I am trying to reproduce the results of the (famous) Myer's paper "Dielectric Branes" https://arxiv.org/abs/hep-th/9910053. I am struggling a bit to obtain the numerical factors in the equations. When he explains the D0-D2 bound state in terms of a stack of D0 branes, thus the non-abelian prescription, he decomposes the Chern-Simons term as follows in his equation (67a), $$ {\rm Tr}\left(\Phi^j\Phi^i[\Phi^k~\partial_k C_{ijt}^{(3)}(t)+ C_{ijk}^{(3)}(t)~D_t\Phi^k]\right) \tag{67a} $$ I am assuming that the $Tr(...)$ symbol here is actually Tsetlyn's symmetric trace description, $$ \text{STr}(A_1...A_n)=\frac1{n!}(A_1...A_n+\text{all possible permutations}), $$ which is supposed to capture the $\alpha'$ corrections in the non-abelian D$p$-brane action up to order four or so. My problem is the following: Equation (67) is supposed to be equal to $$ \frac{1}{3} {\rm Tr}(\Phi^i\Phi^j\Phi^k)~ F_{tijk}^{(4)}(t), \tag{67b} $$ To obtain (67b), you have to realize that $\partial_k C_{ijk}(t)=0$ since $k$ are spatial coordinates and $C^{(3)}$ only depends on time, and then you do a partial integration over the other term, which (says me) leads to $$ \text{Tr}(\Phi^i\Phi^j\Phi^k)F_{tijk}^{(4)}. $$ So, then I say that this $1/3$ probably comes from the symmetrized trace prescription, but I do not see how, since I believe that $$ \text{STr}[\Phi^i\Phi^j\Phi^k F_{tijk}^{(4)}]=0, $$ because of the indexes $F_{tijk}$ is antisymmetric in $i,j,k$. Maybe this factor $1/3$ is coming from somewhere else?


Tsetlyn's symmetric trace prescription is not used here. We basically only need to use (i) that a form field $C^{(3)}$ is totally antisymmetric in its indices, and (ii) that the trace ${\rm Tr}$ over the $N\times N$ matrix-valued scalars $\Phi^{\ell}$ is cyclic: $$\begin{align} &~{\rm Tr}\left(\Phi^j\Phi^i[\Phi^k~\partial_k C_{ijt}^{(3)}(t)+ C_{ijk}^{(3)}(t)~D_t\Phi^k]\right) \tag{67a} \cr ~=~~&-{\rm Tr}\left(\Phi^i\Phi^j[\Phi^k~\partial_k C_{ijt}^{(3)}(t)+ D_t\Phi^k~C_{ijk}^{(3)}(t)]\right) \cr ~=~~&-{\rm Tr}\left(\Phi^i\Phi^j\Phi^k~\partial_i C_{jkt}^{(3)}(t)+\frac{1}{3}D_t[\Phi^i\Phi^j\Phi^k]~C_{ijk}^{(3)}(t)\right)\cr ~\stackrel{\begin{array}{c}\text{int. by}\cr\text{parts}\end{array} }{\sim}& \frac{1}{3} {\rm Tr}(\Phi^i\Phi^j\Phi^k)~[-3\partial_i C_{jkt}^{(3)}(t)+ \partial_tC_{ijk}^{(3)}(t)] \cr ~=~~&\frac{1}{3} {\rm Tr}(\Phi^i\Phi^j\Phi^k)~[-3\partial_{[i} C_{jk]t}^{(3)}(t)+ \partial_tC_{[ijk]}^{(3)}(t)]\cr ~=~~&\frac{1}{3} {\rm Tr}(\Phi^i\Phi^j\Phi^k)~4 \partial_{[t}C_{ijk]}^{(3)}(t)\cr ~\stackrel{(B.4.3)}{=}&\frac{1}{3} {\rm Tr}(\Phi^i\Phi^j\Phi^k)~ (\mathrm{d}C^{(3)}(t))_{tijk}\cr ~=~~&\frac{1}{3} {\rm Tr}(\Phi^i\Phi^j\Phi^k)~ F_{tijk}^{(4)}(t). \tag{67b} \end{align}$$


  1. J. Polchinski, String Theory Vol. 2, 1998; eq. (B.4.3).
| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.