Differential operator on a manifold in Geometric Calculus

Question

In the context of Geometric Calculus, as stated in book Clifford Algebra to Geometric Calculus (pag. 142), let

$M$ be a differentiable vector manifold,
$F$ be a field on $M$ and
$a$ be a tangent vector at $x$ of $M$, then

the derivative of $F$ in the direction $a$ is defined by:
$$\lim_{\tau \to 0} \frac{F(x(\tau))-F(x)}{\tau} $$ and usually denoted by $\partial_a F$.

The central concept of differential calculus on manifold in this context is the derivative with respect to a point on a manifold, denoted by differential operator $\partial_x$. How can it be defined, provided that $x$ is not a tangent vector of $M$ at $x$?

Answer 1

This uses a typical abuse of notation to talk about $x$ as a point on the manifold as well as a one-parameter function $x(\tau)$ that generates a curve on the manifold.

Your edition seems to be a bit different from mine. Mine says

The derivative of $F$ in the direction $a$ is defined by $$\begin{equation} a \cdot \partial F = a(x) \cdot \partial_x F(x) = \left. \frac{dF}{d\tau} (x(\tau)) \right|_{\tau=0} = \lim_{\tau \to 0} \frac{F(x(\tau)) - F(x)}{\tau} \tag{4.1.5} \end{equation}$$

In any event, it should be understood that $x(\tau)$ denotes a curve such that $x'(0) = a$.

The concepts of vector manifold theory are not very different from conventional differential geometry. The main difference is that points are typically vectors in some embedding space (often, this space is the universal geometric algebra, which has infinite dimensions and takes away the arbitrary nature of embedding).

For this reason, all the results of differential geometry that apply to manifolds in an embedding are valid here. Sometimes tangent vectors are defined as equivalence classes of curves that pass through a given point. These have to be walked back into a flat space using a chart to talk about their derivatives in a formal setting; all vector manifolds have done is eliminated the need to use a chart.