Derive formula for Dirac operator using its coordinate free definition I was looking at this article about Dirac operator where a coordinate-free definition is given by this
How to derive from this abstract definition to the usual one given in text, i.e.

 A: The coordinate-free definition is explained in the sentence following it:

where  the  first  map  is  the  connection,  the  second  map  is  the  isomorphism $T^∗X \simeq TX$ determined  by  the  Riemannian  structure,  the  third  map  is  the  natural  inclusion,  and  thefourth map is Clifford multiplication.

Let us detail these. The first map is
\begin{align}
\nabla \colon V & \longrightarrow T^*M \otimes V \\
s & \longmapsto \nabla s 
\end{align}
and the second map is
\begin{align}
^{\sharp}\colon  T^*X\otimes V & \longrightarrow TX \otimes V \\
\alpha \otimes s &\longmapsto \alpha^{\sharp}\otimes s
\end{align}
extended by linearity. If $(e_1,\ldots,e_n)$ is an orthonormal basis and $\alpha^{\sharp} = \sum_{i=1}^n\alpha(e_i)e_i$. The third map is the inclusion map, and is quite explicit. Finally, the last map is the Clifford action:
\begin{align}
c \colon \mathrm{Cliff}(TX) \otimes V & \longrightarrow V \\
e_i \otimes s & \longmapsto e_i \cdot s  
\end{align}
Hence, combining these maps gives:
$$
D s = c \left(\left(\nabla s\right)^{\sharp} \right) = c\left(\sum_{i=1}^n e_i \otimes \nabla_{e_i}s \right)= \sum_{i=1}^n e_i\cdot \nabla_{e_i}s.
$$
A: In fact, this is rather a question of multilinear algebra. The covarint derivative in the coordinate-free definition can be interpreted as assigning to $s$ the linear map $\xi\mapsto\nabla_\xi s$. Fixing $s$, this can actually be interpreted point-wise as $\xi(x)\mapsto\nabla_\xi s(x)$. Next you want to convert this into an element of $T_xX\otimes V_x$ using the inner product on the tangent space defined by the Riemannian metric. This can be most easily expressed in terms of an orthnormal basis $\{e_i\}$ and it gives $\sum_i e_i\otimes \nabla_{e_i}s(x)$. Then you just have to apply Clifford multiplication to this expression to get the point-wise expression. Writing things in term of a local orthonormal frame brings you to the expression you are looking for.
