Evaluate a function along a complex curve

Question

Well, I don't know this question is appropriate but I really need to understand it. So, please help me.

In the book Love and Math, The heart of hidden reality of Edward Frenkel, chapter 15, he said :

The equation $x'(t)=\dfrac{nx(t)}{t}$ has a solution $x(t)=t^n$. However, there is a surprise in store: If $n$ is not an integer, then as we evaluate the solution along a closed path on the plane and come back to the same point, the value of the solution at the end point will not necessarily be the value we started with. It will get multiplied by a complex number. In this situation, we say that the solution has undergone a monodromy along this path.

So, what does he really mean by "evaluate"? Could you please explain the above set up mathematically?

Thanks.

Answer 1

Note that $t$ is considered a complex variable now. The solution to the equation $x'(t) = \frac{n x(t)}{t}$ is a function proportional to $t^n$. This function, defined on $\mathbb{C}\backslash\{0\}$ is in fact multivalued. The way it is defined is

$$t^n = |t|^n e^{i n \arg t}$$

The problem is with the argument: it is not uniquely defined, as a function from $\mathbb{C}\backslash\{0\} \to \mathbb{R}$. And you see this as you move around the point $0$. Start from $1$, with argument $0$ and move around counterclockwise. You get back to $1$, and if the argument varies continuously, you should have now the argument $2 \pi$. This multiple choice with difference multiple of $2\pi$ is what the issue is. Now if $n$ is an integer we'll have the same value for $t^n$ no matter what choice for $\arg n$ we take. However, say if $n$ is a rational number with denominator $d$ then for $t^n$ we will have in fact $d$ choices. And if $n$ is irrational then there are infinitely many choices.

Answer 2

"Evaluating the solution along a path" assumes meaning only in a complex environment. We then have a differential equation $$w'(z)=f(z,w)\ ,$$ where both the independent and the dependent variable are assumed to be complex, and furthermore the right hand side is defined in some domain $\Omega\times{\mathbb C}$ and is analytic in both variables. A simple example is given by "complexifying" the ODE in the question, resulting in the ODE $$w'(z)={p\>w(z)\over z}\ .$$ Here $\Omega={\mathbb C}\setminus\{0\}$, and $p$ may be an arbitrary complex constant.

Given the initial condition $w(1)=c$ it is easy to check that the function $$w_0(z)=c\>\exp(p\> {\rm Log}\,z)\tag{1}$$ which parrots the solution $x(t)=c\>t^p$ from the real setup, is a solution of the given IVP in the slit plane $\Omega':={\mathbb C}\setminus{\rm real \ axis}$. When the parameter $p$ happens to be a real integer then this solution can be glued along the negative real axis, and is simply the function $w(z)=w_0 z^p$ defined in all of $\Omega$ (or even of ${\mathbb C}$, when $p\geq0$), but when $p\notin{\mathbb Z}$ such a mending is not possible: The function $(1)$ assumes different values when approaching, say, the point $-1$ from above and from below.

Now what does "evaluating the solution along a closed path" mean? When we pursue the solution $(1)$ counterclockwise along the unit circle then it doesn't stop to exist upon reaching the negative real axis, and we could even write down a formula for it which is valid in the whole left half plane, and then it would continue to exist until we reach again the point $1$. But the end value we see there would not be the starting value $c$. In fact the solution has picked up a factor $e^{2\pi i p}$ during its journey around the origin, so that we arrive with $$w_1(z)=e^{2\pi i p}w_0(z)\ .$$ This surprise stems from the fact that the "complex log" in $(1)$ increases by $2\pi i$ when going around the origin once.