The following is an excerpt from Sharpe's Differential Geometry - Cartan's Generalization of Klein's Erlangen Program.

Now we come to the question of higher derivatives. As usual in modern differential geometry, we shall be concerned only with the skew-symmetric part of the higher derivatives. In essence, what we shall be doing is taking the partial derivatives with respect to the base (i.e., manifold) variables and skew-symmetrizing the result, thus forgetting about the part of the higher derivatives that vanish under this procedure. However, this will not be made explicit in our treatment. The part of the higher derivative that disappears has not been studied much in differential geometry since Elie Cartan showed how useful it is to consider only the skew-symmetric part, that is, the exterior derivative. The old masters did use the symmetric part...

"Partial derivatives with respect to the base" must be the covariant derivative of the connection. If I correctly understand what's written in this answer, then we have for any torsion free connection on a manifold the equality $ \mathrm dw=\operatorname{Alt}(\nabla w)$.

This already seems rather remarkable since the exterior derivative is intrinsic.

Question 1. What is the geometric meaning of the above fact for torsion-free connections? How does taking anti-symmetrization "forget" the structure of a horizontal bundle (connection)?

Question 2. Still for a torsion-free connection, what is the geometric meaning of "the part of the higher derivative that vanish under this procedure" of anti-symmetrization (i.e the symmetric part)?

Question 3. What's the conceptual picture for an arbitrary connection on a manifold (possibly with torsion)? Is the anti-symmetrization still the exterior derivative? What is the symmetric part?

Remark on Q3. I vaguely recall being told that torsion measures the failure of the fundamental theorem of calculus, and also that in torsion-free connections the parallel transport commutator is given by the Lie bracket (again, the latter is intrinsic).

  • $\begingroup$ The torsion-free condition is exactly what is required for the anti-symmetric part of the covariant derivative to be the exterior derivative. It is one reason why we usually restrict to torsion-free connections. The other one is that there is a unique torsion-free connection compatible with a Riemannian manifold. $\endgroup$ – Deane Yang May 27 '19 at 20:59

The meaning of higher-order derivatives in differential geometry is better understood through jet bundles. The covariant derivative $\nabla\phi$ of (say) a smooth section $\phi:M\rightarrow E$ of a vector bundle $\pi:E\rightarrow M$, being a derivative, can be seen as a section of the so-called first-order jet bundle $\pi^1_0:J^1 E\cong\mathrm{End}(TM,E)\rightarrow E$ of $E$. Likewise, derivatives of higher order (say, $k\in\mathbb{N}$) of smooth sections of $\pi$ are described in a coordinate-free fashion by the so-called jet bundle of order $k$ $\pi^k_0:J^k E\rightarrow E$. This hierarchy of (vector) bundles over $E$ is endowed with vector bundle projections $\pi^k_l:J^kE\rightarrow J^l E$, $k>l$ such that $\pi^l_m\circ\pi^k_l=\pi^k_m$ for all integers $k>l>m\geq 0$. Roughly speaking, one may identify elements of $J^k E$ locally with Taylor polynomials of order $k$, and the projections $\pi^k_l$ discard derivatives of order above $l$.

The key point is that the leading-order derivatives are, in a sense, always symmetric: the kernel of $\pi^k_{k-1}$ is a bundle whose total space is canonically isomorphic to $S^kT^*M\otimes E$ = the tensor product of the symmetrized $k$-fold tensor power of $T^*M$ with $E$. Particularly, any reference to a specific covariant derivative when globally defining highest-order derivatives disappears - more precisely, all yield the same result (differences only affect lower-order terms). One can exploit this fact to write smooth sections of $J^k E$ in terms of symmetrized iterated covariant derivatives (see e.g. this MO question of mine).

This also means that any partial antisymmetrization will always discard highest-order derivatives - this holds irrespectively of $\nabla$ being torsion-free or not (for general vector bundles, this concept usually has no meaning). Hence, if you totally antisymmetrize (as done with $k$-forms), only derivatives of at most first order survive. This is of course related to the nilpotency of the exterior derivative.

Returning to covariant derivatives $\nabla$ on $M$ (that is, on $TM$), the equivalence of the vanishing of the torsion $T_\nabla(X,Y)=\nabla_X Y-\nabla_Y X-[X,Y]$ of the connection $\nabla$ (which, by the way, simply states that $\nabla_X Y-\nabla_Y X$ is the Lie bracket of $X$ and $Y$ = Lie derivative of $Y$ along $X$) to the antisymmetric part of $\nabla\omega$ being equal to $d\omega$ for all 1-forms $\omega$ comes from Henri Cartan's formula relating Lie differentiation and exterior differentiation: $\mathscr{L}_{\!X}\omega=d(i_X\omega)+i_X(d\omega)$. More generally, torsion introduces a sort of "internal rotation" on the frame bundle along parallel transport which cannot be accounted for by pushing forward by the diffeomorphism flow integrating the vector field along which we are differentiating (the latter is the effect measured by Lie derivatives) - hence the failure of the fundamental theorem of calculus. Interestingly, given any covariant derivative $\nabla$ on $M$, there is a unique covariant derivative $\nabla'$ on $M$ which is torsion-free and has the same geodesics as $\nabla$, which means that this "internal rotation" effect can be cancelled by suitably redefining $\nabla$. This is not to say that torsion should be always discarded - for instance, when dealing with foliations, there is a natural connection (the so-called Bott connection) which is only defined along vector fields tangent to the foliation - that is, a so-called partial connection. This connection is flat but has nonvanishing torsion, which therefore is important in dealing with the holonomy of the foliation.


Here's some calculations that may help. (I'll leave it to you to decide if the arguments are valid or not.) We'll work in a two-dimensional manifold with local coordinates $x$ and $y$. Let $d$ be the covariant derivative at a point in some direction and let $dx$ and $dy$ be the components of the directional vector.

We apply $d$ to a scalar function $f(x,y)$ and get $$ df = f_xdx+f_ydy $$ which is the usual differential of $f$. Apply $d$ a second time and we get $$ d^2f = f_{xx}dx^2+2f_{xy}dxdy+f_{yy}dy^2+f_xd^2x+f_yd^2y. $$ This is the second differential of $f$ and is a form on the second order tangent bundle, but since $d$ is to be interpreted as the covariant derivative, the connection gives that the second order differentials are expressed as quadratic forms on the tangent bundle:

\begin{align*} d^2x &= -(\Gamma^1_{11}dx^2+(\Gamma^1_{12}+\Gamma^1_{21})dxdy+\Gamma^1_{22}dy^2)\\ d^2y &= -(\Gamma^2_{11}dx^2+(\Gamma^2_{12}+\Gamma^2_{21})dxdy+\Gamma^2_{22}dy^2) \end{align*} and we have that \begin{align*} d^2f &= (f_{xx}-\Gamma^1_{11}f_x-\Gamma^2_{11}f_y)dx^2\\ &+(2f_{xy}-(\Gamma^1_{12}+\Gamma^1_{21})f_x-(\Gamma^2_{12}+\Gamma^2_{21})f_y)dxdy\\ &+(f_{yy}-\Gamma^1_{22}f_x-\Gamma^2_{22}f_ydy^2. \end{align*}

So the symmetric part of the connection is what comes in to play when we calculate higher order forms, the anti-symmetric part does not affect these calculations.

To calculate an anti-symmetric derivative we let $\partial$ denote the covariant derivative in some other direction and let $\partial x$ and $\partial y$ be the components of this other direction vector. This time we'll investigate the derivative of a one-form so let $\omega=a(x,y)dx+b(x,y)dy$ be some one-form. This one-form can be evaluated on either of the two direction vectors so we let \begin{align*} \omega(d) &= adx+bdy\\ \omega(\partial) &= a\partial x+b\partial y \end{align*} mean that the one-form is evaluated on $(dx,dy)$ and $(\partial x, \partial y)$ respectively.

We now calculate the anti-symmetric derivative of the one-form \begin{align*} \partial\omega(d)-d\omega(\partial) &= a_x\partial xdx + a_y\partial ydx +b_x\partial xdy + b_y\partial ydy + a\partial dx + b\partial dy\\ &- (a_xdx\partial x + a_ydy\partial x +b_xdx\partial y + b_ydy\partial y + ad\partial x + bd\partial y)\\ &= a_y(dx\partial y-\partial xdy) + b_x(\partial xdy-dx\partial y) \\ &+ a(\partial d-d\partial)x + b(\partial d-d\partial)y \end{align*}

Our connection defines that \begin{align*} \partial dx &= -(\Gamma^1_{11}\partial xdx+\Gamma^1_{12}\partial xdy+\Gamma^1_{21}dx\partial y+\Gamma^1_{22}\partial ydy)\\ \partial dy &= -(\Gamma^2_{11}\partial xdx+\Gamma^2_{12}\partial xdy+\Gamma^2_{21}dx\partial y+\Gamma^2_{22}\partial ydy) \end{align*} and we have that \begin{align*} (\partial d-d\partial)x &= -(\Gamma^1_{12}-\Gamma^1_{21})(\partial xdy-dx\partial y)\\ (\partial d-d\partial)y &= -(\Gamma^2_{12}-\Gamma^2_{21})(\partial xdy-dx\partial y) \end{align*} We now replace $\partial xdy-\partial ydx$ with standard notation $dx\wedge dy$ and the anti-symmetric derivative of the one-form is then expressed as \begin{align*} \partial\omega(d)-d\omega(\partial) &= (b_x-a_y-a(\Gamma^1_{12}-\Gamma^1_{21})-b(\Gamma^2_{12}-\Gamma^2_{21}))\,dx\wedge dy \end{align*} The anti-symmetric derivative is also called the covariant exterior derivative and for symmetric connections it coincides with the usual exterior derivative on linear forms.

  • $\begingroup$ In your quadratic form, is there a reason why $dx\,dy = dy\, dx$? $\endgroup$ – Eric Auld Aug 10 '19 at 5:10
  • $\begingroup$ $dx$ and $dy$ (and also $x$, $y$, $\partial x$ and $\partial y$) are variables and multiplication is ordinary scalar multiplication. $\endgroup$ – Jesper Göransson Aug 24 '19 at 9:01
  • $\begingroup$ Oh, I would strongly suggest a change in notation. As it is, it's very easy to get confused and think that dx and dy are covector fields $\endgroup$ – Eric Auld Aug 24 '19 at 11:51

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