Why don't the nuclei have an electric dipole moment? Until yesterday I thought they have. Googling a little bit around the measured values, I've found nothing.
As I've heard on the chat, the nuclei don't have an EDM. Although it would be possible by the symmetry violations of the weak interaction, it doesn't happen.
But I am asking not for the EDM due to weak asymmetry. I am asking for the lack of the EDM due to the asymmetric charge distribution in them.
Nuclei have a lot of elementary particles, interacting in quite complex ways (strong, EM and weakly). Why should the charges be distributed in them always to a 0 EDM? What is the mechanism, or what is the reason, to zero out any EDM in them?
 A: The expectation value of the dipole moment operator for states of definite parity is zero. This is because the dipole moment operator is odd operator.
A: The mystery is really why molecules can have electric dipole moments.
The strong nuclear force and the electromagnetic interaction both conserve energy, angular momentum, and parity, and that's why we usually label states of nuclei with those quantum numbers, e.g., the $2^-$ ground state of a particular odd-odd nucleus. If these quantum numbers are sufficient to identify that particular state, then it follows from parity symmetry that the electric dipole moment must vanish. Under parity, the angular momentum stays the same, but an electric dipole moment would flip. There would then be nothing to determine the direction of the electric dipole moment if, by assumption, the angular momentum is the only vector that could determine it.
In molecular dipoles, you have a nearly degenerate ground state. If there were no interaction between the two states, then you would have one state oriented to the right and one to the left. There is a nice analysis of the ammonia molecule, for example, in the Feynman Lectures. I have a similar analysis in section 14.7.2 of my book Simple Nature, which is free online: http://www.lightandmatter.com/area1sn.html Without interaction, the degeneracy violates the assumption that energy, spin, and parity are sufficient to label a state. When you add in a small interaction, which basically represents tunneling between the two states, the two eigenstates of the Hamiltonian are eigenstates of parity, but they are not states that have a definite orientation. If you put the molecule initially in a state of definite orientation, then it will have a dipole moment, and will not be a state of definite parity.
The reason that we don't get an exactly analogous situation in nuclear physics is the following. Suppose that we consider the possibility of deforming a nucleus into a pear shape, parametrized by a variable $\beta_3$. If you consider the potential energy as a function of $\beta_3$, then for almost all nuclei it looks like a parabola with a minimum at $\beta_3=0$. There is evidence that for a few nuclei, there are minima at $\beta_3=\pm c\ne 0$, so that the potential is shaped like a "W". However, the barrier between the two minima is very shallow, so that the Heisenberg uncertainty principle causes fluctuations in shape that are too big to be confined on one side or the other. This is different from the case of a molecule, for which the barrier is usually very tall, and the tunneling probability through it very small.
