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I have learned that there are 14 distinct Bravais lattices in 3D and any other thought lattice form could be reduced to or expressed in one of these 14 forms. But the primitive unit cell for f.c.c lattice is seen to be a special case of trigonal (rhombohedral) lattice (with angles equal to 60 deg). A similar case is true for b.c.c lattice. So, is f.c.c really distinct while it could be expressed as a special case of trigonal lattice? (I also asked this in mathematics stack.)

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The 14 Bravais lattices are well distinct. Be sure of it. Historically though, a scientist (forgot his name) found out that there were 15 lattices, then Bravais came and proved that two of them were identical, and they were named after him.

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  • $\begingroup$ But how would you refute the claim that FCC lattice is a special case of trigonal lattice? $\endgroup$
    – Ruslan
    Jan 20 '15 at 12:58
  • $\begingroup$ The trigonal lattice does not have the cubic symmetry, so the fcc lattice cannot be reduced to a trigonal one. $\endgroup$
    – mdib
    Jan 20 '15 at 13:20
  • $\begingroup$ In the special case of $\alpha=\beta=\gamma=60^\circ$ it does have the same primitive vectors as FCC lattice, thus it coincides with it. How can their symmetries be different? $\endgroup$
    – Ruslan
    Jan 20 '15 at 13:26
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    $\begingroup$ Because that 'special case' introduces new symmetries into the lattice that are not present in the trigonal case. That is the whole point - special cases are special for a mathematical reason. $\endgroup$
    – Jon Custer
    Jan 20 '15 at 14:47
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    $\begingroup$ Although it is a "special case", at the end it is a special case of "trigonal"; so it is a trigonal. How could FCC be distinct regarding this. I know FCC is a distinct form, because it has been proven mathematically by Bravais. I want an explanation apart from complicated math! $\endgroup$
    – M-J
    Jan 21 '15 at 6:17

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