Feynman Path Integral Formulation

6.5 Invariant Lattice Action 1796.5 Invariant Lattice ActionThe first step in writing down an invariant lattice action, analogous to the continuumEinstein-Hilbert action, is to find the lattice analogue of the Ricci scalar. From theexpression for the Riemann tensor at a hinge given in Eq. (6.36) one obtains bycontractionR(h) =2 δ(h)A C (h) . (6.37)The continuum expression √ gR is then obtained by multiplication with the volumeelement V (h) associated with a hinge. The latter is defined by first joining the verticesof the polyhedron C, whose vertices lie in the dual lattice, with the vertices ofthe hinge h, and then computing its volume.By defining the polygonal area A C as A C (h)=dV(h)/V (d−2) (h), where V (d−2) (h)is the volume of the hinge (an area in four dimensions), one finally obtains for theEuclidean lattice action for pure gravityI R (l 2 )=− k∑hinges hδ(h)V (d−2) (h) , (6.38)with the constant k = 1/(8πG). One would have obtained the same result for thesingle-hinge contribution to the lattice action if one had contracted the infinitesimalform of the rotation matrix R(h) in Eq. (6.32) with the hinge bivector ω αβ ofEq. (6.8) (or equivalently with the bivector U αβ of Eq. (6.31) which differs fromω αβ by a constant). The fact that the lattice action only involves the content of thehinge V (d−2) (h) (the area of a triangle in four dimensions) is quite natural in viewof the fact that the rotation matrix at a hinge in Eq. (6.32) only involves the deficitangle, and not the polygonal area A C (h).An alternative form for the lattice action (Fröhlich, 1981) can be obtained insteadby contracting the elementary rotation matrix R(C) of Eq. (6.32), and not just itsinfinitesimal form, with the hinge bivector of Eq. (6.8),I com (l 2 )=− k∑hinges h12ω αβ (h)R αβ (h) . (6.39)The above construction can be regarded as analogous to Wilson’s lattice gaugetheory, for which the action also involves traces of products of SU(N) color rotationmatrices (Wilson, 1973). For small deficit angles one can of course useω αβ =(d − 2)!V (d−2) U αβ to show the equivalence of the two lattice actions.But in general, away from a situation of small curvatures, the two lattice actionare not equivalent, as can be( seen already in ) two dimensions. Writing the rotationcosδ sinδmatrix at a hinge as R(h)=, expressed for example in terms of−sinδ cosδPauli matrices, and taking the appropriate trace (ω αβ = ε αβ in two dimensions) onefindstr [ 12(−iσ y )(cosδ p + iσ y sinδ p ) ] = sinδ p , (6.40)