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?


21 Answers 21


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 form 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 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.

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    $\begingroup$ This answer is just great, thank you very much! $\endgroup$ – Jan Weidner Apr 7 '10 at 7:29
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    $\begingroup$ 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 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. $\endgroup$ – algori Aug 2 '10 at 5:43
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    $\begingroup$ @ 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. $\endgroup$ – Chris Schommer-Pries Aug 2 '10 at 11:58
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    $\begingroup$ Don't you mean GL_n(C) in GL_{2n}(R)? $\endgroup$ – David Corwin Dec 16 '12 at 1:36
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    $\begingroup$ More generally, it seems to me that the sheaf of torsion free connections should be a (sheaf) torsor under $\rho_1P$. The intrinsic torsion vanishes if and only if this torsor is trivial, i.e. the sheaf has a global section. $\endgroup$ – R. van Dobben de Bruyn May 20 '17 at 20:06

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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    $\begingroup$ This example is indeed very helpful, thanks! $\endgroup$ – 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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    $\begingroup$ 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 Levi-Civita? – $\endgroup$ – Fly by Night Jun 28 '11 at 17:56
  • $\begingroup$ There was no metric in Tom Boardman's answer. $\endgroup$ – Albuquerque Jul 31 '11 at 10:19
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    $\begingroup$ Great answer, of the kind I'd like to see more often on MO. $\endgroup$ – 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 $$T^\nabla = d^\nabla(\mathrm{id}).$$


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.


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 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 :)

  • $\begingroup$ 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! $\endgroup$ – Anton Fetisov Aug 2 '13 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 G-structures and soldering forms) then torsion is the projection of its curvature under $\mathfrak{g}\to \mathfrak{g}/\mathfrak{h}$. $\endgroup$ – Urs Schreiber Dec 18 '14 at 13:32

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.


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.


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.

  • $\begingroup$ Your geometric picture is right. $\endgroup$ – Marc Nieper-Wißkirchen Nov 18 '13 at 11:48
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    $\begingroup$ 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 Levi-Civita connection. But this is not true. $\endgroup$ – Holonomia Jun 1 '15 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 geodesic-preserving (I had not appreciated the distinction), then it must have only skew-symmetric 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!) $\endgroup$ – Bruce Bartlett Jul 1 '15 at 1:09
  • $\begingroup$ See also these notes. $\endgroup$ – Bruce Bartlett Jul 1 '15 at 16:26
  • $\begingroup$ 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? $\endgroup$ – TheQuantumMan Nov 2 '17 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:

Torsion example

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 torsion-free, and they use that figure as justification. I've modified it to show what a connection with torsion looks like.

Torsion after MTW

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 punch-line 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 no-no 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"):

  1. 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.
  2. 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.
  3. 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 picture-based 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.

  • $\begingroup$ Dear Michael, do you know how to go from your explanation to the explanation of frame-rotation? In particular, there are answers that say(or give examples showing that) non-zero torsion means that vectors perpendicular to the tangent vector of a curve will rotate around it. How could the two explanations be connected> $\endgroup$ – TheQuantumMan Nov 2 '17 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 non-zero torsion in 2d. The answer is yes, as you can see. $\endgroup$ – Michael Weiss Nov 3 '17 at 19:32
  • $\begingroup$ 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.) $\endgroup$ – Michael Weiss Nov 3 '17 at 19:37
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    $\begingroup$ @asv Valid point. I thought it merited adding an addendum to my original answer. $\endgroup$ – Michael Weiss Mar 18 '18 at 14:46
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    $\begingroup$ @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. $\endgroup$ – Michael Weiss Apr 9 '18 at 17:30

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.


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:

on the geometrical interpretation of torsion, figure 1 of (Hehl & Obukhov 2007)


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.


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.


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 3-manifold 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 "e-invariant) 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.

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    $\begingroup$ 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. Sep 15 '18 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 :-) $\endgroup$ – ERCÜMENT HASAN ORTAÇGİL Sep 16 '18 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. $\endgroup$ – ERCÜMENT HASAN ORTAÇGİL Sep 16 '18 at 11:40
  • $\begingroup$ More on the nonlinear curvature R: Consider the principal frame bundle P of M and assume that P admits a global crossection e, that is, M is parallelizable. To e we assign a "nonlinear curvature" R which is a highly nontrivial object: For a compact and simply connected 3-manifold M (which is parallelizable) to show the existence of some e whose R vanishes is equivalent to proving Poincare Conjecture!!! $\endgroup$ – ERCÜMENT HASAN ORTAÇGİL Sep 16 '18 at 12:14
  • $\begingroup$ Two intriguing consequences: 1) A topologically trivial principal bundle can be geometrically a highly nontrivial object. 2) The concept of curvature (also torsion!!) does not necessarily belong to any connection but to the geometric structure once it is properly defined. $\endgroup$ – ERCÜMENT HASAN ORTAÇGİL Sep 16 '18 at 12:18

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)


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

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    $\begingroup$ What does this mean? Can you please elaborate? $\endgroup$ – Spiro Karigiannis Aug 28 '12 at 2:33
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    $\begingroup$ 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 anti-symmetric part of the Hessian the "curvature" of the function. $\endgroup$ – Deane Yang Aug 28 '12 at 12:29
  • $\begingroup$ To be more explicit, $R(X,Y)f = \nabla_X \nabla_Y f - \nabla_Y \nabla_X f - \nabla_{[X,Y]} f = XYf - YXf - [X,Y]f$, which is $T(X,Y)f$. In other words, a connection on $TM$ defines connections on all tensor powers $\bigotimes^r TM \otimes \bigotimes^s T^*M$, including the empty product $\underline{\mathbb{R}}$. The curvature the induced connection on $\underline{\mathbb{R}}$ is the torsion of $\nabla$. $\endgroup$ – Jesse Madnick Jan 28 '16 at 9:40

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 non-symmetric 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 $(E, \tilde \nabla)$).

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    $\begingroup$ 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. Sep 19 '18 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 $\endgroup$ – Saal Hardali Sep 19 '18 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. Sep 19 '18 at 17:30
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    $\begingroup$ @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 $X,Y$. $\endgroup$ – Saal Hardali Sep 19 '18 at 17:31

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 non-Riemannian 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.

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    $\begingroup$ 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. $\endgroup$ – Ben McKay Oct 12 '18 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. $\endgroup$ – Aureliano Skirzewski Oct 12 '18 at 15:38
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    $\begingroup$ 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? $\endgroup$ – Ben McKay Oct 12 '18 at 16:04
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    $\begingroup$ 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. $\endgroup$ – Ben McKay Oct 12 '18 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. $\endgroup$ – Aureliano Skirzewski Oct 12 '18 at 17:34

Intutively torsion is the screw-like twist of a manifold. Think of a 2-D sheet, a 2-D 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 2-D 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.

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    $\begingroup$ Yout twisting sheet has NO torsion in its Levi-Civita connection (by definition of Levi-Civita connection). What connection are you using to see torsion emerging on your sheet? $\endgroup$ – Ben McKay Jan 3 '15 at 15:52
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    $\begingroup$ 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. $\endgroup$ – Deane Yang Jan 3 '15 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 Levi--Civita 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. $\endgroup$ – Ben McKay Feb 14 '16 at 10:55

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