I've just spent much too long puzzling over the geometric formalism of a paper of Lazar and Hehl, so I figure I'll write a little something about Riemann-Cartan geometry here. *Note: I will only discuss geometry, and from a very limited perspective at that. For physical applications please see the paper cited above.*

## (pseudo/semi-)Riemannian geometry

First we recall what a semi-Riemannian geometry is (see also Chapter 3 of B.O'Neill, *Semi-Riemannian Geometry*). To begin, let us recall some definitions.

*affine connection*$\nabla$ on a smooth manifold $M$ is a map $C^\infty(M,TM)\times C^\infty(M,TM) \to C^\infty(M,TM)$ taking two smooth vector-fields $V,W$ to a third smooth vector-field denoted $\nabla_VW$ satisfying the following three properties:

*($\mathbb{R}$-linearity in $W$)*$\nabla_V(cW) = c \nabla_VW$ for any real number $c$,*($C^\infty(M,\mathbb{R})$-linearity in $V$)*$\nabla_{fV}W = f\nabla_VW$ for any smooth, real-valued function $f$ on $M$, and*(Leibniz rule for $C^\infty(M,\mathbb{R})$-multiplication in $W$)*$\nabla_V(fW) = V(f) + f \nabla_VW$ for any smooth, real-valued function $f$.

Another way of thinking about the affine connection is that it defines a *derivation* on the tangent bundle. One can naturally extend this derivation to the cotangent bundle and to the tensor bundle of arbitrary rank using the rules of tensor derivation.

*tensor derivation*if

*(Leibniz rule)*$\mathcal{D}(A\otimes B) = (\mathcal{D}A)\otimes B + A\otimes (\mathcal{D} B)$,*(Commute with contractions)*$\mathcal{D}(c(A)) = c(\mathcal{D}A)$ where $c: C^\infty(M,T^{r,s}M)\to C^\infty(M,T^{r-1,s-1}M)$ is an arbitrary tensor contraction.

- By the general Leibniz rule of Definition 2 and the more restricted Leibniz rule of Definition 1, we see that $\nabla_V$ is restricted to act on rank-0 tensors (in other words, smooth functions) as the directional derivative.
- The action on the cotangent bundle can be acquired by using both the Leibniz rule and the contraction rule: \[ \begin{array}{rl} (\nabla_V\omega)(W) & = c[(\nabla_V \omega)\otimes W] = c[ \nabla_V (\omega\otimes W) - \omega\otimes(\nabla_VW)] \newline & = \nabla_V[c(\omega\otimes W)] - \omega(\nabla_VW) = V[\omega(W)] - \omega(\nabla_VW)\end{array} \] where $\omega$ is a smooth one-form and $W$ a vector field, and we note that $\omega(W)$ is a smooth function.
- The action on the higher rank fields can be found analogously using the Leibniz and contraction rules.

The fact that we have the two definitions above is not completely essential. It is just to show that the notion of a connection $\nabla$ acting on the graded algebra of tensor fields such that $\nabla: C^\infty(M,T^{r,s}M)\to C^\infty(M,T^{r,s+1}M)$ can be given independently of the notion of a metric, and that the action of the connection on vector fields is sufficient to determine its action on arbitrary tensors.

Associated to a connection $\nabla$ is an object called the torsion

*torsion*to be the $(1,2)$-tensor $\Theta$ given by $\Theta(V,W) = \nabla_VW - \nabla_WV - [V,W]$ where $[\cdot,\cdot]$ is the Lie bracket.

Now let us introduce a metric

*semi-Riemannian metric*on the manifold $M$ is a smooth $(0,2)$-tensor $g$ which is symmetric, non-degenerate, and whose index is constant.

*compatible*with a metric $g$ if $g$ is parallel with respect to $\nabla$. In other words $\nabla g = 0$.

One of the cornerstones of semi-Riemannian geometry is the following theorem

*Levi-Civita connection*associated to $g$, and can be defined using the Koszul formula.

## Riemann-Cartan (or Einstein-Cartan) geometry

In the geometries named above, we relax the torsion-free requirement on the affine connection. The torsion and the curvature described two different kinds of "defects" a given manifold with a given connection can have that distinguish it from being flat. To see these, imagine we have a very small loop on our manifold, that starts and ends at the same point. Imagine a particle that travels along that loop with unit speed. To travel along the loop with unit speed, the particle is subject to some force (the force maybe zero if, in our particular case, the loop is a geodesic).

Now put this same particle into flat space, and start it moving, and apply some force to the particle so that *from the particle's point of view, the experience of motion is the same*. One can imagine the particle being a small kid riding a roller-coaster with his eyes closed. From our everyday experience, the kid can imagine roughly the trajectory the roller-coaster takes just by feeling how his body is thrown to and fro. Now suppose we move the roller-coaster into space. Without the force of gravity (which, by Einstein's theory of General Relativity we equate with curvature of space) from the Earth, the same roller-coaster will give a very different "feel". In fact, to reproduce the same "feel" in space, the roller-coaster tracks have to be laid in a very different way. ^{1}

At the end of the motion in flat space, following the accelerations imposed, the particle will no longer necessarily end up in the same place where it started. This difference--that in one geometry the curve closes into a loop and in another geometry the curve stays open--can be used to characterize the geometry. Now, for the ease of argument, let us assume our initial loop in curved space is gotten by building a roller-coaster that "only turns left". It does not turn up or down. Nor right. Just left. Intuition tells us that then our initial velocity (forward from the point of view of the little kid on the roller-coaster) and our initial acceleration (to the left from the point of view of the little kid) defines a two-dimensional plane, and our roller-coaster, which stays in the plane, is a boring one indeed. But now consider the same roller-coaster taken to flat space. Again, the initial velocity and acceleration defines a two-dimensional plane, but this time the roller-coaster does not stop where it started. There are two possibilities to how the roller-coaster can end somewhere else: (1) The coaster can have moved vertically in this flat space (I'll explain how this is possible) and (2) The coaster may not have traveled "full circle".

Let us look at (2) first. One way to examine this possibility is by projection. Say it is noon-time and the sun is directly over-head (compared to the starting position of the coaster, where the direction of acceleration is "left" and the direction of motion is "forward"). Then we can look at the shadow of the coaster on the ground. That the coaster have not traveled "full circle" is given by that even the shadow does not form a closed loop. This means that there is an angle defect: the circumference of a circle in the curved manifold is different from the circumference in flat space. And this is a characteristic of the manifold having curvature. In the case of (semi-)Riemannian geometry, this is all we care about.

(1), on the other hand, corresponds to torsion. While it is said that the track can only "turn to the left", there has been no restrictions on the track "twisting" (what is known in coaster-parlance as corkscrews). Imagine the curved manifold being specified in such a way that the only way a curve that constantly turns to the left can only be closed if the curve also steadily twists clockwise. Then the corresponding roller-coaster, when built in flat-space, will have to turn left while having a clockwise twist. A little bit of imagination shows that such a coaster will end up higher (in the vertical direction defined by the initial velocity and acceleration) than where it started. This (roughly speaking) vertical translation is what the torsion tensor measures.

Now, let us examine the geometry in local coordinates. The connection can be expressed in terms of the Christoffel symbols: \begin{equation}\label{eqn7} \nabla_{\frac{\partial}{\partial x^i}}\frac{\partial}{\partial x^j} = \sum_k \Gamma^k_{ij} \frac{\partial}{\partial x^k}. \end{equation} Now, using Definition 3 for the torsion evaluated in the coordinates, we have \[ \Theta(\frac{\partial}{\partial x^i}, \frac{\partial}{\partial x^j}) = \sum_k (\Gamma^k_{ij} - \Gamma^k_{ji})\frac{\partial}{\partial x^k} \] where we used the fact that Lie brackets of coordinate vector fields vanish. So the torsion tensor is the "anti-symmetric" part of the Christoffel symbol.

On the other hand, if the connection is metric compatible, we have that \[ \frac{\partial}{\partial x^k}[ g(\frac{\partial}{\partial x^i}, \frac{\partial}{\partial x^j}) ] = \sum_l \Gamma_{ki}^lg(\frac{\partial}{\partial x^l},\frac{\partial}{\partial x^j}) + \Gamma_{kj}^lg(\frac{\partial}{\partial x^l},\frac{\partial}{\partial x^i}). \] By permuting the indices of the above equation, and writing (for ease of notation) the function $g_{ij} = g(\frac{\partial}{\partial x^i}, \frac{\partial}{\partial x^j})$ and the vector field $\Theta_{ij} = \Theta(\frac{\partial}{\partial x^i}, \frac{\partial}{\partial x^j})$, we have the following expression \[ \partial_i g_{kj} + \partial_j g_{ki} - \partial_k g_{ij} = \sum_l \left[ (\Gamma^l_{ik} - \Gamma^l_{ki})g_{lj} + (\Gamma^l_{jk} - \Gamma^l_{kj})g_{li} + (\Gamma^l_{ij} + \Gamma^l_{ji})g_{lk}\right] . \] So let us write the symmetric part of the Christoffel symbol as \begin{equation}\label{eqn8} \tilde{\Gamma}^l_{ij} = \frac12(\Gamma^l_{ij} + \Gamma^l_{ji}) \end{equation} then we have that \begin{equation}\label{eqn9} \tilde{\Gamma}^l_{ij} = \frac12 \sum_k g^{kl} [ \partial_i g_{kj} + \partial_j g_{ki} - \partial_k g_{ij} +g(\Theta_{ki}, \frac{\partial}{\partial x^j}) + g(\Theta_{kj}, \frac{\partial}{\partial x^i}) ]. \end{equation} So we see that a non-zero torsion contributes both to the symmetric and the anti-symmetric part of the connection. But observe that in a special case the torsion only contributes to the anti-symmetric part: let $\Theta^\flat_{kij} = g(\Theta_{ij},\partial_k)$, then we see that if $\Theta^\flat_{kij}$ is totally anti-symmetric, then the symmetric part of the connection does not depend on the torsion. This particular construction has one further property: in this case the geodesics of the Riemann-Cartan manifold with torsion agree with the geodesics of the Riemannian manifold without torsion. It is immediate from the observations above that the symmetric part of the connection is identical to the Levi-Civita connection. Using that $\Theta(V,V) = 0$ by anti-symmetry, it is clear that if $V$ solves the geodesic equation for the situation with no torsion, it will also solve the geodesic equation when the totally anti-symmetry torsion is added.

## Cartan staircase

The Cartan staircase is a simple case of a three-dimensional Riemann-Cartan manifold with constant torsion. For the Cartan staircase, the base Riemannian manifold is just the Euclidean three-space. But the construction below is more general. In fact, assuming that the torsion contributes only to the anti-symmetric part of the connection, then the construction below gives the only way to have a constant torsion three-manifold.

Consider a Riemannian three-manifold $(M,g)$. Let $\tilde{\nabla}$ denote the Levi-Civita connection. Consider a torsion field $\Theta$ with totally anti-symmetric flattening $\Theta^\flat$. We consider the $g$ compatible affine connection $\nabla$ with torsion $\Theta$. Since $\Theta^\flat$ is totally anti-symmetric, we know that $\Theta^\flat = \Omega \epsilon$ for some smooth function $\Omega$, and where $\epsilon$ denotes the volume three-form of $M$ induced by the metric. As a result of metric compatibility of our affine connection, it is immediate that \[ \nabla \epsilon = 0. \] This implies that \[ \nabla \Theta^\flat = d\Omega \otimes \epsilon. \] In particular, if $\Omega$ is constant, then the corresponding torsion field is parallel.

So, what does this mean? Recall that when $\Theta^\flat$ is anti-symmetric, $\tilde\nabla$ and $\nabla$ have the same geodesics. The difference between the two appear when we consider the normal bundle of a geodesic. The presence of a constant torsion means that the normal bundle will be rotated through an angle proportional to the affine length traveled along the geodesic. Hence the name, "spiral staircase".

*(edit 2019-06-05)*This argument here has been made precise in this later posting.^{^}