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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 $TM$.

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  • $\begingroup$ I couldn't understand exactly your question, but any connection on a principal bundle can be written locally as $d + A$ where $A$ is a Lie algebra valued form. Furthermore you can extend the connection to higher forms by Leibniz rule in order to get a "complex" (not exactly a complex if your connection is not flat). $\endgroup$ – user40276 Oct 7 '16 at 4:51
  • $\begingroup$ By reading better your question, it seems that the term $- \alpha ([X, Y])$ is missing in your $\omega_{\alpha}$. The general expression should be $d_{\nabla} \alpha (X_1, …, X_{k + 1}) = \sum_i (-1)^{i + 1} \alpha (X_1, …, \hat{X_i}, …, X_{k+ 1}) + \sum_{i > j}(-1)^{i + j} \alpha ([X_i, X_j], …, \hat{X_i}, …, \hat{X_j}, …, X_{k + 1})$ . This is actually the differential for the tangent Lie algebroid with values in a vector bundle (i.e, a representation of this Lie algebroid on a vector bundle). $\endgroup$ – user40276 Oct 7 '16 at 6:27
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    $\begingroup$ No, there is nothing missing. By Leibniz we have $(\nabla_X\alpha)(Y) = \nabla_X(\alpha(Y)) - \alpha(\nabla_XY)$. Hence $$(\nabla_X\alpha)(Y) - (\nabla_Y\alpha)(X) = X(\alpha(Y)) - Y(\alpha(X)) - \alpha(\nabla_XY - \nabla_YX).$$ The last term is $\alpha$ evaluated on the commutator, provided that $\nabla$ is torsion free. This also answers the question what is going wrong if there is a torsion... $\endgroup$ – Ivan Izmestiev Oct 7 '16 at 8:54
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    $\begingroup$ @dorebell What you are writing makes perfect sense. You may consult Besse, Einstein manifolds. Section 1.12 defines the exterior derivative of forms with values in a vector bundle by antisymmetrizing the covariant derivative. Section 1.19 provides a connection between $\nabla$ and $d$ in your context. It is a good exercise to check that the $\nabla$-definition of $d$ coincides with a more usual one. $\endgroup$ – Ivan Izmestiev Oct 7 '16 at 9:06
  • $\begingroup$ @IvanIzmestiev Whoops! I misinterpreted the question. The OP actually wants to recover $d$ itself. $\endgroup$ – user40276 Oct 7 '16 at 9:45
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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.

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  • $\begingroup$ What goes wrong when there's torsion? $\endgroup$ – dorebell Oct 7 '16 at 5:18
  • $\begingroup$ I think you mean curvature. In this case, it's possible to correct it and get a representation up to homotopy by using a complex of vector bundles. $\endgroup$ – user40276 Oct 7 '16 at 6:30
  • $\begingroup$ The things is this: using an affine connection gives you a derivative $d_\nabla$ mapping k-forms to k+1-forms. Also it satisfies the Leibniz rule for functions and forms. The important difference is that in general $d_\nabla^2f\neq0,$ as in the remark of Ivan. $\endgroup$ – Sebastian Oct 7 '16 at 9:27
  • $\begingroup$ @user40276: You can play a similar game with a connection on a vector bundle to define what is called the exterior derivative $d^\nabla$ for forms with values in that bundle. What you get in this different situation is $(d^\nabla)^2$ is the curvature. $\endgroup$ – Sebastian Oct 7 '16 at 9:30
  • $\begingroup$ @Sebastian Sorry. I was thinking that you were referring to $d^{\nabla}$ in your answer. This is why I said curvature. $\endgroup$ – user40276 Oct 7 '16 at 9:36

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