Feynman Path Integral Formulation

232 7 Analytical Lattice Expansion Methodsthe hypercube, using the result of Eq. (7.6). This explains the additional factors ofω appearing in the rotation matrices S and T . More importantly, one would expectsuch a combined rotation to be independent of what particular term in the latticeaction one is considering, implying that it can be used to relate other lattice gravitycontributions, such as the cosmological term and higher derivative terms, to theircontinuum counterparts.The lattice action has a local gauge invariance, whose explicit form in the weakfield limit was given in Eq. (7.9). This local invariance has d parameters in d dimensionsand describes lattice diffeomorphisms. In the quantum theory, such localgauge invariance implies the existence of Ward identities for n-point functions. Thechoice of gauge in Eq. (7.22) is of course not the only possible one. Another possiblechoice is the so-called vacuum gauge for which in the continuum h ik,k = 0,h 00 = h 0i = 0. Expressed in terms of the lattice small fluctuation variables such acondition reads in momentum spacee 8 = 0e 9 = 1 2 ω 8 e 1e 10 = 1 2 ω 8 e 2e 12 = 1 2 ω 8 e 4e 11 = 1 3 (1 + ω 8)e 3 − 1 6 (1 − ω 8)(ω 2 e 1 + ω 1 e 2 )e 13 = 1 3 (1 + ω 8)e 5 − 1 6 (1 − ω 8)(ω 4 e 1 + ω 1 e 4 )e 14 = 1 3 (1 + ω 8)e 6 − 1 6 (1 − ω 8)(ω 2 e 4 + ω 4 e 2 ) . (7.28)One can then evaluate the lattice action in such a gauge and again compare to thecontinuum expression. First one expands again the e i ’s in terms of the h ij ’s, as givenin Eq. (7.26), and then expand out the ω’s in powers of k. If one then sets k 4 = 0 onefinds that the resulting contribution can be re-written as the sum of two parts (Hamberand Williams, 2005a,b), the first part being the transverse-traceless contribution14 k2 Tr [ 3 h[P 3 hP − 1 2 PTr (P 3 h)]] = 1 4 k2 ¯h TTij (k) h TTij (k) (7.29)¯h TTij h TTij ≡ Tr [ 3 h[P 3 hP − 1 2 PTr (P 3 h)]] , (7.30)with P ij = δ ij − k i k j /k 2 acting on the three-metric 3 h ij , and the second part arisingdue to the trace component of the metric− 1 4 k2 Tr [PTr (P 3 h)PTr (P 3 h)] = k 2 ¯h T ij(k) h T ij(k) , (7.31)with h T = 1 2 PTr (P 3 h). In the vacuum gauge h ik,k = 0, h ii = 0, h 0i = 0 one canfurther solve for the metric components h 12 , h 13 , h 23 and h 33 in terms of the tworemaining degrees of freedom, h 11 and h 22 ,h 12 = − 12k 1 k 2(h 11 k 2 1 + h 22 k 2 2 + h 11 k 2 3 + h 22 k 2 3)