form and Christoffel Symbols To show: j j Γk = ωk j where the Γk are the Christoffel symbols defined by

k ∇X ei = Γi (X)ek

i j and where ωj = −ωi are the connection one-forms defined by

i i j dθ = ωj ∧ θ

i with ei an orthonormal frame and θ the corresponding dual coframe, so that

i θ (X) = hei,Xi where h·, ·i is the Riemannian metric, and ∇ its Levi-Civita connection. The proof is based on the “S3-lemma” i lemma. Any 3- γjk which is antisymmetric in two indices and symmetric in two other indices is the i zero-tensor: γjk = 0. i j i i Metric compatibility implies that Γj = −Γi . Indeed, Γj(X) = hei, ∇X eji = −h∇X ej, eii = −Γj(X). Torsion freeness asserts that

∇ei ej − ∇ej ei − [ei, ej] = 0 Applying θk to this vector equation yields

k k k Γj (ei) − Γi (ej) − θ ([ei, ej]).

Now k k k k k dθ (ei, ej) = ei[θ (ej)] − ej[θ (ei)] − θ ([ei, ej]) = −θ ([ei, ej]) k k s and the structure equations dθ = ωs ∧ θ , upon application to the pair (ei, ej) yields

k k k dθ (ei, ej) = ωj (ei) − ωi (ej).

Substituting, we find that k k k k Γj (ei) − Γi (ej) + ωj (ei) − ωi (ej) = 0 k k View this equation as one in which the Γj (ei) are given and the ωj (ei) are unknowns. A particular solution is k k ωj (ei) = Γj (ei). k k k Any other solution differs from this one by a 3-tensor γji which must satisfy γji − γij = 0, i.e. which is k k k symmetric in ij. But the γji must be skew symmetric in k and j, since both ωj and Γj are skew symmetric k in these indices. By the S3-lemma, γji = 0 and so our particular solution is the only solution. j i i k Curvature. To show Ωi = dωj − Σkωk ∧ ωj encodes the curvature as a skew-symmetric valued i two-form, Ωj, related in a simple way to the usual curvature tensor of the Levi-Civita connection viewed as an affine connection. We have d∇e = ωe meaning the following: For a given orthonormal frame e = {ei}i=1,...n, the quantity d∇e is the collection of tangent-bundle valued one-forms which maps V ∈ TxM to {∇X ei}. The right hand side it the collection j V 7→ {Σiωi (V )ej}i=1,...n. Then the curvature is given by

d∇d∇e = Ωe

1 j where the left hand side takes X,Y to {R(X,Y )(ei)} while the right hand side yields {ΣjΩi (X,Y )ej}i=1,...n the one-form with values in the . But d∇d∇e = d∇(ωe) = (dω)e − ωωe = (dω − ωω)e). from whence we get that Ω = dω − ωω, the desired result, when ωω is appropriately interpreted using a matrix wedge product of matrix valued one-forms.

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