# Torsion and Parallel Transport

There's a close relationship between curvature and the holonomy group; the holonomy theorem of Ambrose and Singer, for example. It seems to me that there should be an analogous result for torsion. I believe that torsion measures the extent to which certain geodesic parallelograms don't close. Is there a theorem that makes this more precise?

-
I don't understand this question. The Ambrose-Singer theorem applies to connections with torsion and so does the relationship between holonomy and curvature. What do you mean by "analogous result" for torsion? – José Figueroa-O'Farrill Jun 10 '13 at 23:01
@Jose: What I mean is; is there a group associated with a point in the manifold, maybe describable in terms of parallel transport around geodesic parallelograms, which describes torsion? You can have connections with zero curvature but non zero torsion. The zero curvature implies that the restricted holonomy group is trivial and so it tells us nothing about torsion. – Oliver Jones Jun 10 '13 at 23:14

Here is another way to think of the relation between torsion and parallel transport, one that some may find more congenial than many of the other interpretations that have been proposed:

Start with a manifold $M$, and a connection $\nabla$ on $TM$. Consider the bundle $\hat TM=\mathbb{R}\oplus TM$ (where, by $\mathbb{R}$, I really mean the trivial bundle $M\times\mathbb{R}$). Define a connection $\hat\nabla$ on $\hat TM$ by the rule $$\hat\nabla_X\ \begin{pmatrix}a \\\\ Y\end{pmatrix} = \begin{pmatrix}da(X) \\\\ aX + \nabla_XY\end{pmatrix}$$ for any function $a$ on $M$ and any vector fields $X$ and $Y$ on $M$.

I like to think of $\hat TM$ as a sort of 'extended' tangent bundle to $M$. Of course, it canonically contains $TM= 0\oplus TM\subset \mathbb{R}\oplus TM$ as a subbundle, and one sees that, under this identification of $\begin{pmatrix}0\\ Y\end{pmatrix}$ with $Y$, one has $\hat\nabla_X Y = \nabla_XY$. Also, by its definition, $\hat\nabla$-parallel translation in $\hat TM$ along a curve in $M$ will keep the $\mathbb{R}$-component of a section constant, so that the 'affine hyperplane subbundles' $a=const$ in $\hat TM$ are preserved under $\hat\nabla$-parallel translation.

Now, the curvature endomorphism of $\hat\nabla$, namely $$R^{\hat\nabla}(X,Y) = \hat\nabla_X\hat\nabla_Y-\hat\nabla_Y\hat\nabla_X - \hat\nabla_{[X,Y]},$$ is just $$R^{\hat\nabla}(X,Y)\begin{pmatrix}a \\\\ Z\end{pmatrix} = \begin{pmatrix}0&0\\\\T^{\nabla}(X,Y) & R^{\nabla}(X,Y)\end{pmatrix} \begin{pmatrix}a \\\\ Z\end{pmatrix}.$$

Thus, the condition that $T^\nabla(X,Y)=0$ is the condition that $\hat\nabla$-parallel translation around closed loops preserve the splitting $\hat TM = \mathbb{R}\oplus TM$, at least to lowest (i.e., second) order. In particular, if one thinks of the hyperplane subbundle $a=1$ as a sort of 'shadow copy' of the actual tangent plane, the torsion measures how its 'zero', i.e., $\begin{pmatrix}1\\0\end{pmatrix}$, moves in that 'shadow tangent plane' when one uses $\hat\nabla$ to parallel translate around in a loop.

In particular, note that the curvature of $\hat\nabla$ incorporates both the torsion and curvature of $\nabla$, and the vanishing of the torsion of $\nabla$ says precisely that $\hat\nabla$-parallel translation preserves, to lowest order, the origins in the 'shadow copies' of the tangent plane.

NB: It is a notational accident that, in this interpretation, $T(X,Y)$ represents 'infinitesimal translation' while $R(X,Y)$ represents 'infinitesimal rotation'.

NB: In the earlier version of this answer, I foolishly stated that if the torsion vanishes then the section $\begin{pmatrix}1\\0\end{pmatrix}$ is $\hat\nabla$-parallel, which, of course, is nonsense. My apologies for making such a careless error.

Remark: Finally, going back to the OP's original request for how to think about torsion in terms of failure of 'quadrilaterals' to close, I checked on the coordinate expressions and looked at the Taylor series. This is, of course, very classical; it's in Schouten's work, but it might be worth making explicit in the following way: Let $\nabla$ be a connection on $TM$, fix a point $p$ and $p$-centered coordinates $x=(x^i)$. Define the connection coefficients $\Gamma^i_{jk}$ by the usual rule $$\nabla_{\partial_i}\partial_j = \Gamma^k_{ij}\ \partial_k\ ,$$ where $\partial_i$ are the dual vector fields, i.e., $dx^i(\partial_j) = \delta^i_j$.

Choose $v,w\in T_pM$ be tangent vectors and write $v=v^i\ \partial_i(p)$ and $w=w^i\ \partial_i(p)$. Let $a(t)=\exp_p(tv)$ be the $\nabla$-geodesic starting at $p$ with initial velocity $v$, let $b(t)\in T_{a(t)}M$ be the $\nabla$-parallel translate along $a$ of $w$, and set $c(s,t) = \exp_{a(t)}(sb(t))$. Then the functions $c^i= x^i(c(s,t))$ have Taylor expansions in $s$ and $t$ of the form $$c^i(s,t) = tv^i+sw^i - \tfrac12\Gamma^i_{jk}(0)\bigl(t^2v^jv^k+2st\ v^jw^k+s^2w^jw^k\bigr) + R^i_3(s,t).$$ Thus, if switching $(t,v)$ and $(s,w)$ in this construction is to yield the same result up to second order in $s$ and $t$ for all $v$ and $w$, one must have $\Gamma^i_{jk}(0) = \Gamma^i_{kj}(0)$. In particular, if all the 'attempted parallelograms' close to second order at all points, the torsion of $\nabla$ must vanish.

-
Very nice. I never saw this before. I plan to teach this next semester when the students inevitably ask me this. – Spiro Karigiannis Jun 13 '13 at 0:16
@Robert: Thanks for your answer; that's an interesting approach. So no one has a comment on Nakahara's heuristic description? – Oliver Jones Jun 13 '13 at 23:07
@Oliver: I woke in the middle of the night and realized that there's a serious error in my answer, so I'm fixing it. I don't know of Nakahara's heuristic description; I just saw it for the first time in your comment to Peter's answer. Looking at your description, I have to ask: How do you use the connection to parallel translate a pair of points (such as $pq$ in your comment)? I know how to parallel translate vectors, but not pairs of points. Perhaps you meant something else? – Robert Bryant Jun 14 '13 at 12:02
I think by the notation $pq$ he meant that you take a tangent vector $v$ at $p$ so that the geodesic through $p$ with initial tangent vector $v$ reaches $q$ at time $1$. Since $q$ is assumed close to $p$, we could assume that there is a unique such vector $v$. – Ben McKay Jun 14 '13 at 12:44
@Robert: Beautiful and simple. Is this connection associated to the Cartan connection on the $\mathbb R^n \rtimes SO(n)$-bundle which involves the soldering form as the part with values in the translations? – Peter Michor Jun 14 '13 at 13:17

See this question and its answers.

My view is the following: Suppose that $\nabla$ is a linear connection on a vector bundle $E\to M$, and that there is $\sigma\in \Omega^1(M;E)$, a 1-form on $M$ with values in $E$ such that $\sigma_x:T_xM\to E_x$ is a linear isomorphism. This is called a soldering form. It identifies $E$ with $TM$.

The torsion is then $d^{\nabla}\sigma\in\Omega^2(M;E)$. The usual formula of torsion is then $T(X,Y)=\sigma^{-1}((d^\nabla\sigma)(X,Y))$. Torsion is an obstruction against the soldering form being parallel for $\nabla$. Maybe this explains, that space is twisting along geodesics if the torsion is non-zero. So torsion can be viewed either as a property of the soldering form (choose it better if you want to get rid of torsion), or as a property of $\nabla$ (if you identify $TM$ with $E$ with the given soldering form).

This works also with $G$-structures on $M$. Consider a principal $G$-bundle $P\to M$ and a representation $\rho:G\to GL(V)$ where $\dim(V)=\dim(M)$. A soldering form is now a $G$-equivariant and horizontal 1-form $\sigma\in\Omega^1(P,V)^G_{hor}$ which is fiberwise surjective. This induces a form $\bar\sigma\in\Omega^1(M,P\times_G V)$ which is a soldering form in the sense above. You can compute torsion either on $P$ or on $M$ and they correspond to each other.

Edit: Parallel transport and torsion. In 24.2 (page 299ff) of [2] it is stated that The parallel transport along the flow lines of a vector field $X$ are the flow lines of the horizontal lift $C(X)$ of the vector field: $Pt(Fl^X,t) = Fl^{C(X)}_t$. Thus by 3.16, parallel transporting along a parallelogram consisting of flow lines and differentiating twice is: $$\partial_t^2|_0 Pt(Fl^X,-t) Pt(Fl^Y,-t) Pt(Fl^X,t) Pt(Fl^Y,t) = \partial_t^2|_0 Fl^{C(X)}_{-t} Fl^{C(Y)}_{-t} Fl^{C(X)}_t Fl^{C(Y)}_t = [C(X),C(Y)]$$ By 24.6 we have $[C(X),C(Y)](Z) - C([X,Y])(Z) = vl(Z, R(X,Y)Z)$, where $vl$ is the vertical lift, so you get curvature and not torsion.

The only way to get torsion is using again 24.2: $\nabla_XY = \partial_t|_0 Pf(Fl^X,t)^* Y$ and building torsion out of this.

A global formula for torsion which runs over the second tangent bundle you find in 5 of [3].

[2] Peter W. Michor: Topics in Differential Geometry. Graduate Studies in Mathematics, Vol. 93 American Mathematical Society, Providence, 2008. (pdf)

[3] Peter W. Michor: The Jacobi Flow. Rend. Sem. Mat. Univ. Pol. Torino 54, 4 (1996), 365-372 (pdf)

-
@Peter: I just found a related question on MO. Here's the link: mathoverflow.net/questions/20493. I’m looking for something more geometric. – Oliver Jones Jun 11 '13 at 7:44
@Peter: I'm wondering if there isn't a simpler characterization of torsion. On page 258 of Geometry, Topology, and Physics by Mikio Nakahara, torsion is described as the failure of certain parallelograms to close. Fix a point $p$ in the manifold and choose two other points $q$ and $s$ close by. Now, parallel transport $pq$ to $s$ along the geodesic through $p$ and $s$ to get $r_1$. Similarly, transport $ps$ to $q$ along the geodesic through $p$ and $q$ to get $r_2$. In general, $r_1\ne r_2$ and torsion somehow measures the difference. Can't this argument be made more formal? – Oliver Jones Jun 11 '13 at 23:45
@Oliver: I do not know this source; but I doubt it, because I just computed parallel transport along a parallelogram in the answer. – Peter Michor Jun 14 '13 at 13:36
@Peter: I'm not clear why you doubt the result. Here's a better reference; a paper by Hehl and Obukhov <a href="arxiv.org/pdf/0711.1535.pdf"; title="here">here</a> , page 2. – Oliver Jones Jun 14 '13 at 22:18
@Oliver: Robert gave the concise answer. Sorry, I could not see this as parallel transport along a parallelogram, but as exchanging two Fermi coordinates. But this picking hairs. – Peter Michor Jun 16 '13 at 8:18