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

146 submissions , 123 unreviewed
3,953 questions , 1,403 unanswered
4,889 answers , 20,761 comments
1,470 users with positive rep
506 active unimported users
More ...

BRST quantization of point particle

+ 1 like - 0 dislike
172 views

Suppose we have Lie algebra $\mathfrak{g}$ with basis $t_a$ with a representation $$ t_a \mapsto K_a: V \to V. $$ Denote by $c^a$ the dual basis. Chevalley differential is defined as $$ Q = c^i K_i - \frac{1}{2} f_{ab}^c c^a c^b \frac{\partial}{\partial c^c}. $$ If I try to write BRST transformation for the theory of free relativistic point, $\mathfrak{g}$ will be the algebra of vector fields on the line, which is infinite-dimensional. I don't want to compute structure constants in some basis. Note that the first part of the differential is the action of geberal odd element of the Lie algebra $v = c^i t_i$. Its square is $$ v^2 = c^i t_i c^j t_j = \frac{1}{2} c^i c^j f_{ij}^k t_k. $$ Than to compute the action of $Q$ to $c$-ghost we should take the square of the universal odd element with minus sign. In the case of the relativistic particle the universal odd element is $$ v = c(t) \frac{d}{dt}, $$ its square is $$ v^2 = c \dot c \frac{d}{dt}. $$ Then we get $$ Q X = c \dot X, Qc = -c \dot c, $$ but the sign is wrong. I can't figure out mistake. This $Q$ is not nilpotent. Are there other ways to copmute BRST-transformation not introducing the basis of the Lie algebra?

This post imported from StackExchange Physics at 2014-12-09 15:14 (UTC), posted by SE-user user46336
asked Nov 28, 2014 in Theoretical Physics by user46336 (5 points) [ no revision ]
$Q$ is an anti-derivation see here en.wikipedia.org/wiki/Antiderivation#Graded_derivations also $Qc = \frac{i}{2}[c,c]$

This post imported from StackExchange Physics at 2014-12-09 15:14 (UTC), posted by SE-user Autolatry
Comment to the question (v1): Consider adding references in order to receive useful and focused answers.

This post imported from StackExchange Physics at 2014-12-09 15:14 (UTC), posted by SE-user Qmechanic
Clearly $Q$ is an odd derivation. I don't have any $i$. Commutation relations are $[t_i,t_j] = f_{ij}^k t_k$. Then $Qc^a$ should be $1/2 f^a_{bc} c^b c^c$. I have looked through a lot of references about BV-BRST, and did not get an answer to my question.

This post imported from StackExchange Physics at 2014-12-09 15:14 (UTC), posted by SE-user user46336

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$\hbar$ysicsOve$\varnothing$flow
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
...