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I am trying to reproduce the results of the (famous) Myer's paper "Dielectric Branes" https://arxiv.org/abs/hep-th/9910053. In eq (33), when he expands the determinant factor for a flat-space background, there is a 1/4 in the first term that, says me, should be 1/2. It is rather straightforward to compute the determinant in the flat background limit. This is my computation:

$$ Q^i_j=\delta^i_j+i\lambda[\phi^i,\phi^k]\eta_{kj},\\ \Rightarrow\det Q^i_j=1-\lambda^2[\phi^i,\phi^j][\phi^j,\phi^i]\\ \sqrt{\det Q^i_j}\approx1-\frac{\lambda^2}{2}[\phi^i,\phi^j][\phi^j,\phi^i].\\ $$ But the end result of Myer's is $$ \sqrt{\det Q^i_j}\approx1-\frac{\lambda^2}{4}[\phi^i,\phi^j][\phi^j,\phi^i]\tag{33}. $$ The steps are so straightforward that I feel I am really missing some fundamental content here. Does somebody knows where this extra 1/2 factor comes from?

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Hint: If we write

$$ Q^i{}_j ~=~\delta^i_j -C^i{}_j\qquad\Leftrightarrow\qquad Q~=~{\bf 1} -C , \tag{A}$$

then

$$\ln \det Q ~=~ {\rm tr}\ln Q~\stackrel{(A)}{=}~{\rm tr}\ln ({\bf 1} -C)~=~-\sum_{n=1}^{\infty}\frac{1}{n}{\rm tr}(C^n)~=~-\underbrace{{\rm tr}(C)}_{ = 0 } -\frac{1}{2}{\rm tr}(C^2) + {\cal O}(C^3),\tag{B} $$

so that

$$ \sqrt{\det Q}~=~ \exp\left(\frac{1}{2}\ln \det Q \right) ~\stackrel{(B)}{=}~1-\color{red}{\frac{1}{4}}{\rm tr}(C^2) + {\cal O}(C^3) .\tag{C}$$

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  • $\begingroup$ Awesome! Clear and nice derivation. Thank you very much! $\endgroup$ – DB. Rane Jul 28 '18 at 13:51

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