# sub-variety of (P^1)^4

Let $S$ be the sub-algebra generated by the set $\{x_1x_2y_3y_4, x_1x_3y_2y_4, x_1x_4y_2y_3, x_2x_3y_1y_4, x_2x_4y_1y_3, x_3x_4y_1y_2\}$ of homogeneous polynomials of degree $4$. I need to study the projective variety $Proj(S)$. The relation among the above set of polynomials are $f_1f_6=f_2f_5=f_3f_4$ in the above order. So the projective dimension is $3$. Looks like this is a $3$ dimensional sub-variety of $\mathbb P^1 \times \mathbb P^1 \times \mathbb P^1 \times \mathbb P^1$ but what is this variety ?

You can think of this ring as the semigroup ring of the semigroup $S$ generated by $$(1,0,0,1),(-1,0,0,1),(0,1,0,1),(0,-1,0,1),(0,0,1,1),(0,0,-1,1).$$ The above semigroup elements correspond to $f_1,f_6,f_2,f_5,f_3,f_4$ respectively.
I believe that after saturation at the origin the semigroup ring is the polytope ring of the octahedron. Then the corresponding toric variety has a fan based on the cube of size $2$. This is a rather singular Fano variety of Picard number 5. It has $12$ lines worth of $A_1$ singularities and $6$ rather complicated Gorenstein singularties. I don't see it as lying inside $(\mathbb P^1)^4$, because the algebra is not a quotient of the homogeneous ring of the latter, but rather is a subalgebra of it.
• It's the closure of a generic $T^4$-orbit on the Grassmannian $Gr(2,4)$, so there's something it lies inside. Jun 12 '15 at 11:50