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: \[ \det \begin{bmatrix} 0 & R_{12} & R_{13} & R_{14} & 1\\ R_{12} & 0 & R_{23} & R_{24} & 1\\ R_{13} & R_{23} & 0 & R_{34} &1\\ R_{14} & R_{24} & R_{34} & 0 &1 \\ 1&1&1&1&0 \end{bmatrix} =0 \] What about when D=11 and E=32 ? Simplest is D=1, E=2 with (?): \[ R^2 = (x-y)^2 + (x-y)(\theta \zeta + \zeta^* \theta^* ) \] with for example: \[ R^4 = (x-y)^4 + 2 (x-y)^3(\theta \zeta + \zeta^* \theta^* ) + 2(x-y)^2( \theta \theta^* \zeta \zeta^* ) \] \[ R^6 = (x-y)^6 + 3 (x-y)^5(\theta \zeta + \zeta^* \theta^* ) + 6(x-y)^4( \theta \theta^* \zeta \zeta^* ) \] Then the area of the triangle is 0 (or and polynomial with 0 constant term). We should have an identity, something like (?): \[ R^6 - 3 R^4 (x-y)^2 + R^2 (x-y)^4 + (x-y)^6 = 0\] So we would have: \[ \Delta ( L_{AB} ) = 0 \land F(L_{AB},R_{AB}) = 0 \rightarrow ? G(R_{AB})=0\] \[ (R_{xy})^2(R_{xz})^2 = (x-y)^2(x-z)^2 + (x-y)^2(x-z)(\theta \zeta + \zeta^* \theta^* ) + 2 (x-z)^2(x-y)(\theta \beta + \beta^* \theta^* )\] \[ + 2(x-y)(x-z) \theta \theta^* (\zeta \beta^*+\beta\zeta^* ) \] So (?) \[ \Delta \equiv 0 + (x-y)(z-x)(z-y)(\theta \zeta + \zeta^* \theta^*+\zeta \beta^*+\beta\zeta^*+... )\] \[+2(x-y)^2( \theta \theta^* \zeta \zeta^* ) +(x-y)(x-z) \theta \theta^* (\zeta \beta^*+\beta\zeta^* ) +...\] So very simply we can say that the distance geometry is compataible with D real and 4 grassman dimensions if (?): \[ (\Delta_D) ^ 2 =0 \]