Covariant derivative and $\nabla X$ If I consider the covariant derivative of a vector field $Y$ with respect to $X$ this is usually indicated as $\nabla_X Y$, but what's about $\nabla X$? This is the covariant derivative of $X$ and I have read it is  tensor field of type (1,1) such that if $Y\in \chi(\cal{M})$ (vector field) and $w\in \chi^*(\cal{M})$ (covector)
$$\nabla X(Y,w)=w(\nabla_XY)$$
Why it is true? I only know that given a smooth function $f$ then $\nabla_{f X} Y=f\nabla_X Y$, but really I can't understand the relationship between $\nabla X$ and $\nabla_X Y$ and so why the expression above holds.
Can you help me please?
 A: First off, it should be
\begin{equation}
\nabla X(Y,\omega)=\omega(\nabla_YX)
\end{equation}
Over a general vector bundle $E\rightarrow M$ we have that the connection is a map
$$
\Gamma(E)\xrightarrow{\nabla}\Gamma(T^*M\otimes E)
$$
where $\Gamma(-)$ denotes smooth sections. In your case $E=TM$:
$$
\Gamma(TM)\xrightarrow{\nabla}\Gamma(T^*M\otimes TM)
$$
i.e. $\nabla$ eats a vector field and spits out a (1,1) tensor field. We'll work in coordinates for these calculations. So if $\omega$ is a covector field and $Y$ a vector field, then
\begin{align}
\nabla X(Y,\omega)&=\nabla(X^i\partial_i)(Y,\omega) \\
&=(\nabla X^i\otimes\partial_i)(Y,\omega)+(X^i\nabla\partial_i)(Y,\omega) \\
\end{align}
This is since $\nabla$ satisfies a sort of Leibnitz rule. Continuing:
\begin{align}
&=(d(X^i)\otimes\partial_i)(Y,\omega)+(X^i\nabla\partial_i)(Y,\omega)
\end{align}
Since by definition, $\nabla(f)=df$ for any smooth function. Then:
\begin{align}
&=(\partial_m X^idx^m\otimes\partial_i)(Y,\omega)+(X^i\Gamma^{l}_{mi}dx^m\otimes\partial_l)(Y,\omega) \\
\end{align}
Where the $\Gamma$'s are the usual connection coefficients: $\nabla\partial_i=\Gamma^l_{mi}dx^m\otimes\partial_l$. Hence:
\begin{align}
&=(\partial_mX^i)Y^p\omega_\alpha dx^m(\partial_p)\partial_i(dx^\alpha)+X^i\Gamma^l_{mi}Y^p\omega_\alpha dx^m(\partial_p)\partial_l(dx^\alpha) \\
&=(\partial_m X^i)Y^m\omega_i+X^i\Gamma^l_{mi}Y^m\omega_l
\end{align}
On the other hand,
\begin{align}
\omega(\nabla_YX)&=\omega(\nabla_{Y^m\partial_m}X^i\partial_i) \\
&=\omega[Y^m((\partial_m X^i)\partial_i+X^i \nabla_{\partial_m}\partial_i)] \\
&=\omega[Y^m((\partial_m X^i)\partial_i+X^i\Gamma^s_{mi}\partial_s)] \\
&=\omega_ldx^l[Y^m((\partial_m X^s)+X^i\Gamma^s_{mi})\partial_s] \\
&=\omega_l[(\partial_mX^s)Y^m+X^i\Gamma^s_{mi}Y^m]dx^l(\partial_s) \\
&=(\partial_mX^s)Y^m\omega_s+X^i\Gamma^s_{mi}Y^m\omega_s
\end{align}
where I leave the justifications to you. And then replacing dummy indices yields the result.
A: This is really not the right way to look at it. $\nabla X$ is a vector-field valued $1$-form $\omega$ whose value  on $Y$ is $\omega(Y)=\nabla_Y X$. That is, it's giving you a (tensorial) map from vector fields to vector fields. It's a section of $T^*M\otimes TM$, and as such is a $(1,1)$-tensor.
