Vector Fields as Differential Operators Given a manifold $\mathcal{M}$, the notion of a vector field $\xi$ on $\mathcal{M}$ can be interpreted as a collection of arrows on the manifold. In the book The Road to Reality by Roger Penrose, Chapter 12, Section 4, he mentions that the vector field $\xi$ can act on any (smooth) scalar field $\Phi$ to produce another scalar field $\xi(\Phi)$ in the manner of a differential operator. He uses the following definition:

The interpretation of $\xi(\Phi)$ is to be the 'rate of increase' of $\Phi$ in the direction indicated by the arrows that represent $\xi$ ...

The book would like the reader to show that the scalar product
\begin{align*}
\alpha\cdot\xi=\alpha_1\xi^1+\alpha_2\xi^2+\cdots+\alpha_n\xi^n
\end{align*}
is consistent with $d\Phi\cdot\xi=\xi(\Phi)$, in the particular case where $\alpha=d\Phi$ using the chain rule, but unfortunately without knowing the precise definition of $\xi(\Phi)$, I have no way to prove this. There seems to be a similarity between this definition and the directional derivative, and I was always taught that the directional derivitive in multivariable calculus was $\nabla_\mathbf{v}f=\nabla f\cdot\mathbf{v}$ by definition. But if that were true, there would be nothing to prove.
 A: We begin by defining the scalar product of a covector and a vector as
\begin{align*}
\alpha\cdot\xi=\alpha_1\xi^1+\alpha_2\xi^2+\cdots+\alpha_n\xi^n
\end{align*}
Since $\alpha=d\Phi$, we have
\begin{align*}
\alpha\cdot\xi&=d\Phi\cdot\xi
\end{align*}
The covector $d\Phi$ is equal to
\begin{align*}
d\Phi=\left(\frac{\partial\Phi}{\partial x^1},\frac{\partial\Phi}{\partial x^2},\ldots,\frac{\partial\Phi}{\partial x^n}\right)
\end{align*}
Using the expression for the scalar product defined earlier,
\begin{align*}
d\Phi\cdot\xi&=\frac{\partial\Phi}{\partial x_1}\xi^1+\frac{\partial\Phi}{\partial x_2}\xi^2+\cdots+\frac{\partial\Phi}{\partial x_n}\xi^n
\end{align*}
The vector field $\xi$ can be represented by its set of components in terms of partial differential operators, and we will of course choose the coordinates to be the same as the previously used coordinates for $\Phi$:
\begin{align*}
\xi=\xi^1\frac{\partial}{\partial x^1}+\xi^2\frac{\partial}{\partial x^2}+\cdots+\xi^n\frac{\partial}{\partial x^n}
\end{align*}
Therefore,
\begin{align*}
\xi(\Phi)&=\left(\xi^1\frac{\partial}{\partial x^1}+\xi^2\frac{\partial}{\partial x^2}+\cdots+\xi^n\frac{\partial}{\partial x^n}\right)\Phi\\
&=\xi^1\frac{\partial\Phi}{\partial x^1}+\xi^2\frac{\partial\Phi}{\partial x^2}+\cdots+\xi^n\frac{\partial\Phi}{\partial x^n}
\end{align*}
Hence, the definition of the scalar product implies $d\Phi\cdot \xi=\xi(\Phi)$ as required.
