Talk:Braid group

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[edit] Infinite?

The article mentions that the Braid group for n > 1 is infinite. As far as I can tell, it is possible to enumerate all possible Braids in a given Braid group. So would someone please elaborate on this point?

Hi. By "enumerate", you mean "list" them, right? Does this list ever end? For example, with B_2, you can keep twisting the two strands in one direction and keep getting different elements of the braid group (n twists is not the same as m twists for n not equal to m). So "enumerate" here means that you can list the braid elements, but you can never stop. That means the group is infinite. --C S (Talk) 12:43, 23 April 2006 (UTC)

Ok, thanks! Now I understand.

The braid group is countably infinite.

[edit] unencyclopedic tone

This article is written like a math lesson, not an encyclopedia ("We" "you", etc.) Someone needs to rewrite the examples into a neutral third person. Night Gyr 02:55, 1 May 2006 (UTC) Meh! It's readable, what do you want?

This is standard terminology used in mathematical texts, so, using terms such as "we" make the tone of the article more appropriate, not less. See also "Tone", pronoun use, etc. in math articles at Wikipedia talk:WikiProject Mathematics. Dysprosia 22:26, 1 May 2006 (UTC)

My bigger problem with this article is not that it uses the mathematical "we" but that it's written like a math lesson. You've already excised the single most blatant example "(Verify these on a piece of paper!)" The "You are given four strands" example could easily be rewritten without the second-person voice, though, and I don't think second person is even acceptable in most math works either. Night Gyr 23:03, 1 May 2006 (UTC)

Obviously I'm happy with the tone of the article since I wrote most of it, but if you like "one" better than "you", please feel free and change it. Personally I think that math can be intimidating enough even without formalistic language. AxelBoldt 06:34, 2 May 2006 (UTC)

Well, I see that the "(Verify these on a piece of paper!)" was deleted as too informal. I won't fight it but I'm sad about it: nobody really understands the braid group unless they have actually verified these relations on a piece of paper. AxelBoldt 06:38, 2 May 2006 (UTC)

Wikipedia articles aren't supposed to contain exercises, as we intend to be a reference work. What would be really nice and appropriate is a verification in the article. Clearly that skips the point of the reader performing the exercise, but we're not an educational work. Dysprosia 00:54, 3 May 2006 (UTC)

Well I reinstated the "draw it yourself" bit. As a non (pure) mathematician interested in braids, it helped me no end. As for the "tone", I reckon it's perfectly appropriate.

best wishes, Robinh 07:30, 2 May 2006 (UTC)

I've attempted a minor re-wording of the main sections of the article to make the tone a little less informal, but hopefully without losing the excellent clarity of the article. I have also added a MathWorld reference. Gandalf61 12:50, 4 May 2006 (UTC)
Thanks for doing that. I moved the MathWorld link to the External links section. It's probably best to restrict references to peer reviewed work or standard texts in the field; I don't think that MathWorld is any more authoritative than we are, although they certainly are better with listing relevant references. AxelBoldt 17:05, 4 May 2006 (UTC)

[edit] Braid subgroups

Firstly, I would like to say that I like this article a lot, as a mathematician who is rather unfamiliar with this area (despite having John Horton Conway tell me about it informally in 1979). Being realistic, it is doubtful if many readers with very little mathematical knowledge are going to be even looking at this article, so comments above that it is not aimed at a general reader are rather spurious.

The issue I want to bring up is the ways in which the Braid groups are subgroups of bigger Braid groups - the article mentions one monomorphism, but there are clearly many others. The question is "how many?". Intuitively, I see 2(n-1) ways in which Bn-1 is a subgroup of Bn, one for each of the braids on the ends and 2 for each of the ones not on the end (with the other braids being linked in a way that is either all "on top" or all "underneath" the unused braid. I doubt this is the final answer. Is anyone aware of what can be said about the monomorphisms between Bn-1 and Bn beyond the existence of one (as currently described)? Elroch 15:25, 15 May 2006 (UTC)

I don't know the answer, but I feel there are a lot more subgroups isomorphic to Bn-1 in Bn. One could start with the examples you have above, and then let automorphisms of Bn (which I also don't know, except for the inner ones) act on them. AxelBoldt 16:21, 16 May 2006 (UTC)
True. I was thinking only of the ones where the subgroup corresponded to a subset of the braids in the most obvious way, but there may well be others. The most obvious automorphism is the one which inverts the order of the braids, but this just swaps around the examples I mentioned. The braid group seems more rigid than the symmetric group, because the order of the braids is significant, whereas the order of the permuted elements isn't. Despite this, in the situation with symmetric groups, it's clear (or seems so at 2:30 am) that there are n analogous subgroups, the obvious ones. Elroch 01:32, 17 May 2006 (UTC)

[edit] Reminded of Lin Alg

This is the first time I've come across Braid groups, so hopefully someone can enlighten me. Reading the "Generators and relations" section reminded me of basis vector sets in linear algebra. In fact, quite a few properties of braid groups reminded me of matrices. Though I should note that all I know about lin alg is a undergrad course I took. I'd like to add "This is analogous to the basis vector sets in linear algebra." under that section, but, never having seen this material before, I'm really not confident that I really know what I'm talking about. So, to anyone who does know, is there some analogous connection? -- 69.139.198.89 18:24, 20 May 2006 (UTC)

There is a loose analogy between generators of a group and basis elements of a linear space, but it is a very tenuous one for non-abelian groups like Braid groups (but of some relevance to abelian groups), and is not worth mentioning here. Braid groups, like a large class of other groups, may be represented as matrix groups, which is a quite separate topic to generators and relations, and could justify a separate section if anyone knows enough about it. Elroch 23:19, 22 May 2006 (UTC)

[edit] Braid group; presentation of modular group

conversation moved here from User talk:Linas

Hello,

in your addition to Braid group you give a presentation of the modular group that doesn't look right to me. The presentation you give doesn't seem to admit any elements of finite order. Looking at the matrices, you have s4 = t6 = 1, but in the braid group that's not true. Also, (even though that doesn't mean much), the modular group article gives a different presentation. Cheers, AxelBoldt 19:05, 13 May 2007 (UTC)

I re-edited braid group so as to make the statement clear; your confusion was due to my fuzzy presentation of the topic. linas 20:46, 13 May 2007 (UTC)
Yes, now I get it. I was missing the 1 in the presentation I guess. Do we know that <c> is the center of B3? I took that out for now. AxelBoldt 21:54, 13 May 2007 (UTC)
Yes, its in the center, and this is very easy to prove:
σ1c = σ1(σ1σ2σ1)(σ1σ2σ1)
= σ1(σ2σ1σ2)(σ1σ2σ1)
= σ1σ2σ1(σ2σ1σ2)σ1
= (σ1σ2σ1)(σ1σ2σ1)σ1
= cσ1
and similarly easily, σ2c = cσ2; ergo c commutes with all elements.
Yes, but you used the fact that <c> is equal to the center, which isn't yet clear to me.AxelBoldt 16:10, 14 May 2007 (UTC)
The only properties used in the proof above are the definition of c, which is c = (σ1σ2σ1)(σ1σ2σ1), and the braid relationship on B3, which is σ1σ2σ1 = σ2σ1σ2. The parenthesis have no significance, except to aid the reader in grouping the symbols, that is all. The proof shows that c, defined in this way, commutes with σ1 and σ2. Since σ1 and σ2 generate B_3, this means that every element of B_3 may be written as a product of σ1 and σ2. Therefore, c commutes with every element of B_3. Next, it is clear that cn for any finite integer n also commutes with every element of B_3. Now, the group <c> is defined to be the set of elements cn with finite interger n. Thus, every element of the group <c> commutes with every element of B_3. Ergo, one concludes that <c> is in the center of B_3. Is this acceptable? linas 19:02, 14 May 2007 (UTC)
BTW, v and p are sometimes common notations for the elements of order 2 and 3 in PSL(2,Z); its what Rankin uses. S and T are have several different conflicting conventions. linas 04:07, 14 May 2007 (UTC)
Ok, then we shouldn't use them for the elements in the braid group, which have infinite order. I'll switch the notation. AxelBoldt 16:10, 14 May 2007 (UTC)
I notice you removed the statements about the center. Do you believe that B_3 has a center that is larger than the one generated by c? linas 18:49, 14 May 2007 (UTC)

Now I still have another question about the isomorphism Sn = Bn/Fn with Fn free. Is it clear that Fn is normal? Because we already know that Sn = Bn/Pn, where Pn is the pure braid group, which isn't free, and Fn is contained in Pn. It seems to me that Pn is the normal subgroup generated by Fn. AxelBoldt 22:05, 13 May 2007 (UTC)

I'll have to read the article more carefully. If something contains the free group as a subgroup, then, almost by definition, the free group will be a normal subgroup... how could it be otherwise? Right? (since you are "free" to tack on any elements to the left or right hand side, as you wish....) However, bedtime for me, I'll look tommorow, perhaps. linas 04:07, 14 May 2007 (UTC)
Well, B3 contains for example the free group <σ1> as a subgroup, but that isn't a normal subgroup. I'll take the Fn claim out for now. AxelBoldt 16:10, 14 May 2007 (UTC)
Ohh, I reviewed the statements about the free group, and, yes, indeed, they were insane. According to the article history, it seems I added these, so um, yes, oooops. Must of been in a fog, not sure what I was thinking. linas 23:13, 14 May 2007 (UTC)

Actually <c> is the center of B_3. This is because the modular group PSL(2,Z) is centerless (being the free product of two nontrivial groups). Any element in the center of B_3 must map to something in the center of PSL(2,Z) since the homomorphism B_3 → PSL(2,Z) is surjective. It follows that the center of B_3 is contained in the kernel of the map B_3 → PSL(2,Z). -- Fropuff 20:42, 14 May 2007 (UTC)

Thanks, that makes sense. I'll add it to the article. AxelBoldt 22:03, 15 May 2007 (UTC)

[edit] Antihomomorphism

With the definitions given, I believe that the braid group is anti-isomorphic to the mapping class group of the punctured disk. Similarly, the map from the braid group to the symmetric group is an antihomomorphism, not a homomorphism. Sam nead (talk) 21:53, 10 February 2008 (UTC)