Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,353 answers , 22,789 comments
1,470 users with positive rep
820 active unimported users
More ...

  Why is the Yang-Mills gauge group assumed compact and semi-simple?

+ 8 like - 0 dislike
2306 views

What is the motivation for including the compactness and semi-simplicity assumptions on the groups that one gauges to obtain Yang-Mills theories? I'd think that these hypotheses lead to physically "nice" theories in some way, but I've never, even from a computational perspective. really given these assumptions much thought.

This post imported from StackExchange Physics at 2015-01-19 14:11 (UTC), posted by SE-user joshphysics
asked Jan 29, 2013 in Theoretical Physics by joshphysics (835 points) [ no revision ]
retagged Jan 19, 2015
Compactness is needed for the bilinear form on the adjoint representation to be positively definite. For example, $SO(2,1)$ would be no good because the signature on the adjoint is ${+}{-}{-}$. If we had an indefinite form, the norm of the different colorful polarizations of the gauge bosons would have different signs (ghosts, negative probabilities). In a similar way, some Lie algebras (not semisimple etc.) have "zero norm" directions. Ultimately, we decompose the gauge group to simple compact pieces - the factors behave independently and decouple.

This post imported from StackExchange Physics at 2015-01-19 14:11 (UTC), posted by SE-user Luboš Motl

3 Answers

+ 6 like - 0 dislike

It's because you want the kinetic part of the Yang Mills action $$ \int Tr({\bf{F^2}}) dV$$ to be positive definite. To guarantee this, the Lie algebra inner product you're using (Killing form) needs to be positive definite. This is guaranteed if the gauge group is compact and semi-simple. (I'm not sure if it's only if G is compact and semi simple though. Maybe someone else could fill in this detail).

This post imported from StackExchange Physics at 2015-01-19 14:11 (UTC), posted by SE-user twistor59
answered Jan 29, 2013 by twistor59 (2,500 points) [ no revision ]
for non-compact groups, the Killing form is indefinite; for compact ones, the Killing form is negative definite or negative semi-definite, depending on whether the Lie algebra is semi-simple or reductive, respectively

This post imported from StackExchange Physics at 2015-01-19 14:11 (UTC), posted by SE-user Christoph
@Christoph OK thanks for the clarification.

This post imported from StackExchange Physics at 2015-01-19 14:11 (UTC), posted by SE-user twistor59
@Christoph : I think it's worthwhile to add here that you require the Killing Form to be (semi-) negative-definite since you are implicitly using a Russian metric (+,-,-,...,-). $\\$ I just spent a while getting confused between your answer and that above, which is implicitly using a mostly-plus metric (-,+,+,...,+). Please correct me if this is wrong.

This post imported from StackExchange Physics at 2015-01-19 14:11 (UTC), posted by SE-user Flint72
+ 6 like - 0 dislike

As Lubos Motl and twistor59 explain, a necessary condition for unitarity is that the Yang Mills (YM) gauge group $G$ with corresponding Lie algebra $g$ should be real and have a positive (semi)definite associative/invariant bilinear form $\kappa: g\times g \to \mathbb{R}$, cf. the kinetic part of the Yang Mills action. The bilinear form $\kappa$ is often chosen to be (proportional to) the Killing form, but that need not be the case.

If $\kappa$ is degenerate, this will induce additional zeromodes/gauge-symmetries, which will have to be gauge-fixed, thereby effectively diminishing the gauge group $G$ to a smaller subgroup, where the corresponding (restriction of) $\kappa$ is non-degenerate.

When $G$ is semi-simple, the corresponding Killing form is non-degenerate. But $G$ does not have to be semi-simple. Recall e.g. that $U(1)$ by definition is not a simple Lie group. Its Killing form is identically zero. Nevertheless, we have the following YM-type theories:

  1. QED with $G=U(1)$.

  2. the Glashow-Weinberg-Salam model for electroweak interaction with $G=U(1)\times SU(2)$.

Also the gauge group $G$ does in principle not have to be compact.

This post imported from StackExchange Physics at 2015-01-19 14:11 (UTC), posted by SE-user Qmechanic
answered Jan 29, 2013 by Qmechanic (3,120 points) [ no revision ]
First link dead: In lack of better, use e.g. this link instead.

This post imported from StackExchange Physics at 2015-01-19 14:11 (UTC), posted by SE-user Qmechanic
+ 5 like - 0 dislike

I recommend you to read the chapter 15.2 in "The Quantum Theory of Fields" Volume 2 by Steven Weinberg, he answers precisely your question.

Here a short summary
In a gauge theory with algebra generators satisfying $$ [t_\alpha,t_\beta]=iC^\gamma_{\alpha\beta}t_\gamma $$ it can be checked that the field strength tensor $F^\beta_{\mu\nu}$ transforms as follows $$ \delta F^\beta_{\mu\nu}=i\epsilon^\alpha C^\beta_{\gamma\alpha} F^\gamma_{\mu\nu} $$ We want to construct Lagrangians. A free-particle kinetic term must be a quadratic combination of $F^\beta_{\mu\nu}$ and Lorentz invariance and parity conservation restrict its form to $$ \mathcal{L}=-\frac{1}{4}g_{\alpha\beta}F^\alpha_{\mu\nu}F^{\beta\mu\nu} $$ where $g_{\alpha\beta}$ may be taken symmetric and must be taken real for the Lagrange density to be real as well. The Lagrangian above must be gauge-invariant thus it must satisfy $$ \delta\mathcal{L}=\epsilon^\delta g_{\alpha\beta}F^\alpha_{\mu\nu}C^\beta_{\gamma\delta}F^{\gamma\mu\nu}=0 $$ for all $\epsilon^\delta$. In order not to impose any functional restrictions for the field strengths $F$ the matrix $g_{\alpha\beta}$ must satisfy the following condition $$ g_{\alpha\beta}C^\beta_{\gamma\delta}=-g_{\gamma\beta}C^\beta_{\alpha\delta} $$ In short, the product $g_{\alpha\beta}C^\beta_{\gamma\delta}$ is anti-symmetric in $\alpha$ and $\gamma$.
Furthermor the rules of canonical quantization and the positivity properties of the quantum mechanical scalar product require that the matrix $g_{\alpha\beta}$ must be positive-definite. Finally one can prove that the following statements are equivalent

  1. There exists a real symmetric positive-definite matrix $g_{\alpha\beta}$ that satisfies the invariance condition above.
  2. There is a basis for the Lie algebra for which the structure constants $C^\alpha_{\beta\gamma}$ are anti-symmetric not only in the lower indices $\beta$ and $\gamma$ but in all three indices $\alpha$, $\beta$ and $\gamma$.
  3. The Lie algebra is the direct sum of commuting compact simple and $U(1)$ subalgebras.

The proof for the equivalence of these statements as well as a more in-detail presentation of the material can be found in the aforementioned book by S. Weinberg.

A proof for the equivalence for $g_{\alpha\beta}=\delta_{\alpha\beta}$ (actually the most common form) was given by M. Gell-Mann and S. L. Glashow in Ann. Phys. (N.Y.) 15, 437 (1961)


This post imported from StackExchange Physics at 2015-01-19 14:11 (UTC), posted by SE-user Stan

answered Sep 18, 2013 by Stan (60 points) [ revision history ]
edited Mar 3, 2015 by Arnold Neumaier
+1: Thanks Stan I'll definitely take a close look at Weinberg as well.

This post imported from StackExchange Physics at 2015-01-19 14:11 (UTC), posted by SE-user joshphysics

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOverfl$\varnothing$w
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...