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3 votes
0 answers
59 views

Global properties of the gauge group

In this very good P.E. answer, it is explained precisely what it means for a quantum system/theory to have a symmetry group $G$ (where $G$ is a Lie group): going back to first principles, it means ...
SolubleFish's user avatar
  • 5,989
3 votes
1 answer
58 views

Why semi-simple and compact Gauge Group in YM Theory? [duplicate]

I'm studying the Yang-Mills theory, with the Action: $$ S=-\frac{1}{2}\int\mathrm{tr}_{\rho}(\mathcal{F}\wedge\star\mathcal{F}) $$ where $\mathcal{F}:=\mathrm{d} \mathcal{A}+\frac{1}{2}[\mathcal{A},\...
BoyanLiu's user avatar
0 votes
0 answers
39 views

Wilson loop is not an element of $\mathrm{SU}(3)$ in color deconfinement

The center symmetry in QCD comes from the $$a\ \mathcal{P}\mathrm{exp}\left(ig_s \int_C dx^\mu \ A_\mu(x)\right) a^{-1} = \mathcal{P}\mathrm{exp}\left(ig_s \int_C dx^\mu \ A_\mu(x)\right),$$ where $C$ ...
Joao Vitor's user avatar
0 votes
1 answer
78 views

Why we only have one gauge coupling constant for $\rm SU(2)$ gauge symmetry?

I know that in the case of a Yang-Mills $\rm SU(2)$ gauge symmetry, the covariant derivative is written as: $$D^{\mu} = \partial^{\mu} + i\frac{g}{2}W^{\mu}_{a}\tau^{a}$$ With $g$ is the coupling ...
ياسين المهتدي's user avatar
0 votes
2 answers
122 views

Is there any physical reason behind the choice of Lie group in a Yang-Mills theory?

A Yang-Mills theory can be constructed for any Lie group that is compact and semisimple. The motivation behind this is discussed in this question. Is there any physical reason we choose $SU(3)$ or $U(...
CBBAM's user avatar
  • 3,350
1 vote
1 answer
83 views

(Anti-)Fundamental Representation of $SU(5)$ GUT

In many places, it has been mentioned that the sum of electrical charges of the particles present in $\overline{5}$ of $SU(5)$ is zero since the trace of $SU(5)$ generators is zero. I do not ...
Alex's user avatar
  • 85
0 votes
0 answers
73 views

Do we know a reason for why exactly $\rm U(1)$, $\rm SU(2)$ and $\rm SU(3)$? [duplicate]

I always found it a curiousity that in the symmetry groups of the known fundamental forces we find the nice arithmetic progression $1,2,3$: first there is $\DeclareMathOperator{\U}{U}\...
M. Winter's user avatar
  • 531
0 votes
0 answers
63 views

$U(n)$ vs $SU(n)$ Symmetry and Tracelessness - Chargeless Gauge Bosons

I've recently been studying the symmetries that give rise to the bosons of the standard model. I have taught myself just enough group theory to kind of understand what $U(1)$, $SU(2)$, and $SU(3)$ ...
18th Shard's user avatar
-1 votes
1 answer
183 views

How do I see the connections between abelian/non-abelian group and their gauge transformations?

I'm learning QED in my QFT class without too much background in group theory. Recently I'm introduced to the Abelian gauge transformation $$ D_\mu\psi(x)\rightarrow e^{i\alpha(x)}D_\mu\psi(x)\quad A_\...
IGY's user avatar
  • 1,783
2 votes
1 answer
594 views

What does $SU(2)\times U(1)$ really mean?

I know that $SU(2)\times U(1)$ is the product of each element of $SU(2)$ by an element of $U(1)$. But looking at the Lagrangian, such a product exists only in covariant derivatives of Higgs or fermion ...
Peter's user avatar
  • 343
1 vote
1 answer
137 views

Trace of $SU(N)$ generators and number of spin components

I am reading Peskin's and Schroeder's book "An Introduction to Quantum Field Theory". At Chapter 16.6, the authors write down an expression for the trace of of two generators of $SU(N)$ $$\...
schris38's user avatar
  • 3,992
3 votes
1 answer
130 views

Why does this infinitesimal transformation tell us the boson vectors carry charge?

I'm studying from "Cheng, Li - Gauge theory of elementary particle physics", and in section 8.1, where it talks about non-Abelian gauge symmetry and the theory you can construct imposing ...
Rhino's user avatar
  • 693
4 votes
2 answers
430 views

Why are there no instantons in the gauge group $U(1)$?

I am working my way through Srednicki's QFT book and am in chapter 93. Near the end, Srednicki says "If the gauge group is $U(1)$, there are no instantons, and hence no vacuum angle." I'm ...
Cory's user avatar
  • 143
2 votes
1 answer
294 views

Why standard model uses Lie groups like $SU(2)$ and not $SL(2,\mathbb{C})$?

First of all, the question is written in section $2)$. Also, I known that the $SU(2)$ group do not appears "alone" in standard model, rather, inside the Glashow-Salam-Weinberg model. 1) ...
M.N.Raia's user avatar
  • 3,085
2 votes
0 answers
82 views

Doubt on $SU(2)_{L} \times U(1)_{Y}$ covariant derivative and its action on a fermion

I) Introduction I.1) The mathematical structure is quite clear: given a spacetime $M$, and a Lie group $G$ (the gauge group), we can construct the Principal bundle $P^{G}_{M}$. The connection $1$-form ...
M.N.Raia's user avatar
  • 3,085

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