Tional WZ8040 manufacturer curvature R N (, ) c2 f two f two ,

Tional WZ8040 manufacturer curvature R N (, ) c2 f two f two , ( c2 c1 ) n(ii
Tional curvature R N (, ) c2 f two f two , ( c2 c1 ) n(ii)for any linear independent vectors and ; f satisfies one of the following products: f = a1 t a2 , N = -(n – 1) a2 when c1 = 0; 1 f = a1 exp( f = a1 sin(c1 n t ) a2 exp(- 1 – cn t) a2 cos(( n -1) c1 c1 N 4a1 a2 when c1 0; n t ), = n ( n -1) c1 two c1 ( a1 a2 ) when c1 – n t ), N = two nfor any constants a1 and a2 . Proof. Firstly, by [24,25], I f N n is definitely an Einstein manifold together with the continuous if and only if N n has continual Ricci curvature N and f satisfies the differential equations f ( n – 1) N ( n – 1) f 2 . = and = f n n f2 (11)However, following the notations above, the sectional curvatures of your generalized Robertson alker spacetime are provided by (see [22], Lemma five.2) R(u, ) = f , f R(, ) = R N (, ) – f f(12)for any timelike vectors u on I and any spacelike vectors , on N n . So, the conditions (1) and (2) hold if and only if f c = 1 f n c2 and R N (, ) – f fc2 .(13)Solving the first equation of (13), we acquire the expression of the function f and, substituting f in to the second equation of (11), we obtain the worth of your continual N ; thus, together with (11) and (13), we lastly confirm our proof. For much more difficult examples, we are able to construct other warped solution manifolds or twisted item manifolds. 3. Primary Theorems Within this section, we only present our characterization results of spacelike hypersurfaces n with constant scalar curvature in L1 1 , then presenting their proofs in Sections five and six.Mathematics 2021, 9,5 ofBefore giving our main theorems, we have to have some Methyl jasmonate manufacturer fundamental information and notations. Let us n denote as R AB the elements in the Ricci tensor of L1 1 beneath a suitable local orthonormal n 1 n frame e A A=1 ; employing (1), the scalar curvature R of L1 1 is given byn R=A =R AA =i,j=nRijij 2 Rn1in1i =i =ni,j=nRijij 2c1 .(14)Since the scalar curvature of a Ricci symmetric manifold is continuous, we know from (14) n that i,j=1 Rijij can also be a continual. n Let us contemplate the spacelike hypersurface Mn of L1 1 ; we may perhaps opt for en1 because the standard vector, then the second fundamental type B = i,j hij i j en1 with its square 1 length S = | B|two = i,j h2 along with the imply curvature H = n i hii . Therefore, the Gauss equation ij of Mn is given by Rijkl = Rijkl – (hik h jl – hil h jk ). (15) The elements Rij with the Ricci curvature tensor as well as the normalized scalar curvature R of Mn are provided, respectively, by Rij =Rikjk – nHhij hik hkj ,k kn ( n – 1) R =Rikik – n2 H2 S.i,k(16)n If we assume the normalized scalar curvature R of Mn in L1 1 is usually a constant and defineP := R – then P is usually a continuous and (16) becomes1 n ( n – 1)Rijij c,i,jn(n – 1) P = n(n – 1)c – n2 H two S.(17)n In distinct, if L1 1 can be a Lorentz space kind with constant sectional curvature c, then i,j Rijij = n(n – 1)c and P = R; then, (17) is just the Gauss Equation (16). 2 Let be a symmetric tensor on Mn defined by ij = hij – Hij with ||2 = i,j ij . It follows, from (17), that||2 = S – nH 2 = n(n – 1)( H 2 P – c).A well-known reality is the fact that ||two = 0 if and only if Mn is completely umbilical. 1 Now, with c := 2c2 – cn , we’re in the position to state our main benefits.(18)Theorem 1. Let Mn (n 3) be a full spacelike hypersurface with continual normalized scalar n curvature R inside a Ricci symmetric manifold L1 1 satisfying (1) and (two). Let us suppose that H is n and c 0. bounded on M (i) (ii) If n P c, then Mn is entirely umbilical and Mn is entirely geodesic if and only if P = c; ( n -2) c If 0 P n , then either sup ||2 = 0 and Mn i.