Commutator of Lorentz boost generators : visual interpretation I have always struggled to visualize the correctness of the commutation relation for the generators of the boost in the Lorentz group. We have $$[K_i,K_j] = i \epsilon_{ijk} L_k$$ I fail to picture this. For definiteness' sake, let's take a point $\vec{x}$ in my coordinate system, lying in the $O_{xy}$ plane. The difference between boosting first in the $x$ and then in the $y$ direction, and boosting first in the $y$ and then in the $x$ direction ammounts simply to a rotation in the $Oxy$ plane, since $$[K_1,K_2] = iL_3$$  I must be doing something wrong when I draw this on a paper. I realize that this holds at the infinitesimal level, but a simple drawing should be able to realize this principle 'at first order' shouldn't it ? My problem is that when I draw it, I end up with the same situation regardless of which direction I boost into first.
 A: Visualizing this is, in my opinion, pretty tricky because boosts generally act in four dimensions (time included!) and it's hard to draw 4D on a piece of paper.  Fortunately, for this question, you can get around this artistic limitation.  
Here's what I did right after reading your question to attempt to visualize what's going on.  I encourage you to try to draw each step.  The picture I ended up drawing is at the end.
If we order our coordinates in the standard way $(t,x,y,z)$, then boosts with boost parameter $\beta$ in the $x$ and $y$ directions are given by
$$
  \Lambda_x(\beta) = \left(
\begin{array}{cccc}
 \gamma & -\gamma\beta & 0 & 0 \\
 -\gamma\beta & \gamma & 0 & 0 \\
 0 & 0 & 1 & 0 \\
 0 & 0 & 0 & 1 \\
\end{array}
\right), \qquad
\Lambda_y(\beta)=\left(
\begin{array}{cccc}
 \gamma & 0 & -\gamma\beta & 0 \\
 0 & 1 & 0 & 0 \\
 -\gamma\beta & 0 & \gamma & 0 \\
 0 & 0 & 0 & 1 \\
\end{array}
\right)
$$
We can now attempt to apply these transformations to some point in the $x$-$y$ plane in different orders and draw the results.  The key to visualizing the math I'm about to do is to note that both of these boosts do nothing to the $z$ coordinate of any spacetime point because there is a $1$ in the lower right hand corner of each transformation matrix with zeros to its left and above.  As a result, we can ignore the $z$ axis and draw spacetime diagrams in which the time axis is vertical and the $x$-$y$ plane is horizontal.
For simplicity, let's see what happens to the point $e_x=(0,1,0,0)^t$.  Here I will use a superscript $t$ for transpose since I want to emphasize that points in spacetime will be written as column vectors.  The point $e_x$ is simply the three-dimensional unit vector along the $x$-axis at time zero.  What happens when we apply a small boost in the $y$-direction followed by a small boost in the $x$-direction?  Well, let's compute;
$$
  \Lambda_x(\beta)\Lambda_y(\beta)e_x\approx e_x+(-\beta, \beta^2/2, 0, 0)^t
$$
The $\approx$ symbol means that I have simply dropped lower order terms in the Taylor expansion in $\beta$ about $\beta = 0$.  The point is translated backward in time and forward in $x$.  Now what happens when we apply the transformations in the other order and then expand the result?
$$
  \Lambda_y(\beta)\Lambda_x(\beta)e_x\approx e_x+(-\beta, \beta^2/2, \beta^2, 0)^t
$$
The same thing happens to the time and $x$ components as in the first order, but now the point is also shifted forward in $y$ a bit.  When we subtract these two, we therefore get
$$
  \Big(\Lambda_x(\beta)\Lambda_y(\beta)-\Lambda_y(\beta)\Lambda_x(\beta)\Big)e_x \approx (0,0,-\beta^2,0)^t
$$
Does this agree with what a rotation in $z$ would do?  Well a rotation in $z$ by an angle $\phi$ looks like this:
$$
  R_z(\phi)=\left(
\begin{array}{cccc}
 1 & 0 & 0 & 0 \\
 0 & \cos  \phi  & \sin  \phi  & 0 \\
 0 & -\sin  \phi  & \cos  \phi  & 0 \\
 0 & 0 & 0 & 1 \\
\end{array}
\right)
$$
If you Taylor expand this about $\phi=0$, you'll get
$$
  R_z(\phi) \approx I + \left(
\begin{array}{cccc}
 0 & 0 & 0 & 0 \\
 0 & 0 & \phi  & 0 \\
 0 & -\phi  & 0 & 0 \\
 0 & 0 & 0 & 0 \\
\end{array}
\right)
$$
So a rotation in $z$ acts on the point $e_x$ to give
$$
  R_z(\phi) e_x \approx e_x + (0,0,-\phi,0)^t
$$
In other words, it subtracts a bit from the $y$-coordinate.  But this subtraction is precisely what happened in the case of of the two boosts acting in different orders once we identify $\phi =\beta^2$!

I know that this was all rather mathematical, but I can't see a way of getting around doing some math so as to be able to draw what happens to a point you choose under a given small boost or rotation.
