The Math Thread

[lpetrich]

Key to this finite-number theorem is the "crystallographic restriction theorem". A rotation group for two lattice dimensions can only do rotations by multiples of 2*pi/n, where n = 1, 2, 3, 4, or 6. This explains why  Quasicrystals seemed like such an oddity at first. They seemed impossible at first, but it was eventually shown that those are amorphous solids with short-range order that violates the crystallographic restriction theorem, like having fivefold symmetry or icosahedral symmetry.

-

Let's first consider 1D. There are two groups, lattice-only and lattice + reflections:
(1,k) for k in Z
(1,k) and (-1,k+a) for k in Z and some a

We can set a = 0 to put the reflection planes onto the lattice, and also ignore the lattice: (1,0) and (-1,0)

For a 1D lattice and two overall dimensions, one gets the "frieze groups". Their elements are (rr along the lattice, rr perpendicular to the lattice, offset relative to the lattice). The elements (1,1,0) and (-1,1,0) are similar to their pure 1D counterparts, but let us look at (1,-1,a). Multiplying two of them together gives (1,1,2a), meaning a = 0 or a = 1/2. Thus the possible group elements are
(1,1,0), (-1,1,0), (1,-1,0) or (1,-1,1/2), (-1,-1,0) or (-1,-1,1/2).

There are thus 7 possible frieze groups:

1. p1: (1,1,0)
2. p11m: (1,1,0), (1,-1,0)
3. p11g: (1,1,0), (1,-1,1/2)
4. p1m1: (1,1,0), (-1,1,0)
5. p2: (1,1,0), (-1,-1,0)
6. p2mm: (1,1,0), (1,-1,0), (-1,1,0), (-1,-1,0)
7. p2mg: (1,1,0), (1,-1,1/2), (-1,1,0), (-1,-1,1/2)

m = mirror, g = glide (mirror + shift). There are various other notations for these groups.

In ASCII form:

p1:
b b b b b b
p11m:
b b b b b b
p p p p p p
p11g:
b p b p b p
p1m1:
b d b d b d
p2:
b q b q b q
p2mm:
b d b d b d
p q p q p q
p2mg:
b d b d b d
q p q p q p

These groups are all related to the axial 3D point groups. That is because those groups can be understood as the result of wrapping a frieze group around a cylinder. Those groups are finite, based on a Z lattice rather than a Z lattice, but they are otherwise constructed the same.

 

[lpetrich]

For 2D rotoreflections in a 2D lattice, one gets 17 "wallpaper groups".

These have five lattice types: oblique (general), rectangular, rhombic (centered rectangular), square, and hexagonal.

For 3D rotoreflections in a 1D lattice, one gets 13 infinite families of 3D "line groups". This is because of a lack of lattice restrictions on two of the dimensions. These infinite families can be understood as wallpaper groups wrapped in cylinders. Imposing lattice restrictions on those two dimensions gives the 75 "rod groups".

For 3D rotoreflections in a 2D lattice, one gets 80 "layer groups".

For 3D rotoreflections in a 3D lattice, one gets 230 "space groups", or 219 if one counts the 11 pairs of mirror-image ones together.

-

All these groups are subgroups of some continuous groups, the "Euclidean groups", E or Euc for n dimensions. These groups are constructed like the lattice space groups, but their translations are continuous -- Rⁿ (R = real numbers under addition) -- and their rotoreflections are O.

Also like the lattice groups, the translations Rⁿ form a normal subgroup of Euc, with rotoreflections O as their quotient group.

If the space has a metric with a signature with opposite signs, n1 +'s and n2 -'s, then O gets replaced by O(n1,n2), making Euc(n1,n2). The Poincaré group of space-time symmetries is Euc(3,1).

That is for flat space-time, and curved space-time usually has less overall symmetry. However, there are some cosmological solutions with equal amounts of overall symmetry, the De Sitter and Anti De Sitter ones. These have a cosmological constant but no other sources. The cosmological constant can be interpreted as a constant density ρ and a pressure - ρ*c² where the negative sign is a feature and not a bug. In the De Sitter case, ρ > 0, while in the Anti De Sitter case, ρ < 0. Here are their symmetry groups:

De Sitter: O(4,1)
Anti De Sitter: O(3,2)

An additional kind of symmetry is "conformal symmetry", symmetry under stretching and squeezing. For space-time, the maximum possible conformal symmetry is SO(4,2).

 

[lpetrich]

These continuous groups seem very horrible to work with, but there is a simplification that was discovered by mathematician Sophus Lie and some others.

A continuous group may have disconnected parts, parts that are internally continuous. The rotoreflection groups do, with pure rotations being disconnected from reflections. The pure rotations are a normal subgroup, with {rotations, reflections} as its quotient group: Z2. The Lorentz group has four parts (no reversals), (space reversed), (time reversed), (both space and time reversed). The quotient group with respect to normal subgroup (no reversals) is Z2 (space reversal) * Z2 (time reversal).

The part continuous with the identity I will call the identity part. The parts are given by (all a part's elements) = (one of them) * (all the identity part's elements).

This leaves us with the task of understanding the identity parts. Fortunately, there is an elegant theory of these parts. It's the theory of Lie ("Lee") groups, named after mathematician Sophus Lie. These groups can be constructed from elements arbitrarily close to the identity element, elements that form a "Lie algebra". In effect,
D = I + e*L + O(e²)

where D is the group element, e is arbitrarily small, and L is the algebra element.

This gives rise to a general expression for Lie-group elements:
D(a) = exp(sum over i of a_i*L_i)

The L's are closed under commutation, and their commutators' values define the algebra:
[L_i,L_j] = L_i.L_j - L_j.L_i = f_ij^kL_k
summed over k, where the f values are the "structure constants".

The commutators are antisymmetric, of course: [L_j,L_i] = - [L_i,L_j] and there is a relation that is easily proved called the Jacobi identity:

[[L_i,L_j], L_k] + [[L_j,L_k], L_i] + [[L_k,L_i], L_j] = 0

These relations put constraints on the possible structure-constant values: f_ji^k = - f_ij^k and f_ij^a*f_ka^l + f_jk^a*f_ia^l + f_ki^a*f_ja^l = 0

Using the algebra is often much easier than using the group to find what one wants to know. Things like subgroups and representations and combinations of them. For representations, one usually finds only their basis spaces and not their matrices, but their basis spaces usually have what one wants to know.

Lie algebras are very important for particle physics. They appear in space-time symmetries, in gauge symmetries, and in approximate flavor symmetries.

 

[lpetrich]

For rotations, the generators of their Lie aigebra can be expressed in a simple form. The generators of rotations in two of the dimensions. The generator L_ab generates a rotation in the plane of dimensions a and b. It can be expressed in matrix form as

(L_ab)ⁱ_j = δ_aⁱ*δ_bj - δ_a^j*δ_bi

Its commutator is thus
[L_ab,L_cd] = δ_bc*L_ad - δ_bd*L_ac - δ_ac*L_bd + δ_ad*L_bc

This can be generalized to the Lorentz group and Lorentz-like groups with metric g as
(L_ab)ⁱ_j = δ_aⁱ*g_bj - δ_a^j*g_bi
[L_ab,L_cd] = g_bc*L_ad - g_bd*L_ac - g_ac*L_bd + g_ad*L_bc

For the Euclidean and Poincaré and similar groups, one adds the generators of translations, P_a. Rotations interchange them, as one would expect, and one gets
[P_a,P_b] = 0
[L_ab,P_c] = g_bc*P_a - g_ac*P_b

Overall,
[P,P] = 0
[L,P] = P
[L,L] = L

Or [algebra,P] = P
Thus, the P's form an "ideal" of the Euclidean-group algebra and its relatives. Ideals also appear in discussions of the structure of abstract-algebra rings, though defined with multiplication: (ideal)*(ring) = (ideal) and (ring)*(ideal) = (ideal). For the ring of integers, the sets a*Z, a an integer, are ideals.

A Lie-algebra ideal corresponds to a normal subgroup in the algebra's group.

-

Now for "simple" Lie algebras. The commutators of its elements must span the space of those elements, and it must contain no ideals other than itself. Much like a simple group. A "semisimple" Lie algebra is a sum of simple ones, each of which commutes with all the other ones. Much like a product group.

One can define a "derived series" from commutation: L1 = [L,L], L2 = [L1,L1], L3 = [L2,L2], ... If it ends in the empty algebra, then the algebra is solvable.

Likewise, one can define a "lower central series": L1 = [L,L], L2 = [L,L1], L3 = [L,L2], ... If it ends in the empty algebra, then the algebra is nilpotent.

 

[lpetrich]

Let's now look at some small Lie algebras.

A one-element algebra with element L has [L,L] = 0. This is clearly U(1).

A two-element algebra, (L1, L2), has [L1,L2] = a1*L1 + a2*L2. It has two solutions:
[L1,L2] = 0 ... abelian: R²
[L1,L2] = L1 ... from redefinition of L1 and L2 for at least one of a1 and a2 nonzero
L1 = {{0,1},{0,0}} and L2 = -(1/2){{1,0},{0,-1}}

Three-element algebras are more challenging. There are some 9 or 11 types of them, depending on how one counts them. Their most general commutator is
[L_i,L_j] = ε_ijk*(m_k*L_k) + (v_i*L_j - v_j*L_i)
where
m_i*a_i = 0 component by component.

Here is Bianchi's classification, by (m1,m2,m3,v) values:
I: (0,0,0,0) ... abelian: R³
II: (1,0,0,0) ... Heisenberg
III: (1,-1,0,1) ... nonabelian 2D * 1D
IV: (1,0,0,1)
V: (0,0,0,1)
VI: (1,-1,0,a)
VI0: (1,-1,0,0) ... 2D Poincare group Euc(1,1)
VII: (1,1,0,a)
VII0: (1,1,0,0) ... 2D Euclidean group Euc(2)
VIII: (1,1,-1,0) ... SO(2,1) ~ SL(2,R)
IX: (1,1,1,0) ... SO(3) ~ SU(2)

Note: all but the last two have the form [L1,L2] = 0, [L3,L_i] = N_ijL_j for i,j = 1,2 and some arbitrary matrix N. Here is that matrix:
I: {{0,0},{0,0}}
II: {{0,1},{0,0}}
III: {{1,1},{1,1}}
IV: {{1,1},{0,1}}
V: {{1,0},{0,1}}
VI: {{a,1},{1,a}}
VI0: {{0,1},{1,0}}
VII: {{a,-1},{1,a}}
VII0: {{0,-1},{1,0}}

One can choose for the generator matrices L1 = ..1 ... ..., L2 = ... ..1 ..., L3 = N11,N12,. N21,N22,. ... (.=0) for all of these.

 

[lpetrich]

First, for the Euclidean and Poincaré groups and algebras, one can define them using (n+1)*(n+1) matrices:

(P_a)ⁱ_j = δ_aⁱ*δ_n+1,j

Note that P has only one term on the right, while L has two terms on the right. The Euclidean/Poincaré group is thus a degenerate case of SO(n1,n2) with n1+n2=n+1, one where g_n+1,n+1 = 0 for g assumed to be a diagonal matrix. That's like how a line is a degenerate case of an circle or that a plane is a degenerate case of a sphere.

Back to Lie algebras in general, one can define a "Killing form" or metric for them from their structure constants:
K_ij = f_ia^b * f_jb<sup>a[/sub]
summed over a and b.

K is symmetric, and thus has well-defined eigenvalues. If some of them are 0, then the algebra is not simple or semisimple.

For 1 generator, K = {{0}}.

For 2 generators and the nonabelian case, K has eigenvalues 0 and a12+a22.

For 3 generators and the non-simple cases (all but VII and IX), K has eigenvalues 0, 0, and Tr(N.N) = N112 + N222 + 2*N12*N21

For VII, it has eigenvalues proportional to (1, 1, -1) -- different signs
For IX, it has eigenvalues proportional to (1, 1, 1) -- all the same sign
I should note that the algebras are here real-valued, because for complex-valued ones, one can analytically continue case IX to case VIII.

For all-nonzero, same-sign K, the algebra is "compact", meaning that its parameter space effectively has a finite range. SO(3) is compact, but none of the other algebras here are compact.</sup>

 

[George S]

The essence of Linear Algebra. I thought I knew it; now I know it much better.

https://www.youtube.com/playlist?list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab

 

[lpetrich]

George S's link titled: Essence of linear algebra - YouTube -- lots of videos for visualizing linear-algebra objects and operations. Matrices as transformations. Matrix multiplication as composition. Determinants as areas. I'd like to see the videos' author's take on eigenvalues and eigenvectors.

I'll try to guess. A transform's eigenvector is a vector that stays in the same direction after applying the transform, or else reverses direction. Its eigenvalue is (transformed length) / (original length), where reversing direction makes a negative sign.

 

[lpetrich]

I won't do 4-generator algebras, because there does not seem to be any easy way to simplify their structure constants, as there is for 2 and 3 generators.

But there is an elegant theory of simple Lie algebras. They fall into four infinite families with five exceptional algebras.

• A: SU(n+1) -- n(n+2)
• B: SO(2n+1) -- n(2n+1)
• C: Sp(2n) -- n(2n+1)
• D: SO(2n) -- n(2n-1)
• G2: 14
• F4: 52
• E6: 78
• E7: 133
• E8: 248

The lowest-order ones have some isomophisms:

• SO(2) ~ U(1)
• SO(3) ~ SU(2) ~ Sp(2)
• SO(4) ~ SU(2) * SU(2)
• SO(5) ~ Sp(4)
• SO(6) ~ SU(4)

There are even more isomorphisms if one counts analytic continuations, like SO(3) -> SO(2,1) ~ SL(2,R), SO(4) -> SO(3,1) ~ SL(2,C), and SO(4) -> SO(2,2) ~ SL(2,R) * SL(2,R).

There is a nice graphical representation of these algebras' structure called Dynkin diagrams, but some of them are rather difficult to represent as ASCII art.

 

[beero1000]

Solved this problem today and enjoyed it. It's tricksome, but with a nice elementary solution.


Find x and y.

 

[lpetrich]

This problem is incompletely constrained, or so it seems to me. If one changes the lengths of EH and GH while keeping the angles at E and G fixed, one can change x and y. However, x+y will remain fixed.

 

[beero1000]

This problem is incompletely constrained, or so it seems to me. If one changes the lengths of EH and GH while keeping the angles at E and G fixed, one can change x and y. However, x+y will remain fixed.

The lengths of EH and GH are determined by the given angles and the length of EG, so it is well-specified. This is one of those problems where the straightforward approaches don't get anywhere, though. Frustrating.

 

[Bomb#20]

Solved this problem today and enjoyed it. It's tricksome, but with a nice elementary solution.
View attachment 8153

Find x and y.

Piece of cake. x = 23, y = 25.

Full disclosure...

Show

x = 22.99965, y = 25.00035. Thank you Descartes. But you probably had in mind a more tricksome and less elementary solution.

 

[beero1000]

Solved this problem today and enjoyed it. It's tricksome, but with a nice elementary solution.
View attachment 8153

Find x and y.

Piece of cake. x = 23, y = 25.

Full disclosure...

Show

x = 22.99965, y = 25.00035. Thank you Descartes. But you probably had in mind a more tricksome and less elementary solution.

More tricksome and more elementary, I think. Doesn't need any of that new-fangled trigonometry the kids are doing these days.

 

[beero1000]

For those interested, here's my solution:

Show

Draw circle FGH and extend G and H through E to intersect the circle at G' and H', respectively. Fill in the above angles via judicious use of the vertical and inscribed angle theorems. Then, congruent triangles mean that angle EHF is equal to angle G'HH'. Inscribed angles mean that 2(x + 67°) = 180°, or x = 23°.

An added bonus of this approach is that is works for a variety of choices for the angles α and β, with the other angles changed appropriately. We get that x = 90° - α and y = 90° - β.

 

[Bomb#20]

For those interested, here's my solution:

Show

View attachment 8177

Draw circle FGH and extend G and H through E to intersect the circle at G' and H', respectively. Fill in the above angles via judicious use of the vertical and inscribed angle theorems. Then, congruent triangles mean that angle EHF is equal to angle G'HH'. Inscribed angles mean that 2(x + 67°) = 180°, or x = 23°.

An added bonus of this approach is that is works for a variety of choices for the angles α and β, with the other angles changed appropriately. We get that x = 90° - α and y = 90° - β.

No doubt it's because I'm way too rusty at this, but I'm not seeing how the vertical and inscribed angle theorems imply angle EH'F is 67°.

 

[beero1000]

For those interested, here's my solution:

Show

View attachment 8177

Draw circle FGH and extend G and H through E to intersect the circle at G' and H', respectively. Fill in the above angles via judicious use of the vertical and inscribed angle theorems. Then, congruent triangles mean that angle EHF is equal to angle G'HH'. Inscribed angles mean that 2(x + 67°) = 180°, or x = 23°.

An added bonus of this approach is that is works for a variety of choices for the angles α and β, with the other angles changed appropriately. We get that x = 90° - α and y = 90° - β.

No doubt it's because I'm way too rusty at this, but I'm not seeing how the vertical and inscribed angle theorems imply angle EH'F is 67°.

It implies that angles HH'F and HGF are supplementary as they are on opposite sides of the same chord FH.

 

[lpetrich]

I'm still not convinced. That proposed solution does not resolve the arbitrariness of the HG / HE ratio. I'll do an analytic-geometry sort of solution.

H = origin
HG = {1,0}
HE = a*{1,t}
where t = tan(angle GHE)

Now shoot lines from G and E, lines l1*{1,t1} and l2*{1,t2}. Their intersection is point F.

The solution:
l1 = (a*t + t2 - a*t2)/(t1 - t2)
l2 = (a*t - t1 + a*t1)/(t1 - t2)
F position = {(a*t + t1 - a*t2), (a*t + t2 - a*2)} / (t1 - t2)

In our problem, t = tan(48d), t1 = tan(67d), and t2 = - tan(17d), all different, and a is undetermined.

Or is there some problem condition that I did not include?

 

[beero1000]

I'm still not convinced. That proposed solution does not resolve the arbitrariness of the HG / HE ratio. I'll do an analytic-geometry sort of solution.

H = origin
HG = {1,0}
HE = a*{1,t}
where t = tan(angle GHE)

Now shoot lines from G and E, lines l1*{1,t1} and l2*{1,t2}. Their intersection is point F.

The solution:
l1 = (a*t + t2 - a*t2)/(t1 - t2)
l2 = (a*t - t1 + a*t1)/(t1 - t2)
F position = {(a*t + t1 - a*t2), (a*t + t2 - a*2)} / (t1 - t2)

In our problem, t = tan(48d), t1 = tan(67d), and t2 = - tan(17d), all different, and a is undetermined.

Or is there some problem condition that I did not include?

I'm not familiar with your {a,b} notation, but it doesn't seem like you've fixed all four of the given angles. I see 67d, and I think you're using 65d and 50d to get 48d and -17d, but where is the 46d angle in your equations?

My argument is that if EG is fixed, then by ASA there is exactly one triangle EGF and exactly one triangle EGH with the given angles. That completely determines the quadrilateral EFGH, and thereby the angles.

 

[lpetrich]

I used rectangular coordinates: {x,y}, where the origin is at H and where the x-axis is along HG. From the angles that you gave, I could find angle GHE from the angles that you gave. Your angle y (angle GHF) is arctan(y(F)/x(F)) in my coordinates.