## Hot answers tagged presentations-of-groups

35

The fact that giving a presentation at the black board slows down the speed of presentation and helps the audience to digest the stuff is related to the fact that mathematicians use a language that has a high information density (formulas, diagrams) compared to other fields. Moreover the symbols and other structures used in that language CAN be written on a ...

33

For me it depends on the type of talk. But if I'm giving either a lecture or a seminar that involves presenting proofs (or sketches of proofs) then I like blackboards best. There are many reasons for this, but three important ones are (i) it forces me to understand the material well enough to memorize it (I don't use notes for talks, and what I say applies ...

26

When you write something on the blackboard, you have to think and verify that what you are writing is correct. A good speaker shares this chain of thought with the audience.
At every pause where the speaker has to think, the (good) audience will automatically try to understand what the problem could be.
An example would be in what order one might want to ...

20

After my first Beamer talk, a member of the audience came up and politely suggested that I should repeat things from one frame to another rather than jumping backwards to remind the audience of what I had previously said. I adopted that piece of advice and my subsequent presentations (I was giving a series of three talks) were much less irritating.

18

Not only these subgroups are finitely presented, they are all finite free products of cyclic groups; most of them (for sufficiently large $n$) are actually free of finite rank (once congruence subgroup contains no elements of order 2 and 3). For instance, you can easily check that $\Gamma(n)$ is torsion-free for all $n\ge 2$ by looking at traces for $n\ge 3$ ...

16

A major reason for giving live talks is that you can react to the audience and adjust your talk. This is difficult with slides. Even experienced speakers overestimate how much information they can convey in a lecture and are stuck with flashing through slides that almost no one is understanding. I have heard good talks using slides, but they are rare.
On ...

14

I think using Beamer well is hard. Even just getting the layout right is hard - I spend hours moving things around a bit so that the layout aids comprehension. I agree that one should not insert pauses "just because one can", but I think one should insert pauses because it helps people understand what is going on, as if one were writing it gradually on a ...

13

First of all, I completely agree with Tim Gowers' first sentence. However, I put the dividing line between presentations and board-talks much further over towards presentations than everyone else.
As a speaker:
Giving a beamer talk means that I have to prepare the talk properly. I have to think about the audience that I am speaking to, the level at ...

12

An algorithm for computing the hyperbolic "thinness" constant $\delta$ is described in the paper:
Epstein, David B. A.; Holt, Derek F
Computation in word-hyperbolic groups.
Internat. J. Algebra Comput. 11 (2001), no. 4, 467–487.
and I did implement it. It works OK on reasonably straightforward examples like surface groups, hyperbolic triangle groups, etc, ...

11

At the risk of being subjective (and possibly even argumentative), I feel like I should offer an answer to the implicit question, answered piquantly by Derek Holt:
'a large proportion of researchers in Geometric Group Theory appear to be interested only in the existence of algorithms and their complexity, rather than in actually using them to do ...

11

I agree with the comments from gowers, javier etc. on why blackboards are preferable to PowerPoint and similar media, but I thought that I might try to cover in a little more depth why mathematicians might prefer them specifically to whiteboards: (i) material written on a blackboard is more easily visible, especially at long ranges, and especially compared ...

11

It is still an open problem to find a short and simple way to derive a finite presentation for the mapping class group. The book by Farb and Margalit (in the recent preliminary version 4.00) gives a clear sketch of the known derivations, which are rather long and complicated. For the details one must consult the original papers cited in the book. There are ...

10

This problem is undecidable. If it were not, you could use it to construct an algorithm for testing if a given f.p. group is trivial or not, which is well known to be undecidable:
Input: f.p. group G
Test if G is residually solvable.
If it is not, output "non-trivial".
If it is, find the abelianization of G.
If the abelianization is trivial, output ...

9

My personal preference as a speaker and a listener is for the minimal: none of the stuff at the bottom or the top, and no navigation icons (those things at the bottom-right of the screen). Thus, if a slide contains no text, it's a totally blank white rectangle.
Enough people have asked me how to do this that I put instructions on a web page.
(You'll see ...

9

Here is an answer from the viewpoint of an adviser of graduate students. A graduate student can give a talk on Beemer without actually understanding what they are saying. It is impossible to give a decent blackboard talk without so understanding. I am coming around to the position that all graduate student talks (thesis defenses, but we also have a sort ...

9

Finitely generated torsion-free nilpotent groups are polycyclic. Therefore, their cohomological dimension equals their Hirsch length.This is a result of Gruenberg. One can find it in Gruenberg's book 'Cohomological topics in group theory' in section 8.8 or in Robert Bieri's Book on 'homological dimension of discrete groups' as Th. 7.14.
On the other hand, ...

8

I think Lee's and Steve's comments pretty much answer this question. Let me try to summarize, and clear up a couple of misconceptions that seem to be lurking. For convenience, I'll denote your group by $G$.
The map $G\to F_2\times F_2$.
Actually, I don't think the map $G\to F_2\times F_2$ that you describe is an injection. The images of $x$ and $z$ ...

8

My number one tip is read the beamer user guide from cover to cover. As well as telling you how to do stuff, it also contains an amazing amount of good advice on how to do good presentations.
(And if you're using graphics, read the PGF/TikZ manual as well - even if you aren't using that package to do your graphics.)

8

In my experience, "messy" blackboard lectures can mostly been blamed on lack of preparation. Using beamer forces you to prepare your lecture in advance, but I think that a well prepared lecture on a blackboard is easier to digest. Some reasons for this are:
Since one has to write everything down, the natural tendency is to make short, concise statements.
...

7

As the authors explain in the 2010-04-22 version of their text, an explicit presentation was first obtained by Harer, using the method developed by Hatcher and Thurston. The latter defined a 2-dimensional simply connected polyhedral complex on which Mod(S) acts :
cocompactly (i.e. finitely many orbits of 0,1, and 2 faces)
with explicitly finitely presented ...

7

A subgroup of finite index in a finitely presented group is finitely presented (see Exercise 6.1.6 in Robinson: A course in the theory of groups), so all congruence subgroups of the modular group are finitely presented. I cannot answer your second question.

7

This discussion: http://www.math.niu.edu/~rusin/known-math/95/rubik seems to culminate in a presentation (due to Dan Hoey). I did not read it carefully, I must admit. The presentation is quite complicated. For the 2x2x2 group there is this:
http://cubezzz.dyndns.org/drupal/?q=node/view/177

6

My answer to your question here
included a proof of this. Indeed, such a group has a non-abelian 2-nilpotent quotient. In the discrete case, it even shows that there is a non-abelian torsion-free 2-nilpotent quotient. In the profinite case, it shows that for all $p$ (although I gave a full proof only for odd $p$) there is a non-abelian 2-nilpotent $p$-group ...

6

You probably already noticed that, but $B_{p,q}$ is the fundamental group of
$$
X_n/(S_p \times S_q)
$$
where $X_n$ is the configuration space of $n$ points in the complex plane. Ths may help to guess some facts about these groups.
So far I know these group are usually called "mixed braid groups" in the litterature, though this name is sometimes used for ...

6

Unless you have a specific type of subgroup, like one that acts cocompactly on something, you're in deep trouble. Braid groups are incoherent (see Artin groups, 3-manifolds and coherence by Cameron Gordon). That is, there are finitely generated subgroups that are not finitely presented.

6

I prefer blackboards/whiteboards to any other option (even OHP) for the following two reasons:
1) I find nowadays (having been giving talks for about 15 years) that I only semi-prepare them now, and never use notes, and just say what comes out: I have a "plan" and talk around it. This way I'm guaranteed to never give the same talk twice, which is ...

5

GAP has the following:
http://www.gap-system.org/Gap3/Manual3/C023S010.htm
(But note GAP4 exists now I think.)
(The following were comments, but should be parts of the `answer'.)
(i) There is on the GAP site, the following: ftp.gap-system.org/pub/gap/…
(ii) There was a lovely book that listed lots of this sort of material presentations, Group Tables by ...

5

Related to your question, perhaps... there's the Atlas of finite group representations. Although the atlas itself might not be what you're after, the above page contains links that might be more relevant.

5

Yes. In fact, slightly more is true: the simple homotopy type of the presentation 2-complex is preserved under Nielsen transformations. For a proof of this fact, see
Micheal N. Dyer and Allan J. Sieradski, Trees of homotopy types of two-dimensional CW-complexes. I., Comment. Math. Helv. 48 (1973), 31–44. MR0377905

5

If I understand your presentation correctly, you have determined that $\Gamma(4)$ is a free group on $5$ generators. This is not surprising. The modular curve $X(4)$ is $\mathbb P^1$ with six cusps and no elliptic points, so its fundamental group is the free group on $5$ generators. You appear to have found one set of $5$ generators. There is nothing wrong ...

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