springer resolution over $\wedge^3 \mathbb{C}^6$

Question

The action of $GL_6$ on $P(\wedge^3 \mathbb{C}^6)=P^{19}$ has 4 orbits (of dim 19, 18, 14, 9). Can you describe how the springer resolution applies to each of these orbits? It should have positive dimensonal fibers over the 14 and 9 dimensional orbits (probably some flag variety?).

Answer 1

A representative of the quartic orbit is $e_{156} + e_{246} + e_{345}$ (I write $e_{ijk}$ for $e_i\wedge e_j\wedge e_k$). A resolution of this orbit is given by the projective closure of the tangent bundle to the Grassmannian. Indeed, if $V = \mathbb{C}^6$ and $U$ is the tautological bundle on $X = Gr(3,V)$ then $$ T_X \cong U^*\otimes V/U \cong \Lambda^2U\otimes V/U\otimes O(1), $$ so $T_X(-1) \cong \Lambda^2U \otimes V/U$. Note that the tautological filtration $U\subset V$ induces a filtration of $\Lambda^3 V$ with factors $\Lambda^3U$, $\Lambda^2U\otimes V/U$, $U\otimes\Lambda^2(V/U)$, and $\Lambda^3(V/U)$. Consequently, there is a canonical exact sequence $$ 0 \to T^+ \to \Lambda^3 V\otimes O_X \to T^- \to 0, $$ where $T^+$ is the extension of $\Lambda^2U\otimes V/U$ by $\Lambda^3U$ and $T^-$ is the extension of $\Lambda^3(V/U)$ by $U\otimes\Lambda^2(V/U)$. In particular, $T^+$ fits into exact sequence $$ 0 \to O_X(-1) \to T^+ \to T_X(-1) \to 0. $$ The embedding $T^+\to \Lambda^3 V\otimes O_X$ induces a map $f:P_X(T^+) \to P(\Lambda^3 V)$. Its image is the invariant quartic hypersurface.

EDIT: Concerning the fibers of the map $f$. It is easy to show that the fiber of $f$ over a point $\lambda \in P(\Lambda^3V)$ is the subvariety of $Gr(3,V)$ consisting of those $U \subset V$ such that $$ \lambda \wedge \Lambda^2U = 0. $$ Using this it is easy to describe the fibers over representatives of the orbits: $$ f^{-1}(e_{156}+e_{246}+e_{345}) = \langle e_4,e_5,e_6 \rangle \in Gr(3,V), $$ $$ f^{-1}(e_{123}+e_{145}) = \{ U\ |\ e_1 \in U \subset \langle e_1,e_2,e_3,e_4,e_5 \rangle,\ (e_{23}+e_{45})\wedge \Lambda^2(U/e_1) = 0 \}, $$ so this is isomorphic to $Q^3 \subset Gr(2,4) \subset Gr(3,V)$, and $$ f^{-1}(e_{123})=\{ U\ |\ \dim (U \cap \langle e_1,e_2,e_3 \rangle) \ge 2 \}, $$ which is isomorphic to a $P^3$ bundle over $P^2$ with a section contracted to a point.

Answer 2

The dimension of the nilpotent orbits of $\mathfrak{gl}_{6}$ can be described using the corresponding partition $\pi: d_{1}+d_{2}+\ldots + d_{k} =6$ associated to a nilpotent orbit - so $\pi$ is the partition corresponding to the Jordan representative of the nilpotent orbit - as follows:

consider the 'dual partition of $\pi$', $\pi': e_{1}+\ldots +e_{l}=6$. If we consider the Young diagram of $\pi$ then $\pi'$ is the partition of 6 corresponding to the transpose Young diagram. Then, the dimension of a nilpotent orbit is $6^{2} - \sum_{i=1}^{l}e_{i}^{2}$. A quick check of the 11 partitions of 6 shows that there can't exist nilpotent orbits of dimension 9,14 or 19 (furthermore, orbits are always even dimensional). Perhaps you could give more information as to why these orbits given in the question are expected to be nilpotent orbits?

There are two orbits of dimension 18 (corresponding to the partitions $3+1+1+1$ and $2+2+2$, with dual partitions $4+1+1$ and $3+3$, respectively). In this case, more information as required.