Let $\alpha$ be a differential form on a smooth manifold $M$. For simplicity, let's suppose that it is a $1$-form. Then we can think of $\alpha$ as a smooth map from $M$ to $T^* M$, the cotangent bundle.
The exterior derivative $d\alpha$ is a $2$-form on $M$ that somehow "differentiates" $\alpha$. On the other hand, a $2$-form is a special kind of (i.e. alternating) bundle map from $TM$ to $T^* M$. Thinking of $\alpha$ as a smooth map, we obtain a map $D \alpha$ from $TM$ to $TT^* M$. Given a linear Ehresmann connection on the vector bundle $T^* M$, there is a linear map $\phi$ from $TT^* M$ to $T^* M$, with the latter viewed as the sub-bundle of $TT^* M$ which projects to $0$ in $M$. So, $\phi \circ D \alpha$ is a bundle map from $TM$ to $T^* M$, also known as a section of $T^*M \otimes T^* M$. This has a canonical projection $P$ to its skew-symmetrization $T^* M \wedge T^* M$, so $P \circ \Phi \circ D \alpha$ gives a section of $T^* M \wedge T^* M$, also known as a $2$-form.
Perhaps more transparently, we can write this in terms of the covariant derivative associated to the Ehresmann connection. This is an operator $\nabla$ which maps sections $\alpha$ of $T^*M$ to a "$T^*M$-valued $1$-form" $\nabla \alpha: X \mapsto \nabla_X \alpha$, where $\nabla_X \alpha$ is a $1$-form, such that the map is tensorial in $X$ and follows the Leibniz rule in $\alpha$. This is related to $\phi$ precisely by the formula $\phi \circ D \alpha(X) = \nabla_X \alpha$, so this is the same as the previous definition, up to the skew-symmetrization. This should replace the contravariant $2$-tensor $\nabla \alpha(X, Y) = (\nabla_X \alpha)(Y)$ with the $2$-form $\omega_\alpha(X, Y) = (\nabla_X \alpha)(Y) – (\nabla_Y \alpha)(X)$. By the Leibniz rule for $\nabla$, we have \begin{align*}\omega_{f\alpha}(X, Y) &= \nabla_X (f\alpha)(Y) – \nabla_Y (f\alpha)(X)\\
&= f \omega_\alpha(X, Y) + X(f)\alpha(Y) – Y(f) \alpha(X)\\
&= f\omega_\alpha(X, Y) + (df \wedge \alpha)(X, Y)
\end{align*}
So the map $\alpha \mapsto \omega_\alpha$ satisfies the usual derivation property of the exterior derivative and produces an honest $2$-form. However, it uses the entirely non-intrinsic Ehresmann connection $\nabla$. What gives?
Does something like this work for higher $k$-forms? Replacing the bundle $T^* M$ with $\wedge^k T^* M$ everywhere in the arguments gives an analogous definition of an "exterior derivative" of an arbitrary $k$-form, but it is not so clear that this now satisfies $d(\omega_1 \wedge \omega_2) = d\omega_1 \wedge \omega_2 + (-1)^k \omega_1 \wedge d \omega_2$.
EDIT I realized this was confusing – when I say an Ehresmann connection, I'm referring to one that is linear, so it is equivalent to the usual notion of covariant derivative for a vector bundle. However, I wanted the definition to include connections that aren't necessarily determined by an affine connection on $TM$.
Best Answer
An affine connection induces the exterior derivative by taking the ant symmetrization of the covariant derivative if and only if the torsion of the connection vanishes. This can be computed directly.