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 geometric picture behind this definition—what does torsion measure intuitively?

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 Gstructures. 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 antisymmetrization. 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 Gstructure 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 Gcompatible connection is a section of this third space. Suppose that we have two compatible connections. Then their 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 Gstructure. 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:
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. 


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 LeviCevita connection. 


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 LeviCivita 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 nonzero nature perhaps we get nonzero 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. 


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 1form 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 $$T^\nabla = d^\nabla(\mathrm{id}).$$ 


Perhaps, the following two facts help to understand torsion: 


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. 


Torsion is easy to understand but this knowledge seems to be lost. I had to go back to Elie Cartan's articles to find an intuitive explanation (for example, chapter 2 of http://www.numdam.org/numdambin/fitem?id=ASENS_1923_3_40__325_0). Let $M$ be a manifold with a connection on its tangent bundle. The basic idea is that any path $\gamma$ in $M$ starting at $x\in M$ can be lifted as a path $\tilde\gamma$ in $T_xM$, but is the $\gamma$ is a loop $\tilde \gamma$ need not be a loop. The resulting translation of the end point is the torsion (or its macroscopic version). The situation is easy in a Lie group $G$ (which I imagine Cartan had in mind). $G$ has a canonical flat connection for which the parallel vectors fields are left invariant vectors fields. For this connection the parallel transport is simply the left translation. The MaurerCartan form $\alpha$ is then the parallel transport to the tangent space $T_1G$ at the identity $1\in G$. If $\gamma:[0,1]\to G$ is a path in $G$ starting at $1$. $\gamma'$ is a path in $TG$ and $\alpha(\gamma')$ is a path in $T_1M$. $\alpha(\gamma')$ can be integrated to another path $\tilde \gamma$ in $T_1M$. Let $\gamma_{\leq x}$ be the path $\gamma:[0,x]\to G$, then we define $$ \tilde \gamma(x) = \int_0^x\alpha(\gamma'(t))dt = \int_{\gamma_{\leq x}}\alpha. $$ In the sense given by the connection, $\gamma$ and $\tilde\gamma$ have the same speed and the same starting point, so they are the same path (but in different spaces). If $\gamma$ is a loop and $D$ a disk bounding $\gamma$, $\tilde\gamma$ is a loop iff $\tilde\gamma(1)=0\in T_1G$. We have $$ \tilde\gamma(1) = \int_\gamma\alpha = \int_Dd\alpha. $$ $\tilde\gamma$ is a loop iff this integral is zero. Now, $\alpha$ can be viewed as the solder form for $TG$, so the torsion is the covariant differential $T=d^\nabla\alpha$. As the connection is flat $T$ reduces to $T=d\alpha$. The MaurerCartan equation gives an explicit formula: $T=d\alpha = \frac{1}{2}[\alpha,\alpha]$. The previous integral is then the integral of the torsion $$ \tilde\gamma(1) = \int_Dd\alpha = \frac{1}{2}\int_D[\alpha,\alpha] $$ and may not be zero. The situation is the same for a general manifold, but the parallel transport is not explicit and formulas are harder. The notion behing this is that of affine connection. As I understand it, an affine connection is a data that authorize to picture the geometry of $M$ inside the tangent space $T_xM$ of some point $x$. If I move away from $x$ in $M$, there will be a corresponding movement away from the origin in $T_xM$ (this is the above lifting of path). If I transport in parallel a frame with me, the frame will move in $T_xM$. Globally the movement of my point and frame is encoded by a family of affine transformations in $T_xM$. Of course this picture of the geometry of $M$ in $T_xM$ is not faithful. Because of the torsion, if I have two paths in $G$ starting at $x$ and ending at the same point, they may not end at the same point in $T_xM$. Because of curvature, even if my two lifts end at the same point, my two frames may not be parallel. The picture is faithful if $M$ is an affine space iff both torsion and curvature vanish (Cartan's structural equations for affine space). I think torsion is beautiful :) 


Let me expand a little the answer of José FigueroaO'Farrill. 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 1form 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 $d^{\nabla}\sigma\in\Omega^2(M;E)$. It is an obstruction against the soldering form being parallel for $\nabla$. Maybe this explains, that space is twisting along geodesics if the torsion is nonzero. 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 1form $\sigma\in\Omega^1(P,V)^G_{hor}$ 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. This ties in with the answer of Chris SchommerPries. 


My geometric picture of torsion is as follows. Perhaps I am wrong? Let $M$ be a Riemannian manifold and let $\nabla$ be a connection on it which is compatible with the metric, so that parallel transport preserves orthonormal frames. Let $\exp: T_p (M) \rightarrow M$ be the exponential map, given by by sending a tangent vector $v \in T_pM$ to the endpoint $\sigma(1)$ of the paralleltransported curve $\sigma$ in $M$ with initial velocity vector $v$. So we are regarding $T_p M$ as a geodesic coordinate system on $M$. Let $v \in T_p M$, and upgrade it to a frame $v, e_2, \ldots e_n$ at $0 \in T_pM$ by choosing $n1$ vectors orthogonal to it. The radial line proceeding from the origin with initial velocity vector $v$ is a geodesic. Consider the moving frame $v(t), e_2(t), \ldots, e_n(t)$ along this line. Since it is a geodesic, $v(t) = v$ remains constant, but the frame can rotate around $v$. Claim: the torsion of the connection measures the extent to which the moving frame is rotating around the axis $v$ along this straight line. That's why it's "torsion"... it measures the twisting of the frame. 


I'm afraid that the torsion is not motivated by any picture. It's just the skewsymmetric 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 nonsymmetry of $\nabla$. In general (for noncommuting vector fields), the formula $\nabla_XY\nabla_YX$ does not define a tensor and the term $[X,Y]$ fixes this problem. 


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 LeviCivita connection very powerful and useful. 


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 (figure1) 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 antisymmeterized spin density and vorticity respectively. In harmonic analysis on (vector bundles over) homogeneous spaces G/H, the LeviCivita 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 Hconnection which is not torsion free whose Laplacian is diagonal. The explanation of this result is that the information about the inducing Hrepresentation defining the vector bundle is contained in the torsion tensor. 


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). 


Intutively torsion is the screwlike twist of a manifold. Think of a 2D sheet, a 2D real manifold. Imagine that it ripples in one direction, say uniformly in a sinusoidal curve, but not in the second, orthogonal, direction  all geodesics are straight lines along the second axis, and sine curves along the first. This is a manifold with curvature but no torsion. Now imagine the sheet instead twisted about some axis like a screw. This is a manifold with torsion. The generaization of this simple picture for the 2D sheet to more general forms of torsion is probably fairly obvious, but (for me at least) visualizing it in higher dimensions is quite a challenge. 


Here's a simple picture for connections on the tangent bundle from Kock's synthetic geometry of manifolds. Let $x$ and $y$ be infinitesimally close points in a manifold, and let $\nabla(x,y)$ denote the parallel transport map which takes the infinitesimal neighbourhood $\mathcal{N}(x)$ of $x$ into the infinitesimal neighbourhood $\mathcal{N}(y)$ of $y$. If we have a third point $z$ which lies in $\mathcal{N}(x)$, then we can transport it along the infinitesimal line segment between $x$ and $y$ to get a point $\nabla(x,y)z$ in $\mathcal{N}(y) \cap \mathcal{N}(z)$. But we could instead transport $y$ along the infinitesimal line segment from $x$ to $z$ to get a potentially different point $\nabla(x,z)y$ in $\mathcal{N}(y) \cap \mathcal{N}(z)$. Thus, we have two different ways to complete the infinitesimal wedge $zxy$ to an infinitesimal parallelogram. The torsion measures the extent to which these two completions differ. 


One more interpretation: The torsion is the curvature of the smooth functions (as a vector bundle over your manifold). 


Here is a simple physical example of a line with no curvature, but with torsion. This is a cartesian coordinate system with unit vectors zhat, yhat and xhat. I am using a Frenet frame. t = xhat n = cos(x) yhat  sin(x) zhat b =  sin(x) yhat  cos(x) zhat This is a straight line along the xaxis which is constantly "spinning". 

