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Given a Lie algebra $[K_i,K_j]=f_{ij}^k K_k$, and ghost fields satisfying the anticommutation relations $\{c^i,b_j\}=\delta_j^i$, the ghost number operator is then $U=c^ib_i$ (duplicate indices are summed).

The question raised in proving the BRST operator raising the ghost number by 1, given in the Example 6.1 on the page 116-118 of the book, String theory demystified.


On the second and the third lines of the formula derivation of $UC^iK_i$


where we need $c^r U K_r =-c^r K_r U$

The change of sign above is not manifestly obvious to me.

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There's not enough context to answer the question. Nothing in the first two paragraphs has anything to do with parity. In the first paragraph you wrote down two unrelated (anti)commutation relations, and in the second paragraph you want to derive mixed commutation relations that don't follow from the separate relations (at least not without further context). Then in the third paragraph you try to argue with parity, which hadn't occurred anywhere before. Either you're mixing things up, or you're assuming a lot of physics background that you can't assume here. –  joriki Sep 30 '11 at 23:49
Thanks for comments. The link to the Google book is added. –  Osiris Xu Oct 2 '11 at 2:38
I suggest you delete this question. It appears you found the answer, but neither the question nor the answer is self-contained, and it's not clear (to me) what the problem or the solution was, so I don't see much value in it for anyone other than you. If you disagree, please write the answer up as an answer (not as part of the question) such that it's accessible to everyone, and accept it so that the question doesn't remain unanswered. –  joriki Oct 2 '11 at 5:42

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Never mind. I find the second line cheated me. It should be $UC^iK_i=...=c^iK_i+c^i\displaystyle\sum_rc^rb_rK_i=c^iK_i+c^iK_i\displaystyle\sum_rc^rb_r=...$

Furthermore, in the proof followed in this example, there a few typos need to be corrected. But the final result is right. The BRST operator raises the ghost number by 1!

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