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?).

2$\begingroup$ Perhaps you could clarify what is meant by 'Springer resolution' in this scenario? To me the Springer resolution is a resolution of singularities of the nilpotent cone of a simple Lie algebra  I'm interested in how this is related to the setup given. $\endgroup$ – George Melvin Oct 4 '12 at 16:44

$\begingroup$ well in fact I guess that the the 4 orbits I have pointed out are nilpotent orbits. in fact the 9 dim is the G(3,6)  a representative for this orbit is $e_1\wedge e_2 \wedge e_3$ , the 14 dim is the orbit of $e_1\wedge e_2 \wedge e_3 + e_1\wedge e_4 \wedge e_5$ and it is the singular locus of the 18dim orbit, which is a quartic hypersurface  I don't know the shape of the elements of this orbit. The 19 dimensional is all $P^19$ and the elements are of type $e_1\wedge e_2 \wedge e_3 + e_4\wedge e_5 \wedge e_6$. I would like to understand the Springer resolution of the quartic. $\endgroup$ – IMeasy Oct 4 '12 at 18:08

$\begingroup$ This resolution should involve some  I guess  flag variety over the 14 dimensional orbit. $\endgroup$ – IMeasy Oct 4 '12 at 18:09
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.

$\begingroup$ OK thank you, great answer!this is now clear to me. And I guess that on the complement of the 14 dimensional orbit this map is 1:1. Can one see easily from this description what are the fibers of the map $P_X(T^+)\rightarrow P(\wedge^3V)$ over this orbit and over the grassmannian? $\endgroup$ – IMeasy Oct 5 '12 at 12:59

1$\begingroup$ @MBeasy: I added a description of fibers to the answer. $\endgroup$ – Sasha Oct 5 '12 at 15:26

$\begingroup$ Dear Sasha, this is an amazing answer, and very clear too. Thank you. Do you have general references to suggest on these topics? $\endgroup$ – IMeasy Oct 5 '12 at 21:09
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.

$\begingroup$ I guess that the discrpancy between your dimension computations and mines is that I take projective dimensions and not linear. the dimension 19 is the full, projetivized, space. The description of this orbits is classical and developed for instance in Donagi's "on the geometry of grassmannians". in that paper though the springer resolution is not developed. $\endgroup$ – IMeasy Oct 5 '12 at 9:01

$\begingroup$ @MBeasy: Of course, I had overlooked this. Thanks for pointing that out and for the reference. However, this means that it is only possible for the 9dim orbit to be a nilpotent orbit  there are no nilpotent orbits of dimension 15,19 or 20. The 9dim orbit must correspond to the minimal orbit in this case. I'm still unsure where the Springer reaolution appears in your setup. In any case, it appears that @Sasha has given a complete description of this problem. $\endgroup$ – George Melvin Oct 5 '12 at 16:58