Surfaces: from generalised Weierstrass representation to Cartan moving frame

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Claude Pierre Massé

Laboratoire de Physique Fondamentale

11 February 2017

Konopelchenko’s generalised Weierstraß representation of a surface immersed in the three dimensional Euclidean space is derived by the way of the Maurer-Cartan form, and then expressed in the framework of the moving frame. The data of surface theory are given each time. This is then applied to solving the non-linear Liouville equation by the inverse scattering transform. Connection and symmetries are briefly mentioned.

Latest version: http://phy.clmasse.com/surfaces-weierstrass-cartan-v2.html

Introduction

The Enneper-Weierstraß representation allows to represent a minimal surface immersed in $\Bbb R^3$ by a meromorphic function and a holomorphic $1$-form. Recently, Konopelchenko [1] proposed a similar representation with two functions given by a solution of a two-dimensional Dirac equation. In this article, that formalism is presented in a transparent and systematic way, so that the Cartan’s moving frame can be introduced naturally. In section §1, we first derive the generalised Weierstraß representation by the way of the Maurer-Cartan $1$-form of the Dirac spinor, and then switch to the vector representation of a spinor that leads to Konopelchenko’s formula. Up to there, everything is expressed in an arbitrary, fixed reference frame. Then in section §2, the representation is recast along the lines of the Cartan’s moving frame, so that it becomes unnecessary. The mathematical expressions of the traditional data of a surface are obtained each time. The geometrical representation of integrable non-linear differential equations follows directly, and in section §3 it is shown how this can be applied to solving the non-linear Liouville equation by the inverse scattering transform. Finally in the last section §4, other points of view are touched upon.

§1. The generalised Weierstraß representation

The Dirac equation can be written in a Euclidean two-dimensional space using the Pauli matrices: $$\DeclareMathOperator{\Tr}{Tr}\DeclareMathOperator{\d}{d\!}\DeclareMathOperator{\e}{e} \sigma_1=\pmatrix{0& 1\\1& 0},\quad\sigma_2=\pmatrix{0& -i\\i& 0},\quad\sigma_3=\pmatrix{1& 0\\0& -1}\tag{1.1}$$ as follow: $$\{i\sigma_1\partial_\tau+i\sigma_2\partial_\xi+m\}\psi=0.\tag{1.2}$$ Here the mass parameter $m$ is generalised to a real function of $\tau$ and $\xi.$ We shall work with complex numbers for compactness, and denote the complex conjugate by an overbar. We first define the Wirtinger differential operators $$\partial=\frac12(\partial_\tau-i\partial_\xi),\quad\bar\partial=\frac12(\partial_\tau+i\partial_\xi),\tag{1.3}$$ or using the complex variable: $$\zeta=\tau+i\xi,\tag{1.4}$$ $$\partial=\partial_\zeta,\quad\bar\partial=\partial_{\bar\zeta}.\tag{1.5}$$ Writing the wave function as: $$\psi=\binom\varphi\chi,\tag{1.6}$$ the Dirac equation (1.2) is now decomposed into a system of two simple and handy equations: $$\left\{\eqalign{ \bar\partial(\varphi,-\bar\chi)& =\frac i2m(\chi,\phantom-\bar\varphi)\\ \partial(\chi,\phantom-\bar\varphi)& =\frac i2m(\varphi,-\bar\chi). }\right.\tag{1.7a,b}$$ If $m$ vanishes identically, $\varphi$ is a holomorphic and $\chi$ an antiholomorphic function.

We first focus on the spinor. It is uniquely extended as follow: $$\Phi=\pmatrix{\varphi& -\bar\chi\\\chi& \phantom-\bar\varphi}.\tag{1.8}$$ It is a $SU(2)$ matrix multiplied by the factor $\sqrt\rho,$ with the definitions: $$\rho=\e^u:=\varphi\bar\varphi+\chi\bar\chi,\tag{1.9}$$ so that $\Phi\Phi^\dagger=\rho{\rm I}.$ We then consider the associated (right-invariant) Maurer-Cartan $1$-form: $$\d\Phi\Phi^{-1}=\rho^{-1}\d\Phi\Phi^\dagger.\tag{1.10}$$ Its entries can be calculated from the Dirac equation, with the result $$\rho^{-1}\d\Phi\Phi^\dagger= \frac12\pmatrix{2\partial u& iQ\\im& 0}\d\zeta+ \frac12\pmatrix{0& im\\i\bar Q& 2\bar\partial u}\d\bar\zeta,\tag{1.11}$$ where this defining relation has been used: $$\bar\chi\partial\varphi-\varphi\partial\bar\chi=:\frac i2\rho Q,\quad \frac i2\rho\bar Q:=\bar\varphi\bar\partial\chi-\chi\bar\partial\bar\varphi.\tag{1.12}$$ In addition, the following differential constraints have been found: $$\varphi\partial\bar\varphi+\bar\chi\partial\chi=0=\bar\varphi\bar\partial\varphi+\chi\bar\partial\bar\chi,\tag{1.13a}$$ $$\bar\varphi\partial\chi-\chi\partial\bar\varphi=\frac i2\rho m=\bar\chi\bar\partial\varphi-\varphi\bar\partial\bar\chi,\tag{1.13b}$$ $$\bar\varphi\partial\varphi+\chi\partial\bar\chi=\rho\partial u,\quad \rho\bar\partial u=\varphi\bar\partial\bar\varphi+\bar\chi\bar\partial\chi.\tag{1.13c}$$ For completeness, let us list two further constraints: $$\bar\varphi\partial\bar\varphi-\bar\chi\bar\partial\bar\chi=0=\chi\partial\chi-\varphi\bar\partial\varphi,\tag{1.14a}$$ $$\bar\varphi\partial\chi+\varphi\bar\partial\bar\chi=0=\chi\partial\bar\varphi+\bar\chi\bar\partial\varphi,\tag{1.14b}$$

Remark that if $Q(\tau,~\xi)$ is given and $\rho$ is constant, $\psi$ is fully determined if it is known at a given point. The function $Q$ is further invariant under a rotation in the $(\tau,~\xi)$ plane, it is a scalar. The related quantity $J$ used by Bracken [3] (eq. (3.7)) is $J=\frac12\rho Q.$ Taking the derivative of the defining relation (1.12) and using the Dirac equation, we directly get $$(\bar\partial+\bar\partial u)Q=(\partial-\partial u)m.\tag{1.15}$$ The quantity $Q$ could be called the mirror mass, and by solving (1.12) and (1.13c) we have the mirror Dirac equation $$\left\{\eqalign{ \partial(\rho^{-1}\varphi,-\rho^{-1}\bar\chi)& =\frac i2Q(\rho^{-1}\chi,\phantom-\rho^{-1}\bar\varphi)\\ \bar\partial(\rho^{-1}\chi,\phantom-\rho^{-1}\bar\varphi)& =\frac i2\bar Q(\rho^{-1}\varphi,-\rho^{-1}\bar\chi). }\right.\tag{1.16a,b}$$ It is actually the one satisfied by $\Phi^{-1}.$

Now we use the well known vector representation. A matrix such as $\Phi$ represents a rotation times a scaling in $\Bbb R^3$ as follow: $$\boldsymbol V=v^i\sigma_i\to\boldsymbol V'=\Phi\boldsymbol V\Phi^\dagger.\tag{1.17}$$ The matrix $\Phi$ can thus be represented by a multiple of an orthogonal $3\times3$ real matrix, whose elements are easily found by using the inner product in the space of $\boldsymbol V$: $$\langle\boldsymbol V_1,~\boldsymbol V_2\rangle=\frac12\Tr[\boldsymbol V_1\boldsymbol V_2],\tag{1.18}$$ then $$e^j_i=\frac12\Tr[\sigma_i\Phi\sigma_j\Phi^\dagger],\tag{1.19}$$ and after some straightforward calculations: $$e={\small\pmatrix{\boldsymbol e_+:={\sqrt2\over2}(\boldsymbol e_1-i\boldsymbol e_2)\\ \boldsymbol e_-:={\sqrt2\over2}(\boldsymbol e_1+i\boldsymbol e_2)\\ \boldsymbol e_3}} ={\small\pmatrix{{\sqrt2\over2}(\varphi\varphi-\bar\chi\bar\chi)& -i{\sqrt2\over2}(\varphi\varphi+\bar\chi\bar\chi)& \sqrt2~\varphi\bar\chi\\ {\sqrt2\over2}(\bar\varphi\bar\varphi-\chi\chi)& i{\sqrt2\over2}(\bar\varphi\bar\varphi+\chi\chi)& \sqrt2~\bar\varphi\chi\\ -(\varphi\chi+\bar\varphi\bar\chi)& i(\varphi\chi-\bar\varphi\bar\chi)& \varphi\bar\varphi-\chi\bar\chi}.\tag{1.20}}$$ It is an $SO(3)$ matrix multiplied by a factor $\rho,$ so that $ee^\dagger=\rho^2{\rm I}.$

According to the properties of the inner product, we can write $$\boldsymbol E_i:=e^j_i\sigma_j=\Phi^\dagger\sigma_i\Phi,\tag{1.21}$$ with which we calculate $$\d\boldsymbol e_i\cdot\boldsymbol e_j=\langle\d\boldsymbol E_i,~\boldsymbol E_j\rangle =\frac12\Tr[\d\,(\Phi^\dagger\sigma_i\Phi)\Phi^\dagger\sigma_j\Phi].\tag{1.22}$$ Differentiating, this gives $$\d\boldsymbol e_i\cdot\boldsymbol e_j= \frac12\rho\Tr[\d\Phi\Phi^\dagger\sigma_j\sigma_i+(\d\Phi\Phi^\dagger\sigma_j\sigma_i)^\dagger]= \rho~{\frak Re}\{\Tr[\d\Phi\Phi^\dagger\sigma_j\sigma_i]\},\tag{1.23}$$ from which the vector representation of the Maurer-Cartan form is deduced: $$\rho^{-2}\d ee^\dagger={\small \pmatrix{2\partial u& 0& -i{\sqrt2\over2}Q\\0& 0& i{\sqrt2\over2}m\\ -i{\sqrt2\over2}m& i{\sqrt2\over2}Q& \partial u}}\d\zeta+{\small \pmatrix{0& 0& -i{\sqrt2\over2}m\\0& 2\bar\partial u& i{\sqrt2\over2}\bar Q\\ -i{\sqrt2\over2}\bar Q& i{\sqrt2\over2}m& \bar\partial u}}\d\bar\zeta.\tag{1.24}$$

Remark: There is a two-ways fast lane for switching between both representations. If the generators of $SU(2)$ associated with $\Phi$ are $\frac i2\sigma_i,$ the corresponding generators of $SO(3),$ whose Lie algebra is the same, associated with $e$ are $$\Sigma_1=\frac{\sqrt2}2{\small\pmatrix{0& 0& -i\\0& 0& i\\-i& i& 0}},~ \Sigma_2=\frac{\sqrt2}2{\small\pmatrix{0& 0& -1\\0& 0& -1\\1& 1& 0}},~ \Sigma_3={\small\pmatrix{i& 0& 0\\0& -i& 0\\0& 0& 0}.\tag{1.25}}$$ Similarly, the generator of the scaling is $\frac12{\rm I}$ for he spinor representation, and ${\rm I}$ for the vector representation, where ${\rm I}$ is the unit matrix of the appropriate size. The Maurer-Cartan form belongs to the respective Lie algebra. One passes from one to the other just by replacing the basis of the Lie algebra representation. Denoting this abstract basis by $S_i,$ $i$ running from $0$ to $3,$ the form reads $$\e^{-2uS_0}\d {\cal II}^\dagger=\d u~S_0 -i(\partial u\d\zeta-\bar\partial u\d\bar\zeta)S_3+\\ +\frac12[(m+Q)\d\zeta+(m+\bar Q)\d\bar\zeta]S_1 -\frac i2[(m-Q)\d\zeta-(m-\bar Q)\d\bar\zeta]S_2,\tag{1.26}$$ where ${\cal I}$ stands either for $\Phi$ or $e.$ The same works as well for $\Phi$ and $e$ using the exponential map of the Lie algebra, from which the Maurer-Cartan form can be derived. It is still simpler using the step generators $S_\pm=S_1\pm iS_2$ and $\tilde S_\pm=S_0\pm iS_3,$ but I shall not spoil the fun of the reader, who may want to prove the claim too.

The surface is constructed by using the family of orthogonal frames parameterised by $(\zeta,~\bar\zeta)$ that we get through the spinor function. Its normal is taken to be $\boldsymbol e_3$ so that it is a conformal immersion in $\Bbb R^3.$ By mere inspection of the expression of $\d ee^\dagger$ (1.24), we read the simple equation $$(\bar\partial,~\partial,~0)e=\partial\boldsymbol e_-+\bar\partial\boldsymbol e_+=0.\tag{1.27}$$ Each component of $\boldsymbol e_-$ and $\boldsymbol e_+$ is then conserved, the third one is just the Dirac (electromagnetic) current. So there are three linearly independent conserved currents: $$\pmatrix{\phantom-\varphi\varphi\\-\chi\chi},\quad \pmatrix{-\bar\chi\bar\chi\\\phantom-\bar\varphi\bar\varphi},\quad \pmatrix{\varphi\bar\chi\\\bar\varphi\chi}.\tag{1.28}$$ The $\Bbb R^3$-valued differential form $$\d\boldsymbol r=i\frac{\sqrt2}2\boldsymbol e_+\d\zeta-i\frac{\sqrt2}2\boldsymbol e_-\d\bar\zeta\tag{1.29}$$ is then closed, and in addition it is normal to $\boldsymbol e_3,$ and real because $\boldsymbol e_-=\bar{\boldsymbol e}_+.$ Moreover, it is exact in virtue of the Poincaré’s lemma since every loop in $\Bbb C$ can be shrunk to a point. Therefore, integrating it along a path $\Gamma$ in the $(\zeta,~\bar\zeta)$ plane from a fixed point $\zeta_0,$ we arrive at the wanted equation of the surface: $$\eqalignno{ x+iy& =i\int_\Gamma(\varphi^2\d\zeta'+\chi^2\d\bar\zeta')+x_0+iy_0&(\text{1.30a})\\ x-iy& =i\int_\Gamma(-\bar\chi^2\d\zeta'-\bar\varphi^2\d\bar\zeta')+x_0-iy_0&(\text{1.30b})\\ z& =i\int_\Gamma(\varphi\bar\chi\d\zeta'-\bar\varphi\chi\d\bar\zeta')+z_0&(\text{1.30c}) }$$ for the point $\boldsymbol r$ of coordinates $(x,y,z)$ in $\Bbb R^3.$ That is the same as in [1] (equation 2) up to some cosmetic rearrangements.

From this derivation, it is obvious that the Gauß map is directly given by: $$\boldsymbol\phi=\boldsymbol e_3/\rho.\tag{1.31}$$ The north pole corresponds to $\chi=0,$ and the south pole to $\varphi=0.$ To go further, let us remark that in the vector representation, we use row vectors while the spinor is a column. If we consider the row spinor $(\varphi,-\bar\chi)$ and the associated stereographic projection, or fundamental field: $$w=-\bar\chi/\varphi,\tag{1.32}$$ then, dividing the numerator and the denominator by $\bar\varphi\varphi$ in the expression of $\boldsymbol\phi$ (1.31), we obtain $$\boldsymbol\phi=(w+\bar w,~i(w-\bar w),~1-w\bar w)/(1+w\bar w),\tag{1.33}$$ that is a function of $w$ alone. In other words, $w$ is also the stereographic projection of the Gauß map from the south pole. Moreover, for a minimal surface ($m=0,$) it is readily shown that $w$ is an analytic function of $\zeta,$ as is known for a long time. Actually, $w$ is precisely the holomorphic function of the Enneper-Weierstraß representation. That generalises for any non-minimal surface with $w$ being any function. On the other hand, if $Q=0,$ the definition of $Q$ (1.12) implies $\partial w=0,$ that is, $w$ is an antiholomorphic function. The use of the differential constraints (1.12) and (1.13b) yields further $$\eqalign{ m^2& =4{\bar\partial w~\partial\bar w\over(1+w\bar w)^2},\\ mQ& =4{\partial w~\partial\bar w\over(1+w\bar w)^2},\quad m\bar Q& =4{\bar\partial w~\bar\partial\bar w\over(1+w\bar w)^2}\\ Q\bar Q& =4{\partial w~\bar\partial\bar w\over(1+w\bar w)^2}.\tag{1.34a,b,c} }$$

We now have all what is necessary to get the traditional data of a surface. Since $\boldsymbol e_\pm^2=0$ and $\boldsymbol e_+\cdot\boldsymbol e_-=\rho^2,$ the first fundamental form is $${\rm I}:=\d\boldsymbol r\cdot\d\boldsymbol r=\rho^2\d\zeta~\d\bar\zeta\tag{1.35}$$ (the coordinates $\tau,~\xi$ are therefore isothermal.) The second fundamental form can be directly calculated too, using $\boldsymbol e_3\cdot\d\boldsymbol r=0$: $$\eqalign{ {\rm II}:=& -\d\boldsymbol\phi\cdot\d\boldsymbol r =-\rho^{-1}\d\boldsymbol e_3\cdot\d\boldsymbol r\\ =& -\frac12\rho(Q\d\zeta\d\zeta+2m\d\zeta\d\bar\zeta+\bar Q\d\bar\zeta\d\bar\zeta).\tag{1.36} }$$ $${\rm III}:=\d\boldsymbol\phi\cdot\d\boldsymbol\phi\ =mQ\d\zeta\d\zeta+(m^2+Q\bar Q)\d\zeta\d\bar\zeta+m\bar Q\d\bar\zeta\d\bar\zeta\\ =(Q\d\zeta+m\d\bar\zeta)(\overline{Q\d\zeta+m\d\bar\zeta}).\tag{1.36'}$$ Finally the usual formulæ gives the mean curvature $H$ and the gauß curvature $K$: $$H=\frac12\Tr[{\rm II}\cdot{\rm I}^{-1}]=-\rho^{-1}m,\tag{1.37}$$ $$K=\det({\rm II}\cdot{\rm I}^{-1})=\rho^{-2}(m^2-Q\bar Q)=H^2-\rho^{-2}Q\bar Q.\tag{1.38}$$ Konopelchenko [1] (equation 4) gives a different expression: $$K=-4\rho^{-2}\partial\bar\partial u,\tag{1.39}$$ which is known as the Gauß-Riemann curvature. But Ferapontov and Grundland [2] (equation 2.6) proved that they are equivalent, in accordance with Gauß’ Theorema Egregium. Then the modulus of $Q$ is a function of $\rho$ and $m$: $$Q\bar Q=m^2+4\partial\bar\partial u.\tag{1.40}$$ From (1.34) can still be given the alternative expression: $$\rho^2K=4{\bar\partial w\partial\bar w-\partial w\bar\partial\bar w\over(1+w\bar w)^2} =\partial\Big(2{\bar w\bar\partial w-w\bar\partial\bar w\over1+w\bar w}\Big) +\bar\partial\Big(2{w\partial\bar w-\bar w\partial w\over1+w\bar w}\Big),\tag{1.41} $$ where the curvature nature is manifest.

It is interesting to note that since the difference between the principal curvatures is given by $$(\Delta k)^2=4(H^2-K),\tag{1.42}$$ it holds $$|\Delta k|=2\rho^{-1}|Q|,\tag{1.43}$$ that is, the modulus of $Q$ is a measure of the local deformation from a spherical surface, like $m$ is a measure of the local deformation from a minimal surface. Indeed $$k=-\rho^{-1}(m\pm|Q|).\tag{1.44}$$

§2. The Cartan moving frame

So far, we implicitly used a fixed referential frame in $\Bbb R^3.$ Indeed, by varying this frame, with the same solution of the Dirac equation we get a whole set of surfaces that are deduced from each other by a rigid motion. This way has been deemed awkward by Cartan, so he developed the powerful method of the moving frame [4]. To an infinitesimal displacement in the $(\zeta,~\bar\zeta)$ plane is now associated an infinitesimal change of the origin and of the basis vectors, but this time expressed with respect to the moving frame itself. By using differential $1$-forms (also called Pfaffian forms,) this is written down as $$\left\{\eqalign{ \d\boldsymbol r& =\omega^j\boldsymbol e_j\\ \d\boldsymbol e_i& =\omega^j_i\boldsymbol e_j }\right.,\quad i,j=+,-,3,\tag{2.1a,b}$$ which is called the moving frame equations. The vectors $\boldsymbol e_i$ satisfy the orthonormality conditions $$\boldsymbol e_\pm^2=0,\quad\boldsymbol e_+\cdot\boldsymbol e_-=\rho^2, \quad\boldsymbol e_\pm\cdot\boldsymbol e_3=0,\quad\boldsymbol e_3^2=\rho^2.\tag{2.2}$$ Differentiating them, and using the moving frame equations (2.1), we get relations among the differential forms, e.g.: $$\d\boldsymbol e_+^2=0\Rightarrow 2\boldsymbol e_+\cdot\d\boldsymbol e_+=0\Rightarrow \omega^-_+=0.\tag{2.3}$$ By a similar method, we collect the following relations: $$\omega^+_-=\omega^-_+=0,\tag{2.4a}$$ $$\omega^-_3+~\omega^3_+=\omega^+_3+~\omega^3_-=0.\tag{2.4b}$$ $$\omega^-_-+~\omega^+_+=2\d u,\tag{2.4c}$$ $$\omega^3_3=\d u,\tag{2.4d}$$ As $\bar{\boldsymbol e}_+=\boldsymbol e_-$ and $\bar{\boldsymbol e}_3=\boldsymbol e_3,$ with the complex conjugates of the same equations we moreover find $$\eqalignno{ \bar\omega^3_+& =\omega^3_-,&\text{(2.5a)}\\ \bar\omega^+_+& =\omega^-_-.&\text{(2.5b)} }$$ Then assuming the moving frame equations are integrable, by differentiating them and substituting $\d\boldsymbol e_i$ (2.1b) wherever possible, we get compatibility equations, called the structure equations by Cartan: $$\left\{\eqalign{ \d\omega^i& =\omega^k\wedge\omega^i_k\\ \d\omega^i_j& =\omega^k_j\wedge\omega^i_k. }\right.\tag{2.6a,b}$$ The second equality (2.6b) is always true as long as the frames are given, and the first one (2.6b) is the equivalent, expressed in the moving frame language, of the requirement that the form $\d\boldsymbol r$ be exact.

From the expression of $\d\boldsymbol r$ (1.29) we immediately get $$\omega^+=i\frac{\sqrt2}2d\zeta,\quad\omega^-=-i\frac{\sqrt2}2\d\bar\zeta,\quad\omega^3=0,\tag{2.7}$$ and $$\bar\omega^+=\omega^-.\tag{2.8}$$

The expression of the remaining $1$-forms are easily obtained by writing the second moving frame equation (2.1b) as a matrix: $$\d e=\Omega e,\tag{2.9}$$ where $$\Omega=\pmatrix{ \omega^+_+& \omega^-_+& \omega^3_+\\ \omega^+_-& \omega^-_-& \omega^3_-\\ \omega^+_3& \omega^-_3& \omega^3_3\\ }.\tag{2.10}$$ Multiplying (2.9) by $e^\dagger$ from the right, there results $$\Omega=\rho^{-2}\d ee^\dagger\tag{2.11}$$ which is already known, it is the Maurer-Cartan form (1.24), the preceding relations (2.4), (2.5) are incidentally satisfied. Further writing $$\Theta=(\omega^+,~\omega^-,~0),\quad \d\Theta=0,\tag{2.12}$$ the structure equations (2.6) become $$\left\{\eqalign{ \d\Theta& =\Theta\wedge\Omega=0\\ \d\Omega& =\Omega\wedge\Omega. }\right.\tag{2.13a,b}$$ If we decompose $\Omega$ as follow: $$\Omega={\cal Z}\d\zeta+\tilde{\cal Z}\d\bar\zeta,\tag{2.14}$$ with $${\cal Z}={\small\pmatrix{ 2\partial u& 0& -i{\sqrt2\over2}Q\\ 0& 0& i{\sqrt2\over2}m\\ -i{\sqrt2\over2}m& i{\sqrt2\over2}Q& \partial u }},\quad\tilde{\cal Z}={\small\pmatrix{ 0& 0& -i{\sqrt2\over2}m\\ 0& 2\bar\partial u& i{\sqrt2\over2}\bar Q\\ -i{\sqrt2\over2}\bar Q& i{\sqrt2\over2}m& \bar\partial u }},\tag{2.15}$$ the moving frame equation (2.1b) is then $$\partial e={\cal Z}e,\quad\bar\partial e=\tilde{\cal Z}e,\tag{2.16}$$ which is known in the theory of surfaces as the Gauß-Weingarten equations. The second structure equation (2.6b) reads $$\bar\partial{\cal Z}-\partial\tilde{\cal Z}+[{\cal Z},~\tilde{\cal Z}]=0,\tag{2.17}$$ also known as the Gauß-Codazzi-Mainardi equations. We already met them, here they are gathered: $$Q\bar Q=m^2+4\partial\bar\partial u.\tag{Gauß}$$ $$(\bar\partial+\bar\partial u)Q=(\partial-\partial u)m.\tag{Codazzi-Mainardi}$$ Many integrable non-linear partial differential equations are expressed within this formalism. The last equation (2.17) is then called the zero curvature condition, and is the non-linear equation represented as a compatibility condition of a linear system.

In the spinor representation, the corresponding matrices are readily obtained from the Maurer-Cartan form (1.11), they are $$Z=\frac12\pmatrix{ 2\partial u& iQ\\ im& 0},\quad \tilde Z=\frac12\pmatrix{ 0& im\\ i\bar Q& 2\bar\partial u },\tag{2.18}$$ and the equations (2.16), (2.17) become $$\partial\Phi=Z\Phi,\quad\bar\partial\Phi=\tilde Z\Phi,\tag{2.19}$$ $$\bar\partial Z-\partial\tilde Z+[Z,\tilde Z]=0.\tag{2.20}$$

Spelling out the structure equations (2.6), we easily derive the already known equations (Gauß) and (Codazzi-Mainardi) again. In addition, as $\omega^3=0,$ from the structure equation for $\d\omega^3$ we have $$0=\omega^+\wedge\omega^3_++\omega^-\wedge\omega^3_-.\tag{2.21}$$ According to the Cartan’s lemma, $\omega^3_+$ and $\omega^3_-$ are then equal to a linear combination of $\omega^+$ and $\omega^-,$ like $$\eqalign{ \omega^3_+& =h_{++}\omega^+& +h_{+-}\omega^-\\ \omega^3_-& =h_{-+}\omega^+& +h_{--}\omega^-, }\tag{2.22a,b}$$ and $$h_{-+}=h_{+-},\tag{2.23}$$ as can be seen by substituting $\omega^3_+$ and $\omega^3_-$ in the equation (2.21). Finally by identification we get $$h_{++}=-Q,\quad h_{--}=-\bar Q,\tag{2.24a}$$ $$h_{+-}=h_{-+}=m,\tag{2.24b}$$ or $$h=\pmatrix{ -Q& m\\ m& -\bar Q }.\tag{2.25}$$

Now using this matrix together with the moving frame equations (2.1), we are able to express the data of the surface in term of the Pfaffian forms: $${\rm I}=2\rho^2\omega^+\omega^-,\tag{2.26}$$ $$\eqalign{ {\rm II}& =-\rho(\omega^+\omega^3_++\omega^-\omega^3_-)\\ & =-\rho(h_{++}\omega^+\omega^++(h_{+-}+h_{-+})\omega^+\omega^-+h_{--}\omega^-\omega^-).\tag{2.27} }$$ The surface element is $$\d S=i\rho^2\omega^+\wedge\omega^-,\tag{2.28}$$ and the corresponding surface element on the Gauß map $$\d\sigma=i\omega^3_-\wedge\omega^3_+=i(h_{+-}h_{-+}-h_{++}h_{--})\omega^+\wedge\omega^-.\tag{2.29}$$ The total curvature is the ratio of these two surfaces: $$K={\d\sigma\over\d S}=\rho^{-2}(h_{+-}h_{-+}-h_{++}h_{--}),\tag{2.30}$$ and the mean curvature is $$\eqalign{ H& =\frac i2\rho{\omega^-\wedge\omega^3_--\omega^+\wedge\omega^3_+\over\d S}\\ & =-\frac i2\rho{(h_{+-}+h_{-+})\omega^+\wedge\omega^-\over \d S}=-\frac12\rho^{-1}(h_{+-}+h_{-+}). }\tag{2.31}$$ The results are the sames as in the end of the previous section (1.35-38), as expected.

§3. Application

This application illustrates the use of the equations (2.16), (2.17) or (2.19), (2.20) to express a non-linear partial differential equation and its representation by a linear problem. As particular case we shall take a spherical surface, then $Q=0.$ Equation (1.38) simplifies to $$K=\rho^{-2}m^2,\tag{3.1}$$ where $K$ is now a constant, and equation (Gauß), which is deduced from (2.17) or (2.20), becomes $$4\partial\bar\partial u+m^2=0.\tag{3.2}$$ For the special choice $K=1,$ as $m=\rho,$ substituting $$m=\e^u,\tag{3.3}$$ it gives $$4\partial\bar\partial u+\e^{2u}=0,\tag{3.4}$$ and the only remaining variable is $u.$ It is the non-linear Liouville equation as well known. The other equation (Codazzi-Mainardi) is trivially satisfied.

Because of the spinor representation of the Maurer-Cartan form (1.11) from which $Z$ and $\tilde Z$ are deduced (2.18), for any non singular complex matrix function $A$ there is the gauge transformation $$\eqalign{ \Phi& \to A\Phi\\ Z& \to AZA^{-1}+\partial AA^{-1}\\ \tilde Z &\to A\tilde ZA^{-1}+\bar\partial AA^{-1}, \tag{3.5}}$$ for which the non-linear equation (2.20) still holds. Following Lund and Regge [5], (see also [2]) we choose the matrix $$A=\pmatrix{\bar p^{1/2}& 0\\0& p^{1/2}}\pmatrix{\e^{u/2}& 0\\0& \e^{-u/2}}\e^{-u/2}.\tag{3.6}$$ The purpose of each factor, from right to left, is to:

The result is a linear system that can be used to solve the non-linear equation by the inverse scattering transform: $$\partial\Phi_p'=\frac12\pmatrix{2\partial u& 0\\ip& -2\partial u}\Phi_p',\quad \bar\partial\Phi_p'=i\frac{\bar p}2\pmatrix{0& \e^{2u}\\0& 0}\Phi_p'.\tag{3.8}$$ where $$\Phi_p'=\pmatrix{\bar p^{1/2}\varphi& -\bar p^{1/2}\bar\chi\\p^{1/2}\e^{-u}\chi& p^{1/2}\e^{-u}\bar\varphi},\quad \det \Phi_p'=1.\tag{3.9}$$ A system similar to the AKNS one [6] is obtained by using a further factor $$\frac{\sqrt2}2\pmatrix{1& -1\\1& \phantom-1},\tag{3.10}$$ that is $$\partial\Phi_p''=\frac14\pmatrix{-ip& 4\partial u-ip\\ 4\partial u+ip& ip}\Phi_p'',\quad \bar\partial\Phi_p''=i\frac{\bar p}4\pmatrix{-\e^{2u}& \e^{2u}\\-\e^{2u}& \e^{2u}}\Phi_p''.\tag{3.11}$$

Our aim is not to work out the full inverse scattering transform, we shall instead proceed backwards and insert a general solution of the Liouville equation. It is given by some arbitrary antiholomorphic map $g$ of a surface with a conformal metric to the sphere [7]: $$\e^u=2{\sqrt{\bar\partial g\partial\bar g}\over1+g\bar g}.\tag{3.12}$$ From the formula (1.34a) is immediately recognised that $w,$ which is antiholomorphic for $Q=0,$ is identical to $g$: $$g(\bar\zeta) = -\bar\chi/\varphi,\tag{3.13}$$ hence $g$ is the stereographic projection of the Gauß map. Thus $\bar\chi$ can be written as follow: $$-\bar\chi=\varphi g.\tag{3.14}$$ The easiest way to determine $\varphi$ is to use the differential constraint (1.13b). With the given expression of $\bar\chi,$ the derivatives of $\varphi$ cancel out, and using equation (3.3) it simplifies to $$\varphi^2\bar\partial g=\frac i2\e^{2u},\tag{3.15}$$ or $$\varphi=\pm\frac{1+i}2\e^u/\sqrt{\bar\partial g}.\tag{3.16}$$ The condition $\varphi\bar\varphi+\chi\bar\chi=\e^u$ is then fulfilled. As an example, the choice $$g(\bar\zeta)=i\bar\zeta\tag{3.17}$$ gives the complete solution $$m=e^u={2\over 1+\zeta\bar\zeta},\quad \Phi={\sqrt2\over 1+\zeta\bar\zeta}\pmatrix{1& -i\bar\zeta\\i\zeta& 1}.\tag{3.18}$$ It is an easy exercice to verify that it is actually the case. This $\Phi$ solution stands for all the solutions of the spectral problem because there is only one spherical surface of curvature $1.$ Backpedalling the introduction of the spectral parameter, all the $\Phi_p$ then degenerate into the same function.

What is shown is that this linear system is derived from the Dirac equation used to represent the surface. That’s the explanation of the link between non-linear evolution equations and geometry. In fact, it is not the usual representation with a surface of constant negative gaußian curvature and isometry group $SL(2\Bbb R),$ but the AKNS system can be derived the same way.

There is an additional symmetry, since if $\Phi$ is multiplied on the right by a constant $SU(2)$ matrix, it is still a solution of the Dirac equation, and its determinant remains identical to $\e^u.$ Writing this matrix as $$\pmatrix{a& -\bar b\\b& \phantom-\bar a},\quad a\bar a+b\bar b=1,\tag{3.19}$$ we get the transformation $$\bar\varphi\to\bar a\bar\varphi-\bar b\chi,\quad \chi\to a\chi+b\bar\varphi,\tag{3.20}$$ and accordingly the non-linear transformation $$-\bar\chi/\varphi=g\to{\bar ag-\bar b\over bg+a},\tag{3.21}$$ which is a (proper) isometry of the Gauß sphere, and then of the spherical surface. That is to say, the isometry group is $SU(2)$ instead of $SL(2\Bbb R).$

Because of the mirror symmetry, it is natural to wonder whether the rôles of $m$ and $Q$ can be swapped. And that is indeed true, it is a recently discovered duality [7] that is more transparent in this framework. So, setting $m=0,$ for the choice $K=-\e^{-4u}$ the non-linear equation is now $$4\partial\bar\partial u-\e^{-2u}=0.\tag{3.22}$$ The antiholomorphic functions are replaced by holomorphic ones, and the surface is minimal. Since by substituting $u=-v$ the non-linear equation is the same as in the previous case, the general solution is $$\e^{-u}=2{\sqrt{\partial g\bar\partial\bar g}\over1+g\bar g},\tag{3.23}$$ while our example becomes $$\e^{-i\theta}Q=\e^{-u}={2\over 1+\zeta\bar\zeta},\quad \Phi_\theta=\frac{\sqrt2}2\pmatrix{\e^{i\theta/2}& 0\\0& \e^{-i\theta/2}}\pmatrix{1& i\zeta\\-i\bar\zeta& 1},\tag{3.24}$$ where $\theta$ is a real constant. No further details are given since the reasoning is strictly parallel. Let us only remark that these are two constant mean curvature surfaces, and that they have the same isometry group, the Gauß map being merely inverted. The improper isometries are generated by $g(\zeta)\to g(\bar\zeta),$ which transform a surface into its mirror image (the mirror is not plane, as it were.)

§4. Odds and ends

1 - Connection

The differential form $\Omega$ (2.11) is a connection, and actually an affine connection of $\Bbb R^3$ that is flat on the surface, which is the meaning of the structure equations. We have thus the stunning result that the (two-dimensional) Dirac equation is but a way of writing an affine connection. Its restriction to the surface is $$\Omega_{/2}=2\pmatrix{ \partial u\d\zeta& 0\\ 0& \bar\partial u\d\bar\zeta }\tag{4.1}$$ and its intrinsic scalar curvature is precisely the Riemann-Gauß curvature of the surface. The restriction of the structure equations doesn’t necessarily hold, and indeed we have: $$\d\omega_+^+=\omega_+^+\wedge\omega_+^++\omega_+^-\wedge\omega_-^++\omega_+^3\wedge\omega_3^+=\omega_+^3\wedge\omega_3^+.\tag{4.2}$$ But $\omega_+^3$ isn’t defined on the surface, thus the term $\omega_+^3\wedge\omega_3^+=\omega^3_-\wedge\omega^3_+$ stands for the affine curvature form, whose explicit expression taken in three dimensions, that is extrinsically, is the Gauß curvature. That’s still another way of looking at the Theorema Egregium. There is no torsion from the sheer fact that $\omega^3=0,$ and thus it is indeed the Levi-Civita connection.

2 - Symmetries

For investigating the different global symmetries, let us first remark that if we make the substitution $$\matrix{ \varphi& \to& -\bar\chi\\ \chi& \to& \phantom-\bar\varphi, }\tag{4.3}$$ which is in fact the charge conjugaison, the Dirac equation (1.7) still holds. That is, we get the same solutions if we put $\Phi$ instead of $\psi$ in equation (1.2). Then, $\Phi$ multiplied on the right by any constant non singular matrix is a solution of the equation if $\Phi$ is so. Thus, the full symmetry group is $GL(2\Bbb C).$ The substitution above belongs to this group, which makes charge conjugaison a continuous transformation (i.e. continuously connected to identity.) The surface is determined by the functions $\rho$ and $Q$ alone, its transformation then depends only on the transformation properties of these functions. For example, the substitution (4.3) leaves them invariant, so the surface doesn’t change. More generally, the function $\rho$ is invariant under the unitary subgroup $SU(2),$ in this case only the function $Q$ is to be examined. As both curvatures remain invariant under this subgroup since they are independent of $Q$, the surfaces are identical up to a rigid motion. Actually, in virtue of equation (Gauß) only the global phase of $Q$ varies.


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