Exterior Derivative of One-Form vs Torsion of Connection

Question

Let $\omega$ be a $1$-form. Then $d\omega$ may be defined by the formula $$ d\omega(X,Y) = \frac{\partial}{\partial X}\iota_Y\omega - \frac{\partial}{\partial Y}\iota_X\omega-\omega([X,Y]) $$ where $X,Y$ are vector fields. This formula bears a resemblance to the formula for the torsion of a connection $\nabla$: $$ \nabla_XY - \nabla_YX - [X,Y] $$ Is there a geometric explanation for this resemblance?

Answer 1

I don't have an answer to your question, but it leads to a simple coordinate-free definition of the exterior derivative of a $1$-form $\theta$:

Let $\nabla$ be a torsion-free connection. It defines the covariant derivative of a vector field. The covariant derivative of a $1$-form is uniquely determined by the product rule $$ d\langle \theta, V\rangle = \langle \nabla\theta, V\rangle + \langle \theta,\nabla V\rangle. $$ Since the left side of the equation above does not depend on the connecton, it follows that if $\tilde\nabla$ is another torsion-free connection, then \begin{equation}\label{change} \langle \tilde\nabla\theta, V\rangle + \langle \theta,\tilde\nabla V\rangle = \langle \nabla\theta, V\rangle + \langle \theta,\nabla V\rangle. \end{equation}

The exterior derivative of $\theta$ can be defined by $$ \langle d\theta, V\otimes W\rangle = \langle \nabla_V\theta,W\rangle - \langle V,\nabla_W\theta\rangle, $$ where $V, W$ are tangent vectors at a point. This definition makes it obvious that $d\theta$ is a well-defined exterior $2$-tensor. Using the equations above and the torsion-free property, it is easy to show that this definition does not depend on the connection.

The better known coördinate-free formula follows from the equations above and the torsion-free property, because \begin{align*} \langle d\theta, V\otimes W\rangle &= \langle \nabla_V\theta,W\rangle - \langle V,\nabla_W\theta\rangle\\ &= \langle V,d\langle\theta,W\rangle\rangle - \langle \theta, \nabla_VW\rangle - \langle W,d\langle\theta,V\rangle\rangle + \langle \theta,\nabla_WV\rangle\\ &= \langle V,d\langle\theta,W\rangle\rangle - \langle W,d\langle\theta,V\rangle\rangle - \langle \theta,[V,W]\rangle\\ \end{align*} If you choose local coordinates and use the flat connection with respect to those coordinates, then you get the usual formula for the exterior derivative.

Answer 2

Note a very deep result, but I found a common pattern and that the third term cancels some terms.

You gave two expressions and I found a third on Wikipedia: $$\begin{align} d\omega(X,Y) &= \partial_X \iota_Y \omega - \partial_Y \iota_X \omega - \omega([X,Y]) \\ T(X,Y) &= \nabla_X Y - \nabla_Y X - [X,Y] \\ R(X,Y) &= \nabla_X \nabla_Y - \nabla_Y \nabla_X - \nabla_{[X,Y]} \end{align}$$

They all have a common form, $$ \Omega(X,Y) = D(X) A(Y) - D(Y) A(X) - A([X,Y]), $$ where $D(X)=\partial_X+\Gamma_X$ for some automorphism $\Gamma_X$ (possibly vanishing). Below, $D_j=\partial_j+\Gamma_j$ will be used for $D(\partial_j).$ Also, $A_j=A(\partial_j).$

For the three cases we have $$\begin{align} d\omega(X,Y) &: D(X)=\partial_X,\ A(X)=\iota_X\omega = \omega(X) \\ T(X,Y) &: D(X)=\nabla_X,\ A(X)=X \\ R(X,Y) &: D(X)=\nabla_X,\ A(X)=\nabla_X \end{align}$$

With $X=X^j\partial_j,\ Y=Y^k\partial_k$ we have $$\begin{align} D(X) A(Y) &= D(X^j\partial_j) A(Y^k\partial_k) = X^j D(\partial_j) (Y^k A(\partial_k)) \\ &= X^j (\partial_j+\Gamma_j) (Y^k A_k) \\ &= X^j ((\partial_j Y^k) A_k + Y^k (\partial_j A_k) + Y^k \Gamma_j A_k) \\ &= X^j (\partial_j Y^k) A_k + X^j Y^k (D_j A_k) \\ D(Y) A(X) &= Y^k (\partial_k X^j) A_j + X^j Y^k (D_k A_j) \\ A([X,Y]) &= A(X^j(\partial_j Y^k)\partial_k - Y^k(\partial_k X^j)\partial_j) \\ &= X^j(\partial_j Y^k) A_k - Y^k (\partial_k X^j) A_j \end{align}$$ and when we then simplify $\Omega(X,Y)$ we see that $A([X,Y])$ cancels two terms from $D(X) A(Y) - D(Y) A(X)$: $$ D(X) A(Y) - D(Y) A(X) - A([X,Y]) \\ = \left( X^j (\partial_j Y^k) A_k + X^j Y^k (D_j A_k) \right) - \left( Y^k (\partial_k X^j) A_j + X^j Y^k (D_k A_j) \right) - \left( X^j(\partial_j Y^k) A_k + Y^k (\partial_k X^j) A_j \right) \\= X^j Y^k (D_j A_k - D_k A_j) $$

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EDIT 15 July 2021

A deeper explanation is that all three of these are outer derivatives.

We are used to handle scalar-valued forms, but what if the form is vector-valued? Then we use the exterior covariant derivative, which I assume (without thorough investigation) can be written $$ D\omega(X,Y) = \nabla_X\omega(Y) - \nabla_Y\omega(X) - \omega([X,Y]). $$

Taking $\omega(X)=X$ makes $D\omega$ be the torsion, and taking $\omega_Z(X)=\nabla_X Z$ makes $D\omega$ be the curvature.