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I am trying to understand the computation of the monodromy group of the differential equation given by $$x^2f''(x)-f(x)=0.$$

The differential equation has one singular point in $\mathbb C$, namely, $0$.

It is a homogeneous equation, and the vector space of solutions is $V:=\langle f_1,f_2\rangle_{\mathbb C},$ given by $$f_1:=x^{(1+\sqrt{5})/2},\qquad f_2:=x^ {(1-\sqrt{5})/2}.$$

From my understanding, one needs to take a loop $\eta$ around the singularity at $0$ with base point say $z_0=1$, and extend both $f_1$ and $f_2$ holomorphically around $\eta$. This will give rise to a new basis $g_1,g_2$, and the change of basis matrix will generate the monodromy group.

I would like to see an explicit example of what happens when one extends $f_i$ holomorphically around $\eta$, which wraps around the singularity at $0$ and understand how to write $g_i$ as a linear combination of $f_i$. Any reference on this or a similar example would be much appreciated. More generally, if one has a good reference to learn about monodromies starting from an undergraduate background in Complex Analysis, that would be very useful.

I came across this other post: How to compute the monodromy group for a given differential equation., but I find it difficult to understand how they computed the corresponding $k_1,k_2,...$.

did
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1 Answers1

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As you have explicitly determined a basis for the ODE, the problem is relatively easy now:

$$x^a=e^{a\log(x)} = e^{a\log(re^{i\theta})}=e^{a(\log(r)+i\theta)}$$

So when the solution runs a full circle around $0$, nothing is changed, except $\theta$ becomes $\theta+2\pi$, therefore the new solution is

$$e^{a(\log(r)+i\theta)}e^{(2\pi i)a}$$

So the monodromy group is generated by the diagonal matrix:

$$\begin{pmatrix} e^{(2\pi i)a_1} & \\ & e^{(2\pi i)a_2}\end{pmatrix}$$

where $a_1=\frac{1+\sqrt 5}{2}$ and $a_2=\frac{1-\sqrt 5}{2}$.

Just a user
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