Why is Rieman Curvature tensor a (1,3) tensor?

Question

First for the covariant derivative in local coordinates:

$$\nabla_XY = (X^i\partial_i Y^k + X^i Y^j \Gamma_{ij}^k)\partial_k$$

So the basis $\frac{\partial}{\partial x_k}$ requires $f : M \to \mathbb{R}$ as in put.

The Riemann Curvature tensor has term say $\nabla_X \nabla_Y Z$. The map is $(T_pM \times T_pM \times T_pM) \to T_p(M)$, There are only three arguments it doesn't send to $\mathbb{R}$, so how can it be a $(3,1)$ tensor?

Answer 1

Reading your comments and your question, I believe there are two primary sources for your confusion.

1. The Riemann curvature is not a tensor, it is a tensor field. (Often people use the word tensor when they mean tensor field.)
2. In fact, the Riemann curvature is not technically a $(1,3)$-tensor field either, we require an identification $$\hom(V\times V \times V \times V^*,\mathbb R) \cong \hom(V\times V \times V, V).$$

I will now explain these in more detail. A tensor is an element of $$V\otimes\cdots\otimes V\otimes V^*\otimes\cdots \otimes V^*.$$ Tensors only make sense given some vector space $V$. However, the Riemann curvature is defined in terms of a connection! It doesn't just exist on a single tangent space, it exists on the entire manifold. The Riemann curvature is not a tensor, instead, it is a tensor field. In other words, it is a smooth section of a tensor bundle. While a tensor takes vectors and dual vectors as inputs and real numbers as outputs, a tensor field takes vector fields as inputs and outputs smooth real valued functions. That is, if the base manifold is $M$, $$T:\Gamma(TM) \times \cdots \times \Gamma(TM)\times\Gamma(T^*M)\times\cdots\times\Gamma(T^*M)\to C^{\infty}(M).$$ Now let's look at the specific case of the Riemann curvature. We find that the domain and codomain are $$R:\Gamma(TM) \times \Gamma(TM)\times \Gamma(TM) \to \Gamma(TM).$$ You might say - hey that's not a tensor field! The codomain is wrong! But we now have access to a natural isomorphism as stated above. Specifically, on each tangent space, the isomorphism is \begin{align*}\varphi :\hom(V\times V \times V, V) &\to \hom(V\times V \times V \times V^*, \mathbb R)\\ L &\mapsto [(u,v,w,\omega)\mapsto \omega(L(u,v,w))] \end{align*} This isomorphism can be used to show that there is a natural identification of the Riemann curvature with some tensor field $\widehat R$: $$\widehat R:\Gamma(TM) \times \Gamma(TM)\times \Gamma(TM) \times \Gamma(T^*M)\to C^{\infty}(M).$$ Because we have access to this natural isomorphism, many maps (e.g. linear endomorphisms) can be identified with tensors/tensor fields, and we refer to all of them as tensors.