TITLE: What is the precise relationship between connections in differential geometry and Kähler differentials?
QUESTION [6 upvotes]: I've been reading some differential geometry at my leisure, and I couldn't help but getting a very familiar feeling when I've read the definition of a connection:
A derivation of $M$ (or in some notations, the derivation of the identity map $M\rightarrow M$) is defined as: a function, $D$, that takes tensor fields to tensor fields of the same type, such that $D(C \otimes A)=D(C) \otimes A + C\otimes D(A)$ for any two tensor fields $C$ and $D$, and such that $D(aA+bB)=aD(A)+bD(B)$ for any two tensor fields $A$ and $B$ and (real) scalars $a$ and $b$.
A connection is defined as a function $\nabla$ that takes a vector field $X$ to a derivation $\nabla _X$, such that $\nabla$ satisfies: If $f$ is a function on $M$ then $\nabla_X(f)=Xf$, and $\nabla$ is linear (for the module of vector fields over the ring of $C^{\infty}$-functions), and such that $\nabla_X$ commutes with contraction.
This was the first time I saw the definition of a connection formulated in this way, and it is very reminiscent of themes in Kähler differentials. I wonder if there is a rigorous relationship between the two notions.

REPLY [2 votes]: Indeed there is!  A more abstract algebraic definition of a connection is:
Let $A$ be a commutative algebra over a field $k$.  Let $\Omega^1(A)$ be the module of Kähler differentials on $A$.  Let $d:A\rightarrow\Omega^1(A)$ be the universal derivation.
A connection $\bigtriangledown$ on an $A$-module $M$ is a $k$-linear map,
$$\bigtriangledown:M\rightarrow M\otimes\Omega^1(A)$$ such that, for $a\in A$ and $m \in M$, $$\bigtriangledown(am)=m \otimes da+a \bigtriangledown m$$