# Help with Rotations

1. Mar 14, 2006

### Oxymoron

DISCLAIMER: I would like to post a few things about what I am studying. Hopefully by writing all this I will get a better idea of it. There will probably be a few mistakes. Please feel free to comment on anything I said or add your own point of view, I'd love to hear it!

I have three vector fields in $\mathbb{R}^3$ given by

$$U=y\partial_z - z\partial_y$$
$$V=z\partial_x - x\partial_z$$
$$W=x\partial_y - y\partial_x$$

I wish to find the local one-parameter group generated (or flow if you prefer) by $U,V,W$.

Now I thought that you could do this by solving an ODE for each vector field. For example, for U, we should have a coupled ODE...

$$\frac{dy}{dt} = -z$$
$$\frac{dz}{dt} = y$$

Similarly for V, we have

$$\frac{dz}{dt} = -x$$
$$\frac{dx}{dt} = z$$

and for W we have

$$\frac{dx}{dt} = -y$$
$$\frac{dy}{dt} = x$$

So we have three coupled first order linear differential equations.

Assuming we have the following initial conditions,

$$x(0) = A, y(0) = B, and z(0) = C$$

the general solutions are (for U)

$$z(t) = C\cos t - B\sin t$$
$$y(t) = C\sin t + B\cos t$$

and it is easy to see what they are for V and W.

Now this is all easy stuff to do (assuming I got it right! :) ), it is basic ODE stuff. But I want to relate all this to the rotation group...

Last edited: Mar 14, 2006
2. Mar 14, 2006

### Oxymoron

I believe that I have found what are known as the maximal integral curve for each vector field U,V, and W.

Arbitrary points in $\mathbb{R}^3$ can be written as column vectors:

$$\left(\begin{array}{c} x\\y\\z\end{array}\right)$$

Lets take the vector field U for the moment. The maximal integral curve is given by the solution to the couple ODE, which is

$$z(t) = C\cos t - B\sin t$$
$$y(t) = C\sin t + B\cos t$$

Now this is a parametrization of a circle in the y-z axis in $\mathbb{R}^3$. In other words the integral curve of the flow generated by the vector field U is a rotation in the y-z axis. So the curve rotates the y and z points but leaves the x coordinate alone. In other words, this is a transformation and can be written as a 3x3 rotation matrix, denoted by $\bold{R}_U$,

$$[\bold{R}_U] = \left(\begin{array}{ccc} \cos t & -\sin t & 0 \\ \sin t & \cos t & 0 \\ 0 & 0 & 1 \\ \end{array}\right)$$

Last edited: Mar 14, 2006
3. Mar 14, 2006

### Oxymoron

Now, for V and W we have similar rotation matrices

$$[\bold{R}_V] = \left(\begin{array}{ccc} \cos t & 0 & -\sin t \\ 0 & 1 & 0 \\ \sin t & 0 & \cos t \end{array}\right)$$

$$[\bold{R}_W] = \left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & \cos t & -\sin t \\ 0 & \sin t & \cos t \end{array}\right)$$

4. Mar 14, 2006

### Oxymoron

What I want to investigate now is the link between the commutator (or Lie bracket) of two vector fields and the commutator of their respective flows.

I was taught that a flow is a map

$$\sigma\,:\,\mathbb{R}\times M\rightarrow M$$

generated by $X\in\mathcal{X}(M)$, which satisfies

$$\sigma(t,\sigma^{\mu}(s,x_0)) = \sigma(t+s,x_0)[/itex] which I imagine emphasizes the fact that if a particle is observed in a fluid (such as a flowing river) at a point x at t=0, then it will be found at $\sigma(t,x)$ at a late time t. Now, given my vector field [tex]U = z\partial_y - y\partial_z$$

it is easy to see that

$$\sigma(t,(x,y)) = (x\cos t - y \sin t, x \sin t + y \cos t)$$

is a flow generated by U (obviously not written as a column vector). The flow through some point $(x,y)$ is a circle with center at the origin.

Exactly the same can be said about V and W.

5. Mar 14, 2006

### Oxymoron

If we fix $t\in\mathbb{R}$ then a flow $\sigma$ is actually a diffeomorphism from M to M. But most importantly, the flow can be made into a commutative group by enforcing the following rules

$$\sigma_t(\sigma_s(x)) = \sigma_{t+s}(x)$$
$$\sigma_0$$ is the identity map
$$\sigma_{-t} = (\sigma_t)^{-1}$$

This commutative group is called the one-parameter group of transformations. It is isomorphic to $\mathfrak{SO}_2$, which is the multiplicative group of 2x2 real matrices of the form

$$\left(\begin{array}{cc} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{array}\right)$$

Now lets consider two vector fields at the same time, say U and V. U and V are vector fields on a differentiable manifold M, and they generate flows $\psi_{\sqrt{t}}$ and $\phi_{\sqrt{t}}$. Let $C_t$ be the curve through $p\in M$ defined by

$$C_t = \psi_{-\sqrt{t}} \circ\phi_{-\sqrt{t}} \circ \psi_{\sqrt{t}} \circ \phi_{\sqrt{t}}$$

which I take to mean travelling along the four sides of a parallelogram via two push-forward maps and two pullback maps.

However, it can easily happen that by traversing such curves you dont always end up exactly where you started! That is, the parallelogram may not close by taking such a curve.

My question here is, why then is $\sigma$ a curve whose initial tangent is the commutator? I.e.

$$\dot{C}_0(p) = [U,V]_p$$