The Riemann curvature tensor is defined as:
$
R(X,Y) = [\nabla_X, \nabla_Y]
$
when there is no curvature (no loss of generality in the question). If we expand this to coordinate notation, we get the following expression involving the connection:
$
R^\rho_{\sigma \mu \nu} = \partial_\mu \Gamma^\rho_{\sigma \nu} + \partial_\nu \Gamma^\rho_{\sigma \mu} + \Gamma^\rho_{\mu \lambda} \Gamma^\lambda_{\nu \sigma} – \Gamma^\rho_{\nu \lambda} \Gamma^\lambda_{\mu \sigma}
$
My question is, why is it a rank $4$ tensor, or even a tensor? As the connection is not a tensor, but a Lie algebra valued one-form, it would be more logical if the Riemann curvature tensor were not a tensor, as it is composed of the Christoffel symbols. Is the fact that the Riemann curvature tensor is a tensor just pure coincidence, or is there a reason behind this?
Best Answer
The notation $R(X,Y)=[\nabla_X,\nabla_Y]$ is somewhat vague. In particular, $\nabla_X$ refers to a family of differential operators (since covariant derivatives can be applied to all kinds of objects), and the definition of the Riemann tensor has a much more specific case in mind:
Let $\mathfrak{X}M$ be the space of vector fields on a smooth manifold $M$. The Riemann tensor $R$ of an affine connection $\nabla$ is a multilinear map $(\mathfrak{X}M)^3\to \mathfrak{X}M$ defined by $$ R(X,Y)Z=\nabla_X(\nabla_YZ)-\nabla_Y(\nabla_XZ)-\nabla_{[X,Y]}Z $$ Where the last argument $Z$ is left outside of the parenthesis as a matter of convention. It's a basic result that a multilinear map between copies $\mathfrak{X}M$ (or its various tensor spaces) is tensorial (i.e. equivalent to a contraction with a particular tensor field) if and only if it is $C^\infty(M)$-linear. This is (somewhat surprisingly) the case for $R$ defined as above; even though it is built out of differential operators, the differential parts "cancel out" and result in a tensorial map.
As to your comment about Lie algebra valued forms, as it is formulated in elementary Riemannian geometry, connections are not Lie algebra valued forms. Principal connections on principal $G$-bundles can be defined as Lie algebra-valued $1$-forms, but here $\nabla$ refers instead to a affine connection on a vector bundle ($TM$ in this case), which is a differential operator and not a Lie algebra-valued form. We can still interpret $R$ as a Lie algebra-valued $2$-form, though.