqnarray} \tag{ 4.18 }$$

Proof. First of all, we have from the definition of $E(p)$ that $C={\zeta }_{2}+\tfrac{({\zeta }_{3}-{\zeta }_{2})}{K(p)}{\int }_{0}^{K(p)}{\mathrm{sn}}^{2}\ u\ {du}$. Secondly, we have from (4.4) that

$$\begin{eqnarray*}&&{\mathfrak{P}}(\sqrt{2}\sqrt{-1}\hat{y})+\displaystyle \frac{a}{3}=Y(\hat{y}+{\omega }^{{\prime} }{}^{0})={\zeta }_{2}-({\zeta }_{2}-{\zeta }_{3}){\mathrm{sn}}^{2}\ (\sqrt{2}\sqrt{{\zeta }_{2}-{\zeta }_{1}}\hat{y}+\sqrt{-1}{K}^{{\prime} }(p),p).\end{eqnarray*}$$

We calculate

$$\begin{eqnarray*}\begin{array}{rcl}{\displaystyle \int }_{0}^{K(p)}{\mathrm{sn}}^{2}\ u\ {du} & = & \sqrt{2}\sqrt{{\zeta }_{2}-{\zeta }_{1}}{\displaystyle \int }_{-{\omega }^{{\prime} }{}^{0}}^{{\omega }^{0}-{\omega }^{{\prime} }{}^{0}}{\mathrm{sn}}^{2}(\sqrt{2}\sqrt{{\zeta }_{2}-{\zeta }_{1}}\hat{y}+\sqrt{-1}{K}^{{\prime} }(p),p)\ d\hat{y}\\ & = & \displaystyle \frac{\sqrt{2}\sqrt{{\zeta }_{2}-{\zeta }_{1}}}{{\zeta }_{3}-{\zeta }_{2}}{\displaystyle \int }_{-{\omega }^{{\prime} }{}^{0}}^{{\omega }^{0}-{\omega }^{{\prime} }{}^{0}}\left({\mathfrak{P}}(\sqrt{2}\sqrt{-1}\hat{y})+\displaystyle \frac{a}{3}-{\zeta }_{2}\right)\ d\hat{y}\\ & = & \displaystyle \frac{\sqrt{2}\sqrt{{\zeta }_{2}-{\zeta }_{1}}}{{\zeta }_{3}-{\zeta }_{2}}{\left[-\displaystyle \frac{1}{\sqrt{2}\sqrt{-1}}{\zeta }_{w}(\sqrt{2}\sqrt{-1}\hat{y})+\left(\displaystyle \frac{a}{3}-{\zeta }_{2}\right)\hat{y}\right]}_{-{\omega }^{{\prime} }{}^{0}}^{{\omega }^{0}-{\omega }^{{\prime} }{}^{0}}\\ & = & \displaystyle \frac{\sqrt{2}\sqrt{{\zeta }_{2}-{\zeta }_{1}}}{{\zeta }_{3}-{\zeta }_{2}}\left(-\displaystyle \frac{1}{\sqrt{2}\sqrt{-1}}{\zeta }_{w}({\omega }^{{\prime} })+\left(\displaystyle \frac{a}{3}-{\zeta }_{2}\right){\omega }^{0}\right)\\ & = & \displaystyle \frac{K(p)}{{\zeta }_{3}-{\zeta }_{2}}\left(-\displaystyle \frac{{\zeta }_{w}({\omega }^{{\prime} })}{{\omega }^{{\prime} }}+\displaystyle \frac{a}{3}-{\zeta }_{2}\right),\end{array}\end{eqnarray*}$$

where we used (4.2), (4.3) and (4.10). □

We gave a formula of Blaschke immersion in terms of elliptic functions in (4.6). In this section, we rewrite the formula (4.6) in terms of Riemann theta function. For this purpose, we consider the elliptic curve ${{ \mathcal C }}^{0}$ as a Prym variety of a compact Riemann surface $\hat{{ \mathcal C }}$ of genus 2. When $u=u(\hat{y})$ is given by (4.1), the spectral curve is defined by the equation $\det \ (U(\lambda )+V(\lambda )-\mu I)=0$, where $U(\lambda )$ and $V(\lambda )$ are as in (2.5). Since the spectral curve is independent of the parameter $\hat{y}$, if we take $\hat{y}=0$ then the initial condition of $u=u(\hat{y})$ implies that the defining equation of the spectral curve is ${\mu }^{3}-2a\mu ={\lambda }^{3}+{\lambda }^{-3}$. The unramified covering map $\lambda \to \nu ={\lambda }^{3}$ yields ${\mu }^{3}-2a\mu =\nu +{\nu }^{-1}$. Projectivizing this affine plane curve, we obtain a compact Riemann surface $\hat{{ \mathcal C }}$ of genus 2. Since ${(\nu -{\nu }^{-1})}^{2}={(\nu +{\nu }^{-1})}^{2}-4={({\mu }^{3}-2a\mu )}^{2}-4$, setting $\widetilde{\nu }=\nu -{\nu }^{-1}$, we see that $\hat{{ \mathcal C }}$ may be regarded as a hyperelliptic curve ${({\mu }^{3}-2a\mu )}^{2}-4={\widetilde{\nu }}^{2}$. We may observe that $\hat{{ \mathcal C }}$ admits holomorphic involution σ and an anti-holomorphic involution ρ defined by $\sigma (\mu ,\nu )=(-\mu ,-\nu )$ and $\rho (\mu ,\nu )=(\overline{\mu },{\overline{\nu }}^{-1})$. If we transform the parameters by

$$\begin{eqnarray*}&&{\mu }^{2}=-2\zeta +2a,\quad \nu -{\nu }^{-1}=\widetilde{B},\end{eqnarray*}$$

then we obtain the elliptic curve ${\hat{{ \mathcal C }}}^{0}:{\widetilde{B}}^{2}=-8(\zeta -{\zeta }_{1})(\zeta -{\zeta }_{2})(\zeta -{\zeta }_{3})$. We may use ${\hat{{ \mathcal C }}}^{0}$ as a Prym variety of $\hat{{ \mathcal C }}$. Let $\varphi :\hat{{ \mathcal C }}\longrightarrow {\hat{{ \mathcal C }}}^{0}$ be a covering map defined as above. We may choose the canonical homology basis $\{{\hat{a}}_{1},{\hat{a}}_{2},{\hat{b}}_{1},{\hat{b}}_{2}\}$ of $\hat{{ \mathcal C }}$ so that

$$\begin{eqnarray*}&&{\varphi }^{* }{a}_{1}^{0}={\hat{a}}_{1}+{\hat{a}}_{2},\quad {\varphi }^{* }{b}_{1}^{0}={\hat{b}}_{1}+{\hat{b}}_{2},\quad \sigma ({\hat{a}}_{1})=-{\hat{a}}_{2},\quad \sigma ({\hat{b}}_{1})=-{\hat{b}}_{2}\end{eqnarray*}$$

hold (see Figure 3).

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Let $\{{w}_{1},{w}_{2}\}$ be the basis of ${{\rm{H}}}^{1}(\hat{{ \mathcal C }},{\bf{C}})$ such that $(2\pi \sqrt{-1}\ {\rm{I}}\ \hat{T})$ is the Riemann period matrix for $\hat{{ \mathcal C }}$, where $\hat{T}=({\hat{T}}_{{ij}})$ and ${\hat{T}}_{{ij}}={\int }_{{\hat{b}}_{i}}{w}_{j}$ for i, $j=1$, 2. It is well known that the Jacobian variety $J(\hat{{ \mathcal C }})$ of $\hat{{ \mathcal C }}$ is a complex torus ${{\bf{C}}}^{2}/{\rm{\Lambda }}$, where ${\rm{\Lambda }}={\mathrm{Span}}_{{\bf{Z}}}\{2\pi \sqrt{-1}{\rm{I}},\hat{T}\}$. The basis $\{{w}_{1},{w}_{2}\}$ can be constructed as follows. Set ${u}_{1}=\displaystyle \frac{d\mu }{\widetilde{\nu }},{u}_{2}=\displaystyle \frac{\mu d\mu }{\widetilde{\nu }}$. We then see that ${\varphi }^{* }{w}_{1}^{0}=-\tfrac{\pi \sqrt{-1}}{{\omega }^{0}}\ {u}_{2}$. We want to look for ${w}_{1},{w}_{2}$ which satisfy ${w}_{1}+{w}_{2}=-\tfrac{\pi \sqrt{-1}}{{\omega }^{0}}\ {u}_{2}$. Note that $\{{u}_{1},{u}_{2}\}$ is a basis of ${{\rm{H}}}^{1}(\hat{{ \mathcal C }},{\bf{C}})$ and ${\sigma }^{* }{u}_{1}={u}_{1}$ and ${\sigma }^{* }{u}_{2}=-{u}_{2}$. If we set ${C}_{{ij}}={\int }_{{\hat{a}}_{i}}{u}_{j}$ for i, $j=1$, 2 then we have ${C}_{11}=-{C}_{21}$ and ${C}_{12}={C}_{22}=-2{\omega }^{0}$. The last equation follows from ${\int }_{{\hat{a}}_{j}}{\varphi }^{* }{w}_{1}^{0}={\int }_{\varphi ({\hat{a}}_{j})}{w}_{1}^{0}={\int }_{{a}_{1}^{0}}{w}_{1}^{0}=2\pi \sqrt{-1}$ for $j=1$, 2. The Riemann bilinear relation means that ${C}_{11}\ne 0$. Thus, we find ${w}_{1},{w}_{2}$ :

$$\begin{eqnarray*}&&{w}_{1}=\displaystyle \frac{\pi \sqrt{-1}}{{C}_{11}}\left({u}_{1}-\displaystyle \frac{{C}_{11}}{2{\omega }^{0}}{u}_{2}\right),\quad {w}_{2}=\displaystyle \frac{\pi \sqrt{-1}}{{C}_{11}}\left(-{u}_{1}-\displaystyle \frac{{C}_{11}}{2{\omega }^{0}}{u}_{2}\right),\end{eqnarray*}$$

with the properties ${\varphi }^{* }{w}_{1}^{0}={w}_{1}+{w}_{2}$ and ${\sigma }^{* }{w}_{1}=-{w}_{2}$. If we set ${T}_{j1}={\int }_{{\hat{b}}_{j}}({w}_{1}-{w}_{2})$ then we have ${T}_{11}=-{T}_{21}=: T$. We then know that

$$\begin{eqnarray}&&{\hat{T}}_{11}={\hat{T}}_{22}=\displaystyle \frac{1}{2}({\rm{\Pi }}+T),\quad {\hat{T}}_{12}={\hat{T}}_{21}=\displaystyle \frac{1}{2}({\rm{\Pi }}-T)\end{eqnarray} \tag{ 4.19 }$$

for ${\rm{\Pi }}={\rm{\Pi }}(p)$, which is as in (4.9). Moreover, since $\overline{{\rho }^{* }{u}_{j}}=-{u}_{j}$ for $j=1$, 2 we have that

$$\begin{eqnarray}&&\rho ({\hat{a}}_{j})=-{\hat{a}}_{j},\quad \rho ({\hat{b}}_{j})={\hat{b}}_{j}\quad {\rm{for}}\ j=1,2.\end{eqnarray} \tag{ 4.20 }$$

In fact, since $\hat{{ \mathcal C }}$ is an M-curve (see [27]), there are 3 real ovals, which are connected components of fixed points of ρ, and we may take ${\hat{b}}_{j}\ (j=1,2)$ as these ovals. In this case, the number C₁₁ above is real and $\overline{{\rho }^{* }{w}_{j}}={w}_{j}$ for $j=1,2$ hold, which is compatible with the reality of Π. We define a map ${ \mathcal B }:\hat{{ \mathcal C }}\longrightarrow \mathrm{Prym}(\hat{{ \mathcal C }})$ by ${ \mathcal B }(\hat{P})={\int }_{{\hat{P}}_{0}}^{\hat{P}}({w}_{1}+{w}_{2})$. If we change $\hat{P}\to \hat{P}+{m}_{j}{\hat{a}}_{j}+{n}_{j}{\hat{b}}_{j}$, where ${m}_{j},{n}_{j}\in {\bf{Z}}$ and $j=1$, 2, then ${ \mathcal B }(\hat{P})$ changes into ${ \mathcal B }(\hat{P})+2\pi \sqrt{-1}{m}_{j}\,+{\rm{\Pi }}{n}_{j}$, hence it is a well-defined map into $\mathrm{Prym}(\hat{{ \mathcal C }})\cong {{ \mathcal C }}^{0}={\bf{C}}/{\rm{\Gamma }}$, where ${\rm{\Gamma }}={\mathrm{Span}}_{{\bf{Z}}}\{2\pi \sqrt{-1},{\rm{\Pi }}\}$. We have the relations as follows.

$$\begin{eqnarray}&&\overline{{ \mathcal B }(\rho (\hat{P}))}={ \mathcal B }(\hat{P}),\quad { \mathcal B }(\sigma (\hat{P}))=-{ \mathcal B }(\hat{P})\quad (\mathrm{mod}\quad {\rm{\Gamma }})\end{eqnarray} \tag{ 4.21 }$$

We may consider the function f on $\hat{{ \mathcal C }}$ defined by $f(\hat{P})=\theta ({ \mathcal B }(\hat{P})-{\bf{e}})$ for ${\bf{e}}\in {\bf{C}}$, where θ is the Riemann theta function $\theta (z)=\theta (z;{\rm{\Pi }}(p))$ as in section 4.3. Let ${\hat{{ \mathcal C }}}_{0}$ be the simply-connected domain obtained by removing all $\hat{a}$- and $\hat{b}$-cycles from $\hat{{ \mathcal C }}$. The boundary of ${\hat{{ \mathcal C }}}_{0}$ is denoted by $\partial {\hat{{ \mathcal C }}}_{0}={\hat{a}}_{1}{\hat{b}}_{1}{\hat{a}}_{1}^{-1}{\hat{b}}_{1}^{-1}{\hat{a}}_{2}{\hat{b}}_{2}{\hat{a}}_{2}^{-1}{\hat{b}}_{2}^{-1}$. We then know that f is single-valued on ${\hat{{ \mathcal C }}}_{0}$. However, since the theta function satisfies the relations $\theta (z+2\pi \sqrt{-1})=\theta (z),\theta (z+{\rm{\Pi }})\,=\exp \left(-\tfrac{1}{2}{\rm{\Pi }}-z\right)\theta (z)$, it is no problem to consider f on $\hat{{ \mathcal C }}$ when we search for the zeros of the function f. With respect to this problem, there is a study of J. Fay.

Lemma 4.5 If $f(\hat{P})=\theta ({ \mathcal B }(\hat{P})-{\bf{e}})\not\equiv 0$ then the zeros of f is a degree 2 divisor $\hat{{ \mathcal D }}$. Moreover, we have ${ \mathcal B }(\hat{{ \mathcal D }})\equiv {\bf{K}}+2{\bf{e}}\ (\mathrm{mod}\ {\rm{\Gamma }})$, where ${\bf{K}}$ is given by

$$\begin{eqnarray*}&&{\bf{K}}=\displaystyle \frac{1}{2\pi \sqrt{-1}}\displaystyle \sum _{j=1}^{2}{\int }_{{\hat{a}}_{j}}{ \mathcal B }({w}_{1}+{w}_{2})-\displaystyle \sum _{j=1}^{2}{ \mathcal B }({\hat{b}}_{j}(0)),\end{eqnarray*}$$

and ${\hat{b}}_{j}(0)$ is the initial point of the path ${\hat{b}}_{j}$ in the boundary $\partial {\hat{{ \mathcal C }}}_{0}$.

This is, of course, a special one of the higher genus case, which is stated in the next section. Fay states the property of the divisor $\hat{{ \mathcal D }}$ in terms of the Abelian map ${ \mathcal A }:\hat{{ \mathcal C }}\longrightarrow J(\hat{{ \mathcal C }})$. Therefore, we here address the outline of the proof. When $\hat{P}\in {\hat{a}}_{j}({\rm{resp.}}\ {\hat{b}}_{j})$, we denote by ${\hat{P}}^{-}$ the corresponding point of ${\hat{a}}_{j}^{-1}({\rm{resp.}}\ {\hat{b}}_{j}^{-1})$. Set ${f}^{-}(\hat{P})=f({\hat{P}}^{-})$. If f is not identically zero, then the number n of the zeros of f is given by

$$\begin{eqnarray*}\begin{array}{rcl}n & = & \displaystyle \frac{1}{2\pi \sqrt{-1}}{\displaystyle \int }_{\partial {\hat{{ \mathcal C }}}_{0}}d\mathrm{log}f\\ & = & \displaystyle \frac{1}{2\pi \sqrt{-1}}\displaystyle \sum _{j=1}^{2}\left\{{\displaystyle \int }_{{\hat{a}}_{j}}(d\mathrm{log}f-d\mathrm{log}{f}^{-})+{\displaystyle \int }_{{\hat{b}}_{j}}(d\mathrm{log}f-d\mathrm{log}{f}^{-})\right\}.\end{array}\end{eqnarray*}$$

But, since we may observe that ${ \mathcal B }({\hat{P}}^{-})={ \mathcal B }(\hat{P})+{\rm{\Pi }}$ when $\hat{P}\in {\hat{a}}_{j}$ and ${ \mathcal B }({\hat{P}}^{-})={ \mathcal B }(\hat{P})-2\pi \sqrt{-1}$ when $\hat{P}\in {\hat{b}}_{j}$, the property of θ yields $n=\tfrac{1}{2\pi \sqrt{-1}}{\sum }_{j=1}^{2}{\int }_{{\hat{a}}_{j}}d{ \mathcal B }=2$. Thus, the divisor $\hat{{ \mathcal D }}$ is of degree 2 and expressed as $\hat{{ \mathcal D }}=\{{\hat{p}}_{1},{\hat{p}}_{2}\}$. By the residue theorem, we have ${ \mathcal B }(\hat{{ \mathcal D }})=\displaystyle \frac{1}{2\pi \sqrt{-1}}{\int }_{\partial {\hat{{ \mathcal C }}}_{0}}{ \mathcal B }(\hat{P})d\mathrm{log}f$. Using the same calculation as in above, we obtain ${ \mathcal B }(\hat{{ \mathcal D }})\equiv \displaystyle \frac{1}{2\pi \sqrt{-1}}{\sum }_{j=1}^{2}{\int }_{{\hat{a}}_{j}}{ \mathcal B }({w}_{1}+{w}_{2})+{\sum }_{j=1}^{2}{\int }_{{\hat{b}}_{j}}d\mathrm{log}f$, $(\mathrm{mod}\ {\rm{\Gamma }})$. Finally, if we denote by ${\hat{b}}_{j}(1)$ the end point of the path ${\hat{b}}_{j}$ in the boundary $\partial {\hat{{ \mathcal C }}}_{0}$, then the property of θ yields $f({\hat{b}}_{j}(1))=\exp (-\tfrac{1}{2}{\rm{\Pi }}-{ \mathcal B }({\hat{b}}_{j}(0))+{\bf{e}})f({\hat{b}}_{j}(0))$, which implies the formula of ${ \mathcal B }(\hat{{ \mathcal D }})$ stated in lemma 4.5.

In our case, take ${\bf{e}}=\pi \sqrt{-1}$. Then, since the zeros of $\theta (z)$ are given by $z=\pi \sqrt{-1}\pm \tfrac{1}{\ 2\ }{\rm{\Pi }}\ (\mathrm{mod}\ {\rm{\Gamma }})$, it follows from (4.32) below that $\hat{{ \mathcal D }}=\{{\hat{P}}_{2},{\hat{P}}_{-2}\}$, where ${\hat{P}}_{2}=({\mu }_{2},1),{\hat{P}}_{-2}=(-{\mu }_{2},-1)$ in terms of affine coordinate system $(\mu ,\nu )$. Moreover, we see that ${ \mathcal B }(\hat{{ \mathcal D }})\equiv 0\ (\mathrm{mod}\ {\rm{\Gamma }})$.

Next, we look for the solution $\hat{{\rm{\Psi }}}=\hat{{\rm{\Psi }}}(x,t,\hat{\nu })$ of the Schrödinger equation ${\partial }_{x}{\partial }_{t}\hat{{\rm{\Psi }}}={e}^{u}\hat{{\rm{\Psi }}}$ on $\hat{{ \mathcal C }}$ of which the integrability condition ensures that ${e}^{u}$ coincides with the one in (4.17). Moreover, the evaluation of $\hat{{\rm{\Psi }}}$ at $\nu =1$ gives the Blaschke immersion $\psi (\hat{x},\hat{y})$ in (4.6). For this purpose, we want to describe $\hat{{\rm{\Psi }}}(x,t,\hat{\nu })$ in terms of the theta function θ by the following conditions : Denote by ${\hat{P}}_{0}$ and ${\hat{P}}_{\infty }$ the points on $\hat{{ \mathcal C }}$ which corresponds to the values $\hat{\nu }=0$ and $\hat{\nu }=\infty$, respectively, where $\hat{\nu }$ and ${\hat{\nu }}^{-1}$ are the local coordinates around ${\hat{P}}_{0}$ and ${\hat{P}}_{\infty }$ of $\hat{{ \mathcal C }}$, respectively.

• (1)  
  $\hat{{\rm{\Psi }}}$ is a meromorphic function on $\hat{{ \mathcal C }}\setminus \{{\hat{P}}_{0},{\hat{P}}_{\infty }\}$ and the divisor of the poles is nonspecial and given by ${\hat{{ \mathcal D }}}_{\infty }=\{{\hat{p}}_{1},{\hat{p}}_{2}\}$ which is independent of the parameters x and t,
• (2)  
  $\hat{{\rm{\Psi }}}$ has the following asymptotic expansions.

  $$\begin{eqnarray*}\hat{{\rm{\Psi }}}=\left\{\begin{array}{ll}\exp (x{\hat{\nu }}^{-1})\left(1+\displaystyle \sum _{j=1}^{\infty }{\hat{\xi }}_{j}{\hat{\nu }}^{j}\right) & {\rm{near}}\ \ {\hat{P}}_{0},\\ \exp (-t\hat{\nu })\left(1+\displaystyle \sum _{j=1}^{\infty }{\hat{\eta }}_{j}{\hat{\nu }}^{-j}\right) & {\rm{near}}\ \ {\hat{P}}_{\infty }.\end{array}\right.\end{eqnarray*}$$

Consider the Abelian differential ${\omega }_{\hat{{\rm{\Psi }}}}$ defined by ${\omega }_{\hat{{\rm{\Psi }}}}=d\mathrm{log}\hat{{\rm{\Psi }}}$. Then, the principal parts of ${\omega }_{\hat{{\rm{\Psi }}}}$ around $\hat{\nu }=0$ and $\hat{\nu }=\infty$ are given by $-x{\hat{\nu }}^{-2}d\hat{\nu }$ and $t{\hat{\nu }}^{2}d{\hat{\nu }}^{-1}$, respectively. Since the residue of ${\omega }_{\hat{{\rm{\Psi }}}}$ is zero, it follows from the condition (1) that there are two points of zeros, which are denoted by ${\hat{q}}_{1}(x,t)$ and ${\hat{q}}_{2}(x,t)$. Therefore, ${\omega }_{\hat{{\rm{\Psi }}}}$ may be described as follows.

$$\begin{eqnarray}&&{\omega }_{\hat{{\rm{\Psi }}}}=x{\hat{{\rm{\Omega }}}}_{\infty }+t{\hat{{\rm{\Omega }}}}_{0}+\displaystyle \sum _{j=1}^{2}\omega ({\hat{q}}_{j},{\hat{p}}_{j})+\displaystyle \sum _{j=1}^{2}{m}_{j}{w}_{j},\end{eqnarray} \tag{ 4.22 }$$

where $\omega ({\hat{q}}_{j},{\hat{p}}_{j})$ is the normalized Abelian differential of 3rd kind with the principal part ${(\hat{\nu }-{\hat{q}}_{j})}^{-1}d\hat{\nu }$ and $-{(\hat{\nu }-{\hat{p}}_{j})}^{-1}d\hat{\nu }$, and ${\hat{{\rm{\Omega }}}}_{\infty }$ and ${\hat{{\rm{\Omega }}}}_{0}$ are the normalized Abelian differentials of second kind of the forms ${\hat{{\rm{\Omega }}}}_{\infty }=(-{\hat{\nu }}^{-2}+O(1))d\hat{\nu }\quad ({\rm{near}}\quad \ \hat{\nu }=0)$ and ${\hat{{\rm{\Omega }}}}_{0}=({\hat{\nu }}^{2}+O(1))d{\hat{\nu }}^{-1}\quad ({\rm{near}}\quad \ \hat{\nu }=\infty )$, respectively. We may define ${\hat{{\rm{\Omega }}}}_{\infty }$ and ${\hat{{\rm{\Omega }}}}_{0}$ as follows.

$$\begin{eqnarray}&&{\hat{{\rm{\Omega }}}}_{\infty }=-{\varphi }^{* }{{\rm{\Omega }}}_{\infty }^{0}+\displaystyle \frac{1}{\ 2\ }d\mu ,\ \ {\hat{{\rm{\Omega }}}}_{0}={\varphi }^{* }{{\rm{\Omega }}}_{\infty }^{0}+\displaystyle \frac{1}{\ 2\ }d\mu .\end{eqnarray} \tag{ 4.23 }$$

If we set $\mu ={\hat{\nu }}^{-1}$, we then see that

$$\begin{eqnarray*}&&{\hat{{\rm{\Omega }}}}_{\infty }=(-{\hat{\nu }}^{-2}+O(1))d\hat{\nu },\quad {\hat{{\rm{\Omega }}}}_{0}=(C+O({\hat{\nu }}^{2}))d\hat{\nu },\end{eqnarray*}$$

near $\hat{\nu }=0$, where C is as that in (4.18). It then follows from (4.11) that ${\sigma }^{* }{\hat{{\rm{\Omega }}}}_{\infty }=-{\hat{{\rm{\Omega }}}}_{\infty },{\sigma }^{* }{\hat{{\rm{\Omega }}}}_{0}=-{\hat{{\rm{\Omega }}}}_{0}$ and $\overline{{\rho }^{* }{\hat{{\rm{\Omega }}}}_{\infty }}={\hat{{\rm{\Omega }}}}_{0}$. Now, since ${\int }_{{\hat{a}}_{j}}{\omega }_{\hat{{\rm{\Psi }}}}=2\pi \sqrt{-1}{m}_{j}$, the integration of (4.22) over the cycle ${\hat{a}}_{j}$ yields that ${m}_{j}\in {\bf{Z}}$. Set ${\bf{U}}=({U}_{1},{U}_{2})=\left({\int }_{{\hat{b}}_{1}}{\hat{{\rm{\Omega }}}}_{\infty },{\int }_{{\hat{b}}_{2}}{\hat{{\rm{\Omega }}}}_{\infty }\right)$ and ${\bf{V}}=({V}_{1},{V}_{2})=\left({\int }_{{\hat{b}}_{1}}{\hat{{\rm{\Omega }}}}_{0},{\int }_{{\hat{b}}_{2}}{\hat{{\rm{\Omega }}}}_{0}\right)$. Then, the various properties of ${\hat{{\rm{\Omega }}}}_{\infty },{\hat{{\rm{\Omega }}}}_{0}$ under the involutions σ, ρ and ${\int }_{{\hat{b}}_{j}}{\hat{{\rm{\Omega }}}}_{\infty }=-{\int }_{\varphi ({\hat{b}}_{j})}{{\rm{\Omega }}}_{\infty }^{0}=-{\int }_{{b}_{1}^{0}}{{\rm{\Omega }}}_{\infty }^{0}=-{{\bf{U}}}^{0}$ imply that

$$\begin{eqnarray}&&{U}_{1}={U}_{2}=-{{\bf{U}}}^{0},\quad {V}_{1}={V}_{2}=\overline{-{{\bf{U}}}^{0}}={{\bf{U}}}^{0}.\end{eqnarray} \tag{ 4.24 }$$

Remark 4.6. Since ${w}_{1}+{w}_{2}=\displaystyle \frac{\pi \sqrt{-1}}{{\omega }^{0}}(1+O({\hat{\nu }}^{2}))d\hat{\nu }$, we have

$$\begin{eqnarray*}&&{\left.\displaystyle \frac{d}{d\hat{\nu }}\right|}_{\hat{\nu }=0}{ \mathcal B }(\hat{P})=\displaystyle \frac{\pi \sqrt{-1}}{{\omega }^{0}}=2{{\bf{U}}}^{0}=-2{U}_{j}\,\quad (j=1,2)\end{eqnarray*}$$

which is nothing but the consequence of the reciprocity law for differentials of 1st and 2nd kinds.

The integration of (4.22) over the cycle ${\hat{b}}_{j}$ and the reciprocity law gives

$$\begin{eqnarray}&&\displaystyle \sum _{j=1}^{2}{\int }_{{\hat{p}}_{j}}^{{\hat{q}}_{j}}({w}_{1}+{w}_{2})\equiv 2(x-t){{\bf{U}}}^{0}\quad \mathrm{mod}\ {\rm{\Gamma }}.\end{eqnarray} \tag{ 4.25 }$$

Define $F(z)$ by

$$\begin{eqnarray*}&&F(z)=-\displaystyle \frac{1}{\sqrt{2}\sqrt{-1}}{\zeta }_{w}(z)+\displaystyle \frac{z}{\sqrt{2}\sqrt{-1}}\displaystyle \frac{{\zeta }_{w}({\omega }^{{\prime} })}{{\omega }^{{\prime} }}.\end{eqnarray*}$$

Let $\{{z}_{1},{z}_{2},{z}_{3}\}$ be the set of points where ${{\mathfrak{P}}}^{{\prime} }(z)=0$. We then assume that ${z}_{1}={\omega }^{{\prime} },{z}_{2}=\omega ,{z}_{3}=\omega +{\omega }^{{\prime} }$. Note that ${\mathfrak{P}}({z}_{1})={\eta }_{1}$, ${\mathfrak{P}}({z}_{2})={\eta }_{2}$, ${\mathfrak{P}}({z}_{3})={\eta }_{3}$. We denote by ${\hat{P}}_{1}$, ${\hat{P}}_{2}$, ${\hat{P}}_{3}$ the points on $\hat{{ \mathcal C }}$ corresponding to ${\eta }_{1}$, ${\eta }_{2}$, ${\eta }_{3}$,respectively. Precisely, ${\hat{P}}_{1}=({\mu }_{1},1)$, ${\hat{P}}_{2}=({\mu }_{2},1)$, ${\hat{P}}_{3}=({\mu }_{3},1)$ by the affine coordinate $\hat{P}=(\mu ,\nu )$. We choose a path γ from ${\hat{P}}_{0}$ to ${\hat{P}}_{\infty }$ as in Figure 3 and we fix a path from ${\hat{P}}_{0}$ to ${\hat{P}}_{1}$ in the following. Since we have ${\int }_{\gamma }({w}_{1}+{w}_{2})\,\equiv 0\ (\mathrm{mod}\quad {\rm{\Gamma }})$ and we also have ${\int }_{\gamma }({w}_{1}+{w}_{2})$ is purely imaginary by $\rho (\gamma )=-\gamma$ and $\overline{{\rho }^{* }{w}_{j}}={w}_{j}$, we may fix a path from ${\hat{P}}_{1}$ to ${\hat{P}}_{-1}=\sigma ({\hat{P}}_{1})$ which passes through only $\hat{a}$-cycles. It then follows from

$$\begin{eqnarray*}\begin{array}{rcl}{\displaystyle \int }_{{\hat{P}}_{0}}^{{\hat{P}}_{1}}({w}_{1}+{w}_{2}) & = & -{\displaystyle \int }_{{\hat{P}}_{0}}^{{\hat{P}}_{-1}}({w}_{1}+{w}_{2}),\\ {\displaystyle \int }_{{\hat{P}}_{0}}^{{\hat{P}}_{-1}}({w}_{1}+{w}_{2}) & = & {\displaystyle \int }_{{a}_{1}^{0}}{w}_{1}^{0}+{\displaystyle \int }_{{P}_{0}}^{{P}_{1}}{w}_{1}^{0}=2\pi \sqrt{-1}+{\displaystyle \int }_{{\hat{P}}_{0}}^{{\hat{P}}_{1}}({w}_{1}+{w}_{2}),\end{array}\end{eqnarray*}$$

that

$$\begin{eqnarray*}&&{\int }_{{\hat{P}}_{0}}^{{\hat{P}}_{1}}({w}_{1}+{w}_{2})\equiv \pi \sqrt{-1}\quad \mathrm{mod}\ 2\pi \sqrt{-1}{\bf{Z}}.\end{eqnarray*}$$

In fact, this is verified directly in (4.32) below. We have

$$\begin{eqnarray*}&&{\int }_{{\hat{P}}_{1}}^{\hat{P}}{\hat{{\rm{\Omega }}}}_{\infty }+\displaystyle \frac{1}{\ 2\ }{\mu }_{1}\equiv F(z)+\displaystyle \frac{1}{\ 2\ }\mu ,\quad {\int }_{{\hat{P}}_{1}}^{\hat{P}}{\hat{{\rm{\Omega }}}}_{0}+\displaystyle \frac{1}{\ 2\ }{\mu }_{1}\equiv -F(z)+\displaystyle \frac{1}{\ 2\ }\mu \quad (\mathrm{mod}\ {\rm{\Gamma }}).\end{eqnarray*}$$

We have ${\int }_{{\hat{P}}_{1}}^{\hat{P}}{\hat{{\rm{\Omega }}}}_{\infty }+\tfrac{1}{\ 2\ }{\mu }_{1}={\hat{\nu }}^{-1}+O(\hat{\nu })$ and ${\int }_{{\hat{P}}_{1}}^{\hat{P}}{\hat{{\rm{\Omega }}}}_{0}+\displaystyle \frac{1}{\ 2\ }{\mu }_{1}=C\hat{\nu }+O({\hat{\nu }}^{3})\ ({\rm{near}}\quad \hat{\nu }=0)$. We put

$$\begin{eqnarray}&&\left\{\begin{array}{l}{{\rm{\Phi }}}_{\theta }(x,t,\hat{P})=\displaystyle \frac{\theta ({ \mathcal B }(\hat{P})-(x-t){{\bf{U}}}^{0}-{\bf{e}})}{\theta ({ \mathcal B }(\hat{P})-{\bf{e}})},\\ {{\rm{\Phi }}}_{e}(x,t,\hat{P})=\exp \left(x\left({\displaystyle \int }_{{\hat{P}}_{1}}^{\hat{P}}{\hat{{\rm{\Omega }}}}_{\infty }+\displaystyle \frac{1}{\ 2\ }{\mu }_{1}\right)+t\left({\displaystyle \int }_{{\hat{P}}_{1}}^{\hat{P}}{\hat{{\rm{\Omega }}}}_{0}+\displaystyle \frac{1}{\ 2\ }{\mu }_{1}\right)\right),\end{array}\right.\end{eqnarray} \tag{ 4.26 }$$

where ${\bf{e}}\in {\bf{C}}$ is chosen so that $f(\hat{P})\not\equiv 0$ and the divisor of the poles of ${{\rm{\Phi }}}_{\theta }$ is $\hat{{ \mathcal D }}=\{{\hat{p}}_{1},{\hat{p}}_{2}\}$. We observe that ${{\rm{\Phi }}}_{\theta }{{\rm{\Phi }}}_{e}$ is invariant under the translation $\hat{P}\longrightarrow \hat{P}+{m}_{j}{\hat{a}}_{j}+{n}_{j}{\hat{b}}_{j}$ by the property of θ and (4.24). Therefore, it is a meromorphic function on $\hat{{ \mathcal C }}$. It then follows from lemma 4.5 and (4.25) that $\hat{{\rm{\Psi }}}{{\rm{\Phi }}}_{\theta }^{-1}{{\rm{\Phi }}}_{e}^{-1}$ is a holomorphic function on $\hat{{ \mathcal C }}$,hence it is a constant. Evaluating it at $\hat{\nu }=0$ we see that the constant is equal to $\displaystyle \frac{\theta ({\bf{e}})}{\theta ((x-t){{\bf{U}}}^{0}+{\bf{e}})}$. We thus obtain the following.

Lemma 4.7.  $\hat{{\rm{\Psi }}}(x,t,\hat{P},{\bf{e}})$ is given by

$$\begin{eqnarray*}&&\hat{{\rm{\Psi }}}(x,t,\hat{P},{\bf{e}})=\displaystyle \frac{\theta ({ \mathcal B }(\hat{P})-(x-t){{\bf{U}}}^{0}-{\bf{e}})\ \theta ({\bf{e}})}{\theta ({ \mathcal B }(\hat{P})-{\bf{e}})\ \theta ((x-t){{\bf{U}}}^{0}+{\bf{e}})}\ {{\rm{\Phi }}}_{e}(x,t,\hat{P}),\end{eqnarray*}$$

where ${{\rm{\Phi }}}_{e}(x,t,\hat{P})$ is as that in (4.26), and we fix the path from ${\hat{P}}_{0}$ to ${\hat{P}}_{1}$.

Moreover, using the expression of $\hat{{\rm{\Psi }}}$ we see that the following reality condition holds:

$$\begin{eqnarray}&&({\hat{\sigma }}^{* }\hat{{\rm{\Psi }}})(x,t,\hat{P},{\bf{e}}):= \hat{{\rm{\Psi }}}(t,x,\sigma (\hat{P}))=\overline{\hat{{\rm{\Psi }}}(x,t,\sigma \rho (\hat{P}),{\bf{e}})}\end{eqnarray} \tag{ 4.27 }$$

As we see later, we may prove that $\hat{{\rm{\Psi }}}$ satisfies the Schrödinger equation ${\partial }_{t}{\partial }_{x}\hat{{\rm{\Psi }}}={e}^{u}\hat{{\rm{\Psi }}}$. For $U(\lambda )$, $V(\lambda )$ given in (2.5), let ${F}_{\lambda }:{{\bf{R}}}^{2}\longrightarrow {{\bf{C}}}^{3}$ be a solution of the differential equation

$$\begin{eqnarray*}&&{{dF}}_{\lambda }={F}_{\lambda }\ (U(\lambda ){dt}+V(\lambda ){dx}).\end{eqnarray*}$$

To simplify this equation, define ${\hat{F}}_{\nu }$ by

$$\begin{eqnarray*}{\hat{F}}_{\nu }={F}_{\lambda }\left(\begin{array}{ccc}{\lambda }^{-2}{e}^{\tfrac{u}{2}} & 0 & 0\\ 0 & {\lambda }^{-1}{e}^{-\tfrac{u}{2}} & 0\\ 0 & 0 & 1\end{array}\right).\end{eqnarray*}$$

We then obtain

$$\begin{eqnarray*}&&d{\hat{F}}_{\nu }={\hat{F}}_{\nu }(\hat{U}(\nu ){dt}+\hat{V}(\nu ){dx}),\end{eqnarray*}$$

where

$$\begin{eqnarray}\hat{U}(\nu )=\left(\begin{array}{ccc}{u}_{t} & 0 & \nu \\ 1 & -{u}_{t} & 0\\ 0 & 1 & 0\end{array}\right),\quad \hat{V}(\nu )=\left(\begin{array}{ccc}0 & {e}^{-2u} & 0\\ 0 & 0 & {e}^{u}\\ {\nu }^{-1}{e}^{u} & 0 & 0\end{array}\right).\end{eqnarray} \tag{ 4.28 }$$

If we express ${\hat{F}}_{\nu }$ as ${\hat{F}}_{\nu }=({\psi }_{0}\ {\psi }_{1}\ {\psi }_{2})$ then we write down the system of differential equations which we must solve as follows.

$$\begin{eqnarray}&&\left\{\begin{array}{l}{\partial }_{t}{\psi }_{0}={u}_{t}{\psi }_{0}+{\psi }_{1}\\ {\partial }_{t}{\psi }_{1}=-{u}_{t}{\psi }_{1}+{\psi }_{2}\\ {\partial }_{t}{\psi }_{2}=\nu {\psi }_{0}\end{array}\right.,\quad \left\{\begin{array}{l}{\partial }_{x}{\psi }_{0}={\nu }^{-1}{e}^{u}{\psi }_{2}\\ {\partial }_{x}{\psi }_{1}={e}^{-2u}{\psi }_{0}\\ {\partial }_{x}{\psi }_{2}={e}^{u}{\psi }_{1}\end{array}\right.\end{eqnarray} \tag{ 4.29 }$$

Lemma 4.8.  $\{{\psi }_{0}={\nu }^{-1}{\partial }_{t}\hat{{\rm{\Psi }}},{\psi }_{1}={e}^{-u}{\partial }_{x}\hat{{\rm{\Psi }}},{\psi }_{2}=\hat{{\rm{\Psi }}}\}$ is a solution of the system of differential equations (4.29) above.

Set ${\hat{\psi }}_{0}={\hat{\nu }}^{6}{e}^{-\tfrac{u}{2}}{\psi }_{0},{\hat{\psi }}_{1}={\hat{\nu }}^{3}{e}^{\tfrac{u}{2}}{\psi }_{1},{\hat{\psi }}_{2}=\psi$ with $\nu ={\hat{\nu }}^{3}$. We define $\hat{W}(x,t,\hat{P})$ by

$$\begin{eqnarray}&&\hat{W}(x,t,\hat{P})=\displaystyle \sum _{j=0}^{2}{\hat{\psi }}_{j}(\hat{P})\cdot \overline{({\hat{\sigma }}^{* }{\hat{\psi }}_{j})(\rho (\hat{P}))}.\end{eqnarray} \tag{ 4.30 }$$

It follows from (4.29) that $\hat{W}$ is independent of the parameters x and t. Thus, we may write $\hat{W}(x,t,\hat{P})=\hat{W}(\hat{P})$.

We will arrive at the following conclusion.

Theorem 4.9. For $\hat{{\rm{\Psi }}}$ in lemma 4.7, the Blaschke immersion $\psi (\hat{x},\hat{y})$ given in (4.6) can be described in terms of Riemann theta functions as

$$\begin{eqnarray*}&&\psi (\hat{x},\hat{y})={}^{t}\left({\hat{c}}_{1}\displaystyle \frac{\hat{{\rm{\Psi }}}(x,t,{\hat{P}}_{1},\pi \sqrt{-1})}{\sqrt{\hat{W}({\hat{P}}_{1})}},\ {\hat{c}}_{2}\displaystyle \frac{\hat{{\rm{\Psi }}}(x,t,{\hat{P}}_{2},\pi \sqrt{-1})}{\sqrt{\hat{W}({\hat{P}}_{2})}},\ {\hat{c}}_{3}\displaystyle \frac{\hat{{\rm{\Psi }}}(x,t,{\hat{P}}_{3},\pi \sqrt{-1})}{\sqrt{\hat{W}({\hat{P}}_{3})}}\right),\end{eqnarray*}$$

which is a solution of (2.1), where ${\hat{c}}_{1}{\hat{c}}_{2}{\hat{c}}_{3}=1$.

Moreover, the equation ${\partial }_{x}\ {\partial }_{t}\ \hat{{\rm{\Psi }}}={e}^{u}\ \hat{{\rm{\Psi }}}$ holds, where ${e}^{u}$ coincides with one given by (4.17), that is, a solution of the Tzitzèica equation (1.1).

Proof. Proof. Since $\hat{{\rm{\Psi }}}$ is single-valued, we carry out the calculations of integrations using the paths ${\gamma }_{+},{\hat{b}}_{1+},{\hat{a}}_{1+}$ by the parts of the disjoint unions $\gamma ={\gamma }_{+}\cup {\gamma }_{-},{\hat{b}}_{1}={\hat{b}}_{1+}\cup {\hat{b}}_{1-},{\hat{a}}_{1}={\hat{a}}_{1+}\cup {\hat{a}}_{1-}$

$$\begin{eqnarray*}&&{\hat{P}}_{0}\mathop{\longrightarrow }\limits^{{\gamma }_{+}}{\hat{P}}_{1}\mathop{\longrightarrow }\limits^{-{\hat{b}}_{1+}}{\hat{P}}_{3}\mathop{\longrightarrow }\limits^{{\hat{a}}_{1+}}{\hat{P}}_{2}\end{eqnarray*}$$

Let $\{{\hat{P}}_{1},{\hat{P}}_{2},{\hat{P}}_{3}\}$ be the points of $\hat{{ \mathcal C }}$ as above, which are also points of $\nu =1$. It follows from (4.10) and (4.11) that

$$\begin{eqnarray*}&&\left\{\begin{array}{l}{\displaystyle \int }_{{\hat{P}}_{1}}^{{\hat{P}}_{j}}{\hat{{\rm{\Omega }}}}_{\infty }+\displaystyle \frac{1}{\ 2\ }{\mu }_{1}=F({z}_{j})+\displaystyle \frac{1}{\ 2\ }{\mu }_{j},\ \ {\displaystyle \int }_{{\hat{P}}_{1}}^{{\hat{P}}_{j}}{\hat{{\rm{\Omega }}}}_{0}+\displaystyle \frac{1}{\ 2\ }{\mu }_{1}=-F({z}_{j})+\displaystyle \frac{1}{\ 2\ }{\mu }_{j},\\ F({z}_{1})=0,\ F({z}_{2})=\displaystyle \frac{1}{2}{{\bf{U}}}^{0},\ F({z}_{3})=\displaystyle \frac{1}{2}{{\bf{U}}}^{0},\end{array}\right.\end{eqnarray*}$$

where $j=1,2,3$. Therefore, we obtain

$$\begin{eqnarray}{{\rm{\Phi }}}_{e}(x,t,{\hat{P}}_{j})=\left\{\begin{array}{ll}\exp \left(\displaystyle \frac{x+t}{2}{\mu }_{1}\right) & {\rm{for}}\ j=1,\\ \exp \left(\displaystyle \frac{1}{2}{{\bf{U}}}^{0}(x-t)+\displaystyle \frac{x+t}{2}{\mu }_{2}\right) & {\rm{for}}\ j=2,\\ \exp \left(\displaystyle \frac{1}{2}{{\bf{U}}}^{0}(x-t)+\displaystyle \frac{x+t}{2}{\mu }_{3}\right) & {\rm{for}}\ j=3.\end{array}\right.\end{eqnarray} \tag{ 4.31 }$$

On the other hand, since ${\varphi }^{* }\displaystyle \frac{d\zeta }{\widetilde{B}}={\varphi }^{* }\displaystyle \frac{{dz}}{\sqrt{2}\sqrt{-1}}$ we obtain

$$\begin{eqnarray}\begin{array}{rcl}{ \mathcal B }({\hat{P}}_{j}) & = & \displaystyle \frac{\pi \sqrt{-1}}{{\omega }^{0}}{\displaystyle \int }_{{\hat{P}}_{0}}^{{\hat{P}}_{j}}{\varphi }^{* }\displaystyle \frac{{dz}}{\sqrt{2}\sqrt{-1}}=\displaystyle \frac{\pi \sqrt{-1}}{\sqrt{2}\sqrt{-1}{\omega }^{0}}\ {z}_{j}\\ & = & \left\{\begin{array}{cl}\pi \sqrt{-1} & {\rm{for}}\ j=1,\\ -\displaystyle \frac{1}{2}{\rm{\Pi }} & {\rm{for}}\ j=2,\\ -\displaystyle \frac{1}{2}{\rm{\Pi }}+\pi \sqrt{-1} & {\rm{for}}\ j=3.\end{array}\right.\end{array}\end{eqnarray} \tag{ 4.32 }$$

We here remark that if $u=\sqrt{2}\sqrt{{\zeta }_{2}-{\zeta }_{1}}\ \hat{y}$ then $v=\displaystyle \frac{\ u\ }{2K(p)}=\displaystyle \frac{\ {{\bf{U}}}^{0}(x-t)\ }{2\pi \sqrt{-1}}$ by (4.2) and (4.12). We now set $u=\sqrt{2}\sqrt{{\zeta }_{2}-{\zeta }_{1}}\ \hat{y}$. It then follows from (4.13), (4.14), (4.31) and (4.32) that

$$\begin{eqnarray*}\begin{array}{rcl}\mathrm{dn}\ (u,p) & = & \displaystyle \frac{\theta (\pi \sqrt{-1};{\rm{\Pi }})\ \theta ({{\bf{U}}}^{0}(x-t);{\rm{\Pi }})}{\theta (0;{\rm{\Pi }})\ \theta ({{\bf{U}}}^{0}(x-t)+\pi \sqrt{-1};{\rm{\Pi }})}\\ & = & \hat{{\rm{\Psi }}}(x,t,{\hat{P}}_{1},\pi \sqrt{-1})\exp \left(-\displaystyle \frac{x+t}{2}{\mu }_{1}\right),\\ \mathrm{sn}\ (u,p) & = & -\sqrt{-1}\ \displaystyle \frac{\theta (0;{\rm{\Pi }})\ \theta ({{\bf{U}}}^{0}(x-t)+\pi \sqrt{-1}+\tfrac{1}{2}{\rm{\Pi }};{\rm{\Pi }})}{\theta \left(\tfrac{1}{2}{\rm{\Pi }};{\rm{\Pi }}\right)\ \theta ({{\bf{U}}}^{0}(x-t)+\pi \sqrt{-1};{\rm{\Pi }})}\exp \left(\displaystyle \frac{1}{2}{{\bf{U}}}^{0}(x-t)\right)\\ & = & \displaystyle \frac{1}{\hat{{\rm{\Psi }}}({\omega }^{0},-{\omega }^{0},{\hat{P}}_{2},\pi \sqrt{-1})}\hat{{\rm{\Psi }}}(x,t,{\hat{P}}_{2},\pi \sqrt{-1})\exp \left(-\displaystyle \frac{x+t}{2}{\mu }_{2}\right),\\ \mathrm{cn}\ (u,p) & = & \displaystyle \frac{\theta (\pi \sqrt{-1};{\rm{\Pi }})\ \theta ({{\bf{U}}}^{0}(x-t)+\tfrac{1}{2}{\rm{\Pi }};{\rm{\Pi }})}{\theta \left(\tfrac{1}{2}{\rm{\Pi }};{\rm{\Pi }}\right)\ \theta ({{\bf{U}}}^{0}(x-t)+\pi \sqrt{-1};{\rm{\Pi }})}\exp \left(\displaystyle \frac{1}{2}{{\bf{U}}}^{0}(x-t)\right)\\ & = & \hat{{\rm{\Psi }}}(x,t,{\hat{P}}_{3},\pi \sqrt{-1})\exp \left(-\displaystyle \frac{x+t}{2}{\mu }_{3}\right).\end{array}\end{eqnarray*}$$

Take a function $C(\hat{P})$ on $\hat{{ \mathcal C }}$ and define $\widetilde{{\rm{\Psi }}}$ by

$$\begin{eqnarray*}\left\{\begin{array}{ll}\widetilde{{\rm{\Psi }}}(x,t,\hat{P},\pi \sqrt{-1}) & =\,C(\hat{P})\hat{{\rm{\Psi }}}(x,t,\hat{P},\pi \sqrt{-1})\\ \mathrm{with}\quad & \overline{C(\sigma \rho (\hat{P}))}=C(\hat{P}),\end{array}\right.\end{eqnarray*}$$

where $C({\hat{P}}_{1})=C({\hat{P}}_{3})=1,C({\hat{P}}_{2})={(\hat{{\rm{\Psi }}}({\omega }^{0},-{\omega }^{0},{\hat{P}}_{2},\pi \sqrt{-1}))}^{-1}=0$. Let $\widetilde{P}$ be the point near ${\hat{P}}_{2}$ along ${\hat{a}}_{1+}$. We may write ${ \mathcal B }(\widetilde{P})=-\displaystyle \frac{1}{2}{\rm{\Pi }}-(a-\pi )\sqrt{-1}$, where $a\in {\bf{R}}$. We then have ${ \mathcal B }({\hat{P}}_{2})={\mathrm{lim}}_{a\to \pi }{ \mathcal B }(\widetilde{P})$. We define $\widetilde{W}$ using $\widetilde{{\rm{\Psi }}}$ as well as the way we defined $\hat{W}$ using $\hat{{\rm{\Psi }}}$. We then see that

$$\begin{eqnarray*}&&\sqrt{\widetilde{W}(\widetilde{P})}=\sqrt{C(\widetilde{P})\overline{C(\rho (\widetilde{P}))}}\sqrt{\hat{W}(\widetilde{P})}=\left|\displaystyle \frac{\theta (\tfrac{1}{2}{\rm{\Pi }}+a\sqrt{-1})\theta (0)}{\theta (\tfrac{1}{2}{\rm{\Pi }}+(a-\pi )\sqrt{-1})\theta (\pi \sqrt{-1})}\right|\sqrt{\hat{W}(\widetilde{P})}.\end{eqnarray*}$$

Here, we note that

$$\begin{eqnarray*}\begin{array}{rcl}\overline{\left(\displaystyle \frac{\theta \left(\tfrac{1}{2}{\rm{\Pi }}+(a-\pi )\sqrt{-1}\right)}{\theta \left(\tfrac{1}{2}{\rm{\Pi }}+a\sqrt{-1}\right)}\right)} & = & \displaystyle \frac{\theta \left(-\tfrac{1}{2}{\rm{\Pi }}-(a-\pi )\sqrt{-1}+{\rm{\Pi }}\right)}{\theta \left(-\tfrac{1}{2}{\rm{\Pi }}-a\sqrt{-1}+{\rm{\Pi }}\right)}\\ & = & -\displaystyle \frac{\theta \left(\tfrac{1}{2}{\rm{\Pi }}+(a-\pi )\sqrt{-1}\right)}{\theta \left(\tfrac{1}{2}{\rm{\Pi }}+a\sqrt{-1}\right)}.\end{array}\end{eqnarray*}$$

Therefore, we see that

$$\begin{eqnarray*}&&\begin{array}{l}\displaystyle \frac{\widetilde{{\rm{\Psi }}}(x,t,{\hat{P}}_{2})}{\sqrt{\widetilde{W}({\hat{P}}_{2})}}=\mathop{\mathrm{lim}}\limits_{a\to \pi }\displaystyle \frac{C(\widetilde{P})\hat{{\rm{\Psi }}}(x,t,\widetilde{P})}{\sqrt{\hat{W}(\widetilde{P})}}\left|\displaystyle \frac{\theta \left(\tfrac{1}{2}{\rm{\Pi }}+(a-\pi )\sqrt{-1}\right)\theta (\pi \sqrt{-1})}{\theta \left(\tfrac{1}{2}{\rm{\Pi }}+a\sqrt{-1}\right)\theta (0)}\right|\\ =\,\mathop{\mathrm{lim}}\limits_{a\to \pi }\displaystyle \frac{\hat{{\rm{\Psi }}}(x,t,\widetilde{P})}{\sqrt{\hat{W}(\widetilde{P})}}\displaystyle \frac{\exp (-\pi \sqrt{-1}/2)\theta \left(\tfrac{1}{2}{\rm{\Pi }}+a\sqrt{-1}\right)\theta (0)}{\theta \left(\tfrac{1}{2}{\rm{\Pi }}+(a-\pi )\sqrt{-1}\right)\theta (\pi \sqrt{-1})}\left|\displaystyle \frac{\theta \left(\tfrac{1}{2}{\rm{\Pi }}+(a-\pi )\sqrt{-1}\right)\theta (\pi \sqrt{-1})}{\theta \left(\tfrac{1}{2}{\rm{\Pi }}+a\sqrt{-1}\right)\theta (0)}\right|\\ =\,\pm \mathop{\mathrm{lim}}\limits_{\widetilde{P}\to {\hat{P}}_{2}}\displaystyle \frac{\hat{{\rm{\Psi }}}(x,t,\widetilde{P})}{\sqrt{\hat{W}(\widetilde{P})}}.\end{array}\end{eqnarray*}$$

Since we may obtain the explicit expression of $\widetilde{{\rm{\Psi }}}({\hat{P}}_{j}),(j=1,2,3)$ using the Jacobi elliptic functions, we obtain the following.

$$\begin{eqnarray*}&&\widetilde{W}({\hat{P}}_{1})=2+\displaystyle \frac{{\zeta }_{3}}{{\zeta }_{2}},\quad \widetilde{W}({\hat{P}}_{2})=1-\displaystyle \frac{{\zeta }_{1}}{{\zeta }_{2}},\quad \widetilde{W}({\hat{P}}_{3})=2+\displaystyle \frac{{\zeta }_{1}}{{\zeta }_{2}},\end{eqnarray*}$$

which implies that $\sqrt{\widetilde{W}({\hat{P}}_{1})\widetilde{W}({\hat{P}}_{2})\widetilde{W}({\hat{P}}_{3})}=\sqrt{2}\sqrt{{\zeta }_{2}-{\zeta }_{1}}({\zeta }_{3}-{\zeta }_{1})$.

Therefore, $\psi (\hat{x},\hat{y})$ in (4.6) can be described as those forms stated in the theorem. Next, we show that the $\hat{{\rm{\Psi }}}$ satisfies the Schrödinger equation ${\partial }_{x}{\partial }_{t}\hat{{\rm{\Psi }}}={e}^{u}\hat{{\rm{\Psi }}}$. Near ${\hat{P}}_{0}$, $\hat{{\rm{\Psi }}}(x,t,\hat{\nu })=\exp (x{\hat{\nu }}^{-1})(1+{\sum }_{j=1}^{\infty }{\hat{\xi }}_{j}{\hat{\nu }}^{j})$. We set ${e}^{u}={\partial }_{t}{\hat{\xi }}_{1}$. We then see that

$$\begin{eqnarray*}&&\hat{{\rm{\Phi }}}:= ({\partial }_{x}{\partial }_{t}\hat{{\rm{\Psi }}}-{e}^{u}\hat{{\rm{\Psi }}}){\hat{{\rm{\Psi }}}}^{-1}\ \longrightarrow 0\quad {\rm{as}}\quad \hat{\nu }\to 0.\end{eqnarray*}$$

Since the poles of $\hat{{\rm{\Psi }}}$ are independent of the parameters $x,t$, we see that the poles of $({\partial }_{x}{\partial }_{t}\hat{{\rm{\Psi }}}-{e}^{u}\hat{{\rm{\Psi }}})$ coincides with the zeros of ${\hat{{\rm{\Psi }}}}^{-1}$. But, since the zeros of $({\partial }_{x}{\partial }_{t}\hat{{\rm{\Psi }}}-{e}^{u}\hat{{\rm{\Psi }}})$ changes with $x,t$, if $\hat{{\rm{\Psi }}}\not\equiv 0$ then it follows from lemma 4.5 that the zeros of $\hat{{\rm{\Psi }}}$ are the degree 2 divisor $\{{\hat{q}}_{1}(x,t),{\hat{q}}_{2}(x,t)\}$. We then have $\hat{{\rm{\Phi }}}\in {{\rm{H}}}^{0}(\hat{{ \mathcal C }},{{ \mathcal O }}_{\hat{{ \mathcal C }}}({\hat{q}}_{1}(x,t)+{\hat{q}}_{2}(x,t)))$. Since $\{{\hat{q}}_{1}(x,t),{\hat{q}}_{2}(x,t)\}$ is a non-special divisor at the origin $(x,t)=(0,0)$, it remains non-special near the origin. Thus, $\hat{{\rm{\Phi }}}$ must be a constant on $\hat{{ \mathcal C }}$. Evaluating it at $\hat{\nu }=0$ we obtain $\hat{{\rm{\Phi }}}\equiv 0$, hence we have ${\partial }_{x}\ {\partial }_{t}\ \hat{{\rm{\Psi }}}-{e}^{u}\hat{{\rm{\Psi }}}\equiv 0$. Finally, we show that the ${e}^{u}$ above coincides with one in (4.17). Near ${\hat{P}}_{0}$, we may write ${\int }_{{\hat{P}}_{1}}^{\hat{P}}{\hat{{\rm{\Omega }}}}_{0}+\displaystyle \frac{1}{\ 2\ }{\mu }_{1}\,=C\hat{\nu }+O({\hat{\nu }}^{3})$, where $C=\tfrac{a}{3}-\tfrac{{\zeta }_{w}({\omega }^{{\prime} })}{{\omega }^{{\prime} }}$ as in (4.18). Differentiating $\mathrm{log}\hat{{\rm{\Psi }}}$ by $\hat{\nu }$ and setting $\hat{\nu }=0$, we obtain from lemma 4.7.

$$\begin{eqnarray*}&&{\hat{\xi }}_{1}={\left.\displaystyle \frac{d}{d\hat{\nu }}\right|}_{\hat{\nu }=0}(\mathrm{log}\theta ({ \mathcal B }(\hat{P})-(x-t){{\bf{U}}}^{0}-{\bf{e}})-\mathrm{log}\theta ({ \mathcal B }(\hat{P})-{\bf{e}}))+{Ct}+{C}_{0}(x),\end{eqnarray*}$$

where ${C}_{0}(x)$ is a function of the parameter x only. It then follows from ${e}^{u}={\partial }_{t}{\hat{\xi }}_{1}$ that ${e}^{u}=C\,+{\partial }_{t}{\left.\displaystyle \frac{d}{d\hat{\nu }}\right|}_{\hat{\nu }=0}\mathrm{log}\theta ({ \mathcal B }(\hat{P})-(x-t){{\bf{U}}}^{0}-{\bf{e}})$. It follows from the reciprocity law for 1st and 2nd kinds that

$$\begin{eqnarray*}&&{\left.\displaystyle \frac{d}{d\hat{\nu }}\right|}_{\hat{\nu }=0}{ \mathcal B }(\hat{P})=-\displaystyle \sum _{j=1}^{2}{\int }_{{\hat{b}}_{j}}{\hat{{\rm{\Omega }}}}_{\infty }=2{{\bf{U}}}^{0}\end{eqnarray*}$$

which, together with ${e}^{u}={\partial }_{t}\ {\hat{\xi }}_{1}$, yields

$$\begin{eqnarray*}&&{e}^{u}=C-2\ {\partial }_{x}\ {\partial }_{t}\ \mathrm{log}\theta \ (\ (x-t){{\bf{U}}}^{0}+{\bf{e}}).\end{eqnarray*}$$

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