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There is a lot of normal subgroups in braid groups (for example there is an action of $B_n$ on unitriangular bilinear forms on $R^n$ over arbitrary commutative ring $R$: $b_i\colon e_j\mapsto e_j$, $j\ne i, i+1$, $b_i\colon e_{i+1}\mapsto e_i$, $b_i\colon e_i\mapsto e_{i+1}-(e_i, e_{i+1})e_i$ and set $R=\mathbb Z_m$).

Is there any classification (with no conditions on terms of classification) of normal subgroups of $B_n$?

update: the interesting case of this classification for me is the case of finite index normal subgroups. For example I don't even know what is the kernel of the action described above. The answer may be useful in algebraic geometry (see A.L. Gorodentsev, TRANSFORMATIONS OF EXCEPTIONAL BUNDLES ON $\mathbb P^n$)

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@zroslav - I think I found the answer. Take a look at my final edit. –  HJRW Mar 19 '11 at 0:54
    
@HW: I'm not sure that I understand this classification. Neither geometric nor the "listing" meaning are clear for me. –  zroslav Mar 19 '11 at 18:31

3 Answers 3

It depends what you mean by classification. Let's start with the example of the free group of rank $n$, $F_n$. The normal subgroups of $F_n$ correspond to $n$-generator groups with marked generating sets, which of course is a hopelessly complicated set. So in some sense, the normal subgroups of $F_n$ are not classifiable. On the other hand, we have:

Greenberg's Theorem: If $N\lhd F_n$ then precisely one of the following holds.

  1. $N$ is trivial.
  2. $N$ is of finite index in $F_n$.
  3. $N$ is infinitely generated.

Of course, the second option is pretty complicated, and the third even more so. But you might call this a sort of 'classification'.

Now, the pure braid group $PB_n$ admits a natural map onto $PB_{n-1}$, by forgetting a strand. As observed above, $B_3$ surjects $\mathbb{Z}/2*\mathbb{Z}/3$, which has a free subgroup of finite index. So every $B_n$ virtually surjects a free group. This indicates that the normal subgroups of $B_n$ are almost as complicated as the normal subgroups of $F_m$.


The OP has now indicated that he is interested in the normal subgroups of finite index. The answer above shows that we at least have the following:

Let $n\geq 3$. For every finite group $Q$ there is a subgroup $N\lhd B_n$ of finite index with $Q\hookrightarrow B_n/N$.

That said, there are probably many more normal subgroups of finite index. Here's a question in this direction to which I don't know the answer.

Are braid groups residually finite simple?


Further edit:

I think this is what you're looking for. Here's an article by McReynolds, giving a proof that pure braid groups have the congruence subgroup property. This can be thought of as a classification of the subgroups of finite index.

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I suppose what I'm driving at is that there could be a theorem of the form: every finitely generated normal subgroup of $PB_n$ is either of finite index or contains the kernal or a 'forgetting a strand' map. I would guess that this is an open question, but am no expert. Where's Andy Putman when you need him? –  HJRW Mar 18 '11 at 19:01
    
This is a nice classification! But not detailed as it can be desired. Maybe I should give some more examples of normal subgroups of $B_n$? I want to have the complete classification of all the finite index normal subgroups. This may be useful in algebraic geometry (and the study of exceptional bundles) where the described in my question action of $B_n$ came from. –  zroslav Mar 19 '11 at 0:22
    
Well, that's the first time that you've mentioned that you're interested in the finite-index normal subgroups. You should edit your question to say so. –  HJRW Mar 19 '11 at 0:30
    
@HW: Ok, it's done! –  zroslav Mar 19 '11 at 0:52
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@zroslav - that shows that, for every finite $Q$ , there is an $n$ such that $Q$ is a subgroup of a quotient of $B_n$. But my statement is uniform in $n$. Anyway, the third edit, with the congruence subgroup property, is what you're looking for. –  HJRW Mar 19 '11 at 1:16

If you're willing to consider "any classification" you might consider the covering space theory of configuration spaces.

Equivalence classes of regular (finite) connected coverings of the configuration space $C_{n}(\mathbb{R}^{2})$ of n unordered points in $\mathbb{R}^{2}$ completely classify (finite index) normal subgroups of $B_n=B_n(\mathbb{R}^{2})=\pi_{1}(C_{n}(\mathbb{R}^{2}))$ via the usual Galois correspondence.

This may not be very enlightening but may offer a practical approach to studying the lattice of normal subgroups of $B_n$. Hansen's polynomial covering theory might also help to identify interesting subgroups of $B_n$.

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Yes, but is this "classification-by-coverings" saying anything about indexes of these subgroups? I know that this is a number of points in the fiber. Can we list all these coverings? –  zroslav Mar 19 '11 at 18:29
    
The index is equal to the degree of the covering map. In general, counting covering spaces is extremely complicated. For instance, the normal subgroups of index d in free group correspond to the regular coverings of a degree d of a rose. Counting them is a non-trivial exercise... –  HJRW Mar 20 '11 at 1:18
    
I mean: any combinatorial or geometrical realisations of these coverings? –  zroslav Mar 20 '11 at 11:09

This does not seem like a realistic hope. When you say a lot of normal subgroups we can be more precise. A group is conjugacy separable if given two elements which are not conjugate then there is a finite quotient group such that the images of the two elements are not conjugate in this finite group. This is stronger than residually finite which is the special case where one of the group elements is the identity.

The point is I think the braid groups are conjugacy seperable, For $B_3$ we have the short exact sequence $$0 \rightarrow \mathbb{Z} \rightarrow B_3 \rightarrow \mathbb{Z}_2 * \mathbb{Z}_3 \rightarrow 0 $$ and the free product of two cyclic groups is conjugacy seperable.

I don't know the argument for $B_n$.

Edit (in response to comments). The result that braid groups are residually finite is stated in Magnus (Property III) and Rolfsen (Theorem 2.5) . The argument is that free groups are residually finite and the automorphism group of a residually finite group is residually finite.

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I'm not sure that I've understood you correctly –  zroslav Mar 18 '11 at 16:54
    
This is not a complete proof that $B_3$ is conjugacy separable. (Although it's true that it is.) In general, you have to be careful about separating the kernel. Indeed, Deligne gave an example of a central extension of a residually finite group which is not itself residually finite. –  HJRW Mar 18 '11 at 18:28
    
I'd be very interested to see a reference for the fact that all braid groups are conjugacy separable. –  HJRW Mar 18 '11 at 18:29
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I thought this was an unsolved problem; surely they are residually finite, and have solvable conjugacy problem, but I've never seen they are conjugacy separable. –  Steve D Mar 18 '11 at 18:41
    
Steve - that sounds plausible to me. $B_3$ happens to be the fundamental group of a compact Seifert-fibred space, and these are known to be conjugacy separable. –  HJRW Mar 18 '11 at 18:45

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