Choose unit quaternions $q_0, q_1, q_2, q_3, q_4$ that form the vertices of a regular 4-simplex in the quaternions. Assume $q_0 = 1$. Let the other four generate a group via quaternion multiplication. Is this a free group on 4 generators?

I heard from Adrian Ocneanu that the answer is yes, but I don't know a proof.

Here's why I care. As shown in this image by Greg Egan, you can inscribe a cube in a regular dodecahedron:

twin dodecahedra

If you rotate the cube 90 degrees about an axis of 4-fold symmetry, the dodecahedron will be mapped to a different dodecahedron. Ocneanu calls this a twin of the original dodecahedron. For example, the red dodecahedron above has the blue one as a twin, and vice versa. Despite the name, a regular dodecahedron actually has 5 different twins, one for each cube that can be inscribed in it.

You can create a graph as follows. Start with a node for our original dodecahedron. Draw nodes for all the dodecahedra you can get from this one by repeatedly taking twins. Connect two nodes with an edge if and only if they are twins of each other.

Ocneanu claims the resulting graph is a tree! In other words, if you start at your original dodecahedron, and keep walking along edges of this graph by taking twins, you’ll never get back to where you started except by undoing all your steps.

Ocneanu didn't tell me the proof, but he said the key to the proof was this:

Claim: if you take unit quaternions at the vertices of a regular 4-simplex, one of them equal to 1, the remaining four are generators of a free group.

Indeed, Egan and I were able to use this claim to prove that the graph is a tree:

So now I want to know why Ocneanu's claim is true --- and indeed, I want to know that it is true.

If it helps, you can assume the regular 4-simplex has these vertices:

$$ q_0 = 1 $$

$$q_1 = -\frac{1}{4} + \frac{\sqrt{5}}{4} i + \frac{\sqrt{5}}{4} j + \frac{\sqrt{5}}{4} k $$

$$ q_2 = -\frac{1}{4} + \frac{\sqrt{5}}{4} i -\frac{\sqrt{5}}{4} j -\frac{\sqrt{5}}{4} k $$

$$ q_3 = -\frac{1}{4} -\frac{\sqrt{5}}{4} i + \frac{\sqrt{5}}{4} j -\frac{\sqrt{5}}{4} k $$

$$ q_4 = -\frac{1}{4} -\frac{\sqrt{5}}{4} i -\frac{\sqrt{5}}{4} j +\frac{\sqrt{5}}{4} k $$

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    $\begingroup$ Yes, it would imply that. The unit quaternions form the group isomorphic to $\mathrm{SU}(2)$ under multiplication, and I believe that if you randomly choose $n$ elements of $\mathrm{SU}(2)$, they generate a free group with probability 1. (Here I'm using Haar measure to define probability.) I'm having trouble finding a reference for that, though. $\endgroup$
    – John Baez
    May 1, 2015 at 17:30
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    $\begingroup$ Okay, here is an embedding of the free group on 2 generators into $\mathrm{SO}(3)$, and here are some proofs that you can embed a free group of any countable number of generators in the free group on 2 generators. So, $\mathrm{SU}(2)$ (double covering $\mathrm{SO}(3)$) contains free groups of arbitrary countable rank. $\endgroup$
    – John Baez
    May 1, 2015 at 17:39
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    $\begingroup$ Existence implies that a random set with $n$ generators is free: Because there are countably many words in the $n$ generators, it is sufficient to check that each one is nontrivial with probability $1$. But each word defines an analytic function $SU(2)^N \to SU(2)$ which is trivial with probability $0$ unless it is the constant function with value the identity. But it is not the constant function with value the identity, because it is sometimes not the identity. $\endgroup$
    – Will Sawin
    May 1, 2015 at 17:42
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    $\begingroup$ @John: I was a little thrown off by your usage of the word "put"; at first it sounded to me like I could arbitrarily choose some quaternions to put at the vertices of a regular $4$-simplex, but now I think you're talking about the vertices of a regular $4$-simplex in the quaternions, so once I've selected a regular $4$-simplex I can't choose anything else. Can you clarify the wording here? $\endgroup$ May 1, 2015 at 17:44
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    $\begingroup$ Thinking of these quaternions as elements in $SU(2)$, they have trace $-\frac12$. This indicates that there ought to be a non-trivial action on a Bass-Serre tree. It might be possible to show that the action on this tree is free, in which case the group is free. $\endgroup$
    – Ian Agol
    May 1, 2015 at 17:48

5 Answers 5


Edit: The previous answer had an error, which I realized from a comment of @Will Sawin, and I've completely revised it.

This group is a subgroup of an S-arithmetic lattice, which acts discretely on finite-valence Serre tree associated to $SL_2$ (really, a Bruhat-Tits building associated to $SL_2(F)$, where $F$ is a local field), hence is virtually free.

The rational quaternions is a quaternion algebra with Hilbert symbol $\binom{\underline{-1,-1}}{\mathbb{Q}}$. We may tensor with $F=\mathbb{Q}(\sqrt{5})$ to get the quaternion algebra $A=\binom{\underline{-1,-1}}{F}$. The given elements $q_i$ are unit norm elements in the quaternion algebra $A$. Since the real quaternions $\binom{\underline{-1,-1}}{\mathbb{R}}$ is ramified (i.e. a division algebra), the algebra $\binom{\underline{-1,-1}}{F}$ is ramified at both real places (tensoring $A$ with $\mathbb{R}$ over the two embeddings of $F$ into $\mathbb{R}$.

For all odd places (i.e. tensoring $A$ with $F_\mathcal{P}$, the $\mathcal{P}$-adic completion of $F$), the quaternion algebra $A_\mathcal{P}=A\otimes_F F_\mathcal{P}$ splits, i.e. is isomorphic to a matrix algebra $M_2(F_\mathcal{P})$. Since $2$ does not split over $F$, the algebra $A_{(2)}$ must also split, since $A$ must split at an even number of places by Hilbert's Reciprocity Law.

The given elements lie in an order $\mathbb{Z}[\sqrt{5}][\frac12][1,i,j,k] \subset A$. For each odd prime $\mathcal{P}$, this lies in a compact subgroup of $A_\mathcal{P}$, and lies in a compact subgroup of the real places. So it must be a lattice in $A_{(2)}^1\cong SL_2(F_{(2)})$. Therefore, it acts on the tree associated to $SL_2(F_{(2)})$, described in Serre's book Trees Chapter II.1 (this is the Bruhat-Tits building associated to $SL_2(F_{(2)})$). Thus, the group is virtually free.

The residue field of $\mathcal{O}_{F_{(2)}}$ is $\mathbb{Z}[(1+\sqrt{5})/2]/(2)=\mathbb{F}_4$, the field with 4 elements, so the Serre tree has degree 5 ($=|\mathbb{P^1F}_4|$). It is tempting to guess that vertices of the Serre tree will correspond to dodecahedra, and neighbors to twins, but I haven't checked this. However, it's clear that the automorphism group $A_5$ of the dodecahedron stabilizes a vertex of the Serre tree, and the twin dodecahedra should have automorphism group stabilizing adjacent vertices of the tree.

  • $\begingroup$ What would I need to read to understand the phrase "the Bass-Serre tree associated to the prime $2$"? $\endgroup$ May 2, 2015 at 6:01
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    $\begingroup$ @QiaochuYuan, you should start with Serre's book Trees. $\endgroup$
    – HJRW
    May 2, 2015 at 6:31
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    $\begingroup$ @Ian: Sorry, I should have been more specific. My question is not about the "Bass-Serre tree" part but about the "associated to the prime $2$" part (e.g. as far as I can tell the Wikipedia article doesn't tell me anything about this). $\endgroup$ May 2, 2015 at 7:47
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    $\begingroup$ All $4$ elements written down in the original question lie in the usual quaternions tensored with $\mathbb Q(\sqrt{5})$ But that's a different algebra from the one you wrote down - the original quaternion algebra is ramified only over $2$ and $\infty$, so this is ramified only over primes lying over $\infty$. Why do they lie in two different quaternion algebras? $\endgroup$
    – Will Sawin
    May 2, 2015 at 14:30
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    $\begingroup$ @QiaochuYuan: Ok, I was using the term "Bass-Serre tree" incorrectly. I really meant the tree associated to $SL_2$ of a local field, which is a special case of a Bruhat-Tits building. Serre's book Chapter II section 1 is the usual reference. Shalen's notes (linked in my other comment) Chapter 3 also has a nice discussion. $\endgroup$
    – Ian Agol
    May 3, 2015 at 15:00

As Ian Agol said, the group is virtually free.

Now, let us note that all your quaternions are congruent to $1$ modulo $\sqrt 5$. This implies your group is torsion free (a kind of Minkowski theorem), and hence is free.

[ Proof : Let $\Gamma$ be the group of quaternions of norm $1$ in $\mathbf Z[\frac{1}{2},\sqrt{5}]$, and let $\Gamma_0$ be the subgroup of those who are congruent to $1$ mod $\sqrt{5}$.

Assume $q_0^p=(1+\sqrt 5 q_0')^p=\sum_k C_p^k \sqrt 5^kq_0^k=1$. Looking at the $\sqrt 5$ valuation of the summands shows that one has to have $q_0'=0$.]

Added later :

Let us use the following notations :

$\phi:=\frac{1+\sqrt 5}{2}$.

$A=\mathbf Z[\phi]$ is the ring of integers in $K:=\mathbf Q[\sqrt 5]$.

$\mathbf H$ is the algebra of Hamilton quaternions over $\mathbf Q$. It's underlying quadratic space is just the euclidean $4$-dimensional space over $\mathbf Q$, which is positive definite at $\mathbf R$, anisotropic at $\mathbf Q_2$, and split at $\mathbf Q_p$ for $p$ odd.

$\mathbf H_K$ is the extension $\mathbf H\otimes K$. It is obviously totally positive definite (i.e. $\mathbf H_K\otimes \mathbf R$ is positive definite for each of the two embeddings $K\to\mathbf R$), and split at all non-dyadic primes. At the prime $2$, there are isotropic quaternions, e.g. $\sqrt{-7}+i+j+\sqrt 5 k$ (note : $-7$ is a square in $\mathbf Q_2$), hence (since an isotropic quaternion algebra is split) $\mathbf H_K\otimes \mathbf Q_2[\sqrt 5]$ is split (as it had to be by the law saying that the number of ramified primes has to be even).

$\mathbf H_A$ is the set of quaternions that are integral, i.e. have trace and norm in $A$. It obviously contains the $A$-subalgebra $H_0$ generated by $i$ and $j$ (i.e. $A.1+A.i+A.j+A.k$). This subalgebra contains 8 vectors of norm 1. Also, it contains the bigger $A$-subalgebra $A_1$ generated by Hurewicz quaternions, i.e. ($A+A.i+A.j+A.\frac{1+i+j+k}{2}$). This subalgebra contains 24 vectors of norm 1, and is still contained in a bigger one, obtained by appending the integral quaternion $\frac{1+\phi i+\bar \phi j}{2}$ to the generators. This last algebra is $\mathbf H_A$ and contains 120 vectors of norm 1. Let us call $Sp(A)$ this group of order 120.

Let $L_A$ be the orthogonal complement of $1$ in $\mathbf H_A$. Since (as a bilinear module) $H_A$ has determinant $1$ (because it's the case at each place), and $1$ has norm $1$, $L_A$ (as a bilinear module) has to have determinant $2$. It contains at leat $30$ vectors of length 1 (all the quaternions of the form $qi\bar q$ for $q$ in $Sp(A)$. One checks that there are no other vectors of length 1 in $L_A$, and that these vectors generate $L_A$.

For an $A$-algebra $k$, let us write $Sp(k)$ for the set of norm $1$ quaternions in $\mathbf H_A\otimes k$, write $L_k$ for the extension $L_A\otimes k$, and write $SO(k)$ for the special orthogonal group of $L_k$. We have the usual morphism $f : Sp(k)\to SO(k)$ obtained by making the quaternions act by conjugation on elements of $L_k$.

Example 1 : one checks that the following is an exact sequence $$1\to\{\pm 1\}\to Sp(A)\to SO(A)\to 1\ \ \ .$$

Example 2 : for the reduction mod $\sqrt 5$ : the following is an exact sequence $$1\to\{\pm 1\}\to Sp(\mathbf F_5)\to SO(\mathbf F_5)\to \mathbf F_5^\times/(\mathbf F_5)^{\times 2}\to 1\ \ \ .$$

Example 3 : the group $SO(A[\frac 12])$ acts transitively on the maximal simplices (edges) of the building for $SO(\mathbf Q_2[\sqrt 5])$, which is a tree. One deduces a decomposition as an amalgamated sum $$ SO(A[\frac 12])\simeq \mathfrak{A}_5\ast_{\mathfrak{A}_4} \mathfrak S_4\ \ \ .$$ The image of $f$, that we denote by $SO'(A[\frac 12])$ also acts on the building, but this time it has $2$ orbits of maximal vertices, $2$ orbits of edges, and $1$ orbit of minimal simplices. One deduces a decomposition as an amalgamated sum $$ SO'(A[\frac 12])\simeq \mathfrak{A}_5\ast_{\mathfrak{A}_4} \mathfrak A_5\ \ \ .$$ It follows that the cokernel of $f$ has order $2$.

Finally we have a commutative diagram in which all rows and columns are exact :

So the groups $\Gamma_0$ and $\Gamma_0'$ are isomorphic. Now $SO(A[\frac 12])$ has virtual Euler-Poincaré characteristic $$\chi_v( SO(A[\frac 12]))=\frac{1}{60}+\frac{1}{24}-\frac{1}{12}=-\frac{1}{40}$$ and $SO(\mathbf F_5)$ has order $120$, so we obtain $$\chi_v( \Gamma_0')=-{3}\ \ \ .$$ Since $\Gamma_0'$ has no torsion, we deduce that it is a free group on $4$ generators ... the ones that correspond to the quaternions (different from $1$) you gave in the OP.

Further edit :

@IanAgol asked for a justification of the last claim, i.e. that the 4 non-trivial quaternions given in the OP generate $\Gamma_0$. Sorry if it seems raw.

Here it is :

In $\mathbf H_A$, there are 2520 quaternions of norm 4. Among them, 120 are of the form $2q$ with $q$ in $Sp(A)$. The other 2400 quaternions, once divided by two, furnish elements of $Sp(A[\frac 12])$. The group $Sp(A)$ acts faithfully on them, and gives 20 left cosets. Using $f$, two elements of a same coset will send $L_A$ on the same image, which is one of the vertices of the tree at distance $2$ of $L_A$. This proves the affirmation that the quotient of the tree by $SO'(A)$ is as indicated above.

Each one of these cosets contain exactly one element of $\Gamma_0$. These elements are ($y=\sqrt 5$):

$$X_1=\left\{\begin{array}{} (&-1/4& ,& 1/4*y& ,& -1/4*y& ,& 1/4*y& )\\ (&-1/4& ,& -1/4*y& ,& -1/4*y& ,& -1/4*y& )\\ (&-1/4& ,& 1/4*y& ,& 1/4*y& ,& -1/4*y& )\\ (&-1/4& ,& -1/4*y& ,& 1/4*y& ,& 1/4*y& )\\ \end{array}\right.$$

$$X_2=\left\{\begin{array}{} (&-1/4& ,& 0& ,& 1/8*(y + 5)& ,& 1/8*(y - 5)& )\\ (&-1/4& ,& 1/8*(-y - 5)& ,& 1/8*(y - 5)& ,& 0& )\\ (&-1/4& ,& 1/4*y& ,& -1/4*y& ,& -1/4*y& )\\ (&-1/4& ,& 1/8*(-y + 5)& ,& 0& ,& 1/8*(y + 5)& )\\ \end{array}\right.$$

$$X_3=\left\{\begin{array}{} (&-1/4& ,& 1/8*(y + 5)& ,& 1/8*(y - 5)& ,& 0& )\\ (&-1/4& ,& 1/8*(y - 5)& ,& 0& ,& 1/8*(-y - 5)& )\\ (&-1/4& ,& -1/4*y& ,& -1/4*y& ,& 1/4*y& )\\ (&-1/4& ,& 0& ,& 1/8*(y + 5)& ,& 1/8*(-y + 5)& )\\ \end{array}\right.$$

$$X_4=\left\{\begin{array}{} (&-1/4& ,& -1/4*y& ,& 1/4*y& ,& -1/4*y& )\\ (&-1/4& ,& 1/8*(y + 5)& ,& 1/8*(-y + 5)& ,& 0& )\\ (&-1/4& ,& 1/8*(y - 5)& ,& 0& ,& 1/8*(y + 5)& )\\ (&-1/4& ,& 0& ,& 1/8*(-y - 5)& ,& 1/8*(y - 5)& )\\ \end{array}\right.$$

$$X_5=\left\{\begin{array}{} (&-1/4& ,& 0& ,& 1/8*(-y - 5)& ,& 1/8*(-y + 5)& )\\ (&-1/4& ,& 1/8*(-y - 5)& ,& 1/8*(-y + 5)& ,& 0& )\\ (&-1/4& ,& 1/8*(-y + 5)& ,& 0& ,& 1/8*(-y - 5)& )\\ (&-1/4& ,& 1/4*y& ,& 1/4*y& ,& 1/4*y& ) \end{array}\right.$$

Given an edge $E_i$ with source $L_A$ in the tree (there are 5 of them), the quaternions $q$ such that the geodesic between $L_A$ and $f(q)L_A$ begins with $E_i$ have been isolated in the set $X_i$.

Note that $X_1$ is the set of inverses of the quaternions of the OP. Moreover, all the elements in the $X_i's$ are of the form $q_1^{-1}q_2$ with $q_1,q_2\in X_1\cup\{1\}$. The claim follows.

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    $\begingroup$ Nice description. One thing I still don't quite get is why $\Gamma_0$ is generated by the 4 given elements? I don't see how you can deduce this without a bit more information (a free group on 4 elements has many 4 generator subgroups). $\endgroup$
    – Ian Agol
    May 5, 2015 at 4:00
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    $\begingroup$ Ok, excellent, that shows it is a lattice! $\endgroup$
    – Ian Agol
    May 5, 2015 at 18:20

By Stalling's theorem any torsion free virtually free group is free see here. By Ian Agol's answer the group is virtually free.

All four quaternions are congruent to $-1/4$ mod $\sqrt 5$ and since $5/4-1/4=1$ as few-reps has noted in his or her answer they are all congruent to $1$. So their finite products are always congruent to a non-zero real number mod $\sqrt 5$. So the sum of their squares is a fraction with relatively prime denominator and numerator,with the numerator divisible by 5.

Now the real term of any quaternion is the cosine of the angle through which quaternion rotates the plane consisting of the two vectors with zero real terms which are perpendicular to the non-real part of the quaternion. For this to have a prime power equal to one the angle of rotation must be rational.

Now here we have a classification of all rational and all quadratic values for cosines of rational angles they are $0$, $\pm 1/2$, $\pm 1$ and also some terms involving square roots of two or three which can be ignored and finally $\pm 1/4 \pm (\sqrt 5)/4)$.

Now we can look at these cases individually for the real part equals $1$ that is the identity and that does not contribute to torsion, for $1/2$ the sum of the squares of the non-real components will not equal a fraction with relatively prime denominator and numerator,with the numerator divisible by $5$, for zero the real component is zero when by the above it must be non zero.

For the real part equals $\pm 1/4 \pm (\sqrt 5)/4)$ the squares of the non=real components will not equal a fraction with relatively prime denominator and numerator,with the numerator divisible by $5$.

Finally if we have real part equal to $-1$ we note it does not equal $1$ mod square root of 5.

So we have covered all cases and there is no torsion and if the group is virtually free it is free.


Edit: Added proof that the 4-simplex generates a free group.

Here's a fairly simple proof that the graph of twin dodecahedra is a tree, which is then used to prove that the 4-simplex generates a free group.

We'll start by working with a graph of twin 4-simplexes, rather than twin dodecahedra. (This is an approach that John Baez suggested.) Given any 4-simplex with five unit quaternions $q_i$ as its vertices, we will define five "twins" of this 4-simplex, relative to each of its five vertices, as the 4-simplexes with vertices:

$$q^{(m)}_i = q_m q_i^{-1} q_m$$

When $q_m = q_i$ we have:

$$q^{(m)}_m = q_m$$

So the simplex and its twin always share a vertex. If we then "twinned" the twin simplex relative to that shared vertex, we would be taken back to the original simplex:

$$q^{(m)}_m (q^{(m)}_i)^{-1} q^{(m)}_m = q_m q_m^{-1} q_i q_m^{-1} q_m = q_i$$

We will define a map $G: \mathbb{Q}^4 \to \mathbb{Q}(\sqrt{5})^4$ as:

$$G(x) = \left(x_0, x_1 \sqrt{5}, x_2 \sqrt{5}, x_3 \sqrt{5}\right)$$

Now suppose we have unit quaternions:

$$q_a = G(a) \qquad q_b = G(b)$$

where $a, b \in \mathbb{Q}^4$, and $q_a$ and $q_b$ are distinct vertices of the same 4-simplex, so that $q_a \cdot q_b = -\frac{1}{4}$. Then the quaternion multiplications of the twinning operation simplify to an extraordinary extent, and we have:

$$q_a q_b^{-1} q_a = -\frac{1}{2} G(a+2b)\qquad q_a \ne q_b$$

(If $q_a = q_b$, then of course the twin itself is simply equal to $q_a$.)

Let's further suppose that all $a_i, b_i$ are equal to integers divided by powers of 2, and that we write each unit quaternion $q_a$ and $q_b$ with a common denominator across all four components:

$$q_a = \frac{1}{2^{n_a}} G(r)$$ $$q_b = \frac{1}{2^{n_b}} G(s)$$

where $r, s \in \mathbb{Z}^4$, and at least one of the $r_i$ and at least one of the $s_i$ are odd.

We then have:

$$q_a q_b^{-1} q_a = -\frac{1}{2^{n_a+n_b+1}} G(2^{n_b}r+2^{n_a+1}s)$$

If $n_a+1 \gt n_b$, we can cancel a factor of $2^{n_b}$, giving us:

$$q_a q_b^{-1} q_a = -\frac{1}{2^{n_a+1}} G(r+2^{n_a-n_b+1}s)$$

Since at least one of the $r_i$ is odd, no further cancellation is possible.

If $n_a+1 \lt n_b$, we can cancel a factor of $2^{n_a+1}$, giving us:

$$q_a q_b^{-1} q_a = -\frac{1}{2^{n_b}} G(2^{n_b-n_a-1}r+s)$$

Since at least one of the $s_i$ is odd, no further cancellation is possible.

If $n_a+1 = n_b$, we have:

$$q_a q_b^{-1} q_a = -\frac{1}{2^{n_b}} G(r+s)$$

Here it's possible that there will be further cancellation, depending on the details of the $r_i$ and $s_i$. However, it will turn out that we'll never need to make use of this case, because it only shows up if we move backwards through the graph.

So $n_t$, the power of 2 in the denominator of the twin's vertex, will be:

$$n_t=\begin{cases} max(n_a+1,n_b) & n_a+1 \ne n_b \\ n_b - ? & n_a+1 = n_b\end{cases}$$

In the original simplex, one of the vertices has $n=0$ as the power of 2 in its denominator, and the other four vertices have $n=2$. In its twins, we have the cases:

$$\begin{array}{cccc} n_a & n_b & q_a = q_b & n_t \\ 0 & 0 & T & 0 \\ 0 & 2 & F & 2 \\ 2 & 0 & F & 3 \\ 2 & 2 & F & 3 \\ 2 & 2 & T & 2 \end{array}$$

So the twin relative to the vertex $q_0=(1,0,0,0)$ has $n_t=0,2,2,2,2$ again, while those relative to the other vertices have $n_t=2,3,3,3,3$. But we only get $n_t=2$ when $q_a=q_b$, so as we extend the graph outwards we will never twin these twins relative to that vertex, since it would just take us back to the original simplex. This means that when we take twins of the twins, we don't encounter the case $n_b = n_a+1$, since in the first level twins we never set $q_a$ equal to the vertex with $n_t=2$.

As we take twins of twins, any sequence of powers we find starting from the twin with powers $0,2,2,2,2$ will just match one we find starting from the original simplex, so wlog we can assume we extend the graph through a twin with powers $2,3,3,3,3$. The possibilities are then:

$$\begin{array}{cccc} n_a & n_b & q_a = q_b & n_t \\ 3 & 2 & F & 4 \\ 3 & 3 & F & 4 \\ 3 & 3 & T & 3\end{array}$$

Once again, we will never take higher-level twins with the vertex we get from $q_a=q_b$, so again we avoid the case $n_b = n_a+1$. And it's clear now that the pattern will continue like this: every $n$th level twin (that we did not reach via $q_0$) will have powers $n+1,n+2,n+2,n+2,n+2$, while those that we did reach via $q_0$ will just look like $(n-1)$th level twins, with powers $n, n+1, n+1, n+1, n+1$.

If we choose a sequence $m_1, m_2, ... , m_M$ of vertices, with $m_k \ne m_{k+1}$, and apply the twinning operation repeatedly using the vertex with index $m_k$, for even $M$ we end up with:

$$q^{(m_1 m_2 ... m_M)}_i = q_{m_1} q_{m_2}^{-1} ... q_{m_M}^{-1} \: q_i \: q_{m_M}^{-1} ... q_{m_2}^{-1} q_{m_1}$$

while for odd $M$ we get:

$$q^{(m_1 m_2 ... m_M)}_i = q_{m_1} q_{m_2}^{-1} ... q_{m_M} \: q_i^{-1} \: q_{m_M} ... q_{m_2}^{-1} q_{m_1}$$

Now suppose that our graph was not a tree. Then there would be two different sequences of indexes like this, say $m_1, m_2, ... , m_M$ and $p_1, p_2, ... , p_P$ that resulted in the same simplex. We will define the rotation:

$$R(x) = q_{m_M}^{\pm 1} ... q_{m_2} q_{m_1}^{-1} \: x \: q_{m_1}^{-1} q_{m_2} ... q_{m_M}^{\pm 1}$$

where $q_{m_M}^{\pm 1} = q_{m_M}$ if $M$ is even, and $q_{m_M}^{-1}$ if $M$ is odd.

Applying $R$ to the first simplex will reduce it to either the original simplex, $q_i$, if $M$ is even, or to the twin relative to $q_0$, with vertices $q_i^{-1}$, if $M$ is odd.

If $M$ is even, applying $R$ to the second simplex will map it into the simplex that arises from the sequence $m_M, ..., m_1, p_1, ..., p_P$. If the two original sequences agree at the beginning, then $q_{m_1}^{-1}$ will cancel with $q_{p_1} = q_{m_1}$ in the string of vertex products, and so on for as long as the sequences are the same, but since the sequences are not identical, some non-empty sequence will remain.

If $M$ is odd, applying $R$ to the second simplex will map it into the inverse of the simplex that arises from the sequence $m_M, ..., m_1, p_1, ..., p_P$, again with some possible cancellations.

We then have the result that either the original simplex (if $M$ is even) or its inverse (if $M$ is odd), both of which have powers $0,2,2,2,2$, is equal to some other simplex in the graph (if $M$ is even), due to some non-empty sequence of twinning operations, or the inverse of that simplex (if $M$ is odd).

The only simplex in the graph, besides the original, with powers $0,2,2,2,2$ is the one due to the index sequence $0$: the twin of the original with respect to $q_0$. But we know this is the inverse of the original simplex, rather than being equal to it. Whether $M$ is even or odd, the claim ends up being that the original simplex and its inverse are equal, which is false. So our supposition that the graph is not a tree must be false.

Finally, we can relate this to the graph of dodecahedra as follows. Each simplex with vertices $q_i$ can be used to construct a dodecahedron, whose 20 vertices are:

$$d_{i,j} = q_i q_j^{-1} \qquad i \ne j$$

The dodecahedron constructed from a twin simplex has vertices:

$$d^{(m)}_{i,j} = q^{(m)}_i (q^{(m)}_j)^{-1} = q_m q_i^{-1} q_m (q_m q_j^{-1} q_m)^{-1} = q_m q_i^{-1} q_j q_m^{-1}$$

These share the 8 vertices where $i=m, j \ne m$ or $j=m, i \ne m$:

$$d^{(m)}_{m,j} = q_j q_m^{-1} = d_{j,m}$$ $$d^{(m)}_{i,m} = q_m q_i^{-1} = d_{m,i}$$

which makes them twin dodecahedra. So the tree of twin simplexes gives rise to a graph of twin dodecahedra.

The reason we can't immediately claim that the graph of dodecahedra is also a tree is that two non-identical simplexes can be used to construct the same dodecahedron. If we right-multiply all the vertices of a given simplex by the same unit quaternion $q_R$, rotating it into another simplex, then $q_R$ cancels out in the formula for the dodecahedron vertices.

So, could two of the simplexes in our tree be the same up to right multiplication by some quaternion $q_R$? We can show that this is impossible by exploiting the fact that all of these simplexes have their vertices in a particular subset of the golden field, $G(\mathbb{Q}^4) \subset \mathbb{Q}(\sqrt{5})^4$, in which the first coordinate is purely rational and all the other coordinates are rational multiples of $\sqrt{5}$.

Suppose we have three linearly independent, unit quaternions $q_i \in G(\mathbb{Q}^4)$, each of which is described in terms of a 4-vector of rationals, $a_i \in \mathbb{Q}^4$:

$$q_i = G(a_i) \qquad i=1,2,3$$

And suppose we pick some fourth quaternion, $p \in G(\mathbb{Q}^4)$, that we wish to map $q_1$ into by right multiplication, with $p = G(b)$ for some $b \in \mathbb{Q}^4$. The quaternion we need to right-multiply with to achieve this will be:

$$q_R = q_1^{-1} p$$

In general, $q_R$ will not belong to $G(\mathbb{Q}^4)$. The question we want to answer is: what restrictions are imposed on $p$ by requiring the images of the other $q_i$ under right multiplication:

$$q_i q_R = q_i q_1^{-1} p \qquad i=2,3$$

to lie in $G(\mathbb{Q}^4)$. Multiplying this out and setting the appropriate rational and irrational parts of the components of the product to zero, we obtain a set of eight linear equations in the rational parameters $b_j, j=0,1,2,3$:

$$\begin{array}{lcr} b_3 \left(a_{1,2} a_{2,1}-a_{1,1} a_{2,2}\right)+b_2 \left(a_{1,1} a_{2,3}-a_{1,3} a_{2,1}\right)+b_1 \left(a_{1,3} a_{2,2}-a_{1,2} a_{2,3}\right) & = & 0\\ b_3 \left(a_{1,0} a_{2,2}-a_{1,2} a_{2,0}\right)+b_2 \left(a_{1,3} a_{2,0}-a_{1,0} a_{2,3}\right)+b_0 \left(a_{1,2} a_{2,3}-a_{1,3} a_{2,2}\right) & = & 0\\ b_3 \left(a_{1,1} a_{2,0}-a_{1,0} a_{2,1}\right)+b_1 \left(a_{1,0} a_{2,3}-a_{1,3} a_{2,0}\right)+b_0 \left(a_{1,3} a_{2,1}-a_{1,1} a_{2,3}\right) & = & 0\\ b_2 \left(a_{1,0} a_{2,1}-a_{1,1} a_{2,0}\right)+b_1 \left(a_{1,2} a_{2,0}-a_{1,0} a_{2,2}\right)+b_0 \left(a_{1,1} a_{2,2}-a_{1,2} a_{2,1}\right) & = & 0\\ b_3 \left(a_{1,2} a_{3,1}-a_{1,1} a_{3,2}\right)+b_2 \left(a_{1,1} a_{3,3}-a_{1,3} a_{3,1}\right)+b_1 \left(a_{1,3} a_{3,2}-a_{1,2} a_{3,3}\right) & = & 0\\ b_3 \left(a_{1,0} a_{3,2}-a_{1,2} a_{3,0}\right)+b_2 \left(a_{1,3} a_{3,0}-a_{1,0} a_{3,3}\right)+b_0 \left(a_{1,2} a_{3,3}-a_{1,3} a_{3,2}\right) & = & 0\\ b_3 \left(a_{1,1} a_{3,0}-a_{1,0} a_{3,1}\right)+b_1 \left(a_{1,0} a_{3,3}-a_{1,3} a_{3,0}\right)+b_0 \left(a_{1,3} a_{3,1}-a_{1,1} a_{3,3}\right) & = & 0\\ b_2 \left(a_{1,0} a_{3,1}-a_{1,1} a_{3,0}\right)+b_1 \left(a_{1,2} a_{3,0}-a_{1,0} a_{3,2}\right)+b_0 \left(a_{1,1} a_{3,2}-a_{1,2} a_{3,1}\right) & = & 0\end{array}$$

This system has a 1-parameter family of solutions:

$$b_i = \lambda a_{1,i}$$

In other words, the only vectors we can map $q_1$ into by right multiplication are scalar multiples of itself, if we want the images of $q_2$ and $q_3$ to lie in $G(\mathbb{Q}^4)$. So right multiplication can't map one of the simplexes in the tree into another, and each distinct simplex belongs in a distinct equivalence class modulo right multiplication.

This means the dodecahedra created from distinct simplexes in the tree are distinct, and the graph of dodecahedra is a tree.

To prove that the 4-simplex generates a free group, suppose we have any finite product of positive or negative powers of the elements $\{q_1, q_2, q_3, q_4\}$. By inserting the element 1 in the form $q_0$ or $q_0^{-1}$ where necessary, we can write this product as:

$$p=q_{m_1} q_{m_2}^{-1} q_{m_3} ... q_{m_M}^{\pm 1}$$

for some sequence $m_1, m_2, ... , m_M$ with $m_k \ne m_{k+1}$. We then have the vertices of the dodecahedron constructed from the 4-simplex associated with this sequence being, for even $M$:

$$d^{(m_1 m_2 ... m_M)}_{i,j} = p q_i q_j^{-1} p^{-1}$$

and for odd $M$:

$$d^{(m_1 m_2 ... m_M)}_{i,j} = p q_i^{-1} q_j p^{-1}$$

If $p=1$ for some non-empty sequence, then this dodecahedron would also appear elsewhere in the tree, either as the dodecahedron constructed from the original simplex, or the one constructed from its twin relative to $q_0$. But this is impossible, so $p \ne 1$.


Have you seen the preprint by Adrian Ocneanu? It has been available since May 13; he has these proofs worked out in full detail. See http://arxiv.org/abs/1505.03248.


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