Harrison-construction in Solid-State Physics I have a question concerning the Fermi-surfaces of a square primitive cell with 2 and 4 valence electrons.
Consider these two plots, where the upper plot is for 2 valence electrons and the lower for 4:


The way I understand it is that in the upper plot, the white area in the middle circle is for 1st zone electrons. However, why does this area for the lower plot (which is smaller and has another geometric form in the lower plot) correspond to 2nd zone holes?
 A: I am not an expert in solid-state physics. As far as understand, you call some region in the $k$ space a hole, in case it is unfilled in the lowest unfilled level. And the electrons correspond to the filled domains, in the zones above the lowest unfilled.
In the first case, the square is not completely filled - so the blue domain corresponds to 1st zone holes, but there are some domains, already filled twice - there are electrons of the 2nd zone.
On the second picture, the first level is completely filled, but the second is not yet, and the orange corresponds thus to the 2nd zone holes. The domain, not painted with any colored, corresponds to filled 2nd zone, but empty 3rd.
A: A picture might also help.  Here is a plot of the band structure running from the center of the zone to the zone boundaries of the first zone.  Since this plot is in the reduced zone scheme, all bands are in this zone.
If we imagine the red line to be the fermi level, we see that the second band has holes in the center of the zone.  The filled levels are to either side.

This image comes from here.
