User talk:Gauge
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feel free to place any constructive criticism or comments here. i am a friendly guy, so please no yelling. - Gauge 00:23, 4 Aug 2004 (UTC)
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Cauchy-Schwarz inequality
My faithful bot asked me to tell you that he is much happier now with downcased "which". :) Actually I do remember that particular article (I double check every article) but I was too lazy to actually downcase "which" myself. I think the style of the article is better now. And thanks for not reverting on me. :) Oleg Alexandrov 04:28, 22 Mar 2005 (UTC)
- I'm glad the bot is happy; it was annoying me too :-) I don't revert except in cases of vandalism. These bots are a great service, so thanks! - Gauge 05:30, 26 Mar 2005 (UTC)
Hi
Just noticed that you're editing articles on my watchlist, so I thought I'd say hi! linas 04:35, 20 May 2005 (UTC)
- Hi :-) I recently became interested in analytic number theory so I've been browsing and editing those pages quite a bit. In the modular forms article I was wondering if F could somehow be interpreted as a functor. Would you know anything about this? - Gauge 04:43, 20 May 2005 (UTC)
- I've never had any formal education on category theory, and whenever I try to read a book on it, I fall asleep on page 3 because it always seems too simple. User:Charles Matthews would know; but he's also quite busy as he's the math overlord here on Wikipedia.
Color charge
I can't quite give an accurate definition for color charge off the top of my head. I think you can say that the color charge of a Lie algebra (any Lie algebra) is the same as the root system of the algebra. To be more correct, we should really talk about a given irreducible and/or semi-simple rep. The reason that this is so is understood by working with the principal bundle of the associated Lie group. The connection (fiber bundle) is Lie-algebra valued. The Hamiltonian mechanics of a classical (not quantum!) particle coupled to the principal bundle is such that the momentum can be written as
- <math>p = d+qA<math>
where d is the exterior derivative on the base space (or the covariant derivative if the base space is curved) and A is the connection on the principal G-bundle. Here q is the charge, it is a vector that multiplies the Lie-algebra-valued A so that p becomes a vector in the space of the Lie algebra as well. It should be immediately recognized that p is the covarient derivative on the fiber bundle. When the underlying manifold (the base space) is physical space-time, then p can be understood to be the momentum of a particle with a given color charge, being acted on by the gauge field A.
The term charge comes from analogy to the electromagnetic field. Mathematicaly speaking, the elecromagentic field can be understood to be a line bundle or a principle bundle with fibre U(1). The strength of the electromagnetic field (literally, the strength of the electric field and the magnetic field) is given by the curvature <math>F=dA<math> of the connection A. (In Minkowski spacetime, one of the four components of A is called the electric potential and the other three are called the magnetic vector potential). (More generally, the curvature of a principle bundle is given by <math>F=dA+[A,A]<math> but the second term vanishes when the Lie algebra is Abelian. In the non-Abelian case, F is known as the Yang-Mills field.) The traditional coupling of an electrically charged particle is given thorough it's momentum, p=d+qA as above, the momentum being important as it gives the particle's motion through space, and, in the fourth component, through time. The momentum can essentially be understood to be the covariant derivative on the electromagnetic line bundle.
The root system for U(1) is trivial, it is just the scalar 1. Thus, the fiber-bundle interpretation immediately gives some insight to the problem of the quantization of the electric charge: the charge is not just any value, it is a particular value,and that value is 1.
In quantum chromodynamics, the relevant Lie group is SU(3) and the Lie algebra is the smallest irreducible representation su(3). Actually, su(3) has a pair of conjugate representations; they look identical except that they are complex conjugates. The root system can best be understood to be two pairs of three vectors each, oriented 120 degrees apart, forming a Star of David. The vectors are of length 1/2; one triangle belongs to one representation, the other to its conjugate. Quarks transform as one representation, anti-quarks transform as the conjugate representation. That is, the color charge of a quark is thus associated with one of the three vectors, and the charge of the anti-quark is associated with one of the vectors of the conjugate representation.
Generalizations to supersymmetry and to curved base spaces follow the same pattern laid out here: one defines a principle bundle over some group, and couples fields by means of the covariant derivative on the bundle. For good measure, one recognizes that four-dimensional space-time can be represented by a pair of spinors, each spinor transforming under su(2) or its complex conjugate. Spinors naturally have an anti-symmetric algebra, by means of the Pauli exclusion principle. One fundamental problem of supersymmetry is that there are so many different Lie groups and couplings and representations one can choose from, leading to a bewildering number of fields and charges and the like.
Phew. Heh, reasonable for an imprecise definition? I'm gonna copy this into the article,I guess ... linas 05:24, 21 May 2005 (UTC)
- Outstanding work! More than I expected, and certainly welcome. I will ponder this for a while. - Gauge 05:34, 21 May 2005 (UTC)
Crap. Actually, there is a glaring, embarassing error in there. I need to think it through. Wham-bam job, leads to errors. Dohh. Unfortunately I already copied into the color-charge page, so I fully expect to get whacked in a few hours when others notice. linas 06:53, 21 May 2005 (UTC)
Err, well, at the hand-waving level, I guess maybe it can pass muster. I'll fix the wording in the main article. For quantum fields, the field carries the charge, not q, so q becomes this dead scalar, and the field becomes a vector. linas 06:58, 21 May 2005 (UTC)
Proof of the Sylow Theorems
The following proofs are based on combinatorial arguments of Wielandt and give much shorter proofs of the Sylow theorems than those found in most texts. In the following, we use a | b as notation for "a divides b" and a <math>\nmid<math> b for the negation of this statement.
Theorem 1: A finite group G whose order |G| is divisible by a prime power pk has a subgroup of order pk.
Proof: Let |G| = pkm, and let pr be chosen such that no higher power of p divides m. Let Ω denote the set of subsets of G of size pk and note that |Ω| = <math>{p^km \choose p^k}\mathrm{,}<math> and furthermore that pr+1 <math>\nmid<math> <math>{p^km \choose p^k}<math> by the choice of r. Let G act on Ω by left multiplication. It follows that there is an element A ∈ Ω with an orbit θ = AG such that pr+1 <math>\nmid<math> |θ|. Now |θ| = |AG| = [G : GA] where GA denotes the stabilizer subgroup of the set A, hence pk | |GA| so pk ≤ |GA|. Note that the elements ga ∈ A for a ∈ A are distinct under the action of GA so that |A| ≥ |GA| and therefore |GA| = pk. Then GA is the desired subgroup.
Lemma: Let G be a finite p-group, let G act on a finite set Ω, and let Ω0 denote the set of points of Ω that are fixed under the action of G. Then |Ω| ≡ |Ω0| mod p.
Proof: Write Ω as a disjoint sum of its orbits under G. Any element x ∈ Ω not fixed by G will lie in an orbit of order |G|/|CG(x)| (where CG(x) denotes the centralizer), which is a multiple of p by assumption. The result follows immediately.
Theorem 2: If H is a p-subgroup of a finite group G and P is a Sylow p-subgroup of G then there exists a g ∈ G such that H ≤ gPg−1. In particular, the Sylow p-subgroups for a fixed prime p are conjugate in G.
Proof: Let Ω be the set of left cosets of P in G and let H act on Ω by left multiplication. Applying the Lemma to H on Ω, we see that |Ω0| ≡ |Ω| = [G : P] mod p. Now p <math>\nmid<math> [G : P] by definition so p <math>\nmid<math> |Ω0|, hence in particular |Ω0| ≠ 0 so there exists some gP ∈ Ω0. It follows that hgP = gP so g−1hgP = P, g−1hg ∈ P, and thus h ∈ gPg−1 ∀ h ∈ H, so that H ≤ gPg−1 for some g ∈ G. Now if H is a Sylow p-subgroup, |H| = |P| = |gPg−1| so that H = gPg−1 for some g ∈ G.
Theorem 3: The number of Sylow p-subgroups of a finite group G divides the order of G and is congruent to 1 mod p.
Proof: By Theorem 2, the number of Sylow p-subgroups in G is equal to [G : NG(P)], where P is any such subgroup, and NG(P) denotes the normalizer of P in G, so this number is a divisor of |G|. Let Ω be the set of all Sylow p-subgroups of G, and let P act on Ω by conjugation. Let Q ∈ Ω0 and observe that then Q = xQx−1 for all x ∈ P so that P ≤ NG(Q). By Theorem 2, P and Q are conjugate in NG(Q) in particular, and Q is normal in NG(Q), so then P = Q. It follows that Ω = {P} so that, by the Lemma, |Ω| ≡ |Ω0| = 1 mod p.