Nelson's car and Lie bracket

Question

I just read Stillwell's Naive Lie Theory, had a very basic understanding about Lie bracket, then I ran into Prof Edward Nelson's book Tensor Analysis, where from 32 to 36, it discussed how to drive a car with the help of Lie bracket.

First, on page 32, it introduced a formula

$$e^{t[x,y]} = e^{\sqrt{t}x} e^{\sqrt{t}y} e^{-\sqrt{t}x} e^{-\sqrt{t}y} + o(t)$$

Then started on a car model:

Consider a car. The configuration space of a car is the four dimensional manifold parameterized by $(x,y,\phi, \theta)$, where $(x,y)$ are the Cartesian coordinates of the center of the front axle, the angle $\phi$ measures the direction in which the car is headed, and $\theta$ is the angle made by the ront wheels with the car. (More realistically, the configuration space is the open submanifold $-\theta_{max} < \theta < \theta_{max}$.)

There are two distinguished vector fields, called Steer and Drive, on $M$ corresponding to the two ways in which we can change the configuration of a car. Clearly (15) $$\operatorname{Steer} = \frac{\partial}{\partial \theta}$$

....

Then (16)

$$\operatorname{Drive} = \cos (\phi+\theta) \frac{\partial}{\partial x} + \sin (\phi + \theta) \frac{\partial}{\partial y} + \sin \theta \frac{\partial}{\partial \phi}$$

So far so good, then it said,

By (15) and (16)

$$[\operatorname{Steer}, \operatorname{Drive}] = -\sin(\phi+\theta) \frac{\partial}{\partial x} + \cos (\phi+ \theta) \frac{\partial}{\partial y} + \cos \theta \frac{\partial}{\partial \phi}$$ named (17)

-- I'm lost here, how could I arrive to this result?

The first thought is this is just a calculation of the partial differentiation according to the chain rule:

$$[\operatorname{Steer}, \operatorname{Drive}] = \operatorname{Steer} \operatorname{Drive} - \operatorname{Drive} \operatorname{Steer} $$ $$ = \frac{\partial}{\partial \theta} \left[\cos (\phi+\theta) \frac{\partial}{\partial x} + \sin (\phi + \theta) \frac{\partial}{\partial y} + \sin \theta \frac{\partial}{\partial \phi}\right] - \left[\cos (\phi+\theta) \frac{\partial}{\partial x} + \sin (\phi + \theta) \frac{\partial}{\partial y} + \sin \theta \frac{\partial}{\partial \phi}\right] \frac{\partial}{\partial \theta} $$ $$ = -\sin(\phi+\theta) \frac{\partial}{\partial x} + \cos (\phi+ \theta) \frac{\partial}{\partial y} + \cos \theta \frac{\partial}{\partial \phi}$$

But... how could I map these to what I read from Naive Lie Theory?

As I understand, the essential idea of Lie theory is that given a curve in Lie group space $A(t)$, given $A(0) = \mathbf{1}$ where $\mathbf{1}$ is the unit matrix, its corresponding Lie algebra element would be the derivative $\left.\frac{d A(t)}{dt} \right|_{t=0}$ , while at the same time for any element $A$ in the Lie group there's a corresponding Lie algebra $X$ such that $A = e^X$.

I tried to map as

• the configuration space is $(x,y,\phi, \theta)$

• the Lie group is the transforming matrix $A(t)$ at time $t$ so that $$ \begin{pmatrix} x\\ y\\ \phi\\ \theta \end{pmatrix} = A(t) \begin{pmatrix} x_0\\ y_0\\ \phi_0\\ \theta_0 \end{pmatrix} $$

• the change introduced by Steer and Drive, are the vectors in the Lie algebra, namely

$$\operatorname{Steer} = \begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}, \quad \operatorname{Drive} = \begin{pmatrix} \cos (\phi + \theta) & 0 & 0 & 0\\ 0 & \sin (\phi + \theta) & 0 & 0\\ 0 & 0 & 0 & \sin \theta\\ 0 & 0 & 0 & 0 \end{pmatrix} $$

• The Steer formula makes sense: if the driver only turns the steering wheel, after time $t$ the status is $$ \begin{pmatrix} x\\ y\\ \phi\\ \theta \end{pmatrix} = \begin{pmatrix} x_0\\ y_0\\ \phi_0\\ \theta_0 + \dot{\theta} t \end{pmatrix} = A(t) \begin{pmatrix} x_0\\ y_0\\ \phi_0\\ \theta_0 \end{pmatrix} $$ so $$ A(t) = \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 + \frac{\dot{\theta}}{\theta} t \end{pmatrix}, \quad \frac{d A(t)}{dt} = \frac{\dot{\theta}}{\theta} \begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix} $$ Normalize it, we can say $$\operatorname{Steer} = \begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}$$ Now if we denote $\begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}$ as $\frac{\partial}{\partial \theta}$, then we can write as $$\operatorname{Steer} = \frac{\partial}{\partial \theta}$$ So here I'm understanding $\frac{\partial}{\partial \theta}$ as a notation of a unit in the Lie algebra, it's not a real partial differential that can calculate

• Similarly, when the driver only drives at speed $v$, denote the distance between the front and rear wheels as $L$, after time $t$, $$ \begin{pmatrix} x\\ y\\ \phi\\ \theta \end{pmatrix} = \begin{pmatrix} x_0 + vt \cos(\phi + \theta) \\ y_0 + vt \sin(\phi + \theta) \\ \phi_0 + \frac{vt}{L} \sin(\theta) \\ \theta_0 \end{pmatrix} = A(t) \begin{pmatrix} x_0\\ y_0\\ \phi_0\\ \theta_0 \end{pmatrix} $$ so $$ A(t) = \begin{pmatrix} 1 + \frac{vt}{x_0} \cos(\phi + \theta) & 0 & 0 & 0\\ 0 & 1 + \frac{vt}{y_0} \sin(\phi + \theta) & 0 & 0\\ 0 & 0 & 1 + \frac{vt}{L\phi_0} \sin(\theta) & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}, \quad \frac{d A(t)}{dt} = \begin{pmatrix} \frac{v}{x_0} \cos(\phi + \theta) & 0 & 0 & 0\\ 0 & \frac{v}{y_0} \sin(\phi + \theta) & 0 & 0\\ 0 & 0 & \frac{v}{L\phi_0} \sin(\theta) & 0\\ 0 & 0 & 0 & 0 \end{pmatrix} $$ Normalize it, we can say $$\operatorname{Drive} = \begin{pmatrix} \cos(\phi + \theta) & 0 & 0 & 0\\ 0 & \sin(\phi + \theta) & 0 & 0\\ 0 & 0 & \sin(\theta) & 0\\ 0 & 0 & 0 & 0 \end{pmatrix}$$ Now if we denote $$\frac{\partial}{\partial x}:=\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix}, \quad \frac{\partial}{\partial y}:=\begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix}, \quad \frac{\partial}{\partial \phi}:=\begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix}$$ then we can write as $$\operatorname{Drive} = \cos(\phi + \theta) \frac{\partial}{\partial x} + \sin(\phi + \theta) \frac{\partial}{\partial y} + \sin(\theta) \frac{\partial}{\partial \phi} $$ Again here I'm understanding $\frac{\partial}{\partial x}$, $\frac{\partial}{\partial y}$, and $\frac{\partial}{\partial \phi}$ as notations of unit vectors in the Lie algebra, it's not a real partial differential that can calculate

But then I'm stuck.

First I tried to compute $[\operatorname{Steer}, \operatorname{Drive}]$ via matrix commuter, but I got $$[\operatorname{Steer}, \operatorname{Drive}] = \operatorname{Steer} \operatorname{Drive} - \operatorname{Drive} \operatorname{Steer} = \begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \cos (\phi + \theta) & 0 & 0 & 0\\ 0 & \sin (\phi + \theta) & 0 & 0\\ 0 & 0 & \sin \theta & 0\\ 0 & 0 & 0 & 0 \end{pmatrix} - \begin{pmatrix} \cos (\phi + \theta) & 0 & 0 & 0\\ 0 & \sin (\phi + \theta) & 0 & 0\\ 0 & 0 & \sin \theta & 0\\ 0 & 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix} $$ This doesn't tally with (17) $$[\operatorname{Steer}, \operatorname{Drive}] = -\sin(\phi+\theta) \frac{\partial}{\partial x} + \cos (\phi+ \theta) \frac{\partial}{\partial y} + \cos \theta \frac{\partial}{\partial \phi}$$

Then I thought I have to use the formula $$e^{t[x,y]} = e^{\sqrt{t}x} e^{\sqrt{t}y} e^{-\sqrt{t}x} e^{-\sqrt{t}y} + o(t)$$ So I got

$$e^{\sqrt{t}\operatorname{Steer}} = \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & e^{\sqrt{t}} \end{pmatrix}, \quad e^{-\sqrt{t}\operatorname{Steer}} = \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & e^{-\sqrt{t}} \end{pmatrix}, \quad $$ $$ e^{\sqrt{t}\operatorname{Drive}} = \begin{pmatrix} e^{\sqrt{t} \cos (\phi+ \theta)} & 0 & 0 & 0\\ 0 & e^{\sqrt{t} \sin (\phi+ \theta)} & 0 & 0\\ 0 & 0 & e^{\sqrt{t} \sin \theta} & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}, \quad e^{-\sqrt{t}\operatorname{Drive}} = \begin{pmatrix} e^{-\sqrt{t} \cos (\phi+ \theta)} & 0 & 0 & 0\\ 0 & e^{-\sqrt{t} \sin (\phi+ \theta)} & 0 & 0\\ 0 & 0 & e^{-\sqrt{t} \sin \theta} & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} $$ Multiply them together I still got

$$e^{\sqrt{t}\operatorname{Steer}} e^{\sqrt{t}\operatorname{Drive}} e^{-\sqrt{t}\operatorname{-Steer}} e^{-\sqrt{t}\operatorname{-Drive}} = \mathbf{1} $$ Then $$[\operatorname{Steer}, \operatorname{Drive}] = \frac{d}{dt}\left. e^{\sqrt{t}\operatorname{Steer}} e^{\sqrt{t}\operatorname{Drive}} e^{-\sqrt{t}\operatorname{-Steer}} e^{-\sqrt{t}\operatorname{-Drive}} \right|_{t=0} = \mathbf{0} $$ Also wouldn't tally with (17) $$[\operatorname{Steer}, \operatorname{Drive}] = -\sin(\phi+\theta) \frac{\partial}{\partial x} + \cos (\phi+ \theta) \frac{\partial}{\partial y} + \cos \theta \frac{\partial}{\partial \phi}$$

Pls enlighten me how the (17) formula shall be interpreted in the Lie algebra?

Answer 1

There is some confusion going on when you represent elements of your Lie algebra as matrices. If we work in the basis $\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial \phi}, \frac{\partial}{\partial \theta}$, we get:

$\begin{align} \text{Steer} &= \frac{\partial}{\partial \theta} \\ \text{Drive} &= \cos(\phi + \theta) \frac{\partial}{\partial x} + \sin(\phi + \theta) \frac{\partial}{\partial y} + \sin(\theta)\frac{\partial}{\partial \phi} \\ \vdots \end{align}$

So, in this basis $\text{Steer}, \text{Drive}, \text{Wriggle}, \text{Slide}$ should not be written as diagonal matrices, but rather vectors with components given above:

$\begin{align*} \text{Steer} &= \begin{pmatrix} 0\\ 0\\ 0\\ 1\end{pmatrix}, \quad \text{Drive} = \begin{pmatrix} \cos(\phi+\theta)\\ \sin(\phi+\theta)\\ \sin(\theta)\\ 0\end{pmatrix} \end{align*}, \ldots$

From the book on Naive Lie Theory you may already know how to differentiate a Lie group into a Lie algebra, by taking the tangent space at the identity. It is also possible to integrate a Lie algebra into a Lie group: we can take any Lie algebra (like the one spanned by $\text{Steer}, \text{Drive}, \text{Wriggle}, \text{Slide}$) and find a (local) Lie group whose Lie algebra is the same as the one we started with.

One way to do this is via Ado's theorem, which involves representing the vectors in our Lie algebra as matrices. But we cannot do this in the same way you did; we must make sure that the commutator is preserved! If we represent elements of our Lie algebra as diagonal matrices like you did, then all the commutators become 0, since diagonal matrices commute. So we have represented the trivial Lie algebra, and all the calculations work out to 0.

What we have to do is find matrices $\text{Steer}, \text{Drive}, \text{Wriggle}, \text{Slide}$ satisfying (in what follows, the bracket is now a matrix commutator, and not the composition of derivation operators):

$\begin{align} [\text{Steer}, \text{Drive}] &= \text{Wriggle}\\ [\text{Steer}, \text{Wriggle}] &= -\text{Drive}\\ [\text{Wriggle}, \text{Drive}] &= \text{Slide}\\ \text{all other commutators} &= 0. \end{align}$

How do we do this? It's not so easy in general. The proof of Ado's theorem gives a general construction. I can post how the steps work, if you're interested. But the bottom line is that you can't represent elements in your Lie algebra as diagonal matrices corresponding to the vectors and then hope that the commutator is non-zero (it will always be zero): you have to choose your matrices carefully to make the commutator non-zero.

Answer 2

Short answer: You are confusing specialized techniques that are specific to the narrow context of Lie groups/Lie algebras with more general commutator formulas that apply on arbitrary manifolds.

Nelson's book starts with a general discussion of commutators of smooth vector fields on general manifolds (which are not necessarily Lie groups). The commuter of two smooth vector fields, also known as their Lie bracket, can be defined in this very general context as follows.

Definition. In local coordinates if the vector field $A$ is written as a first-order differential operator $A=\sum_j a^j(x) {\partial x_j}$ and similarly $B=\sum_k b^k(x) {\partial x_k}$ then their Lie bracket is another vector field given by the operator commutator formula $[A,B]= \sum_{j,k} (a_j b^k_j \partial x_k - b_k a^j_k \partial x_j)$.

Here e.g $a^k_j= \frac{\partial a_k}{\partial x_j}$. As you noted in your post, this method for computing the commutator gives the result mentioned in the book.

P.S. Later in Nelson's discussion he points out the remarkable fact that the fields $A=$ Steer and $B=$ Drive can be augmented with two other vector fields to obtain a four- dimensional Lie algebra. That is, the commutators of these four fields with one another can be expressed as linear combinations of one another that have constant coefficients. (The property that these coefficients are constant is the distinctively "rigid" property of a true finite-dimensional Lie algebra).

P.P.S. In the typical case of arbitrary smooth vector fields on manifolds one would expect the commutators of four arbitrary linearly-independent basis fields to be expressible as linear combinations of these basis fields, but the expansion of each commutator in terms of this basis would have variable coefficients. This is a much less rigid property than the former case of true Lie algebra.)

Answer 3

In order to define the concept of vector fields in abstract, non embedded manifolds, one has to take a more abstract viewpoint, which is indipendent from the coordinates (because coordinates are defined only locally); one way is to think tangent vector as derivations.

For example, if you want to think the vector field $\mu:\mathbb{R}^n \rightarrow \mathbb{R}^n, x = (x_1,\dots,x_n)\mapsto v_x= (v_1(x_1,\dots, x_n),\dots,v_n(x_1,\dots,x_n))$ without using coordinates, you can consider the map

$$ C^\infty(\mathbb{R}^n) \rightarrow C^\infty(\mathbb{R}^n), f(x)\mapsto \frac{\partial f}{\partial v_x}(x), $$ in this way, you look at the effect that $\mu$ has on functions, rather than at its coordinates.

Observe that the map above, as any derivative, is linear and verifies the Liebnitz rule. So, in an abstract manifold $M$, you define vector fields as derivations of $C^\infty$ functions: $$V(M) :=\{\mu: C^\infty(M) \rightarrow C^\infty(M) : \mu \ is \ linear \ and \ \mu(fg) = \mu(f)g +f\mu(g)\}.$$

Now $V(M)$ has a natural structure of Lie algebra, where the bracket of $\mu$ and $\eta$ is defined by $$[\mu,\eta](f) := \mu(\eta(f))-\eta(\mu(f)).$$

If you have local coordinates $(x_1,\dots,x_n)$, then you can prove that there is a unique form $\mu = \sum a_i \frac{\partial}{\partial x_i}$, with $a_i$ are $C^\infty$ where defined, so you can apply chain rule, as you did in your calculation.