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  Writing the Yang-Mills topological charge using differential forms

+ 3 like - 0 dislike

I have a very pedestrian knowledge of differential forms and I am having some trouble in a derivation. The topological charge $Q$ in Yang-Mills theories is supposed to be
My text https://munsal.files.wordpress.com/2014/10/marino-lectures2014.pdf equation (4.48) says that
I am trying to prove this. My attempt is the following
From here on I am stuck. I figure I am missing some key identity. Any tip would be appreciated.

asked Jun 14, 2016 in Theoretical Physics by Dmitry hand me the Kalashnikov (735 points) [ no revision ]

shouldn't $d^\beta$ be $dx^\beta$? should the last two formulas be preceded by an equal sign?

There is some mistake in the definition of your charge which is a scalar quantity. The indices $\rho,\sigma$ are not contracted.

2 Answers

+ 5 like - 0 dislike

I am not sure about the normalisations and your contraction is wrong since the charge should be a scalar quantity but you can play with the indices using the fact that 

$$ F_{ab}  F_{mn} dx^a \wedge dx^b \wedge dx^m \wedge dx^n = F_{ab}  F_{mn} \, \epsilon_{efgh} \epsilon^{pqrs}dx^edx^fdx^gdx^g \delta_p^a \delta_{q}^b\delta_{r}^m \delta_s^n$$

possibly times some normalisations that you can fix later. I trust this is correct but I would advise to check it out in books like Freedman's and Van Proyen's Supergravity or similar. After you do all the horrifying horrible contractions you should probably get what you want.

answered Jun 15, 2016 by conformal_gk (3,625 points) [ no revision ]
+ 3 like - 0 dislike

Hint: The $dx^\mu$ anticommute in the wedge product, and $d^4x$ is the wedge product of the $dx^\mu$ in increasing order.

answered Jun 14, 2016 by Arnold Neumaier (15,787 points) [ no revision ]

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