97
$\begingroup$

This is perhaps a vague question, but hopefully there exists literature on the subject. The question is motivated by an answer I gave to this question. I also asked a related question on MO, although hopefully this question should be easier.

There exists a rather remarkable relationship between the 5 platonic solids, and the factor groups of the $n$-string braid groups $B_n$ by adjoining the relation $\sigma_i^k=1$ (here the $\sigma_i$, $1\leq i\leq n-1$ are the usual generators of $B_n$). We'll call these groups $B_n(k)$ the truncated braid groups of type $(n,k)$ where $B_n(k)=B_n/\langle \sigma_i^k\rangle$.

Theorem *: For $n\geq 3$, the group $B_n(k)$ is finite if and only if $k=2$ or $(n,k)$ is the Schläfli type of one of the 5 platonic solids. For these cases, $$|B_n(k)|=\left(\frac{f(n,k)}{2}\right)^{n-1}n!$$ where $f(n,k)$ is the number of faces of the platonic solid of type $(n,k).$

The 5 platonic solids correspond to the pairs $(n,k)\in\{(3,3),(3,4),(4,3),(3,5),(5,3)\}$. This is equivalent to the pair $(n,k)$ being a solution to the inequality $$\frac{1}{n}+\frac{1}{k}>\frac{1}{2}.$$

*It appears that this theorem was proved by Coxeter in

H. S. M. Coxeter, Factor groups of the braid group, Proceedings of the Fourth Can. Math. Cong., Banff 1957, University of Toronto Press (1959), 95–122.

although it is proving difficult for me to find a copy of this online or in my institution's library. From what I can gather, the proof is rather algebraic/combinatorial, although without access to a copy I can't say for sure. My question is:

Question Can one view the finite truncated braid groups in a geometric way as some action (in the vague, not necessarily strict group-action, sense) on the corresponding platonic solid or related objects?

Some of the approaches I have taken have included:

  • considering the isometry group on the corresponding solid (group orders don't match up),
  • considering paths on the surface which 'remember' the side that a face was entered from (relations don't match up),
  • considering labellings of the edges of the faces of the solid so that no face has a pair of edges with the same label, with group elements being permutations of edges which preserve this property (not sure if order matches up - seems difficult to calculate but may be the best approach so far),
  • considering 'rolling' the solid along a surface until it reaches its starting point again (no obvious way of making this a group via some kind of homotopy)

It seems that there should be some nice geometric interpretation of these groups, especially as every face has $n$ edges and every vertex is shared by $k$ faces.

It would be especially nice if the usual generators can be realised in a nice geometric way. Ultimately it would be nice to extend such a geometric interpretation of these finite truncated braid groups to the infinite cases as well (which correspond to regular tilings of the hyperbolic plane in most cases, and the complex plane in the case $(n,k)\in\{(3,6),(4,4),(6,3)\}$).

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    $\begingroup$ This is a really interesting problem. I know it has been a long time since you posted this, but I found this. Perhaps it has what you were looking for. $\endgroup$ – N. Owad Feb 12 '14 at 18:07
  • $\begingroup$ For n=3 there's definitely a link to the Platonic solids; I can probably give an answer tying it together on a representation-theoretic level. For n>3 you might be able to find a link but it's probably going to require you do something like cut out a particular slice; consider k=2, where the resulting group is the good old symmetric group, and the dihedral symmetry that comes from the corresponding "solid" is much smaller than the full symmetric group. $\endgroup$ – W. Cadegan-Schlieper Sep 29 '15 at 17:52
  • $\begingroup$ Here is a minor observation -- for $k$ even, there is a surjection $B(n,k) \to S_n$ sending $\sigma_i$ to $(i \ i+1)$. So maybe we should ask why the kernel of this map (the "pure truncated praid group") has size $\left( \tfrac{f(n,k)}{2} \right)^{n-1}$ $\endgroup$ – David E Speyer Jun 2 '20 at 12:04
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+250
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An idea to search for a structural answer would be as follows. The five groups are complex reflection groups, i.e. generated by reflection symmetries $s$ in a finite dimensional vector space $V$ over $\Bbb C$. So we may (want to effectively) realize each of these five truncated braid groups geometrically as a group $G(n,p)$ acting on some vector space $V$ over $\Bbb C$. Each reflection symmetry has an $1$-eigenspace of codimension one. (By definition.) As in the real case, we obtain a separation of the space $V$ by an arrangement $A$ of hyperplanes, and the symmetry is captured by a regular complex polytope. Regular complex polytopes were introduced by Shephard in 1950 (in an article with the same name), and some years later (1973) we also have a book (with the same name, well,) written by H.S.M Coxeter. I will refer to this book as [Coxeter1973]. The front cover of this book is a picture due to McMullen of the four dimensional complex polytope with $240$ vertices and $2160$ "edges" associated to the most complicated group among our five, the one of type $$ \underset 3\circ\overset {\color{blue}3}{-\!\!-\!\!-}\! \underset 3\circ\overset {\color{blue}3}{-\!\!-\!\!-}\! \underset 3\circ\overset {\color{blue}3}{-\!\!-\!\!-}\! \underset 3 \circ\ , $$ written symbolically as $ 3{\color{blue}{[3]}} 3{\color{blue}{[3]}} 3{\color{blue}{[3]}} 3$. (There is a convention to omit the $\color{blue}3$ upper decoration for "joined generators" $s,t$, which then defaults to thebraid relation $sts=tst$, but let us show this $\color{blue}3$ to suggest that it becomes a $\color{blue}{[3]}$ when writing liniearly.) Chapter 12 of the book is describes the groups and the polytopes for the types

  • $3\color{blue}{[3]}3$,
  • $3\color{blue}{[3]}3\color{blue}{[3]}3$, and
  • $3\color{blue}{[3]}3\color{blue}{[3]}3\color{blue}{[3]}3$.

Now one may try to exhibit starting from $G(n,p)$ acting on $V$ also an action of the "associated polyhedral symmetry group" $Q(n,p)$ on some real vector space $V'$, so that we may establish a commutative diagram of the shape: $$ \require{AMScd} \begin{CD} 1 \\ @VVV \\ K(n,p) \\ @VVV \\ \color{red}{G(n,p)} @>>> \operatorname{U}(n)\\ @VVV @VVV \\ \color{red}{Q(n,p)} @>>> \operatorname{SO}(n)\\ @VVV \\ 1 \end{CD} $$ Here, $K(n,p)$ should be the kernel of the surjective group homomorphism $\color{red}{G(n,p)\to Q(n,p)}$. In fact, for the purposes of the question in the OP, for the construction of a polyhedral action of $G(n,p)$, it is enough to have such a homomorphism only, realizing the corresponding group of polyhedral symmetries as a quotient group $Q(n,p)$ of $G(n,p)$.

Our point of view here is to construct this map at the level of presentations. In the cases $G(3,p)\to Q(3,p)$ the kernel is the center of $G(3,p)$, i.e. $K(3,p)=ZG(3,p)$, we have an explicit description of the center as being cyclic, generated by an element $z$ of order $\frac 12 V$, and using $z$ the presentations for the two groups $G(3,p)$ and $Q(3,p)$ are related.

For the setting, the framework of the OP, there are tons of related material in the mentioned book [Coxeter1973]. For instance:

  • Chapter 13 considers regular complex polytopes and their symmetry groups. The combinatorial structure is revealed. It contains in 13.44 a simpler presentations of the finite groups of the shape $p_1[q_1]p_2[q_2]\dots$ (in case this leads to a finite group). After 13.44, the former article [Coxeter1957] is cited (page 148).
  • §7.2 is dedicated to the binary tetrahedral group, related to $G(3,3)$. Space symmetries can be described using quaternions, an explicit representation inside the quaternion algebra is given.
  • §7.3 is dedicated to the binary octahedral group, related to $G(3,4)$. Quaternion realization.
  • §7.4 is dedicated to the binary icosahedral group, related to $G(3,5)$.
  • §13.8 gives a (simple) presentation of the simple group of order $25920$. This is related to a "rigidity property" of the group $G(5,3)$.

I am trying to give an answer that follows the spirit of the question, forcing most of the dream to become true. First, i will start some discussion on the metamathematics related by explaining a combinatorial formula joining numbers as an identiy joining structural objects, followed by counting. Yes, it is often a human wish to prove combinatorial identities (e.g. ones where on the LHS and on the RHS we have some quantities joined by addition and product) by arranging them in a given set theoretical setting, then showing that two sets are in a (more or less canonical bijection). The sets involved may be obtained in pieces by specific construction in a more or less functorial way. (Then we would like to translate operations appearing in the LHS and RHS of the identity in terms of such functorial constructions, so an addition would correspond after counting to a disjoint sum, and a product after a cartesian product.) Often, the combinatorial relation has to be reshaped for this process.

In our case, in the link posted by N. Owad, here in a separated line:

Coxeter1957 :: H.S.M. Coxeter, Factor Groups of the Braid Groups, Two lectures delivered at Banff, September 5 and 6, 1957

gives the formula for the cardinality of the five finite groups, that appears as quotients of the braid groups as, compared with the cardinality of the symmetric groups...

... we find (in §12) that the order is changed from $n!$ to $$\left(\frac 12 V\right)^{n-1}\cdot n!\ , $$ where $V$ is the number of faces of the regular polyhedron or tesselation $\{p,n\}$.

Now the question is if this handy formula, managed to work for five special cases, is in the "right combinatorial form". We would like to see then inside the category of groups a corresponding construction delivering the above formula. So we would need to search for a group of order $n!$ that can be brought in relation to the five quotients of the braid groups. I need now a handy notation for the five groups, and since i could not find a handy one, let us use $G(n,k)$ with $n,k$ as in the OP, so using "G" also as a short cut for "truncated braid group"... $$ \begin{aligned} G(3,3) &=\langle \ s,t\ :\ sts=tst\ ,\ s^3=t^3=1\ \rangle\\ &=G\left(\underset 3\circ\overset 3{-\!\!-\!\!-}\!\underset 3\circ\right) =G(\ 3[3]3\ ) \ , \\[2mm] G(3,4) &=\langle \ s,t\ :\ sts=tst\ ,\ s^4=t^4=1\ \rangle\\ &=G\left(\underset 4\circ\overset 3{-\!\!-\!\!-}\!\underset 4\circ\right) =G(\ 4[3]4\ ) \ , \\[2mm] G(3,5) &=\langle \ s,t\ :\ sts=tst\ ,\ s^5=t^5=1\ \rangle\\ &=G\left(\underset 5\circ\overset 3{-\!\!-\!\!-}\!\underset 5\circ\right) =G(\ 5[3]5\ )\ , \\[2mm] G(4,3) &=\langle \ s,t,u\ :\ sts=tst\ ,\ tut=utu\ ,\ \ s\leftrightarrows u\ ,\ s^3=t^3=u^3=1\ \rangle\\ &=G \left( \underset 3\circ\overset 3{-\!\!-\!\!-}\!\underset 3\circ\overset 3{-\!\!-\!\!-}\!\underset 3 \circ\right) =G(\ 3[3]3[3]3\ )\ , \\[2mm] G(5,3) &=\langle \ s,t,u,v\ :\ sts=tst\ ,\ tut=utu\ ,\ uvu = vuv\ , \\ &\qquad\qquad \ s\leftrightarrows u\ ,\ \ s\leftrightarrows v\ ,\ \ t\leftrightarrows v\ ,\ \\ &\qquad\qquad \ s^3=t^3=u^3=v^3=1\ \rangle\\ &=G \left( \underset 3\circ\overset 3{-\!\!-\!\!-}\! \underset 3\circ\overset 3{-\!\!-\!\!-}\! \underset 3\circ\overset 3{-\!\!-\!\!-}\! \underset 3 \circ\right) =G(\ 3[3]3[3]3[3]\ )\ . \end{aligned} $$ (I am using $s,t,u,v$ instead of $R_1,R_2,R_3,R_4$. It makes typing simpler)


Let us start with the simplest group, $G(3,3)$. In [Coxeter1957], §3, page 99, Coxeter, citing Moser, already gives the structure of $G(3,3)$ as a binary tetrahedral group $2T=\langle 2,3,3\rangle$ with $24$ elements. It has as quotient the tetrahedral group $T=(2,3,3)$, and there is a short exact sequence: $$ 1\longrightarrow \pm1\longrightarrow 2T\longrightarrow T\longrightarrow 1\ . $$ So we have an action of $2T\cong G(3,3)$, factorizing via the quotient morphism $2T\to T$, on the tetrahedron. This simplest example shows that we may rewrite the formula for $|G(3,3)|$ in the form $$ |G(3,3)|=24=\underbrace{2}_{=|ZG(3,3)|}\cdot \underbrace{|T|}_{=|A_4|=4!/2} $$ to connect it with a polyhedral group, here $A_4$, which corresponds to the orientation preserving symmetries of the tetrahedron, the polyhedron with Schläfli symbol $\{3,3\}$. Here, $K(3,3)=ZG(3,3)$ is the center of $G(3,3)$, a group with two elements.

The wiki page mentions the following situation: $$ \require{AMScd} \begin{CD} 1 \\ @VVV \\ K(n,p) \\ @VVV \\ \color{red}{G(n,p)} @>>> \operatorname{Spin}(3)\\ @VVV @VVV \\ \color{red}{Q(n,p)} @>>> \operatorname{SO}(3)\\ @VVV \\ 1 \end{CD} $$

Can we do "the same" for the other groups?


Related to the above question, there is also an observation in [Coxeter1957] regarding $G(3,p)$, $p=3,4,5$. Let $R$ be the element $$R=st\in G(3,p)\ .$$ Then $R^3=(Rs)^2$. Because of $R^3=st\; st\;st=sts\; tst=sts\;sts=(sts)^2=(Rs)^2$. The observation is, citing loc. cit.:

The element $R^3=(Rs)^2$ of $G(3,p)=G(p[3]p)$ generates the center, whose quotient group: $$\langle \ s,R\ :\ s^p=R^3=(Rs)^2=1\ \rangle$$ is polyhedral. This central quotient group is of order $\displaystyle\frac{12p}{6-p}$ if $p=2,3,4$, or $5$.

So in these three cases we obtain an action of $G(3,p)$ via a quotient group.

Let us see why $z:=R^3=(Rs)^2$ generates the center of $G(3,p)$, $p=3,4,5$, and compute its order. It commutes with $Rs$, and with $R$, so also with $s$, so also with $t$ (because of $R=st$.)

For $p=3,4,5$ we have $z=(st)^3=ststst=s\;sts\;st=s^2ts^2t$, and similarly $z=t^2st^2s$. On the other side, $z=(st)^3=ststst=st\;tst\;t=st^2st^2$, and similarly $z=ts^2ts^2$.

  • Then in case of $p=3$, $$z^2=(s^2ts^2t)(t^2st^2s)=s^2(t(s^2(tt^2)s)t^2)s=1\ ,$$ using $s^3=t^3=1$.

  • For $p=4$ we have $s^4=t^4=1$, and as above $z=s^2ts^2t=ts^2ts^2$, and we need a longer dance around the fire to show $z^4=1$. $$ \begin{aligned} z^2 &=(ts^2ts^2)(s^2ts^2t)\\ &=ts^2t^2s^2t \ ,\text{ and similarly}\\ z^2 &= st^2s^2t^2s\ ,\\ z^4 &=(ts^2t^2s^2t)(st^2s^2t^2s)\\ &=ts^2t^2s^2\; tst\; ts^2t^2s\\ &=ts^2t^2s^2\; sts\; ts^2t^2s\\ &=ts^2t^2s^3\; tst\; s^2t^2s\\ &=ts^2t^2s^3\; sts\; s^2t^2s\\ &=ts^2t^2s^4\; t\;s^3t^2s\\ &=ts^2t^2\; t\;s^3t^2s\\ &=ts^2t^3\;s^3t^2s\\ &=ts^{-1}\; s^{-1}t^{-1}s^{-1}\; t^2s\\ &=ts^{-1}\; t^{-1}s^{-1}t^{-1}\; t^2s\\ &=t\;s^{-1}t^{-1}s^{-1}\; ts\\ &=t\;t^{-1}s^{-1}t^{-1}\; ts\\ &=1\ . \end{aligned} $$

  • For $p=5$ we delegate the work to sage. (It is a matter of time, i have to submit and switch to the real life job.)

Let us make a table of the data for the five groups: $$ \begin{array}{|c|c|c|c|c|} \hline (n,p) & |G(n,p)| & \text{ST} & \text{struct} & |ZG(n,p)| & K & Q(n,p) & |Q(n,p)|\\\hline (3,3) & 24 = 2^2\cdot 3! & G_4 & 2T\cong\operatorname{SL}(2,3) & 2 & Z(3,3) =\langle z\rangle & T=(2,3,3)=A_4 & 12\\\hline (3,4) & 96 = 4^2\cdot 3! & G_8 & C_4\rtimes\operatorname{SL}(2,3) & 4 & Z(3,4) =\langle z\rangle & O=(2,3,4)=S_4 & 24\\\hline (3,5) & 600 = 10^2\cdot 3! & G_{16} & 2I\times C_5=\operatorname{SL}(2,5)\times C_5 & 10 & Z(3,5) =\langle z\rangle & I=(2,3,5)=A_5 & 60\\\hline (4,3) & 648 = 3^3\cdot 4! & G_{25} & \text{see wiki link of ST-types} & 3 & H_{27} =C_3\rtimes(C_3\times C_3) & \text{not }C=O=(2,3,4)=S_4 & 24\\\hline (5,3) & 155520 = 6^4\cdot 5! & G_{32} & \text{see wiki link of ST-types} & 6 & - & \text{not }D=I=(2,3,5)=A_5 & 60 & \\\hline \end{array} $$ The column ST shows the Shephard-Todd number of the corresponding complex reflexion group, wiki :: Complex reflexion groups, see also the table List of complex reflexion groups inside this wiki link.

The cube and the octahedron are dual, so the corresponding symmetry groups $C$ and $O$ coincide.

The dodecahedron and the icosahedron are dual, so the corresponding symmetry groups $D$ and $I$ coincide.

The symmetry groups of the regular polyhedra above are $$ \begin{aligned} T &=(2,3,3)=\langle\ a,b,c\ :\ a^2=b^3=c^3=1\ ,\ abc=1\ \rangle=\langle\ a,b\ :\ a^2=b^3=(ab)^3=1\ \rangle\ ,\\ O &=(2,3,4)=\langle\ a,b,c\ :\ a^2=b^3=c^4=1\ ,\ abc=1\ \rangle=\langle\ a,b\ :\ a^2=b^3=(ab)^4=1\ \rangle\ ,\\ I &=(2,3,5)=\langle\ a,b,c\ :\ a^2=b^3=c^5=1\ ,\ abc=1\ \rangle=\langle\ a,b\ :\ a^2=b^3=(ab)^5=1\ \rangle\ . \end{aligned} $$ (Adding one more relation coming from the central element to the $(s,t)$ group presentation we can exhibit in the new presented group the elements $a,b,c$. This is done implicitly in many cases in [Coxeter1973]. Alternatively, a computer search for elements $a,b,c$ in the groups $T,O,I$, and of lifts shows which choices of words for $a,b,c$ in terms of $s,t$ are simplest.)

In the case of the group of order $648$ we do not find a faithful action on the cube. In fact, the group $G(4,3)=G_{25}$ has exactly one normal group of order $27$, but the quotient has the structure $\operatorname{SL}(2,3)$, which is not the expected structure $S_4$. See also the structure in the wiki list of the Shephard-Todd types.

In the case of the group of order $155520$ we have a too big situation for the computation. It turns out that $R=stuv$ is an element of order $30$ in $G(5,3)$, and $R^5$ generates the center. The structure of $G(5,3)$ is described in the list of the Shephard-Todd types as $G_{32}\cong W(L_4)=C_3\times \operatorname{Sp}_4(3)$. The number $155520/6$ is $25920$, and this is the order of the simple group $S_{25920}$ say, which is $G(5,3)/C_6$, [Coxeter1973], (13.83), page 155. $$ 1\to C_6 \to G(5,3) \to S_{25920}\to 1\ . $$ It is this major rigidity that makes it impossible to exhibit a quotient of order $60$, or to have related constructions joining this truncated braid group with the symmetry group of the dodecahedron or icosahedron.

The above table is supported by sage code delivered in the sequel. Sage is a CAS for all mathematical purposes, done to respect mathematical structure and thinking. The commands are close to the mathematical writing, so i hope the code is readable also for mathematicians who never saw a programming language before. The (pythonic) list comprehension in a line like [ H for H in G.normal_subgroups() if H.order() == 27 ] is literally almost the same as the meaning $\{\ H\ :\ H\triangleleft G\text{ and }|H|=27\ \}$.

A final observation (already done): There is a coincidence of the order of the center with the expression $\frac 12V$ from the OP, half of the number of faces of the associated polyhedron of Schläfli type $\{n,p\}$. The center is related to the element $R=st\dots$, a suitable power of this element generates the center. The (groupal) combinatorics of the situations is for the five truncated braid groups reach, but this may be an important open point that needs structural support.


Note: Searching for information to prepare the above, (well, i am far, far away from home,) i also noticed some sources that may be interesting in the context.

  • Sage also provides support for the Assion groups. The commands AssionGroupS and AssionGroupU are available.
  • Assion groups are introduced and studied in [Joachim Assion, A proof of a theorem of Coxeter].
  • Truncated braid groups occur also in physics. See for instance the article in [Inside Particles and Fields, Discrete Gauge Theories, Mark de Wild Propitius and F Alexander Bais].

Sage code supporting the computations for $G(3,3)=G(3[3]3)$:

The group and the corresponding elements are initialized by:

F.<S,T> = FreeGroup()
G333 = F / [S^3, T^3, S*T*S*(T*S*T)^-1]
R = G333(S*T)
s, t = G333(S), G333(T)
z = R^3

Then we can ask in sage for:

sage: G333.is_finite()
True
sage: G333.order()
24
sage: G333.structure_description()
'SL(2,3)'
sage: G333.center()
Group([ T*S^2*T^-2*S^-1 ])
sage: G333.center().order()
2
sage: z.order()
2

And the quotient $Q(3,3)$ of $G(3,3)$ w.r.t. $\langle z\rangle$ is:

sage: QG333 = F / [S^3, T^3, S*T*S*(T*S*T)^-1, (S*T)^3]
sage: QG333.structure_description()
'A4'

This shows the claimed structure for $G(3,3)$.

Note that we can also introduce this group as cubic braid group in sage. For instance, alternatively:

sage: CB3 = CubicBraidGroup(3)
sage: CB3
Cubic Braid group on 3 strands
sage: CB3.structure_description()
'SL(2,3)'
sage: CB3.gens()
(c0, c1)
sage: CB3.inject_variables()
Defining c0, c1
sage: CB3.relations()
(c0*c1*c0*c1^-1*c0^-1*c1^-1, c0^3, c1^3)
sage: c0^3 == CB3(1), c1^3 == CB3(1), c0*c1*c0 == c1*c0*c1
(True, True, True)
sage: R = c0*c1
sage: z = R^3
sage: z in C3.center()
True
sage: z.order()
2

This will be the way to construct $G(4,3)$ and $G(5,3)$.


Sage code supporting the computations for $G(4,3)=G(4[3]4)$:

The group and the corresponding elements are initialized by:

F.<S,T> = FreeGroup()
G434 = F / [S^4, T^4, S*T*S*(T*S*T)^-1]
R = G434(S*T)
s, t = G434(S), G434(T)
z = R^3

Then we can ask in sage for:

sage: G434.is_finite()
True
sage: G434.order()
96
sage: G434.structure_description()
'SL(2,3) : C4'
sage: G434.center().order()
4
sage: z.order()
4

And the quotient $Q(4,3)$ of $G(4,3)$ w.r.t. $\langle z\rangle$ is:

sage: QG434 = F / [S^4, T^4, S*T*S*(T*S*T)^-1, (S*T)^3]
sage: QG434.structure_description()
'S4'

This shows the claimed structure for $G(4,3)$.


Sage code supporting the computations for $G(5,3)=G(5[3]5)$:

The group and the corresponding elements are initialized by:

F.<S,T> = FreeGroup()
G535 = F / [S^5, T^5, S*T*S*(T*S*T)^-1]
R = G535(S*T)
s, t = G535(S), G535(T)
z = R^3

Then we can ask in sage for:

sage: G535.is_finite()
True
sage: G535.order()
600
sage: G535.structure_description()
'C5 x SL(2,5)'
sage: G535.center()
Group([ T^3*S^-1*T^-2*S^-1, S*T^2*S^3*T*S^-3*T^-2, T*S^2*T^3*S^3*T*S^-3*T^-2*S^-2 ])
sage: G535.center().order()
10
sage: z.order()
10
sage: ZG535 = G535.center()
sage: z in ZG535
True

And the quotient $Q(5,3)$ of $G(5,3)$ w.r.t. $\langle z\rangle$ is:

sage: QG535 = F / [S^5, T^5, S*T*S*(T*S*T)^-1, (S*T)^3]
sage: QG535.order()
60
sage: QG535.structure_description()
'A5'

This shows the claimed structure for $G(5,3)$.


Sage code supporting the analysis of the case $G(4,3)=G(3[3]3[3]3)$:

The group and the corresponding elements are initialized by:

sage: CB4 = CubicBraidGroup(4)
sage: CB4.inject_variables()
Defining c0, c1, c2
sage: s, t, u = CB4.gens()
sage: R = s*t*u
sage: R.order()
12
sage: CB4.center().order()
3
sage: R^4 in CB4.center()
True

We need to work with the associated permutation group to have more available methods.

sage: PCB4 = CB4.as_permutation_group()
sage: PCB4.center().order()
3

sage: for H in PCB4.normal_subgroups():
....:     print(H.order(), H.structure_description())
....:
648 (((C3 x C3) : C3) : Q8) : C3
216 ((C3 x C3) : C3) : Q8
54 ((C3 x C3) : C3) : C2
27 (C3 x C3) : C3
3 C3
1 1

sage: z = PCB4.center().gens()[0]
sage: H27 = [ H for H in PCB4.normal_subgroups() if H.order() == 27 ][0]
sage: z in H27
True
sage: H27.structure_description()
'(C3 x C3) : C3'

sage: QPCB4 = PCB4.quotient( H27 )
sage: QPCB4.structure_description()
'SL(2,3)'

Sage code supporting the analysis of the case $G(5,3)=G(3[3]3[3]3[3]3)$:

The group and the corresponding elements are initialized by:

CB5 = CubicBraidGroup(5)
CB5.inject_variables()
s, t, u, v = CB5.gens()
R = s*t*u*v
print("R = stuv has order {}".format(R.order()) )
print("G(5,3) has order {} = {}".format(CB5.order(), CB5.order().factor()))
print("The center ZG(5,3) has order {}".format(CB5.center().order()))
print("Is R^5 in the center ZG(5,3)? {}".format( R^5 in CB5.center() ))

And we obtain:

Defining c0, c1, c2, c3
R = stuv has order 30
G(5,3) has order 155520 = 2^7 * 3^5 * 5
The center ZG(5,3) has order 6
Is R^5 in the center ZG(5,3)? True
$\endgroup$
6
  • $\begingroup$ This answer is a true tour de force and your efforts are very much appreciated. This question has received a surprising amount of attention over the years with respect to bounties that various people have offered, and so it nice to see that at least one of those bounties has attracted further attention in the form of a detailed, well-researched and carefully-written answer. $\endgroup$ – Dan Rust Jun 3 '20 at 12:17
  • $\begingroup$ A cursory read looks like the question isn't perhaps fully answered, but it is maybe at least answered as well as it reasonably could be, given the fairly loose parameters with which it was asked. When I have the time, I will definitely give your answer the undivided and detailed attention that it deserves $\endgroup$ – Dan Rust Jun 3 '20 at 12:17
  • $\begingroup$ @DanRust Thanks for reading the answer, this is certainly a hard experience, i was facing a concrete, very rich structure, and i'm afraid the structure of the answer and the ingrediends are not optimally presented. I'm coming from some project on Hecke algebras, reflection groups, Weyl groups, etc., it was the way i found this post. After only five minutes i was convinced that the combinatorial quantity $$(V/2)^{n-1}$$ has a deep structural meaning, two days i worked hard to discover some new connection. All i have so far is the connection of $(V/2)$ with the center of the truncated braid... $\endgroup$ – dan_fulea Jun 3 '20 at 15:21
  • $\begingroup$ ... group. Reading [Coxeter1973], i am almost convinced that the same question was also leading the research on the topic, which finally gives geometrical shapes for the symmetry groups, on the geometrical part, and connections between the presentations of groups on the algebraic part. Citing [Coxeter1973], page 140, the start and preface of Chapter 13... <<Theorem (13.44) is the climax, for which most of the earlier chapters were preparing the way so as to make it seem natural and inevitable. However it is only 'verified' by consideration of separate cases. Perhaps some reader...>> This... $\endgroup$ – dan_fulea Jun 3 '20 at 15:29
  • $\begingroup$ ... is definitively a late conclusion after a decade of search for the "unifying structure". I have the same feeling when reading this OP, when reading the book. I tried to give a group theoretical meaning for $(V/2)$, best in the form of $(V/2)^{n-1}$ on the one side, and $n!$ on the other side. But the $n!$ is the "problem" somehow, there is not always a symmetric (permutation) group related to the symmetries of the regular polyhedra. The question found me unprepared, i had no gap3+chevie on the box - but i may well install them next days and start experimenting... $\endgroup$ – dan_fulea Jun 3 '20 at 15:35
2
$\begingroup$

Here is a suggestion towards a geometric interpretation of the group $B_3/(\sigma_1^k)$ with $3\le k\le 5$.

First note that $B_3$ is generated by $\sigma_1$ and $x=\sigma_1\sigma_2$. The braid relation is equivalent to $(\sigma_1x)^2=x^3$, and $\Delta=x^3$ is the Garside element. It generates the center of $B_3$, but I will not use this fact.

Let $K$ be the set of oriented edges of the platonic solid. Each such edge $e$ determines two vertices (source $s(e)$ and target $t(e)$) and two faces ($l(e)$ on the left of $e$ and $r(e)$ on the right of $e$). For any oriented edge e, let $\sigma_1\cdot e$ and $x\cdot e$ be the oriented edges with $$ s(\sigma_1\cdot e)=s(e),\quad r(\sigma_1\cdot e)=l(e) $$ and $$ s(x\cdot e)=t(e),\quad l(x\cdot e)=l(e). $$ It follows that $$ x^3\cdot e=e,\quad \sigma_1\cdot (x\cdot (\sigma_1\cdot (x\cdot e)))=e, $$ and $k$-fold action of $\sigma_1$ on $e$ gives $e$ for each oriented edge $e$. It is not very difficult to show that this defines a free and transitive action of $B_3/(\sigma_1^k,\Delta)$ on $K$.

Since $\Delta $ is central in $B_3$, we may extend this action to an action of $B_3/(\sigma_1^k)$ on $K\times C_N$ with $N$ the order of $\Delta $ in $B_3/(\sigma_1^k)$ and $C_N$ the cyclic group of order $N$. Note that $x^3=\Delta $. In order to implement this action, let us fix a set $L$ of oriented edges such that each face is adjacent to exactly one edge in $L$ in
counterclockwise orientation. (There may be other edges in $L$ adjacent to the face in clockwise orientation.) Moreover, we add to each oriented edge an integer $l'(e)$. Now we define $$ \sigma_1\cdot (e,m)=(\sigma_1\cdot e,m),\quad x\cdot (e,m)=(x\cdot e,m+l'(e)+\varepsilon_e),$$ where $\varepsilon_e\in \{0,1\}$, and $$ \varepsilon_e=1 \quad \Leftrightarrow e\in L. $$ It turns out that this defines an action of $B_3/(\sigma_1^k)$ on $K\times C_N$ if and only if for each face $F$, $$ \sum_{e\in K,l(e)=F}l'(e)=0 $$ and for each oriented edge $e$, $$ l'(e)+l'(-e)+\varepsilon_e+\varepsilon_{-e}=1. $$ Here $-e$ is the edge $e$ with opposite orientation. Such a labeling exists if and only if $N|f/2$ (with $f$ the number of faces).

It follows that $N=f/2$ and $$ |B_3/(\sigma_1^k)|=\frac f2|K|=6\left(\frac f2\right)^2 $$ since $|K|=3f$.

$\endgroup$

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