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.


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!

  • $\begingroup$ The ghost number operator $U$ is manifestly Grassmann-even, and commutes with the Lie algebra generators, $UK_r=+K_rU$. Therefore, the sought-for minus sign in the last equation of the question(v1) is not possible (unless, of course, the Lie algebra is just $u(1)$, in which case both sides are actually zero because the ghost $c^1c^1=0$ is nilpotent). $\endgroup$
    – Qmechanic
    Oct 1 '11 at 16:53

The book has a minus sign typo in the previous equation. The commutation in the step you are looking at does not introduce a minus sign, but the previous line had a term of the form

$$ - c^r c^i b_r K_i$$

with a minus sign in front, which requires you to commute the two c's past each other. The next step writes

$$ - c^i c^r b_r K_i $$

incorrectly, it should be a plus sign, because the c's are commuted, and then the next step says

$$ c^i K_i c^r b_r $$

Which fixes the error. You assumed that the minus sign came from moving the K, which does nothing, when it actually comes from commuting the c's past each other. Thank you for linking the book, it would have been impossible to find the error otherwise.

Given the unmotivated and needlessly formal introduction of BRST in this book, I would recommend that you read the introduction to BRST in the appendices of Polchinski. The exercise in question does not require formal symbol manipulation gymnastics in an operator calculus, it is a simple question.

  • $\begingroup$ Just found this typo by myself too. Yes, there are many typos in the current version of this book. I will read Polchinski's. Thanks a lot. $\endgroup$
    – Osiris Xu
    Oct 2 '11 at 4:44

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