Exterior derivative of 1-form and derivatives of sections of $T^*M$

Question

Let $M$ be a compact manifold and consider a differential form $\alpha\in\Omega^1(M)$, which we can think of as a map $\alpha:M\to T^*M$. Since $T^*M$ is a smooth manifold we can compute the differential of this map $$D\alpha:TM\to T(T^*M)$$ How does this differential differ from the exterior derivative on 1-forms? I have a feeling that $D\alpha(X) = d(\alpha(X))$, but I'm not sure how to show this.

Answer 1

As Ted points out, your guess isn't exactly right.

I explain in this answer that $d\alpha$ is basically the "antisymmetric part" of $D \alpha$. Things are a little bit awkward to formulate viewing $\alpha$ as a map $M \to T^*M$, so let's view it as a fiberwise linear map $TM \to \mathbb{R}$ instead. Then the derivative is $D\alpha : TTM \to T\mathbb{R}$. There is a canonical identification $T \mathbb{R} = \mathbb{R} \times \mathbb{R}$, so that $D \alpha$ viewed this way just does $\omega \circ \pi$ on the first factor. Let $k : T\mathbb{R} \to \mathbb{R}$ be the projection onto the second factor. (If you want to be extremely fancy, $k$ is the connector of the canonical affine connection on the vector space $\mathbb{R}$.)

Using $\operatorname{flip} : TTM \to TTM$ to denote the canonical flip on the double tangent bundle, the difference $k \circ D \alpha - k \circ D\alpha \circ \operatorname{flip} : TTM \to \mathbb{R}$ factors through the projection $(D\pi, \pi) : TTM \to TM \times_M TM$ to yield a bilinear map $TM \times_M TM \to \mathbb{R}$---this bilinear map is alternating for the reason I explain in the linked answer, and is in fact the exterior derivative $d \alpha$.

---

If we fix an affine connection $\nabla$ on $M$ and $C : TM \times_M TM \to TTM$ is the associated horizontal lift map, then $\nabla \alpha = k \circ D \alpha \circ C$. All of the information in $D\alpha$ is contained in $\nabla \alpha$, because on the so-called vertical subbundle of $TTM$ the derivative $D\alpha$ depends only on $\alpha$.

The condition that $\nabla$ is torsionfree turns out to be equivalent to the requirement that $\operatorname{flip} \circ\ C(X, Y) = C(Y, X)$. Thus in this situation $$ \operatorname{Alt}(\nabla \alpha)(X, Y) = (\nabla_X \alpha)(Y) - (\nabla_Y \alpha)(X) = (k \circ D\alpha \circ C - k \circ D\alpha \circ \operatorname{flip} \circ \ C)(X,Y). $$ By definition $C$ is a section of $(D\pi, \pi) : TTM \to TM \times_M TM$, so this difference of covariant derivatives again recovers the exterior derivative $d\alpha$.

In summary given a torsionfree connection $\nabla$, $d\alpha$ is the alternating part of $\nabla \alpha$, which turns out to be invariant of the choice of $\nabla$. In contrast the symmetric part of $\nabla \alpha$ does depend on $\nabla$.