# How I can get the numerical factor in the relation between string coupling and YM coupling?

I'm trying to understand some references about Wilson loops being used to test AdS/CFT. Some of them are

It seems that they use

$$g_{ym}^2=4\pi g_s \tag{A}$$

But, it is not clear for me how to get that identification. I tried to expand the D3-brane action and compare to the Yang-Mills action. This is done for example in

but I got $$g_{ym}^2=2\pi g_s$$.

Is (A) correct? How can I get that relation?

Recall that a Lie algebra valued field strength is written $$$$F_{\mu\nu}=F_{\mu\nu}^{a}T^a$$$$ and you choose a normalisation constant, $$c$$, for the Lie algebra generators in the fundamental representation: $$$$\mathrm{Tr}\left[T^a,T^b\right]=c\delta^{ab}.$$$$ Deriving the D3-brane Yang-Mills action from the lowest DBI action gives an action of the form $$$$S_{\mathrm{YM}}=-\frac{1}{4(2\pi g_s)}\int \mathrm{d}^4x\,\mathrm{Tr}\left[F_{\mu\nu}F^{\mu\nu}\right]=-\frac{c}{4(2\pi g_s)}\int \mathrm{d}^4x\,F_{\mu\nu}^a F^{a,\mu\nu},$$$$ and with canonical normalisation we therefore have that $$$$g_{\mathrm{YM}}^2=\frac{2\pi g_s}{c}.$$$$ The most often used choice is $$c=1/2$$ giving the relation you referenced in the papers: $$$$g_{\mathrm{YM}}^2=4\pi g_s.$$$$
• Are you sure? I did that actually for just one brane (N=1), in other words I obtained $U(1)$ Yang-Mills action. So there is no dependence of the choice of normalization of the Lie algebra generator, because we dont need even to define that. I think the dependence is in the definition of what we call $g_{YM}$ and not your constant $c$. – ConfusedPhDstudent Sep 22 '19 at 6:30