Conical Semi-Groups Versus Torroidal Semi-Groups, Displacements

Let us say that one were to consider one conical-shaped orbifold that holomorphically struck a torroidal-shaped orbifold, at a spot other than at the general locus of its annulus, to where the latter general format of orbifold -- of which were to here be approaching the so-eluded-to conical-shaped orbifold, at the same velocity, while in the process, was moving in the directly antiholomorphic directoral-based Hamiltonian mappable tracing, over time.  Let us, furthermore, say that both the so-stated conical-shaped orbifold and the torroidal-shaped orbifold that I have mentioned here were to bear the same Hodge-Index of Hamiltonian-based composition -- both in terms of the overall intrinsic rest energy quanta, and, also in terms of their intrinsic Lorentz-Four-Contraction, in so that the degree of compactification of both of the orbifolds that I have mentioned here were to be the same -- in terms of the resultant of the Lorentz-Four-Contractions of both of the said orbifolds, as both of the so-eluded-to semi-groups are to here approach each other in a Ward-Caucy manner that relates to a tense of 180 degrees.  So, here, the resultant genus of Polyakov-Action of both of the so-stated orbifolds are to be of the same parity -- as each of the said orbifolds are to here approach each other on a common Real Reimmanian plane, that curves as to the intrinsic curvature of space-time-fabric, and not of a Wilson Line-based plane of binary approach.  Here, the conical-based orbifold structure that I have eluded-to will tend to scatter the torroidal-based orbifold structure in a Rayleigh scattering -- as the two so-eluded-to orbifolds are to here strike each other upon the general gauge-metric, in which the two said semi-groups are to then work to bear a viable Gliossi-based Yakawa Coupling.  Yet, if everything here were to be the same, except that both of the so-eluded-to orbifold eigensets that are to here be involved were to be of a torroidal-based nature, then, the scattering will, instead, be of a Reimman scattering.  To Be Continued!  I will continue with the suspense later!!! Sam.