Question

Hi, given a connection on the tangent space of a manifold, one can define its torsion: $$T(X,Y):=\triangledown_X Y - \triangledown_Y X - [X,Y]$$ What is the geoemtric picture behind this definition, what does torsion meassure intuitively?

Answer 1

The torsion is a notoriously slippery concept. Personally I think the best way to understand it is to generalize past the place people first learn about torsion, which is usually in the context of Riemannian manifolds. Then you can see that the torsion can be understood as a sort of obstruction to integrability. Let me explain a little bit first.

The torsion really makes sense in the context of general G-structures. Here $G \subseteq GL_n(\mathbb{R}) = GL(V)$ is some fixed Lie group. Typical examples are $G = O(n)$ and $G = GL_n(\mathbb{C})$. We'll see that these will correspond to Riemannian metrics and complex structures respectively. Now given this data, we have an exact sequence of vector spaces,

$$0 \to K \to \mathfrak{g} \otimes V^\ast \stackrel{\sigma}{\to} V \otimes\wedge^2 V^\ast \to C \to 0 $$

Here $\sigma$ is the inclusion $\mathfrak{g} \subseteq V \otimes V^\ast$ together with anti-symmetrization. K and C are the kernel and cokernel of $\sigma$.

If we are given a manifold with $G$-structure, we then get four associated bundles, which fit into an exact sequence:

$$ 0 \to \rho_1P \to ad(P) \otimes T^*M \to \rho_3P \to \rho_4P \to 0$$

Now the difference of two connections which are both compatible with the G-structure is a tensor which is a section of the second space $\rho_2P = ad(P) \otimes T^*M$. This means that we can write any connection as $$\nabla + A$$ where $A$ is a section of $\rho_2(P)$.

Now the torsion of any G-compatible connection is a section of this third space. Suppose that we have two compatible connections. Then there torsions are sections of this third space. However since we can write the connections as $\nabla$ and $\nabla + A$, the torsion differ by $\sigma(A)$. Thus they have the same image in the fourth space $\rho_4(P)$.

The section of this fourth space is the intrinsic torsion of the G-structure. It measures the failure of our ability to find a torsion free connection. If this obstruction vanishes, then the torsion free connections are a torsor over sections of the smaller bundle $\rho_1P$. Now some examples:

1. $G = O(n)$. This is the case of a Riemannian structure. In this case $\sigma$ is an isomorphism so that the there is always a unique torsion free connection. The Levi-Civita connection.
2. $G = GL_m(\mathbb{C})$. This is the case of a complex structure. More precisely a $GL_m(\mathbb{C})$-structure is the same as an almost complex structure. In this case the intrinsic torsion can be identified with the Nijenhuis tensor. So it vanishes precisely when the almost-complex structure is integrable (i.e. a ordinary complex structure).
3. $G = Sp(n)$. Having an $Sp(n)$-structure on a manifold for which the intrinsic torsion vanishes is equivalent to having a symplectic manifold.

From these examples you can see that the vanishing of torsion can be viewed as a sort of integrality condition. In these latter two cases the space of torsion free connections consists of more then a single point. There are many such connections. That's one reason why we don't see them popping up more often.

Answer 2

Here is an example which I found useful when learning about torsion. Consider $\mathbb{R}^3$. Let $X$, $Y$ and $Z$ be the coordinate vector fields, and take the connection for which $$\begin{matrix} \nabla_X(Y)=Z & \nabla_Y(X)=-Z \\ \nabla_X(Z)=-Y & \nabla_Z(X)=Y \\ \nabla_Y(Z)=X & \nabla_Z(Y)=-X \end{matrix}$$

A body undergoing parallel translation for this connection spins like an American football: around the axis of motion with speed proportional to its velocity. So the geodesics are straight lines, and this connection preserves the standard metric, but it has torsion and is thus not the Levi-Cevita connection.

Answer 3

Here's another reinterpretation of the torsion tensor which seems perhaps more natural.

Consider the identity endomorphism $\mathrm{id}:TM \to TM$, but thought of as a 1-form with values in $TM$; that is, $$\mathrm{id} \in \Omega^1(M;TM).$$ The connection $\nabla$ defines an exterior covariant derivative: $$d^\nabla : \Omega^1(M;TM) \to \Omega^2(M;TM)$$ and the torsion of $\nabla$ is precisely $$T^\nabla = d^\nabla(\mathrm{id}).$$

Answer 4

Almost too basic an approach to give, but I think the only way to intuitively get under the hood of torsion (at least in the Levi-Civita sense) is to really understand the ideas of Lie bracket and connection:

We're used to the fact, working on $\mathbb{R}^n$, that partial derivatives commute: $\frac{\partial}{\partial x_i}\circ\frac{\partial}{\partial x_j}=\frac{\partial^2}{\partial x_ix_j}=\frac{\partial}{\partial x_j}\circ\frac{\partial}{\partial x_i}$. But not only is this untrue in the setting of general $C^2$ manifolds, it also makes no sense- with no global coordinates to turn to, we need some other way of defining a 'direction of differentiation' globally. Fortunately that's exactly what vector fields do, so now our updated equation $\frac{\partial}{\partial X}\circ\frac{\partial}{\partial Y}=\frac{\partial}{\partial Y}\circ\frac{\partial}{\partial X}$ makes sense (modulo some issues of notation)- our only problem being its falsehood in general, which we measure with the Lie bracket.

Now it might be tempting to blame our vector fields for the Lie bracket's general non-zero nature- perhaps we get non-zero Lie brackets just when we pick a really weird vector field... but close examination (of, say, the image of the coordinate vector fields under the differential of your faourite chart map) reveals this is not the case. In fact the $C^2$ness of the vector field ensures that on an infinitessimal level our vector fields are never really very pathological: what the Lie bracket is measuring is something much more intrinsic about our manifold- about how vector fields must locally twist as they move along each other to keep time with the metric.

But telling us how vector fields do move along one another is the job of a connection- which, by giving us $\nabla_X Y$, prescribes $\frac{\partial}{\partial X}Y$, but $Y$ is really $\frac{\partial}{\partial Y}$ so this 'prescribes a value' for the Lie bracket as $\nabla_X Y-\nabla_Y X $.

Subrtracting the former from the latter gives the actual infinitessimal twist minus the neccessary infinitessimal twist to give the 'unneccessary twist' of the connection.

Answer 5

Perhaps, the following two facts help to understand torsion:
1. Two connections are equal if and only if they have the same geodesics and equal torsions.
2. For any connection there is a unique torsion-free connection with the same geodesics.
This is proved in Spivak, volume II, page 249.

Answer 6

Similar to José's answer, one can consider the following: for each connection $\nabla$ on the tangent bundle (or its dual), one can consider the induced connection $\nabla\colon\Gamma(M;\Lambda^k T^* M)\to\Gamma(M; T^* M\otimes \Lambda^k T^* M).$ Denote by $\Lambda\colon T^* M \otimes \Lambda^k T^* M\to \Lambda^{k+1} T^* M$ the antisymmetrising map, and by $d_\nabla=\Lambda\circ\nabla$ some kind of exterior derivative. Then $d_\nabla$ is the exterior derivative if and only if $\nabla$ is torsion free. Moreover $d_\nabla^2=0$ if and only if $\nabla$ is torsion free. This is very similar to the equation of the curvature of a connection $\tilde\nabla$ of an arbitary bundle in terms of its absolute exterior derivative $d^{\tilde\nabla}.$

The torsion of a connection is the obstruction to the induced calculus of the connection to be the usual/natural calculus on a manifold.

Answer 7

I'm afraid that the torsion is not motivated by any picture. It's just the skew-symmetric part of $\nabla$.

Let $M$ be your manifold and $p\in M$. Consider two tangent vectors $v,w\in T_pM$. You can extend them to commuting vector fields $V$ and $W$ in a neighborhood of $p$. Then $$ T(v,w) = \nabla_vW-\nabla_wV , $$ so in this case $T$ measures non-symmetry of $\nabla$. In general (for non-commuting vector fields), the formula $\nabla_XY-\nabla_YX$ does not define a tensor and the term $[X,Y]$ fixes this problem.

Answer 8

ere is a review article by Hehl and Obukhov about the role of torsion in geometry and physics. The article contains the intutive geometric explanation of the torsion tensor as stated by Deane Yang as a measure (figure-1) of the failure to close an infinitesimal parallel transport parallelogram.

The article also contains an interpretation of the torsion tensor in three dimensions as the dislocation density of a dislocated crystal.

Here are a few additional properties of the torsion tensor. In dynamically generated gravity theories and fluid dynamics, the generated torsion tensor is proportional to the anti-symmeterized spin density and vorticity respectively.

In harmonic analysis on (vector bundles over) homogeneous spaces G/H, the Levi-Civita Lagrangian, based on the torsionless connection is not diagonal in the spaces of sections beonging to irreducible G representations (except for the trivial representation). On the other hand there exists an H-connection which is not torsion free whose Laplacian is diagonal. The explanation of this result is that the information about the inducing H-representation defining the vector bundle is contained in the torsion tensor.

Answer 9

Well, one should think in term of Euclidean motions, i.e. rotations AND translations (see Cartan connections) - hence the name affine connection. The (Cartan) curvature of this (Cartan) connection splits into two parts: one measuring infinitesimal rotations (i.e. the ordinary Riemannian curvature) and one measuring infinitesimal translations ("slipping") (i.e. the torsion).
Maybe one should elaborate on this in more detail. (This explanation is related to Jose's)

Answer 10

I'm convinced that there is geometric explanation analogous to curvature measuring infinitesimal holonomy, but I haven't been able to work it out yet.

In any case, at least in the context of Riemannian geometry, what's geometrically natural is zero torsion, so it's not surprising that a geometric interpretation of nonvanishing torsion is a little elusive.

Here are some things that are implied by (and are essentially equivalent to) zero torsion:

1) The ability to define the Hessian of a function as a symmetric tensor

2) A parameterized curve is a constant speed geodesic if and only if its velocity curve is parallel along the curve

This extends some useful properties of Euclidean space to a Riemannian manifold. These properties (and probably some others) along with its uniqueness make the Levi-Civita connection very powerful and useful.