I am confused by what the Leibniz rule means here. In my understanding, Leibniz rule is only for connection on tensor field bundle in the following form $$\begin{align}
\nabla_X\left(T(\omega^1,\dots,\omega^r,X_1,\dots, X_s)\right)&=(\nabla_XT)(\omega^1,\dots, \omega^r,X_1,\dots, X_s)\\
&+\sum_{i=1}^rT\left(\omega^1,\dots, \nabla_X(\omega^i),\dots, \omega^r,X_1,\dots, X_s\right)\\
&+ \sum_{j=1}^sT\left(\omega^1,\dots, \omega^r,X_1,\dots,\nabla_X(X_j),\dots, X_s\right).
\end{align}$$
But for the one in the link, it does not treat the curvature operator as the (1,3) type tensor. I wonder how that formula is obtained.
Edit: I understand that the curvature operator can be thought as a linear mapping over smooth functions from $\Gamma(TM)\times\Gamma(TM)\times\Gamma(TM)$ to $\Gamma(TM)$. But the formula above for Leibniz rule only works for tensors; how can one obtain the one in the link for linear mapping from $\Gamma(TM)\times\Gamma(TM)\times\Gamma(TM)$ to $\Gamma(TM)$.
Best Answer
Well, you’re going to need a definition of how $\nabla$ acts on such guys. And the definition is pretty much to enforce the desired Leibniz rule: \begin{align} (\nabla_{\xi} R)(X,Y)[Z]&:=\nabla_{\xi}\left(R(X,Y)[Z]\right)-R(\nabla_{\xi}X,Y)[Z]-R(X,\nabla_{\xi}Y)[Z]-R(X,Y)[\nabla_{\xi}Z]. \end{align}
For clarity, let me denote the curvature operator as done here, $R(X,Y)[Z]$, and let me call the associated $(1,3)$ tensor field $r$, so that for all covector fields $\omega$ and vector fields $X,Y,Z$, we have \begin{align} r(\omega,Z,X,Y):=\omega(R(X,Y)[Z]). \end{align} You can then check that with the above definition, $\nabla_{\xi}R$ ‘corresponds’ to $\nabla_{\xi}r$.
Notice that the way I presented things, you first need a definition for how $\nabla$ acts on $R$; only then can you ask how it is related to the (‘originally defined’) action of $\nabla$ on $r$ (and notice that the same notation $\nabla$ is being used for multiple different (though related) purposes).
Alternatively, you can define the action of $\nabla$ on $R$ to be such that $\nabla_{\xi}R$ ‘corresponds’ to $\nabla_{\xi}r$ (or more formally, a certain square diagram commutes), and then as a consequence, you can derive the above formula.
But in either case, the answer is that it’s almost by definition.
Finally, I should mention that whenever you deal with some new ‘flavour’ of tensor fields, you should really ask yourself which vector bundle you’re working with, and whether/how a connection in your original vector bundle induces a connection in the new vector bundle. For instance, above, we viewed the curvature operator as a map taking 3 vectors and spitting out a vector; but really the ‘better’ way to look at it is to think of it as eating two vectors and spitting out an endomorphism (which aligns itself nicely with the more general viewpoint that the curvature of a linear connection $\nabla$ in a vector bundle $(E,\pi,M)$ is an $\text{End}(E)$-valued $2$-form on $M$… and now one can ask about how a linear connection on $E$ induces one on $\text{End}(E)$ and so on).