Riemann-Hilbert problem via quiver description

Question

The moduli space of Fuchsian systems over $\mathbb{P}^1$ with prescribed adjoint orbits conditions at poles a.k.a. additive Deligne-Simpson problem can be presented under purely quiver description.The quiver is give by $\{A_i\in \mathcal{O}^i_{\mathfrak{gl}_n(\mathbb{C})}, i=1,2,…,m|\sum A_i=0\}/\mathrm{GL}_n(\mathbb{C})$.

Meanwhile, the solution of multiplicative Deligne-Simpson problem is subset of character varieties $\mathrm{Hom}(\pi_1(\mathbb{P^1}-\{p_1,…,p_m\}),\mathrm{GL}_n(\mathbb{C}))/\mathrm{GL}_n(\mathbb{C})$

which can be described by multiplicative quiver variety directly, i.e. the space $\{M_i\in \mathcal{O}_{\mathrm{GL}_n(\mathbb{C})}^i,i=1,…,m|M_1M_2…M_m=1\}/\mathrm{GL}_n(\mathbb{C})$.

To get a map from additive one to multiplicative one via quiver ,my naive hope is to translate the Riemann-Hilbert correspondence into quiver language. More precisely, Let’s assume $A_i$ are nonresonant, consider the monodromy of connection $d+\sum \frac{A_i}{z-i}dz$, then $e^{2i\pi A_i}$ will conjugate to the monodromy matrix around point $z=i$, so we can define a map from additive Deligne-Simpson problem to the multiplicative one by taking exponential $(A_1,…,A_m)\to (e^{2i\pi A_1},…,e^{2i\pi A_m})$ and then conjugate each $e^{2i\pi A_j}$ to monodromy matrix around $z=i$.

I don’t know if these conjugacy matrices can be expressed by $A_i$, But at least by looking at monodromy we have a well-defined map from additive one to multiplicative one. If $A_i$ are not nonresonant, I’m even not sure taking monodromy will send conjugacy class of $A_i$ to conjugacy class of $M_i$, any counterexample will also help.

And I wonder in general, can we set up an isomorphism of these two quiver varieties by Riemann-Hilbert correspondence? Or do we have an isomorphism in purely quiver description?

Answer 1

I don't think the map you are considering is surjective in general. Let us consider the case where where you have $m=4$ points and all the $A_i$ lie in the conjugacy class of $$\begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix},$$ and so all the $M_i$ live in the conjugacy class of $$\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}$$ There is a canonical local system on $\mathbb{P}^1-\{x_1, \cdots, x_4\}$, called the uniformizing local system, with these local monodromies; its (global) monodromy depends on the $x_i$. One can construct it via non-abelian Hodge theory as follows: it is a complex variation of Hodge structure, with corresponding graded Higgs bundle $\mathscr{O}(1)\oplus \mathscr{O}(-1)$, where the Higgs field $$\mathscr{O}(1)\to \mathscr{O}(-1)\otimes \Omega^1_{\mathbb{P}^1}(x_1+\cdots+x_4)=\mathscr{O}(1)$$ is a non-zero isomorphism.

But this implies the flat bundle $(\mathscr{E}, \nabla)$ on $\mathbb{P}^1$ with regular singularities at the $x_i$ and residues $A_i$ in our nilpotent conjugacy class has underlying vector bundle $\mathscr{E}=\mathscr{O}(1)\oplus \mathscr{O}(-1)$, i.e. the corresponding ODE is not "Fuchsian" (if I understand correctly, these are the ODEs you are considering -- please correct me if I'm wrong). Since this bundle is unique by the Riemann-Hilbert correspondence, there is no Fuchsian ODE with the required monodromy, so the monodromy of the uniformizing local system is not in the image of your map.

EDIT: The OP asks for a non-resonant example, i.e. one where the $A_i$ have distinct eigenvalues which do not differ by integers. I think one can make one as follows. Let $A_1, A_2, A_3$ lie in the conjugacy class of $$\begin{pmatrix} 1/4 & 0 \\ 0 & 3/4 \end{pmatrix}$$ and let $A_4$ lie in the conjugacy class of $$\begin{pmatrix} -3/4 & 0 \\ 0 & -9/4 \end{pmatrix}$$ We first construct a stable graded parabolic Higgs bundle with weights $(1/4, 3/4)$ at each of $x_1, \cdots, x_4$. Namely, let $$\mathscr{E}=\mathscr{O}(-1)\oplus \mathscr{O}(-3),$$ with Higgs field $$\theta:\mathscr{O}(-1)\to \mathscr{O}(-3)\otimes \Omega^1(x_1+\cdots+x_4)=\mathscr{O}(-1)$$ non-zero, and parabolic weights $(1/4, 3/4, 1/4, 3/4)$ on $\mathscr{O}(-1)$ at $x_1, \cdots, x_4$, and weights $(3/4, 1/4, 3/4, 1/4)$ on $\mathscr{O}(-3)$ at $x_1,\cdots, x_4$. This bundle is parabolically stable of (parabolic) slope zero, since $\mathscr{O}(-1)_\star$ has parabolic degree $1$ and $\mathscr{O}(-3)_\star$ has parabolic degree $-1$. The corresponding ODE is not Fuchsian for the same reason as in the previous example; its residues all lie in the conjugacy class of $$\begin{pmatrix} 1/4 & 0 \\ 0 & 3/4 \end{pmatrix}.$$ The OP might reasonably argue that this is unfair, since the sum of the traces of the $A_i$ is non-zero; but twisting up by $\mathscr{O}(2x_4)$ makes the residue at $x_4$ lie in the conjugacy class of $$\begin{pmatrix} -3/4 & 0 \\ 0 & -9/4 \end{pmatrix}$$ and doesn't change the fact that the ODE is still not Fuchsian.