How to define orbital angular momentum in other than three dimensions? In classical mechanics with 3 space dimensions the orbital angular momentum is defined as
$$\mathbf{L} = \mathbf{r} \times \mathbf{p}.$$
In relativistic mechanics we have the 4-vectors $x^{\mu}$ and $p^{\mu}$, but the cross product in only defined for 3 dimensions. So how to define orbital angular momentum e.g. in special relativity in terms of 4-vectors? Or more generally in $d$ dimensions?
 A: Dear asmaier, you shouldn't view $\vec L = \vec x \times \vec p$ as a primary "definition" of the quantity but rather as a nontrivial result of a calculation.
The angular momentum is defined as the quantity that is conserved because of the rotational symmetry - and this definition is completely general, whether the physical laws are quantum, relativistic, both, or nothing, and whether or not they're mechanics or field theory.
To derive a conserved charge, one may follow the Noether's procedure that holds for any pairs of a symmetry and a conservation law:

http://en.wikipedia.org/wiki/Noether_charge

In particular, the angular momentum has no problem to be evaluated in relativity - when the background is rotationally symmetric. The fact that you write $\vec L$ as a vector is just a bookkeeping device to remember the three components. More naturally, even outside relativity, you should imagine
$$ L_{ij} = x_i p_j - x_j p_i $$
i.e. $L_{ij}$ is an antisymmetric tensor with two indices. Such a tensor, or 2-form, may be mapped to a 3-vector via $L_{ij} = \epsilon_{ijk} L_k$ but it doesn't have to be. And in relativity, it shouldn't. So in relativity, one may derive the angular momentum $L_{\mu\nu}$ which contains the 3 usual components $yz,zx,xy$ (known as $x,y,z$ components of $\vec L$) as well as 3 extra components $tx,ty,tz$ associated with the Lorentz boosts that know something about the conservation of the velocity of the center-of-mass.
Incidentally, the general $x\times p$ Ansatz doesn't get any additional "gamma" or other corrections at high velocities. It's because you may imagine that it's the generator of rotations, and rotations are translations (generated by $\vec p$) that linearly depend on the position $x$. So the formula remains essentially unchanged. In typical curved backgrounds which still preserve the angular momentum, the other non-spatial components of the relativistic angular momentum tensor are usually not preserved because the background can't be Lorentz-boost-symmetric at the same moment.
