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?

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    $\begingroup$ a-priori I'm not sure about a geometric explanation, but you can view the torsion of the connection as the covariant exterior derivative of the identity mapping $\text{id}_{TM}:TM\to TM$ (which we can consider as a $TM$-valued 1-form on $M$), $T:= d_{\nabla}(\text{id}_{TM})$, and this is a $TM$-valued 2-form on $M$. And if you look at the coordinate-free definition of exterior covariant derivative, it's analogous to that of the normal exterior derivative, except all the Lie-derivatives are replaced by covariant derivatives. $\endgroup$
    – peek-a-boo
    Commented Jul 2, 2021 at 21:52
  • $\begingroup$ I'm trying to understand the formula for $d\omega.$ How is $\frac{\partial}{\partial X}$ of a 0-form (like $\iota_Y\omega$) defined? Is it just the directional derivative? $\endgroup$
    – md2perpe
    Commented Jul 4, 2021 at 10:02
  • $\begingroup$ @md2perpe yes, its just the directional derivative, and an abuse of notation. $\frac{\partial}{\partial X} f$ means the same thing as $X(f)$. $\endgroup$
    – namsos
    Commented Jul 11, 2021 at 14:03
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    $\begingroup$ Another similar formula is that for curvature: $$R(X,Y)Z=\nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[X,Y]} Z.$$ $\endgroup$
    – md2perpe
    Commented Jul 11, 2021 at 20:22
  • $\begingroup$ I just added an update to my answer, connecting the three formulas into all being exterior covariant derivatives. $\endgroup$
    – md2perpe
    Commented Jul 15, 2021 at 13:29

3 Answers 3


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.

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    $\begingroup$ If you add $\langle \theta, T(V, W)\rangle$ to the definition of $d\theta$ where $T$ is the torsion tensor, you don't even need $\nabla$ to be torsion free. $\endgroup$
    – Paul Bryan
    Commented Jul 4, 2021 at 0:26

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 $D(X) A(Y) - D(Y) A(X)$: $$ D(X) A(Y) - D(Y) A(X) - A([X,Y]) \\ = \left( X^j (\partial_j Y^k) A_k + X^j Y^k (D_j A_k) \right) - \left( Y^k (\partial_k X^j) A_j + X^j Y^k (D_k A_j) \right) - \left( X^j(\partial_j Y^k) A_k + Y^k (\partial_k X^j) A_j \right) \\= X^j Y^k (D_j A_k - D_k A_j) $$

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.


An underlying geometric relation is that the condition that $\nabla$ be torsion-free is precisely the condition that its action on differential forms reduces to the exterior derivative upon antisymmetrization. In particular, the tensorial Leibiz rule for $\nabla$ together with invariance under contraction yield the general coordinate-free formula for the action of $\nabla$ on a $(0,k)$-tensor $\omega$:

\begin{align} (\nabla \omega)(X_0, X_1, \dots, X_k) & = (\nabla_{X_0} \omega)(X_1, \dots, X_k) \\ & = X_0 \left( \omega(X_1,\dots,X_k) \right) - \omega(\nabla_{X_0} X_1, \dots, X_k) - \cdots - \omega(X_1, \dots, \nabla_{X_0} X_k) \end{align}

Meanwhile, if we take $\omega$ to be a differential $k$-form, Cartan's magic formula can be used to show that we may obtain a similar coordinate-free formula for the action of the exterior derivative:

$$(d \omega)(X_0, X_1, \dots, X_k) = \sum_{i=0}^k (-1)^i X_i \left( \omega(X_0,\dots, \widehat{X_i}, \dots,X_k) \right) + \sum_{i<j} (-1)^{i+j} \omega([X_i,X_j],X_0,\dots, \widehat{X_i}, \dots, \widehat{X_j}, \dots,X_k),$$ where hats denote omission. I'll leave it up to the reader to convince themselves that, if $\nabla$ is torsion-free, then antisymmetrizing the action of $\nabla$ on a $k$-form yields the action of $d$ (up to a factor of $k!$).

For example, in your $1$-form case, the former is $$(\nabla \omega)(X,Y) = X(\omega(Y)) - \omega(\nabla_X Y) $$ Antisymmetrizing yields $$\text{Alt}(\nabla \omega)(X,Y) = (\nabla \omega)(X,Y) - (\nabla \omega)(Y,X) = X(\omega(Y)) - Y(\omega(X)) - \omega( \nabla_X Y - \nabla_Y X)$$ If $\nabla$ is torsion-free, this is equivalent to your formula $$(d \omega)(X,Y) = X(\omega(Y)) - Y(\omega(X)) - \omega([X, Y]). $$


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