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Artin's presentation of braid group on three strands is: $$ B_3 = \langle l,r : lrl = rlr \rangle $$ where you should think of "$l$" as the positive crossing between the left and middle strands and "$r$" as the positive crossing of the right and middle strands:

     |   |   |         |   |   |
      \ /    |         |    \ /
 l =   \     |     r = |     \
      / \    |         |    / \
     |   |   |         |   |   |

Then there is a surjection $B_3 \to S_3$ given by $l \mapsto (12)$ and $r \mapsto (23)$. ($S_3$ is the symmetric group on three letters: it is generated by $(12)^2 = 1 = (23)^2$ and the braid relation above.) The pure braid group $PB_3$ is the kernel of this surjection.

The six-crossing braid $b = lr^{-1}lr^{-1}lr^{-1}$ is an element of the pure braid group. Let $N$ be the minimal normal subgroup of $B_3$ that contains $b$. Certainly $N \subseteq PB_3$.

Question: Do we have $N = PB_3$?

Motivation

The motivation for my question comes from a neat trick that Conway showed us years ago. It leads to a more nuanced question than what I asked that I will pose as its own question if the answer above is "no". My memory is that at the time Conway did not know the answer to the more nuanced question, which suggests that the answer above cannot be "yes".

Take a long and reasonably thin rectangle of paper, and score it with two cuts, so that you have three strips of paper that are connected at both ends, so that you end up with a (framed, oriented, ...) "theta graph":

 |--------|
 |        |
 |  |  |  |
 |  |  |  |
 |  |  |  |
 |  |  |  |
 |  |  |  |
 |  |  |  |
 |        |
 |--------|

Then with some finagling, you can in fact "tie" the six-crossing braid in those three strands, without ripping the paper further, by passing the bottom through itself a few times. (The trick is that it's easier to unbraid than to braid, so make your paper long enough that you can put $bb^{-1}$ into it, and then unbraid the $b^{-1}$.) Put another way: you can first put a hair-tie at the end of your ponytail, and then braid your hair.

The harder question is to characterize all braids that you can put on the above "theta graph". The following facts are essentially obvious:

  • Any "braiding" of the theta graph is pure.
  • The set of braidings of the theta graph is a subgroup of $B_3$.
  • The set of braidings of the theta graph is closed under conjugating by arbitrary braids.

Therefore, the set $T$ of braidings of the theta graph is a normal subgroup of $B_3$, with $N \subseteq T \subseteq PB_3$. In particular, a positive answer to the question above characterizes $T$.

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5 Answers

up vote 15 down vote accepted

No.

Let $P_n$ be the pure braid group on $n$ strands. Forgetting the first strand gives a projection $P_3\to P_2$. The kernel is the free group $F_2=\pi_1(\mathbb{C}\setminus \{2,3\},1)$ given by moving the first strand around the other two while they are held in place. Since $P_2$ is just $\mathbb{Z}$, we have a short exact sequence

$1\to F_2\to P_3\to \mathbb{Z}\to 1$

Your element $b$ is contained in the kernel of this sequence (see the picture below) so its normal closure is contained in $F_2\lt P_3$. For an appropriate basis $\langle x,y\rangle=F_2$, the element $b$ represents $xyx^{-1}y^{-1}$.

\ |  |
 \   |
  \  |
   \ |
  |  |
  |  |\
  |  | |
  |   /
  |  /
  | / 
   / |
  /  |
 /   |
| |  |
 \|  |
  |  |
  | \
  |  \
  |   \
  |  | |
  |  |/
  |  |
  |/ |
  |  |
  |  |
/ |  |

In fact, removing any of the three strands results in a trivial 2-strand braid (as can be seen from the picture). Closing up the ends of the braid thus results in a 3-component link so that if any one component is removed, the other two are unlinked. This is the well-known Borromean rings. There are three different surjections $P_3\to \mathbb{Z}$, given by forgetting each strand separately, and the group of Brunnian braids is exactly the intersection of their kernels. Since $xyx^{-1}y^{-1}$ normally generates the commutator subgroup of $F_2$, the normal closure $N$ of $b$ will be exactly the group of Brunnian braids. In particular, $N$ is an infinite-rank free group.

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Any idea what the rank of N is? Is it infinite? –  Dan Margalit Apr 13 '10 at 0:37
1  
Great answer, Tom! –  Dan Margalit Apr 13 '10 at 10:05
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By the way, what is the minimal normal generating set for the Brunnian braid group on n strands? –  Dan Margalit Apr 13 '10 at 12:29
    
I'm not sure a minimal generating set is known -- but it is generated inside $F_{n-1}$ by all "basic commutators" that involve all the generators. See Johnson "Towards a characterization of smooth braids" journals.cambridge.org/action/displayAbstract?aid=2086212 or Stanford "Brunnian braids and some of their generalizations" arxiv.org/abs/math.GT/9907072. –  Tom Church Apr 13 '10 at 16:21
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Let $\epsilon: F(2)\rightarrow \mathbb{Z}$ be the map that sends a word to its total exponential sum. It is a homomorphism. Notice that $lrlr^{-1}l^{-1}r^{-1}$ is sent to zero, as is any conjugate of it, so that the map descends to $\epsilon:PB_3\rightarrow \mathbb{Z}$. The map is nonzero, as $l^2$ is a pure braid but $\epsilon(l^2)=2$. Notice that $(lr^{-1})^3$ is in the kernel of $\epsilon$ so that the normal subgroup generated by $(lr^{-1})^3$ is in the kernel of $\epsilon$, meaning that it cannot generate $PB_3$ normally.

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Oddly enough I was just thinking about this question a few days ago and nearly posted it to MO. From an old Martin Gardner paper (which I'd vaguely remembered from Malia telling me about it) I found a reference to: MR0133125 J. Shepperd "Braids which can be plaited with their threads tied together at each end." That paper answers the unframed version of your question for arbitrary B_n. The framed case is more complicated and not addressed in that paper. (Jim Tanton gave a Mathcamp class about some open questions related to the framed 4-strand case last summer, unfortunately I didn't attend that class.)

My motivation was some research I'm doing on relatives of $U_q(\mathfrak{g}_2)$, where I needed to understand all 1-dimensional representations of B_3/T_3. This was easy enough to work out directly, but it made me realize I'd really like to know about all small representations of B_n/T_n for small n. Essentially this is because "low weight spaces" in ribbon planar algebras (i.e. ribbon categories with a fixed favorite object) always have the structure of a representation of B_n/T_n (where you're looking at low-weight vectors in the n-box space, i.e. invariants of the nth tensor power of your fixed object).

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Do you have a generating set for $T_n$? –  Charlie Frohman Apr 13 '10 at 13:55
    
In the unframed case there's a generating set for T_n given in the above reference and an (impractical) recognition algorithm for elements of T_n. The basic idea is that the generators correspond to taking the bottom part and pushing it through one of the slots. –  Noah Snyder Apr 13 '10 at 16:32
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Just a comment/question too large for comment section, a related issue that is bugging me: it is not mentioned here that $PB_3$ is isomorphic to $F_2\times \mathbb Z$, perhaps it is not.

The configuration space of 3 labeled points $\mathbb R^2$ is a fiber bundle with base the config space of the first 2 points and fiber the position of the 3rd point. Let a "model fiber" be the plane minus $\{0,1\}$, and given a configuration on 3 labeled points $\{v_1,v_2,v_3\}$ in the plane, apply the unique orientation preserving affine linear map $f$ which translates, rotates and scales to send $v_1$ to $0$, $v_2$ to $1$. Then $f$ is defined globally, and continuously, as a projection to this typical fiber, making the bundle trivial.

A generator for the base is $(rl)^3$ (in your notation given above), generators for the fiber are, say, $r^2$ and $rl^2r$. Yet we should have, then, that $\[(rl)^3,r^2]=1$, let alone that the subgroup generated by $\{r^2,rl^2r\}$ is normal in $PB_3$, or even that, say, $l^2$ is a product of these 3 generators listed, which doesn't seem to be the case.

I'm missing something subtle or something obvious.

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It's mentioned above that we have an extension $F_2 \to PB_3 \to F_1$, which is in fact split. To me anyway, the obvious generators for the $F_2$ part are $l^2$ and $r^2$, and for the $F_1$ part $(lr)^3=(rl)^3$. This generating set does present $PB_3$ as a product: the braid relation has $l^2(rl)^3=llrlrlrl=l(rlrlrl)l=(lr)^3l^2$. So, yes, I agree that $PB_3=F_2\times F_1$. I'm not sure about your claimed generators --- maybe you mean to conjugate these generators by $r$? –  Theo Johnson-Freyd Nov 24 '10 at 0:14
    
ah! I had lost track of the braid relations, so it works after all. Incidentally, $l^{-2}(lr)^3=l^{-1}(rlr)lr=rllr$ so the sets of generators are equivalent. I had chosen $rl^2r$ only because it effectively leaves the first two strands fixed, while moving the 3rd around them both. –  AndrewLMarshall Nov 25 '10 at 19:27
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Does anyone have a reference from the literature for the fact that $PB_3=F_2\times F_1$?

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It follows immediately from Fadell and Neuwirth's fibre bundle. In this particular case, $C_3 \mathbb R^2$ is diffeomorphic to $C_2 \mathbb R^2 \times (\mathbb R^2 \setminus \{0,1\})$. So apply the fundamental group and you have the result. –  Ryan Budney Jan 22 '11 at 23:49
    
Thanks. I have an algebraic proof. I was actually looking for a citation from the literature. –  Rick Kubelka Jan 23 '11 at 5:29
    
I can't remember if we stated this exactly, but if you combine isomorphisms given on pages 101 and 252 of "A Primer on Mapping Class Groups", you get this statement. –  Dan Margalit Aug 20 '12 at 14:16
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