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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?

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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 $T(f_2) = f_2+2i\pi f_1$

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This is very helpful! Thank you! – Samuel Reid Nov 11 '13 at 22:04

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