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Complex geometry studies spaces locally described by complex coordinates with holomorphic transition maps. A complex $n$-dimensional complex manifold has real dimension $2n$, but its structure is more restrictive than a smooth atlas: the transition maps satisfy the Cauchy-Riemann equations. The complex projective line is a compact example, described by two complex charts related by $w=1/z$. Hermitian metrics and Kähler manifolds connect this analytic structure to Riemannian geometry, while complex algebraic varieties connect it to algebraic geometry.

A rotation of a circular annulus through an angle $\theta \notin 2\pi \mathbb{Z}$ is a continuous map with no fixed point. In complex coordinates it is $z\mapsto e^{i\theta }z$. Since an annulus excludes the center, $z\ne 0$, and the fixed-point equation $(e^{i\theta }-1)z=0$ has no solution. Thus the annulus lacks the fixed-point property.

The condition is nonzero modulo a full turn: an angle such as $2\pi$ is nonzero as a real number but gives the identity map, which fixes every point.

On the unit sphere, use $\zeta =(x+iy)/(1-z)$ away from the north pole and $\eta =(x-iy)/(1+z)$ away from the south pole. Their overlap relation $\eta =1/\zeta$ is a holomorphic map, giving a holomorphic atlas for the Riemann sphere. The opposite sign of $iy$ in the second manifold chart ensures a holomorphic rather than antiholomorphic transition.

Writing the sphere coordinates as $(X,Y,Z)$ avoids confusing their height with the complex coordinate $z$. The two manifold charts cover the whole unit sphere:On their overlap, $X^2+Y^2=1-Z^2$ givesThe coordinate transformation and its inverse are holomorphic, so these manifold charts make the sphere a Riemann surface. The sign in $X-iY$ is essential. Geometric projection from the south pole with the same planar axis labels would give $v=(X+iY)/(1+Z)=1/\overline{z}$. The complex conjugation in that transition prevents it from being holomorphic. For $z=i$, the correct second coordinate is $w=-i$: positive $Y$ corresponds to negative imaginary part in this manifold chart.

Robert Laughlin's 1983 variational method uses pairwise zeros to suppress close electron encounters within the lowest Landau level. For aligned spin, odd $m$ gives fermionic exchange symmetry.^[ref]

Here $\mathcal{N}$ enforces wavefunction normalization, ${\ell }_B$ is the magnetic length, and complex coordinates $z_j=x_j-i{y}_j$ describe electrons in a magnetic field $B\hat{\mathbf{z}}$, $B>0$, with the symmetric gauge. Reversing the orientation conjugates the polynomial without changing probability densities. The Gaussian function confines the wavefunction.

The exchange check is especially simple for two electrons. Writing $w=z_1-z_2$, exchange sends $w\mapsto -w$, so $w^m\mapsto (-1{)}^m{w}^m$. An aligned two-electron spin state is symmetric, requiring the spatial wavefunction to change sign. Hence $m=1,3,5,\dots$ works, whereas $m=2$ would give the wrong sign. At a close encounter, the probability density vanishes as $|w{|}^{2m}$: $m=3$ imposes a sixth-order zero, stronger than the quadratic zero at $m=1$. This is the local geometric mechanism for avoiding a large repulsive Coulomb potential energy.

A useful large-$N$ counting outline follows directly from the polynomial. Each complex coordinate has maximum degree $m(N-1)$. A lowest Landau level orbital proportional to $z^k{e}^{-|z{|}^2/(4{\ell }_B^2)}$ has its areal probability density maximum at $r^2=2k{\ell }_B^2$, by differentiating its logarithm. This suggests droplet area $A\simeq 2\pi mN{\ell }_B^2$, neglecting the boundary correction. Dividing $N/A$ by the degeneracy of a Landau level per area, $1/(2\pi {\ell }_B^2)$, gives filling factor $\nu \simeq 1/m$. This counting is a motivation for the fraction, not an independent proof that the wavefunction minimizes the interacting Hamiltonian.

Choose the positive Green function conventionAway from its pole, $G$ is harmonic, and the maximum principle for harmonic functions makes it positive inside $D$. The sign of its logarithmic singularity is important: it is the convention for $-\Delta$, whereas a Green function for $+\Delta$ has the opposite sign.

Green's second identity, justified by removing a small disk about the pole and taking its radius to zero, givesSince $\Delta u=-f$, $\Delta G=-{\delta }_{\xi }$ and $G$ vanishes on the outer boundary, this rearranges toThe local logarithmic singularity is integrable, so continuous $f$ gives a well-defined volume integral.

In complex coordinates $z=x+iy$, $\zeta =\xi +i\eta$ for the unit disk, useThe numerator has no zeros inside the disk, so its logarithm is harmonic there. On $|z|=1$, the numerator and denominator have equal modulus, giving the required zero boundary value. Differentiating in the outward radial direction at $z=e^{i\theta }$ givesWith zero forcing the representation therefore becomes the Poisson integralFinally replace $\eta$ by $-\eta$ and change variable $\theta \mapsto 2\pi -\theta$. Cosine stays the same, sine changes sign and the assumed boundary symmetry leaves $g$ unchanged. The integral is therefore invariant, so $\boxed{u(\xi ,-\eta )=u(\xi ,\eta )}$. Equivalently, reflection produces a harmonic function with the same Dirichlet data, and uniqueness identifies it with the original solution.

The equation of motion in a rotating frame with constant angular velocity $\mathbf{\omega }$ isWith $\mathbf{F}=-4m{\omega }^2\mathbf{x}$ and $\mathbf{\omega }=(0,0,\omega )$, this becomesThe given initial data make $z=0$. For the complex coordinate $\zeta =x+iy$, the planar equations combine intoIts characteristic roots are $i\omega$ and $-3i\omega$. Applying $\zeta (0)=1$, $\dot{\zeta }(0)=0$ gives $\zeta =\frac{3}{4}{e}^{i\omega t}+\frac{1}{4}{e}^{-3i\omega t}$. The triple-angle identities reduce this astroid motion of a harmonic oscillator in a rotating frame toIt traces an astroid. Differentiating gives the relative speedFor $\omega >0$, its maxima occur at $t=(2n+1)\pi /(4\omega )$, withThis is speed measured in the specified rotating frame. The same physical motion has inertial complex coordinate $e^{i\omega t}\zeta =\cos (2\omega t)+(i/2)\sin (2\omega t)$, an ellipse; its initial velocity is the frame-rotation contribution even though the initial relative velocity is zero.

A Dirichlet Green function satisfies ${\Delta }_{\mathbf{x}}{G}_D(\mathbf{x},\mathbf{y})={\delta }^{(2)}(\mathbf{x}-\mathbf{y})$ in the domain and $G_D=0$ when $\mathbf{x}$ lies on its boundary. For a source $\mathbf{y}\ne 0$ in the unit disc, put its opposite image at ${\mathbf{y}}^{\ast }=\mathbf{y}/|\mathbf{y}{|}^2$. The method of images givesThe image lies outside the unit disc, so its logarithm is a harmonic function inside. The original logarithm has precisely the unit source strength found in part (a). On $|\mathbf{x}|=1$,so the boundary value is zero. Equivalently, in complex coordinates, the Dirichlet Green function inside the unit disk isThis also covers a source at the origin, where $G_D(z,0)=(2\pi {)}^{-1}\log |z|$.

A Riemann surface is a connected one-complex-dimensional manifold with local complex coordinates whose transition maps are holomorphic. It is locally a piece of the complex plane, but its global shape can be a sphere, a torus or a surface with more handles. The coordinate compatibility makes holomorphic functions and meromorphic functions meaningful independently of the chosen chart.

The relation $w^2=z$ illustrates why these surfaces are useful. Away from zero, each $z$ has two square-root values; on the surface parametrized by $w$, the function $w$ is single-valued and the projection to $z=w^2$ records the two sheets. The apparent branch cut in the plane is replaced by the geometry of the projection.

Stereographic projection identifies a sphere minus one pole with the plane and supplies the standard charts of the Riemann sphere.

In the equatorial-plane convention of the Riemann sphere, draw the straight line through $N=(0,0,1)$ and a point $P=(X,Y,Z)\ne N$ on the unit sphere. Its intersection with the plane $Z=0$ has complex coordinateThe inverse function, for $z=x+iy$ and $r^2=x^2+y^2$, isHere positive $x$ gives positive $X$, and positive $y$ gives positive $Y$. Thus $1$ and $i$ lie on the positive $X$ and $Y$ ends of the equator. The omitted north pole is assigned $z=\infty$.^[ref]

For original worked points, the real numbers $1/2$ and $2$ lift to points with the same $X$ coordinate but opposite $Z$ coordinates:Projecting back gives $(4/5)/(1+3/5)=1/2$ and $(4/5)/(1-3/5)=2$. A point off the real axis checks the other sign:The coordinates satisfy $4/9+1/9+4/9=1$, as a point on the unit sphere must. Also $P(0)=S$, $P(-1)=(-1,0,0)$ and $P(i)=(0,1,0)$.

A short derivation starts with the line $N+t(P-N)=(tX,tY,1+t(Z-1))$. Setting its last coordinate to zero gives $t=1/(1-Z)$, hence the projection formula. Conversely the line from $N$ to $(x,y,0)$ is $(tx,ty,1-t)$. Substitution in the unit sphere equation givesThe root $t=0$ is $N$; the other root $t=2/(1+r^2)$ gives $P(z)$. These line intersections establish both the formula and its bijection property without guessing the inverse.

The same algebra explains how large planar distances become small spherical ones:Consequently $|z|\to \infty$ brings $P(z)$ to $N$ independently of direction. The last expression is twice the chordal metric distance $\chi (z,\infty )$. A whole unbounded exterior in the complex plane therefore becomes a small cap on the unit sphere.

Consider the original two-electron example obtained by taking $N=2$ in the Laughlin wavefunction, with no additional polynomial for the centre of mass. Introduce $Z=(z_1+z_2)/2$, relative complex coordinate $w=z_1-z_2$, separation $r=|w|$, and dimensionless separation $u=r/{\ell }_B$. Since $|z_1{|}^2+|z_2{|}^2=2|Z{|}^2+|w{|}^2/2$, the probability density factorizes:Integrating out $Z$ leaves an isotropic relative probability density. Define $D_m(u)$ per unit dimensionless area and $p_m(u)$ per unit dimensionless separation:The physical areal probability density is $D_m(r/{\ell }_B)/{\ell }_B^2$. The probability of a separation between $r$ and $r+dr$ is $p_m(r/{\ell }_B)dr/{\ell }_B$. The extra $u$ comes from the Jacobian determinant in polar coordinates: a thin annulus has area $2\pi rdr$. Omitting that factor changes the question being answered.

Both wavefunction normalization and the peak shift can be checked exactly. The substitution $t=u^2/4$ gives the Gamma function integralFor $m>0$, logarithmic differentiation givesConsequently the areal maximum lies at $u=2\sqrt{m}$, while the most probable separation lies farther out, at $u=\sqrt{4m+2}$. For $m=1$ these are $2$ and $\sqrt{6}\simeq 2.449$; for $m=3$, $\sqrt{12}\simeq 3.464$ and $\sqrt{14}\simeq 3.742$. The same substitution gives $⟨u^2⟩=4(m+1)$, so the root mean square separations are $\sqrt{8}$ and $4$ respectively. The distinction between a probability density maximum and a radial maximum therefore has a directly calculable geometric origin.

This two-electron model explains exchange and pair avoidance. It does not establish a nonzero many-particle spectral gap, an extended quantum Hall plateau, or the topological properties of a fluid in the thermodynamic limit. Those require additional many-electron reasoning and experimental tests.