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  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?

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