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Let $Q=(I,\Omega)$ be the $D_4$ affine quiver. We choose as dimension vector $(2,1,1,1,1)$ (where $2$ is on the central vertex). As this dimension vector is indivisible, we can choose a generic $\theta \in \mathbb{Z}^I$. We have the usual moment map $$\mu: R(\overline{Q},v) \to \mathbb{C}^I $$ from representations of the doubled quiver to $\mathbb{C}^I$.

We have different quiver varieties associated to these data (I do not know whether the notation is standard): $$\mathfrak{M}_{0,\theta}(v)=\mu^{-1}(0)//_{\theta}G(v)$$ and $$\mathfrak{M}_{\theta,\theta}(v)=\mu^{-1}(\theta)/G(v) $$

A computation shows that these varieties are actually of dimension $2$: is there an explicit geometric presentation of such objects? If we choose $\theta$ in an appropriate way, we can think of $\mathfrak{M}_{0,\theta}(v),\mathfrak{M}_{\theta,\theta}(v)$ as certain moduli spaces of parabolic Higgs bundles/parabolic connections over trivial vector bundle over $\mathbb{P}^1$.

I know that in this case we have a concrete description of the full moduli space (the so-called Hausel toy model). Also the character variety side is well understood as affine Del Pezzo surfaces. What about quiver side?

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  • $\begingroup$ Aren't they just the simple surface singularity $\mathbb{C}^2/\Gamma$, where $\Gamma \subset \mathrm{SL}(2,\mathbb{C})$ is the dihedral group $D_4$, and its minimal resolution? $\endgroup$
    – Sasha
    Commented Sep 20, 2021 at 10:51
  • $\begingroup$ I think this should be true for $\mathfrak{M}_{0,\theta}(v)$. The other one should be diffeomorphic to the former,but not algebraically isomorphic (I think) $\endgroup$
    – Tommaso Scognamiglio
    Commented Sep 20, 2021 at 12:58

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It is Kronhimer's result that $\mathfrak M_{\zeta_{\mathbb R},\zeta_{\mathbb C}}(\mathbf v)$ is $\mathbb C^2/\Gamma$ ($\zeta_{\mathbb R}=\zeta_{\mathbb C}=0$), its deformation ($\zeta_{\mathbb R}=0$), and the minimal resolution of the deformation (in general). Here $\Gamma$ is the binary dihedral group of type D4.

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    $\begingroup$ Perhaps just to extend on this a little bit, as the OP asked for "explicit geometric presentation": the singularity $Y=\mathfrak M_{0,0}(\mathbf v) =\mathbb C^2/\Gamma$ is affine, and well known to be a hypersurface in ${\mathbb A}^3$ given by an explicit equation; its deformation $Y_t = \mathfrak M_{0,\zeta_{\mathbb C}}(\mathbf v)$ is still affine, and is given by a generic deformation of the equation of the hypersurface; the minimal resolution $X = \mathfrak M_{\zeta_{\mathbb R},0}(\mathbf v)$ of $Y$ is not affine of course but is a composite of blowups of $Y=Y_0$... $\endgroup$
    – Balazs
    Commented Nov 9, 2021 at 11:23
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    $\begingroup$ ...finally indeed $X$ and $Y_t$ for $t\neq 0$ are diffeomorphic, in fact algebraic deformation equivalent, as they sit in the Grothendieck resolution family. $\endgroup$
    – Balazs
    Commented Nov 9, 2021 at 11:24

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