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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 '19 at 20:59
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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.

(EDIT 2) To illustrate this, let us consider the concept of iterated covariant derivatives, mentioned in the MO question linked in the second paragraph. To that end, we combine a covariant derivative on the vector bundle $\pi$ with a covariant derivative on $TM$ in order to define covariant derivatives on the tensor bundles $\otimes^k T^*M \otimes E\rightarrow M$ for all $k\in\mathbb{N}$ by means of Leibniz's rule. We denote all these covariant derivatives collectively by $\nabla$. Once this is done, the iterated covariant derivatives $\nabla^k\phi$ of all orders $k\in\mathbb{N}$ of smooth sections $\phi$ of $\pi$ are defined recursively as follows: $$\nabla^0\phi=\phi\ ,\,\nabla^1\phi=\nabla\phi\ ,\,\nabla^{k+1}\varphi=\nabla(\nabla^k\phi)\ ,$$ so that $\nabla^k\phi$ is a smooth section of $\otimes^k T^*M \otimes E\rightarrow M$ for all $k\in\mathbb{N}$. Particularly, for $k=2$, one can write $$\nabla^2\phi(X,Y)=\nabla_X(\nabla_Y\phi)-\nabla_{\nabla_X Y}\phi\ .$$ If we antisymmetrize $\nabla^2\phi$, we obtain $$\nabla^2\phi(X,Y)-\nabla^2\phi(Y,X)=\nabla_X(\nabla_Y\phi)-\nabla_Y(\nabla_X\phi)-\nabla_{\nabla_X Y-\nabla_Y X}\phi$$ $$=\text{Riem}_\nabla(X,Y)\phi-\nabla_{T_\nabla(X,Y)}\phi\ ,$$ where $$\text{Riem}_\nabla(X,Y)\phi=\nabla_X(\nabla_Y\phi)-\nabla_Y(\nabla_X\phi)-\nabla_{[X,Y]}\phi$$ is the Riemann curvature tensor of the covariant derivative on $\pi$ and $$T_\nabla(X,Y)=\nabla_X Y-\nabla_Y X-[X,Y]$$ is the torsion tensor of the covariant derivative on $TM$. The above expression clearly depends on derivatives of $\phi$ of at most first order. Notice that the above antisymmetrization is closely related to the operation of exterior covariant differentiation $d^\nabla$ of $E$-valued differential forms, as mentioned in Jesper Gรถransson's answer: since $\phi$ is a $E$-valued 0-form and therefore $d^\nabla\phi=\nabla\phi$, we have that $$d^\nabla(d^\nabla\phi)(X,Y)=\nabla_X(\nabla_Y\phi)-\nabla_Y(\nabla_X\phi)-\nabla_{[X,Y]}\varphi=\text{Riem}_\nabla(X,Y)\varphi$$ $$=\nabla^2\phi(X,Y)-\nabla^2\phi(Y,X)+\nabla_{T_\nabla(X,Y)}\varphi\ .$$ Particularly, if $E=M\times\mathbb{R}$ (i.e. sections of $\pi$ are just real-valued functions on $M$) and $\nabla$ is just ordinary differentiation of functions on $M$ (so that $\text{Riem}_\nabla=0$ and $T_\nabla=0$), we thus recover the nilpotency of the (ordinary) exterior derivative.

A similar phenomenon happens if we antisymmetrize the vector field entries of the tensor field $\nabla^k\phi$ for $k>2$ - any antisymmetrization with respect to (say) $j$ entries leaves us with an expression which depends on $\nabla^i\phi$ for $i$ only up to $k-(j-1)=k-j+1$ for all $2\leq j\leq k$. If $j<k$, this is what is called "partial antisymmetrization" above.

Returning to covariant derivatives $\nabla$ on $M$ (that is, on $TM$), the equivalence of the vanishing of the torsion tensor $T_\nabla(X,Y)$ 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 $\omega$, mentioned by Deane Yang in his comment to the OP's question, comes from Henri Cartan's formula relating Lie differentiation and exterior differentiation: $\mathscr{L}_{\!X}\omega=d(i_X\omega)+i_X(d\omega)$.

(EDIT 1) To see this, suppose that the torsion tensor of $\nabla$ vanishes, i.e. $$\nabla_X Y-\nabla_Y X=[X,Y]=\mathscr{L}_{\!X}Y\ .$$ Recall that the action of $\nabla$ on an 1-form $\omega$ is given by Leibniz's rule$$(\nabla_X\omega)(Y)=X(\omega(Y))-\omega(\nabla_X Y)$$ and so is the Lie derivative of $\omega$: $$(\mathscr{L}_{\!X}\omega)(Y)=X(\omega(Y))-\omega(\mathscr{L}_{\!X}Y)=X(\omega(Y))-\omega([X,Y])\ .$$ The antisymmetric part of $\nabla\omega$ is then given by $$(\nabla_X\omega)(Y)-(\nabla_Y\omega)(X)=X(\omega(Y))-Y(\omega(X))-\omega([X,Y])\ .$$ Notice that I used the vanishing of the torsion of $\nabla$ in the rhs. On the other hand, Henri Cartan's formula tells us that $$d\omega(X,Y)=(\mathscr{L}_{\!X}\omega)(Y)-Y(\omega(X))=X(\omega(Y))-Y(\omega(X))-\omega([X,Y])\ ,$$ hence indeed $(\nabla_X\omega)(Y)-(\nabla_Y\omega)(X)=d\omega(X,Y)$. Conversely, if one assumes this last identity and backtracks the above calculations, one obtains that $\omega(T_\nabla(X,Y))=0$ for all 1-forms $\omega$, hence $T_\nabla(X,Y)=0$ for all vector fields $X,Y$.

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.

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  • $\begingroup$ "The equivalence...comes from Henri Cartan's formula relating Lie differentiation and exterior differentiation: โ„’๐‘‹๐œ”=๐‘‘(๐‘–๐‘‹๐œ”)+๐‘–๐‘‹(๐‘‘๐œ”)." Could you maybe add 1-2 more sentences to clarify why this equivalence comes from Cartan's formula? It's not obvious to me. $\endgroup$ – Eric Auld Mar 22 at 5:32
  • $\begingroup$ Also, could you explain a bit more the statements: "This also means that any partial antisymmetrization will always discard highest-order derivatives - this holds irrespectively of โˆ‡ being torsion-free or not (for general vector bundles, this concept usually has no meaning). Hence, if you totally antisymmetrize (as done with ๐‘˜-forms), only derivatives of at most first order survive. This is of course related to the nilpotency of the exterior derivative." What do you mean by "first order" here for k-forms, and what do you mean by "partial antisymmetrization"? $\endgroup$ – Eric Auld Mar 29 at 15:19
  • $\begingroup$ Hi Eric, sorry for the delay. I've expanded my answer in order to account for your first question. As for your second question, it may take a little while since I need to think about how to explain this without drowning too much on notation. I'll come back to you about it as soon as I can. $\endgroup$ – Pedro Lauridsen Ribeiro Apr 1 at 23:25
  • $\begingroup$ Ok, done. I tried to illustrate how antisymmetrization kills higher-order derivatives in terms of iterated covariant derivatives in my answer. $\endgroup$ – Pedro Lauridsen Ribeiro Apr 2 at 0:19
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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 '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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Although higher derivatives are best thought through with jet bundles where we actually differentiate a bundle and not just a manifold, there is a useful description of second derivatives using secondary tangent bundles, this is just iterating the tangent bundle, ie $TTM$. This is described in one of the early chapters in Kolรกr, Michor & Slovรกks book on natural bundles. (They also describe a generalisation of this in one of the later chapters which they call section forms - but they don't do anything with it).

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?

On the last part of this question you may be interested in this paper, A Description of the Derivations of the Algebra of Symmetric Tensors, by A. Heydari, N. Boroojerdian, E. Peyghan and published in Archivum Mathematicum. They introduce symmetric forms, bracket, Lie derivative and differential. Whilst the symmetric differential is not nilpotent, ie $(d^s)^2$ does not vanish, they show

proposition 5:

Let $(M, g)$ be a Riemmanian manifold with Levi-Civita connection $โˆ‡$. The 1-form $ฯ‰$ is Killing if and only if $d^sฯ‰ = 0$.

and

proposition 7:

Every derivation of degree 1 on symmetric forms, whose value for a function is the differential of the function, is of the form of a symmetric differential of a connection which is also unique.

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