How can crystals be isotropic? In cubic crystals where $a=b=c$, there are rotational invariances that leave the system unchanged. If some of the electrons are responsible for many properties of solids and that they are free to move (like in a good metal), it makes no difference if one applies a current along any of the crystallographic axis, I get this part. However if one applies a current along a non crystallographic axis, I would expect a different response from the material because the distance between atoms in that direction is not the same than the distance between atoms along a crystallographic axis. However these crystals can be isotropic. How is this possible?
In my mind isotropic means that no matter the angle of the rotation of the crystal, all its properties are the same. But from a microscopic point of view I notice differences in the crystal if the rotation is arbitrary, hence I expect its properties not to be isotropic, yet they are. I know I am missing something but I don't see it.
So, in short, how can a crystal have isotropic properties?
 A: If you look at the energy diagram of a crystal you generally see a complex band structure which represents all the discrete symmetries of the crystal. However, when you look at the region near the origin (low energy limit), you'll see that the structure becomes very simple. For a cubic crystal this region is completely isotropic.
So what it means is that if the energy at which you observe the crystal is lower than a certain threshold, the crystal will behave completely isotropic.
A: Here is a counterexample: calcium fluoride and barium fluoride are birefringent despite their cubic lattices. The birefringence is only observable for short wavelengths and is very small (1e-6), but it is actually a problem for high-performance UV lenses.
Common birefringence in non-cubic lattices has two or three axes at perpendicular angles - the light propagating along one of the axes with the polarization aligned to the other axes is not affected. The birefringence in CaF$_2$ and BaF$_2$ is between the Cartesian axes of the cube unit cell and the diagonals, as you would expect.
