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 ...

  Yang-Mills potential and principal bundles

+ 4 like - 0 dislike
1656 views

In section 2.7.2 of Bertlmann's "Anomalies in quantum field theory", it is stated that since a non-trivial principal bundle (based on a Lie group $G$) does not admit a global section, the Yang-Mills potential (the pullback of the connection on the total space) exists locally but not globally.
Covering the base space with various charts, we get different Yang-Mills potentials and we can identify two such potentials in the overlapping region of two charts, this give a transformation rule for going from one chart to the other chart. This is the geometrical interpretation of the gauge transformation.

Later, in section 6.1, the principal bundle of QED is defined based on $U(1)$, with the 4-dimensional Minkowski space being the base space. It is then stated that "the principle bundle is just trivial, $P=\mathbb{R}^4\times U(1)$, since the base space is contractible". From that, I infer that a single Yang-Mills potential can be defined globally on the base space and therefore there should not be any gauge transformation, which seems in contradiction with the usual textbook formulation of QED. What am I missing? Are things different for a non-Abelian Lie group?

This post imported from StackExchange Physics at 2016-07-07 18:16 (UTC), posted by SE-user BeJ
asked Jul 6, 2016 in Theoretical Physics by BeJ (20 points) [ no revision ]
This is a tricky little issue, which arises due to differences in math and physics terminology. I could answer in the weekend; no time right now. I think that everything you need is contained in these notes.

This post imported from StackExchange Physics at 2016-07-07 18:16 (UTC), posted by SE-user Danu

1 Answer

+ 3 like - 0 dislike

The term gauge transformation refers to two related notions in this context. Let $P$ be a principal $G$-bundle over a manifold $M$, and let $\cup_i U_i$ be a cover of $M$. A connection on $P$ is specified by a collection of $\mathfrak{g}=\mathrm{Lie}(G)$ valued 1-forms $\{A_i\}$ defined in each patch $\{U_i\}$, together with $G$-valued functions $g_{ij} : U_i \cap U_j \to G$ on each double overlap, such that overlapping gauge fields are related by

$$A_j = g_{ij} A_i g_{ij}^{-1} + g_{ij} \mathrm{d} g_{ij}^{-1}.\tag{1}$$

The transition functions must also satisfy the cocycle condition on triple overlaps, $g_{ij}g_{jk}g_{ki} =1$. This is the first notion of a gauge transformation, relating local gauge fields on overlapping charts.

Second, there is a notion of gauge equivalence on the space of connections. Two connections $\{ A_i, g_{ij} \}$ and $\{A_i',g_{ij}'\}$ are called gauge-equivalent if there exist $G$-valued functions $h_i : U_i \to G$ defined on each patch such that $$A_i' = h_i A_i h_i^{-1} + h_i \mathrm{d}h_i^{-1} ~~\text{and}~~ g_{ij}' = h_j g_{ij} h_i^{-1}\tag{2}$$

In terms of the globally defined connection 1-form $\omega$ on $P$, the local gauge fields $\{A_i\}$ are defined by choosing a collection of sections $\{\sigma_i\}$ on each patch of $M$. The local gauge fields are obtained by pulling back the global 1-form, $A_i = \sigma_i^* \omega$. On overlapping patches, such pullbacks are related by (1). On the other hand, the choice of sections was arbitrary; a different collection of sections $\{\sigma'_i\}$ related to the first by $\sigma'_i = \sigma_i h_i$ leads to the gauge-equivalence (2).

Given a map $f: M \to M'$ between two manifolds and a bundle $P'$ over $M'$, we obtain a bundle over $M$ by pullback, $f^* P'$. Moreover, the pullback bundle depends only on the homotopy class of $f$. Suppose we have a contractible manifold $X$. By definition, there exists a homotopy between the identity map $\mathbf{1}:X \to X$ and the trivial map $p: X \to X$ which takes the entire manifold to a single point $p\in X$. Let $P$ be a bundle over $X$. The identity pullback of course defines the same bundle, $\mathbf{1}^* P = P$. On the other hand, the pullback $p^* P$ is a trivial bundle; it maps the same fiber above $p$ to every point on $X$. But the bundles $\mathbf{1}^*P$ and $p^*P$ are equivalent since $\mathbf{1}$ and $p$ are homotopic maps. Thus, a bundle over a contractible space is necessarily trivial (i.e. a direct product).

In particular, a $G$-bundle over $\mathbb{R}^4$ is trivial, whether $G$ is abelian or non-abelian. The cover $\cup_i U_i$ has a single chart, $\mathbb{R}^4$ itself. There is a single gauge field $A$, which is a globally defined $\mathfrak{g}$-valued 1-form. It is obtained from the 1-form $\omega$ on $P$ by pullback, $A = \sigma^* \omega$, where $\sigma$ is a globally defined section. Picking another section $\sigma' = \sigma g(x)$ produces a gauge-equivalent connection, related to $A$ by the usual gauge transformation law given above.

For more details, see e.g. Nakahara "Topology, Geometry, and Physics," chapter 10.

This post imported from StackExchange Physics at 2016-07-07 18:16 (UTC), posted by SE-user user81003
answered Jul 6, 2016 by user81003 (130 points) [ no revision ]

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$\hbar$ysicsO$\varnothing$erflow
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
...