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…

Additional Reading

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