Misplaced Pages

Revision as of 01:33, 11 June 2002

The axiom of closure:

(Closure) for all a and b in G, a * b belong to G.

is superfluous, by definition of a binary operation. It's worth mentioning that closure follows from the definition, though.

The test of closure in the examples is in fact a test that the described mapping is inded a binary operation.

Any thoughts before I wade on in and make changes?

"This was our first example of a non-abelian group, because the operation o here is not commutative as the table shows. " If the table did show commutativity, would it be symmetrical about the diagonal from top left to bottom right? TimJ 5 Feb 2002

Yes. The group is abelian if and only if the table is symmetric about the main diagonal. --Zundark, 2002 Feb 5

Could someone put up a good description of Sylow's Theorem?

See Sylow theorems.

Also, it'd be nice to see a page dedicated to examples of groups.

Nice exercise: Classify all (isomorphism classes of) groups of order <=60. I'd like to see a page on that.

Do you realise how complicated this is, especially for order 32? Doing order <= 15 might be feasible, however. --Zundark, 2002 Feb 22

I started it at list of small groups. AxelBoldt

It simply is not true that the translation group is "our first example of a Lie group" (as the article is currently arranged). (Z,+) is also a Lie group; it is simply discrete, or 0 dimensional. 0 dimensional Lie groups (discrete groups) are studied in ordinary group theory rather than Lie theory, but they are still technically Lie groups. Hence the necessity for the adjective "nondiscrete". (If you want to change "nondiscrete" to "nontrivial", then I won't fight that, although I won't advocate it either.) -- Toby, 2002/04/03

Is it common to allow 0-dimensional manifolds? What could possibly be gained by that? Every set is a 0-dimensional manifold. In EDM they don't specify what n is in an n-dimensional manifold, but from their definition of "manifold with boundary" it is clear that they implicitly assume n>0. Furthermore, would you typically find or expect Z_p in a list of simple Lie groups? AxelBoldt

Since the upper half plane of R is all of R, every 0D manifold with boundary is actually a boundaryless manifold. So it is safe to ignore n = 0 in that case. But I certainly hope that EDM allows for n = 1 for manifolds with boundary. In that case, what dimension is the boundary of this manifold with boundary? If you don't believe in 0D manifolds, you'll have to put an annoying "unless n = 1" into your statement of the theorem that the boundary of an nD manifold with boundary is an (n-1)D boundaryless manifold. And that answers your question "What could possibly be gained by that?" -- you gain the rest of the just at night knowing that you don't have to hunt through your theorems for the odd "unless" and "except". This is just one example of my favourite principle in mathematics: Don't ignore the trivial case! The trivial case may seem uninteresting (that's why we call it "trivial"), but if you slight it, then it will come back to haunt you with a thousand paper cuts.

As for lists of simple Lie groups, I wouldn't expect to find Z_p because the classification of simple discrete groups is part of another subject. (The only things that Lie theorists need to classify are simple connected Lie groups.) But I certainly wouldn't ignore that other subject when classifying semisimple Lie groups (unless I were only interested in the connected ones, again). For example, the Lorentz group is not connected and can't be decomposed into connected simple Lie groups. But it is a semidirect product of a discrete group and a connected semisimple Lie group. So it can be decomposed into simple Lie groups, where some of these simple Lie groups are discrete and some are connected. (The only Lie group that is both discrete and connected, of course, is the trivial group, which is too simple to be simple.) Again we see that the trivial dimension cannot be profitably ignored.

Nobody working on Lie groups is ignorant of their need for the classification of simple discrete groups. But they're so used to the fact that simple Lie groups come in 2 families (the discrete ones and the connected ones) and the fact that only 1 of those families is classified by their own field of mathematics that they naturally consider only their family to be simplie as Lie groups. And this is easy to make rigorously true by redefining the term "simple Lie group". Instead of saying that a simple Lie group is a Lie group that has exactly 2 normal subgroups, say that its normal subgroups fall into exactly 2 dimensions, or that its Lie algebra is simple. (The classification of simple Lie groups, after all, is generally studied on the level of Lie algebras.)

So to sum up: No, I would not expect to find a discrete group on a list of simple Lie groups. But yes, I would expect to find -- and have found -- semisimple Lie groups in use that have nontrivial discrete normal subgroups, and I would expect the Lie theorists that use them to know and understand this. Lie theorists don't study discrete groups, but they ignore them at their peril, and the latter is true of anyone that would ignore the trivial case.

-- Toby Bartels

PS: Could change "nondiscrete" to "connected" as well as "nontrivial", assuming that the trivial group is never discussed above the word, but I think that that gets less at what we want to say. Could also put the word in parentheses to indicate its status as a technicality. -- Toby

I am convinced. Allowing dimension 0 gives a much nicer category of Lie groups. For instance, the kernel of a map between "ordinary" Lie groups may very well end up to be a discrete group. We should probably add this to Lie group. AxelBoldt

That's probably a better argument than anything that I said. Heh. (Certainly related to what I said but better expressed.) -- Toby

The article contained a characterization of GL(n,R) as consisting of rotations, reflections, dilations of R that keep the origin fixed; SL(n,R) was likewise explained as the rotations and reflections. I added those originally, then they were removed and readded. But they are indeed wrong. Even SL(2,R) contains lots of transformation different from rotations and reflections. There are skew transformations like ], dilations like ], and combinations of these like ]. AxelBoldt, Thursday, May 23, 2002

I put them back because I thought that I saw the error — you didn't specify that the origin was fixed, so affine transformations were included. Of course, the errors were more extensive, so I've now gone over everything more carefully. What we have now should be correct; I hope that you like it. — Toby Bartels, Thursday, May 23, 2002

Axel, you say that there are elements of SO(n) that are not rotations. Did you have an example in mind? I'm 100% certain that this is wrong, and I checked with 2 other mathematical physicists (Miguel and John Baez if you want credentials), who agree that I am not having an acid flashback or something. (Heck, it's even called the "rotation group" — which would make a nice name for an article on it.) I'm not sure that I like the description of GL(n) yet, and I'll think about that some more, but in any case, surely we can agree that the rotation group consists precisely of rotations? — Toby Bartels, Tuesday, May 28, 2002

The example I had in mind is reflection at the origin in R: ]. Or can that be interpreted as a rotation somehow? I don't even know what a rotation is :-) AxelBoldt, Tuesday, May 28, 2002

When I talked to John, he remarked "How obvious it is depends on how you define a rotation. I define it as an element of SU(n).". Your matrix has infinitesimal generator (say) a := ]; we can transform R continuously (even smoothly — analytically if you get right down to it ^_^) along the path t → exp(ta); at each stage, globally and infinitesimally, we preserve lengths and angles; at t = π, we reach your matrix. (Of course, there are all kinds of curvier paths that we could take, but I think that this is a minimal geodesic, so it must ^_^ be best.) This seems intuitively like a rotation to me. (And this talk of preserving lengths and angles isn't entirely begging the question — we get orientation preserving for free. The same thing shows why only the proper orthochronous part of SO(3,1) is the Lorentz group of rotations on Minkowski spacetime.) I don't know whether or not this cuts it as an intuitive argument, but I'm still certain that both mathematicians and physicists would count your matrix as a rotation — and who besides them ever thinks about rotations in R anyway? -- Toby Bartels, Friday, June 7, 2002

I like this notion of rotation. I had some muddy image in mind where you stake a line through the space and then rotate around that line, but this is a lot cleaner. I will officially give up my resistance to the term "rotation" and retract everything I said before. AxelBoldt, Friday, June 7, 2002

Ah, but in 4D a rotation is about a plane... :) BTW, has anyone written about Jean-Pierre Petit? He does great vulgarization of maths in comic book form, his book on topology (which covers Klein bottles & Boy's surface etc) has a preface which recommends readers have a packet of aspirins & a glass of water to hand... same applies here I think ;) Tarquin

And in 2D, a rotation is about a point. Also, in 4D, a rotation need not be about a single plane (Axel's −1 is an example of such) but may instead be around a sort of linear combination of planes. This is related to the fact that not every 2form in 4D can be factored as a wedge product of 1forms.

In general, the direction of the "axis" of a rotation in nD can be described by an (n−2)form, specified up to a scalar multiple. This (n−2)form is another way of looking at the infinitesimal generator of a rotation. You can see how the infinitesimal generator that I gave for Axel's rotation is a linear combination of 2 blocks, which correrspond to 2 factorable 2-forms and hence 2 planes. The rotation is about both of those planes simultaneously. -- Toby Bartels, Sunday, June 9, 2002

So it is true that any element of SO(n) is a product of rotations about n-2 dimensional hyperplanes? AxelBoldt

I'm pretty sure that that's true. If it isn't, then the analogy between axes of rotation and 2forms doesn't work like I think that it does (but perhaps that's exactly what you're getting at). I can check it out if you want. — Toby Bartels, Tuesday, June 11, 2002