I am interested in a group of transformations that leave the Plucker embedding of complex Grassmannian $G(2,4)$ into $CP^5$ given by $\lambda\_{12}\lambda\_{34}-\lambda\_{13}\lambda\_{24}+\lambda\_{14}\lambda\_{23}=0$ invariant. ($\lambda\_{ij}$ are natural coordinates in $C^4\wedge C^4$). Naively I would say that it is $SU(3,3)$ because this quadratic form has signature (1,1,1,-1,-1,-1). However, $SU(3,3)$ does not contain $SU(4)$, the isometry group of the Grassmannian, and I believe the embedding should be invariant at least under that.



Transformations that leave the Plucker embedding of G(2,4) invariant
**Corrected after Noam's comments below**


The subgroup of $\mathrm{PSL}(6,\mathbb{C})$ acting in its usual way on $\mathbb{CP}^5$ that preserves the quadric hypersurface that is $G(2,4)\subset \mathbb{CP}^5$ is $\mathrm{PO}(6,\mathbb{C}) = \mathrm{O}(6,\mathbb{C})/\{\pm I\_6\}$, which has two components. Its identity component, by the usual exceptional isomorphism (i.e., $A\_3=D\_3$), is $$\mathrm{PSO}(6,\mathbb{C}) = \mathrm{SO}(6,\mathbb{C})/\{\pm I\_6\}=\mathrm{PSL}(4,\mathbb{C}).$$ 


[See Noam's comments below, which point out that my original answer of $\mathrm{PSO}(6,\mathbb{C})$ was not correct and that my first attempted fix also had an error. I'm leaving them there because they might help some else avoid making my mistake in the future. ]


