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3.6 Theorem. (Levi-Civita) Given a Riemannian manifold $M$, there exists a unique affine connection $\nabla$ on $M$ satisfying the conditions:
 a) $\nabla$ is symmetric.
 b) $\nabla$ is compatible with the Riemannian metric.

 – From Riemannian Geometry, do Carmo

I understand that, if we get rid of a) above, the connection is no longer uniquely defined. But I don't know what sorts of structures we'd get.

For example, on $\Bbb R^n$, what happens when you give it a nonsymmetric connection? Are geodesics still straight lines, or are there some connections (still compatible with the metric) for which the geodesics are curved?

If geodesics are always straight lines, I imagine something weird happens with parallel transport instead. Is there any nonsymmetric, metric-compatible connection on $\Bbb R^2$, or do I need to go to dimensions $3$ and up to find them? I know parallel transport is an isometry on the tangent space, so this means that it would have to twist the tangent space somehow if it's not the Levi-Civita connection, and I think that only makes sense if you have three or more dimensions.

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    $\begingroup$ A nice question. It is a good exercise to check that geodesics will not change if you replace a L-C connection with another compatible connection. The key is that the difference between them is a differential form with values in the bundle of skew-symmetric matrices (just a random reference: users.math.msu.edu/users/parker/GA/GeometryPrimer.pdf) and skew-symmetric matrices have zeros on the diagonal. $\endgroup$ Mar 8 '18 at 1:02
  • $\begingroup$ Not a direct answer to your specific questions, but this MO page has a lot of interesting discussion of torsion. $\endgroup$ Mar 8 '18 at 2:30
  • $\begingroup$ @MoisheCohen I didn't go through the reference or attempt the exercise yet, but- Ted Shifrin's answer seems to have a counterexample, if his math checks out. (After the "EDIT") $\endgroup$ Mar 8 '18 at 5:14
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    $\begingroup$ Yes, I think Moishe's comment is mistaken: while the metric-preserving connections and the geodesic-matching connections are both uniquely determined by their torsion, they do not coincide. You can read about this in section 2.1 of this article, which was linked from the MO question I mentioned above. $\endgroup$ Mar 8 '18 at 7:49
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One of the simplest examples I can suggest is this (although it's not on Euclidean space). A compact Lie group (e.g., $SO(3)$) has a unique (up to scaling) bi-invariant metric. For the associated (symmetric) Riemannian connection the geodesics through the identity element are one-parameter subgroups, and in general they're just group translates of one-parameter subgroups. The connection is given by $\nabla_XY = -\frac12[X,Y]$ for left-invariant vector fields $X,Y$, and the curvature tensor can be computed algebraically from the Lie bracket.

However, we can easily define an invariant flat connection by setting the covariant derivative $\nabla_XY=0$ for any left-invariant vector fields $X,Y$. Then the torsion tensor $\tau$ will be given by the Lie bracket, i.e., $\tau(X,Y) = -[X,Y]$. So now all the Lie algebra is in the torsion.

You probably can't detect the difference with geodesics (since any invariant vector field is parallel along its integral curves either way), but you sure can see the difference with general parallel translation.

EDIT: Here's a connection $\nabla$ on $\Bbb R^2$ so that the parabolas $y=x^2+c$ are geodesics. Let $X_1 = \dfrac1{\sqrt{1+4x^2}}(1,2x)$ and $X_2 = (0,1)$, and set $\nabla X_1 = 0$, $\nabla X_2 = \left(\dfrac 2{\sqrt{1+4x^2}} X_1 - \dfrac{4x}{1+4x^2} X_2\right)\otimes dx$. The connection is metric-compatible and definitely not symmetric.

If we write $g_{ij} = X_i\cdot X_j$ (using the usual $\Bbb R^2$ dot product) and $\nabla X_i = \sum_j \omega_i^j\otimes X_j$, metric compatibility is given by the equations $$dg_{ij} = \sum_k g_{ik}\omega_j^k + g_{kj}\omega_i^k.$$ Here we have $\omega_1^1 = \omega_1^2 = 0$, $\omega_2^1 = \dfrac2{\sqrt{1+4x^2}}dx$, and $\omega_2^2 = -\dfrac{4x}{1+4x^2}dx$, and you can check that compatibility follows. Note that the connection is obviously not symmetric, as \begin{align*} \nabla_{X_2}X_1 &= 0, \quad \text{and yet} \\ \nabla_{X_1}X_2 &= \omega_2^1(X_1)X_1 + \omega_2^2(X_1)X_2 = \frac2{1+4x^2}X_1-\frac{4x}{(1+4x^2)^{3/2}}X_2. \end{align*} Last, since $\nabla_{X_1}{X_1} = 0$, the integral curves of $X_1$ are geodesics. These are the parabolas $y=x^2+c$.

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  • $\begingroup$ I don't understand how you got the formula for $\nabla_{X_1}X_2$ from the definition of $\nabla X_2$. What happened to the square root, and where did the 3/2 power come from? $\endgroup$ Mar 8 '18 at 18:49
  • $\begingroup$ I feed $X_1$ into the $dx$. $\endgroup$ Mar 8 '18 at 19:04
  • $\begingroup$ Oh, $X_1$ and $X_2$ aren't the unit horizontal and vertical vector fields. I see. $\endgroup$ Mar 8 '18 at 19:04
  • $\begingroup$ Well, $X_2$ is. $\endgroup$ Mar 8 '18 at 19:06
  • $\begingroup$ Right, it's engineered so that $X_1$ is the unit tangent vector for the curves we want to be geodesics. $\endgroup$ Mar 8 '18 at 19:07
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I asked the question. However, I think I have a simpler example than the currently accepted answer.

Let $X=(1,0)$ and $Y=(0,1)$ be the constant unit vector fields in the $x$ and $y$ directions. Define a connection such that: $$\begin{matrix}\nabla_XX=Y&\nabla_XY=-X\\ \nabla_YX=0&\nabla_YY=0\end{matrix}$$ Equivalently, $\Gamma_{12}^1=-1$, $\Gamma_{11}^2=1$, and all other Christoffel symbols are zero. Clearly, it is not symmetric.

Parallel transport of a vector along a curve is equivalent to rotating the vector by an amount equal to the horizontal displacement of the curve. Since rotation is an isometry of tangent spaces, this property tells us that the connection is compatible with the metric.

Since $\nabla_YY=0$, vertical lines are geodesics. The other geodesics each look like a component of the graph of $\ln(\cos(x))$, but translated some amount horizontally and vertically. (This can be derived from the aforementioned property of parallel transport.) In other words, the nonvertical geodesics look like components of the graph of $\ln(\cos(x+C_0))+C_1$ for some $C_0,C_1$.

To check this, we can take $C_0=C_1=0$ without loss of generality. We need a unit-speed parametrization of $y=\ln(\cos(x))$. It can be checked that the following is such a parametrization: \begin{align}x(t)&=\tan^{-1}(\sinh(t))\\ y(t)&=-\ln(\cosh(t))\end{align}

The equations for a geodesic, in this case, demand: \begin{align}\frac{{\rm d}^2x}{{\rm d}t^2}-\frac{{\rm d}x}{{\rm d}t}\frac{{\rm d}y}{{\rm d}t}&=0\\ \frac{{\rm d}^2y}{{\rm d}t^2}+\left(\frac{{\rm d}x}{{\rm d}t}\right)\!^2&=0\end{align} Calculation shows that $\frac{{\rm d}x}{{\rm d}t}=\operatorname{sech}(t)$ and $\frac{{\rm d}y}{{\rm d}t}=-\tanh(t)$, and the result follows.


EDIT: There's another interesting property of this connection. If we define $\nabla'$ such that ${\Gamma'}_{ij}^k=\frac12(\Gamma_{ij}+\Gamma_{ji})$, then we end up with a symmetric connection with the same geodesics. However, parallel transport changes, and it is no longer compatible with the metric. In fact, $\nabla'$ is not compatible with any metric!

To see this, note that there is no geodesic that goes between $(0,0)$ and $(\pi,0)$, or between any pair of points whose horizontal distance is more than or equal to $\pi$. If $\nabla'$, which is symmetric, were compatible with any metric, then it would be the Levi-Civita connection on the corresponding Riemannian manifold. This would contradict the Hopf–Rinow theorem, which states that, for any complete Riemannian manifold, there exists a geodesic between any two points.

(This also means that the Hopf–Rinow theorem crucially uses the symmetry of the connection, since it fails with our asymmetric $\nabla$ as well.)

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  • $\begingroup$ The geodesics look like $\bigcap$, essentially. $\endgroup$ Mar 9 '18 at 0:25
  • $\begingroup$ Notice how there's no rotation when you parallel transport around a closed loop. The same is true of Ted's connection. I don't know if there's any compatible connection on Euclidean space for which this is not true... $\endgroup$ Mar 9 '18 at 0:27
  • $\begingroup$ Does the second answer to this question give what you want? $\endgroup$ Mar 9 '18 at 6:09
  • $\begingroup$ @JesseMadnick Yeah, I didn't fully understand the significance of that example when I first saw it, partly because I didn't have as much of an intuition about $\nabla$ as I do now. So there, essentially, if $V$ and $W$ are constant vector fields, then $\nabla_VW=V\times W$. That means that the geodesics are lines, which I now see is the exception rather than the rule. Parallel transport is twisting. It does give rotation when you parallel transport around a closed loop—to see this, take an equilateral triangle of sidelength $\pi$, and transport a vector normal to the triangle around it. $\endgroup$ Mar 9 '18 at 13:15
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    $\begingroup$ If you modify your example slightly, setting $\nabla_X X = yX$ and $\nabla_X Y = -yX$ (with $\nabla_Y X = \nabla_Y Y = 0$ still holding), then you'll get constant curvature $1$ for this connection (well, up to a sign I'm too lazy to check). $\endgroup$ Mar 9 '18 at 19:37

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