In case of hypersurfaces why we take { D_X\Xi}=0 ?
where D is a normal connection,\XI is a normal vector field.
For a hypersurface, DX ζ=0 holds if and only if the normal vector field ζ is of constant length. Hence you cannot always take DX ζ=0, unless you assume that the normal vector field ζ is of constant length.
An atomic surface within projection method is assumed to be flat as it derives as orthogonal space decomposition of embedding space into parallel or physical space resp. in perpendicular or internal space. The atomic surface then is nothing but a finite realm within that perp space component.
You might like to point invert your hyper surface and attach that around the cut. This then becomes equivalent, i.e. whether the cut would cross some hypersurface or whether that thickened cut (the acceptance domain) would contain some lattice point (where the hypersurfaces formerly were attached to).
Isn't that the most natural choice? The normal bundle is one-dimensional and trivial (provided the hypersurface is orientable), hence the normal connection has zero curvature, thus you take a basis (just one vector in this case) of parallel sections. If you have an arbitrary section (normal vector field), it is f\Xii for some function f and D_X(f\Xi) = (d_X f)\Xi. Thus the D-derivative is the ordinary derivative.
Best regards
Jost
For a hypersurface, DX ζ=0 holds if and only if the normal vector field ζ is of constant length. Hence you cannot always take DX ζ=0, unless you assume that the normal vector field ζ is of constant length.
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