# Can we define exterior derivatives using pushforwards and connections?

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 $TM$.

• 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). – user40276 Oct 7 '16 at 4:51
• 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). – user40276 Oct 7 '16 at 6:27
• 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... – Ivan Izmestiev Oct 7 '16 at 8:54
• @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. – Ivan Izmestiev Oct 7 '16 at 9:06
• @IvanIzmestiev Whoops! I misinterpreted the question. The OP actually wants to recover $d$ itself. – user40276 Oct 7 '16 at 9:45

## 1 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.

• What goes wrong when there's torsion? – dorebell Oct 7 '16 at 5:18
• 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. – user40276 Oct 7 '16 at 6:30
• 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. – Sebastian Oct 7 '16 at 9:27
• @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. – Sebastian Oct 7 '16 at 9:30
• @Sebastian Sorry. I was thinking that you were referring to $d^{\nabla}$ in your answer. This is why I said curvature. – user40276 Oct 7 '16 at 9:36