I am asking this in the context of differential geometry (specifically Riemannian).

When the Levi-Civita Connection is defined, we require that the torsion tensor is 0, which in local coordinates translates to the requirement that $\Gamma_{ij}^{k} = \Gamma_{ji}^{k}$; which is the covariant derivative version of saying partial derivatives commute: $\nabla_{\partial_i}(\partial_j)=\nabla_{\partial_j}(\partial_i)$.

This is obviously true in the Euclidian settings, and I understand all the details of the proofs. But why is this such an essential property? Why does this capture our intuitive sense of derivatives?

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    $\begingroup$ A related question might be: What uses, if any, are there for connections that aren't torsion free? $\endgroup$ – Harald Hanche-Olsen Feb 23 '13 at 15:42
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    $\begingroup$ There are nice interpretations of torsion in mathoverflow.net/questions/20493/… Especially, check out Tom Boardman's answer. $\endgroup$ – Claudio Gorodski Feb 23 '13 at 20:24
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    $\begingroup$ So that exterior differentiation makes a chain complex! $\endgroup$ – David Corwin Feb 24 '13 at 16:11
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    $\begingroup$ Isn't this a repeat of the thread Claudio links to above? $\endgroup$ – Ryan Budney Feb 24 '13 at 19:44
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    $\begingroup$ "Which is the covariant derivative version of saying partial derivatives commute" not quite. It only says that for scalar functions. $\endgroup$ – Willie Wong Feb 25 '13 at 9:51

Here is another way of obtaining the Christoffel symbols with the symetry imposed by the torsion free condition

$$ \Gamma^i_{k\ell}=\Gamma^i_{\ell k}. $$

This goes back to Riemann's Habillitation.

Suppose that $(M,g)$ is a Riemann manifold of dimension $N$, $p\in M$. By fixing an orthonormal frame of $T_pM$ we can find local coordinates $(x^1,\dotsc, x^N)$ near $p$ such that, $\newcommand{\pa}{\partial} $

$$ x^i(p)=0, \;\; g=\sum_{i,j} g_{ij}(x) dx^i dx^j, $$

$$g_{ij}(x)= \delta_{ij} +\sum_{i,j}\left(\sum_k\pa_{x^k}g_{ij}(0) x^k\right) dx^i dx^j + O(|x|^2). $$

In other words, in these coordinates,

$$ g_{ij}(x)=\delta_{ij} +O(|x|). $$

Riemann was asking whether one can find new coordinates near $p$ such that in these coordinates the metric $g$ satisfies $g_{ij}=\delta_{ij}$.

As a first step, we can ask whether we can find a new system of coordinates such that, in these coordinates the metric $g$ is described by

$$ g=\sum_{ij}\hat{g}_{ij} dy^idy^j, $$


$$\hat{g}(y)=\delta_{ij}+ O(|y|^2). \tag{1} $$

The new coordinates $(y^j)$ are described in terms of the old coordinates $(x^i)$ by a family of Taylor approximations

$$y^j= x^j + \frac{1}{2}\sum_{ij}\gamma^j_{\ell k} x^\ell x^k + O(|x|^3),\;\; \gamma^j_{\ell k}=\gamma^j_{k\ell}. $$

The constraint (1) implies

$$ \gamma^j_{\ell k}=\frac{1}{2}\left(\pa_{x^\ell}g_{jk}+\pa_{x^\ell}g_{jk}-\pa_{x^j}g_{\ell k}\right)_{x=0}. $$

We see that, in the $x$ coordinates

$$ \Gamma^i_{k\ell}(p)=\gamma^i_{k\ell}, $$

because $g^{ij}(p)=\delta^{ij}$.

It took people several decades after Riemann's work to realize that the coefficients $\Gamma^i_{k\ell}$ are related to parallel transport, and ultimately, to a concept of connection.

Ultimately, to my mind, the best explanation for the torsion-free requirement comes from Cartan's moving frame technique. The clincher is the following technical fact: given a connection $\nabla$ on $TM$ and a $1$-form $\alpha\in \Omega^1(M)$ then for any vector fields $X,Y$ on $M$ we have

$$d\alpha(X,Y)= X\alpha(Y)-Y\alpha(X)-\alpha([X,Y]) $$

$$= (\nabla_X\alpha)(Y)-(\nabla_Y\alpha)(X)+\alpha(\nabla_XY-\nabla_YX)-\alpha([X,Y]) $$

$$= (\nabla_X\alpha)(Y)-(\nabla_Y\alpha)(X)+\alpha\bigl(\;T_\nabla(X,Y)\;\bigr). $$

If the torsion is zero, the above equality looses a term, and one obtains rather easily Cartan's structural equations of a Riemann manifold.


To me, a Riemannian metric and the Levi-Civita connection associated with the metric represent the intrinsic geometric properties of a submanifold in Euclidean space induced by the inner product and natural flat connection on Euclidean space. Since they are intrinsic, their definitions can be extended from submanifolds of Euclidean space to abstract manifolds.

If you don't assume the connection is torsion-free, then there are an infinite number of connections that are compatible with the metric (instead of exactly one), so the link between the geometric properties of the metric and that of the connection is much weaker.

  • $\begingroup$ Thanks. This makes sense, although in a way it kind of cancels the entire point of differential geometry as I see it so far (which is not a lot): I thought the motivation is to define calculus again in an intrinsic, coordinate-free way on general smooth manifolds; specifically, since we believe our universe is modeled well by one. Reducing it all back to the Euclidian setting when we stumble along some complication does not seem to follow the same spirit. Perhaps this means we should think our universe is some complicated embedding in a larger Euclidian space, unreachable to us? $\endgroup$ – R S Feb 23 '13 at 15:59
  • $\begingroup$ Maybe it's circular in this setting, but there is Nash's embedding theorem. $\endgroup$ – horse with no name Feb 23 '13 at 16:09
  • $\begingroup$ I have two responses: Designing an intrinsic geometric structure modeled on submanifolds of Euclidean space does not mean we are restricting our attention to submanifolds of Euclidean space, notwithstanding the Nash isometric embedding theorem. Nor does it mean that the geometric structure is not co-ordinate-free. The study of Riemannian geometry does not depend on co-ordinates. Of course, even the study of submanifolds in Euclidean space does not, too. $\endgroup$ – Deane Yang Feb 23 '13 at 16:52

The covariant derivative version of trying to commute partial derivatives is: $\nabla_{\partial_i}\nabla_{\partial_j}-\nabla_{\partial_j}\nabla_{\partial_i} - 0 = R(\partial_i,\partial_j)$.

Torsion is measuring something different: It is the covariant derivative of the soldering form $\sigma\in\Omega^1(M,E)$ which you use to identify the vector bundle $E$ with $TM$, where $E$ is the bundle you are considering your covariant derivative on.

  • $\begingroup$ Thanks. It is possible to explain this in simpler terms ("soldering form" seems to be advanced from the basic place in diff. geometry in which I now stand)? Or do you think a true understanding of the torsion tensor requires more advanced concepts? $\endgroup$ – R S Feb 23 '13 at 16:05
  • $\begingroup$ There are lots of nice explanations of torsion here mathoverflow.net/questions/20493/… although you may consider them to be advanced also. $\endgroup$ – Paul Reynolds Feb 23 '13 at 19:12

The Levi-Civita connection is just a very special one - torsion free. It is interesting that the same geometry may be described by switching to another, non-torsion-free connection. E.g. there is such a version of general relativity which is called teleparallel formulation. While the curvature tensor (based on the new connection) vanishes, all the deviation from flatness has been shifted to the torsion tensor (better: vector-valued 2-form). Einstein exchanged his ideas with Cartan in the 1920 about that. Torison has also an equivalent in physics as dislocation density (disclocations are defects in crystals). The theory has been developed in the 1950's by Kondo, Bilby and Kröner. See also the book Ricci calculus by J.A. Schouten. To summarize, it is not that important that the connection is symmetric, it is merely a matter of choice. The metric, for instance, is independent of the connection.

  • $\begingroup$ You put the cart before the horse. First, you fix a metric. Then you fix a connection compatible with the metric. If you are interested only in geodesic then, as E. Cartan observed, there are several connections compatible with the metric that give the same geodesics. If you are interested in more than geodesics, then curvature matters, and the curvature does depend on the choice of connection. $\endgroup$ – Liviu Nicolaescu Feb 25 '13 at 13:14
  • $\begingroup$ I don't think that contradicts what I said, but it is surely more precise to start the discussion with the metric, yes. Thanks for clarifying. $\endgroup$ – ClassicalPhysicist Feb 27 '13 at 10:45

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