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. $w_{i}=\delta_{i,k}$)?


1 Answer 1


We, at least, have the explicit answer for their Betti numbers. For an example, let $Q$ be type $E_8$, and $k$ be the triplet vertex. Take $\mu = 0$. The normalized Poincare polynomial of $\mathfrak M(\mathbf v,\mathbf w)$ was computed in https://arxiv.org/pdf/math/0606637.pdf as $$1357104 + 2232771t^2 + 2002423t^4 + 1317308t^6 + 716312t^8 + 342421t^{10} + 148512t^{12} + 59490t^{14} + 22162t^{16} + 7687t^{18} + 2463t^{20} + 726t^{22} + 192t^{24} + 44t^{26} + 8t^{28} + t^{30}.$$ Here the constant term corresponds to the middle degree, while the highest term corresponds to the degree 0.


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