# What is the precise relationship between connections in differential geometry and Kähler differentials?

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.

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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 $m \in M$, $$\bigtriangledown(am)=m \otimes da+a \bigtriangledown m$$
I wanted to put this as a comment, but I don't have any reputation points yet. I'm a little confused by John's answer. It is true (in differential geometry) that you can think of $\nabla$ as a map from $M$ to $M \otimes T*(M)$ (the cotangent bundle). How does the cotangent bundle relate to the module of Kahler differentials? I'm finding it hard to connect the dots.. –  Nicole Jul 5 '11 at 22:00
Notice,though, that if $M$ is a smooth manifold, then the module of Kähler differentials of the ring $C^\infty(M)$ is not the same thing as the $C^\infty(M)$-module of $1$-forms on $M$. In particular, when $M=\mathbb R$, in the former $\mathrm d(\mathrm e^t)\neq\mathrm e^t\mathrm dt$. (One needs to construct a "module of Kähler differentials" with more care —taking into account topologies when constructing tensor products— in order to obtain an equality) –  Mariano Suárez-Alvarez Jul 5 '11 at 22:07
@Mariano - I've been thinking about this some more. Since $d(e^t) \neq e^tdt$ in the module of Kahler differentials, if we use such differentials in our definition of the connection (rather than the usual differential 1-forms), do we get something essentially different? I think not, since the role of the 1-forms in the theory of connections is simply to pair with a derivation. To continue your above example, if the derivation is $\partial/\partial t$, then its natural pairing with $e^tdt$ gives $e^t$, but its pairing with $d(e^t)$ also gives $e^t$, so everything stays consistent. –  John M Aug 26 '11 at 21:46