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Sasha
Sasha
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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 $P_X(T^+) \to P(\Lambda^3 V)$$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.

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 $P_X(T^+) \to P(\Lambda^3 V)$. Its image is the invariant quartic hypersurface.

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.

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Sasha
Sasha
  • 39.3k
  • 2
  • 54
  • 104

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 $P_X(T^+) \to P(\Lambda^3 V)$. Its image is the invariant quartic hypersurface.