Why is the Riemann curvature tensor a rank 4 tensor?

Question

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?

Answer 1

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.

Answer 2

The proof of this went like :

$$(\nabla _a \nabla _b -\nabla_b \nabla )(fw^c)=f((\nabla _a \nabla _b-\nabla _b \nabla _a) w^c))$$

$f$ is a scalar field. Take a special case of this identity. Take $f$ to be equal to $1$ at just one manifold point $p$, and $\neq 1$ at every other point.

So when you compute $fw^c$, the value of the vector field $w^c$ changes at every other point except at $p$. The RHS says that $R(fw^c(p))=f(p)R(w^c(p))=R(w^c(p))$.

This means that, for fixed $p_0$, $Rw^c(p_0)$ is solely a function of $w^c(p_0)$. It doens't depend on $w^c(p)$ at any other point $p$. This means $R$ is a linear map from dual vectors to tensors, and is hence a tensor.

Answer 3

Since you're not even convinced it has to be a true tensor, we'll work throughout with covariant derivatives, which by definition map tensors to tensors. In fact, it's generally a good idea to use covariant derivatives as much as you can early in a calculation like this; that way you'll need use Christoffel symbols not only less, with better understanding.

Unlike partial derivatives, covariant derivatives don't commute. But although $\nabla_a\nabla_bV_c$ is a second-order derivative of a vector field $V_c$, $[\nabla_a,\,\nabla_b]V_c$ is easily shown to have no second-order part. As it's the difference of two tensors, it must be of the form $R_{abcd}V^d+X_{abcde}\nabla^dV^e$ for some tensors $R,\,X$. And a more general tensor will have$$[\nabla_a,\,\nabla_b]T_{c_1\cdots c_q}{}^{f_1\cdots f_p}=\sum_{I=1}^qR_{abc_Id}T_{c_1\cdots c_{I-1}}{}^d{}_{c_{I+1}\cdots c_q}{}^{f_1\cdots f_p}-\sum_{J=1}^pR_{ab}{}^{f_J}{}_{g}T_{c_1\cdots c_q}{}^{f_1\cdots f_{J-1}gf_{J+1}\cdots f_p}+\cdots$$(where $X$s appear in the last $\cdots$), so as to satisfy the product rule, contraction etc. It's not hard to see $X=0$ in a torsion-free theory, but either way the rank-$4$ Riemann tensor has naturally arisen.

Then we can compute it from Christoffel-based formulae for covariant derivatives. "Why is it a tensor?" then makes about as much sense as asking why $\partial_bV^c+\Gamma_{bd}{}^cV^d$ is a tensor: that's how Christoffel symbols are defined. Similarly, the formula you quoted for the Riemann tensor's components describes a tensor because it emerges from an obviously tensorial definition of same.