Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given a differential equation, say $\frac{d^2f}{dz^2}+\frac{f}{z-5}=0$. We know that this equation has two linearly independent solutions $f_1(z), f_2(z)$. By analytic continuation, the solution $f_i$ is taken to $g_i$, $i=1, 2$, and $g_i=c_{i1}f_1+c_{i2}f_2$. The set of matrices $(c_{ij})$ form a group called monodromy group. My question is how to compute the monodromy group explicitly?

share|cite|improve this question

For the sake of simplicity, say you are talking about the equation $\frac{d^2f}{dz^2} = f/z$. For each nonzero $z_0$, one can search for power series at $z_0$ : we look for sequences $(a_n)$ such that $f(z) = \sum a_i (z-z_0)^i$ is a solution near $z_0$. Given the first two coefficients, you can quickly determine all the others. This gives you a linear space of dimension $2$ of solutions near $z_0$ with a radius of convergence probably bounded by $|z_0|$. Suppose for the sake of the argument that nothing especially bad happens away from $0$ and that this radius is always $|z_0|$

Then, if $|z_1 - z_0| \le |z_0|$, you should have a linear map from the solutions near $z_0$ to the solutions near $z_1$, simply by taking a power series at $z_0$ and looking at what it's like at $z_1$

For example, starting from the set of power series at $1$ that are solution to the differential equation near $1$, you get a map to the set of power series at $\exp(2i\pi/8)$, and from there to the set of power series at $\exp(4i\pi/8)$, and so on. In $8$ steps you can do a full loop around $0$ and return back to $1$. This gives you a way to approximate the monodromy map even if you don't have an exact map for each step.

For the case at hand, keeping only $100$ coefficients to compute the maps, and doing those $8$ steps we find that the power series $a + b(z-1) + \ldots$ is sent to $[a+i(k_1a+k_2b)] + [b+i(k_3a-k_1b)](z-1) + \ldots$ where $k_1 = 22.7827821414\ldots, k_2 = -15.8972480162\ldots, k_3 = 32.6506299438\ldots$.

We find that the if $T$ is the corresponding matrix, $\operatorname{tr}(T) = 2$ and $\det(T) = 1$. Its eigenvalues are $1$, and $T$ is equivalent to $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ .

In particular, $k_1^2+k_2k_3=0$, and if $f_1(z) = k_1 + k_3(z-1)\ldots$ then $T(f_1) = f_1$, and if $f_2(z) = k2-(1+i)k_1(z-1) + \ldots$ then $T(f_2) = f_2+k_2f_1$.

If nothing bad happens elsewhere, $f_1$ has a zero at $0$ and is actually an entire function, while $f_2$ is "a function of $\log(z)$", by which I mean that there is an entire function $g_2$ (which is a solution of a corresponding differential equation) such that $g_2(z) = f_2(\exp z)$ on a neighbourhood of $0$. In this setting, $g_1$ is a $2i\pi$-periodic entire function, and the monodromy is just translating by $2i\pi$ so $g_2(z+2i\pi) = g_2(z)+k_2f_1(z)$. And actually, for every solution $g$ there is a constant $k \in \Bbb C$ such that $g(z+2i\pi) = g(z) + kg_1(z)$

From this newfound knowledge, we can look for solutions of the form $f(z) = \sum_{k \ge 0} a_k z^k + \sum_{k \ge 1} b_k z^k \log(z)$ around $0$, and we find some solutions $f_1(z) = \sum_{n \ge 1} \frac {z^n}{n!(n-1)!}$ and $f_2(z) = \sum_{n \ge 1} \frac {z^n (\log(z) + 1 - H_n - H_{n-1})}{n!(n-1)!}$, which are entire (as functions of $\log(z)$). It is quick to check that the monodromy group acts on them by $T(f_1) = f_1$ and $T(f_2) = f_2+2i\pi f_1$

share|cite|improve this answer
This is very helpful! Thank you! – Samuel Reid Nov 11 '13 at 22:04

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.