Trace-free curvature-type tensors in dimension three vanish

Let T(x, y, z, w) be a quadrilinear form on a free module of rank 3 which is antisymmetric in its first pair and in its last pair of arguments, as a Riemann curvature tensor Rm(x, y, z, w) = ⟨R(x, y)z, w⟩ is. If its Ricci-type contraction ∑ᵢ T(eᵢ, y, z, eᵢ) vanishes for some basis (eᵢ), then T = 0.

Applied to the Weyl tensor, which is trace-free and has these antisymmetries, this shows that the Weyl tensor of a three-dimensional Riemannian manifold vanishes, so that its Riemann curvature is determined by its Ricci curvature.

The scalars form a commutative ring in which 2 is left regular, that is, multiplication by 2 is injective. In characteristic 2 the statement fails: T(x, y, z, w) = x₁ y₁ z₂ w₂ on 𝔽₂³ is a counterexample.

Main declarations

• LinearMap.eq_zero_of_sum_basis_eq_zero_of_card_eq_three: a quadrilinear form on a module with a basis of three elements, antisymmetric in both pairs of arguments and with vanishing contraction against that basis, is zero.

References

• A. L. Besse, Einstein Manifolds, Springer (1987), Section 1.G.

theorem LinearMap.eq_zero_of_sum_basis_eq_zero_of_card_eq_three {R : Type u_1} {M : Type u_2} {ι : Type u_3} [CommRing R] [AddCommGroup M] [Module R M] [Fintype ι] (h2 : IsLeftRegular 2) (b : Module.Basis ι R M) (hι : Fintype.card ι = 3) {T : M →ₗ[R] M →ₗ[R] M →ₗ[R] M →ₗ[R] R} (h₁ : ∀ (x y z w : M), (((T x) y) z) w = -(((T y) x) z) w) (h₂ : ∀ (x y z w : M), (((T x) y) z) w = -(((T x) y) w) z) (hT : ∀ (y z : M), ∑ i : ι, (((T (b i)) y) z) (b i) = 0) : T = 0

A quadrilinear form T on a module with a basis (bᵢ) of three elements which is antisymmetric in its first and in its last pair of arguments, and whose contraction ∑ᵢ T(bᵢ, y, z, bᵢ) vanishes, is zero, provided multiplication by 2 is injective on the scalars. For an orthonormal basis of an inner product space this contraction is the Ricci contraction of T.