D O C . 1 5 8 O N R I E M A N N C U RVAT U R E T E N S O R 1 7 7 Here, yields the same curvature in the surface elements ( and ( ), whereas yields the surface curvature of the same magnitude but opposing directions. If for the sake of brevity one sets (5a) , this means that for any choice of surface element, the equations[13] (5b) should be satisfied. With these conditions the decomposition is uniquely deter- mined. We will now see that the ࣔantisymmetrical” component is closely related to the tensor , to the extent that the vanishing of one has the vanishing of the other as a consequence and vice versa. For, (6) , where, to abbreviate, we set (7) . For the proof we first introduce the tensor (8) where or stands for the values ±1, depending on whether iklm is an even or an odd permutation of (1, 2, 3, 4). Then (9) holds, since it is easy to prove that then the equations (5a) and equation (10) are satisfied. Furthermore, we introduce the notation (11) , and analogously for the tensors and . If we multiply (4) by , we obtain on the basis of (11): (12) , while on the basis of (9) and (11) the relations (5b) assume the form Sik lm , f ik) f ik Aik lm , gilgkm gimgkl)f – ( ikf lm gilgkm gimgkl)f – ( ikf lm 1 = = Sik lm , f ikf lm Sik lm , f ikf lm = Aik lm , f ikf lm Aik, lm f ikf lm = Aik, lm Rik 1 4 --gikR - – Aik lm , 1 2 --( - gilGkm gkmGil gimGkl – gklGim) – + –= Gil Rik 1 4 -- - gikR –= Δik αβ 1 2 --------δαβστgσigτk - 1 g - 1 2 -- - gδikστgσαgτβ = = δiklm δiklm f ik Δαβ ik f αβ = gilgkm gimgkl)f – ( ikf lm 0= R ik lm , Δik λρΔlm στRλρ στ , = [p. 102] Sik lm , Aik lm , f ukf lm R ik lm , S ik lm , A ik lm , –=