to the four derivative terms in both the field theory and the supergravity descriptions.

In the case where there are no flavor multiplets, the perturbative beta function does not vanish and the field theory is asymptotically free. We will focus attention on the perturbative contributions to the two and four derivative terms appearing in the low energy effective action. The one loop results for the Kähler metric and real function ${\cal H}$ are

$$K_{\varphi\bar{\varphi}}\sim 
{\log (\varphi\bar{\varphi}/\Lambda^{2})\over\varphi\bar{\varphi}},$$

$${\cal H}(A,\bar{A})\sim \log \Big({A\over\Lambda}\Big)
\log\Big({\bar{A}\over\Lambda}\Big).$$

This leads to the following four derivative term for the scalars, after performing the field redefinition $$\eqalign{S&=\int d^{4}x 
   (\partial^{m}\varphi\partial_{m}\varphi)
   (\partial^{n}\bar{\varphi}\partial_{n}\bar{\varphi})
   {8+4\log\Big({\varphi\bar{\varphi}\over\Lambda^{2}}\Big)
   +\Big[\log\Big({\varphi\bar{\varphi}\over\Lambda^{2}}\Big)\Big]^{2}\over
   \varphi^{2}\bar{\varphi}^{2}
   \Big[\log\Big({\varphi\bar{\varphi}\over\Lambda^{2}}\Big)\Big]^{2}}\Big]\cr
  &=\int d^{4}x (\partial^{m}\varphi\partial_{m}\varphi)
  (\partial^{n}\bar{\varphi}\partial_{n}\bar{\varphi})
  \Big[{1\over (\varphi\bar{\varphi})^{2}}+
  O\Big({1\over (\varphi\bar{\varphi})^{2}log|\varphi|}\Big)\Big].}$$

The $|\varphi|^{-4}$ fall off at large separations (large $|\varphi|$) suggests that the dominant interaction between the branes is again due to the exchange of massless supergravity modes propagating in the six dimensional space transverse to the three branes.

The problem of finding the relevant background geometry corresponding to the asymptotically free gauge theory is considerably more complicated. The Laplace equation  becomes (we will show all $N$ dependence in this section)

$$\Big[{N\over\lambda }
\Big({8\pi\over g_{cl}^{2}}+{6\over\pi}+{2\over\pi}
\log \Big({a\bar{a}\over\Lambda^{2}}\Big)
\Big){\partial^{2}\over\partial y^{2}}
+{\partial\over\partial a}{\partial\over\partial \bar{a}}\Big]f=
-(2\pi )^{4}N\delta^{(4)}(y)\delta^{(2)}(a).$$

Performing a Fourier transform 