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.
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1$\begingroup$ Over the complex numbers there's no signature; it's just PGO(6). $\endgroup$– Noam D. ElkiesCommented Jul 20, 2015 at 18:44
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2 Answers
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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. ]
answered Jul 20, 2015 at 18:42
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2$\begingroup$ It's actually twice as big, right? There are automorphisms of determinant $-1$ that come from an outer automorphism of PSL(4). $\endgroup$ Commented Jul 20, 2015 at 20:25
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$\begingroup$ Oh, yes, Noam, good point. The automorphism group should be $\mathrm{PO}(6,\mathbb{C}) = \mathrm{O}(6,\mathbb{C})/\{\pm I_6\}$, which, as you surmise, is a (nontrivial) extension of $\mathrm{PSO}(6,\mathbb{C})$. [However, technically, what I wrote is correct, because I was only considering the subgroup of $\mathrm{PSL}(6,\mathbb{C})$ that preserves the hypersurface, and that is $\mathrm{PSL}(4,\mathbb{C})$. Instead, I should have considered the subgroup of $\mathrm{PGL}(6,\mathbb{C})$ that preserves the quadric hypersurface.] $\endgroup$ Commented Jul 20, 2015 at 22:38
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2$\begingroup$ PGL and PSL are the same over C, though. (Multiply the transformation of determinant $-1$ by $i$ or any other sixth root of $-1$; the resulting transformation does not quite preserve the quadric form, but does preserve its zero-locus.) $\endgroup$ Commented Jul 20, 2015 at 23:12
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$\begingroup$ Ach! Thanks, Noam, for pointing out the true nature of my mistake (again). I'm sorry for being so careless. I should have remembered that, even though $\mathrm{O}(6,\mathbb{C})$ is not a subgroup of $\mathrm{SL}(6,\mathbb{C})$, the group $\mathrm{PO}(6,\mathbb{C})$ is a subgroup of $\mathrm{PSL}(6,\mathbb{C})$, precisely because this latter group is, as you say, the same as $\mathrm{PGL}(6,\mathbb{C})$. $\endgroup$ Commented Jul 21, 2015 at 0:53
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If you denote by $W$ the 6D space $\mathbb{C}^4\wedge\mathbb{C}^4$, then the (conformal) quadratic form you've written down in terms of the $\lambda_{ij}$'s is intrinsically defined by the natural symmetric bilinear form $$ W\times W\ni (\omega_1,\omega_2)\longmapsto q(\omega_1,\omega_2):=\omega_1\wedge\omega_2\in (\mathbb{C}^{4})^{\wedge 4}\equiv \mathbb{C}\, . $$ As a subset of $\mathbb{CP}^5=\mathbb{P}(W)$, the complex Grassmannian $G(2,4)$, is the zero locus of $q$ and, as Robert Bryant pointed out, such a locus is preserved by $\mathrm{PSO}(6,\mathbb{C})$. But, as you said, as an abstract object, $G(2,4)$ admits $\mathrm{SU}(4)$ as a symmetry group.
In answer to your query, let me just notice that, for $\omega_i=v_i\wedge w_i$, we have $$ q(\phi(\omega_1)\wedge \phi(\omega_2))=q(\phi(v_1\wedge w_1)\wedge (\phi(v_2\wedge w_2))=q(\phi(v_1)\wedge \phi(w_1),\phi(v_2)\wedge \phi(w_2))=\phi(v_1)\wedge \phi(w_1)\wedge\phi(v_2)\wedge \phi(w_2)=\det\phi\, \cdot\, v_1\wedge w_1 \wedge v_2\wedge w_2=\det\phi \, \cdot\, q(\omega_1,\omega_2) $$ for any $\phi\in \mathrm{SU}(4)$, i.e., $\mathrm{SU}(4)$ acts on $\mathbb{P}(W)$ by (conformal) isometries. Hence, there is a group homomorphism $\mathrm{SU}(4)\longrightarrow \mathrm{PSO}(6,\mathbb{C})$, and the Plucker embedding in invariant along it.
