Distance Geometry in Superspace

We can define a distance invariant under superspace transformations: $$R^2=|x-y{|}^2+\theta {\gamma }_{\mu }(x-y{)}^{\mu }\zeta$$ For D space-time dimensions and E Grassman dimensions, does this satisfy a Cayley-Menger type equation? In 2 dimensions we know the volume of a tetrahedron is zero. Or: $$det\left[\begin{array}{ccccc}0& R_{12}& R_{13}& R_{14}& 1\\ R_{12}& 0& R_{23}& R_{24}& 1\\ R_{13}& R_{23}& 0& R_{34}& 1\\ R_{14}& R_{24}& R_{34}& 0& 1\\ 1& 1& 1& 1& 0\end{array}\right]=0$$ What about when D=11 and E=32 ? Simplest is D=1, E=2 with (?): $$R^2=(x-y{)}^2+(x-y)(\theta \zeta +{\zeta }^{\ast }{\theta }^{\ast })$$ with for example: $$R^4=(x-y{)}^4+2(x-y{)}^3(\theta \zeta +{\zeta }^{\ast }{\theta }^{\ast })+2(x-y{)}^2(\theta {\theta }^{\ast }\zeta {\zeta }^{\ast })$$ $$R^6=(x-y{)}^6+3(x-y{)}^5(\theta \zeta +{\zeta }^{\ast }{\theta }^{\ast })+6(x-y{)}^4(\theta {\theta }^{\ast }\zeta {\zeta }^{\ast })$$ Then the area of the triangle is 0 (or and polynomial with 0 constant term). We should have an identity, something like (?): $$R^6-3R^4(x-y{)}^2+R^2(x-y{)}^4+(x-y{)}^6=0$$ So we would have: $$\mathrm{\Delta }(L_{AB})=0\wedge F(L_{AB},R_{AB})=0\to ?G(R_{AB})=0$$ $$(R_{xy}{)}^2(R_{xz}{)}^2=(x-y{)}^2(x-z{)}^2+(x-y{)}^2(x-z)(\theta \zeta +{\zeta }^{\ast }{\theta }^{\ast })+2(x-z{)}^2(x-y)(\theta \beta +{\beta }^{\ast }{\theta }^{\ast })$$ $$+2(x-y)(x-z)\theta {\theta }^{\ast }(\zeta {\beta }^{\ast }+\beta {\zeta }^{\ast })$$ So (?) $$\mathrm{\Delta }\equiv 0+(x-y)(z-x)(z-y)(\theta \zeta +{\zeta }^{\ast }{\theta }^{\ast }+\zeta {\beta }^{\ast }+\beta {\zeta }^{\ast }+...)$$ $$+2(x-y{)}^2(\theta {\theta }^{\ast }\zeta {\zeta }^{\ast })+(x-y)(x-z)\theta {\theta }^{\ast }(\zeta {\beta }^{\ast }+\beta {\zeta }^{\ast })+...$$ So very simply we can say that the distance geometry is compataible with D real and 4 grassman dimensions if (?): $$({\mathrm{\Delta }}_D{)}^2=0$$