Monopole and dipole dynamics on Riemann surfaces

Posted on 30 July, 2022 by jesperg

This blog has ended and is replaced by the blog Postmödern Mathematics where you can read

Monopole and dipole dynamics on Riemann surfaces

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The Symmetries of the Quantum Kepler Problem

Posted on 22 January, 2022 by jesperg

This blog has ended and is replaced by the blog Postmodern Mathematics where you can read

The Symmetries of the Quantum Kepler Problem (prime version)

The Symmetries of the Quantum Kepler Problem (differential version)

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The geometry of a second order differential equation, part 2

Posted on 4 July, 2020 by jesperg

We start over with a general second order ordinary differential equation

\displaystyle y''=\phi(x,y,y').

and this time we won’t impose any restrictions on the right-hand side. We express y'=dy/dx and y''=(d^2ydx-d^2xdy)/dx^3 which gives that

\displaystyle d^2ydx-d^2xdy=\phi\left(x,y,\frac{dy}{dx}\right)dx^3.

The variables dx and dy denote velocity and are tangent space coordinates. The variables d^2x and d^2y denote acceleration and are jet space (of curves) coordinates. Since the right-hand side is simultaneous homogeneous of degree 3 in dx and dy we instead consider equations of the form

\displaystyle d^2ydx-d^2xdy=f(x,y,dx,dy),

where f is homogeneous of degree 3 in dx and dy, that is f(x,y,\lambda dx,\lambda dy)=\lambda^3f(x,y,dx,dy). Euler’s theorem then gives that we can express

\displaystyle f(x,y,dx,dy)=Adx^3+3Bdx^2dy+3Cdxdy^2+Ddy^3

with

\displaystyle \begin{aligned} A&=\tfrac{1}{6}f_{dxdxdx}(x,y,dx,dy)\\B&=\tfrac{1}{6}f_{dxdxdy}(x,y,dx,dy)\\C&=\tfrac{1}{6}f_{dxdydy}(x,y,dx,dy)\\D&=\tfrac{1}{6}f_{dydydy}(x,y,dx,dy)\end{aligned}

and A,B,C,D are homogeneous of degree 0 in dx and dy. We put p=dy/dx and consider A,B,C,D to be functions in x,y and p:

\displaystyle \begin{aligned} A(x,y,p)&=\tfrac{1}{6}f_{dxdxdx}(x,y,1,p)\\B(x,y,p)&=\tfrac{1}{6}f_{dxdxdy}(x,y,1,p)\\C(x,y,p)&=\tfrac{1}{6}f_{dxdydy}(x,y,1,p)\\D(x,y,p)&=\tfrac{1}{6}f_{dydydy}(x,y,1,p)\end{aligned}

We have that

\displaystyle d^2ydx-d^2xdy=Adx^3+3Bdx^2dy+3Cdxdy^2+Ddy^3

Polarisation gives that

\displaystyle\begin{aligned} d^2x&=-Bdx^2-2Cdxdy-Ddy^2\\d^2y&=\quad Adx^2+2Bdxdy+Cdy^2,\end{aligned}

which are geodesic equations of a connection of the two-dimensional tangent spaces in the three-dimensional projectivised tangent bundle (also called the manifold of elements). Each tangent space has coordinates dx and dy and the projectivised tangent bundle has coordinates x, y and p. We can read off the Christoffel symbols of this connection as

\displaystyle \begin{aligned}\Gamma^1_{11}=B\quad\Gamma^1_{12}=\Gamma^1_{21}=C\quad\Gamma^1_{22}=D\\\Gamma^2_{11}=-A\quad\Gamma^2_{12}=\Gamma^2_{21}=-B\quad\Gamma^2_{22}=-C.\end{aligned}

We collect the Christoffel symbols in a connection one-form

\displaystyle \mathbf{\Gamma}=\begin{pmatrix}B&C\\-A&-B\end{pmatrix}dx+\begin{pmatrix}C&D\\-B&-C\end{pmatrix}dy.

We calculate the curvature two-form \mathbf{\Omega} by applying the exterior covariant derivative D to the connection one-form

\displaystyle \begin{aligned}\mathbf{\Omega}&=D\mathbf{\Gamma}=\delta\mathbf{\Gamma}+\mathbf{\Gamma}\wedge\mathbf{\Gamma}\\&=\begin{pmatrix}b&c\\-a&-b\end{pmatrix}dx\wedge dy+\begin{pmatrix}kp&-k\\kp^2&-kp\end{pmatrix}(dy-pdx)\wedge dp\end{aligned}

where \delta means ordinary exterior derivative and

\displaystyle \begin{aligned}a&=B_x-A_y+2AC-2B^2\\b&=C_x-B_y+AD-BC\\c&=D_x-C_y+2BD-2C^2\\k&=-\frac{A_p}{p^3}=\frac{B_p}{p^2}=-\frac{C_p}{p}=D_p=\tfrac{1}{6}\phi_{y'y'y'y'}(x,y,p).\end{aligned}

We extend the tangent spaces of the base manifold to projectives planes by adding a line at infinity in each tangent space. The points in the affine part of the projective tangent spaces have homogenous coordinates (dx:dy:1) and points on the line at infinity have homogenous coordinates (dx:dy:0). We also extend the connection into two projective connections \mathbf{\Pi}_1 and \mathbf{\Pi}_2:

\displaystyle  \mathbf{\Pi}_1=\begin{pmatrix}B&C&1\\-A&-B&0\\0&0&0\end{pmatrix}dx+\begin{pmatrix}C&D&0\\-B&-C&1\\0&0&0\end{pmatrix}dy,

\displaystyle  \mathbf{\Pi}_2=\begin{pmatrix}B&C&1\\-A&-B&0\\-a&-b&0\end{pmatrix}dx+\begin{pmatrix}C&D&0\\-B&-C&1\\-b&-c&0\end{pmatrix}dy.

At an element (x_0,y_0,p_0) a connection in the bundle defines an osculating connection in the base manifold by regarding p=p_0 as constant. Also an osculating connection is defined in the dual manifold by regarding (x,y)=(x_0,y_0) as constant. The curvature of the osculating connection of \mathbf{\Pi}_2 in the base manifold is

\displaystyle D\mathbf{\Pi}_2|_{\text{primal}} = \begin{pmatrix}0&0&0\\0&0&0\\ \alpha&\beta&0\end{pmatrix}dx\wedge dy

with

\displaystyle \begin{aligned}\alpha&=A_{yy}-2B_{xy}+C_{xx}-3AC_y+2AD_x+6BB_y-3BC_x-3CA_y+DA_x\\\beta&=B_{yy}-2C_{xy}+D_{xx}-AD_y+3BD_x+3CB_y-6CC_x-2DA_y+3DB_x.\end{aligned}

The curvature of the osculating connection of \mathbf{\Pi}_1 in the dual manifold is

\displaystyle D\mathbf{\Pi}_1|_{\text{dual}} =k\begin{pmatrix}p&-1&0\\p^2&-p&0\\0&0&0\end{pmatrix}(dy-pdx)\wedge dp

The curvature form of the primal osculating connection is characterised by that all lines through the origin are invariant. The curvature form of the dual osculating connection is characterised by that all points on the line dy-pdx=0 are invariant. These two relations are dual and it follows that the two osculating connections are dual.

From the curvatures we get two invariants in the bundle:

\displaystyle \begin{aligned} k(x,y,p) &= \tfrac{1}{6}\phi_{y'y'y'y'}(x,y,p)\\ \gamma(x,y,p)&=\alpha(x,y,p)+\beta(x,y,p) p\end{aligned}

which are dual to each other.

 

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The geometry of a second order differential equation, part 1

Posted on 3 July, 2020 by jesperg

We start with a general second order ordinary differential equation

\displaystyle y''=\phi(x,y,y').

and express y'=dy/dx and y''=(d^2ydx-d^2xdy)/dx^3 which gives that

\displaystyle d^2ydx-d^2xdy=\phi\left(x,y,\frac{dy}{dx}\right)dx^3.

The variables dx and dy denote velocity and are tangent space coordinates. The variables d^2x and d^2y denote acceleration and are jet space (of curves) coordinates. Since the right-hand side is simultaneous homogeneous of degree 3 in dx and dy we instead consider equations of the form

\displaystyle d^2ydx-d^2xdy=f(x,y,dx,dy),

where f is homogeneous of degree 3 in dx and dy, that is f(x,y,\lambda dx,\lambda dy)=\lambda^3f(x,y,dx,dy). Euler’s theorem then gives that we can express

\displaystyle f(x,y,dx,dy)=Adx^3+3Bdx^2dy+3Cdxdy^2+Ddy^3

with

\displaystyle \begin{aligned} A&=\tfrac{1}{6}f_{dxdxdx}(x,y,dx,dy)\\B&=\tfrac{1}{6}f_{dxdxdy}(x,y,dx,dy)\\C&=\tfrac{1}{6}f_{dxdydy}(x,y,dx,dy)\\D&=\tfrac{1}{6}f_{dydydy}(x,y,dx,dy)\end{aligned}

and A,B,C,D are homogeneous of degree 0 in dx and dy.

We now make the simplifying assumption that A,B,C,D do not depend on dx and dy and hence are functions in x and y only. (This means that we are restricting to equations with right-hand side that is a third degree polynomial in the first derivative, \phi(x,y,y')=A+3By'+3C(y')^2+D(y')^3.)

\displaystyle \begin{aligned} A(x,y)&=\tfrac{1}{6}f_{dxdxdx}\\B(x,y)&=\tfrac{1}{6}f_{dxdxdy}\\C(x,y)&=\tfrac{1}{6}f_{dxdydy}\\D(x,y)&=\tfrac{1}{6}f_{dydydy}\end{aligned}

We have that

\displaystyle d^2ydx-d^2xdy=Adx^3+3Bdx^2dy+3Cdxdy^2+Ddy^3

Polarisation gives that

\displaystyle\begin{aligned} d^2x&=-Bdx^2-2Cdxdy-Ddy^2\\d^2y&=\quad Adx^2+2Bdxdy+Cdy^2,\end{aligned}

 which are geodesic equations and we can read off the Christoffel symbols as

\displaystyle \begin{aligned}\Gamma^1_{11}=B\quad\Gamma^1_{12}=\Gamma^1_{21}=C\quad\Gamma^1_{22}=D\\\Gamma^2_{11}=-A\quad\Gamma^2_{12}=\Gamma^2_{21}=-B\quad\Gamma^2_{22}=-C.\end{aligned}

We collect the Christoffel symbols in a connection one-form

\displaystyle \mathbf{\Gamma}=\begin{pmatrix}B&C\\-A&-B\end{pmatrix}dx+\begin{pmatrix}C&D\\-B&-C\end{pmatrix}dy.

We calculate the curvature two-form \mathbf{\Omega} by applying the exterior covariant derivative D to the connection one-form

\displaystyle \begin{aligned}\mathbf{\Omega}&=D\mathbf{\Gamma}=\delta\mathbf{\Gamma}+\mathbf{\Gamma}\wedge\mathbf{\Gamma}\\&=\begin{pmatrix}b&c\\-a&-b\end{pmatrix}dx\wedge dy\end{aligned}

where \delta means ordinary exterior derivative and

\displaystyle \begin{aligned}a&=B_x-A_y+2AC-2B^2\\b&=C_x-B_y+AD-BC\\c&=D_x-C_y+2BD-2C^2.\end{aligned}

We read off the Riemann tensor components

\displaystyle \begin{aligned}&R^1_{112}=b\quad &R^1_{121}=-b\\&R^1_{212}=c\quad &R^1_{221}=-c\\&R^2_{112}=-a\quad &R^2_{212}=a\\&R^2_{212}=-b\quad &R^2_{222}=b.\end{aligned}

By contracting the Riemann tensor we get the components of the Ricci quadratic form

\displaystyle \begin{gathered}R_{ij}=R^k_{ikj}\\R_{11}=a\quad R_{12}=b\quad R_{21}=b\quad R_{22}=c\\R=adx^2+2bdxdy+cdy^2.\end{gathered}

The Ricci quadratic form is the two-dimensional analogue of the Schwarzian derivative and allows us to construct a projective structure for each geodesic. Set

\displaystyle \mathbf{v}=\begin{pmatrix}u\\v\\w\end{pmatrix}

and let

\displaystyle d^2\mathbf{v}=-R(x,y,dx,dy)\mathbf{v}.

The vector \mathbf{v} will trace out a curve in \mathbf{R}^3. Since the acceleration d^2\mathbf{v} is parallel to \mathbf{v} the curve will be planar. The containing plane defines homogeneous coordinates on the geodesic and this gives the geodesic a projective structure.

Since the Ricci quadratic form defines a projective structure on every solution curve we might wonder if these projective structures fit together to also give a two-dimensional projective structure on the x,y-plane. To investigate this we reinterpret the Ricci form as a one-form valued one-form and express it as a row vector of one-forms

\displaystyle R=\begin{pmatrix}a&b\end{pmatrix}dx+\begin{pmatrix}b&c\end{pmatrix}dy.

We apply the exterior covariant derivative to this form

\displaystyle \begin{aligned}DR&=\delta R+R\wedge \mathbf{\Gamma}\\&=\begin{pmatrix}\alpha&\beta\end{pmatrix}dx\wedge dy,\end{aligned}

with

\displaystyle \begin{aligned}\alpha&=A_{yy}-2B_{xy}+C_{xx}-3AC_y+2AD_x+6BB_y-3BC_x-3CA_y+DA_x\\\beta&=B_{yy}-2C_{xy}+D_{xx}-AD_y+3BD_x+3CB_y-6CC_x-2DA_y+3DB_x.\end{aligned}

DR is a one-form valued two-form, which perhaps can be called the Bianchi form, and is the obstruction of the integrability of the one-dimensional projective structures into a two-dimensional projective structure. This can, perhaps, be understood geometrically in the following way. Parallel transport a plane around a loop in \mathbf{R}^3. After the parallel transport the plane might have become tilted compared to its initial position. The form DR which when evaluated on a bivector at a point gives a covector. The null direction of this covector is a rotation axis and the magnitude of the covector is the rotational velocity around this axis in \mathbf{R}^3.

When DR vanishes, a plane when parallel transported in a tangential direction in \mathbf{R}^3 will always be the tangent plane of an integral surface. The Ricci quadratic form is the centroaffine fundamental form of this surface. The solution curves will be curves on the surface cut by planes through the origin and \mathbf{R}^3 will be homogeneous coordinates thus giving a projective structure to the x,y-plane.

If the surface is projected from the origin onto a plane not passing through the origin, the solution curves are projected onto straight lines. This gives a diffeomorphism that maps the differential equation to the trivial one.

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The geometry of y”+y=0

Posted on 2 July, 2020 by jesperg

In the differential equation y''+y=0 we rewrite y''=(d^2ydx-d^2xdy)/dx^3 and get

\displaystyle d^2ydx-d^2xdy=-ydx^3

and with polarisation we get

\displaystyle\begin{aligned} d^2x&=0\\d^2y&=-ydx^2.\end{aligned}

We read off the Christoffel symbols to be

\displaystyle \begin{aligned}\Gamma^1_{11}=0\quad\Gamma^1_{12}=\Gamma^1_{21}=0\quad\Gamma^1_{22}=0\\\Gamma^2_{11}=y\quad\Gamma^2_{12}=\Gamma^2_{21}=0\quad\Gamma^2_{22}=0.\end{aligned}

The connection one-form becomes

\displaystyle \mathbf{\Gamma}=\begin{pmatrix}0&0\\y&0\end{pmatrix}dx

and the curvature two-form is

\displaystyle \begin{aligned}\mathbf{\Omega}&=D\mathbf{\Gamma}=\delta\mathbf{\Gamma}+\mathbf{\Gamma}\wedge\mathbf{\Gamma}\\&=\begin{pmatrix}0&0\\-1&0\end{pmatrix}dx\wedge dy\end{aligned}

where D is the covariant exterior derivative and \delta means the ordinary exterior derivative.

The Ricci quadratic form is

\displaystyle R = dx^2

which is the centroaffine fundamental form of a centered elliptical cylinder in \mathbf{R}^3. The coordinates x and y parameterise the cylinder, x is the angular coordinate and y is the longitudinal coordinate on the cylinder. The solution curves are curves that are given by intersecting the cylinder with planes through the origin.

In this way \mathbf{R}^3 becomes homogenous coordinates on the x,y-plane and realises the projective structure inherent in the differential equation. The differential equation thus has the symmetry group PGL(3).

We can extend the connection to a connection in \mathbf{R}^3. We define cylindrical coordinates (x,y,r), where x is the angular coordinate, y is the longitudinal coordinate and r is the radial coordinate. In these coordinates we define the connection one-form as

\displaystyle \mathbf{\Pi}=\begin{pmatrix}0&0&dx\\ydx&0&dy\\-dx&0&0\end{pmatrix},

where the upper left 2\times 2-submatrix is \mathbf{\Gamma}, the rightmost column is the standard column one-form and the bottom row is the Ricci quadratic form expressed as a row one-form. A geodesic of this connection that starts parallel to a coordinate cylinder (dr=0) will stay on the cylinder and has its acceleration vector parallel to the position vector and so will trace out an ellipse that is the intersection of the cylinder and a plane through the origin.

sin_sin3d_v3

 

The curvature of this connection vanishes:

\displaystyle \begin{aligned}D\mathbf{\Pi}&=\delta\mathbf{\Pi}+\mathbf{\Pi}\wedge\mathbf{\Pi}\\&=\begin{pmatrix}0&0&0\\0&0&0\\0&0&0\end{pmatrix}\end{aligned},

confirming that the connection defines a projective structure on the x,y-plane.

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Vortex dynamics

Posted on 17 January, 2020 by jesperg

(Updated on 26/4-20)

We will investigate the motion of point vortices and dipole vortices on a Riemann surface. We’ll start with the point vortex case and will mainly follow the well-written exposition given by Björn Gustafsson in Vortex motion and geometric function theory: the role of connections.

monopole

The setup is that we have a Riemann surface M of genus g and equipped with a conformal metric ds=\lambda(z)|dz|. On this surface we specify n point vortices located at points z_1,\dots,z_n with strengths given by real numbers \Gamma_1,\dots,\Gamma_n such that \sum_k \Gamma_k=0 and a steady background flow defined by 2g periods a_1,\dots,a_g and b_1,\dots,b_g.

We construct a Green function G(z,a) for the Riemann surface which is a solution to

\displaystyle \Delta G=\delta_a-\frac{1}{\text{area}(M)},

where \Delta is the Laplace operator, \delta_a is a point source at z=a and -1/\text{area}(M) is a constant function acting as a background sink. 

The Green function can be expanded as

\displaystyle G(z,a)=\frac{1}{2\pi}\left(-\log|z-a|+h_0(a)+\frac{1}{2}\left(h_1(a)(z-a)+\overline{h_1(a)}(\bar{z}-\bar{a})\right)\right.

\displaystyle \left.+\frac{1}{2}\left(h_2(a)(z-a)^2+\overline{h_2(a)}(\bar{z}-\bar{a})^2\right)+h_{11}(a)(z-a)(\bar{z}-\bar{a})+O(|z-a|^3)\right)

The stream function is constructed by adding together the individual Green functions

\displaystyle \psi=\sum_k\Gamma_kG(z,z_k)+U^*

where U^* is the harmonic conjugate of the background flow defined by the given periods a_1,\dots,a_g and b_1,\dots,b_g. The complex potential \Phi=\varphi+i\psi consists of the velocity potential \varphi and the stream function \psi related by -d\varphi=*d\psi. 

The complex potential can be expanded at z=z_k as

\displaystyle \Phi(z) = \frac{\Gamma_k}{2\pi i}\left(\log(z-z_k)-c_0(z_k)-c_1(z_k)(z-z_k)-c_2(z_k)(z-z_k)^2+O((z-z_k)^3)\right)

and we have that

\displaystyle c_0(z_k)=h_0(z_k)+\sum_{j\neq k}\frac{2\pi\Gamma_j}{\Gamma_k}G(z_k,z_j)+\frac{2\pi}{\Gamma_k}U^*(z_k)

\displaystyle c_1(z_k)=h_1(z_k)+\sum_{j\neq k}\frac{4\pi\Gamma_j}{\Gamma_k}\frac{\partial G(z_k,z_j)}{\partial z_k}+\frac{4\pi}{\Gamma_k}\frac{\partial U^*(z_k)}{\partial z_k}

\displaystyle c_2(z_k)=h_2(z_k)+\sum_{j\neq k}\frac{2\pi\Gamma_j}{\Gamma_k}\frac{\partial^2 G(z_k,z_j)}{\partial z_k^2}+\frac{2\pi}{\Gamma_k}\frac{\partial^2 U^*(z_k)}{\partial z_k^2}

We note that the expansion is dependent on the coordinate system. This means that -2c_1(z)dz is not a proper one-form but rather an affine connection. If \tilde{z}=f(z) is a holomorphic coordinate change then -2c_1(z)dz transforms as -2\tilde{c}_1(\tilde{z})d\tilde{z}=-2c_1(z)dz-\{\tilde{z},z\}_1dz, where \{\tilde{z},z\}_1=f''/f' is the affine derivative of f.

The differential of \Phi is then expanded as

\displaystyle d\Phi=\frac{\Gamma_k}{2\pi i}\left(\frac{1}{z-z_k}-c_1(z_k)+O(z-z_k)\right)dz

The complex flow vector field is given by using the metric to convert the conjugate of the differential component of \Phi to the component of the vector field

v(z) = \displaystyle \frac{1}{\lambda(z)^2}\overline{\frac{d\Phi}{dz}}.

It is reasonable to expect a vortex to flow along the fluid, but since the velocity field is singular at z=z_k we investigate this further by expanding the conjugated derivative of \Phi at z=z_k

\displaystyle \frac{1}{\lambda(z)^2}\overline{\frac{d\Phi}{dz}} =\frac{\Gamma_k}{4\pi i\lambda(z)^2}\left(-\frac{1}{\bar{z}-\bar{z}_k}+\overline{c_1(z_k)}+O(\bar{z}-\bar{z}_k)\right)

We might to try to renormalise this expression by simply discarding the complex pole term and then letting z\to z_k, keeping only the finite part to be the velocity of the vortex point

\displaystyle \dot{z}_k\stackrel{?}{=}\frac{\Gamma_k}{4\pi i\lambda(z_k)^2}\overline{c_1(z_k)}

But since c_1(z)dz is not a proper one-form this won’t do, the calculation will be dependent on the coordinate system. We remedy this by noting that there is a special coordinate system: the normal coordinate system in which \lambda(z)=1. We thus choose the renormalisation to be the c_1 in the normal coordinate system, treat this as the component of a one-form and translate it to our original coordinate system. We then get

\displaystyle \dot{z}_k=\frac{\Gamma_k}{4\pi i\lambda(z_k)^2}\left(\overline{c_1(z_k)}+\frac{\partial\log \lambda(z_k)}{\partial \bar{z}_k}\right)

This can be written as

\displaystyle \dot{z}_k=\frac{\Gamma_k}{8\pi i\lambda(z_k)^2}\left(\overline{r_{\mathrm{metric}}(z_k)}-\overline{r_{\mathrm{robin}}(z_k)}\right)

where  r_{\mathrm{metric}}(z)=2\partial\log \lambda(z)/\partial z and r_{\mathrm{robin}}(z)=-2c_1(z). These are both affine connection component functions and since the difference between two affine connections is a proper one-form we see that our renormalised expression for the vortex motion is well-defined.

We now continue to investigate the motion of dipole vortices on the Riemann surface M.

dipole

On the surface we specify n dipole vortices located at points z_1,\dots,z_n with strengths and orientations given by complex numbers dz_1,\dots, dz_n. We also specify a steady background flow defined by 2g periods a_1,\dots,a_g and b_1,\dots,b_g.

The absolute strength of a dipole vortex at z_k is \lambda(z_k)|dz_k| and the orientation is \arg dz_k. If \tilde{z}=f(z) is a holomorphic coordinate change then the strength-and-orientation numbers transform according to d\tilde{z}_k=f'(z_k)dz_k.

We let d_a mean the exterior derivative with respect to a. The a-differential of a Green function d_aG=d_aG(z,a,da) is then a stream function on the Riemann surface depending on a and da where da is a complex variable denoting the strength-and-orientation of the stream field.

We have that

\displaystyle \Delta d_a G=d_a\Delta G=d_a\left(\delta_a-\frac{1}{\text{area}(M)}\right)=d_a\delta_a=-\frac{\partial\delta_a}{\partial z}da-\frac{\partial\delta_a}{\partial\bar{z}}d\bar{a}

where d\bar{a} is the conjugate value of da. This means that -d_aG can be understood as a “Green function” of a dipole. We also see that -d_aG is harmonic for z \neq a. The decomposition of -d_aG

\displaystyle -d_aG=-\frac{\partial G}{\partial a}da-\frac{\partial G}{\partial \bar{a}}d\bar{a}

is then the decomposition of -d_aG into a meromorphic function -(\partial G/\partial a)da and an antimeromorphic function -(\partial G/\partial\bar{a})d\bar{a}=-\overline{(\partial G/\partial a)da}. This means that -2i(\partial G/\partial a)da is the complex potential of the flow.

The dipole complex potential has a pole at z=a and is expanded as

\displaystyle -2i\frac{\partial G}{\partial a}da=-\frac{da}{2\pi i}\left(\frac{1}{z-a}-2\frac{\partial h_0(a)}{\partial a}+h_1(a)-\left(\frac{\partial h_1(a)}{\partial a}-2h_2(a)\right)(z-a)+O((z-a)^2)\right)

The complex potential \Phi(z) of the total flow is constructed from the differentiated Green function and the periods by

\displaystyle\Phi(z)=\sum_{k=1}^n-2idz_k\frac{\partial G(z,z_k)}{\partial z_k}+\hat{U}(z),

where \hat{U}(z) is the potential of the background flow defined by the given periods a_1,\dots,a_g and b_1,\dots,b_g. At z=z_k the complex potential can be expanded as

\displaystyle\Phi(z) = -\frac{dz_k}{2\pi i}\left(\frac{1}{z-z_k}-2\frac{\partial c_0(z_k)}{\partial z_k}+c_1(z_k)-\left(\frac{\partial c_1(z_k)}{\partial z_k}-2c_2(z_k)\right)(z-z_k)+ O((z-z_k)^2)\right)

where

\displaystyle c_0(z_k)=h_0(z_k)+\sum_{j\neq k}\frac{2\pi dz_j}{dz_k}G(z_k,z_j)+\frac{2\pi}{dz_k}U^*(z_k)

\displaystyle c_1(z_k)=h_1(z_k)+\sum_{j\neq k}\frac{4\pi dz_j}{dz_k}\frac{\partial G(z_k,z_j)}{\partial z_k}+\frac{4\pi}{dz_k}\frac{\partial U^*(z_k)}{\partial z_k}

\displaystyle c_2(z_k)=h_2(z_k)+\sum_{j\neq k}\frac{2\pi dz_j}{dz_k}\frac{\partial^2 G(z_k,z_j)}{\partial z_k^2}+\frac{2\pi}{dz_k}\frac{\partial^2 U^*(z_k)}{\partial z_k^2}.

The differential of \Phi is then expanded as

\displaystyle d\Phi=-\frac{dz_k}{2\pi i}\left(-\frac{1}{(z-z_k)^2}-\frac{\partial c_1(z_k)}{\partial z_k}+2c_2(z_k)+O(z-z_k)\right)dz

The expansion is dependent on the coordinate system. So the form -6(\partial c_1(z)/\partial z-2c_2(z))dz^2 is not a proper differential quadratic form but rather a projective connection. If \tilde{z}=f(z) is a holomorphic coordinate change then -6(\partial c_1(z)/\partial z-2c_2(z))dz^2 transforms as -6(\partial c_1(\tilde{z})/\partial\tilde{z}-2c_2(\tilde{z}))d\tilde{z}^2=-6(\partial c_1(z)/\partial z-2c_2(z))dz^2-\{\tilde{z},z\}_2dz^2, where \{\tilde{z},z\}_2=f'''/f'-(3/2)(f''/f')^2 is the Schwarzian derivative of f.

The velocity field component at z is then

\displaystyle v(z)=\frac{1}{\lambda(z)^2}\overline{\frac{d\Phi}{dz}}=\frac{d\bar{z}_k}{2\pi i\lambda(z)^2}\left(-\frac{1}{(\bar{z}-\bar{z}_k)^2}-\overline{\frac{\partial c_1(z_k)}{\partial z_k}}+2\overline{c_2(z_k)}+O(\bar{z}-\bar{z}_k)\right)

where d\bar{z}_k is the conjugate value of dz_k. We renormalise the velocity at z=z_k by choosing the expansion to be in the normal coordinate system, discarding the pole, treating \partial c_1/\partial z-2c_2 to be the component of a differential quadratic form and then translating it to the original coordinate system. We get

\displaystyle \dot{z}_k=\frac{d\bar{z}_k}{2\pi i\lambda(z_k)^2}\left(-\overline{\frac{\partial c_1(z_k)}{\partial z_k}}+\overline{2c_2(z_k)}-\frac{1}{3}\frac{\partial^2\log \lambda(z_k)}{\partial \bar{z}_k^2}+\frac{1}{3}\left(\frac{\partial\log \lambda(z_k)}{\partial \bar{z}_k}\right)^2\right)

which can be written as

\displaystyle \dot{z}_k=\frac{d\bar{z}_k}{12\pi i\lambda(z_k)^2}\left(\overline{q_{\mathrm{robin}}(z_k)}-\overline{q_{\mathrm{metric}}(z_k)}\right)

where q_{\mathrm{metric}}(z)=2\partial^2\log \lambda(z_k)/\partial z_k^2-2\left(\partial\log \lambda(z_k)/\partial z_k\right)^2 and q_{\mathrm{robin}}(z)=-6(\partial c_1(z)/\partial z-2c_2(z)). These are both projective connection component functions and since the difference between two projective connections is a differential quadratic form we see that the dipole vortex motion is indeed well-defined.

It is reasonable to expect the strength-and-orientation numbers dz_k to be affected by the flow. The dz_k is equivalent to the component of a tangent vector and so its time evolution should be governed by the acceleration vector at the point, \dot{dz}_k=\ddot{z}_k. To investigate this further we specialise to the case where there is only one dipole vortex at z=a with strength-and-orientation number da and no background flow. The complex potential is then

\displaystyle\Phi(z) = -\frac{da}{2\pi i}\left(\frac{1}{z-a}-2\frac{\partial h_0(a)}{\partial a}+h_1(a)-\left(\frac{\partial h_1(a)}{\partial a}-2h_2(a)\right)(z-a)+O((z-a)^2)\right)

and the motion of the dipole is given by

\displaystyle \dot{a}=d\bar{a}\Lambda(a)

where

\displaystyle \Lambda(a)=\frac{1}{2\pi i\lambda(a)^2}\left(-\overline{\frac{\partial h_1(a)}{\partial a}}+\overline{2h_2(a)}-\frac{1}{3}\frac{\partial^2\log \lambda(a)}{\partial \bar{a}^2}+\frac{1}{3}\left(\frac{\partial\log \lambda(a)}{\partial \bar{a}}\right)^2\right)

By setting \dot{da}=\ddot{a} and substituting \dot{a}=d\bar{a}\Lambda(a) we get the equation

\displaystyle \dot{da}=\dot{d\bar{a}}\Lambda+d\bar{a}(\Lambda_a \dot{a}+\Lambda_{\bar{a}}\dot{\bar{a}})=

=\dot{d\bar{a}}\Lambda+d\bar{a}(\Lambda_a d\bar{a}\Lambda+\Lambda_{\bar{a}}da\bar{\Lambda})

=\dot{d\bar{a}}\Lambda+\Lambda\Lambda_a d\bar{a}^2+\overline{\Lambda}\Lambda_{\bar{a}}dad\bar{a},

where \Lambda_a=\partial\Lambda/\partial a and \Lambda_{\bar{a}}=\partial\Lambda/\partial\bar{a}. Solving for \dot{da} gives

\displaystyle \dot{da}=\frac{\Lambda(\overline{\Lambda}\overline{\Lambda_a} da^2+\Lambda\overline{\Lambda_{\bar{a}}}dad\bar{a})+\Lambda\Lambda_a d\bar{a}^2+\overline{\Lambda}\Lambda_{\bar{a}}dad\bar{a}}{1-|\Lambda|^2}=

\displaystyle =\frac{|\Lambda|^2\overline{\Lambda_a}}{1-|\Lambda|^2}da^2+\frac{\Lambda^2\overline{\Lambda_{\bar{a}}}+\overline{\Lambda}\Lambda_{\bar{a}}}{1-|\Lambda|^2}dad\bar{a}+\frac{\Lambda\Lambda_a}{1-|\Lambda|^2}d\bar{a}^2

If we instead substitute da=\dot{\bar{a}}/\overline{\Lambda} we can rewrite this as

\displaystyle \ddot{a}=\frac{\Lambda_a}{\Lambda(1-|\Lambda|^2)}\dot{a}^2+\frac{\Lambda^2\overline{\Lambda_{\bar{a}}}+\overline{\Lambda}\Lambda_{\bar{a}}}{|\Lambda|^2(1-|\Lambda|^2)}\dot{a}\dot{\bar{a}}+\frac{\Lambda\overline{\Lambda_a}}{\overline{\Lambda}(1-|\Lambda|^2)}\dot{\bar{a}}^2

This is a geodesic equation with Christoffel symbols

\displaystyle \Gamma_{11}=\frac{\Lambda_a}{\Lambda(|\Lambda|^2-1)}

\displaystyle \Gamma_{1\bar{1}}=\Gamma_{\bar{1}1}=\frac{\Lambda^2\overline{\Lambda_{\bar{a}}}+\overline{\Lambda}\Lambda_{\bar{a}}}{2|\Lambda|^2(|\Lambda|^2-1)}

\displaystyle \Gamma_{\bar{1}\bar{1}}=\frac{\Lambda\overline{\Lambda_a}}{\overline{\Lambda}(|\Lambda|^2-1)}

We conclude that the dipole moves on a geodesic of a connection generated by the Green function.

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The Gelfand-Fuchs cocycle

Posted on 10 July, 2016 by jesperg

Let d be a vector field on the Euclidean line \mathbf{E}^1. Expressed in a coordinate x the vector field is d=v(x)d/dx. The logarithm of the component value at each point of the vector field is the fundamental Euclidean invariant

\displaystyle i_0 = \ln|v(x)|.

We now let dx denote the component of the vector field, dx=v(x) and so d=dx\,d/dx. Let y=f(x) be a diffeomorphism of the Euclidean line. The vector field is then pushed-forward by dy=f'dx. By taking the difference between the fundamental invariants of the pushed-forward vector field and the original vector field,

\displaystyle I_0 = \ln|dy|-\ln|dx|=\ln|f'|,

we recover the logarithm of the derivative as the fundamental Euclidean invariant of a diffeomorphism.

From the invariant i_0 further invariants can be constructed by differentiating with respect to the vectorfield

\displaystyle i_1 = di_0 = d\,\ln|dx| = \frac{d^{2}x}{dx}\, (\, =\, v' )

Let \partial=\partial x\, \partial/\partial x be another vector field. Then we have that

\displaystyle\begin{aligned} i_1([d,\partial ]) &= i_1(d\partial-\partial d)=\frac{(d\partial-\partial d)}{(d\partial-\partial d)x}(d\partial-\partial d)x=\\ &=\frac{d}{dx}d\partial x-\frac{\partial}{\partial x}\partial dx =\\ &= \frac{d}{dx}dx\frac{\partial}{\partial x}\partial x-\frac{\partial}{\partial x}\partial x\frac{d}{dx} dx =\\ &=d\frac{\partial^2x}{\partial x}-\partial\frac{d^2x}{dx}=\\ &=d\, i_1( \partial)-\partial\, i_1(d),\end{aligned}

which shows that i_1 is a cocycle on \mathrm{Vect}(\mathbf{E}^1).

The invariant i_1 is also an invariant for vector fields on the affine line \mathbf{A}^1. For a diffeomorphism y=f(x) of the affine line we get an invariant one-form by taking the difference of the invariants of the pushed-forward and original vector fields:

\displaystyle I_1= \frac{d^{2}y}{dy}-\frac{d^{2}x}{dx}=\frac{f''}{f'}dx.

By differentiating i_1 with respect to the vector field we get another affine invariant

\displaystyle i_2=di_1=d\frac{d^{2}x}{dx}=\frac{d^{3}x}{dx}-\frac{d^2x^2}{dx^2}\, (\, =v''dx\, )

This is also a cocycle but instead of showing this by direct calculation, we assume that the diffeomorphism f is generated by the flow the of the vector field

\displaystyle f(x) = e^{\lambda d}x = x+\lambda\, dx+O(\lambda^2),

where \lambda is a fixed parameter of the flow. Then

\displaystyle\begin{aligned}I_1(f)&=i_1(df)-i_1(dx)=i_1(dx+\lambda\, d^2x+O(\lambda^2))-i_1(dx)=\\ &=\lambda\, di_1(dx)+O(\lambda^2)=\lambda\, i_2(dx)+O(\lambda^2),\end{aligned}

which shows that i_2 is the first order term of I_1 and since I_1 is a diffeomorphism group cocycle, it follows that i_2 is its corresponding vector field algebra cocycle.

We now consider a vector field d=v(x)d/dx=dx\, d/dx on the projective line \mathbf{P}^1 and lift it to homogeneous coordinates (x_1:x_2) by

\displaystyle\begin{aligned} x_1&=\frac{x}{\sqrt{|dx|}}\\ x_2&=\frac{1}{\sqrt{|dx|}}\end{aligned}

This gives a constant area-velocity vector field which means that the acceleration vector field is a central field. The ratio between the acceleration vector and the position vector then gives the fundamental projective invariant of this vector field:

\displaystyle s_0=\frac{d^2x_1}{x_1}=\frac{d^2x_2}{x_2}=\frac{d^3x}{dx}-\frac{3}{2}\frac{d^2x^2}{dx^2}\, \left(\, =v''v-\frac{v'^2}{2}\, \right)

Again, by pushing forward the vector field by a diffeomorphism y=f(x) and taking the difference of the invariants we get an invariant of the diffeomorphism, the Schwarzian derivative:

\displaystyle S_0(f)=s_0(dy)-s_0(dx)=\frac{d^3y}{dy}-\frac{3}{2}\frac{d^2y^2}{dy^2}-\frac{d^3x}{dx}+\frac{3}{2}\frac{d^2x^2}{dx^2}=\left(\frac{f'''}{f'}-\frac{3}{2}\left(\frac{f''}{f'}\right)^2\right)dx^2.

Differentiating s_0 gives another projective invariant of the vector field

\displaystyle s_1=ds_0=\frac{d^4x}{dx}-4\frac{d^3xd^2x}{dx^2}+3\frac{d^2x^3}{dx^3}\, \left(\,=\frac{v'''}{2}dx^2\,\right)

And this invariant then has the interpretation that it gives the intrinsic change of the acceleration/position ratio of the lifted vector field.

Since s_1 is related to the cocycle S_0 in the same way as the cocycle i_2 is related to the cocycle I_1 (and also i_1 to I_0) it follows that s_1 is a cocycle. It is the vector field algebra cocycle counterpart of the Schwarzian diffeomorphism group cocycle, the Gelfand-Fuchs cocycle.

 

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The Schwarzian derivative

Posted on 14 September, 2015 by jesperg

For a diffeomorphism f:\mathbf{E}^1\rightarrow\mathbf{E}^1 between two Euclidean lines, the derivative f'(x)=\lim_{\Delta x\rightarrow 0}\Delta f/\Delta x is an invariant in the sense that to calculate the derivative we fix Euclidean coordinate systems on the domain and on the range and the calculated value will not depend on the choice of coordinate systems.

For a diffeomorphism f:\mathbf{A}^1\rightarrow\mathbf{A}^1 between affine lines, things are not that simple, the calculated value of the derivative f' will be dependent on the affine coordinate systems it is computed in. To find an invariant expression we look at the fundamental invariant of the affine line, the three-point ratio:

\displaystyle [x,x_1,x_\alpha]=\frac{x_\alpha-x}{x_1-x}

We put x_1=x+dx and x_\alpha=x+\alpha dx. Then [x,x_1,x_\alpha]=\alpha and the ratio of the images of the three points is then

\displaystyle \begin{aligned}[] [f(x),&f(x_1),f(x_\alpha)] =\frac{f(x+\alpha dx)-f(x)}{f(x+dx)-f(x)}\\ &=\frac{\alpha f'(x)dx+(\alpha^2/2)f''(x)dx^2+O(dx^3)}{f'(x)dx+(1/2)f''(x)dx^2+O(dx^3)}\\ &= \alpha+\frac{((\alpha^2-\alpha)/2)f''dx^2+O(dx^3)}{f'dx+O(dx^2)}\\ &= \alpha+\frac{\alpha^2-\alpha}{2}\frac{f''}{f'}dx+O(dx^2)\end{aligned}

and so the one-form \omega=(f''/f')dx is an invariant that measures how much the diffeomorphism deviates from being an affine transformation.

Now let y be a coordinate on the range and thus y=f(x). We can express the derivatives of f as

\begin{aligned} f'&=\frac{dy}{dx}\\f''&=\frac{d^2ydx-d^2xdy}{dx^3}\end{aligned}

and the invariant one-form can be reexpressed as \omega=(f''/f')dx=d^2y/dy-d^2x/dx. Each of the two quotients is invariant under affine coordinate changes so in fact any function g(d^2y/dy,d^2x/dx) would define an invariant of the diffeomorphism, but it is only by taking the difference that the second order differentials cancel and we get an invariant that is a one-form on the tangent space.

Let’s carry out this calculation, which is in a sense the last calculation carried out backwards. Since the second differential of f is d^2f=f''dx^2+f'd^2x, we get

\displaystyle \frac{d^2y}{dy}-\frac{d^2x}{dx}=\frac{f''dx^2+f'd^2x}{f'dx}-\frac{d^2x}{dx}=\frac{f''}{f'}dx

as expected.

Moving on to instead study a diffeomorphism f:\mathbf{P}^1\rightarrow\mathbf{P}^1 between projective lines, we now want to see how the cross ratio

\displaystyle [x,x_1,x_a,x_{a^2}]=\left.\frac{x_a-x}{x_1-x}\middle/\frac{x_a-x_{a^2}}{x_1-x_{a^2}}\right.

changes under the diffeomorphism. We put x_1=x+dx, x_a=x+adx, x_{a^2}=x+a^2dx and calculate that [x,x_1,x_a,x_{a^2}]=1+a. On the range space we want to make use of the availability of homogenous coordinates and define a lift of f to a vector \mathbf{v}=(u,v) by

\displaystyle \begin{aligned}u&=\frac{f}{\sqrt{f'}}\\v&=\frac{1}{\sqrt{f'}}\end{aligned}.

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This lift has the property that the curve has constant angular momentum \mathbf{v}'\times\mathbf{v}=u'v-uv'=1 and thus the acceleration vector is parallel to the position vector \mathbf{v}''=\lambda(x)\mathbf{v}. The cross ratio can be calculated in homogenous coordinates as

\displaystyle [\mathbf{v},\mathbf{v}_1,\mathbf{v}_a,\mathbf{v}_{a^2}]=\frac{(\mathbf{v}_a\times\mathbf{v})(\mathbf{v}_1\times\mathbf{v}_{a^2})}{(\mathbf{v}_1\times\mathbf{v})(\mathbf{v}_a\times\mathbf{v}_{a^2})}.

We have that

\displaystyle \begin{aligned} \mathbf{v}_1&=\mathbf{v}(x+dx)=\mathbf{v}+\mathbf{v}'dx+\frac{1}{2}\mathbf{v}''dx^2+\frac{1}{6}\mathbf{v}'''dx^3+O(dx^4)\\ &=\left(1+\frac{\lambda}{2}dx^2+\frac{\lambda'}{6}dx^3\right)\mathbf{v}+\left(dx+\frac{\lambda}{6}dx^3\right)\mathbf{v}'+O(dx^4)\\ \mathbf{v}_a&=\mathbf{v}(x+adx)\\ &=\left(1+\frac{\lambda}{2}a^2dx^2+\frac{\lambda'}{6}a^3dx^3\right)\mathbf{v}+\left(adx+\frac{\lambda}{6}a^3dx^3\right)\mathbf{v}'+O(dx^4)\\ \mathbf{v}_{a^2}&=\mathbf{v}(x+a^2dx)\\ &=\left(1+\frac{\lambda}{2}a^4dx^2+\frac{\lambda'}{6}a^6dx^3\right)\mathbf{v}+\left(a^2dx+\frac{\lambda}{6}a^6dx^3\right)\mathbf{v}'+O(dx^4)\end{aligned}

The cross products become

\displaystyle \begin{aligned}\mathbf{v}_1\times\mathbf{v}&=dx+\frac{1}{6}\lambda dx^3+O(dx^4)\\ \mathbf{v}_a\times\mathbf{v}&=adx+\frac{a^3}{6}\lambda dx^3+O(dx^4)\\ \mathbf{v}_1\times\mathbf{v}_{a^2}&=(1-a^2)dx+\left(\frac{1}{6}-\frac{a^2}{2}+\frac{a^4}{2}-\frac{a^6}{6}\right)\lambda dx^3+O(dx^4)\\ \mathbf{v}_a\times\mathbf{v}_{a^2}&=(a-a^2)dx+\left(\frac{a^3}{6}-\frac{a^4}{2}+\frac{a^5}{2}-\frac{a^6}{6}\right)\lambda dx^3+O(dx^4)\end{aligned}

and the cross ratio of the image points is

\displaystyle \begin{aligned}[][\mathbf{v},\mathbf{v}_1,\mathbf{v}_a,\mathbf{v}_{a^2}]&=\frac{1-a^2+\left(\frac{1}{6}-\frac{a^2}{3}+\frac{a^4}{3}-\frac{a^5}{6} \right) \lambda dx^2+O(dx^3)}{1-a+\left(\frac{1}{6}-\frac{a}{6}+\frac{a^2}{6}-\frac{a^3}{2}+\frac{a^4}{2}-\frac{a^5}{6}\right) \lambda dx^2+O(dx^3)}\\&=1+a+\left( -\frac{a^2}{3}+\frac{3a^4-2a^5}{6-6a} \right) \lambda dx^2+O(dx^3)\end{aligned}.

So we get a quadratic form, the Schwarzian derivative, S(f)=-2\lambda dx^2 that tells us how much the diffeomorphism deviates from being a projective transformation. The coefficient \lambda is calculated as

\displaystyle\lambda = \frac{u''}{u}=\frac{v''}{v}=-\frac{1}{2}\frac{f'''f'-\frac{3}{2}(f'')^2}{(f')^2}

and the Schwarzian derivative when directly expressed in derivatives of f becomes

\displaystyle S(f)=\left(\frac{f'''}{f'}-\frac{3}{2}\left(\frac{f''}{f'}\right)^2\right)dx^2.

We have already given above the first and second derivative of f expressed as differentials of x and y. The third derivative then can also be reexpressed as

\displaystyle f'''=\frac{d^3ydx^2-d^3xdydx-3d^2yd^2xdx+3d^2x^2dy}{dx^5}

and the Schwarzian derivative then becomes

\displaystyle S(f)=\frac{d^3y}{dy}-\frac{3}{2}\frac{d^2y^2}{dy^2}-\frac{d^3x}{dx}+\frac{3}{2}\frac{d^2x^2}{dx^2}.

Thus we see, just as in the affine case, that the invariant form for the diffeomorphism is a difference between two separately invariant higher order forms, one on the range and one on the domain.

We now define s(x)=d^3x/dx-(3/2)d^2x^2/dx^2 and s(y|x)=s(y)-s(x). For a composition of diffeomorphisms f=g\circ h, we get

S(f)=s(y|x)=s(y)-s(x)=s(y)-s(h)+s(h)-s(x)=s(y|h)+s(h|x)=S(g)+S(h)

and we have shown the cocycle condition for the Schwarzian derivative.

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The Lie bracket is torsion

Posted on 20 June, 2015 by jesperg

We start with a Lie group G. An element g\in G acts on an external vector x_0 by x=gx_0. Let g be generated by a vector de at the identity e, g(t)=e^{tde}. By differentiating, with dt=1, the velocity of g is

dg=e^{tde}de=gde.

This can also be interpreted that dg is the parallel transport of de to g. The velocity of x can then be expressed as

dx=dgx_0=gdex_0=degx_0=dex,

so we see that the action of an element of the Lie algebra on a vector generates a velocity of this vector.

Now let \partial be differentiation in some independent direction and let \partial e be a corresponding vector in the Lie algebra. The \partial-derivative of de is then multiplication with \partial e:

\partial de=\partial ede.

Swapping the order of differentiation gives

d\partial e=de\partial e

and combining these gives

(d\partial-\partial d)e=de\partial e-de\partial e.

The operator d\partial-\partial d is the exterior (covariant) derivative squared D^2 and computes the torsion of the action (i.e. the connection) of the Lie group. The righthand side is of course the usual commutator in the Lie algebra. We can thus reexpress the last expression as

D^2e=[de,\partial e].

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The cross ratio

Posted on 10 May, 2015 by jesperg

The purpose of this post is to show how the cross ratio can be understood as a generalisation of the corresponding invariants in Euclidean and affine geometry and how it is closely related to the concept of a point frame. For the Euclidean line \mathbf{E}^1, a point frame is a single point x_0 and it allows us to specify a Euclidean invariant of another point x with respect to x_0:

f(x,x_0)=x-x_0.

So if we consider x and x_0 to be coordinates in a Euclidean coordinate system, then f(x,x_0) is simply the coordinate of x in the coordinate system with x_0 as its origin. This is invariant under non-reflection Euclidean coordinate changes.

Moving on to the affine line \mathbf{A}^1, we want to make a corresponding construction that is invariant under affine coordinate changes. Thus we extend the point frame into an affine point frame which consists of two points \{x_0,x_1\} and define an invariant of a point x with respect to this frame to be

\displaystyle f(x,x_0,x_1)=\frac{x-x_0}{x_1-x_0}.

This can be interpreted to be the coordinate of x in the coordinate system with x_0 as its origin and x_1 as its unit point.

For the projective line \mathbf{P}^1 we add a third point x_\infty which makes the frame into a projective frame \{x_0,x_1,x_\infty\}. The point x_\infty is to be understood as the point at infinity and allows us to define an affine coordinate system with x_0 as its origin and x_1 as its unit point. The coordinate of some point x in this coordinate system is

\displaystyle f(x,x_0,x_1,x_\infty)=\left.\frac{x-x_0}{x_1-x_0}\middle/\frac{x-x_\infty}{x_1-x_\infty}\right. .

Thus the cross ratio of four points is the coordinate value of one of the points with respect to the projective frame defined by the other three points. The projective frame gives the additional structure missing in the projective geometry needed to be able to compute an invariant coordinate value and specialises the projective line into an Euclidean line.

A projective frame with three points which we now denote by \{\mathbf{x}_1,\mathbf{x}_2, \mathbf{x}_\text{mid}\} uniquely determines an element of PGL(2). This matrix is constructed by choosing homogenous coordinates of the three points (x_1,y_1), (x_2,y_2), (x_\text{mid},y_\text{mid}) rescaled such that

\displaystyle \begin{pmatrix}x_1\\y_1\end{pmatrix}+\begin{pmatrix}x_2\\y_2\end{pmatrix}=\begin{pmatrix}x_\text{mid}\\y_\text{mid}\end{pmatrix}.

Then the corresponding element in PGL(2) is

\displaystyle \begin{pmatrix}x_1 & x_2\\y_1 & y_2\end{pmatrix}.

Whether we interpret the third point in a projective frame as a point at infinity or as a midpoint of the other two points depends on the circumstance and is a matter of what is most convenient. The two interpretations are equivalent since the midpoint and the infinity point are harmonic conjugates with respect to the two other points of the frame. If we want to emphasis the equivalence of the midpoint and the infinity point we can include both in the frame and thus define a projective frame to be four points with a harmonic cross ratio.

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