# Riemann Hilbert Problems and Wall Crossing Formulae – for Calc I Students

Holomorphic functions are basically functions in terms of $z$ and not the complex conjugate $\overline{z}$. The function $f(z) = x = z + \overline{z}$ is not holomorphic since it depends on the conjugate.

Meromorphic functions are allowed to blow up to infinity in a specific way. Obviously, $f(z) = \frac{1}{z}$ is divergent at $z = 0$ and it’s our prototype for a “pole”. Using the geometric series formula we can show it’s holomorphic everywhere else.

$\displaystyle \frac{1}{z} = \frac{1}{z_0 + (z z_0)} = \frac{1/z_0}{1 + \frac{z-z_0}{z_0}} = z_0 \left( 1 - \left(\frac{z-z_0}{z_0}\right) + \left(\frac{z-z_0}{z_0}\right)^2 + \dots \right)$

Infinite series tend to work in some regions and ‘break’ in others.

$\displaystyle \frac{1}{1-z} = 1 + z + z^2 + \dots$

in the region |z| < 1 and otherwise.

$\displaystyle \frac{1}{1-z} = \frac{-1/z}{1-1/z} = - \frac{1}{z} + \frac{1}{z^2} - \frac{1}{z^3} + \dots$

So maybe the theory of analytic functions is the study of where 'long division' produces a legitimate infinite series expansion. If we do this in two variables, we are led to the theory of amoebas…

Then we'd like to say that $\ln (z)$ is singular at 0 in the same way that $\frac{1}{z}$ is. Indeed if we zoom very closely around the origin $z = r e^{i\theta}$ we get $\ln z = \ln r + i\theta$ where $\theta \in \mathbb{R}$. This is kind of bizarre… $e^{0} = e^{2\pi I } = 1$ but if we move continuously around a circle from 1 back to itself, $\ln z$ moves from $0 \to 2\pi$.

Integrals of complex-valued functions are better-behaved then real-valued functions. Cauchy's integral formula lets us deform integrals into small circles around the poles.

$\displaystyle \oint_{|z|=1} \frac{dz}{\sin z} = \oint_{|z|=\epsilon} \frac{dz}{\sin z} = \int_0^{2\pi} i\, d\theta = 2 \pi i$

If we let $z = \epsilon e^{i\theta}\approx \sin z = \epsilon e^{i\theta}, dz = \epsilon i \theta d\theta$

### Solving a differential equation

I remember once time in class we solved y’ = y(y-1)(y+1). We can separate the y and t into two sides and integrate.

$\displaystyle \int \frac{dy}{y (y-1)(y+1)} = \int dt$
The left side is a short partial fractions problem.

$\displaystyle \frac{1}{y(y-1)(y+1)} = \frac{A}{y-1} + \frac{B}{y} + \frac{C}{y+1}$

Residue calculus is like a great big partial fractions problem where we have to guess the coefficients on both sides. Let’s multiply both sides by y-1 and set y = 1

$\displaystyle \frac{1}{y(y+1)} = A + (y-1) \left(\frac{B}{y} + \frac{C}{y+1}\right)$

and get A = 1/2. If we multipled by y+1 instead and plugged in y = -1 we get C = 1/2. If we multiply by y and set y = 0 we get B = -1. We can do the integrals

$\displaystyle \ln \left|\frac{y^2-1}{y} \right| = t + c$
and exponentiate
$\displaystyle \left|\frac{y^2-1}{y} \right| = k e^t$

We don’t know whether inside the absolute value sign is + or -. It depends on where y lies on the real line

• y < -1
• -1 < y < 0
• 0 < y < 1
• 1 < y

In fact, y(t) can’t cross these lines since the log would be infinite, so this would take infinite time. So the equilibrium solutions y = -1,0,1 are walls so to speak.

### Jumping Across Contours

The funciton $e^z$ is holomorphic everywhere and has no zeros, but if we chop off it’s Taylor series $1 + z + \frac{z^2}{2!} + \dots + \frac{z^n}{n!}$ a polynomial of degree n must have exactly n roots in the complex plane $\mathbb{C}$ so what happened?

This problem was solved by Szego and later reviewed by Dieudonne in the 1930’s (long before Mathematica could plot the roots!).

Let $\gamma$ any circle around the origin.

$\displaystyle \frac{1}{2\pi i} \oint_\gamma \frac{s^{-n}e^{n(s-1)}ds}{s-z} = \frac{1}{2\pi i} \oint_\gamma \frac{e^{n(s-1 - \ln s)}ds}{s-z} = \left\{ \begin{array}{rl} -(ez)^n \left(1 + (nz) + \frac{(nz)^2}{2!} + \dots + \frac{(nz)^{n-1}}{(n-1)!} \right) & z\text{ inside }\gamma \\ z^{-n}e^{n(z-1)}-(ez)^n \left(1 + (nz) + \frac{(nz)^2}{2!} + \dots + \frac{(nz)^{n-1}}{(n-1)!} \right) & z\text{ outside }\gamma \end{array} \right.$

We can estimate the contour integral using steepest descent. It can be approximated by a Gaussian around the critical point $s=1$, if we can pick the right contour of integration.

$\displaystyle \frac{1}{2\pi i} \oint_\gamma \frac{s^{-n}e^{n(s-1)}ds}{s-z} = \frac{1 + O(\frac{1}{n})}{\sqrt{2\pi n } (1-z)}$

This error term is uniform in $\mathbb{C} \backslash (1 + \epsilon \mathbb{D})\cup \gamma$. Where could the roots lie, inside the contour

$\displaystyle \frac{1 + O(\frac{1}{n})}{\sqrt{2\pi n } (1-z)} = \left\{ \begin{array}{rl} 0 & z\text{ inside }\gamma \\ z^{-n}e^{n(z-1)} & z\text{ outside }\gamma \end{array} \right.$

and indeed if we take nth roots we get no solution the first case and $|ze^{1-z}| = 1$ for the second.

### Wall-Crossing Formulae

So what have we learned?

• contour integrals localize around a small circle
• residue theory is just partial fractions
• analytic expansions is just geometric series

I admit this is a very gimmicky (grossly oversimplified) way of learning things but maybe it can pave the way for a new approach.

In a way, this blog entry was a bust, since I was looking to for a simplification of the Kontesevich-Soibelman Wall-Crossing formula as proven by Gaiotto Moore and Neitzke. I’ll try again soon…