Returning from a trip to New Mexico to explore some Puebloan ruins, I picked up this beautiful chunk of labradorite in the town of Quartzsite. This mineral creates an eerie blue shimmer in the sunlight: a phenomenon called ‘labradorescence’. Reading up on it, I discovered it’s a form of feldspar.
60% of the Earth’s crust is feldspar, and I know so little about this stuff! It turns out there are 3 fundamental kinds:
• orthoclase is potassium aluminosilicate
• albite is sodium aluminosilicate
• anorthite is calcium aluminosilicate
Then there are lots of feldspars that contain different amounts of potassium, sodium and calcium. We get a triangle of feldspars with orthoclase, albite and anorthite at the corners. You can find labradorite on this triangle:
But not all points in this triangle are possible kinds of feldspar! There’s a big region called the ‘miscibility gap’, where as you cool the molten mix it separates out. Apparently this is because the radius of calcium is too much bigger than that of potassium for them to get along. Sodium has an intermediate radius so it gets along with either calcium or potassium.
And there are also subtler issues. When you cool down the feldspar called labradorite, it separates out a little, forming tiny layers of two different kinds of stuff. When the thickness of these layers is the wavelength of visible light, you get a weird optical effect: labradorescence! You really need a movie to see the strange shimmer as you turn a piece of labradorite in the sunlight.
In fact there are 3 kinds of feldspar that separate out slightly as they cool and harden, forming thin alternating layers of two substances:
• The ‘peristerite gap’ produces layers in feldspars with 2-16% anorthite and the rest albite: these layers create the beauty of moonstone!
• The ‘Bøggild gap’ produces layers in feldspars with 47-58% anorthite and the rest albite: these are labradorites!
• The ‘Huttenlocher gap’ produces layers in feldspars with 67-90% anorthite and the rest albite: these are called ‘bytownites’. For some reason these layers do not seem to produce an interesting visual effect. Maybe their thickness is too far from the wavelength of visible light.
All these gaps are ‘miscibility gaps’: that is, feldspars with these concentrations of anorthite and albite are unstable: they want to separate out. That’s why they form layers.
The physics and math of all this stuff is fascinating. Crystals try to do whatever it takes to minimize free energy, which is energy minus entropy times temperature. That’s why many feldspars have different high- and low-temperature forms. But sometimes when molten rock cools quickly, it doesn’t have time to reach its free energy minimizing state.
For feldspar all of these issues are complex, because feldspar crystals are complicated structures:
Aluminum and silicon have to be distributed among the corners of the tetrahedra here, and there are various ways to do this. The distribution is determined by the relative amounts of potassium, sodium and calcium, which are the white balls. The distribution of aluminum and silicon in turn controls the symmetry of the crystal, which can be either ‘monoclinic’ or the less symmetrical ‘triclinic’.
The picture here shows the difference between monoclinic and triclinic crystals:
But the picture doesn’t fully capture the symmetry group of an actual crystal—because there’s more to a crystal than just a shape of a parallelipiped! There may be the same atoms at all corners of the parallelipiped, or not, and there may also other atoms not on the corners.
Let’s get into a bit of the math.
The symmetry group G of a crystal, called its ‘space group’, fits into a short exact sequence:
0 → T → G → P → 1
where T ≅ ℤ³ is the group of translational symmetries and P is the group of symmetries that fix a point, called the ‘point group’. This sequence may or may not split! It splits iff G is a semidirect product of P and T.
For a triclinic crystal, there are only two possible space groups G, and both are semidirect products. P is either trivial or ℤ/2, acting by negation.
For a monoclinic crystal, there are 3 choices of the point group P as a subgroup of O(3):
• P = ℤ/2 (a single 2-fold rotation)
• P = ℤ/2 (a single reflection)
• P = ℤ/2 × ℤ/2 (generated by a 2-fold rotation and inversion
(𝑥,𝑦,𝑧) ↦ -(𝑥,𝑦,𝑧): their product is a reflection).
For each choice of P there are 2 fundamentally different choices of lattice T ≅ ℤ³ it can act on. One is made up of copies of the parallelipiped I showed you. The other is twice as dense; then we call the lattice ‘base-centered monoclinic’:
So, we get 3 × 2 = 6 space groups G that are semidirect products.
But there are 7 other non-split extensions! These other 7 give nontrivial elements of the cohomology group H²(P, T). It’s not obvious that there are just 7 options. Thus, the hardest part of the classification of all 13 monoclinic space groups is essentially the computation of H²(P, ℤ³) for all 6 choices of groups P and their actions on ℤ³.
I knew that cohomology rocks. But it turns out cohomology helps classify rocks!
Now, which of these various groups are symmetry groups of feldspars?
Apparently all the feldspars in the triangle have just two different symmetry groups:
• For the monoclinic feldspars (including sanidine, orthoclase, and high-temperature albite), the crystal has a 2-fold rotational symmetry, a mirror plane, and inversion symmetry
(𝑥,𝑦,𝑧) ↦ -(𝑥,𝑦,𝑧).
The point group is the Klein four-group ℤ/2 × ℤ/2. The lattice is base-centered monoclinic, so there’s an extra translational symmetry shifting by half a cell diagonally across one face of the parallelipiped.
• For the triclinic feldspars (including microcline, low-temperature albite, and anorthite), the only symmetry beyond translation is inversion. So the point group is just ℤ/2. And there are no extra generators of translation symmetry beyond the three edges of the parallelipiped.
Alas, each of these space groups G is the semidirect product of their point group P and their translation symmetry group T ≅ ℤ³. So, no interesting cohomology classes show up!
Nontrivial cohomology classes show up only in crystals where you can’t cleanly separate the translations from the symmetries that fix a point of the crystal. This happens when your crystal has ‘screw axes’ or ‘glide planes’. A screw axis is an axis where you’ve got a symmetry of translating along that axis, but only if you also rotate around it:
A glide plane is a plane where you’ve got a symmetry of translating along that plane, but only if you also reflect across it:
But wait! There’s a rarer kind of feldspar made with barium. It’s called celsian, after Anders Celsius, the guy who invented the temperature scale. Chemically it’s barium aluminosilicate. And its crystal structure has both screw axes and glide planes! So its space group G is not a semidirect product! It’s an extension of ℤ³ by the point group P = ℤ/2 × ℤ/2 that gives a nonzero element of H²(P, ℤ³). See the end of this post for some details.
All this is lots of fun to me: you start with a pretty rock, and before long you’re doing group cohomology. But the classification of symmetry groups is just the start. For mathematical physicists, one fun thing about feldspars is their phase transitions, especially the symmetry-breaking phase transition from the more symmetrical monoclinic feldspars to the less symmetrical triclinic ones! There’s a whole body of work—by Salje, Carpenter, and others—applying Landau’s theory of symmetry-breaking phase transitions to map out the space of different possible feldspar crystals! Here’s one way to get started:
• Ekhard Salje, Application of Landau theory for the analysis of phase transitions in minerals, Physics Reports 215 (1992), 49–99.
Even if you don’t particularly care about feldspars, there are a lot of good general principles of physics to learn here!
Details
Let me sketch out why barium aluminosilicate, or celsian, has a space group G that’s described by a non-split short exact sequence:
0 → T → G → P → 1
Its point group is P = {e, r, m, i} ≅ ℤ/2 × ℤ/2, where we can take r to be a 180° rotation about the y axis and m to be a reflection that negates the y coordinate, so that i = rm is inversion. In coordinates:
r acts as (x, y, z) ↦ (−x, y, −z)
m acts as (x, y, z) ↦ (x, −y, z)
i acts as (x, y, z) ↦ (−x, −y, −z)
We can take the translation lattice T ≅ ℤ³ to be the lattice generated by
f₁ = (1,0,0), f₂ = (0,1,0), f₃ = (½,½,½)
Note that (0,0,½) is not in T.
To compute the 2-cocycle we need a set-theoretic section s: P → G. We choose
s(e) = identity
s(m) = a glide reflection: (x, y, z) → (x, −y, z + ½)
s(i) = inversion: (x, y, z) → (−x, −y, −z)
s(r) = s(i)·s(m): (x, y, z) → (−x, y, −z + ½)
As usual, the 2-cocycle c: P2 → G is defined by
c(g,h) = s(g)·s(h)·s(gh)⁻¹
The interesting value is c(m, m): the glide composed with itself gives (x, y, z) → (x, −y, z+½) → (x, y, z+1), so s(m)² = translation by (0, 0, 1), while s(m²) = s(e) is the identity. Thus c(m, m) = (0, 0, 1). The other values are trivial: c(i, i) = 0, c(r, r) = 0.
Now, is this cocycle nontrivial in H²(P, T)? It would be trivial if we could find a different section that makes the cocycle zero—that is, find a function b: P → T such that replacing s(g) with s'(g) = s(g) + b(g) makes
c'(g,h) = s'(g)·s'(h)·s'(gh)⁻¹
be the identity for all g,h. I will spare you the calculation proving this is impossible. The idea is simply this: the reflection m squares to the identity in the point group, but no matter how we choose b, s'(m) is a glide reflection, so it squares to a nontrivial translation. On the other hand, s'(m2) is trivial since m2 is, so
c'(m,m) = s'(m)·s'(m)·s'(m2)⁻¹
is nontrivial.
