Tor: the map N1 R defined as dxi dx j dxk xl – ( xl , , ,) e(dxi , dx j , dxk ,).s Even so, as any tensor in N1 is symmetric in the 1st two indices, it readily follows that this map is identically zero. 1 1 If dim X = 3, let us initially describe the Sl-equivariant endomorphisms T3 T3 . To this finish, let x1 , x2 and x3 be coordinates centered at p such that = dx1 dx2 dx3 is positively oriented, and let e be its dual 3-vector. Working with and e, we can construct 16 generators, and they will all be expressed as a 1 permutation with the things of T3 PX-478 manufacturer followed by one of those 4 maps:(a) (b) (c)dxi dx j dxk xl – e(dxi , dx j , dxk) xl , dxi dx j dxk xl – e(dxi , dx j , dxk) ( xl , , l dxi dx j dxk xl – (i) e(dx( j) , dx(k) , ,s N)I ,) , S3 ,(d) dxi dx j dxk xl – ( xl ,) dx(i) e(dx( j) , dx(k) ,), S3 . Because the initial two covariant indices of are symmetric, the following maps are identically zero: (a), (b), and raising the initial two indices at (c) and (d). That leaves eight non-zero generators. Nevertheless, this symmetry also tends to make raising indices 1, three and 2, three indistinguishable, therefore lowering to just four generators.Mathematics 2021, 9,16 ofThe final step will be to verify which of these maps take values in R. Out of those four generators, only the following two make tensors which are skew-symmetric within the initial two covariant indices: 1 dxi dx j dxk xl two dxi dx j dxk xl= ( xl , =l j,) dx j e(dxi , dxk ,) ,)(9) (ten) e(dxi , dxk ,plus the skew-symmetrization in the remaining two is actually a linear mixture of those. None of those two tensors Compound 48/80 References satisfy the first Bianchi identity, however the linear mixture := 31 – 2 does. Lastly, all that is certainly left to prove is that is R-linearly independent of R, C1 , and C2 . So that you can do that, it is enough to find a symmetric linear connection and an orientation on a 3-manifold X such that the aforementioned tensors on X are R-linearly independent. The following example performs: Let be the linear connection on R3 whose only non-zero Christoffel symbols in cartesian coordinates are 1 = x2 x3 11 , two = two = x1 x2 . 23Assume that dx1 dx2 dx3 is positively oriented, and denote Tij := dxi x j . Direct computation provides the following linearly independent tensors, hence finishing the proof: R = dx1 dx2 (- x3 T11 x2 T32) dx1 dx3 (- x2 T11 x2 T22)two 2 dx2 dx3 x1 T22 – x1 x2 T32 ,C1 = dx1 dx1 1 1 x3 T11 – x3 T22 – x1 T31 – x2 T32 2 2 two 1 1 2 two dx1 dx3 x2 T11 – x1 T21 – x3 T23 x1 x2 T31 – x2 T33 two 2 1 1 1 two two dx2 dx3 x2 T12 – x3 T13 – x1 T22 x1 x2 T32 x1 T33 , 2 two 2 3 3 1 x3 T11 x3 T22 x1 T31 x3 T33 two 2 two 1 1 dx1 dx3 – x1 T21 x3 T23 two 2 1 three three dx2 dx3 – x1 T11 – x3 T13 – x1 T22 – x1 T33 , 2 2C2 = dx1 dx= dx1 dx2 ( x1 T31 – x3 T33) dx1 dx3 (2×1 T21 – 3×2 T22 x3 T23 3×2 T33) dx2 dx3 ( x1 T11 – 3×2 T12 2×3 T13).Definition 13. An endomorphism-valued organic 2-form is said to satisfy the second Bianchi identity if it can be closed inside the sense of Definition 11. Theorem 15. The continuous multiples of your curvature are the only endomorphism-valued natural 2-forms that satisfy both the first and second Bianchi identities.Mathematics 2021, 9,17 ofProof. The curvature tensor R is generally a closed natural 2-form, so, by the earlier Lemma, it really is sufficient to analyze the R-linear span on the differentials of C1 , C2 , and, in dimension three, of . If dim X 3, then dC1 and dC2 are linearly independent by [16] (Thm. three.13), as well as the statement follows. If dim X = three, a direct computation, using the exact same instance as within the previous Lemma, proves that.