Let $B_n$ be the braid group on $n$ strands. As is well known, if $\sigma_i$ is the operation of crossing the string in position $i$ over the string in position $i+1$, then the elements $\sigma_1,\dots,\sigma_{n-1}$ generate $B_n$ and the relations $\sigma_i\sigma_j=\sigma_j\sigma_i$ ($|i-j|\geq 2$) and $\sigma_i\sigma_{i+1}\sigma_i=\sigma_{i+1}\sigma_i\sigma_{i+1}$ give a presentation of the group.

I am interested in a group that can be obtained from $B_n$ by adding another set of relations. For each $k$, define $T_k$ to be the "twist" of the first $k$ strands. Geometrically, you take hold of the bottom of the first $k$ strands and rotate your hand through 360 degrees in such a way that strands to the left go over strands to the right. In terms of the generators, $T_k=(\sigma_1\dots\sigma_{k-1})^k$, since $\sigma_1\dots\sigma_{k-1}$ takes the $k$th most strand and lays it across the next $k-1$ strands. Let us also define $S_k$ to be the right-over-left twist of the strands from $k+1$ to $n$. That is, $S_k=(\sigma_{k+1}\dots\sigma_{n-1})^{-(n-k)}$. (Also, let us take $T_1$ and $S_{n-1}$ to be the identity.)

I am interested in the group you get if you start with $B_n$ and add in the relation $T_kS_k=1$ for every $k$. Let me make some utterly trivial observations.

If $n=2$ then we get the cyclic group $C_2$. That's because $\sigma_1$ is the only generator and $T_2S_2=\sigma_1^2$. If $n=3$ then we get $S_3$. That's because $T_1S_1=\sigma_2^{-2}$ and $T_2S_2=\sigma_1^{-2}$, so the relations we are adding are $\sigma_1^2$ and $\sigma_2^2$, and it is well known that those, together with the braid relations, give a presentation of the symmetric group.

Beyond that I don't know what to say, though I've convinced myself (without a proof) that when $n=4$ the group is infinite: in general, it seems that the extra relations can be used to do only a limited amount of untwisting. (I do have a proof that there are pure braids that cannot be reduced to the identity once we have four strands. It's a fairly easy exercise and I won't give it here.)

What exactly is my question? Well, I'd be interested to know whether the word problem in this group is soluble in reasonable time. In the service of that, I'd like to know whether this group is one that people have already looked at, or whether it at least belongs to a class of groups that people have already looked at. (E.g., perhaps the solubility of the word problem follows from some general theory.) And is there some nice way of characterizing the subgroup of $B_n$ that we are quotienting by? That is, which braids belong to the normal closure of the set of braids $T_kS_k$? (One way of answering this would be to characterize their normal forms.)

The motivation for the question comes from part of an answer that Thurston gave to a question I asked about unknots. It seems to me that this question ought to be relevant to the untying of unknots, but easier.

One final remark: the word problem in $B_n$ can be solved in polynomial time. (If my understanding is correct, this is a result of Thurston that built on work of Garside.) Since adding more relations makes more braids equal to the identity but also gives more ways of converting a word into another, it is not clear whether the problem I am asking should be easier or harder than the word problem for braid groups. However, my hunch is that it is harder (for large $n$, that is).

• Just a comment on your last paragraph: There are now many ways of solving the word problem for braids quite quickly. Practically, the quickest way is Dehornoy's handle reduction. As we know have explicit faithful linear representations (thanks to Bigelow and Krammer), one can also solve the word problem by multiplying matrices... www.math.unicaen.fr/~dehornoy/Surveys/Dhu.pdf is a relevant survey. May 11, 2011 at 19:05
• I have seen that survey and I like it very much. Perhaps you can answer another question: is the list of methods for solving the word problem given in that paper essentially complete (in the sense that all methods are small variants of one of the methods mentioned there)? May 11, 2011 at 19:36
• Is there something else you want to know about this group? May 25, 2011 at 4:19

If I understand your group correctly, your quotient $G_n = B_n / \langle T_kS_k : k = 1,2,\cdots, n\rangle$ can be thought of as a fairly geometric object -- it's the mapping class group of a sphere with $n$ marked points.

Let $\pi_0 Diff(S^2,n+1)$ be the mapping class group of $S^2$ that preserves $n+1$ points. There is an epi-morphism $\pi_0 Diff(S^2,n+1) \to \Sigma_{n+1}$, so we can talk about the subgroup of $\pi_0 Diff(S^2,n+1)$ that fixes one of the $n+1$ permuted points, call it $A_n$. I believe

$$A_n \simeq B_n / \langle T_n \rangle$$

and I believe Ian Agol mentioned this. I'm using angle brackets to indicate "normal subgroup generated by".

But we still have to account for the other relators you want to mod out by, $T_1S_1, T_2S_2, \cdots, T_{n-1}S_{n-1}$.

Let $X_n$ be the subgroup of $Diff(S^2,n+1)$ that stabilizes one of the $n+1$ points. Then there is a map $X_n \to Diff(S^2,n)$ given by forgetting that point. This map is the fibre of a locally-trivial fibre bundle with base-space a $n$-times punctured sphere. Moreover, this expresses $\pi_0 Diff(S^2,n)$ as a quotient of $A_n$, and the relators are precisely the remaining ones you list (this requires a bit of a computation).

So I believe the above is an argument your group $G_n$ is $\pi_0 Diff(S^2,n)$. This explains why you get $\mathbb Z_2$ in the $n=2$ case, and $\Sigma_3$ in the $n=3$ case. This group is linear for all $n$. There's two published proofs of this, one by Bigelow and myself and another by Korkmaz. The representation is obtained as a tweaking of the Lawrence-Krammer representation.

• I think the kernel is normally generated by $T_{n-1}^{-1}T_n=\sigma_{n-1}\cdots \sigma_1\sigma_1\cdots \sigma_{n-1}$, since this corresponds to moving the rightmost strand across infinity. So you need only check that this is killed by Gowers' relators. But changing the $\sigma_1^2=T_1$ in the middle to $S_1$, I think you get the twist on the last $n-1$ strands, which is conjugate to $T_{n-1}$. May 11, 2011 at 20:33

Your group indeed will be infinite for $n>3$. The $n$-strand braid group is actually the mapping class group of the $n$-punctured disk, which has a natural homomorphism to a finite index subgroup of the mapping class group of the $n+1$-punctured sphere. This is the subgroup that fixes a strand (the strand "at infinity" for the braid group, or corresponding to the boundary of the disk). This group is obtained from the braid group by quotienting by a full twist (this corresponds to twisting about the boundary of the disk, which is trivial in the mapping class group since it just fixes a puncture). Now, we have a further epimorphism to the $n$-punctured sphere mapping class group, by "filling in" the puncture at infinity. I claim that the image of $T_k$ and $S_k^{-1}$ in this group are the same. $T_k$ corresponds to a Dehn twist about a curve surrounding the strands $1,\ldots,k$. But on $S^2$, this curve is equivalent to the curve surrounding strands $k+1,\ldots, n$, and so the corresponding Dehn twists are equivalent. If you can show that the full twist is killed by your relators, then this must be the quotient group. Otherwise, I'm not quite sure what you get. I think I can show in $B_4$ that the full twist is killed by your relators, so the quotient of $B_4$ is the mapping class group of the 4-punctured sphere.

Edit: Ryan Budney's answer shows what I expected, that you get the $n$-punctured sphere mapping class group.

• Mapping class groups were shown by Mosher to be automatic, so this gives a polynomial-time solution to the word problem (in fact, the conjugacy problem is also solvable, which is more akin to the unknot question). May 11, 2011 at 22:12
• @Agol : I think that it is still open whether or not the conjugacy problem is solvable in an automatic group. The algorithm I know uses a biautomatic structure (but Hammenstadt proved a few years ago that mapping class groups were biautomatic). May 19, 2011 at 15:40
• Yes, Andy, this follows from Hamenstadt's result, but was known earlier by work of Hemion. Another algorithm was given by Mosher. I didn't mean to imply that automaticity implied that the conjugacy problem is solvable. For braids, the conjugacy problem was solved by Garside: I wouldn't be surprised if some version of his method applied in the punctured sphere mapping class group too. May 19, 2011 at 15:59

THIS ANSWER IS WRONG. See the comment by Gowers. The mapping class group of the n-punctured sphere, which the question was about, is not the same as the spherical braid group. It is obtained from the spherical braid group by quotienting out by a full twist of all the strands. The original answer is below for posterity.

As others have pointed out, this is the braid group on S2. These groups were studied a bit during the 1960's, and then everyone seems to have forgotten all about them. The only book which seems to give braid groups over S2 more than a passing mention is Murasugi and Kurpita's monograph. Only recently does it seem that people have become interested again, because of "braid groups over surfaces" (discussed here for example) and because of the amazing Berrick-Cohen-Wong-Wu paper which connects Brunnian braids over a sphere to homotopy groups of S2! Their result is one of those mysterious alluring connections which makes mathematics worth doing. John Baez wrote a nice blog post about that story.

It is proven by Fadell and van Buskirk that the braid group over S2 is presented by the braid generators $$\sigma_1,\ldots,\sigma_n$$ modulo the relations:

1. $$\sigma_i\sigma_j\sigma_i=\sigma_j\sigma_i\sigma_j$$ if $$\left\vert i-j\right\vert=1$$.
2. $$\sigma_i\sigma_j=\sigma_j\sigma_i$$ if $$\left\vert i-j\right\vert>1$$.
3. $$(\sigma_1\sigma_2\cdots\sigma_{n-1})(\sigma_{n-1}\cdots\sigma_2\sigma_1)=1$$.
Its word problem is solved in a very nice paper by Gillette and van Buskirk by modifying the combing algorithm for braids to work for braids over S2. This has exponential run-time, the same as the combing algorithm for the braid group. But first Thurston ("Word Processing in Groups"), and then Dehornoy, modified the combing algorithm to work in polynomial time, I don't know whether their algorithms can be made to work for the braid group over S2; my intuition is that the answer is surely yes, but surely nobody has bothered doing it yet, because braid groups over S2 are well forgotten.

EDIT: This is part of Open Question 9.3.10 in Word Processing in Groups (see also Open Question 9.3.9). I don't know whether it has been solved in the two decades since that book's publication.
• I'm not sure why you keep saying these groups have been forgotten. The mapping class groups of punctures spheres have been studied pretty steadily. "Forgotten" because people don't call them braid groups? May 19, 2011 at 15:09
• This is the impression I got from looking at the literature- all the papers which seriously study these groups seem to be from the 1960's. Do you know of much between, say, 1970 and 2000? If so, I'll edit that in! May 19, 2011 at 15:14
• @Daniel : The mapping class group literature (say, starting from Thurston's work in the late '70's) is enormous. A substantial portion of it is general enough to cover the case of the punctured sphere (ie the spherical braid group). May 19, 2011 at 15:39
• Ian Agol says in a comment below that Mosher showed that mapping class groups are automatic and hence that there is a polynomial-time algorithm for the word problem. Am I missing something here, or does that mean that the problem you mention was solved in 1995 (the date of Mosher's paper)? May 23, 2011 at 10:01
• The two are almost the same groups. The key relation is the fibre bundle $$Diff(S^2,n) \to Diff(S^2) \to C_n(S^2)/\Sigma_n$$ which means that you have a short exact sequence where the base group is $\pi_0 Diff(S^2,n)$, the middle group is the braid group of the sphere, and the fiber/kernel is $\mathbb Z_2$. So it's just a $\mathbb Z_2$-central extension between the two groups. May 25, 2011 at 4:01