Is there a linear differential equation on $\mathbb{C}$ with singularities at $0$ and $1$ whose monodromy group represents the fundamental group of $\mathbb{C}-\{0,1\}$? If so, can someone give a specific example?



A specific linear differential equation on $\mathbb{C}-\{0,1\}$ whose monodromy group represents the fundamental group of $\mathbb{C}-\{0,1\}$
Yes. they exist. The problem breaks up into two steps:


(1) Embed the free group on two generators into $GL\_n(\mathbb{C})$; see [here](https://sbseminar.wordpress.com/2007/09/17/that-trick-where-you-embed-the-free-group-into-a-lie-group/) or [here](https://math.stackexchange.com/questions/1492799/there-is-a-free-group-f-2-in-so3) among many sources.


(2) Find a connection on $\mathbb{CP}^1 \setminus \{ 0,1,\infty \}$ whose monodromy is the embedding from step (1).


Step (2) is always possible, by the Riemann-Hilbert correspondence, but that result is usually highly non-constructive. I write this answer to advertise that step (2) is not bad for $2 \times 2$ matrices.




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Let $\theta$ be $\cos^{-1}(1/3)$. Let $A$ and $B$ in $SO(3)$ be rotation by $\theta$ around the $x$ and $z$-axes. It is standard (see [here](https://math.stackexchange.com/questions/1492799/there-is-a-free-group-f-2-in-so3)) that $A$ and $B$ generate a free group.


We'd rather work with the preimages of $A$ and $B$ in the double cover $SU(2) \to SO(3)$. These are
$$X := \begin{pmatrix} e^{i \theta/2} & 0 \\ 0 & e^{-i \theta/2} \end{pmatrix} \qquad Y:=\begin{pmatrix} \cos(\theta/2) & \sin(\theta/2) \\ - \sin(\theta/2) & \cos(\theta/2) \end{pmatrix}.$$
Since $A$ and $B$ generate a free group, so do the preimages $X$ and $Y$.


The great thing about $2 \times 2$ matrices is that $X$ and $Y$ are determined up to simultaneous conjugacy by $Tr(X)$, $Tr(Y)$, $Tr(XY)$, $\det(X)$ and $\det(Y)$. In this case, these are $\cos(\theta/2)$, $\cos(\theta/2)$, $\cos(\phi/2)$, where $\phi$ is an angle I could compute explicitly if I had to, and $1$, $1$. So it is enough to write down an ODE where the monodromy is all determinant $1$, and where the traces of loops around $0$, $1$ and $\infty$ are $\cos(\theta/2)$ and $\cos(\theta/2)$, $\cos(\phi/2)$ respectively.


To this end, we take our ODE to be $v'(t) = \frac{1}{(2 \pi i) t} Pv + \frac{1}{(2 \pi i)(t-1)} Qv$, where $P$ and $Q$ are $2 \times 2$ constant matrices with eigenvalues $\pm i \theta/2$, $\pm i \theta/2$ such that $P+Q$ has eigenvalues $\pm i \phi/2$. This is not hard to achieve, taking $P = \left( \begin{smallmatrix} i \theta/2 & 0 \\ 0 & -i \theta/2 \end{smallmatrix} \right)$ and $Q = S P S^{-1}$ for $S$ of the form $\left( \begin{smallmatrix} e^{i \alpha} & 0 \\ 0 & e^{-i\alpha} \end{smallmatrix} \right)$ and solving numerically for what $\alpha$ gives $P+Q$ the correct determinant.




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That said, this answer is kind of wrong headed. One rarely wants to give a specific embedding of the free group into $SU(2)$; one just wants to know that almost any two generators do it. And, while this method will give you an explicit connection, in general one uses the Riemann-Hilbert correspondence non-constructively and notes that almost any connection will have no relations between the monodromy matrices.


