Confusion about tensor fields taking values in vector fields rather than functions

differential-geometrytensors

I'm beginning with tensorial calculus and I have some questions. Let $(M,g)$ a riemannian manifold with $\nabla$ his Levi Civita connection. The curvature tensor $R$ is defined as

\begin{align*}
R : \mathfrak{X}(M) \times \mathfrak{X}(M) \times \mathfrak{X}(M) &\to \mathfrak{X}(M)\\
(X,Y,Z) &\mapsto R(X,Y)Z
\end{align*}

where

$$R(X,Y)Z = \nabla_X \nabla_Y Z – \nabla_Y \nabla_X Z – \nabla_{[X,Y]}Z.$$

In all books of differential geometry, they said that $R$ is a $(1,3)$ tensor but I don't know why, because a $(1,3)$ tensor is a multilinear map

$$\Omega^1(M) \times \mathfrak{X}(M) \times \mathfrak{X}(M)\times\mathfrak{X}(M) \to C^\infty(M).$$

I think that I don't understand any concept or because for example, let $X \in \mathfrak{X}(M)$, then $\nabla X : TM \to TM$, $(\nabla X)(v_p) = \nabla_{v_p}X$, is a $(1,1)$ tensor field and I don't know why.

$\textbf{Remark}$: $\Omega^1(M)$ is the set of all 1-forms, $\alpha : M \to TM^*$.

Best Answer

A $(p, q)$-tensor field on a smooth manifold $M$ is a $C^{\infty}(M)$-multilinear map $T : \Omega^1(M)^p\times\mathfrak{X}(M)^q \to C^{\infty}(M)$.

Given a $C^{\infty}(M)$-multilinear map $S : \Omega^1(M)^p\times\mathfrak{X}(M)^q \to \mathfrak{X}(M)$, there is an associated $(p + 1, q)$-tensor field $T : \Omega^1(M)^{p+1}\times\mathfrak{X}(M)^q \to C^{\infty}(M)$ defined by

$$T(\beta, \alpha^1, \dots, \alpha^p, X_1, \dots, X_q) := \beta(S(\alpha^1, \dots, \alpha^p, X_1, \dots, X_q)).$$

Likewise, given a $C^{\infty}(M)$-multilinear map $S : \Omega^1(M)^p\times\mathfrak{X}(M)^q \to \Omega^1(M)$, there is an associated $(p, q + 1)$-tensor field $T : \Omega^1(M)^p\times\mathfrak{X}(M)^{q+1} \to C^{\infty}(M)$ defined by

$$T(\alpha^1, \dots, \alpha^p, Y, X_1, \dots, X_q) := (S(\alpha^1, \dots, \alpha^p, X_1, \dots, X_q))(Y).$$