Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Allow me to take advantage of your collective scholarliness...

The symmetric group $\mathbb S_n$ can be presented, as we all know, as the group freely generated by letters $\sigma_1,\dots,\sigma_{n-1}$ subject to relations $$ \begin{aligned} &\sigma_i\sigma_j=\sigma_j\sigma_i, && 1\leq i,j<n, |i-j|>1;\\\\ &\sigma_i\sigma_j\sigma_i=\sigma_j\sigma_i\sigma_j, &&1\leq i,j<n, |i-j|=1; \\\\ &\sigma_i^2=1, && 1\leq i<n \end{aligned} $$ If we drop the last group of relations, declaring that the $\sigma_i$'s are involutions, we get the braid group $\mathbb B_n$. Now suppose I add to $\mathbb B_n$ the relations $$ \begin{aligned} &\sigma_i^3=1, && 1\leq i<n \end{aligned} $$ and call the resulting group $\mathbb T_n$.

  • This very natural group has probably shown up in the literature. Can you provide references to such appearances?
  • In particular, is $\mathbb T_n$ finite?
share|improve this question
(One can ask exactly the same last question for the quotients of all Artin braid groups corresponding to finite type Coxeter matrices by the cubes of the simple reflections... I am not greedy) –  Mariano Suárez-Alvarez Dec 9 '10 at 23:36
for n=3, this is the binary tetrahedral group of order 24, a split extension of the quaternion group of order 8 by a cyclic group of order 3 –  Matthew Towers Dec 9 '10 at 23:40
For $n=4$, the group is an extension of the simple group $O(5,3)$ of order $25920$ by a cyclic group of order $6$. –  Mariano Suárez-Alvarez Dec 10 '10 at 0:02
...and for $n=5$ GAP's coset enumeration seems not to stop. –  Mariano Suárez-Alvarez Dec 10 '10 at 0:12
(My two last comments use off-by-one indexing...) For $n=6$, GAP gives up after using 3GiB to build the coset table. –  Mariano Suárez-Alvarez Dec 10 '10 at 0:51

3 Answers 3

up vote 40 down vote accepted

Following up what was mentioned in the comments for $n$ up to $5$. In "Factor groups of the braid group" Coxeter showed that the quotient of the Braid group by the normal closure of the subgroup generated by $\{\sigma_i^k \ | \ 1\le i\le n-1\}$ is finite if and only if $$\frac{1}{n}+\frac{1}{k}>\frac{1}{2}$$ In your case ($k=3$) this translates to this group being infinite for $n\geq 6$.

P.S. For the same question on Artin braid groups one can use the classification of finite complex reflection groups. See for example the first reference there, "On complex reflection groups and their associated braid groups" by Broué, Malle and Rouquier.

share|improve this answer
Which seems to answer the second question. –  Theo Johnson-Freyd Dec 10 '10 at 1:33
Aha! That is precisely what I wanted to hear! :D –  Mariano Suárez-Alvarez Dec 10 '10 at 1:45

There is a way to understand these groups geometrically, as complex reflection groups, which are groups generated by rotations around complex hyperplanes. I believe (but without actually looking up the paper cited by Gjergji Zaimi) that this is what Coxeter was doing.

Let's start with the tetrahedron. Let $A$ be rotation by 120 degrees about one vertex, and $B$ be rotation by 120 degrees about another. $A$ and $B$ both act as even permutations of the vertices, so $AB$ also acts as an even permutation of the vertices; it is a 3-cycle, so they satisfy the relation $(AB)^3 = 1$. As noted by mt in comments, this is not $\mathbb T^3$, but just the quotient by its center. The group $SO(2)$ of orientation preserving isometries of $S^2$ is the quotient of the group of unit quaternions ($S^3$) by its center, and the preimage of the tetrahedral group is the binary tetrahedral group, where the relation is also satisfied. You can think of this as keeping track of how many times you've rotated the tetrahedron, mod 2. The group $S^3$ is the same as $SU(2)$. The 2 lifts of a 120 degree rotations to $S^3$ look like $60$ and $240$ degree rotations about great circles. These great circles are the intersections of complex lines with the unit sphere in $\mathbb C^2$, and the operations are what are called complex reflections. 120 degree rotations about hyperplanes in any dimension making the same angle as these satisfy the braid relation.

In general, to realize the $\mathbb T^n$ geometrically, you need $n-1$ complex hyperplanes $P_i$ corresponding to your transformations $T_i$, so that when $|i-j] > 1$ they are orthogonal, and otherwise they make the same angle as above. To make this work, construct a Hermitian form ($\leftrightarrow$ metric compatible with the complex structure) so normal vectors to these planes have the specified angles. This is a standard process. This won't be a positive definite form in general --- it'll only be positive definite when the group is finite. But, it still has a geometric interpretation. In particular, when there is only one negative direction, it can be interpreted as a complex reflection group acting on complex hyperbolic space.

Quite a lot is known about complex reflection groups, but I'm not very familiar with the literature so I won't try to summarize. I'll just mention a paper of mine Shapes of polyhedra and triangulations of the sphere, in which you can find an interpretation of $\mathbb T^n$ for $n \le 12$ as the modular groups for spaces of convex polyhedra in $R^3$ which have angle defects at their vertices that are multiples of $\pi/3$. When there are no more than 6 points, the polyhedron is non-compact. When $n < 6$, the moduli space is the quotient of complex projective space $\mathbb{CP}^{n-2}$ by your group modulo its center. (The full group is a subgroup of $SU(n-1)$).

In the borderline case $n = 6$, the polyhedron looks like an infinite cylinder at one end, and $\mathbb T^6$ acts as a crystallographic group on $\mathbb{C}^4$ (with its order 2 center acting trivially).

share|improve this answer
Beautiful answer. –  Petya Dec 10 '10 at 4:32

Here is an idea. Fix a commutative ring $R$ and elements $q, z \in R$. Recall that the (Iwahori-)Hecke algebra $H_n(q, z)$ is the $R$-algebra on generators $T_1, ... T_{n-1}$ with relations

$$T_i T_j = T_j T_i, |i - j| \ge 2$$ $$T_i T_{i+1} T_i = T_{i+1} T_i T_{i+1}$$ $$T_i^2 = z T_i + q.$$

It is known that $H_n(q, z)$ is a free $R$-module on elements $T_w, w \in S_n$ which are identified with products of the $T_i$ corresponding to minimal representations of $w$ as a product of transpositions. When $q = 1, z = 0$, we get the group algebra of $S_n$.

When $q = -1, z = -1$, we have $T_i^3 = 1$, hence $\mathbb{T}\_n$ acts on $H_n(-1, -1)$ (via the map sending $\sigma_i$ to multiplication by $T_i$). If we could show that this action is faithful, then it would follow at least that $\mathbb{T}\_n$ has a faithful linear representation, and one might be able to push this to show that $\mathbb{T}\_n$ is finite (for example by showing that the action is faithful when $R$ is a finite field).

share|improve this answer
Isn't $T_i^3=-1$ with those parameters? –  Mariano Suárez-Alvarez Dec 10 '10 at 0:44
You want $q=z=-1$. –  Mariano Suárez-Alvarez Dec 10 '10 at 0:46
Whoops; thanks. –  Qiaochu Yuan Dec 10 '10 at 0:50
This is a great approach. Sadly, according to Gjergji Zaimi's answer, the "push this to show that it's finite" is bound to fail. –  Theo Johnson-Freyd Dec 10 '10 at 1:35

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.