What is torsion in differential geometry intuitively?

Question

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

Answer 1

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

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

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

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

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

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

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

Now the torsion of any G-compatible connection is a section of this third space. Suppose that we have two compatible connections. Then 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.

Answer 2

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

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

Answer 3

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

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

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

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

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

Answer 4

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

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

Answer 5

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 :)

Answer 6

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

Answer 7

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.

Answer 8

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:

Answer 9

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

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

Answer 10

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.

Answer 11

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 torsion-free, 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 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.

Answer 12

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

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

Answer 13

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

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

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

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

Answer 14

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.

Answer 15

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.

Answer 16

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.

Answer 17

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

Answer 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)

Answer 19

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 $(E, \tilde \nabla)$).

Answer 20

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.

Answer 21

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 torsion-free.

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 torsion-free 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 "anti-symmetric part" and if a torsion-free 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".

Answer 22

In page-306 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 $\mathbb{O}(\epsilon^2)$

Answer 23

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.

Answer 24

$ 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.