\section{Gluino condensate and effective superpotential}
\label{gluino_pot}
Let us start our analysis by considering the geometry of the
``internal'' manifold. The bulk solution is---apart from the warp
factor---of the form $\mathbb{R}^{1,3}\times M_6$, where $M_6$ is a 
K\"ahler manifold and geometrically encodes various aspects of the
dual gauge theory. It encodes, first, the effective superpotential 
\begin{equation}
\label{superpot_def}
  \Weff \sim \int_{M_6} F^{(3)} \wedge \Omega~,
\end{equation} 
where $\Omega$ is the holomorphic 3-form of the complex manifold
$M_6$. Since $M_6$ is not compact, $\Weff$ explicitly depends on a
cut-off. The holomorphic 3-form $\Omega$ is given by
\cite{Ivanov:2000fg,Papadopoulos:2000gj}\footnote{We
have chosen the phase so that $W_{\mathrm{eff}}$ is real.}
\begin{equation}
\label{form3}
  \Omega = \frac{\e{2\Phi}}{L^3} 
  \mathcal{E}^1 \wedge \mathcal{E}^2 \wedge \mathcal{E}^3~,
\end{equation}
where the complex 1-forms $\mathcal{E}^i$ are defined by
\begin{align}
\label{E1}
  \mathcal{E}^1 &= (\sigma^3-L A^3) -i d\rho~,\\
  \mathcal{E}^2 &= (\sigma^1-L A^1)+iX \e{h} \sin\tth d\tp -iP
  (\sigma^2-L A^2)~,\\
  \mathcal{E}^3 &= \e{h} d\tth +iX ( \sigma^2-L A^2) +iP \e{h}
  \sin \tth d\tp~,
\end{align}
and 
\begin{equation}
\label{PX}
  P = \frac{\sinh(4\rho+2c)-4\rho}{2\sinh^2(2\rho+c)}~,\qquad 
  X = (1-P^2)^{1/2} = \frac{2 \e{h}}{\sinh(2\rho+c)}~.
\end{equation}
A straightforward calculation yields 
\begin{equation}
\label{superpot}
  \Weff(\rho_0) \sim \frac{16 \pi^3}{L^5} \e{2\Phi_0} f(c)
  \int\limits_{\hr}^{\rho_0} d\rho \left[ 2\rho \coth(2\rho+c) -1
  \right]~.
\end{equation}


The effective superpotential can be recast in terms of a pre-potential after
introducing a canonical basis of homology 3-cycles of $M_6$
\cite{Taylor:1999ii,Mayr:2000hh},
\begin{equation}
  \Weff \sim \int_A F^{(3)} \int_B \Omega - \int_A \Omega \int_B F^{(3)}~.
\end{equation}  
In our case the compact 3-cycle is $A=S^3$, and the non-compact 3-cycle $B$ has
a complicated form. Since we have already found $\Weff$, we shall
consider only the compact $S^3$.
The integral of $F^{(3)}$ over $S^3$
is proportional to the number of D-branes, $N$, see eqn.\
\eqref{charge_quant}, while the integral of $\Omega$ over $S^3$ encodes the gluino condensate,
\begin{equation}
\label{gluino_cond}
  2\pi i |\glue_c| = \tau_5 \int_{S^3} \Omega =
  i \tau_5 \frac{2\pi^2}{L^3} \e{2\Phi} X 
  = i \tau_5 \frac{2\pi^2}{L^3} \e{2\Phi_0} f(c)~. 
\end{equation}
The constant $\tau_5$ is needed for dimensional reasons.
Following our interpretation of the integration constants, we re-write
eqn.\ \eqref{gluino_cond} as 
\begin{equation}
\label{gluino_c}
  |\glue_c| = |\glue_0| f(c)= \Lambda^3 f(c)~,
\end{equation}
where we have used the
convention $f(0)=1$ and the fact that the regular solution is the true
dual of $\mathcal{N}=1$ SYM theory, \emph{i.e.} $\glue_0=\Lambda^3$,
where $\Lambda$ is the dynamically generated mass scale.
Thus, we identify the precise role of $\Phi_0$ relating $\Lambda$ to
the SUGRA parameters by the relation 
\begin{equation}
\label{Phi0_interpret}
  \e{2\Phi_0} = \pi^{-1} \tau_5^{-1} \Lambda^3 L^3 = 
  \frac{2(2\pi)^4}{N^3 g_s^2} \left(\frac{\Lambda}{L}\right)^3~.
\end{equation}
Obviously, for $c=\infty$ we have $|\glue_\infty| = 0$,
in agreement with the fact that a purely perturbative calculation
fails to exhibit the gluino condensate.


\section{Beta function}
\label{betafunc}
In this section we shall analyze and interpret the gauge coupling and
$\tYM$ angle measured by a probe D5-brane. Our main interest lies in the
calculation of the perturbative beta function, but other aspects, such
as the breaking of chiral symmetry from classically $U(1)$ to
$\mathbb{Z}_{2N}$ in the perturbative regime and to $\mathbb{Z}_2$ by
non-perturbative effects, will also become transparent.

Let us start by discussing the chiral symmetry. We argue that the
transformation $\delta\psi=-2\epsilon$, where $\epsilon \in [0,2\pi)$,
corresponds to a chiral transformation of the dual SYM theory. The
perturbative physics is correctly captured by the solution with
$c=\infty$, in which case eqns.\ \eqref{gYM} and \eqref{tYM} simplify
to  
\begin{equation}
\label{YMinf}
  \frac1{\gYM^2} = \frac{N}{4\pi^2} \rho \quad \text{and} \quad 
  \tYM = N\psi \mod 2\pi~.
\end{equation}
Moreover, in this solution the non-abelian gauge field becomes abelian by virtue of $a(\rho)=0$, which removes all terms with
$\sin(\psi-\psi_0)$ and $\cos(\psi-\psi_0)$ from the metric and the
form fields. Hence, the metric and the field strengths $dC^{(2)}$
and $dC^{(6)}$ are symmetric under $\delta\psi=-2\epsilon$ for all
$\epsilon \in [0,2\pi)$. These transformations form the classical
chiral symmetry group $U(1)$. 
However, $\tYM$ is determined by $C^{(2)}$, which is not
invariant under a general chiral transformation. In fact, eqn.\
\eqref{YMinf} is invariant only for $\epsilon=\pi (n-1)/N$ for
$n=1\ldots 2N$, corresponding to the group $\mathbb{Z}_{2N}$ of the
non-anomalous chiral symmetry transformations. 
In contrast, for every solution with $c<\infty$, where a certain
amount of non-perturbative effects are included, terms with
$\sin(\psi-\psi_0)$ and $\cos(\psi-\psi_0)$ appear showing that
the symmetry of the bulk solutions is given by 
$\epsilon=\pi$ only, which represents
the generator of the unbroken $\mathbb{Z}_2$ chiral symmetry of the
quantum theory. We clearly see that the breaking $\mathbb{Z}_{2N} \to
\mathbb{Z}_2$ is a non-perturbative effect.

Let us turn now to the beta function. Considering the case $c=\infty$,
which is given by eqn.\ \eqref{YMinf}, one is tempted to identify
$\e{2\rho} \sim \mu^3$, where the exponent of $\mu$ is chosen such
that the correct coefficient of the one-loop beta function is
reproduced \cite{Apreda:2001qb}. We shall present an 
alternative argument, which is applicable for any solution with
$c<\infty$. To begin, let us combine $\gYM$
and $\tYM$ to the complexified gauge coupling,
\begin{equation}
\label{tau}
  \tau = \frac{\tYM}{2\pi} +i \frac{4\pi}{\gYM^2} 
  =i \frac{N}{2\pi} \left[ 2\rho \coth(2\rho+c) -i\psi -a(\rho)
  \e{-i(\psi-\psi_0)} \right]~.
\end{equation}
In order to identify the energy scale $\mu$, we follow
\cite{Apreda:2001qb,DiVecchia:2002ks} 
and interpret the function $a(\rho)$ as the
gluino condensate measured in units of $\mu$,\footnote{A different
approach to finding a radial/energy relation can be found in
\cite{Wang:2002ka}.} 
\begin{equation}
\label{mudef}
  a(\rho) = \frac{|\glue_c|}{\mu^3}~,
\end{equation}
where $|\glue_c|=\Lambda^3 f(c)$ by virtue of eqn.\ \eqref{gluino_c}.
We shall in the following be able to determine the function $f(c)$. 
The identification \eqref{mudef} is ambiguous in the case
$c=\infty$, but the result of the following arguments will have a
well-defined limit for $c\to\infty$, because any non-zero $|\glue_c|$
drops out when calculating the beta function.  

The complex coupling \eqref{tau} or, alternatively, eqns.\ \eqref{gYM}
and \eqref{tYM}, might be interpreted as exact, non-perturbative
expressions. More precisely, starting at very large $\rho$ and
a certain initial $\psi$, the RG flow proceeds along the direction of
steepest descent of the probe potential $V$ (see Sec.\ \ref{probepot})
towards the 
minimum at $\psi=\psi_0$, $\rho=\hr$. In the regular case, $c=0$, the
gauge coupling, $\gYM$, diverges at $\rho=0$, signalling the
disappearance of the UV degrees of freedom at the vacuum. 
This exact RG flow was analyzed in \cite{Bertolini:2002yr} for the
special case $\psi=\psi_0$.

The exact RG flow ``knows'' about the position of the vacuum at
$\psi=\psi_0$ in the sense that the value of $\tYM$ flows towards
$\tvac=N\psi_0 \mod 2\pi$. In contrast, perturbative calculations
in field theory are typically ignorant about the vacuum state. For
example, an argument as to why a direct perturbative calculation of 
the gaugino condensate yields
zero is that the perturbative analysis averages over all vacua
\cite{Amati:1985uz,Amati:1988ft}.
We wish to confront our results with a perturbative field theory
calculation and, therefore, we must average over all inequivalent vacua, which
can be equivalently expressed as the following change 
of renormalization scheme,\footnote{It is irrelevant whether we
  average over the $N$ discrete values of $\psi_0$ for a given
  $\tvac$, or whether we take the average over the whole interval.}
\begin{equation}
\label{tauredef}
  \tau\to \tau +i\frac{N}{2\pi} 
  \frac{\glue_c}{\mu^3 \e{i\psi}}~,
\end{equation}
where we have written $\glue_c= |\glue_c| \e{i\psi_0}$, leading to the
coupling 
\begin{align}
\label{gYM_new}
  \frac1{\gYM^2} &= \frac{N}{4\pi^2} \rho \coth(2\rho+c)~,\\
\intertext{and the $\theta$-angle}
\label{tYM_new}
  \tYM &= N \psi \mod 2\pi.
\end{align}
It is interesting to observe that, in this renormalization scheme, the
coupling \eqref{gYM_new} does not diverge for $\rho=c=0$, in contrast
to the exact coupling \eqref{gYM}. 

From eqns.\ \eqref{gYM_new} and \eqref{mudef} it is straightforward to
obtain the beta function of the gauge coupling, 
\begin{equation}
\label{beta}
 \beta = \mu \frac{d\gYM}{d\mu} 
  = -\frac{3N}{16\pi^2} \gYM^3 
  \left[ 1-a(\rho)^2\frac{N\gYM^2}{8\pi^2} \right]
  \left[1-\frac{N\gYM^2}{8\pi^2} \right]^{-1}~.
\end{equation}
For $c\to\infty$, where non-perturbative effects are absent, this
coincides with the complete perturbative beta function of NSVZ
\cite{Novikov:1983uc,Novikov:1986rd}. Notice also that in obtaining 
\eqref{beta} no leading order approximation has been made in order to
rewrite $\rho$ in terms of $\gYM$, and thus \eqref{beta} holds at all
energies. Non-perturbative effects are described by $a(\rho)$, which,
in a large $\rho$ expansion, takes the form of instanton
corrections. In fact, $a(\rho) \approx 16\pi^2/(N\gYM^2)
\exp [-8\pi^2/(N\gYM^2)] \e{-c}$. We see that the constant $c$ determines
how much instanton physics is present in the running. 
Notice, however, that these non-perturbative effects have been
obtained via a ``perturbative'' calculation, in that we are
considering the vacuum-averaged coupling.  


By using eqn.\ \eqref{mudef}, eqn.\ \eqref{beta} is recast in terms
of gauge theory quantities only,
\begin{equation}
\label{beta_new}
 \beta  = -\frac{3N}{16\pi^2} \gYM^3 
  \left[1- \frac{|\glue_c|^2}{\mu^6} \frac{N\gYM^2}{8\pi^2} \right]
  \left[1-\frac{N\gYM^2}{8\pi^2} \right]^{-1}~.
\end{equation}
In this way we are also able to determine $f(c)$. In fact, consider
the cases $c>0$, where the beta function diverges for $\gYM^2=8\pi^2/N$
or, equivalently, for $\rho=\rL$, where $\rL$ is determined by 
\begin{equation}
\label{rhoLambda}
  2 \rL \coth( 2\rL +c) =1~.
\end{equation}
Clearly, we have $\rL>\hr$. For the limiting cases, $c=\infty$ and
$c=0$, we find $\rL=1/2$ and $\rL=0$, respectively. (The regular case
$c=0$, in which $\rL=\hr=0$, will be considered below.) Thus, the
perturbative analysis breaks down before the brane reaches the
singularity. Defining $\Lambda$ as the scale
where the beta function diverges, we
obtain from eqn.\ \eqref{mudef} that 
\begin{equation}
\label{f_def}
  f(c) = a(\rL) = \sqrt{1-4\rL^2}~,
\end{equation}
where $\rL$ is determined implicitly by eqn.\ \eqref{rhoLambda}. 
It is straightforward to check that $f(0)=1$ and, for $c\to\infty$,
$f(c)\approx 2\e{-(c+1)}$, as needed. 

In the regular case the beta function is not
singular at $\rho=\rL=0$, because also the numerator is zero. More
precisely, the limit is 
\begin{equation}
\label{beta_limit}
  \beta\to -\frac{3N}{8\pi^2} \gYM^3 
  = -3 \left(\frac{8\pi^2}{N}\right)^{1/2}~.
\end{equation}
Hence, the running changes from NSVZ type for very large $\mu$
to pure one-loop at $\mu=\Lambda$. Notice, however, the factor $2$
with respect to the UV one-loop coefficient.

An interesting observation is that $\rL$ coincides with the location
of the minimum of the effective superpotential, $\Weff$, cf.\
eqn.~\eqref{superpot}. Thus, we are lead to interpret $\Weff$ as a
``perturbative'' expression that averages over all inequivalent vacua. 
 
\documentclass[12pt,a4paper]{JHEP}
\usepackage{amsmath,amssymb}

\title{Perturbative and non-perturbative aspects of pure
  $\mathcal{N}=1$ super Yang-Mills theory from wrapped branes}

\author{W.~M\"uck \\
  Dipartimento di Scienze Fisiche, Universit\`a di Napoli
  ``Federico II'', \\ 
  and INFN, Sezione di Napoli, Via Cintia, 80125 Napoli, Italy\\
  E-mail: \email{mueck@na.infn.it}}

\abstract{The Maldacena-Nu\~nez solution is generalized to include a
  number of integration constants, one of which controls the
  resolution of the singularity of the wrapped D5-brane
  background. Some features of the dual pure $\mathcal{N}=1$
  super Yang-Mills (SYM) theory are calculated, amongst which the gluino
  condensate, the beta function of the gauge coupling and a brane
  probe potential, which is related to the Veneziano-Yankielowicz
  effective potential. Each integration constant has a precise meaning
  in the dual SYM theory, e.g., the amount of non-perturbative SYM
  physics captured by the gravity configuration is described by the
  singularity resolution parameter.}

\keywords{Brane Dynamics in Gauge Theories, D-branes, Nonperturbative Effects}

\preprint{DSF-3-2003\\hep-th/0301171}

%%  begin of definitions
%
\providecommand{\e}{}
\renewcommand{\e}[1]{\mathrm{e}^{{#1}}}
\newcommand{\tth}{\tilde{\theta}}
\newcommand{\tp}{\tilde{\phi}}
\newcommand{\hth}{\hat{\theta}}
\newcommand{\hp}{\hat{\phi}}
\newcommand{\da}{\dot{a}}
\newcommand{\hr}{\hat{\rho}}
\newcommand{\gYM}{g_{\text{YM}}}
\newcommand{\tYM}{\theta_{\text{YM}}}
\newcommand{\tvac}{\theta_0}
\newcommand{\glue}{\langle \lambda^2\rangle} 
\newcommand{\VVY}{V_{\mathrm{VY}}}
\newcommand{\Weff}{W_{\mathrm{eff}}}
\newcommand{\rL}{\rho_\Lambda}
\newcommand{\trace}{\,\mathrm{Tr}\,}
%
% end of definitions

\begin{document}

\input{intro}
\input{revmn}
\input{condens}
\input{probe}
\input{gaugetheory}
\input{probepot}

\acknowledgments
I am indebted to A.~Lerda for his collaboration on this
topic and  careful reading of the manuscript. 
Moreover, thanks go to M.~Bertolini, P.~Di~Vecchia,
A.~Liccardo, R.~Marotta, P.~Merlatti, F.~Nicodemi, C.~Nu\~nez and
R.~Russo, as well as the other participants of the ``gauge/gravity''
workshop during the 2003 RTN Torino winter school, for useful
discussions. Financial support by the European Commission under the
contract RTN-2000-00131 and by a fellowship from INFN is gratefully
acknowledged. 


\input{MN.bbl}


\end{document}

\section{Brane probe analysis}
\label{probe}
Another way of obtaining information about the dual field theory is
by using the probe technique. 
Let us consider a D5-brane embedded in the background
\eqref{metric10}. Its action is given by 
\begin{equation}
\label{D5action}
  S = - \tau_5 \int d^6 \xi\, \e{-\Phi} \sqrt{-\det(G+2\pi\alpha' F)} 
  + \tau_5 \int \left( \sum\limits_n C^{(n)} \wedge
  \e{2\pi\alpha' F} \right)_{\text{6-form}}~.
\end{equation}
We consider a D5-brane wrapping a two-sphere parameterized by two
angles $\hth$ and $\hp$. Expanding the Born-Infeld part of
the action \eqref{D5action} and demanding that the non-abelian gauge
fields $F$ live only in the 4d part of the D5-branes, one finds
\begin{equation}
\label{D5action_exp}
  S = - \int d^4x \left[ V + \frac1{4\gYM^2} F^A_{\mu\nu} F_A^{\mu\nu} 
  -\frac{\tYM+2\pi n}{32\pi^2}  F^A_{\mu\nu} (\star F_A)^{\mu\nu} \right]~,
\end{equation}
where the raising of the indices and the dual of the gauge fields are
taken using the 4d Minkowski metric, and we have used the convention
$\mathrm{tr}(T^A T^B)=\frac12 \delta^{AB}$ for the colour trace over the
non-abelian generators. The potential $V$ is given by 
\begin{equation}
\label{Vdef}
  V = \tau_5 \int d\hth d\hp\, \left( \e{-\Phi} \sqrt{-G} -
    C^{(6)}_{1234\hth\hp} \right)~.
\end{equation}  
For the gauge coupling, $\gYM$, and
the theta angle, $\tYM$, one obtains \cite{DiVecchia:2002ks} 
\begin{align}
\label{gYMdef}
  \frac1{\gYM^2} &= 2 \pi^2 {\alpha'}^2 \tau_5 \int d\hth d\hp\,
  \e{-3\Phi} \sqrt{-\det G}~,\\
\label{tYMdef}
  \tYM &= (2 \pi)^4 {\alpha'}^2 \tau_5 \int d\hth d\hp\,
  C^{(2)}_{\hth\hp} \mod 2\pi~.
\end{align}
In eqns.\ \eqref{D5action_exp} and \eqref{tYMdef} we have used the fact
that the physics of Yang-Mills theory is periodic in the theta angle with
period $2\pi$, and we adopt the convention $\tYM\in [0,2\pi)$. 
The metric, the 2-form and the 6-form are induced from the respective
bulk fields. 

In order to proceed we have to specify how the world volume
coordinates of the D5-brane are related to the bulk coordinates of the
MN solution. The flat 4d part is obvious, but the wrapped $S^2$ needs
some care. In order to use the coordinates $\rho$ and $\psi$ as
parameters, we have to ensure that both of them are trivially fibred
over the world volume \cite{Maldacena:2000yy}. This is done by
imposing the four embedding conditions\footnote{The alternative
  $\theta = -\tth$ is physically equivalent.} 
\begin{equation}
\label{D5cond}
  \theta = \tth=\hth~,\quad \phi =\tp=\hp~,\quad 
  \psi=\text{const}~,\quad \rho=\text{const}~.
\end{equation}
Thus, we have $d\rho=0$ and $\sigma^3-L A^3=0$
on the world volume.
Notice that the first two conditions differ from the ones used in
\cite{DiVecchia:2002ks}, where $\theta$ and $\phi$ are kept constant.
Hence, the induced metric on the world volume of the D5-branes becomes
\begin{equation}
\label{metric6}
  ds_6^2 = \e{\Phi} \left\{ dx_{1,3}^2 + \frac1{4L^2}
  \left[ 4\e{2h} +a(\rho)^2 +1 -2a(\rho)\cos(\psi-\psi_0)\right] 
  \left( d\hth^2 +\sin^2\hth \, d\hp^2 \right) \right\}~,
\end{equation}
so that
\begin{equation}
\label{sqrtG}
  \sqrt{-G} = \e{3\Phi} \sin\hth \frac1{L^2}
  \left[ \rho \coth(2\rho+c) - \frac12 a(\rho) \cos(\psi-\psi_0)\right]~.
\end{equation}
Moreover, the induced 2- and 6-forms are 
\begin{align}
\label{C2ind}
  C^{(2)} &= \frac1{2L^2} \sin\hth \, d\hth \wedge d\hp \left[
  \psi- a(\rho) \sin(\psi-\psi_0)\right]~,\\
\label{C6ind}
   C^{(6)}_{1234\hth\hp} &= -\frac1{L^2} \sin\hth \left[
   \Psi(\rho) +\frac14 \e{2\Phi}\da \cos(\psi-\psi_0) \right]~,
\end{align}
respectively.

Inserting these equations into the general expressions for $V$, $\gYM$
and $\tYM$, we find 
\begin{equation}
\label{V}
  V= \frac{4\pi \tau_5}{L^2} \left[ 
  \e{2\Phi} \rho \coth(2\rho+c) + \Psi(\rho) - \frac14 \e{2\Phi}(
  2a -\da) \cos(\psi-\psi_0) \right]~,
\end{equation}
as well as 
\begin{align}
\label{gYM}
  \frac1{\gYM^2} &= \frac{N}{4\pi^2}   
 \left[ \rho \coth(2\rho+c) -\frac12 a(\rho) \cos(\psi-\psi_0)\right]~,\\
\label{tYM}
  \tYM &= N \left[ \psi - a(\rho) \sin(\psi-\psi_0) \right] \mod 2\pi~.
\end{align}

\section{Probe potential}
\label{probepot}
The vacuum of $\mathcal{N}=1$ SYM is usually described by the
Veneziano-Yankielowicz effective potential, $S \log S$, where $S$ is a
chiral superfield containing the composite operator $\lambda^2$, where
$\lambda$ is the gluino field, as its lowest component. After
integrating out the auxiliary fields, this becomes 
\cite{Veneziano:1982ah},
\begin{equation}
\label{V_VY}
  \VVY \sim |\phi|^{4/3} \left[ \ln^2 \frac{|\phi|}{\Lambda^3} 
  + \left( \alpha -\frac{\tvac+2\pi n}{N} \right)^2 \right]~.
\end{equation}
In the notation of \cite{Veneziano:1982ah},
$\mu^3=\Lambda^3\e{i\tvac/N}$, where $\tvac$ 
is the vacuum angle. The integer $n$ stems from the fact that we may
choose a branch of the logarithm $\ln[(\phi/\mu^3)^N]$
\cite{Kovner:1997im}.
The scalar field $\phi=|\phi|\e{i\alpha}=\lambda^2$ 
describes the effective degrees of freedom at low
energies. Since $\alpha$ has period $2\pi$, there are $N$ inequivalent
values of $n$. Hence, the minimum of $\VVY$ at
$\phi=\Lambda^3\e{i(\tvac+2\pi n)/N}$ describes the $N$ inequivalent
vacua of $\mathcal{N}=1$ SYM. The chirally symmetric minimum at
$|\phi|=0$ is unphysical, since the second derivative of $\VVY$
diverges. 

We would like to find a SUGRA derived quantity that can be compared to
$\VVY$. The effective superpotential \eqref{superpot} is not a good
candidate, because it does not contain an angular variable that could
play the role of $\alpha$. We shall argue in the following that the
probe brane potential $V$ can be compared to a potential derived from
$\VVY$ in the vicinity of the vacuum. 
Consider the potential 
\begin{equation}
\label{V2}
  \tilde{V} = \left(\frac{\Lambda^3}{|\phi|}\right)^\kappa \VVY = C
  \Lambda^{3\kappa} |\phi|^{4/3-\kappa} \left[ \ln^2 \frac{|\phi|}{\Lambda^3} 
  + \left( \alpha -\frac{\tvac+2\pi n}{N} \right)^2 \right]~,
\end{equation}
where $C$ is a dimensionless constant. The potential $\tilde{V}$
possesses the same physical minimum at $|\phi|=\Lambda^3$ as $\VVY$,
whereas the existence and the properties of the unphysical minimum at
$\phi=0$ depend on $\kappa$. In fact, $\phi=0$ is a minimum of
$\tilde{V}$, if $\kappa<4/3$, but for $1/3<\kappa<4/3$ the first
derivative at this minimum is singular. We shall determine in the
following that the probe potential $V$ coincides with $\tilde{V}$
around the vacuum for $C=L/\Lambda$ and $\kappa=1$. 

The brane potential $V$ is given by eqn.~\eqref{V},
\begin{equation}
\label{V_new}
  V= \frac{4\pi\tau_5}{L^2} \left[ 
  \e{2\Phi} \rho \coth(2\rho+c) + \Psi(\rho) - \frac14 \e{2\Phi}(
  2a -\da) \cos(\psi-\psi_0) \right]~,
\end{equation}
where the function $\Psi$ satisfies eqn.~\eqref{dotPsi}
\begin{equation}
\label{dotPsi_new}
  \dot{\Psi}(\rho) =  2 \e{2\Phi}
    \left[\e{2h} +\frac1{16} \e{-2h} (1-a^4) \right]~.
\end{equation}
In order to discuss $V$, we must again distinguish
between the singular solutions, $c>0$, and the regular one, $c=0$,
which are qualitatively very different. Let us
start with the singular cases. We were not able to integrate eqn.\
\eqref{dotPsi_new} except for the case $c=\infty$, where
$\Psi=\e{2\Phi}(\rho-1/2)+\Psi_0$. The integration constant $\Psi_0$
plays the role of a zero-point energy, and we shall set it to zero
ensuring $V=0$ for $\rho=\hr=1/4$. Hence, we obtain
\begin{equation}
\label{Vinf}
  V_{c=\infty} = 
  4L\Lambda^3 \e{2\rho-1} \sqrt{\rho-\frac14}~,
\end{equation}
where we have used eqn.~\eqref{Phi0_interpret} to eliminate $\Phi_0$
and eqn.~\eqref{f_def} for $f(c)$. Eqn.~\eqref{Vinf} captures not only
the large $\rho$ behaviour of $V$ in all cases, but its behaviour at
the singularity is typical for the cases with $c>0$. In fact,
making the ansatz $\Psi= \e{2\Phi}[\rho-1/2+f(\rho)]$, we find from
eqn.~\eqref{dotPsi_new} the following differential equation for $f$,
\begin{equation}
\label{f_eq}
  \frac{d}{d\rho} \left(\frac{\rho f}{a \e{h}}\right) =
  -\frac{\rho}{a\e{3h}} \left(\e{2h} +\frac{a^2-1}4 \right) 
  \left[ 1-\frac1\rho \left( \e{2h} -\frac{a^2-1}4 \right) \right]~.
\end{equation}
It is not difficult to show that $f$ behaves as 
\begin{equation}
\label{fsim}
  f = f_0 \sqrt{\rho-\hr} + \frac{1-4\hr}{\hr} (\rho-\hr) +\cdots
\end{equation}
close to the singularity. Setting the integration constant $f_0$ to
zero we ensure again $V=0$ for $\rho=\hr$. Moreover, using $\e{2\Phi}\sim
(\rho-\hr)^{-1/2}$ and $2a-\da\sim\rho-\hr$, we obtain $V\sim
\sqrt{\rho-\hr}$ with a positive proportionality constant close to the
singularity. Hence, the potential has its absolute minimum at
$\rho=\hr$, but the brane probe feels an infinite attractive force at
the singularity. Furthermore, the last term in eqn.~\eqref{V_new} 
containing $\cos(\psi-\psi_0)$ exactly vanishes at the singularity, which
means that the state of lowest potential energy does not depend on the
choice of $\psi$. Both features of the minimum---invariance under variations of
$\psi$ and a singular first derivative---are in common with the
unphysical minimum of the potential $\tilde{V}$, although
the comparison does not stand up to a more quantitative analysis.

We shall discuss now the regular solution and find a
quantitative agreement between $V$ and $\tilde{V}$ close to the
vacuum for $\kappa=1$ and $C=L/\Lambda$.  
The first feature, which is different from the singular cases, is that
the coefficient in $V$ of $\cos(\psi-\psi_0)$ is strictly
negative. Hence, the vacuum must be found at $\psi=\psi_0$. (Any
multiples of $2\pi$ are irrelevant.) From eqn.~\eqref{tYM} we now
have that $\psi_0= (\tvac + 2\pi n)/N$, so that we can re-write
$\cos(\psi-\psi_0)$ as $\cos[\psi-(\tvac+2\pi n)/N]$, where $\tvac$
is the field theory vacuum angle. 
Expanding the cosine in $V$ to quadratic order and comparing it with
$\tilde{V}$, we find that $\psi=\alpha$, and we can make the following
identifications, which we expect to hold in the vicinity to the
physical minimum, 
\begin{align}
\label{phi_ident}
  C \Lambda^{3\kappa} |\phi|^{4/3-\kappa} &\approx 
  \frac{4\pi\tau_5}{L^2} \e{2\Phi} \frac18 (2a-\da)~,\\
\label{pot_ident}
  C \Lambda^{3\kappa} |\phi|^{4/3-\kappa} \ln^2
  \frac{|\phi|}{\Lambda^3} &\approx 
  \frac{4\pi\tau_5}{L^2} 
  \left[ \e{2\Phi}\left( \rho \coth(2\rho) -\frac12 a +\frac14 \da
  \right) + \Psi \right]~.
\end{align}
It is straightforward to show that the right hand side of eqn.\
\eqref{pot_ident} has exactly one local minimum, 
which is at $\rho=0$. Thus, the potential $V$ has
its minimum at $\rho=0$ and $\psi=\psi_0$, which is not chirally invariant. 

Expanding eqn.~\eqref{phi_ident} about $\rho=0$ and substituting
eqn.~\eqref{Phi0_interpret} for $\Phi_0$, we find
\begin{equation}
\label{phi_exp}
  C \Lambda^{3\kappa} |\phi|^{4/3-\kappa}  = L \Lambda^3 \left( 1
  +\frac23 \rho + \frac29 \rho^2 +\mathcal{O}(\rho^3) \right)~.
\end{equation}
Since $|\phi|=\Lambda^3$ at the minimum, we obtain $C=L/\Lambda$. 
Then, from eqn.~\eqref{phi_exp} follows
\[ \ln \frac{|\phi|}{\Lambda^3} = \frac3{4-3\kappa} \left( \frac23 \rho
+ \mathcal{O}(\rho^3) \right)~, \]
which we use to expand the left hand side of eqn.~\eqref{pot_ident},
thus obtaining 
\begin{equation}
\label{pot_exp}
  \frac{4\rho^2 L \Lambda^3}{(4-3\kappa)^2} \left( 1 +\frac23 \rho
  +\cdots\right) =
  \frac{4\pi\tau_5}{L^2} \e{2\Phi_0} 
  \left( \rho^2 + \frac23 \rho^3 +\cdots \right)~.
\end{equation}
After substituting the constant $\Phi_0$
we find agreement of the second and the third derivatives of
the potentials $V$ and $\tilde{V}$ at the minimum, if $\kappa=1$.
(The fourth derivatives do not agree). Naively one might have expected
to find $\kappa=0$, but one should bear in mind that, while $\VVY$ is
constructed from a holomorphic superfield, the probe potential is not
intrinsically holomorphic. 

Finally, we would like to point out that the result $\kappa=1$ is
supported also by a consideration of the kinetic term.
In fact, it is not difficult to show that, if we allow
for small fluctuations $\rho(x)$ and $\psi(x)$ of the brane positions,
the probe brane action gives rise also to the kinetic term 
\begin{equation}
\label{Skin}
  S_{\mathrm{kin}} = - \frac{2\pi\tau_5}{L^4} \int d^4x\, \e{2\Phi} 
  \left[\rho \coth(2\rho+c) -\frac12 a(\rho) \cos(\psi-\psi_0) \right] 
  \left[ (\partial \rho)^2 +\frac14 (\partial \psi)^2 \right]~.
\end{equation}
We consider the regular case, $c=0$, and evaluate the kinetic term in
the vicinity of the vacuum. Up to quadratic order in $\rho$ and 
$\psi-\psi_0$, we find 
\begin{equation}
\label{Skin_0}
  S_{\mathrm{kin}} \approx -\frac{2\Lambda^3}{L} \int d^4
   x\, \left[ \rho^2 +\frac14(\psi-\psi_0)^2 \right]  
  \left[ (\partial \rho)^2 +\frac14 (\partial \psi)^2 \right]~.
\end{equation}
Using $\kappa=1$ and $C=L/\Lambda$, we find to leading order from eqn.~\eqref{phi_exp}
that $\rho\approx (|\phi|/\Lambda^3 -1)/2$, so that eqn.~\eqref{Skin_0}
can be re-written (again exact to quadratic order) as 
\begin{equation}
\label{Skin_1}
  S_{\mathrm{kin}} \approx -\frac1{8L\Lambda^9} \int d^4
  x\, \left( \phi -\Lambda^3 \e{i\psi_0}\right)
  \left( \phi^\ast -\Lambda^3 \e{-i\psi_0} \right) 
  \partial_\mu \phi \partial^\mu\phi^\ast ~,   
\end{equation}
where $\phi=|\phi|\e{i\psi}$. 
In order to obtain this complexified form the value $\kappa=1$ is
crucial.

It would be very interesting to recast eqn.~\eqref{Skin_1} in
terms of a K\"ahler potential.


\section{Review of the MN solution}
\label{review}
We consider the SUGRA solution corresponding to a system of $N$
D5-branes. One way of finding it is by using $d=7$
gauged SUGRA, which is obtained as a consistent truncation of the
$d=10$ SUGRA by compactification on an $S^3$
\cite{Cvetic:2000dm,Chamseddine:1999uy}. This is a natural setting to
incorporate the twist condition necessary to retain some supersymmetry
\cite{Bershadsky:1996qy,Maldacena:2000mw}. 
The metric in the string frame is \cite{Maldacena:2000yy,DiVecchia:2002ks}
\begin{equation}
\label{metric10}
  ds_{10}^2 = \e{\Phi} \left[ dx_{1,3}^2 + \frac{\e{2h}}{L^2}
  (d \tth^2 +\sin^2 \tth \, d\tp^2) +\frac1{L^2} d\rho^2
  +\frac1{L^2} \sum\limits_{a=1}^3 (\sigma^a-L A^a)^2
  \right]~.
\end{equation}
Here, $\tth \in [0,\pi)$ and $\tp \in [0,2\pi)$ parameterize a
two-sphere, $S^2$, which is part of the gauged SUGRA solution. The 
compactification three-sphere, $S^3$, is parameterized by the
left-invariant one-forms $\sigma^a$ ($a=1,2,3$) and is twisted by the
$SU(2)$ gauge field $A=\tau^a A^a$, where $\tau^a$ denote
the Pauli matrices. For completeness, we give here the
expressions for the $\sigma^a$,
\begin{align}
\label{sigma1}
  \sigma^1 &= \frac12 [ \cos(\psi-\psi_0) \, d\theta 
  + \sin(\psi-\psi_0) \sin\theta \, d\phi ]~,\\
\label{sigma2}
  \sigma^2 &= \frac12 [ -\sin(\psi-\psi_0) \, d\theta 
  + \cos(\psi-\psi_0) \sin\theta \, d\phi ]~,\\
\label{sigma3}
  \sigma^3 &= \frac12 [ d\psi + \cos\theta \, d\phi ]~,
\end{align}
which satisfy
\begin{equation}
\label{dsigma}
  d\sigma^a = - \varepsilon^{abc} \sigma^b \wedge \sigma^c~.
\end{equation}
The angles are defined in the intervals
$\phi\in[0,2\pi)$, $\theta\in [0,\pi)$ and $\psi\in
[0,4\pi)$. In addition to the solution in the literature, 
we have included the constant angle $\psi_0$, 
which arises from a global $U(1)$ gauge transformation. 
More precisely, one could consider the
transformed gauge field $A'= g^{-1} A g +i g^{-1} dg$, where $g\in
SU(2)$. Then, the part of the metric that belongs to the twisted 3-sphere
can be written in the form (apart from the warp factor)
\begin{equation}
  \sum\limits_{a=1}^3 (\sigma^a -L {A'}^a)^2 
  = \frac12 \trace ( \sigma -L A' )^2 
  = \frac12 \trace (g \sigma g^{-1} -iL dg \,g^{-1} -L A)^2~.
\end{equation}
Hence, a \emph{global} gauge transformation corresponds to a pure rotation
of the frame on $S^3$, while \emph{local} transformations will, in
addition to the rotation, contribute to the twisting. Our frame is
obtained from the MN frame by the transformation
$g=\exp (-i\psi_0\tau^3/2)$. 

The dilaton $\Phi$ and the prefactor $\e{2h}$ in the metric are functions of
the radial variable $\rho$ and are given by
\begin{align}
\label{Phi}
  \e{2\Phi} &= \e{2\Phi_0} f(c) \frac{\sinh(2\rho+c)}{2\e{h}}~,\\ 
\label{h}
  \e{2h} &= \rho \coth(2\rho+c) -\frac14 [a(\rho)^2+1]~,\\
\label{a} 
  a(\rho) &= \frac{2\rho}{\sinh(2\rho+c)}~.
\end{align}
In eqn.~\eqref{Phi}, $f(c)$ is part of the overall constant, but we
choose not to absorb it into $\Phi_0$. The reason for this is that we want to
consider $c$ and $\Phi_0$ as \emph{independent} integration constants
related to distict features of the dual gauge theory. We impose
that for the MN solution ($c=0$), $f(0)=1$. Moreover, if the solution
with $c=\infty$ and finite $\Phi_0$ is to make sense, we also need
$f(c)\sim\e{-c}$ for $c\to\infty$. $f(c)$ shall be determined in Sec.\
\ref{betafunc}. 

The $SU(2)$ gauge fields, $A^a$, are given by
\begin{equation}
\label{A}
  A^1 = \frac{a(\rho)}{2L}  d\tth~,\quad
  A^2 = \frac{a(\rho)}{2L}  \sin\tth \, d\tp~,\quad
  A^3 = \frac1{2L} \cos \tth \, d\tp~,
\end{equation}
with the field strengths $F^a = dA^a + L \epsilon^{abc} A^b
\wedge A^c$, 
\begin{equation}
\label{F}
  F^1 = \frac{\da(\rho)}{2L} d\rho \wedge d \tth~,\quad
  F^2 = \frac{\da(\rho)}{2L} d\rho \wedge \sin\tth \,d\tp~,\quad
  F^3 = \frac1{2L} \left[a(\rho)^2-1\right] d\tth \wedge \sin\tth \, d\tp~.
\end{equation}
The dot denotes a derivative with respect to $\rho$.

Furthermore, the solution contains a 2-form potential
\begin{equation}
\label{C2}
\begin{split}
  C^{(2)} &= \frac1{4L^2} \left[ \psi \left(\sin\theta \, d\theta \wedge
  d\phi) + \sin\tth \, d\tth \wedge d\tp \right) + \cos\theta \cos\tth
  \, d\phi \wedge d\tp \right] \\
  &\quad - \frac{a(\rho)}{2L^2} \left( d\tth \wedge \sigma^1 +
  \sin\tth \, d\tp \wedge \sigma^2 \right)~, 
\end{split}
\end{equation}
whose 3-form field strength $F^{(3)} = dC^{(2)}$ is 
\begin{equation}
\label{F3}
  F^{(3)} = \frac2{L^2} (\sigma^1-L A^1) \wedge
  (\sigma^2-L A^2) \wedge (\sigma^3-L A^3) 
  - \frac1{L} \sum\limits_{a=1}^3 F^a \wedge  
  (\sigma^a-L A^a)~.
\end{equation}

The metric \eqref{metric10} is real for $\rho\ge\hr$, where $\hr$ is
defined by $\e{2h(\hr)}=0$. It is not difficult to show that $\hr$ is
implicitly determined by the transcedental equation
\begin{equation}
\label{hatrhodef}
  2\hr\left[ \coth(2\hr+c) +1\right]  = 1~,
\end{equation}
which has a unique solution $0\le \hr \le 1/4$. The limiting cases are 
$\hr=0$ for $c=0$ and $\hr=1/4$ for $c=\infty$.

The constant $L$ is related to the number $N$ of wrapped
D5-branes by the usual charge quantization condition
\begin{equation}
\label{charge_quant}
  \frac1{2\kappa_{10}^2} \int_{S^3} F^{(3)} = N \tau_5~.
\end{equation}
Using $\kappa_{10}=8\pi^{7/2} g_s {\alpha'}^2$ and $\tau_5=(2\pi)^{-5}
g_s^{-1} {\alpha'}^{-3}$ one obtains \cite{DiVecchia:2002ks}
\begin{equation}
\label{lambda}
  L^{-2} = N g_s \alpha'~.
\end{equation}

In addition to the fields listed so far, there is a non-zero 6-form
potential, $C^{(6)}$, defined by $dC^{(6)}=\star F^{(3)}$, where the Hodge
dual is taken with respect to the string frame metric \eqref{metric10}.
Using eqns.~\eqref{metric10}, \eqref{A}, \eqref{F} and \eqref{F3}, it
is straightforward to obtain
\begin{equation}
\label{dC6}
\begin{split}
  dC^{(6)} &= -\frac1{L^2} v^{(4)} \wedge \left\{ 2 \e{2\Phi}
  \left[\e{2h} +\frac1{16} \e{-2h} (1-a^4) \right] d\rho\wedge
  d\tth \wedge \sin\tth\, d\tp \phantom{\frac12} \right. \\
  &\quad \left. + \frac14 d\left[ 
  \e{2\Phi} \da \left( d\tth \wedge \sigma^2 - \sin \tth\, d\tp \wedge
  \sigma^1 \right) 
  - (a^2-1) \e{2\Phi-2h} d\rho \wedge (\sigma^3-L A^3) \right] \right\}~,
\end{split}
\end{equation}   
where we have abbreviated $v^{(4)}=dx^1\wedge dx^2\wedge dx^3\wedge
dx^4$. 
Thus, the 6-form potential is 
\begin{equation}
\label{C6}
\begin{split}
  C^{(6)} &= -\frac1{L^2} v^{(4)} \wedge \left\{ 
  \Psi(\rho)\, d\tth \wedge \sin\tth\, d\tp \right. \\
  &\quad \left. + \frac14 \left[ 
  \e{2\Phi} \da \left( d\tth \wedge \sigma^2 - \sin \tth\, d\tp \wedge
  \sigma^1 \right) 
  - (a^2-1) \e{2\Phi-2h} d\rho \wedge (\sigma^3 -L A^3) \right]
  \right\}~, 
\end{split}
\end{equation}   
where the function $\Psi(\rho)$ satisfies
\begin{equation}
\label{dotPsi}
  \dot{\Psi}(\rho) =  2 \e{2\Phi}
    \left[\e{2h} +\frac1{16} \e{-2h} (1-a^4) \right]~.
\end{equation}
We were not able to integrate this equation, except for the case
$c=\infty$, where $\Psi=\e{2\Phi}(\rho-1/2)+\Psi_0$, with $\Psi_0$
being an integration constant. We shall comment further on the
function $\Psi$ in Sec.~\ref{probepot}.
