1
$\begingroup$

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

$\endgroup$

1 Answer 1

4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.