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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?

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The wikipedia page en.wikipedia.org/wiki/… has some geometric picture of torsion in terms of the "twisting" of reference frames along geodesics. –  José Figueroa-O'Farrill Apr 6 '10 at 14:03
    
See also: mathoverflow.net/questions/122729/… –  Ryan Budney Feb 24 '13 at 20:03

15 Answers 15

up vote 60 down vote accepted

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 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 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.

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This answer is just great, thank you very much! –  Jan Weidner Apr 7 '10 at 7:29
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Chris -- your answer is very interesting but a bit difficult to follow, in particular since you use some notation that you do not define ($P,\rho_1 P,\ldots$), so I'd like to ask: is this notion of torsion the same as in Sebastian's answer below (i.e. the same as the one given e.g. in Milnor-Stasheff, Characteristic classes, appendix C)? More precisely, we take a connection on the (co)tangent bundle compatible with a given $G$-structure on a manifold. In example 1 the answer is yes, so I was wondering about the other two. –  algori Aug 2 '10 at 5:43
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@ algori: P is the G-principal bundle coming from the G-structure (see the link I provided) and $\rho_i(P)$ are the associated bundles induced by the representations $\rho_i$, which come from the short exact sequence I mention. Sorry if that was confusing. It seemed clear enough from context, but perhaps it wasn't. Anyway, I don't have my copy of Milnor and Stasheff handy, but I'm fairly certain this torsion is the same. Most of what I say is explained in more detail in D. Joyce's book "Compact Manifolds with Special Holonomy". I suggest looking there for more details. –  Chris Schommer-Pries Aug 2 '10 at 11:58
    
Thanks Chris! This has helped clarify things. –  algori Aug 2 '10 at 18:59
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Don't you mean GL_n(C) in GL_{2n}(R)? –  David Corwin Dec 16 '12 at 1:36

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.

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This example is indeed very helpful, thanks! –  Jan Weidner Apr 6 '10 at 19:57

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.

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This is a really nice explanation. If someone asks about torsion then the answer should be in terms of things of similar or less complexity than torsion itself, and answers in terms of sections of bundles and Lie algebras add to confusion. Like using the example of a car to explain to a caveman what a wheel is. (Not that Jan is a caveman!) I just wanted to pick up on your comment about the Lie bracket measuring the necessary twist of vector fields to keep time with the metric. What about the general case, where M doesn't necessarily have a metric, and ∇ isn't necessarily Levi-Civita? – –  Fly by Night Jun 28 '11 at 17:56
    
There was no metric in Tom Boardman's answer. –  Albuquerque Jul 31 '11 at 10:19
    
Great answer, of the kind I'd like to see more often on MO. –  Filippo Alberto Edoardo Nov 2 '12 at 1:10

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

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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.

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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.

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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/numdam-bin/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 Maurer-Cartan 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 Maurer-Cartan 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 :)

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So basically zero torsion for some G-structure states that locally our manifold looks like a neighbourhood in some $\mathbb{R}^n$ canonical model of our structure (at least if the connection can be locally chosen flat). I.e. a smooth manifold is locally like $\mathbb{R}^n$, a complex one is like $\mathbb{C}^n$, symplectic is like a symplectic vector space etc. Thank you! Together with answers of Chris Schommer-Pries and Peter Michor this finally completes the geometric puzzle! –  Anton Fetisov Aug 2 '13 at 14:58
    
Further putting this all together, it seems we should say that given an $(H \to G)$-Cartan connection ncatlab.org/nlab/show/Cartan+connection (which subsumes G-structures and soldering forms) then torsion is the projection of its curvature under $\mathfrak{g}\to \mathfrak{g}/\mathfrak{h}$. –  Urs Schreiber 2 days ago

Let me expand a little the answer of José Figueroa-O'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 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 $d^{\nabla}\sigma\in\Omega^2(M;E)$. 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 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. This ties in with the answer of Chris Schommer-Pries.

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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.

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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.

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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.

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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 parallel-transported 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 $n-1$ 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.

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Your geometric picture is right. –  Marc Nieper-Wißkirchen Nov 18 '13 at 11:48

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)

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

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What does this mean? Can you please elaborate? –  Spiro Karigiannis Aug 28 '12 at 2:33
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I believe the point is that if the connection has torsion, then the Hessian of a function is no longer symmetric. So you can call the anti-symmetric part of the Hessian the "curvature" of the function. –  Deane Yang Aug 28 '12 at 12:29
    
Ah, I see. Thanks, Deane. –  Spiro Karigiannis Aug 28 '12 at 13:23

Here is a simple physical example of a line with no curvature, but with torsion. This is a cartesian coordinate system with unit vectors z-hat, y-hat and x-hat. I am using a Frenet frame.

t = x-hat

n = cos(x) y-hat - sin(x) z-hat

b = - sin(x) y-hat - cos(x) z-hat

This is a straight line along the x-axis which is constantly "spinning".

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This answer does not address the torsion of a connection (which is what the question is asking for), but rather the torsion of a space curve. Bruce Bartlett's answer points out the relationship. –  j.c. Jan 20 at 7:59

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