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

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

173 submissions , 136 unreviewed
4,271 questions , 1,618 unanswered
5,069 answers , 21,527 comments
1,470 users with positive rep
623 active unimported users
More ...

  Gauge invariance of Chern-Simons functional integral for a 3-manifold with boundary

+ 4 like - 0 dislike
58 views

I am trying to understand how the functional integral for Chern-Simons theory for a possibly non-compact 3-manifold with boundary is made gauge invariant.

For a compact 3-manifold, $M$, without boundary, it is well known (see, for example, section 2 of this reference), that for a compact simple Lie group $G$ and trivial principal G-bundle $P\rightarrow M$, one may define the Chern-Simons action \begin{equation} S[A]=\frac{k}{4\pi}\int_M\textrm{Tr}\bigg(A\wedge dA+\frac{2}{3}A\wedge A\wedge A\bigg). \end{equation} Here, the group, $\mathcal{G}$, of gauge transformations of $P$, is isomorphic to the group of smooth maps from $M$ to $G$. Under a gauge transformation $g\in \mathcal{G}$, the action changes by the sum of a boundary term and $\textrm{deg}(g)$, which labels the corresponding component of $\mathcal{G}$. The group, $\pi_0(\mathcal{G})$, of components is isomorphic to the group of homotopy classes of maps from $M$ to $G$, which for simply connected $G$ is isomorphic to $\pi_3(G)$. Since $G$ is simple, $\pi_3(G)\cong\mathbb{Z}$. Thus, upon requiring that $k$ is quantized, we find that the integrand of the functional integral, $e^{iS}$, is invariant under gauge transformations.

My question is, how does one extend this to the case where $M$ has a boundary, and is possibly noncompact? The example I have in mind is $M=D\times \mathbb{R}$, where $D$ is the disk. In this paper, it is explained that for gauge invariance, one first chooses one of the boundary components of the connection, $A$, to be zero, and with such a boundary condition the functional integral is invariant only under gauge transformations which are one at the boundary. This requirement is also alluded to below equation 3.18 of this paper by Dijkgraaf and Witten.

It is clear to me that the aforementioned boundary term that arises via gauge transformation will vanish via the boundary condition.

However, it is not clear to me why we require that $g$ be 1 at the boundary for gauge invariance.

Firstly, why do we need to impose another boundary condition on $g$, i.e., in addition to the constraint required to preserve the boundary condition on $A$ under gauge transformations? Secondly, why would another constant value for $g$ at the boundary not suffice? I would think that any common value for $g$ along the boundary would imply that we are effectively studying $M$ with boundary points identified, which is a closed 3-manifold, for which we can apply the arguments of the second paragraph above.

This post imported from StackExchange MathOverflow at 2019-04-13 07:47 (UTC), posted by SE-user Mtheorist
asked Apr 3 in Theoretical Physics by Mtheorist (100 points) [ no revision ]
retagged Apr 13
If you want your gauge transformations to be a group, then your boundary condition should be closed under pointwise multiplication. The only choice for a constant that is closed under multiplication is 1.

This post imported from StackExchange MathOverflow at 2019-04-13 07:47 (UTC), posted by SE-user Danny Ruberman
Sorry, could you elaborate on why the boundary condition ought to be closed under pointwise multiplication for the gauge transformations to be a group? I can only see that one should constrain the gauge transformations such that the boundary condition on A is preserved.

This post imported from StackExchange MathOverflow at 2019-04-13 07:47 (UTC), posted by SE-user Mtheorist
Isn't the gauge group supposed to be a group? So it needs to be closed under multiplication, and needs an identity element. Either of those would require that on a set where the transformation is constant, the only possible constant is 1.

This post imported from StackExchange MathOverflow at 2019-04-13 07:47 (UTC), posted by SE-user Danny Ruberman
Okay, now I understand what you mean, thanks. I think what I don’t understand is why we need the boundary value for g to be a constant in the first place. Is it not possible to show gauge invariance without assuming that g is constant at the boundary?

This post imported from StackExchange MathOverflow at 2019-04-13 07:47 (UTC), posted by SE-user Mtheorist

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).
To avoid this verification in future, please log in or register.




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

Your rights
...