User:LyleRamshaw/sandbox

Metrics on planar domains
Consider a 2-dimensional Riemannian manifold, say with an (x, y) coordinate system on it. The curves of constant x on that surface typically don't intersect the curves of constant y orthogonally. A new coordinate system (u, v) is called isothermal when the curves of constant u do intersect the curves of constant v orthogonally and, in addition, the parameter spacing is the same — that is, for small enough h, the little region with $$a\le u\le a+h$$ and $$b\le v\le b+h$$ is nearly square, not just nearly rectangular. The Beltrami equation is the equation that has to be solved in order to construct isothermal coordinate systems.

To see how this works, let S be an open set in C and let


 * $$ \displaystyle{ds^2=E\,dx^2 + 2F\, dxdy + G\, dy^2}$$

be a smooth metric g on S. The first fundamental form of g


 * $$\displaystyle{g(x,y)=\begin{pmatrix} E & F \\ F & G\end{pmatrix}}$$

is a positive real matrix (E > 0, G > 0, EG − F2 > 0) that varies smoothly with x and y.

The Beltrami coefficient of the metric g is defined to be


 * $$\displaystyle{\mu(x,y)={E-G +2iF\over E+G +2\sqrt{EG-F^2}}}$$

This coefficient has modulus strictly less than one since the identity


 * $$\displaystyle (E-G+2iF)(E-G-2iF)=(E+G+2\sqrt{EG-F^2})(E+G-2\sqrt{EG-F^2})$$

implies that


 * $$\displaystyle{|\mu|^2={E+G - 2\sqrt{EG -F^2}\over E+G + 2\sqrt{EG-F^2}}<1.}$$

Let f(x,y) =(u(x,y),v(x,y)) be a smooth diffeomorphism of S onto another open set T in C. The map f preserves orientation just when its Jacobian is positive:


 * $$\displaystyle u_x v_y-v_x u_y>0.$$

And using f to pull back to S the standard Euclidean metric ds2 = du2 + dv2 on T induces a metric on S given by


 * $$\displaystyle{ds^2=du^2 + dv^2=(u_x^2+v_x^2)\, dx^2 + 2 (u_xu_y + v_xv_y)\, dx dy + (u_y^2 + v_y^2)\, dy^2,}$$

a metric whose first fundamental form is


 * $$\displaystyle \begin{pmatrix} u_x^2+v_x^2 & u_x u_y+v_x v_y \\ u_x u_y+v_x v_y & u_y^2+v_y^2\end{pmatrix}.$$

When f both preserves orientation and induces a metric that differs from the original metric g only by a positive, smoothly varying scale factor r(x, y), the new coordinates u and v defined on S by f are called isothermal coordinates.

To determine when this happens, we reinterpret f as a complex-valued function of a complex variable f(x+iy) = u(x+iy) + iv(x+iy) so that we can apply the Wirtinger derivatives:


 * $$\displaystyle{\partial_z= {1\over 2} (\partial_x-i\partial_y),\,\, \partial_{\overline{z}}={1\over 2} (\partial_x +i\partial_y);\,\,\,\,dz=dx+i\,dy,\,\,d\overline{z}=dx-i\,dy.}$$

Since


 * $$\displaystyle f_z=((u_x+v_y)+i(v_x-u_y))/2$$
 * $$\displaystyle f_{\overline{z}}=((u_x-v_y)+i(v_x+u_y))/2,$$

the metric induced by f is given by


 * $$\displaystyle{ds^2=|f_z dz + f_{\overline{ z}}d\overline{z}|^2= |f_z|^2\, |dz + {f_{\overline{z}}\over f_z} d\overline{z}|^2.}$$

The Beltrami quotient of this induced metric is defined to be $$f_{\overline{z}}/f_z$$.

The Beltrami quotient $$f_{\overline{z}}/f_z$$ of $$f$$ equals the Beltrami coefficient $$\mu(z)$$ of the original metric g just when


 * $$\displaystyle ((u_x+v_y)+i(v_x-u_y))\bigl(E-G+2iF\bigr)$$
 * $$=((u_x-v_y)+i(v_x+u_y))\bigl(E+G+2\sqrt{EG-F^2}\bigr).$$

The real and imaginary parts of this identity linearly relate $$u_x,$$ $$u_y,$$ $$v_x,$$ and $$v_y,$$ and solving for $$u_y$$ and $$v_y$$ gives


 * $$\displaystyle u_y=\frac{F u_x-\sqrt{EG-F^2}\,v_x}{e}\quad\text{and}

\quad v_y=\frac{\sqrt{EG-F^2}\,u_x+F v_x}{e}.$$

It follows that the metric induced by f is then r(x, y) g(x,y), where $$r=(u_x^2+v_x^2)/e,$$ which is positive, while the Jacobian of f is then $$r\sqrt{EG-F^2},$$ which is also positive. So, when $$f_{\overline{z}}/f_z=\mu(z),$$ the new coordinate system given by f is isothermal.

Conversely, consider a diffeomorphiam f that does give us isothermal coordinates. We then have


 * $$\displaystyle \mu(z)=

\frac{(u_x^2+v_x^2)-(u_y+v_y)^2+2i(u_xu_y+v_xv_y)} {(u_x^2+v_x^2)+(u_y^2+v_y)^2+2\sqrt{(u_x^2+v_x^2)(u_y^2+v_y^2)-(u_xu_y+v_xv_y)^2}},$$

where the scale factor r(x, y) has dropped out and the expression inside the square root is the perfect square $$u_x^2v_y^2-2u_xv_xu_yv_y+v_x^2u_y^2.$$ Since f must preserve orientation to give isothermal coordinates, the Jacobian $$u_xv_y-v_xu_y$$ is the positive square root; so we have


 * $$\displaystyle \mu(z)=

\frac{\bigl((u_x+iu_y)+i(v_x+iv_y)\bigr)\bigl((u_x+iu_y)-i(v_x+iv_y)\bigr)} {\bigl((u_x+v_y)+i(v_x-u_y)\bigr)\bigl((u_x+v_y)-i(v_x-u_y)\bigr)}.$$

The right-hand factors in the numerator and denominator are equal and, since the Jacobian is positive, their common value can't be zero; so $$\mu(z)=f_{\overline{z}}/f_x.$$

Thus, the local coordinate system given by a diffeomorphism f is isothermal just when f solves the Beltrami equation for $$\mu(z).$$

Isothermal coordinates for analytic metrics
Gauss proved the existence of isothermal coordinates locally in the analytic case by reducing the Beltrami to an ordinary differential equation in the complex domain. Here is a cookbook presentation of Gauss's technique.

An isothermal coordinate system, say in a neighborhood of the origin (x, y) = (0, 0), is given by the real and imaginary parts of a complex-valued function f(x, y) that satisfies


 * $$\displaystyle{{f_{\overline{z}}\over f_z}=\mu(z)=\frac{E-G+2iF}{E+G+2\sqrt{EG-F^2}}.}$$

Let $$f$$ be such a function, and let $$\psi$$ be a complex-valued function of a complex variable that is holomorphic and whose derivative is nowhere zero. Since any holomorphic function $$\psi$$ has $$\psi_{\overline{z}}$$ identically zero, we have


 * $$\displaystyle \frac{(\psi\circ f)_{\overline{z}}}{(\psi\circ f)_z}

=\frac{(\psi_z\circ f)f_{\overline{z}}+(\psi_{\overline{z}}\circ f)\overline{f_{\overline{z}}}}{(\psi_z\circ f)f_z+(\psi_{\overline{z}}\circ f)\overline{f_z}} =\frac{f_{\overline{z}}}{f_z}=\mu(z).$$

Thus, the coordinate system given by the real and imaginary parts of $$\psi\circ f$$ is also isothermal. Indeed, if we fix $$f$$ to give one isothermal coordinate system, then all of the possible isothermal coordinate systems are given by $$\psi\circ f$$ for the various holomorphic $$\psi$$ with nonzero derivative.

When E, F, and G are real analytic, Gauss constructed a particular isothermal coordinate system $$h,$$ the one that he chose being the one with $$h(0,y)=iy$$ for all y. So the v (imaginary) axis of his isothermal coordinate system coincides with the y axis of the original coordinates and is parameterized in the same way. All other isothermal coordinate systems are then of the form $$\psi\circ h$$ for a holomorphic $$\psi$$ with nonzero derivative. (If you would rather have $$h(x,0)=x$$ for all x, so that your u axis coincides with the original x axis, you can swap x with y, apply the following, and then swap u with v.)

Gauss lets q(t) be some complex-valued function of a real variable t that satisfies the following ordinary differential equation:


 * $$\displaystyle q'(t)=\frac{-F+i\sqrt{EG-F^2}}{G}(t,q(t)),$$

where E, F, and G are here evaluated at x = t and y = q(t). If we specify the value of q(s) for some start value s, this differential equation determines the values of q(t) for t either less than or greater than s. Gauss then defines his isothermal coordinate system h by setting h(x, y) to be $$iq(0)$$ along the solution path of that differential equation that passes through the point (x, y), and thus has q(x) = y.

This rule sets h(0, y) to be $$iy$$, since the starting condition is then q(0)=y. More generally, suppose that we move by an infinitesimal vector (dx, dy) away from some point (x, y), where dx and dy satisfy


 * $$\displaystyle (F-i\sqrt{EG-F^2})\,dx+G\,dy=0.$$

Since $$q'(t)=dy/dx$$, the vector (dx, dy) is then tangent to the solution curve of the differential equation that passes through the point (x, y). Because we are assuming the metric to be analytic, it follows that


 * $$\displaystyle dh =\lambda\bigl((F-i\sqrt{EG-F^2})\,dx+G\,dy\bigr)$$

for some smooth, complex-valued function $$\lambda(x,y).$$ We thus have


 * $$\displaystyle h_z=\lambda(F-i\sqrt{EG-F^2}-iG)/2$$
 * $$\displaystyle h_{\overline{z}}=\lambda(F-i\sqrt{EG-F^2}+iG)/2,$$

from which it follows that


 * $$\displaystyle {h_{\overline{z}}\over h_z}=

\frac{(F-i\sqrt{EG-F^2}+iG)(F+i\sqrt{EG-F^2}+iG)} {(F-i\sqrt{EG-F^2}-iG)(F+i\sqrt{EG-F^2}+iG)}$$
 * $$\displaystyle =\frac{E-G+2iF}{E+G+2\sqrt{EG-F^2}}=\mu(z).$$

The function h thus gives isothermal coordinates.