Today I’m going to talk about parallel parking. We’ve all had to do it at some point (at least, those who drive) and certainly we’ve all noticed how much of a pain it is to get into a small space. Well, as it happens, if your car has length , then for any , it is possible to parallel park, assuming some things like that the driver can make arbitrarily small movements.

So what does this rely upon? Well, we’re going to have to talk about Lie Groups to do this. A Lie Group is a group that is also a manifold, and such that the multiplication and inversion maps are smooth. Lie groups turn out to be incredibly nice geometric objects, as well as a nice group. However, the particularly nice properties of Lie groups aren’t the big feature here.

The important feature here is that if you take the tangent space to the origin of the Lie group’s manifold, you get a natural Lie algebra under commutator of vector fields. It turns out that for “nice” Lie groups, there is a very precise correspondence between the Lie algebra and the Lie group. What this means is that we can take problems about a Lie group and solve them in the Lie algebra. We then get back to the group through the exponential map, where is the Lie algebra of .

Now let’s look at the parallel parking problem, and add more background merely as needed. We first need to translate the car into mathematics. We do this by first assuming that the car is of unit length (for simplicity) and taking to be the point on the center of the rear axle, and to be the point on the center of the front axle. Define to be the coordinates of , to be the angle that the car is facing measured from the horizontal, and to be the angle between the front axle and the line , counterclockwise.

This makes the configuration space of the car into (we can even restrict to the open submanifold given by bounding how much the car can turn, that is .

Now, there are some important vector fields on this manifold (which is also a Lie group, and so we can just work with the vector fields and use the Lie algebra to park). The first one is called Steer, and all it does is change the direction of the front wheels infinitesimally. It can be expressed as . We also want to know what Drive looks like, the vector field which moves the car forward infinitesimally.

So this turns out to be a nasty computation which can be done with the aid of a diagram, a name for the infinitesimal distance, and a LOT of care, but the up side is that Drive . And now we have commutators to compute! . We shall call this Wriggle, following Nelson (this is called Nelson’s Parking Problem, from his 1967 book Tensor Analysis). Wriggle is just what happens if you drive, steer, reverse, and then return to your original direction.

So what we really want to do when we parallel park is to slide sideways. This vector field is given by , so the question is, how can we get this vector field from Steer and Drive? Well, let’s look at the commutator of Wriggle and Drive. Out pops slide, so that in an arbitrarily tight parking spot, we merely have to wriggle, drive, reverse wriggle and reverse drive. Remembering that wriggle is steer, drive, reverse steer, reverse drive, and that reverse wriggle would be reverse steer, reverse drive, steer, drive, this becomes Steer, Drive, Reverse Steer, Reverse Drive, Drive, Reverse Steer, Reverse Drive, Steer, Drive, Reverse Drive, which simplifies to Steer, Drive, Reverse Steer, Reverse Steer, Reverse Drive, Steer.

Now that we’ve shown that you CAN parallel park, I don’t recommend trying to follow this procedure in a tight space, simply because parking in tight spaces is tough, and best avoided. I’ll just continue using public transportation, myself…