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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).

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  • $\begingroup$ 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 at 20:59
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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.

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  • $\begingroup$ In your quadratic form, is there a reason why $dx\,dy = dy\, dx$? $\endgroup$ – Eric Auld Aug 10 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 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 at 11:51

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