JOURNAL Series A - Pure and Applied Mathematics

26 Iva DokuzovaIf (M, g, J) is a locally decomposable space, then R hijk J ij = S ah Jk a . Thus itfollows R hijk J ij J hk = τ. Then from (22) we findτ =1n 2 − tr 2 J ((n2 − n)( ˜P n − P trJ) + n − tr2 J(4˜QtrJ − Qn)). (25)Collecting the system (24), (25), we getFrom (10), (11), (13), and (14) we obtainQn − ˜QtrJ = 4( ˜P n − P trJ). (26)P = ∇ k b k + (1 − n 2 )ϕ˜P = ∇ k˜bk + φ − trJ2 ϕ (27)Q = 2 ˜P − φ + ϕtrJ = 2∇ k˜bk + φ˜Q = 2P + (n − 1)ϕ = 2∇ k b k + ϕ.At first we consider the case φ = ϕ = 0.Remark Obviously, everywhere in our considerations we suppose that g is undefinedmetric. If we accept that g is defined, then we see that (7) is trivial transformation.Thus the vector field b is isotropic one and (27) has the form:P = ∇ k b k , ˜P = ∇k˜bk , Q = 2∇ k˜bk , ˜Q = 2∇k b k . (28)From (21), (26), and the last system we getP n − ˜P trJ = 0˜P n − P trJ = 0.It’s only solution is P = ˜P = Q = ˜Q = 0. So we get R = 0. We prove that ∇ is alocally flat connection if φ = ϕ = 0.Now, let us consider the case ˜b j = b j (φ = ϕ). We have ∇ k˜bj = ∇ k b j and from(27) we findP = ∇ k b k + (1 − n 2 )ϕ˜P = ∇ k b k + (1 − trJ )ϕ + φ (29)2Q = ˜Q = 2∇ k b k + φ.From (21), (26), and the last system we get ˜P = P . Then we have ϕ = 0. This casereduces to the previous case. Analogously if ˜b j = −b j (φ = −ϕ) we get ϕ = 0. Sothe theorem is proved.