“More” because the discussion here is also about regular singularities.

Let $X,Y,Z \in \mathrm{GL}_2$ have $XYZ = 1$.  They determine a rank 2 local system on a 3-times punctured $\mathbf{P}^1$ — $X$ the monodromy around $0$, $Y$ the monodromy around $1$, $Z$ the monodromy around $\infty$.  The simultaneous conjugacy class of the triple $(X,Y,Z)$ is determined by the separate conjugacy classes of $X,Y,Z$, i.e. the corresponding local system is “rigid.”

The analytic source of these local systems is Gauss’s hypergeometric

$$z(z-1) u” + ((a+b+1) z – c) u’ + ab u = 0$$

where $a,b,c$ are complex constants.  See Ahlfors page 316. It has regular singularities, and the monodromy matrices $XYZ$ have characteristic polynomials

$$\begin{array}{c}\chi(X) = (\lambda – 1)(\lambda – e^{2 \pi i (1-c)}), \\ \chi(Y) = (\lambda – 1)(\lambda – e^{2 \pi i (c – a – b)}), \\ \chi(Z) = (\lambda – e^{2 \pi i a})(\lambda – e^{2 \pi i b}) \end{array}$$

The local system is irreducible if and only if none of the quantities $a,b,c-a,c-b$ are integers.

(Rigidity of the system is a good post-hoc reason for studying it.  Katz classifies rigid local systems on $\mathbf{P}^1$ minus $n$ points in his book.  Simpson observes in his ICM talk, about rigid local systems: the corresponding Higgs field is fixed by $\mathbf{C}^*$, and the data of a $\mathbf{C}^*$-fixed Higgs field is a variation of Hodge structures.  And they are used to solve some instances of inverse Galois.)

Question

Is there an exact Lagrangian $L \subset T^* S^2$ which when decorated with some rank one local systems determines these rigid ones under Nadler-Zaslow?  In other words can hypergeometric DEs be “abelianized.”

If it exists, the projection $L \to S^2$ must be two-to-one and (because of the rigidity) $L$ must have genus zero.  Riemann-Hurwitz doesn’t leave many possibilities.  Let $\overline{L} \cong S^2$ be the smooth compactification of $L$; the projection from $L$ to extends to one from $\overline{L}$.  Let $e_0,e_1,e_{\infty}$ be $1$ or $0$ according to whether $\overline{L} \to S^2$ is or is not ramified at $0,1,\infty$  Then $\chi(\overline{L}) = 2\chi(S^2) – e_0 – e_1 – e_{\infty} – n$ where $n$ is the number of critical points of $L \to S^2$.  In other words $n = 2 – e_0 – e_1 – e_{\infty}$.

Since for generic $a,b,c$ the monodromies around the punctures are semisimple, it is tempting to believe that $e_0 = e_1 = e_{\infty} = 0$, i.e. $L$ is the complement of $6$ points on $S^2$ — then $n = 2$.  It is also tempting to take for $L$ the curve

$$z(z-1)w^2 + ((a+b+1)z-c)w + ab = 0$$

i.e. replace the operator $d/dz$ by the coordinate $w$ in the Gauss equation.  That move is dubious, but it gives an $L$ with the right topology.