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?

10$\begingroup$ The wikipedia page en.wikipedia.org/wiki/… has some geometric picture of torsion in terms of the "twisting" of reference frames along geodesics. $\endgroup$– José FigueroaO'FarrillCommented Apr 6, 2010 at 14:03

$\begingroup$ See also: mathoverflow.net/questions/122729/… $\endgroup$– Ryan BudneyCommented Feb 24, 2013 at 20:03

$\begingroup$ See also: math.stackexchange.com/questions/465603/… $\endgroup$– Ziqian XieCommented Apr 12, 2015 at 22:29
24 Answers
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 form a torsor over sections of the smaller bundle $\rho_1P$. Now some examples:
 $G = O(n)$. 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 LeviCivita connection.
 $G = GL_m(\mathbb{C})$. 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 almostcomplex structure is integrable (i.e. a ordinary complex structure).
 $G = Sp(n)$. 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 integrability 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.

5$\begingroup$ This answer is just great, thank you very much! $\endgroup$ Commented Apr 7, 2010 at 7:29

6$\begingroup$ 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 MilnorStasheff, 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. $\endgroup$– algoriCommented Aug 2, 2010 at 5:43

4$\begingroup$ @ algori: P is the Gprincipal bundle coming from the Gstructure (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. $\endgroup$ Commented Aug 2, 2010 at 11:58

4$\begingroup$ Don't you mean GL_n(C) in GL_{2n}(R)? $\endgroup$ Commented Dec 16, 2012 at 1:36

2$\begingroup$ I am amused that a later question tried to clear up the "notoriously slippery concept" of torsion by asking about the Rolling without slipping interpretation of torsion. $\endgroup$– LSpiceCommented Sep 10, 2020 at 14:17
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.

3$\begingroup$ This example is indeed very helpful, thanks! $\endgroup$ Commented Apr 6, 2010 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 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.

16$\begingroup$ 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 LeviCivita? – $\endgroup$ Commented Jun 28, 2011 at 17:56

2$\begingroup$ Great answer, of the kind I'd like to see more often on MO. $\endgroup$ Commented Nov 2, 2012 at 1:10

1$\begingroup$ @AntonioJPan to add to your question, the Lie bracket only depends on the differentiable structure and not the metric, so I too am confused about that sentence. $\endgroup$– BmaCommented Jan 5, 2022 at 2:07

1$\begingroup$ Sorry, what is " the image of the coordinate vector fields under the differential of your faourite chart map"? $\endgroup$– AlexCommented May 17, 2023 at 22:06

1$\begingroup$ "at 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". Can that be true? The Lie bracket only depends on the smooth structure, not the metric. What does it mean "to keep time with the metric"? $\endgroup$ Commented Jan 12 at 16:14
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 $$T^\nabla = d^\nabla(\mathrm{id}).$$
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/item/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 if 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 :)

$\begingroup$ So basically zero torsion for some Gstructure 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 SchommerPries and Peter Michor this finally completes the geometric puzzle! $\endgroup$ Commented Aug 2, 2013 at 14:58

$\begingroup$ 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 Gstructures and soldering forms) then torsion is the projection of its curvature under $\mathfrak{g}\to \mathfrak{g}/\mathfrak{h}$. $\endgroup$ Commented Dec 18, 2014 at 13:32

$\begingroup$ @Mathieu Thank you for your answer! However, it's not quite clear for me what you meant by $\alpha(\gamma ')$. I suppose that is the pullback of $\gamma'$ to $T_1G$ by parallel transport. However, if that's the case, I don't understand how this argument can fit into the following example: consider the standard unit sphere in $R^3$ with standard metric and standard (LC) connection. It is a torsion free connection, so the path $\bar{\gamma}$ should be a loop. Consider $\gamma$ to be a big circle, then $\alpha(\gamma')$ is a nontrivial constant map, preventing $\bar{\gamma}$ from being a loop. $\endgroup$– StudentCommented Jul 5, 2019 at 1:46

$\begingroup$ @Student If I understand well your example $\alpha(\gamma')$ is the constant path with value zero. It is integrated in the constant path with value $\gamma(0)$, which is a loop. $\endgroup$ Commented Jul 25, 2019 at 11:51

$\begingroup$ @MathieuANEL I have to say I don't really understand what you meant by $\alpha(\gamma')$. If it is just the parallel transform of the path in the tangent bundle back to the tangent space at the initial, then the "bigcircle" example I mentioned above gives a contradiction to your claim "$\tilde{\gamma}$ is a loop iff the integral is zero", since then $\alpha(\gamma')$ is a nonzero constant path and thus the integration of it (which gives $\tilde{\gamma}$ by your definition) does not form a loop. $\endgroup$– StudentCommented Jul 25, 2019 at 16:16
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 torsionfree connection with the same geodesics.
This is proved in Spivak, volume II, page 249.
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.

$\begingroup$ Does the soldering form always exist? $\endgroup$– AlexCommented May 17, 2023 at 22:43
The nicely labeled figure on page three of
Friedrich W. Hehl, Yuri N. Obukhov
Elie Cartan's torsion in geometry and in field theory, an essay, arXiv:0711.1535
makes everything intuiitively clear:
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.

$\begingroup$ what is an "absolute exterior derivative" and what's the formula for the curvature? $\endgroup$ Commented Apr 26 at 22:40
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.

$\begingroup$ Your geometric picture is right. $\endgroup$ Commented Nov 18, 2013 at 11:48

2$\begingroup$ Dear Bruce, I do not understand your picture in dimension $n=2$. Namely, $v(t)$ remains constant as you said. But in dimension $n=2$ there are just one orthogonal $e_2(t)$ (up to $\pm$ sign of course). Then it follows that also $e_2(t)$ remains constant. So I do not see that the frame is "rotating" around the axis $v$. Namely, it seems to me that your picture imply (in dimension 2) that any compatible connection is torsion free hence the LeviCivita connection. But this is not true. $\endgroup$ Commented Jun 1, 2015 at 21:35

$\begingroup$ Indeed, the picture is more complicated than I initially thought. I think these notes clears up things. Corollary 2.1 of that paper shows that if a connection is metric and geodesicpreserving (I had not appreciated the distinction), then it must have only skewsymmetric torsion, which means the dimension must be greater than 2. So it seems I was talking about connections which are both compatible with the metric and preserve geodesics. (Still amazed that these notions are different!) $\endgroup$ Commented Jul 1, 2015 at 1:09


$\begingroup$ Dear Bruce, could you please explain to me what you mean by metric and geodesic preserving? For example, metric preserving means compatible with the metric? If so, then what is the analog of geodesic preserving? $\endgroup$ Commented Nov 2, 2017 at 23:36
Here is a 2D example with a picture: \begin{align*} &\nabla_y {\bf e}_x = {\bf e}_y; \quad \nabla_y {\bf e}_y = {\bf e}_x\\ &\nabla_x {\bf e}_x = \nabla_x {\bf e}_y =0 \end{align*} These equations say that the standard xy frame rotates clockwise (according to the connection) as it's transported upwards, and so if we parallel transport a vector upwards, it rotates counterclockwise according to the standard xy frame. Picture:
On the left hand side, I've parallel transported the unit frame from the origin to a bunch of lattice points. The right hand side shows how if we parallel translate the unit x vector in the y direction, and the unit y vector in the x direction, the two tips fail to meet.
In fact, you can think of the torsion as measuring the "failure of quadrilaterals to close under parallel transport", in the same sense as the Lie derivative $[{\bf x},{\bf y}]$ measures the "failure to close" under the Lie flow.
This becomes clear with a second picture. The inspiration comes from Gravitation by Misner, Thorne, and Wheeler, the figure in Box 10.2C (p.250). MTW assume that all their connections are torsionfree, and they use that figure as justification. I've modified it to show what a connection with torsion looks like.
In this picture, ${\bf u}_0$ and ${\bf v}_0$ are vectors at point $P(0)$; we are computing the torsion $\tau({\bf u}_0,{\bf v}_0)$ at that point. We extend ${\bf u}_0$ and ${\bf v}_0$ to smooth vector fields in a neighborhood of $P(0)$. The idea is that the diagram illustrates the situation in an "infinitesimal" neighborhood of $P(0)$, so ${\bf u}_0$ and ${\bf v}_0$ and other vectors must be scaled by $\epsilon$ (or $\epsilon^2$) to fit in the picture. So $\epsilon{\bf u}_\$ is $\epsilon{\bf u}_0$ parallel transported by $\epsilon{\bf v}_0$; $\epsilon{\bf v}_\epsilon$ is the $\epsilon$scaled value of the ${\bf v}$ vector field at $P(\epsilon)$; etc. The punchline is the formula for the torsion: $$\tau( {\bf u} ,{\bf v})= \nabla_{\bf u}{\bf v}\nabla_{\bf v}{\bf u}[{\bf u},{\bf v}]$$ which you can read off the quadrilateral on the upper right corner of the diagram.
The 2d example above has zero curvature. For the geometrical intuition for this, see fig.11.2 (p.278) of Gravitation.
Closing moral: if you are looking for geometrical intuition in differential geometry (rather than rigor), your first port of call should be Misner, Thorne, and Wheeler. I've never seen any other book willing to devote so much paper and ink to that goal.
Addendum: asv makes the valid point that in the diagram I am doing the subtraction $\epsilon u_\\epsilon v_\$, even though $u_\$ and $v_\$ lie in different tangent spaces. Of course this is a nono when doing rigorous proofs. However, the question asked for the intuitive meaning of torsion. (Indeed, no theorem was stated.)
The distinction is discussed in Gravitation in Boxes 8.3 (p.199) and 9.2C (p.238). Box 8.3 ("Three Levels of Differential Geometry") talks about the pictorial, abstract, and component levels. At the pictorial level you draw rough pictures to gain geometric insight. At the abstract level you have precise definitions and rigorous proofs. At the component level you introduce coordinate systems to do computations.
Here's how they put it in Box 9.2C ("Philosophy of Pictures"):
 Pictures are no substitute for computation [or rigor]. Rather, they are useful for (a) suggesting geometric relationships that were previously unsuspected and that one verifies subsequently by computation; (b) interpreting newly learned geometric results.
 This usual noncomputational role of pictures permits one to be sloppy in drawing them. No essential new insight was gained in part B over part A, when one carefully moved the tangent vectors into their respective tangent spaces, and permitted only curves to lie in spacetime. Moreover, the original picture (part A) was clearer because of its greater simplicity.
 This motivates one to draw "sloppy" pictures, with tangent vectors lying in spacetime itself—so long as one keeps those tangent vectors short and
occasionally checks the scaling of errors when the lengths of the vectors are halved.
I'll add that turning a picturebased plausibility argument into a rigorous proof often involves as much work as coming up with the picture in the first place.
Finally, "your mileage may vary": some people are just fine with a purely abstract treatment of differential geometry. Many of us though find it an unsatisfying dish without a pictorial sauce.

1$\begingroup$ Dear Michael, do you know how to go from your explanation to the explanation of framerotation? In particular, there are answers that say(or give examples showing that) nonzero torsion means that vectors perpendicular to the tangent vector of a curve will rotate around it. How could the two explanations be connected> $\endgroup$ Commented Nov 2, 2017 at 23:38

$\begingroup$ The common theme is the failure of quadrilaterals to close under parallel transport. One of the other answers (anonymous, near the top) gives the example of Cartan's spiral staircase; this is a 3d example, so you have more flexibility. I constructed my example when I was wondering if you could have nonzero torsion in 2d. The answer is yes, as you can see. $\endgroup$ Commented Nov 3, 2017 at 19:32

1$\begingroup$ If you want to see a diagram for Cartan's spiral staircase, google that phrase; you'll find an arXiv article by Hehl and Obukhov; look at Fig. 13 (p.22). (Incidentally, they have a figure much like my second diagram; I only discovered this paper after I'd posted my answer. I haven't seen my 2d example elsewhere to date.) $\endgroup$ Commented Nov 3, 2017 at 19:37

1$\begingroup$ @asv Valid point. I thought it merited adding an addendum to my original answer. $\endgroup$ Commented Mar 18, 2018 at 14:46

1$\begingroup$ @asv Yes, that's how you do it for a formal definition. In other words, you compare parallel transport with Lie dragging. MTW discuss this (briefly) in their text; also, if you want a really thorough rigorous discussion of torsion, other books are a better bet. $\endgroup$ Commented Apr 9, 2018 at 17:30
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.
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.
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.
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 $z \sim x \sim y$ to an infinitesimal parallelogram. Torsion measures the extent to which these two completions differ.
The concept of torsion in differential geometry is clarified in the recent book "An Alternative Approach to Lie Groups and Geometric Structures" whose title could be as well "What is Torsion?".
Let me try to briefly explain the picture from the standpoint of this book (following the advice of j.c).
Let P the principal frame bundle of M and assume that P admits a global section e, that is, M is parallelized by e. According to the general theory of connections on principal bundles and their associated vector bundles, e defines an obvious flat connection on P and another obvious linear flat connection on the tangent bundle. However, we assisgn to this geometric structure (=absolute parallelism) a nonlinear curvature R which is a highly nontrivial object: To prove the existence of some e with vanishing R on some compact and simply connected 3manifold M (which is parallelizable) is equivalent to proving the Poincare conjecture! The "linearization" of this picture gives two "connections" on the tangent bundle, one for left and the other for right. One of them is flat and coincides with the above flat linear connection. However, the second is not necessarily flat and differs from the first by "torsion". In fact, the curvature of the second (called the linear curvature of absolute parallelism) vanishes if and only if R vanishes. This is Lie's 3rd Theorem. In this case, "torsion" becomes "constant" over M and coincides with the structure constants (up to sign) of the emerging two Lie algebras (left/right) of vector fields on M. Therefore, "torsion" is the structure functions of certain vector fields on M (those which are "einvariant) and these functions become structure constants of the emerging Lie algebra of vector fields when the nonlinear (or linear) curvature vanishes.
If we generalize the above picture to arbitrary "geometric structures" (including Riemannian), we still have the nonlinear and linear curvatures and Lie's 3rd Theorem. These curvatures belong to the geometric structure and not necessarily to any connection!! Therefore, the illusionary concept of "torsion" is due to the fact that we want to attribute it to some "connection". Clearly, only "very special" connections have torsions. Once we understand clearly what this "very special" means, then the illusion disappears, at least from the standpoint of this book.

5$\begingroup$ As the author of this book, perhaps you could give a few more details about the approach to torsion taken there and what intuitive picture it suggests? $\endgroup$– j.c.Commented Sep 15, 2018 at 16:45

$\begingroup$ Thank you very much for asking this question. We are told that torsion is a property of certain connections (the story begins with connections on the tangent bundle). It is shown in this book that this is a mirror image and the true object is quite different: Torsion is the "analog" of the structure constants of a Lie algebra and measures the deviation from being "abelian". The setup is explained in the introduction of this book which can be reached thru internet. Unfortunately I can't give more technical details here :) $\endgroup$ Commented Sep 16, 2018 at 7:27

$\begingroup$ Let me try a little more with the technical details: The above Lie algebra emerges when the "nonlinear curvature" R of the parallelizable manifold vanishes. Now R is NOT the curvature of any connection on some principle bundle! However its "linearization" turns out to be a linear connection on the tangent bundle whose torsion coincides with those structure constants. This is the simplest example of a geometric structure: Absolute parallelism. $\endgroup$ Commented Sep 16, 2018 at 11:40

1$\begingroup$ Thanks for the details! I recommend editing this material into your answer to make it easier to read. $\endgroup$– j.c.Commented Sep 16, 2018 at 14:00

1$\begingroup$ I see you've figured it out. I'll remove my comment above. I made some small formatting edits  note that "markdown formatting" ignores single line breaks; you have to put in two to create a new paragraph. You can also use LaTeX to insert formulas and symbols if you wish. See more advice here mathoverflow.net/editinghelp $\endgroup$– j.c.Commented Sep 16, 2018 at 18:48
One more interpretation: The torsion is the curvature of the smooth functions (as a vector bundle over your manifold).

1$\begingroup$ What does this mean? Can you please elaborate? $\endgroup$ Commented Aug 28, 2012 at 2:33

4$\begingroup$ 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 antisymmetric part of the Hessian the "curvature" of the function. $\endgroup$ Commented Aug 28, 2012 at 12:29

$\begingroup$ @JesseMadnick Old history, but I believe you meant $R(X,Y) f = \nabla_X \nabla_Y f  \nabla_Y \nabla_X f  \nabla_{\nabla_X Y  \nabla_Y X} f = [X,Y] f  (\nabla_X Y  \nabla_Y X)f$. This is now equivalent to $ T(X,Y) f$, while your expression was identically zero. $\endgroup$– jawheeleCommented Feb 24, 2023 at 19:39

$\begingroup$ @jawheele: Yes, right, my expression was identically zero, which is silly. I've deleted my comment. $\endgroup$ Commented Feb 25, 2023 at 6:05
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)
Let $M$ be a manifold and let $\nabla$ be a connection on its tangent bundle and let $(E, \tilde \nabla )$ be a vector bundle with connection.
These two connections together extended uniquely to connections on all bundle of the form $TM^{\otimes \bullet} \otimes T^*M^{\otimes \bullet} \otimes E^{\otimes \bullet} \otimes E^{*\otimes \bullet}$ via the leibniz rule. One can now try to define a second derivative ("hessian") on $E$ as the composition:
$$\tilde\nabla^2 : \Gamma(M, E)\to \Gamma(M,T^*M \otimes E) \to\Gamma(M,T^*M \otimes T^*M \otimes E)$$
One gets the following expresion for $\tilde\nabla^2$:
$$\tilde\nabla^2_{X,Y}= \tilde\nabla_X \tilde\nabla_Y  \tilde\nabla_{\nabla_X Y}$$
Which is visibly nonsymmetric w.r.t. interchanging $X$ and $Y$. The difference between the two second derivatives is easy to calculate and we get:
$$\tilde\nabla^2_{X,Y}  \tilde\nabla^2_{Y,X} = R_E(X,Y)  \tilde\nabla_{T(X,Y)}$$
Where $R_E$ is the curvature of $\tilde\nabla$ and $T$ is the torsion of $\nabla$. Notice that $R_E$ doesn't depend on $\nabla$ and that $T$ doesn't depend on $E$ or $\tilde \nabla$.
So in summary, torsion can be seen as the intrinsic obstruction to the symmetry of second derivatives for bundles with connection on $M$ (intrinsic in that it depends only on $(TM,\nabla)$ and not on $(E, \tilde \nabla)$).

1$\begingroup$ Interesting answer, because unlike the other answers it talks about connections in arbitrary vector bundles. Unfortunately, I cannot understand it because I have never seen the notation $\tilde \nabla _{X,Y} ^2$; what does it mean? $\endgroup$– Alex M.Commented Sep 19, 2018 at 17:22

$\begingroup$ @AlexM. $\tilde \nabla$ is just the connection on $E$ and $\tilde \nabla^2$ is its square i.e. the composition defined above. You could look here: [en.m.wikipedia.org/wiki/Second_covariant_derivative] for some details $\endgroup$ Commented Sep 19, 2018 at 17:26

$\begingroup$ No, I'm asking about $\tilde \nabla _{X,Y} ^2$: is it $\tilde \nabla ^2 (X,Y) : \Gamma(E) \to \Gamma(E)$? $\endgroup$– Alex M.Commented Sep 19, 2018 at 17:30

1$\begingroup$ @AlexM. Aha sorry. Yeah I think we mean the same thing. Its just the partial evaluation of this operator on the pair of vector fields $X,Y$. $\endgroup$ Commented Sep 19, 2018 at 17:31

$\begingroup$ How does one get that expression for $\tilde{\nabla}^2$? It seems strange that there's any formula relating $\tilde\nabla$ and $\nabla$. $\endgroup$ Commented Apr 26 at 22:37
Define a path as a set of instructions to move, as given in the reference frame of the starting point, i.e. one step forwards, one step to your right, one step back and one step left, all given with respect to the initial reference frame. Of course, in order to follow the instructions, we have to parallel transport the reference frame with the path as we move.
In a flat space, when we parallel transport our reference frame through a closed path (closed in the flat map we parallel transported), we end up at the initial point and the reference frame ends in its initial orientation. If we do this in a torsionless, curved space, we probably will end up somewhere else, unless the path is infinitesimal. Yet, in this case, the reference frame will probably be changed with respect to the initial frame. In Riemannian geometries, the reference frame will simply be rotated with respect to the original but relative angles and the length of the basic vectors is left invariant. In nonRiemannian geometries relative angles and lengths of the basis vectors of the reference frame can vary.
What happens when you go through an infinitesimal closed path if torsion is not zero? The set of instructions close, but we end up in a different point of the space. Torsion is the measure by which an infinitesimal closed path does not result in a closed loop.

1$\begingroup$ This doesn't look much different from the answers above. Could you make clearer how it differs? In particular, David Moshe and Dennis Sullivan seem to have made similar answers. $\endgroup$ Commented Oct 12, 2018 at 15:28

$\begingroup$ I edited after the comment. This makes contact with the notion of torsion given for curves, as the rate of change of the curve's osculating plane. I didn't find an intuitive notion in their answers. $\endgroup$ Commented Oct 12, 2018 at 15:38

2$\begingroup$ You say "If we do this in a torsionless, curved space, we probably will end up somewhere else, unless the path is infinitesimal. However, the reference frame will be changed with respect to the initial frame." How do you measure whether it changed, if they are frames at different points? $\endgroup$ Commented Oct 12, 2018 at 16:04

3$\begingroup$ Moshe said "the failure to close an infinitesimal parallel transport parallelogram," which is more specific than an "infinitesimal closed path does not result in a closed loop", but essentially the same answer. $\endgroup$ Commented Oct 12, 2018 at 16:07

$\begingroup$ @BenMcKay I rewrote once again the explanation. The frames are comparable when the paths are closed curves, that's what I meant. $\endgroup$ Commented Oct 12, 2018 at 17:34
In what follows, summation signs are on repeated indices.
Let $d\ge1$ be an integer. In linear algebra, if $M$ is a $d\times d$ real matrix and $v\in{\mathbb R}^d$, and if we want to calculate $\sum M_{ij} v_i v_j$, then you might as well assume that $M$ is symmetric; if not, and we replace $M$ by its symmetrization $S:=(M+M^t)/2$, then, no matter what $v\in{\mathbb R}^d$ we use, we'll get the same answer whether you use $M$ or $S$. That is, $\sum M_{ij} v_i v_j = \sum S_{ij} v_i v_j$.
So when the goal is to study a quadratic form like $$v\,\,\mapsto\,\,\sum M_{ij} v_i v_j\,\,:\,\,{\mathbb R}^d\,\,\to{\mathbb R},$$ the habit is simply to assume that $M$ is symmetric, because that is often useful, and represents no loss of generality. The symmetry of $M$ can be expressed by saying that, $\forall i,j$, $M_{ij}M_{ji}=0$.
Fix $\varepsilon>0$. Let $I:=(\varepsilon,\varepsilon)$. Let $\Gamma_{ij}^k$ be the Christoffel symbol of a connection on ${\mathbb R}^d$. In differential geometry, if $c:I\to{\mathbb R}^d$ is $C^\infty$, and if $\dot{c}:I\to{\mathbb R}^d$ denotes its derivative, and if we want to calculate, for each $k$, the function $\sum \Gamma_{ij}^k \dot{c}_i \dot{c}_j:I\to{\mathbb R}$, then we might as well assume that, $\forall k$, $\Gamma_{ij}^k$ is symmetric in $i$ and $j$; if not, and we replace $\Gamma$ by its $ij$symmetrization $\Delta_{ij}^k:=(\Gamma_{ij}^k+\Gamma_{ji}^k)/2$, then, no matter what curve $c$ we use, we'll get the same answer, whether you use $\Gamma$ or $\Delta$. That is, $\forall k$, $\sum\Gamma_{ij}^k \dot{c}_i \dot{c}_j= \sum\Delta_{ij}^k \dot{c}_i \dot{c}_j$.
So when the goal is to study, for each $k$, the quadratic map $$c\,\,\mapsto\,\,\sum \Gamma_{ij}^k \dot{c}_i \dot{c}_j \,\,:\,\,C^\infty(I,{\mathbb R}^d)\,\,\to\,\, C^\infty(I,{\mathbb R}),$$ the habit is simply to assume that $\Gamma_{ij}^k$ is $ij$symmetric, because that is often useful, and represents no loss of generality. The $ij$symmetry of $\Gamma_{ij}^k$ can be expressed by saying that, $\forall i,j,k$, $\Gamma_{ij}^k\Gamma_{ji}^k=0$.
The formula for torision is $T_{ij}^k:=\Gamma_{ij}^k\Gamma_{ji}^k$. So $ij$symmetry of $\Gamma_{ij}^k$ is the same as saying $\Gamma$ is torsionfree.
According to the geodesic equation, $c$ is a $\Gamma$geodesic iff, $\forall k$, $\ddot{c}_k+\sum \Gamma_{ij}^k \dot{c}_i \dot{c}_j=0$. Also, $c$ is a $\Delta$geodesic iff, $\forall k$, $\ddot{c}_k+\sum \Delta_{ij}^k \dot{c}_i \dot{c}_j=0$. Then, assuming, as before, that $\Delta_{ij}^k$ is the $ij$symmetrization of $\Gamma_{ij}^k$, we get: $\Gamma$ and $\Delta$ have the same geodesics.
The final upshot of all of this: If you're mainly interested in studying the geodesics of a connection, you can simply subtract off its torsion, and get a torsionfree connection, without changing the collection of geodesics.
I've thought that things would all be a lot clearer if the torsion of a connection were called its "antisymmetric part" and if a torsionfree connection were called a "symmetric" connection. I suppose whoever decided to use "torsion" was aware of examples like the American football spiral described in another answer in this post.
That said, my hunch is that the original motivation was to mimic, in the theory of connections, the representation of a quadratic form by a symmetric matrix. I speculate that, after the definition was made, there was some effort to see what it said geometrically, and that led to the term "torsion".
In page306 of Road to Reality by Roger Penrose, a nice interpretation of torsion is given:
If we consider a parellogram made by two geodesic families of $\lambda$ and $\mu$ then vanishing torsion means the gap closing length of parellogram is of $\mathbb{O}(\epsilon^3)$ and non vanishing Torsion means the gap closing length is of $\mathbb{O}(\epsilon^2)$

1$\begingroup$ How is this different from Michael Weiss's answer? $\endgroup$ Commented Aug 17, 2022 at 14:51

1$\begingroup$ Actually their answer is some what misleading because the parellogram doesn't neccesarily close even if the torsion is 0 @BenMcKay $\endgroup$ Commented Aug 17, 2022 at 14:55
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.

2$\begingroup$ Yout twisting sheet has NO torsion in its LeviCivita connection (by definition of LeviCivita connection). What connection are you using to see torsion emerging on your sheet? $\endgroup$ Commented Jan 3, 2015 at 15:52

2$\begingroup$ This seems to be some kind of 2d notion analogous to the torsion of a curve. I never heard of this before. In any case, as Ben says, it's not the torsion of a connection. $\endgroup$ Commented Jan 3, 2015 at 17:16

$\begingroup$ If the sheet is the graph of $z=sin(x)$ in $(x,y,z)$ coordinates, then it has no curvature or torsion in its LeviCivita connection induced from its metric as a surface in Euclidean space. I would like to understand what you mean. Maybe you could put in some equations to make this clearer. $\endgroup$ Commented Feb 14, 2016 at 10:55
$ T(A,B) $ is just the failure of multiple covariant derivatives to commute on scalar functions. It's $f_{;a;b}  f_{;b;a}$
By the product rule: $$ (f_{;a} A^a)_{;b} = f_{;a;b}A^a + f_{;a} A^a_{;b} $$ So $$ f_{;a;b}A^a = (f_{;a} A^a)_{;b}  f_{;a} A^a_{;b} $$
Now substitute into $(f_{;a;b}  f_{;b;a}) A^aB^b$ $$ = ((f_{;a} A^a)_{;b}B^b  f_{;a} A^a_{;b}B^b)  ((f_{;b} B^b)_{;a}A^a  f_{;b} B^b_{;a}A^a) $$
$$ = B(A(f))  \nabla_BA(f)  A(B(f)) + \nabla_AB(f) $$ $$ = (\nabla_AB  \nabla_BA  [A,B])(f) $$ $$ = T(A,B)(f) $$
PS: The Riemann curvature tensor $R$ measures the failure of failure of $ \nabla $ to commute on vectors. $T$ and $R$ and the product rule can be used to calculate $X_{;a;b}  X_{;b;a}$ for tensors of any rank.