The Math Thread

[beero1000]

Once every now and then, I stumble into discussions with people who apparently don't understand exponentiation. Specifically, people who claim that we need to go to the stars because overpopulation and resulting collapse is inevitable due to unchangeable aspects of human nature. I'm trying to educate them that if that is so, going to the stars will only buy us in the order of centuries at best.

However, the math contains some mistakes, or at least they seem like mistakes to me. I'll address that problem myself. Let's say that the density of stars is ns, and that the population a star's planets and space colonies can support is Ns. The colonies extend out to radius R(t), and the total population is N(t), increasing at a relative rate of p. Then,
\( N(t) = \frac{4\pi}{3} n_s N_s R(t)^3 \)
For
\( \frac{dN(t)}{dt} = p N(t) \)
we get
\( N(t) = N_0 e^{p t} \)
and
\( R(t) = R_0 e^{p t / 3} \)
The velocity of the leading edge of expansion is
\( \frac{dR(t)}{dt} = \frac{p}{3} R(t) \)
Since it must be less than c,
\( R(t) < \frac{3c}{p} \)

I'm not sure I see the problem, the two seem to be in strong agreement to me. In particular, using Jokodo's equations and working out the time of minimum population density gives the time t = 3/ln(R). Using ln(R) as your p (i.e. switching R^t to e^{t ln R}) and remembering to put c back in, the radius is exactly 3c/ln(R) = 3c/p at that point of no return. That corresponds to where the population growth rate finally wins out and the population density starts increasing, requiring ever-increasing superluminal expansion speeds in order to avoid the population density going to infinity.

 

[lpetrich]

I think that I need to show some pretty pictures.


SU(3) rep roots:
Red = fundamental (dimension 3)
Blue = its conjugate or mirror image (also dimension 3)
Black = adjoint (dimension 8, with 2 zero roots)


G2 rep roots:
Black: adjoint (dimension 14, with 2 zero roots)
Green fringe: fundamental (dimension 7, with 1 zero root)

Note that the G2 one looks like the SU(3) ones taken together. That is because SU(3) is a subalgebra of G2, and G2's reps break down into SU(3) ones as follows:

(7 fundamental) = (3 fundamental) + (3* fund conjugate) + (1 scalar)
(14 adjoint) = (8 adjoint) + (3 fundamental) + (3* fund conjugate)

 

[Jokodo]

Considering we don't need a whole body to simulate a consciousness in a brain, perhaps a more efficient configuration of matter would allow a greater amount of "differing personalities per volume"?

Increasing the density will buy us even less than increasing the speed in equal proportion (since volume increases to the cube of the speed).

Increase death rates. War. We need more wars, that kill people only, not destroy infrastructure. And only people without skillsets that maintain the physical infrastructure (kill all the greedy bankers and opportunistic lawyers/lawmakers for a good start).

Death rates are going to increase all by themselves, despite rising life expectancy: The only reason they're currently so low is that globally, many of the people whose turn it is to die weren't born in the first place, because we had a population spurt peaking around 50 years ago, meaning the generation of 70- and 80-somethings (+ even older people) is out of sync with what it should with a constant birth rate.

 

[lpetrich]

SO(5) rep roots:
Red = Spinor (4 it generalizes angular-momentum spin-1/2)
Black = Adjoint (10 with 2 zero roots)
Green Fringe = Vector (5 with 1 zero root)


SO(4) rep roots:
Red = Spinor 1 (2)
Blue = Spinor 2 (2)
Green = Vector (4)
Black = Adjoint (6 with 2 zero roots)

 

[Kharakov]

More from the lighter side of math, here is a Petrie polygon like illusion.

 

[lpetrich]

It is evident that SO(5) is closely related to SO(4):

Spinor (4) = Spinor 1 (2) + Spinor 2 (2)
Vector (5 with 1 zero root) = Vector (4) + Scalar (1 is zero)
Adjoint (8 with 2 zero roots) = Adjoint (6 with 2 zero roots) + Vector (4)


There is a simpler way of working out the size of a Lie-algebra representation than working out all its roots explicitly, and that was worked out by Hermann Weyl:

N(u) = product over positive roots a of (a,u+d)/(a,d)
where d = (1/2)*(sum over positive roots a of a)


For SU(2), N(u) = (u+1/2)/(1/2) = (2u + 1) = (n + 1) -- the well-known formula for quantum-mechanical angular momentum.

For SU, we have (n1+1)/1 * (n2+1)/1 * ... * (n1+n2+2)/2 * (n2+n3+2)/2 * ... * (n1+n2+n3+3)/3 * ...

The other simple Lie algebras have similar expressions, though not as simple.

For products of reps, like for adding angular momentum, it's simple: X1(u1) + X2(u2) have roots v1+v2 with multiplicity m1*m2 (the m's are their original multiplicity). One then looks at its highest root and expands the irrep for it, and subtracts that irrep. One then repeats that process is all gone until the algebra is all gone.

For powers of reps, one can break down by symmetry. For a square of a rep, the symmetric part has roots 2v with multiplicity m^2, and in addition roots v1+v2 with multiplicity m1*m2 where v2 != v1. The antisymmetric part has only the second kind of roots.

Like for SU(3) the 3 rep, (roots) {2/3,1/3} {-1/3,1/3}, {-1/3,-2/3} (weights) {1,0}, {-1,1}, {0,-1}

The antisymmetric part has (roots) {1/3,2/3}, {1/3,-1/3}, {-2/3,-1/3} (weights) {0,1}, {1,-1}, {-1,0}
That is rather plainly the 3* rep, the conjugate of the 3 rep.

The symmetric part has additional roots: (roots) {4/3,2/3}, {-2/3,2/3}, {-4/3,-2/3} (weights) {2,0}, {-2,2}, {0,-2}
That can be shown to be an additional rep: the 6 rep.

So 3 * 3 = 6 + 3*

For SU, one can use a graphical technique called "Young diagrams": SU Multiplets and Young Diagrams I've written a computerized version of the Young-diagram algorithm for products of reps.

 

[bilby]

Increase death rates. War. We need more wars, that kill people only, not destroy infrastructure. And only people without skillsets that maintain the physical infrastructure (kill all the greedy bankers and opportunistic lawyers/lawmakers for a good start).

Death rates are going to increase all by themselves, despite rising life expectancy: The only reason they're currently so low is that globally, many of the people whose turn it is to die weren't born in the first place, because we had a population spurt peaking around 50 years ago, meaning the generation of 70- and 80-somethings (+ even older people) is out of sync with what it should with a constant birth rate.

Yup. The boomers have already started to die in increasing numbers. Add in the fact that they were the first generation with mass media fame, and we would expect a very sharp increase in the number of famous people dying per annum - and that's exactly what has happened over that last couple of years.

People talked about how awful 2016 was, with so many popular and talented actors and musicians dying; but they didn't seem to grasp that it's the new normal - if anything, it's going to get worse for at least the next decade or two.

 

[Kharakov]

Increase death rates. War. We need more wars, that kill people only, not destroy infrastructure. And only people without skillsets that maintain the physical infrastructure (kill all the greedy bankers and opportunistic lawyers/lawmakers for a good start).

Death rates are going to increase all by themselves, despite rising life expectancy: The only reason they're currently so low is that globally, many of the people whose turn it is to die weren't born in the first place, because we had a population spurt peaking around 50 years ago, meaning the generation of 70- and 80-somethings (+ even older people) is out of sync with what it should with a constant birth rate.

Yup. The boomers have already started to die in increasing numbers. Add in the fact that they were the first generation with mass media fame, and we would expect a very sharp increase in the number of famous people dying per annum - and that's exactly what has happened over that last couple of years.

People talked about how awful 2016 was, with so many popular and talented actors and musicians dying; but they didn't seem to grasp that it's the new normal - if anything, it's going to get worse for at least the next decade or two.

I thought death was the norm.

 

[bilby]

Yup. The boomers have already started to die in increasing numbers. Add in the fact that they were the first generation with mass media fame, and we would expect a very sharp increase in the number of famous people dying per annum - and that's exactly what has happened over that last couple of years.

People talked about how awful 2016 was, with so many popular and talented actors and musicians dying; but they didn't seem to grasp that it's the new normal - if anything, it's going to get worse for at least the next decade or two.

I thought death was the norm.

It isn't the norm for a given generation, until that generation starts to approach its life expectancy.

Up until recently, most boomers have been able to imagine that they are immortal - and that dying is for the older people. Now dying is, suddenly and shockingly, the norm for their contemporaries. That's pretty scary for the generation that's dominated the world for their entire lives.

 

[Kharakov]

Yup. The boomers have already started to die in increasing numbers. Add in the fact that they were the first generation with mass media fame, and we would expect a very sharp increase in the number of famous people dying per annum - and that's exactly what has happened over that last couple of years.

People talked about how awful 2016 was, with so many popular and talented actors and musicians dying; but they didn't seem to grasp that it's the new normal - if anything, it's going to get worse for at least the next decade or two.

I thought death was the norm.

It isn't the norm for a given generation, until that generation starts to approach its life expectancy.

Up until recently, most boomers have been able to imagine that they are immortal - and that dying is for the older people. Now dying is, suddenly and shockingly, the norm for their contemporaries. That's pretty scary for the generation that's dominated the world for their entire lives.

Good, they are a generation of lying, manipulative, assholes. To paraphrase an older relative "wow, we really fucked your generation over, didn't we?"

The worse thing is how they selected their successors in corruption- only those who would band with the previous corrupt generation were allowed to gain power and knowledge. All others were deliberately discredited and fucked over, or killed and/or incarcerated if they weren't identified as foolish enough to be useful to the corrupt system.

I, unfortunately, have a programmed in work ethic, caused by the pieces of shit I hate. Mehh, can't kill myself because of it too. Want to make the world a better place... fuck. Anyway.

Fuck those on top of the socioeconomic hierarchy, they are all pieces of shit or fools... and I'm foolish enough to hope that there are some good, intelligent people up there. I suppose I should lol.

 

[lpetrich]

Getting back to Lie algebras, I'll consider their subalgebras. They have some interesting structures that I wish to discuss.

A simple way of making a subalgebra of a (semi)simple algebra is to demote a root from the algebra and turn it into a U(1) factor.

For instance, one gets the Standard Model out of SU(5) Grand Unified Theories by this mechanism.
SU(5) = A4 with Dynkin diagram 1 - 2 - 3 - 4

Demote root 3. It gives us 1 - 2 (3) 4. Roots 1 and 2 stay connected, making A2 = SU(3), while root 3 becomes the U(1) factor and root 4 makes A1 = SU(2). Root 3's number is the U(1) value.

Thus, we get SU(5) = SU(3) * SU(2) * U(1) -- QCD symmetry * weak isospin * weak hypercharge
The latter two are broken even further in electroweak symmetry breaking.

Let's see what a simple rep looks like: 5 = HW's {1,0,0,0} = { {4/5,3/5,2/5,1/5}, {-1/5,3/5,2/5,1/5}, {-1/5,-2/5,2/5,1/5}, {-1/5,-2/5,-3/5,1/5}, {-1/5,-2/5,-3/5,-4/5} }

They break apart into two parts:
{ {4/5,3/5}, {-1/5,3/5}, {-1/5,-2/5} } and 2/5 and { {1/5} }
{ {-1/5,-2/5} } and -3/5 and { {1/5}, {-4/5} }
Subtract {1/3,2/3} times the U(1) value from the first set and {1/2} times it from the second set:
{ {2/3,1/3}, {-1/3,1/3}, {-1/3,-2/3} } and 2/5 and { {0} }
{ {0,0} } and -3/5 and { {1/2}, {-1/2} }
So we get some recognizable irreps of SU(3) and SU(2):
3 and 2/5 and 1 (like spin 0)
1 and -3/5 and 2 (like spin 1/2)

So we get the right-handed part of the down quark and the left-handed parts of the leptons (electron and neutrino).

Likewise, a 10 = HW's {1,1,0,0} and it gives us the left-handed parts of the quarks, the right-handed up quark, and the right-handed electron.

A 24 = HW's {1,0,0,1} = the gluon (QCD particle), the W (WIS particle), the B (WHC particle), and some "leptoquarks" that cause free protons to decay. In fact, proton decay is predicted by most GUT"s.

 

[lpetrich]

Another method is to add a root at some appropriate spot and then remove another root. Here are the prescriptions for making the extended root sets with their Dynkin diagrams:

A: connect the root to both ends, making a loop.
B: make the root long and connect it to the second root from the long-root end, making the ends a short root and a fork.
C: make the root long and connected it to the short-root end, making the ends both long roots.
D: make a fork at the unforked end.
G2: make the root long and connect it to the long root.
F4: make the root long and connect it to the long-root end.
E6: make branching 2 - 2 - 2 (1 end)
E7: make branching 3 - 1 - 3 (2 end)
E8: make branching 2 - 1 - 5 (long end)
The exceptional ones are rather weird, with the extra root jumping around as one moves up in those algebras.

One gets these results:
A gives itself with roots renamed.
B(n1+n2) -> D(n1) + B(n2) ... SO(2n1+2n2+1) -> SO(2n1) + SO(2n2+1)
C(n1+n2) -> C(n1) + C(n2) ... Sp(2n1+2n2) -> Sp(2n1) + Sp(2n2)
D(n1+n2) -> D(n1) + D(n2) ... SO(2n1+2n2) -> SO(2n1) + SO(2n2)
G2 -> SU(3) and SU(2) * SU(2)
F4 -> SO(9) and SU(4) * SU(2) and SU(3) * SU(3) and SU(2) * Sp(6)
E6 -> SU(3)^3 and SU(6) * SU(3)
E7 -> SU(8) among others
E8 -> SO(16) among others


There are several more subalgebra prescriptions.
SU -> SO
SU(2n) -> Sp(2n)
SO(2n1+2n2+2) -> SO(2n1+1) * SO(2n2+1)
With root demotion for SO(2) and with SO(1) dropping out, this completes SO(n1+n2) -> SO(n1) * SO(n2)
SU(n1*n2*...) = SU(n1)*SU(n2)*...
SO(n1*n2*...) = SO(n1)*SO(n2)*...

In the latter one, SO(2) ~ U(1) and SO(1) drops out. Putting in negative even numbers gives the Sp's: SO(-2n) -> Sp(2n). This gives
SO(8) -> SO(4) * SO(2) and Sp(4) * Sp(2)
Sp(8) -> Sp(4) * SO(2) and SO(4) * Sp(2)

There's also a class of subalgebra where the fundamental rep of an infinite-family algebra maps onto some irrep of some smaller algebra. Like SU(6) -> SU(3) because SU(3) has a rep with size 6. I was unable to find a general-case formula for doing the rep reduction, so I have not implemented it, except for a special case.

What I call the "height subalgebra". I take the roots, add their components, then make them roots of SU(2). Then I find out what reps of SU(2) are present.

For the SU(3) one here, { {2/3,1/3}, {-1/3,1/3}, {-1/3,-2/3} } -> {1, 0, -1} -> SU(2) 3 or spin = 1
For the SU(5) one here, { {4/5,3/5,2/5,1/5}, {-1/5,3/5,2/5,1/5}, {-1/5,-2/5,2/5,1/5}, {-1/5,-2/5,-3/5,1/5}, {-1/5,-2/5,-3/5,-4/5} } -> {2, 1, 0, -1, -2} -> SU(2) 5 or spin 2

 

[lpetrich]

In 1949, physicist Giulio Racah attempted to analyze the f electrons of lanthanides and actinides. Starting with 7 slots for electrons, he created a cascade of algebras:
SU(7) -> SO(7) -> G2 -> SU(3) -> SU(2)
The G2 was there as a mathematical convenience without physical meaning.

What he did was equivalent to finding the height subalgebra of SU(7).

He went through that song-and-dance because he had to do his calculations by hand. Computers were in a very primitive state back then.


With my Lie-algebra code, I've derived some other important results, like the light-quark baryon states. I start by giving each quark a color state, a flavor state, and a spin state. I then find the antisymmetric cube of the quark's rep, since quarks must obey Fermi-Dirac statistics. I then select out all colorless states: color singlets. That leaves spin and flavor, with 8 spin-1/2 and 10 spin-3/2 baryons. They are arranged in a hexagon and in a triangle. Which is what one observes. The three quarks are antisymmetric in color, so they are symmetric in flavor and spin combined. The 10 spin-3/2 baryons have both spin and flavor symmetric, while the 8 spin-1/2 baryons have both spin and flavor with mixed symmetry.

I find it satisfying to derive results very quickly that took so long for previous generations of mathematicians and physicists to derive. In fact, most of my time in doing such derivations goes into writing my code and testing it.

 

[Kharakov]

FDS! That's what I was looking for... or just sigmoid functions in general.

 

[lpetrich]

I now turn to an interesting curiosity about Lie algebras' symmetry groups of their roots -- their Weyl groups.

Lie-algebra root a generates a Weyl-group operator on x: T(a,x) = x - 2*a*(a,x)/(a,a)
For short, I'll call it T(a).

T(a) has order 2: T(a).T(a) = identity e. From T(a,T(a,x)) = x.

Let us consider an algebra's simple roots ai. Their corresponding Weyl-group elements ri = T(ai). They satisfy
(ri*rj)^(mij) = e
for positive-integer mij. mii = 1, the other m's > 1.

Those other m's are related to the roots' normalized inner product: (ai,aj)/sqrt((ai,ai)*(aj,aj)) = - sqrt(nij)/2, for nonnegative integer nij.
For nij > 0, longer(a,a)/shorter(a,a) = nij. The m's are related to the n's by:
n = 0: m = 2
n = 1: m = 3
n = 2: m = 4
n = 3: m = 6

So a Weyl group is a Coxeter group, a group given by root elements ri satisfying (ri*rj)^(mij) = e for some matrix m with mii = 1 and other m's integers >= 2.

There are, however, some Coxeter groups that are not the Weyl group of any Lie algebra.

Coxeter groups can be represented diagrammatically much like Lie algebras. Each root element gets a dot, and each pair of roots gets no connection for mij = 2 and a connecting line for mij >= 3. For mij > 4, one writes the value of that number next to the line.

The finite Coxeter groups are:
A: o - o - o - ... - o
B, C: o 4 o - o - ... - o
D: o - (o - o) - o - ... - o
E6: o - o - (o - o) - o - o
E7: o - o - (o - o) - o - o - o
E8: o - o - (o - o) - o - o - o - o
F4: o - o 4 o - o
H3: o 5 o - o
H4: o 5 o - o - o
I2: o n o
A2 = I2(3), B2 = C2 = I2(4), H2 = I2(5), G2 = I2(6)

There are also many infinite ones, with some of them being affine. All but I2(infinity) are related to Lie algebras' extended root systems for doing extension-splitting subalgebras.
A': - o - o - o - ... - o - (loop)
B': o 4 o - o - ... - (o - o) - o
C': o 4 o - o - ... - o - o 4 o
D': o - (o - o) - o - ... - (o - o) - o
E6': o - o - (o - o - o) - o - o
E7': o - o - o - (o - o) - o - o - o
E8': o - o - (o - o) - o - o - o - o - o
F4': o - o 4 o - o - o
G2': o 6 o - o
I2(infinity): o infinity o

Affine ones have an interesting property. From the Coxeter matrix m one can construct a Schläfli matrix c with components cij = - cos(pi/mij).

For a finite Coxeter group, det(c) > 0
For an affine Coxeter group, det(c) = 0
For the remaining groups, "indefinite" or "hyperbolic" Coxeter groups, det(c) < 0

Affine Coxeter groups have (infinite) normal abelian subgroups with finite quotient groups.


I'll close this post by analyzing the simplest Coxeter groups. With one root element r, r² = e, and we get Z2.

Two root elements give the I2 groups, (r*s)ⁿ = e
Multiply by s, then (r*s)^m-n:
s*(r*s)ⁿ = s
s*(r*s)^m = s*(r*s)^m-n = (s*r)^m-n*s = (r*s)^n-m*s = (r*s)^-m*s

Thus, the Coxeter group I2 is the dihedral group, Dih. It has abelian subgroup Z with quotient group Z2: (same number of r's and s's, number of r's and s's different by 1)

 

[lpetrich]

I'll now consider what possible finite Coxeter groups there can possibly be. I'll do this much like I did simple Lie algebras.

The first thing is that every subgroup must also be finite, meaning that an infinite subgroup makes the whole group infinite.

The next thing is what happens when one merges two connected roots. The Schläfli matrix for connection between roots m and m+1 out of n has the following form:
c(1,1) ... c(m,m) -- the Sch matrix for the 1 ... m subgroup
c(m+1,m+1) ... c(n,n) -- the Sch matrix for the m+1 ... n subgroup
c(m,m+1) = c(m+1,m) = - x
All the rest zero (no loops between them).

One can diagonalize each subgroup part, and when one does so, one finds
det(group) = det(sg 1 ... m) * det(sg m+1 ... n) - x² * det(sg 1 ... m-1) * det(sg m+2 ... n)

For roots n-1 and n,
det(group) = det(sg 1... n-1) - x² * det(sg 1 ... n-2)

So for a finite group, det(group) < det(two elements merged)

So if the new group has zero or negative determinant, then the group as a whole is infinite.




Let's now see what is the maximum amount of connection that each root can have. For root 1 connected to roots 2 and 3 with roots 2 and 3 not being connected to each other, the determinant is D = 1 - x12² - x13[/sup]2[/sup].

The minimum value that will make finiteness is x = 1/2. For x12 = x13 = 1/sqrt(2), det = 0

With root merging, that means that there is at most one connection with m > 3.

Now consider branching. For three branches, every branch except for at most one has m = 3, and if that one has m = 4, then det = 0.

For loops, a 3-member loop has det <= 0, with det = 0 only for m = 3 connections. One can calculate the determinants of larger loop subgroups, making all but at most one connection 3. Only all connections 3 makes the determinant 0; all others make it negative.


So a finite Coxeter group's root-element connections have no loops, at most one connection with m > 3, or else at most one branch, a three-way one.

The only branched ones are D and E6, E7, and E8. They all have m = 3, as does their straight-chain relative: A.

By evalulating determinants, the only one with a m = 4 connection away from the ends is F4. By comparison, B = C has its m = 4 connection at one ends.

Likewise, the only ones with a m = 5 connection has that connection on one end and at most 2 additional roots: H3, H4.

The only ones with a m >= 6 connection have only those root elements with that connection: I2(m)

So I've gotten all the finite Coxeter groups.

 

[lpetrich]

Now for expressing the finite Coxeter groups in matrix form.

For A, one starts with the identity matrix with size n+1: ID(n+1).
Root element i: interchange rows i and i+1 of ID(n+1)
This yields the group of all interchanges of rows of ID(n+1) -- Sym(n+1)
Its order is (n+1)!

For B and C, the root elements are those of A(n-1) with one added: ID with last row reversed signs
This yields a group whose elements are diagonal(n +-1's) . interchanges of rows of ID
Its order is 2^n * n!

For D, the root elements are those of A(n-1) with one added: the last one of A(n-1) with last two rows reversed signs
This yields a group that is the B/C ones but with an even number of minus signs.
Its order is 2^(n-1) * n!

For F4, the root elements are those of B(3) with one added: (1/2) * {{1,-1,-1,-1},{-1,1,-1,-1},{-1,-1,1,-1},{-1,-1,-1,1}}
This yields a group that is B(4) with all (1/2) * (4*4 nonsingular matrix of +-1)
Its order is 1152

For I2, its root elements are {{1,0},{0,-1}} and {{cos(2pi/n), sin(2pi/n)}, {sin(2pi/n), -cos(2pi/n}}
Its order is 2n.

For H3, H4, the orders are 120 and 14400

For E6, E7, and E8, the orders are 51840, 2903040, 696729600

Note: these choices of root element are *not* unique.

 

[lpetrich]

A is the symmetry group of the n-simplex (triangle, tetrahedron, 5-cell, ...)

B is the symmetry group of the n-hypercube (square, cube, 8-cell, ...) and also of the n-cross-polytope (square, octahedron, 16-cell, ...)

F4 is the symmetry group of the 24-cell, halfway between an 8-cell and a 16-cell
H3 is the symmetry group of the dodecahedron and the icosahedron
H4 is the symmetry group of the 120-cell and the 600-cell, 4D relatives of the dodecahedron and the icosahedron

I2 is the symmetry group of the n-gon
Special cases:
2: D2
3: A2
4: B2
5: H2
6: G2

The symmetries of a line segment: A1 = Z2 -- identity and end-to-end reflection


To handle the 3D and 4D groups, we introduce the quaternionic forms of 3D and 4D rotation. A quaternion is a 4-vector that can be divided up into scalar and 3-vector parts. For scalar qs and vector qv of some unit quaternion, a 3D rotation is
R_ij = δ_ij*(qs² - qv.qv) + 2*qv_i*qv_j - 2*qs*(sum over k of ε_ijk*qv_k)

A unit quaternion has qs² + qv.qv = 1
To get a reflection, multiply by -1

A 4D rotation with sign factor s = +-1 is
R_ij = δ_ij*qs - (sum over k of ε_ijk*qv_k)
R_i4 = - s*qv_i
R_4i = s*qv_i
R₄₄ = qs

Combining for the two s values gives us a function of a pair of unit quaternions: R({q1,q2}) = R(q1,+1).R(q2,-1)
To get a reflection, multiply by diagonal({1,1,1,-1})

 

[lpetrich]

The quaternionic rotation groups are:

Cyclic: QC = ({cos(a),0,0,sin(a)} where a = 2pi*k/n for k = 0 to n-1)
Dihedral: QD = QC(2n) with ({0,cos(a),sin(a),0} where a = pi*k/n for k = 0 to 2n-1)
Tetrahedral: QT = QD(2) with (1/2)*(4 +-1's)
Octahedral: QO = QT with (1/sqrt(2))*(permutations of {+-1, +-1, 0, 0})
Icosaahedral: QI = QT with (even permutations of {+-(sqrt(5)+1)/4, +-1/2, +-(sqrt(5)+1)/4, 0})


The 3-root Coxeter groups are 3D rotation groups.
A3 = D3 = tetrahedral group
Root elements:
{{0, 1, 0}, {1, 0, 0}, {0, 0, 1}} -- 1/sqrt(2)*{0,1,-1,0}
{{1, 0, 0}, {0, 0, 1}, {0, 1, 0}} -- 1/sqrt(2)*{0,0,1,-1}
{{1, 0, 0}, {0, 0, -1}, {0, -1, 0}} -- 1/sqrt(2)*{0,0,1,1}
Order: 24

B3 = C3 = octahedral group
Root elements:
{{0, 1, 0}, {1, 0, 0}, {0, 0, 1}} -- 1/sqrt(2)*{0,1,-1,0}
{{1, 0, 0}, {0, 0, 1}, {0, 1, 0}} -- 1/sqrt(2)*{0,0,1,-1}
{{1, 0, 0}, {0, 1, 0}, {0, 0, -1}} -- {0,0,0,1}
Order: 48

H3 = icosahedral group
Root elements:
{{-1, 0, 0}, {0, 1, 0}, {0, 0, 1}} -- {0,1,0,0}
{{(sqrt(5)+1)/4, -1/2, (-sqrt(5)+1)/4}, {-1/2, (-sqrt(5)+1)/4, (-sqrt(5)-1)/4}, {(-sqrt(5)+1)/4, (-sqrt(5)-1)/4, 1/2}} -- {0, (sqrt(5)-1)/4, (sqrt(5)+1)/4, 1/2}
{{1, 0, 0}, {0, 1, 0}, {0, 0, -1}} -- {0,0,0,1}
Order: 120

 

[lpetrich]

Now the 4D case.

A4:
Root elements:
{{0, 1, 0, 0}, {1, 0, 0, 0}, {0, 0, 1, 0}, {0, 0, 0, 1}} -- 1/sqrt(2)*{{0,1,-1,0}, {0,-1,1,0}}
{{1, 0, 0, 0}, {0, 0, 1, 0}, {0, 1, 0, 0}, {0, 0, 0, 1}} -- 1/sqrt(2)*{{0,0,1,-1}, {0,0,-1,1}}
{{1, 0, 0, 0}, {0, 1, 0, 0}, {0, 0, 0, 1}, {0, 0, 1, 0}} -- 1/sqrt(2)*{{1,0,0,-1}, {1,0,0,1}}
{{(-sqrt(5)+5)/8, (-sqrt(5)-3)/8, (-sqrt(5)-3)/8, (sqrt(5)-1)/8}, {-sqrt(5)-3)/8, (-sqrt(5)+5)/8, (-sqrt(5)-3)/8, (sqrt(5)-1)/8}, {((-sqrt(5)-3)/8, (-sqrt(5)-3)/8, (-sqrt(5)+5)/8, (sqrt(5)-1)/8}, {(sqrt(5)-1)/8, (sqrt(5)-1)/8, (sqrt(5)-1)/8, (3*sqrt(5)+1)/8}} -- 1/sqrt(2) * { {(-sqrt(5)+3)/4,(-sqrt(5)-1)/4,(-sqrt(5)-1)/4,(-sqrt(5)-1)/4}, {(-sqrt(5)+3)/4,(sqrt(5)+1)/4,(sqrt(5)+1)/4,(sqrt(5)+1)/4} }
Order: 120

The rotation part matches the rotation part of the icosahedral group, and it can be constructed as {q,inv(q)} for all q in QI. inv(q) is the inverse quaternion ({1,-1,-1,-1}*q)/(q.q). The root elements here are a rotation of that construction.

In general, A can be constructed as size-n matrices in addition to size-(n+1) matrices. The construction:
Use the root elements of A(n-1) in size-n form, and add a final root element:
For indices i and j in the matrix:
. if i and j < n, then
. . if i == j then (n-2)(n+1) - 2*sqrt(n+1)
. . else -(n+2) - 2*sqrt(n+1)
. else if one of i and j < n then
. . - 2 + (n-2)*sqrt(n+1)
. else
. . (n-2) + 2(n-1)*sqrt(n+1)
all divided by n²

B4, C4:
Root elements:
{{0, 1, 0, 0}, {1, 0, 0, 0}, {0, 0, 1, 0}, {0, 0, 0, 1}} -- 1/sqrt(2)*{{0,1,-1,0}, {0,-1,1,0}}
{{1, 0, 0, 0}, {0, 0, 1, 0}, {0, 1, 0, 0}, {0, 0, 0, 1}} -- 1/sqrt(2)*{{0,0,1,-1}, {0,0,-1,1}}
{{1, 0, 0, 0}, {0, 1, 0, 0}, {0, 0, 0, 1}, {0, 0, 1, 0}} -- 1/sqrt(2)*{{1,0,0,-1}, {1,0,0,1}}
{{1, 0, 0, 0}, {0, 1, 0, 0}, {0, 0, 1, 0}, {0, 0, 0, -1}} -- {{1,0,0,0}, {1,0,0,0}}
Order: 384

Rotation part: (all dihedral, all dihedral) + (all tetrahedral extra, tetrahedral extra with even number of signs flipped) + (all octahedral, octahedral extra with nonzero in same places xx00 xx00 or in complementary places xx00 00xx))

D4:
Root elements:
{{0, 1, 0, 0}, {1, 0, 0, 0}, {0, 0, 1, 0}, {0, 0, 0, 1}} -- 1/sqrt(2)*{{0,1,-1,0}, {0,-1,1,0}}
{{1, 0, 0, 0}, {0, 0, 1, 0}, {0, 1, 0, 0}, {0, 0, 0, 1}} -- 1/sqrt(2)*{{0,0,1,-1}, {0,0,-1,1}}
{{1, 0, 0, 0}, {0, 1, 0, 0}, {0, 0, 0, 1}, {0, 0, 1, 0}} -- 1/sqrt(2)*{{1,0,0,-1}, {1,0,0,1}}
{{1, 0, 0, 0}, {0, 1, 0, 0}, {0, 0, 0, -1}, {0, 0, -1, 0}} -- 1/sqrt(2)*{{1,0,0,1}, {1,0,0,-1}}
Order: 192

Rotation part: (all dihedral, all dihedral) + (all tetrahedral extra, tetrahedral extra with even number of signs flipped)

F4:
Root elements:
{{0, 1, 0, 0}, {1, 0, 0, 0}, {0, 0, 1, 0}, {0, 0, 0, 1}} -- 1/sqrt(2)*{{0,1,-1,0}, {0,-1,1,0}}
{{1, 0, 0, 0}, {0, 0, 1, 0}, {0, 1, 0, 0}, {0, 0, 0, 1}} -- 1/sqrt(2)*{{0,0,1,-1}, {0,0,-1,1}}
{{1, 0, 0, 0}, {0, 1, 0, 0}, {0, 0, -1, 0}, {0, 0, 0, 1}} -- {{0,0,0,1}, {0,0,0,-1}}
(1/2) * {{1, -1, -1, -1}, {-1, 1, -1, -1}, {-1, -1, 1, -1}, {-1, -1, -1, 1}} -- (1/2) * {{1,1,1,1}, {1,-1,-1,-1}}
Order: 1152

Rotation part: (all tetrahedral, all tetrahedral) + (all octahedral extra, all octahedral extra)

H4:
Root elements:
{{1, 0, 0, 0}, {0, 1, 0, 0}, {0, 0, 1, 0}, {0, 0, 0, -1}} -- {{1,0,0,0},{1,0,0,0}}
(1/2) * {{1, 1, 1, 1}, {1, 1, -1, -1}, {1, -1, 1, -1}, {1, -1, -1, 1}} -- (1/2) * {{1,-1,1,1}, {1,1,-1,-1}}
{{1, 0, 0, 0}, {0, 1, 0, 0}, {0, 0, -1, 0}, {0, 0, 0, 1}} -- {{0,0,0,1},{0,0,0,-1}}
{{1/2, (sqrt(5)-1)/4, (-sqrt(5)-1)/4, 0}, {(sqrt(5)-1)/4, (sqrt(5)+1)/4, 1/2, 0}, {-sqrt(5)-1)/4, 1/2, (-sqrt(5)+1)/4, 0}, {0,0,0,1} -- {{0, 1/2, (-sqrt(5)+1)/4, (sqrt(5)+1)/4}, {0, -1/2, (sqrt(5)-1)/4, (-sqrt(5)-1)/4}}
Order: 14400

Rotation part: (all icosahedral, all icosahedral)

So I've verified the construction of all the finite Coxeter groups except for E6, E7, and E8. I've also verified the quaternion construction of all the 3D and 4D finite Coxeter groups.