Let $Q$ be a quiver of type $ADE$, $I$ is the set of vertices of $Q$. Let $\mathfrak{M}({\mathbf{v}},{\mathbf{w}})$ be a Nakajima quiver variety for such quiver (here ${\mathbf{v}}=(v_i)_{i \in I}$ is the dimension vector and ${\mathbf{w}}=(w_i)_{i \in I}$ is framing). We can associate to $\mathfrak{M}({\mathbf{v}},{\mathbf{w}})$ a pair of weights $\lambda:=\sum_i w_i\omega_i$, $\mu:=\lambda-\sum_i v_i\alpha_i$.

We assume that $\mathfrak{M}({\mathbf{v}},{\mathbf{w}})$ is nonempty (so $\mu$ is a weight of the irreducible representation $L_{\lambda}$). Assume also that $w_i=\delta_{i,k}$ for some $k \in I$ (i.e. all $w_i$ are zero except $w_k$ which is one). In other words we assume that $\lambda=\omega_k$ for some $k$.

If $\mu$ is conjugate to $\omega_k$ via the action of the Weyl group (this for example always holds if $\omega_k$ is minuscule) then the quiver variety $\mathfrak{M}({\mathbf{v}},{\mathbf{w}})$ is one point.

If $\mu$ is not conjugate to $\omega_k$ then $\mathfrak{M}({\mathbf{v}},\mathbf{w})$ has dimension at least two.

Question: do we have a good (explicit) description of the variety $\mathfrak{M}({\mathbf{v}},{\mathbf{w}})$ when $\lambda=\omega_k$ (i.e. $w_{i}=\delta_{i,k}$)?