• 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.


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


(propose a free ad)

Site Statistics

180 submissions , 140 unreviewed
4,526 questions , 1,815 unanswered
5,153 answers , 21,935 comments
1,470 users with positive rep
713 active unimported users
More ...

  One question about "TASI 2003 LECTURES ON ANOMALIES" by Jeffrey A. Harvey

+ 2 like - 0 dislike

I am trying to understand the equation (122)

in  http://xxx.lanl.gov/pdf/hep-th/0509097.pdf

Making my proper computation I am obtaining

$$-\int_{M^4}da \wedge d\omega_{2} ^{1}=-\int_{M^4}[d^2a \wedge\omega_{2} ^{1}-d(da \wedge \omega_{2} ^{1}) ]=$$

$$-\int_{M^4}d^2a \wedge\omega_{2} ^{1}+\int_{M^4}d(da \wedge \omega_{2} ^{1}) =$$

$$-\int_{M^4}d^2a \wedge\omega_{2} ^{1}+\int_{\partial M^4}da \wedge \omega_{2} ^{1}= -\int_{M^4}d^2a \wedge\omega_{2} ^{1}+0= -\int_{M^4}d^2a \wedge\omega_{2} ^{1}$$

My question is:  if my computation is correct and then  a minus sign is missing in the last integral of (122)?

asked Sep 14, 2015 in Theoretical Physics by juancho (1,105 points) [ revision history ]
edited Apr 3, 2016 by juancho

2 Answers

+ 1 like - 0 dislike

You're totally right. Then it seems there may be a sign error in the formula.

answered Apr 3, 2016 by Bootstrapper [ no revision ]
+ 0 like - 0 dislike

It's just a statement about integration by parts, following from

$$d(d a \wedge \omega_2^{(1)}) = d^2 a \wedge \omega_2^{(1)} + d a \wedge d \omega_2^{(1)}$$

(since $a$ is a 0-form) together with $\int d \, (\text{anything}) = 0$.

answered Apr 3, 2016 by Bootstrapper [ no revision ]

Please note that the correct expression is

$$d(d a \wedge \omega_2^{(1)}) = d^2 a \wedge \omega_2^{(1)} - d a \wedge d \omega_2^{(1)}$$

given that $da$ is a 1-form.  Do you agree?

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:
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