Stable Non-BPS Black Holes and

Black Strings in Five Dimensions

Anshul Mishra 111 E-mail -<EMAIL_ADDRESS>and Prasanta K. Tripathy
222E-mail -<EMAIL_ADDRESS>

Centre for Strings, Gravitation and Cosmology,
Department of Physics,
Indian Institute of Technology Madras, Chennai 600 036, India

ABSTRACT

In this paper we study black hole and black string solutions in five
dimensional $N=2$ supergravity theories arising from the compactification of
M-theory on Calabi-Yau manifolds. In particular, we consider explicit examples
of three parameter Calabi-Yau manifolds which are obtained as hypersurfaces in
toric varieties. Using the attractor mechanism, we obtain BPS as well as non-
BPS black holes in these compactified supergravity theories. We also consider
the black string solutions in these models. We analyze the stability of these
extremal black brane configurations by computing the recombination factor. We
find multiple stable non-BPS attractor solutions in some of these models.

##  1 Introduction

It is well known that the compactification of string theory gives rise to a
landscape of consistent low energy effective theories in various dimensions
[1]. This landscape has further been enriched by the swampland program [3, 2]
which contrived a set of general criteria that the low energy effective
theories must hold. One of the key ingredients devised in order to distinguish
the low energy effective theories among each other is the weak gravity
conjecture [4]. This conjecture tells that gravity is the weakest force among
all and as a consequence objects with large charge must decay into their
respective elementary constituents unless they are protected by a symmetry.

In gravity theories, black holes provide simplest objects which can be used to
further analyze the weak gravity conjecture. Of particular interest are black
hole configurations in $N=2$ supergravity theories. For these theories, the
attractor mechanism [5, 6, 7, 8, 9] provides a suitable technique to construct
the extremal black hole solutions. These extremal black holes may or may not
preserve supersymmetry [8, 9]. The attractor mechanism ensures that the scalar
fields in these theories, which in general may take arbitrary values at
spatial infinity, must run into fixed points at the horizon of the
corresponding extremal black hole. Their values at the horizon are completely
determined in terms of the black hole charges. The attractor points in the
moduli space correspond to critical points of a suitably constructed black
hole effective potential [8]. This in turn determines the mass as well as
entropy of these black holes.

Recently Long, Sheshmani, Vafa and Yau have constructed a number of non-BPS
black branes using the attractor mechanism [10]. They have considered $N=2$
supergravity theories in five dimensions [11], obtained upon the
compactification of M-theory on Calabi-Yau manifolds [12, 13]. $M2$ branes
wrapped on two cycles of the Calabi-Yau manifold give rise to black holes and
$M5$ branes wrapped on four cycles give rise to black string configurations.
Depending upon whether the two cycle corresponds to a holomorphic or a non-
holomorphic curve one obtains a BPS or a non-BPS black hole. Likewise, if the
four cycle is a (non)-holomorphic divisor one obtains a (non)-BPS black
string.

BPS objects, being protected by supersymmetry remain stable. Whereas the same
thing may not hold for non-BPS configurations and a separate study is needed
to address the issues of their stability. For non-BPS black holes, the
attractor mechanism determines the black hole entropy. This, in turn is
conjectured [10] to give rise to the asymptotic volume of the non-holomorphic
curve on which the $M2$ brane is wrapped provided the non-BPS black hole is
doubly extremal. Similarly, for doubly extremal non-BPS black strings, the
string tension determines the asymptotic volume of the non-holomorphic divisor
on which the corresponding $M4$ brane is wrapped. These non-holomorphic cycles
correspond to local volume minimizers in a given homology class. If, within
the same homology class there exists a piece-wise calibrated representative
with a smaller volume then the black brane decays into the constituent
brane/anti-brane pairs wrapping this piece-wise representative cycle in
accordance of the weak gravity conjecture. On the other hand, if no such
piece-wise calibrated representative with a smaller volume exists within the
same homology class then the black brane becomes stable against decay into
constituent brane-anti-brane configurations. In the later case, the
constituent brane configurations are said to be undergoing a recombination in
order to form a stable non-BPS black brane [10]. Several examples of Calabi-
Yau manifolds has been considered and it has been observed that the non-BPS
black holes are always unstable [10]. On the other hand, it has been observed
that some of these models admit stable non-BPS black strings [10].

These results have further been extended in [14] to consider an exhaustive
study of all the two parameter Calabi-Yau compactifications of M-theory. It
has been observed that the resulting non-BPS black hole configurations in all
these examples are unstable and decay into the constituent BPS/non-BPS pairs.
Moreover, for a given charge configuration, there exists a unique non-BPS
black hole attractor. In contrast, there exist several non-BPS black string
configurations which are stable against decay into constituent BPS/non-BPS
pairs. In addition, many of these models also admit multiple basin of non-BPS
black string attractors for a given charge configuration. Similar phenomena
have also been observed in [15], in the context of four dimensional $N=2$
supergravity theories arising from the Calabi-Yau compactifications of
type-$IIA$ string theory. The authors have considered non-BPS $D0-D4$ as well
as $D0-D2-D4$ black hole configurations in one and two parameter Calabi-Yau
models and made similar analysis. While many of these configurations are
unstable, there are also examples of several stable non-BPS black holes in
these classes of compactifications.

The authors in [10] considered a few examples of Calabi-Yau models with small
$h_{11}$ and interestingly the non-BPS black holes obtained in all their
examples remained unstable. Surprisingly, this result holds true in all the
two parameter examples considered in [14]. It remains to be seen whether this
result holds for all five dimensional non-BPS black holes resulting in Calabi-
Yau compactifications of M-theory. Further, the analysis of [14] shows the
existence of multiple basin of attractors in some of these two-parameter
models. In models with larger $h_{11}$ there is a possibility of obtaining
more interesting black hole and black string configurations. In the present
work we wish to generalize these results for a number of three parameter
Calabi-Yau manifolds obtained as hypersurfaces in toric varieties. The
cohomology data for these manifolds have been computed. Using these data we
can obtain the black brane configurations and study their properties. In this
paper we consider some of these three parameter models to obtain extremal
black branes in them and analyze their properties.

The plan of the paper is as follows. First we will review the basic results on
extremal black hole and black string attractors in five-dimensional minimal
supergravity in §2. Also in this section we will set up our notations and
conventions and outline the attractor equations that needs to be analyzed
further. In §3, we will shift our focus towards three parameter Calabi-Yau
manifolds where we will outline some useful formulae and introduce some
notations that will be used throughout the paper. Following a comprehensive
treatment of BPS black holes in §4 we turn our attention in particular to the
black hole solutions resulting from the Calabi-Yau manifolds emerging as
hypersurfaces in toric varieties (THCY). Subsequently, our investigation
extends to non-BPS black hole solutions in §5. In §5.1 we take different
examples of three parameter THCY models and investigate non-BPS black hole
configurations. We also show the existence of multiple non-BPS black hole
solutions. In §5.2 we compute the recombination factor for non-BPS extremal
black hole attractors in these THCY models and show that there exists stable
non-BPS black holes. Moving forward, in §6 we outline the essential details
concerning extremal black string attractors in five-dimensional, minimal
supergravity. Once again, we turn our attention into three-parameter Calabi-
Yau manifolds. We take different cases for non-BPS black strings and obtain
multiple solutions. We compute the recombination factor corresponding to these
non-BPS extremal black strings in §6.1 highlighting the stability of these
solutions. Finally we summarize our findings in §7.

##  2 Review of 5D Black Hole Attractor

In this section we will review the basics of black brane attractors in $N=2$
supergravity theory in five dimensions. We consider the five dimensional low
energy effective theory arising upon the compactification of M-theory on a
Calabi-Yau manifold. The corresponding Lagrangian density is given by[17]

$\frac{\mathcal{L}}{\sqrt{-\mathbf{g}}}=-\frac{1}{2}R-\frac{1}{4}G_{IJ}F_{\mu\nu}^{I}F^{J|\mu\nu}-\frac{1}{2}g_{ij}\partial_{\mu}\varphi^{i}\partial^{\mu}\varphi^{j}+\frac{1}{6^{3/2}\sqrt{-\mathbf{g}}}C_{IJK}\varepsilon^{\lambda\mu\nu\rho\sigma}F_{\lambda\mu}^{I}F_{\nu\rho}^{J}A_{\sigma}^{K}\
.$ (2.1)

Here $\mathbf{g}_{\mu\nu}$ is the space-time metric and $\mathbf{g}$ is its
determinant. $R$ is the corresponding Ricci scalar, $\varphi^{i}$ are the real
scalar fields arising from the Kähler moduli $t^{I}(I=1,\cdots,h_{1,1}),$ and
$A_{\mu}^{I}$ are the gauge fields in the vector multiplet with corresponding
field strengths $F^{I}_{\mu\nu}$. The triple intersection numbers of the
Calabi-Yau manifold are denoted by $C_{IJK}$. The volume of the Calabi-Yau
manifold is constrained to satisfy

$C_{IJK}t^{I}t^{J}t^{K}=1\ .$ (2.2)

Thus, only $h_{1,1}-1$ real scalar fields are independent and are denoted by
$\phi^{i}$. The moduli space metric $g_{ij}$ is given by

$g_{ij}=\frac{3}{2}\partial_{i}t^{I}\partial_{j}t^{J}G_{IJ},~{}~{}g^{ij}g_{jk}=\delta_{k}^{i},$
(2.3)

Here $G_{IJ}$ is the pull back of moduli space metric onto the “ambient space”
formed by the Kähler moduli $t^{I}$. The metric $G_{IJ}$ is derived from the
prepotential ${\cal V}=C_{IJK}t^{I}t^{J}t^{K}$ as

$G_{IJ}:=-\frac{1}{3}\left.\frac{\partial^{2}\log
C_{LMN}t^{L}t^{M}t^{N}}{\partial t^{I}\partial
t^{J}}\right|_{\ast}=\left(3C_{ILM}C_{JNP}t^{L}t^{M}t^{N}t^{P}-2C_{IJM}t^{M}\right)|_{\ast},$
(2.4)

where “$\ast$” indicates that we need to impose the constraint (2.2) after
taking the derivatives.

The critical points of this theory have been studied extensively in [18] using
the attractor mechanism. The model admits extremal black hole as well as
extremal black string solutions. We will first review the black hole
configurations. These are electrically charged objects carrying charges
$q_{I}$. They are described in terms the black hole effective potential [18]

$V=G^{IJ}q_{I}q_{J}=Z^{2}+\frac{3}{2}g^{ij}\partial_{i}Z\partial_{j}Z\ .$
(2.5)

Here $Z$ denotes the central charge of the black hole $Z=q_{I}t^{I}$. The
critical points of this potential are given by

$\partial_{i}V=0\ .$ (2.6)

For a comprehensive treatment of the critical points using the so called new
attractor approach see [19, 20]. The above condition is obviously satisfied by
the extrema of the central charge $Z$. All such critical points correspond to
supersymmetry preserving extremal black holes. However, there exist critical
points which do not extremize $Z$. These correspond to non-supersymmetric
extremal black holes. For both these cases, the entropy of the black hole is
determined by the value $V_{0}$ of the effective potential (2.5) at the
critical point. In the present convention, the black hole entropy is given by

$S=2\pi\left(\frac{V_{0}}{9}\right)^{3/4}\ .$ (2.7)

To obtain the equation of motion it is more convenient to express the black
hole effective potential as [14]

$V=\frac{3}{2}Z^{2}-\frac{1}{2}C^{IJ}q_{I}q_{J}\ .$ (2.8)

Here we have introduced the notation $C_{IJ}=C_{IJK}t^{K}$ and $C^{IJ}$ is the
matrix inverse of $C_{IJ}$. Upon extremizing the potential (2.8) with the
constraint (2.2), we obtain

$3Z\big{(}q_{K}-ZC_{KIJ}t^{I}t^{J}\big{)}+\frac{1}{2}\big{(}C^{IL}C^{JM}C_{KLM}-C^{IJ}C_{KLM}t^{L}t^{M}\big{)}q_{I}q_{J}=0\
.$ (2.9)

The supersymmetric critical points $\partial_{i}Z=0$ are given by

$q_{K}-ZC_{KIJ}t^{I}t^{J}=0\ .$ (2.10)

Clearly, the first term of (2.9) vanishes if the above is satisfied. With a
little bit of work it can be shown that the second term of this equation also
vanishes for the supersymmetric critical points. The equation of motion (2.9)
also admits critical points for which (2.10) does not hold. These correspond
to non-BPS critical points of the effective potential (2.8). To find these
non-BPS critical points, we set

$X_{I}:=q_{I}-ZC_{IJK}t^{J}t^{K}$ (2.11)

Using this notation, we will rewrite (2.9) in a suitable form [14]. To this
end, substitute $q_{I}=X_{I}+ZC_{IJK}t^{J}t^{K}$ in (2.9) to obtain

$8ZX_{K}+C_{KLM}\big{(}C^{IL}C^{JM}-C^{IJ}t^{L}t^{M}\big{)}X_{I}X_{J}=0\ .$
(2.12)

Clearly, $X_{I}=0$ solves this equation trivially and gives BPS critical
points. Solutions of (2.12) with $X_{I}\neq 0$ correspond to non-BPS critical
points.

To further analyse the above equation, we note that the constraint (2.2) can
be rewritten in terms of $X_{I}$ as

$t^{I}X_{I}=0\ .$ (2.13)

We can first solve the above constraint to express $X_{I}$ in terms of $t^{I}$
and substitute the resulting values in (2.12) which can then be solved to
obtain the critical points. Care must be taken while solving (2.13) for
$X_{I}$, since it remains unchanged upon multiplying the $X_{I}$s by an
overall factor. The overall multiplicative factor in the $X_{I}$s can be
determined as follows. Multiply both sides of (2.12) with $C^{KN}X_{N}$ and
using (2.13) to obtain

$8ZC^{IJ}X_{I}X_{J}+C_{KLM}C^{KI}C^{LJ}C^{MN}X_{I}X_{J}X_{N}=0\ .$ (2.14)

This equation is not homogeneous in $X_{I}$ and hence will fix the overall
multiplicative factor in it uniquely. Using the above formalism, models with
$h_{1,1}=2$ have been extensively analysed in [14]. In the following we will
consider some explicit examples of Calabi-Yau compactifications with
$h_{1,1}=3$ and analyse the resulting black hole solutions.

Before we turn our discussion on three parameter Calabi-Yau models, we will
outline the Kähler cone condition, which states that the volume of any
effective curve must be positive:

$\int_{C}J>0\ ,$ (2.15)

where $J=t^{I}J_{I}$ is the Kähler form and $C$ is any arbitrary effective
curve in the Calabi-Yau manifold ${\cal M}$. Here $\\{J_{I}\\}$ form an
integral basis of the cohomology class $H_{1,1}({\cal M},\mathbb{Z})$. Thus,
as emphasised in [10], we need to make sure that the resulting solutions for
the attractors must lie within the Kähler cone.

##  3 Three Parameter Model

Our goal in the present work is to analyze black brane attractors in three
parameter Calabi-Yau models. In this section we will set up some notations in
three parameter models that will be used throughout this paper to analyze the
equations of motion. First, for convenience introduce the variables $x,y,z$ to
represent the Kähler moduli $t^{1},t^{2},t^{3}$ respectively. Further,
introduce the parameters $a,b,\cdots,$ to denote the intersection numbers
$C_{IJK}$ such that
$C_{111}=a,C_{112}=b,C_{122}=c,C_{222}=d,C_{113}=e,C_{123}=f,C_{133}=g,C_{223}=h,C_{233}=i,C_{333}=j$.
In addition, we introduce the functions $A_{1},\ldots,A_{6}$ as

$\displaystyle
A_{1}=ax+by+ez\;,\;\;\;\;\;\;\;\;\;A_{2}=bx+cy+fz\;,\;\;\;\;\;\;\;\;\;A_{3}=ex+fy+gz\
,$ $\displaystyle
A_{4}=cx+dy+hz\;,\;\;\;\;\;\;\;\;\;A_{5}=fx+hy+iz\;,\;\;\;\;\;\;\;\;\;A_{6}=gx+iy+jz\
.$ (3.1)

With these notations, the matrix $C_{IJ}=C_{IJK}t^{K}$ reads as

$C_{IJ}=\begin{pmatrix}A_{1}&A_{2}&A_{3}\cr A_{2}&A_{4}&A_{5}\cr
A_{3}&A_{5}&A_{6}\end{pmatrix}\ .$ (3.2)

To express the inverse of the matrix $C_{IJ}$ in an organized way we introduce
the functions $B_{1},\ldots,B_{6}$ as and

$\displaystyle
B_{1}=A_{4}A_{6}-A_{5}^{2}\;,\;\;\;\;\;\;\;\;\;B_{2}=A_{3}A_{5}-A_{2}A_{6}\;,\;\;\;\;\;\;\;\;\;B_{3}=A_{2}A_{5}-A_{3}A_{4}\
,$ $\displaystyle
B_{4}=A_{1}A_{6}-A_{3}^{2}\;,\;\;\;\;\;\;\;\;\;B_{5}=A_{2}A_{3}-A_{1}A_{5}\;,\;\;\;\;\;\;\;\;\;B_{6}=A_{1}A_{4}-A_{2}^{2}\
.$ (3.3)

In terms of these quantities, the matrix inverse $C^{IJ}$ is expressed as

$C^{IJ}=\frac{1}{A_{1}B_{1}+A_{2}B_{2}+A_{3}B_{3}}\begin{pmatrix}B_{1}&B_{2}&B_{3}\cr
B_{2}&B_{4}&B_{5}\cr B_{3}&B_{5}&B_{6}\end{pmatrix}\ .$ (3.4)

We further need the expressions for quantities such as $C_{I}\equiv
C_{IJK}t^{J}t^{K}$ and $C^{IJ}q_{J}$. It is straightforward to obtain

$C_{I}=C_{IJK}t^{J}t^{K}=\begin{pmatrix}A_{1}x+A_{2}y+A_{3}z\cr
A_{2}x+A_{4}y+A_{5}z\cr A_{3}x+A_{5}y+A_{6}z\end{pmatrix}\ ,$ (3.5)

and

$C^{IJ}q_{J}=\frac{1}{A_{1}B_{1}+A_{2}B_{2}+A_{3}B_{3}}\begin{pmatrix}B_{1}q_{1}+B_{2}q_{2}+B_{3}q_{3}\cr
B_{2}q_{1}+B_{4}q_{2}+B_{5}q_{3}\cr
B_{3}q_{1}+B_{5}q_{2}+B_{6}q_{3}\end{pmatrix}\ .$ (3.6)

The central charge $Z=q_{I}t^{I}$ is given as

$Z=(q_{1}x+q_{2}y+q_{3}z)\ ,$ (3.7)

and the black hole effective potential (2.8) becomes

$V=\frac{3}{2}(q_{1}x+q_{2}y+q_{3}z)^{2}-\frac{(B_{1}q_{1}^{2}+2B_{2}q_{1}q_{2}+2B_{3}q_{1}q_{3}+2B_{5}q_{2}q_{3}+B_{4}q_{2}^{2}+B_{6}q_{3}^{2})}{2(A_{1}B_{1}+A_{2}B_{2}+A_{3}B_{3})}\
.$ (3.8)

##  4 BPS Black Holes

We will first analyse the BPS black holes in three parameter Calabi-Yau
models. Thus, we need to solve the equation (2.10) subjected to the constraint
(2.2). For the three parameter models, (2.10) becomes

$\displaystyle Z(A_{1}x+A_{2}y+A_{3}z)=q_{1}\ ,$ (4.1) $\displaystyle
Z(A_{2}x+A_{4}y+A_{5}z)=q_{2}\ ,$ (4.2) $\displaystyle
Z(A_{3}x+A_{5}y+A_{6}z)=q_{3}\ .$ (4.3)

On the other hand the constraint (2.2) becomes

$x(A_{1}x+A_{2}y+A_{3}z)+y(A_{2}x+A_{4}y+A_{5}z)+z(A_{3}x+A_{5}y+A_{6}z)=1\ .$
(4.4)

To further analyze the above conditions, note that the left hand sides in
equations (4.1) and (4.4) are all degree three homogeneous polynomials in
$x,y,z$. We will introduce the following inhomogeneous coordinates $\tau=x/z$
and $t=y/z$. Further we define the charge ratios $\rho=q_{1}/q_{3}$ and
$\sigma=q_{2}/q_{3}$. In terms of these variables, equations (4.1) becomes

$\displaystyle z^{3}(\rho\tau+\sigma t+1)\left(a\tau^{2}+2b\tau
t+ct^{2}+2e\tau+2ft+g\right)=\rho\ ,$ (4.5) $\displaystyle
z^{3}(\rho\tau+\sigma t+1)\left(b\tau^{2}+2c\tau
t+dt^{2}+2f\tau+2ht+i\right)=\sigma\ ,$ (4.6) $\displaystyle
z^{3}(\rho\tau+\sigma t+1)\left(e\tau^{2}+2f\tau
t+2g\tau+ht^{2}+2it+j\right)=1\ .$ (4.7)

The constraint on the volume becomes

$z^{3}(a\tau^{3}+3b\tau^{2}t+3c\tau t^{2}+dt^{3}+3e\tau^{2}+6f\tau
t+3g\tau+3ht^{2}+3it+j)=1$ (4.8)

Using the above equation, we can eliminate $z$ in equations (4.5)-(4.7). This
results in three equations in two variables. However, as we will see below,
only two of them remain independent. We need to solve two of these equations
for $\tau$ and $t$ in terms of the charge ratios $\rho$ and $\sigma$. However,
notice that these equations provide two coupled cubics in two variables. Thus,
it is not possible to obtain an exact analytic solution for arbitrary values
of the intersection numbers. In order to obtain some insight into the problem
we will first study the inverse problem,i.e., we will solve these equations
for the charge rations $\rho$ and $\sigma$ in terms of the scalar moduli
$\tau$ and $t$. Since these equations are linear in $\rho$ and $\sigma$, it is
straightforward to obtain the solution. Consider the first to equations (4.5)
and (4.6). We find

$\displaystyle\rho=\frac{a\tau^{2}+2b\tau
t+ct^{2}+2e\tau+2ft+g}{e\tau^{2}+2f\tau t+2g\tau+ht^{2}+2it+j}\ ,$ (4.9)
$\displaystyle\sigma=\frac{b\tau^{2}+2c\tau
t+dt^{2}+2f\tau+2ht+i}{e\tau^{2}+2f\tau t+2g\tau+ht^{2}+2it+j}\ .$ (4.10)

It can easily be verified that (4.7) holds true upon substitution of the above
values for $\rho$ and $\sigma$ in it.

In this paper we will consider some specific examples of three parameter
Calabi-Yau manifolds. We will consider Calabi-Yau manifolds which are obtained
as hypersurfaces from toric varieties. A toric variety is given in terms of a
reflexive polytope with a specific triangulation of its faces. A polytope is
reflexive provides it is integral (i.e., the coordinates of its vertices are
all integers) and the corresponding dual polytope is also integral. The faces
of a reflexive polytope are in one to one correspondence with the vertices of
the dual polytope. Consider a reflexive polytope with $n$ faces and let the
$n$ vectors $\vec{v}_{i},(i=1,\cdots,n)$ represent the vertices of the dual
polytope. In general, they will not all be linearly independent, and, for a
polytope in $d$ dimensions, we will have $(n-d)$ relations like

$\sum_{i=1}^{n}q^{r}_{i}\vec{v}_{i}=0\ .$ (4.11)

Here the index $r$ takes values $r=1,\cdots,n-d$. We associate each of the
vertices with a homogeneous coordinate $z_{i}\in\mathbb{C}^{n}$. If we view
the coefficients $q^{r}_{i}$ in (4.11) as elements of a $(n-d)\times n$ weight
matrix, then each of its row defines an equivalence relation among these $n$
homogeneous coordinates with the corresponding coefficients as the weights.
Thus there are $(n-d)$ such equivalence relations altogether. Removing the
fixed points and taking quotient with these equivalence relations we obtain a
$d$ dimensional toric variety. A hypersurface of vanishing first Chern class
in this toric variety gives a $(d-1)$ dimensional Calabi-Yau manifold.

We are interested in Calabi-Yau threefolds. Thus we need to construct toric
varieties from four dimensional reflexive polytopes. All such reflexive
polytopes in four dimensions have been classified by Kreuzer and Sakrke[21].
The cohomology data for the corresponding toric Calabi-Yau manifolds have been
computed in [22, 23, 24] and the results are listed in the database [25]. In
the present work, we will use the cohomology data such as the intersection
numbers and Kähler cone conditions from [25] to analyze the black brane
configurations.

We will now consider an explicit example of a toric hypersurface Calabi-Yau
manifold (THCY). The polytope ID associated with this THCY as per the Kreuzer-
Sakrke classification [21] is 260. In the following we will recapitulate some
of the essential cohomology data associated with it from the Calabi-Yau
database [25]. The Hodge numbers for the Calabi-Yau manifold $\mathcal{M}$ are
$h_{1,1}=3$ and $h_{2,1}=123$ and the corresponding Euler number is $-260$.
The resolved weight matrix of $\mathcal{M}$ has the form

$\displaystyle Q=\left(\begin{array}[]{ccccccc}0&0&0&1&1&3&1\\\
0&1&1&2&2&6&0\\\ 1&0&0&0&1&2&0\\\ \end{array}\right)\ .$ (4.15)

The volume of the manifold is given by

$\mathcal{V}=2xy^{2}+2xyz+4y^{3}-4yz^{2}\ .$ (4.16)

From this we can obtain the intersection numbers of the Calabi-Yau manifold
$\mathcal{M}$. We find the non-vanishing intersection numbers to be
$c=2/3,d=4,f=1/3,i=-4/3$.

Each of the homogeneous coordinates $z_{i}$ associated with the toric Calabi-
Yau $\mathcal{M}$ define a holomorphic divisor $D_{i}$. Denoting the basis
elements of $(1,1)$ homology class of $\mathcal{M}$ as
$\\{J_{1},J_{2},J_{3}\\}$, the divisor classes $D_{i}$ in terms of them are
given as
$D_{1}=-2J_{1}+J_{2}-J_{3},D_{2}=J_{1},D_{3}=J_{1},D_{4}=2J_{1}+J_{3},D_{5}=J_{2},D_{6}=2J_{1}+2J_{2}+J_{3},D_{7}=J_{3}$.
The Mori cone matrix associated with $\mathcal{M}$ is

$\displaystyle\mathcal{M}^{i}_{~{}j}=\int_{C^{i}}D_{j}=C^{i}\cdot
D_{j}=\left(\begin{array}[]{ccccccc}0&1&1&0&0&0&-2\\\ 1&0&0&0&1&2&0\\\
-1&0&0&1&0&1&1\\\ \end{array}\right)\ .$ (4.20)

The elements of the Mori cone matrix indicate the intersection of generating
curves $C^{i}\ (i=1,\ldots,3)$ with the toric divisor classes $D_{i}\
(i=1,\ldots,7)$ [23]. In order to ensure the validity of the Kähler cone
condition (2.15) for our attractors, we also need the Kähler cone matrix $K$
corresponding to the Calabi-Yau manifold. This is the matrix whose elements
${K^{i}}_{j}$ are defined by [23]

${K^{i}}_{j}=\int_{C^{i}}J_{j}\ .$ (4.21)

From the expression for the Mori cone matrix we can find the Kähler cone
matrix. In the present case, the Kähler cone matrix is given by

$\displaystyle K=\left(\begin{array}[]{ccc}1&0&-2\\\ 0&1&0\\\ 0&0&1\\\
\end{array}\right)\ .$ (4.25)

This indicates that the Kähler moduli $t^{I}$ for this model will lie inside
the Kähler cone provided $x-2z>0,y>0,z>0$. In terms of the inhomogeneous
coordinates these conditions become $\tau-2>0,t>0$ and $z>0$. Now, consider
the constraint on the volume (2.2) in terms of the inhomogeneous coordinates:

$2z^{3}t(t+1)(\tau+2t-2)=1\ .$ (4.26)

From the above constraint we observe that $z$ becomes positive in the region
$t>0,\tau>2$. Thus, we do not need to impose it as an additional condition for
the solution to remain inside the Kähler cone.

We will now consider the BPS solutions for this model. Substituting the values
of the intersection numbers in (4.10) we can obtain the charge ratios in terms
of the inhomogeneous coordinates. However, in this case the BPS equations
$q_{I}-ZC_{IJK}t^{J}t^{K}=0$ are simple enough to solve directly. Substituting
the values of the intersection numbers and rescaling the charges and variables
appropriately, the equations take the form

$\displaystyle 3\rho-2t(t+1)z^{3}(\rho\tau+\sigma t+1)$ $\displaystyle=$
$\displaystyle 0\ ,$ (4.27) $\displaystyle 3\sigma-2z^{3}\left(6t^{2}+2\tau
t+\tau-2\right)(\rho\tau+\sigma t+1)$ $\displaystyle=$ $\displaystyle 0\ .$
(4.28)

The variable $z$ in the above equations can be eliminated using the constraint
(4.26) to obtain

$\displaystyle\frac{2\rho(3t+\tau-3)-\sigma t-1}{\tau+2t-2}=0\ ,\
\sigma-\frac{\left(6t^{2}+2\tau t+\tau-2\right)(\rho\tau+\sigma
t+1)}{3t(t+1)(\tau+2t-2)}=0\ .$ (4.29)

These equations can easily be solved to obtain a unique solution for $\tau$
and $t$ in terms of the charge ratios $\rho$ and $\sigma$:

$\displaystyle\tau=\frac{24\rho^{2}-4\rho(\sigma-3)-\sigma+1}{\rho(6\rho-\sigma+2)}\
,\ t=-\frac{2\rho+1}{6\rho-\sigma+2}\ .$ (4.30)

The entropy associated with the attractor is

$S=\frac{2\pi}{3\sqrt{3}}|Z|^{3/2}$ (4.31)

Upon substituting the solution (4.30) in the above and simplifying we find

$S=2^{1/3}\sqrt{3}\pi
q_{3}^{2}\Big{(}\rho(2\rho+1)(4\rho-\sigma+1)\Big{)}^{2/3}\ .$ (4.32)

We need to make sure that the solution (4.30) lies within the Kähler cone,
i.e. we must have $\tau>2$ and $t>0$. Expressed in terms of the charge ratios,
these conditions become

$\frac{(2\rho+1)(6\rho-\sigma+1)}{\rho(6\rho-\sigma+2)}>0\ {\rm and}\
\frac{2\rho+1}{6\rho-\sigma+2}<0\ .$ (4.33)

To simplify the above conditions further, note that for the attractor solution
(4.30), we have $\rho^{2}(\tau+2t-2)=\rho(2\rho+1)(t+1)$. Since both
$(\tau+2t-2)$ and $(t+1)$ are positive inside the Kähler cone, we must have
$\rho(2\rho+1)>0$. Thus, we can have $\rho>0$ or $\rho<-1/2$. Now, requiring
$\tau>2$ gives the condition

$\frac{6\rho-\sigma+1}{6\rho-\sigma+2}>0\ .$ (4.34)

Thus, for a given $\rho$ we can either have $6\rho-\sigma+1>0$ or
$6\rho-\sigma+2<0$. In other words, for $\tau>2$ we must have
$\sigma>(6\rho+2)$ or $\sigma<(6\rho+1)$. We will now consider the implication
of these two bounds on $\sigma$ on the $t>0$ condition. Note that for
$\sigma>(6\rho+2)$ we have $(6\rho+2-\sigma)<0$ and hence $t>0$ implies
$2\rho+1>0$ and hence $\rho$ must be positive. For $\sigma<(6\rho+1)$ we have
$(6\rho+1-\sigma)>0$ and hence $(6\rho+2-\sigma)$ is positive. Thus
$2\rho+1<0$ and hence $\rho<-1/2$. To summarize, the solution (4.30) lies
within the Kähler cone provided $\rho>0,\sigma>(6\rho+2)$ or
$\rho<-1/2,\sigma<(6\rho+1)$.

Before we turn our attention to the non-BPS black holes, we will consider one
more example of a three parameter THCY model. The polytope ID associated with
this THCY is 230. It has Hodge numbers $h_{1,1}=3,h_{2,1}=111$ and Euler
number $\chi=-216$. The resolved weight matrix associated with it is given by

$\displaystyle Q=\left(\begin{array}[]{ccccccc}0&0&0&0&0&1&1\\\
0&1&1&1&1&3&0\\\ 1&1&1&0&0&3&0\\\ \end{array}\right)\ .$ (4.38)

The volume of the Calabi-Yau manifold $\mathcal{M}$ is given by

${\mathcal{V}}=2x^{3}+2x^{2}y+x^{2}z-3xz^{2}+9z^{3}$ (4.39)

From the above we can read the respective intersection numbers. We find
$a=2,b=2/3,e=1/3,g=-1,j=9$ and all other intersection numbers are zero. The
Mori cone matrix associated with the manifold $\mathcal{M}$ is

$\mathcal{M}^{i}_{~{}j}=\int_{C^{i}}D_{j}=\left(\begin{array}[]{ccccccc}-1&0&0&1&1&0&0\\\
0&0&0&0&0&1&1\\\ 1&1&1&0&0&0&-3\\\ \end{array}\right)\ ,$ (4.40)

where $\\{C^{i}\\}$ are the Mori cone generators. The divisor classes $D_{j}$
are given in terms of the basis $\\{J_{1},J_{2},J_{3}\\}$ of $(1,1)$ homology
class as
$D_{1}=J_{1}-J_{2},D_{2}=J_{1},D_{3}=J_{1},D_{4}=J_{2},D_{5}=J_{2},D_{6}=3J_{1}-J_{3},D_{7}=J_{3}$.
From (4.40) we find the Kähler cone matrix $K$ for $\mathcal{M}$ to be

$K=\left(\begin{array}[]{ccc}0&1&0\\\ 0&0&1\\\ 1&0&-3\\\ \end{array}\right)\
.$ (4.41)

We can now study the BPS solutions for this model. Substituting the values of
the intersection numbers in (4.10) it can be seen that the charge ratios in
this case take the form

$\sigma=\frac{2\tau^{2}}{\tau^{2}-6\tau+27}\
,\;\;\;\;\;\;\rho=\frac{6\tau^{2}+(4t+2)\tau-3}{\tau^{2}-6\tau+27}\ .$ (4.42)

Now we have much a simpler expression and it can be inverted to find the
inhomogeneous moduli $\tau$ and $t$ in terms of $\sigma$ and $\rho$. We find

$\tau=\frac{3}{\sigma-2}\Big{(}\sigma\pm\sqrt{2(3\sigma-\sigma^{2})}\Big{)}\
,\;\;\;\;\;\;t=\frac{\rho\tau^{2}-6\rho\tau+27\rho-6\tau^{2}-2\tau+3}{4\tau}\
,$ (4.43)

for $\sigma\neq 2$. For $\sigma=2$, we have

$\tau=\frac{9}{2}\ ,\;\;\;\;t=\frac{1}{24}\big{(}27\rho-170\big{)}\ .$ (4.44)

Upon substituting the solution in the expression for the entropy (4.31) and
simplifying, we find

$\displaystyle
S=\frac{\pi}{12}\left(\frac{2q_{3}^{3}}{3\tau^{3}}\right)^{1/2}\frac{\left(\rho^{2}\big{(}\tau^{2}-6\tau+27\big{)}-\rho\big{(}6\tau^{2}+2\tau-3\big{)}+4\tau\big{(}\sigma\tau+1\big{)}\right)^{3/2}}{\left(\rho\tau\big{(}\tau^{2}-6\tau+27\big{)}-\big{(}2\tau^{3}+3\tau-18\big{)}\right)^{1/2}}\
.$ (4.45)

Here for simplicity we have expressed the $\sigma$ dependence through $\tau$
as given in (4.43).

Several comments are in order. First, since $\tau$ and $t$ are real valued we
must have $3\sigma-\sigma^{2}>0$, i.e., $\sigma$ lies in the range
$0\leq\sigma\leq 3$. Further, the values must lie within the Kähler cone. From
the Kähler cone matrix (4.41), we find that $y>0,z>0$ and $x-3z>0$. Thus, we
have $\tau>3$ and $t>0$. We also need to ensure that $z$ remains positive for
both the solutions. We can do so by examining the volume constraint (2.2),
which in our model takes the form:

$z^{3}\left(2\tau^{3}+(2t+1)\tau^{2}-3\tau+9\right)=1\ .$ (4.46)

Notice that the quantity $\left(2\tau^{3}+(2t+1)\tau^{2}-3\tau+9\right)$ in
the left hand side above remains positive for $\tau>3$ and $t>0$. Thus,
imposing the volume constraint (2.2), $z$ becomes automatically positive
provided $\tau>3$ and $t>0$. We do not need to impose this condition
separately for the solutions to lie inside the Kähler cone.

For $\sigma=2$, from (4.44) we find a unique solution lying within the Kähler
cone provided $\rho>170/27$. For $\sigma\neq 2$ we have two possible solutions
as given in (4.43). First consider the following solution for $\tau$:

$\tau=\frac{3}{\sigma-2}\Big{(}\sigma+\sqrt{2\sigma(3-\sigma)}\Big{)}\ .$
(4.47)

It may be observed that, for $0\leq\sigma<2$ the value of $\tau$ becomes
negative and hence lies outside the Kähler cone. On the other hand, for
$2<\sigma\leq 3$ the value of $\tau$ is always greater than $3$ and hence lies
within the Kähler cone. We can substitute this value of $\tau$ in the
expression for $t$ in (4.42). We find that, for $2<\sigma\leq 3$ the value of
$t$ remains positive provided

$\rho>\rho_{-}\equiv\frac{1}{54}\Big{(}(169\sigma-6)-4\sqrt{2\sigma(3-\sigma)}\Big{)}\
.$ (4.48)

For $\sigma=2$, $\rho_{-}$ takes the minimum possible value $\rho_{-}=6$.
Thus, this branch of BPS attractors exists for $2<\sigma\leq 3$ and $\rho>6$.

We will now consider the second solution for $\tau$ in (4.43):

$\tau=\frac{3}{\sigma-2}\Big{(}\sigma-\sqrt{2\sigma(3-\sigma)}\Big{)}\ .$
(4.49)

It is easy to notice that, for $1<\sigma\leq 3$ the value of $\tau$ remains
greater than $3$ and it lies within the Kähler cone. Substituting the above
expression for $\tau$ in the expression for $t$ in (4.43) we find that it
remains positive provided

$\rho>\rho_{+}\equiv\frac{1}{54}\Big{(}(169\sigma-6)+4\sqrt{2\sigma(3-\sigma)}\Big{)}\
.$ (4.50)

Since $\rho_{+}$ monotonically increases with $\sigma$, it takes the minimum
possible value $\rho_{+}=19/6$ at $\sigma=1$. This branch exists for
$1<\sigma\leq 3$ and $\rho>19/6$.

From the above we can see that, while no solution exists for $\sigma<1$, we
have a unique solution described by (4.49) and the corresponding $t$ when
$\rho>\rho_{+}$ and $\sigma$ takes values in the range $1<\sigma\leq 2$. On
the other hand, for $2<\sigma<3$ there is a possibility of existence of both
the solutions. Notice that $\rho_{+}\geq\rho_{-}$, with the equality holding
for $\sigma=3$. Thus, we have a unique solution described by (4.47) (and the
corresponding solution for $t$) if $2<\sigma\leq 3$ and
$\rho_{-}<\rho<\rho_{+}$, whereas both the solutions exist if $\rho>\rho_{+}$.

As a concrete example, take $\sigma=3/2$ and $\rho=5$. In this case the
equations of motion gives rise to a unique solution inside the Kähler cone

$\tau=9\big{(}\sqrt{2}-1\big{)}\
,\;\;\;t=\frac{1}{12}\big{(}19\sqrt{2}-23\big{)}\ ,\;\;{\rm
and}\;\;z=\left(\frac{2}{9\big{(}1887\sqrt{2}-2641\big{)}}\right)^{1/3}\ .$
(4.51)

Similarly, consider the values $\sigma=5/2$ and $\rho=38/5$. We find

$\tau=3\big{(}5+\sqrt{10}\big{)}\
,\;\;\;t=\frac{1}{300}\big{(}13\sqrt{10}-35\big{)}\
,\;\;\;z=\left(\frac{30267}{2}+\frac{46773}{\sqrt{10}}\right)^{-1/3}\ .$
(4.52)

On the other hand, if we take $\sigma=5/2$ and $\rho=8$ we obtain two
solutions both lying inside the Kähler cone:

$\displaystyle\tau=3\big{(}5+\sqrt{10}\big{)}\
,\;\;t=\frac{1}{60}(65+17\sqrt{10})\
,\;\;z=\left(\frac{2}{9(3627+1121\sqrt{10})}\right)^{1/3}\ ,$ (4.53)
$\displaystyle{\rm and}$ $\displaystyle\tau=3\big{(}5-\sqrt{10}\big{)}\
,\;\;t=\frac{1}{60}(65-17\sqrt{10})\
,\;\;z=\left(\frac{2}{9(3627-1121\sqrt{10})}\right)^{1/3}.$ (4.54)

One distinguished feature is the existence of two distinct solutions for a
given value of $\sigma$ and $\rho$ in the range $2<\sigma\leq 3$ and
$\rho>\rho_{+}$. It is rather interesting to find multiple solutions
preserving supersymmetry for the same set of charges in our model. Such
examples were constructed first in the case of five dimensional supergravity
theories arising from the compactification of M-theory on two parameter
Calabi-Yau manifolds. It was later realized that the multiple solutions lie in
different disconnected branches of the moduli space. While the examples
studied in [14] admitted multiple non-BPS attractors, the BPS solutions for a
given set of charges were all unique. Though multiple non-BPS attractors are
found to occur in many supergravity theories, the existence of multiple BPS
attractors seems quite exceptional.

##  5 Non BPS Black Holes

We will now turn our attention to non-BPS black holes in the three parameter
Calabi-Yau models. Thus, we need to analyze the equation of motion

$8ZX_{K}+C_{KLM}\big{(}C^{IL}C^{JM}-C^{IJ}t^{L}t^{M}\big{)}X_{I}X_{J}=0\ ,$
(5.1)

with $X_{I}\neq 0$. We need to keep in mind that $X_{I}$’s satisfy the
constraint (2.13), $t^{I}X_{I}=0$. Setting $X_{I}=\hat{X}\tilde{X}_{I}$ in
(2.14), we obtain

$\hat{X}=-\frac{8ZC^{IJ}\tilde{X}_{I}\tilde{X}_{J}}{C_{KLM}C^{KP}C^{LQ}C^{MN}\tilde{X}_{P}\tilde{X}_{Q}\tilde{X}_{N}}\
.$ (5.2)

Using the above in the definition of $X_{I}$ from (2.11), we find

$q_{I}=Z\left(C_{IJ}t^{J}-\frac{8\tilde{X}_{I}C^{JK}\tilde{X}_{J}\tilde{X}_{K}}{C_{KLM}C^{KP}C^{LQ}C^{MN}\tilde{X}_{P}\tilde{X}_{Q}\tilde{X}_{N}}\right)\
.$ (5.3)

In terms of the charge ratios $\rho=q_{1}/q_{3}$ and $\sigma=q_{2}/q_{3}$ the
above equation can be rewritten as

$\displaystyle\rho$ $\displaystyle=$
$\displaystyle\frac{C_{1J}t^{J}C_{KLM}C^{KP}C^{LQ}C^{MN}\tilde{X}_{P}\tilde{X}_{Q}\tilde{X}_{N}-8\tilde{X}_{1}C^{JK}\tilde{X}_{J}\tilde{X}_{K}}{C_{3J}t^{J}C_{KLM}C^{KP}C^{LQ}C^{MN}\tilde{X}_{P}\tilde{X}_{Q}\tilde{X}_{N}-8\tilde{X}_{3}C^{JK}\tilde{X}_{J}\tilde{X}_{K}}\
,$ (5.4) $\displaystyle\sigma$ $\displaystyle=$
$\displaystyle\frac{C_{2J}t^{J}C_{KLM}C^{KP}C^{LQ}C^{MN}\tilde{X}_{P}\tilde{X}_{Q}\tilde{X}_{N}-8\tilde{X}_{2}C^{JK}\tilde{X}_{J}\tilde{X}_{K}}{C_{3J}t^{J}C_{KLM}C^{KP}C^{LQ}C^{MN}\tilde{X}_{P}\tilde{X}_{Q}\tilde{X}_{N}-8\tilde{X}_{3}C^{JK}\tilde{X}_{J}\tilde{X}_{K}}\
.$ (5.5)

With $\tilde{X}_{I}$ being algebraic functions of $t^{I}$ the right hand sides
of the above equations are only functions of the moduli. A consistent solution
to the above conditions leads a non-BPS black hole. To find whether the
resulting configuration is stable, we need to consider the recombination
factor [10]. The recombination factor $R$ is defined as the ratio of the black
hole mass to the mass of the $M2$-brane wrapping the minimum piece-wise
calibrated curve in the same homology class as the non-holomorphic curve
corresponding to the black hole:

$R=\frac{M_{C}}{M_{C^{\cup}}}\ ,$ (5.6)

where $M_{C}$ is the mass of the $M2$ brane wrapping the non-holomorphic curve
$C$ and $M_{C^{\cup}}$ is the mass of the $M2$ brane wrapping the minimum
volume piece-wise calibrated curve $C^{\cup}$ in the homology class $[C]$. The
mass of the non-BPS black hole is given by $M_{C}=\sqrt{V_{\rm cr}}$, the
square root of the black hole potential at the critical point. On the other
hand, if $C=\sum\alpha_{I}C^{I}$, then $M_{C^{\cup}}=\sum|\alpha_{I}|t^{I}$.
Thus,

$R=\frac{\sqrt{V_{\rm cr}}}{\sum|\alpha_{I}|t^{I}}\ .$ (5.7)

If this quantity is greater than one, the black hole becomes unstable and
decays into the constituent BPS/non-BPS pairs. Whereas, for $R<1$
recombination of the brane/anti-brane pairs takes place and the non-BPS black
hole becomes stable [10].

Let us now focus on the constraint (2.13). In the case of two parameter
Calabi-Yau models [14] it took the form $\tilde{X}_{1}x+\tilde{X}_{2}y=0.$
Thus, for the two parameter case it was possible to obtain the expressions for
$X_{I}$ uniquely upon solving this constraint. However, this is not the case
for models with $h_{1,1}>2$. In the present case the constraint takes the form
$\tilde{X}_{1}x+\tilde{X}_{2}y+\tilde{X}_{3}z=0$. In terms of the rescaled
coordinates we have $\tilde{X}_{1}\tau+\tilde{X}_{2}t+\tilde{X}_{3}=0$, and
hence we can at most express $\tilde{X}_{3}$ in terms of $t^{I}$ and
$\tilde{X}_{1},\tilde{X}_{2}$. Thus, to obtain these quantities we need to
solve the full equations of motion. In the following we will consider examples
of three parameter models where we will explicitly solve the equations of
motion to obtain the non-BPS attractors.

###  5.1 Examples

We will now consider non-BPS attractors for the both the examples considered
in the previous section. Let us first consider the first example, say the
model 1. We need the explicit expression for the black hole effective
potential to obtain the non-BPS solutions. Substituting the intersection
numbers for the model 1 in (3.8) we find

$V=\frac{3}{4y(y+z)(x+2y-2z)}\big{(}v_{11}q_{1}^{2}+v_{22}q_{2}^{2}+v_{33}q_{3}^{2}+2v_{12}q_{1}q_{2}+2v_{23}q_{2}q_{3}+2v_{31}q_{3}q_{1}\big{)}\
,$ (5.8)

where the functions $v_{ij}$ are defined as follows

$\displaystyle v_{11}$ $\displaystyle=$ $\displaystyle
2x^{3}y(y+z)+x^{2}\left(4y^{3}-4yz^{2}+1\right)+8x(y-z)+16\left(3y^{2}+z^{2}\right),$
(5.9) $\displaystyle v_{22}$ $\displaystyle=$ $\displaystyle
y^{2}\left(2xy(y+z)+4y^{3}-4yz^{2}+1\right),$ (5.10) $\displaystyle v_{33}$
$\displaystyle=$ $\displaystyle
2y^{2}\left(xz^{2}+2\right)+y\left(2xz^{3}-4z^{4}+4z\right)+4y^{3}z^{2}+z^{2},$
(5.11) $\displaystyle v_{12}$ $\displaystyle=$ $\displaystyle
y\left(2x^{2}y(y+z)+x\left(4y^{3}-4yz^{2}-1\right)-8y\right),$ (5.12)
$\displaystyle v_{23}$ $\displaystyle=$ $\displaystyle
y\left(2xy^{2}z+y\left(2xz^{2}-4z^{3}-2\right)+4y^{3}z-z\right),$ (5.13)
$\displaystyle v_{31}$ $\displaystyle=$ $\displaystyle
2y^{2}\left(x^{2}z+6\right)+2yz\left(x^{2}z-2xz^{2}+4\right)+4xy^{3}z+z(4z-x)\
.$ (5.14)

The constraint on the volume (2.2) for the Calabi-Yau manifold $\mathcal{M}$
becomes

$2xy^{2}+2xyz+4y^{3}-4yz^{2}=1\ .$ (5.15)

In order to obtain the critical points, we need to extremize the black hole
effective potential (5.8) using the method of Lagrange multiplier to
incorporate the constraint (5.15) on the scalar fields. Upon eliminating the
Lagrange multiplier we obtain two independent equations which after some
simplifications take the form

$\displaystyle
q_{1}^{2}f_{11}+q_{2}^{2}f_{22}+2q_{3}^{2}y(y+z)(2y+z)+f_{12}q_{1}q_{2}+f_{23}q_{2}q_{3}+f_{31}q_{3}q_{1}=0\
,$ (5.16)
$\displaystyle\Big{(}q_{1}\left(2x^{2}y(y+z)+x\left(4y^{3}-4yz^{2}+1\right)+8y\right)+q_{2}y\left(2xy(y+z)+4y^{3}-4yz^{2}-1\right)$
(5.17)
$\displaystyle+q_{3}\left(2xy^{2}z+2y\left(xz^{2}-2z^{3}+1\right)+4y^{3}z+z\right)\Big{)}(q_{1}(x-4z)-q_{3}(y+z))=0\
.$ (5.18)

The functions $f_{ij}$ are defined as

$\displaystyle f_{11}$ $\displaystyle=$ $\displaystyle
2x^{3}y\left(2y^{2}+3yz+z^{2}\right)+x^{2}\left(20y^{4}+16y^{3}z-12y^{2}z^{2}+y\left(2-8z^{3}\right)+z\right)$
(5.19) $\displaystyle+$ $\displaystyle
2x\left(12y^{5}-16y^{3}z^{2}+5y^{2}+4yz\left(z^{3}-1\right)-3z^{2}\right)-8\left(3y^{3}+9y^{2}z+yz^{2}-z^{3}\right),$
(5.20) $\displaystyle f_{22}$ $\displaystyle=$ $\displaystyle y^{2}(y+z)\text{
}\left(2xy(y+z)+4y^{3}-4yz^{2}+1\right),$ (5.21) $\displaystyle f_{12}$
$\displaystyle=$ $\displaystyle
y\left(2x^{2}y^{2}(y+z)+xy\left(16y^{3}+12y^{2}z-8yz^{2}-4z^{3}-1\right)\right.$
(5.22) $\displaystyle+$
$\displaystyle\left.2\left(12y^{5}-16y^{3}z^{2}+5y^{2}+4yz\left(z^{3}+2\right)+z^{2}\right)\right),$
(5.23) $\displaystyle f_{23}$ $\displaystyle=$ $\displaystyle
y(y+z)\left(2xy^{2}z+2xyz^{2}+4y^{3}z-4y\left(z^{3}+1\right)-z\right),$ (5.24)
$\displaystyle f_{31}$ $\displaystyle=$ $\displaystyle
4y^{3}\left(x^{2}z+4xz^{2}-8z^{3}-6\right)-2y^{2}z\left(-3x^{2}z+6xz^{2}+19\right)$
(5.25) $\displaystyle+$ $\displaystyle
2yz\left(x^{2}z^{2}-4xz^{3}-x+4z^{4}-4z\right)+20xy^{4}z+z^{2}(2z-x)+24y^{5}z.$
(5.26)

We can rewrite these equations in terms of the inhomogeneous coordinates and
the charge ratios and eliminate $z$ using the constraint (5.15). The equations
motion simplify remarkably to have the form

$\displaystyle(\rho\tau+4\rho t+t+1)(\rho(\tau-4)-t-1)=0,$ (5.27)
$\displaystyle
t^{3}g_{1}+t^{2}(g_{2}-8\tau\rho^{2})+t(4\rho^{2}+4\rho+1)-2\tau
t\rho^{2}(\tau-2)-\rho^{2}(\tau-2)^{2}=0,$ (5.28)

where $g_{1}$ and $g_{2}$ are functions of the charge ratios $\rho$ and
$\sigma$ with the expression

$\displaystyle g_{1}=12\rho^{2}+\rho(12-8\sigma)+\sigma^{2}-2\sigma+2\ ,\
g_{2}=36\rho^{2}-8\rho(\sigma-2)+\sigma^{2}-2\sigma+3\ .$ (5.29)

It is indeed possible to solve these equations exactly to obtain the
inhomogeneous coordinates in terms of the charge ratios. They admit four
independent solutions. They are given as

$\displaystyle\tau=-\frac{8\rho^{2}+8\rho-\sigma+1}{\rho(2\rho-\sigma)},t=\frac{2\rho+1}{2\rho-\sigma},$
(5.30)
$\displaystyle\tau=\frac{8\rho^{2}+\sigma-1}{\rho(6\rho-\sigma+2)},t=-\frac{2\rho+1}{6\rho-\sigma+2},$
(5.31)
$\displaystyle\tau=\frac{8\rho^{2}-4\rho\sigma+4\rho-\sigma+1}{2\rho^{2}-\rho\sigma},t=\frac{2\rho+1}{2\rho-\sigma},$
(5.32)
$\displaystyle\tau=\frac{24\rho^{2}-4\rho(\sigma-3)-\sigma+1}{\rho(6\rho-\sigma+2)},t=-\frac{2\rho+1}{6\rho-\sigma+2}\
.$ (5.33)

The last of these solutions corresponds to BPS black holes. They have been
analyzed in detail in §4. On the other hand, the first three solutions do not
satisfy the BPS equations. They corresponds to non-BPS attractors.

We need to make sure that the non-BPS solutions lie within the Kähler cone.
From the expression of the Kähler cone matrix we observe that the values of
$t$ and $\tau-2$ must be positive. Let us analyze this condition for the first
solution given in (5.30). We find

$\frac{2\rho+1}{2\rho-\sigma}>0\ {\rm and}\
\frac{(2\rho+1)(6\rho-\sigma+1)}{\rho(2\rho-\sigma)}<0\ .$ (5.34)

Further, we consider the combination $(t+1)(\tau+2t-2)$ which is positive
inside the Kähler cone. For (5.30), this implies $\rho(2\rho+1)<0$. Thus, the
value of $\rho$ must lie in the range $-1/2<\rho<0$. Since $2\rho+1$ is
positive, the first inequality in (5.34) implies $2\rho-\sigma>0$ and hence
$\sigma<2\rho$. The second inequality in (5.34) now implies $\sigma<6\rho+1$.
Thus, the solution (5.30) remains inside the Kähler cone provided
$-1/2<\rho<0$ and $\sigma<{\rm min}\\{2\rho,6\rho+1\\}$.

We will now focus on the Kähler cone condition for the solution (5.31). A
similar analysis tells that once again the values of $\rho$ lies in the range
$-1/2<\rho<0$. Requiring $t>0$ for a given $\rho$ in this range now gives the
bound $\sigma>6\rho+2$. Requiring $\tau-2>0$ gives $\sigma>2\rho+1$. Thus, the
solution (5.31) lies inside the Kähler cone provided $-1/2<\rho<0$ and
$\sigma>{\rm max}\\{2\rho+1,6\rho+2\\}$. Finally, let us consider the solution
(5.32). In this case $\rho$ takes values in the range $\rho>0$ or $\rho<-1/2$.
For $\rho>0$ we must have $\sigma<2\rho$ and for $\rho<-1/2$ the value of
$\sigma$ lies in the range $\sigma>2\rho+1$. For these values of $\rho$ and
$\sigma$ the solution (5.32) remain inside the Kähler cone. Further, these
bounds ensure that all the four branches of solutions as given in
(5.30)-(5.33) are mutually exclusive from each other. Thus, in this model both
the BPS as well as non-BPS attractors are unique.

Let us now turn our attention to the second model consider in §4. The black
hole effective potential (3.8) for this model takes the form

$\displaystyle
V=\frac{3q_{1}}{2x}\left(q_{1}x^{3}+2x^{2}(q_{2}y+q_{3}z)-q_{2}\right)+\frac{1}{8x^{2}(x-9z)}\left(q_{2}^{2}V_{3}+q_{2}q_{3}V_{2}+q_{3}^{2}V_{1}\right)\
.$ (5.35)

where, for easy reading we have introduced the functions

$\displaystyle V_{1}$ $\displaystyle=$ $\displaystyle 4x(3xz^{2}-27z^{3}+1)\
,$ (5.36) $\displaystyle V_{2}$ $\displaystyle=$ $\displaystyle
4x\left(6x^{2}yz-54xyz^{2}-x+3z\right)\ ,$ (5.37) $\displaystyle V_{3}$
$\displaystyle=$ $\displaystyle
12x^{3}y^{2}-x^{2}\left(108y^{2}z-19\right)+3x(2y-55z)-18z(3y+z)\ .$ (5.38)

The constraint on the volume (2.2) becomes

$2x^{3}+x^{2}(2y+z)-3xz^{2}+9z^{3}=1\ .$ (5.39)

The equations of motion is obtained upon extremizing the potential (5.35)
subject to the constraint (5.39). We find

$\displaystyle
24q_{1}^{2}x^{5}(x-9z)^{2}-q_{2}f_{3}(x,y,z)\big{(}4x^{2}(q_{1}x+q_{2}y+q_{3}z)+q_{2}\big{)}+4q_{2}q_{3}x^{2}f_{1}(x,y,z)\
{~{}}$ (5.40)
$\displaystyle+12q_{1}x^{2}(x-9z)^{2}(2q_{2}x^{2}y+q_{2}+2q_{3}x^{2}z)-q_{2}^{2}f_{5}(x,y,z)-4q_{3}^{2}x^{4}=0\
,$ (5.41)
$\displaystyle(q_{2}f_{1}(x,y,z)-2q_{3}x^{2})\big{(}2x^{2}(2q_{1}x(x-9z)^{2}+q_{3}f_{2}(x,y,z))+q_{2}f_{4}(x,y,z)\big{)}=0\
.$ $\displaystyle\ .$ (5.42)

In the above the functions $f_{1},\cdots,f_{5}$ are defined as follows

$\displaystyle f_{1}(x,y,z)$ $\displaystyle=$ $\displaystyle
x^{2}-6xz+27z^{2}\ ,$ (5.43) $\displaystyle f_{2}(x,y,z)$ $\displaystyle=$
$\displaystyle 2x^{2}z-36xz^{2}+162z^{3}+3\ ,$ (5.44) $\displaystyle
f_{3}(x,y,z)$ $\displaystyle=$ $\displaystyle
3(x-9z)^{2}(6x^{2}+2x(2y+z)-3z^{2})\ ,$ (5.45) $\displaystyle f_{4}(x,y,z)$
$\displaystyle=$ $\displaystyle
4x^{4}y-72x^{3}yz+x^{2}(324yz^{2}-1)-18xz+81z^{2}\ ,$ (5.46) $\displaystyle
f_{5}(x,y,z)$ $\displaystyle=$ $\displaystyle
x\big{(}19x^{3}+6x^{2}(2y-55z)-27xz(8y-53z)+324z^{2}(3y+z)\big{)}\ .$ (5.47)

We express these equations in terms of the rescaled variables $\tau=x/z,t=y/z$
and eliminate $z$ using (5.39) to obtain

$\displaystyle
4\tau^{4}\left(2\tau^{3}+(2t+1)\tau^{2}-3\tau+9\right)-12\rho(\tau-9)^{2}\tau^{2}\left(2\tau^{2}(1+\rho\tau)-\sigma\left(4\tau^{3}+\tau^{2}-9\right)\right)$
(5.48)
$\displaystyle-4\sigma\tau^{2}\left(2\tau^{5}+(2t-29)\tau^{4}+(363-24t)\tau^{3}+9(30t-143)\tau^{2}-27(36t+29)\tau+972\right)$
(5.49)
$\displaystyle+\sigma^{2}\left(74\tau^{7}+\left(96t^{2}-3408t+4785\right)\tau^{5}-9\left(192t^{2}-1588t-777\right)\tau^{4}+54\left(144t^{2}+148t\right.\right.$
(5.50)
$\displaystyle\left.\left.-271\right)\tau^{3}-(1259-194t)\tau^{6}-81(174t-263)\tau^{2}+2187(8t+5)\tau-6561\right)=0\
,$ (5.51)

and

$\displaystyle\left(2\tau^{2}-\sigma\left(\tau^{2}-6\tau+27\right)\right)\Big{(}2\tau^{2}\left(6\tau^{3}+(6t+5)\tau^{2}-45\tau+189\right)+4\rho(\tau-9)^{2}\tau^{3}$
(5.52)
$\displaystyle-\sigma\left(2\tau^{5}-(2t-37)\tau^{4}+3(36t-49)\tau^{3}-18(27t+7)\tau^{2}+405\tau-729\right)\Big{)}=0.$
(5.53)

Even though these equations look complicated, it is indeed to solve the
inverse problem exactly to express the charge ratios in terms of the
inhomogeneous coordinates $(\tau,t)$. We find three independent solutions:

$\displaystyle\rho$ $\displaystyle=$
$\displaystyle\frac{6\tau^{2}+(4t+2)\tau-3}{\tau^{2}-6\tau+27}\
,\;\;\;\;\;\;\sigma=\frac{2\tau^{2}}{\tau^{2}-6\tau+27}\ ,$ (5.54)
$\displaystyle\rho$ $\displaystyle=$
$\displaystyle-\frac{2\tau^{3}+2(2t+1)\tau^{2}-9\tau+36}{\tau\left(\tau^{2}-6\tau+27\right)}\
,\;\;\;\;\;\;\sigma=\frac{2\tau^{2}}{\tau^{2}-6\tau+27}\ $ (5.55)

and

$\displaystyle\rho$ $\displaystyle=$
$\displaystyle-\frac{328\tau^{5}+36\left(6t^{2}+4t-63\right)\tau^{3}+(548t+310)\tau^{4}+(9981-1134t)\tau^{2}+81(84t+23)\tau+729}{92\tau^{5}+3(30t-763)\tau^{4}-81(28t-121)\tau^{3}+162(63t+19)\tau^{2}-3645\tau+6561}$
(5.56) $\displaystyle\sigma$ $\displaystyle=$
$\displaystyle-\frac{2\tau^{2}\left(52\tau^{3}+27(2t+3)\tau^{2}-567\tau+1701\right)}{92\tau^{5}+3(30t-763)\tau^{4}-81(28t-121)\tau^{3}+162(63t+19)\tau^{2}-3645\tau+6561}$
(5.57)

The first solution (5.54) corresponds to BPS black holes which has been
considered in §4. On the other hand solutions obtained in (5.55) and (5.56)
give rise to non-BPS black holes. We will now analyze them in detail. Consider
first (5.55). We can solve them to obtain analytic expressions for the
inhomogeneous coordinates $(\tau,t)$. Denoting

$\displaystyle\tau_{\pm}$ $\displaystyle=$
$\displaystyle\frac{3\left(\sigma\pm\sqrt{2\sigma(3-\sigma)}\right)}{(\sigma-2)}\
,$ (5.58) $\displaystyle t_{\pm}$ $\displaystyle=$
$\displaystyle\frac{1}{36(\sigma-2)^{3}\sigma}\Big{(}12(405\rho-11)\sigma^{2}-8(2187\rho+31)\sigma-59\sigma^{4}+222\sigma^{3}+192$
(5.59) $\displaystyle\mp$
$\displaystyle\sqrt{2\sigma(3-\sigma)}\left(972\rho\sigma+5832\rho+55\sigma^{3}-222\sigma^{2}+228\sigma-8\right)\Big{)}\
,$ (5.60)

there are two independent solutions $(\tau_{+},t_{+})$ and $(\tau_{-},t_{-})$.
The expression for $\tau$ in these solutions remain the same as that of BPS
black holes as given in (4.43). Thus, from the discussion followed by (4.49)
we understand that requiring $\tau_{-}$ to lie inside the Kähler cone for the
solution $(\tau_{-},t_{-})$ gives rise to the constraint $1<\sigma\leq 3$ for
the possible values of $\sigma$. Further, introducing $\rho_{\pm}$ as

$\rho_{\pm}=\frac{1}{26244\sigma}\Big{(}\sigma(185\sigma^{2}-948\sigma-396)\pm
2\sqrt{2\sigma(3-\sigma)}(167\sigma^{2}+18\sigma+72)\Big{)}\ ,$

we find that the value $t_{-}$ in the solution $(\tau_{-},t_{-})$ remains
positive if $\rho<\rho_{-}$. At $\sigma=3$, $\rho_{-}$ takes its maximum
possible value $\rho_{-}^{m}=-175/2916$. Thus, the non-BPS configurations
$(\tau_{-},t_{-})$ exist for $1<\sigma\leq 3$ and $\rho<-175/2916$. Let us now
focus on the solution described by $(\tau_{+},t_{+})$. In this case, for
$2<\sigma\leq 3$ the value of $\tau_{+}$ remains greater than $3$. Requiring
$t_{+}>0$ in this solution, we find that $\rho$ must satisfy the upper bound
$\rho<\rho_{+}$. From the expression for $\rho_{+}$ we find that $\rho_{+}\leq
0$ throughout with the equality holding at $\sigma=2$. Thus, this branch
exists when $2<\sigma\leq 3$ and $\rho<0$. Further, we notice that
$\rho_{-}\leq\rho_{+}$ in the range $2<\sigma\leq 3$. Thus, for $2<\sigma\leq
3$ and $\rho_{-}<\rho<\rho_{+}$ only $(\tau_{+},t_{+})$ lies within the Kähler
cone whereas, for $\rho<\rho_{-}$ both the solutions survive.

Before we turn our attention to (5.56), we would like to compare the BPS and
non-BPS solutions we have obtained so far. Though both the set of solutions
exist only when $1<\sigma\leq 3$, the range of $\rho$ is entirely different.
While, for the BPS solutions the minimum possible value of $\rho$ is $19/6$,
for the non-BPS solutions $\rho$ must take negative values. Thus the BPS
solutions obtained from (5.54) and non-BPS solutions derived from (5.55) do
not coexist simultaneously.

We will now consider (5.56). This involves coupled equations of degree five in
$(\tau,t)$ and hence it is not possible to obtain an exact analytical
expression for these variables in terms of $(\rho,\sigma)$. Nevertheless, it
is possible to obtain the range of charge ratios which admit attractor
solutions lying inside the Kähler cone. First thing to notice is that there
exist curves of singularities in the moduli space which do not correspond to
any possible extremal black hole configuration for finite values of black hole
charges. These curves are obtained by setting the denominators in the
expressions for $\rho$ and $\sigma$ in (5.56) to zero. We find

$92\tau^{5}+3(30t-763)\tau^{4}-81(28t-121)\tau^{3}+162(63t+19)\tau^{2}-3645\tau+6561=0\
.$ (5.61)

This equation can be rewritten as

$t=-\frac{92\tau^{5}-2289\tau^{4}+9801\tau^{3}+3078\tau^{2}-3645\tau+6561}{18\tau^{2}\left(5\tau^{2}-126\tau+567\right)}\
.$ (5.62)

The above expression has two poles as well as two zeros. Denote the poles as
$\tau_{p1},\tau_{p2}$ and the zeros as $\tau_{z1},\tau_{z2}$. The values of
$\tau_{p1}$ and $\tau_{p2}$ are

$\tau_{p1}=\frac{9}{5}\big{(}7-\sqrt{14}\big{)}\simeq 5.865\
,\;\;\;\tau_{p1}=\frac{9}{5}\big{(}7+\sqrt{14}\big{)}\simeq 19.335\ ,$

and the zeros $\tau_{z1},\tau_{z2}$ are approximately $\tau_{z1}\simeq 5.859,\
\tau_{z2}\simeq 19.266$. The value of $t$ as given in (5.62) remains positive
in the intervals $(\tau_{z1},\tau_{p1})$ and $(\tau_{z2},\tau_{p2})$. Thus,
there are two curves of singularities. They form a pair of almost straight
lines which start respectively from $\tau_{z1}$ and $\tau_{z2}$ at $t=0$ and
asymptote to $\tau=\tau_{p1}$ and $\tau=\tau_{p2}$ lines as
$z\rightarrow\infty$. Any point inside the Kähler cone which does not lie on
either of these two curves correspond to an extremal non-BPS black hole
configuration for some suitably chosen charges.

The singularity curves naturally divide the moduli space into three regions.
We denote the region bounded by the singularity curve asymptote to the line
$\tau=\tau_{p1}$ and the $\tau=3$ boundary of the Kähler cone as region 1, the
region bounded by the curve asymptote to the line $\tau=\tau_{p2}$ and the
boundary $\tau=\infty$ as region 2 and the region bounded by the two curves of
singularities as region 3. If we treat $\sigma$ and $\rho$ as functions of
$(t,\tau)$, then they do not possess any extremum in regions 1 and 2. In
region 1, $\sigma$ takes a maximum value $-79/257$ at the boundary point
$t=0,\tau=3$. As we move away from this point, $\sigma$ becomes more and more
negative and $\sigma\rightarrow-\infty$ as we approach the singularity curve.
In region 2, $\sigma$ attains a maximum value of $-26/23$ in the limit
$\tau\rightarrow\infty$. It takes a smaller value for any point interior to
this region and as we approach the singularity curve,
$\sigma\rightarrow-\infty$. In region 3, $\sigma\rightarrow\infty$ as we
approach either of the singularity curves from an interior point of this
region. However, it remains positive throughout inside the region 3 and hence
form a valley shaped surface. To find the minimum value of $\sigma$ in this
region we set both the partial derivatives $\frac{\partial\sigma}{\partial t}$
and $\frac{\partial\sigma}{\partial\tau}$ to zero to obtain:

$(\tau-9)^{5}=0$

and

$\displaystyle(\tau-9)\Big{(}729\left(28t^{2}+36t+143\right)\tau^{4}+8(6t+2635)\tau^{6}+144(276t+17)\tau^{5}$
$\displaystyle-1458(30t+217)\tau^{3}+6561(4t+13)\tau^{2}-275562\tau+413343\Big{)}=0\
.$

Thus $\sigma$ has a line of minima given by $\tau=9$, with the minimum value
$\sigma_{m}=3$.

We now focus on the range of values for $\rho$ that admits attractor
solutions. In region 1 as well as region 2 it does not admit an extremum. At
the boundary point $t=0,\tau=3$ of region 1 it takes the maximum value
$\rho=-575/514$. As we move away from this point, it becomes more negative and
diverges as we approach the singularity curve. Similarly, in region 2, it has
a maximum value $\rho=-82/23$ in the limit $\tau\rightarrow\infty$. It becomes
more negative at finite $\tau$ and as we approach the singularity curve,
$\rho$ diverges. In the interior of region 3, $\rho$ takes positive values and
$\rho\rightarrow\infty$ as we approach the singularity curves from any point
in the interior. To find the minimum value of $\rho$ in this region, we set
$\frac{\partial\rho}{\partial t}$ and $\frac{\partial\rho}{\partial\tau}$ to
zero. A numerical computation indicates that these equations do not admit any
solution in the interior of region 3. For any fixed $t$ it admits a minimum
for some $\tau$. The value of this minimum decreases with $t$. Thus, the
functional form of $\rho$ defines a valley which slopes downward towards
smaller values of $t$. To find the smallest allowed value of $\rho$, we set
$t=0$ in it and extremize. We find

$\displaystyle\Big{(}1640\tau^{4}+1240\tau^{3}-6804\tau^{2}+19962\tau+1863\Big{)}\Big{(}92\tau^{5}-2289\tau^{4}+9801\tau^{3}+3078\tau^{2}$
(5.63)
$\displaystyle-3645\tau+6561\Big{)}-\Big{(}460\tau^{4}-9156\tau^{3}+29403\tau^{2}+6156\tau-3645\Big{)}\Big{(}328\tau^{5}+310\tau^{4}$
(5.64) $\displaystyle-2268\tau^{3}+9981\tau^{2}+1863\tau+729\Big{)}=0\ .$
(5.65)

This admits a unique solution $\tau_{m}$ in region 3. Numerically we find
$\tau_{m}\simeq 9.03$. The value of $\rho$ corresponding to this point is
$\rho_{m}\simeq 9.277$.

From the above analysis we find that, for $\sigma>3$ and $\rho>\rho_{m}$ we
have multiple non-BPS solutions in region 3. If $-26/23<\sigma<-79/257$ and
$-82/23<\rho<-575/514$ we have a unique non-BPS attractor in region 1 for a
given set of charges. Finally, for $\sigma<-26/23$ and $\rho<-82/23$ we have
two non-BPS attractors for a given set of charges one of which lies in region
1 and the other in region 2. No attractors exist for charges lying in the
range $-79/257<\sigma<3$ and $-575/514<\rho<\rho_{m}$. Further, we note that
since the above solutions exist only when $\sigma>3$ or $\sigma<-79/257$,
whereas the solutions obtained from (5.54) and (5.55) exist only when
$1<\sigma\leq 3$, all the three set of solutions are mutually exclusive.

As an example, consider the values $\sigma=4,\rho=16$. It gives a pair of
solutions for $(\tau,t,z)$ with approximate values $(7.101,2.231,10.225)$ and
$(12.281,4.102,17.682)$, both lying in region 3. If we take
$\sigma=-2,\rho=-7$ we find one solution in region 1 with $\tau\simeq 4.853$
and $t\simeq 0.61$ and another in region 2 with $\tau\simeq 55.008$ and
$t\simeq 10.363$. In this case $z$ takes the approximate values $6.506$ and
$73.587$ respectively. On the other hand, for $\sigma=-1,\rho=-7/2$ we find a
unique solution in region 1 with $\tau\simeq 4.298,t\simeq 0.383$ and $z\simeq
5.724$.

###  5.2 Stability

Now we will analyze the stability of the non-BPS black holes obtained in the
previous subsection. For this purpose we need to obtain the recombination
factors of the respective black hole configurations. Let the non-BPS black
hole results from wrapping a $M2$ brane on a non-holomorphic curve
$C=\alpha_{1}C^{1}+\alpha_{2}C^{2}+\alpha_{3}C^{3}$ of the Calabi-Yau manifold
$\mathcal{M}$. The charges $q_{I}$ of the black hole are given by

$q_{I}=\int_{C}J_{I}=\alpha_{1}C^{1}\cdot J_{I}+\alpha_{2}C^{2}\cdot
J_{I}+\alpha_{3}C^{3}\cdot J_{I}\ .$ (5.66)

The intersection numbers $C^{I}\cdot J_{L}$ can be obtained from the
expression for the Kähler cone matrix (4.25) corresponding to the THCY
manifold $\mathcal{M}$. We find

$q_{1}=\alpha_{1}-2\alpha_{3}\ ,\ q_{2}=\alpha_{2}\ ,\ q_{3}=\alpha_{3}\ .$
(5.67)

From the above we find the coefficients $\alpha_{I}$ in terms of the black
hole charges $q_{J}$ as
$\alpha_{1}=q_{1}+2q_{3}=(2\rho+1)q_{3},\alpha_{2}=q_{2}=\sigma q_{3}$ and
$\alpha_{3}=q_{3}$. Thus the recombination factor (5.7) corresponding to the
non-BPS black holes in model 1 have the expression

$R=\frac{\sqrt{V_{\rm cr}}}{z|q_{3}|\left(|(2\rho+1)|\tau+|\sigma|t+1\right)}\
.$ (5.68)

Here $V_{\rm cr}$ is the value of the effective black hole potential at the
critical point. Substituting the expression for the effective black hole
potential (5.8), using the volume constraint (5.15) and simplifying, we find

$R=\frac{\sqrt{3\left(\rho^{2}\left(\tau^{2}-4\tau+8\right)+4\rho+t^{2}\left(24\rho^{2}+\rho(12-8\sigma)+\sigma^{2}-2\sigma+2\right)+t\left(4\rho^{2}\tau+8\rho+2\right)+1\right)}}{\left(|(2\rho+1)|\tau+|\sigma|t+1\right)}$
(5.69)

We need to substitute the solutions in the three branches of non-BPS black
holes to obtain the corresponding recombination factors in each of the cases.
Consider the solution (5.30) first. Recall that, the allowed values of $\rho$
and $\sigma$ for this solution are negative, whereas $2\rho+1>0$. Keeping this
in mind and substituting (5.30) in (5.69) we find the recombination factor to
have the form

$R=-\frac{3\rho(2\rho+1)(4\rho-\sigma+1)}{16\rho^{3}+2\rho^{2}(\sigma+11)+10\rho-\sigma+1}\
.$ (5.70)

The black hole becomes unstable when $R>1$. This happens for

$\sigma>\frac{40\rho^{3}+40\rho^{2}+13\rho+1}{4\rho^{2}+3\rho+1}\ .$ (5.71)

However, recall that for the solution (5.30) to remain inside the Kähler cone
we need to impose the bound $\sigma<{\rm min}\\{2\rho,6\rho+1\\}$. This may
not be always compatible with (5.71). We find that (5.71) does not hold when
$\rho$ takes values in the range $(\sqrt{17}-9)/32<\rho<0$ or
$-1/2<\rho<(\sqrt{17}-9)/16$. For these values of $\rho$ the recombination
factor $R$ is always smaller than one and the corresponding black hole
solutions are stable. For $(\sqrt{17}-9)/16<\rho<(\sqrt{17}-9)/32$ we can
choose $\sigma$ to satisfy (5.71). In that case the resulting non-BPS black
holes become unstable. However, we can also choose $\sigma$ not to satisfy
this bound. The non-BPS black holes for those values of $\sigma$ remain stable
against decay into constituent BPS/anti-BPS pairs.

Next consider the second branch of solutions given by (5.31). Here the charge
ratios take values in the interval $-1/2<\rho<0$ and $\sigma>{\rm
max}\\{2\rho+1,6\rho+2\\}$. Hence the values of $\sigma$ and $2\rho+1$ always
remain positive. Substituting (5.31) in (5.69) and simplifying we find

$R=-\frac{3\rho(2\rho+1)(4\rho-\sigma+1)}{16\rho^{3}-2\rho^{2}(\sigma-7)+\sigma-1}\
.$ (5.72)

In order to get $R>1$ the charge ratios need to satisfy the bound

$\sigma>\frac{40\rho^{3}+32\rho^{2}+3\rho-1}{8\rho^{2}+3\rho-1}\ .$ (5.73)

However, for $-1/2<\rho<0$, the rhs of the above equation is always smaller
than ${\rm max}\\{2\rho+1,6\rho+2\\}$. Since the Kähler condition implies
$\sigma>{\rm max}\\{2\rho+1,6\rho+2\\}$, all such values of $\sigma$ satisfy
the bound (5.73). Thus, the recombination factor is always greater than one
and hence the resulting non-BPS black holes in this branch are all unstable.

Finally, for the branch (5.32), the recombination factor takes the form

$R=\left|\frac{3\rho(2\rho+1)(4\rho-\sigma+1)}{|2\rho+1|\left(8\rho^{2}-4\rho(\sigma-1)-\sigma+1\right)+\rho(2\rho|\sigma|+|\sigma|+2\rho-\sigma)}\right|\
.$ (5.74)

For this solution to be inside the Kähler cone we can have either
$\rho>0,\sigma<2\rho$ or $\rho<-1/2,\sigma>2\rho+1$. We will treat the
stability conditions for $\rho>0,0<\sigma<2\rho$ and $\rho>0,\sigma<0$
separately. Consider first the case $\rho>0,0<\sigma<2\rho$. Analyzing the
formula (5.74) we find that for
$\rho>\left(2-\sqrt{2}\right)^{1/3}\left.\left(2^{1/3}+\left(2+\sqrt{2}\right)^{2/3}\right)\right/4$
the value of the recombination factor always becomes greater than one
irrespective of the value of $\sigma$ and the black holes become unstable.
However, for
$0<\rho<\left(2-\sqrt{2}\right)^{1/3}\left.\left(2^{1/3}+\left(2+\sqrt{2}\right)^{2/3}\right)\right/4$
the black holes become unstable if $\sigma>1-8\rho^{3}/(3\rho+1)$, and stable
if $\sigma<{\rm min}\\{2\rho,1-8\rho^{3}/(3\rho+1)\\}$. In the case
$\rho>0,\sigma<0$, black holes become stable if $\sigma<{\rm
min}\\{0,(1+3\rho-8\rho^{3})/(1+5\rho+4\rho^{2})\\}$ and unstable otherwise.
Similarly, for $\rho<-1/2,\sigma>0$ we find stable non-BPS black holes
provided $\sigma>{\rm
max}\\{0,(1+3\rho-4\rho^{2}-8\rho^{3})/(1+3\rho+4\rho^{2})\\}$. For
$\rho<-1/2,2\rho+1<\sigma<0$ there are no stable attractors if $\rho<\rho_{0}$
where $\rho_{0}\simeq-0.776$ is a root of $8\rho^{3}+4\rho^{2}-3\rho-1=0$. For
$\rho_{0}<\rho<-1/2$ we have stable attractors if $\sigma$ takes values in the
interval ${\rm max}\\{2\rho+1,-(8\rho^{3}+4\rho^{2}-3\rho-1)/(\rho+1)\\}$.

We will now analyze the stability of the non-BPS black holes in model 2. Let
the corresponding $M2$ brane wraps a non-holomorphic curve
$C=\sum\alpha_{I}C^{I}$. Using the expression for the Kähler cone matrix in
(4.41), we find the charges $q_{I}$ of the black hole to be

$q_{1}=\alpha_{3}\ ,\;\;q_{2}=\alpha_{1}\ ,\;\;{\rm
and}\;\;q_{3}=\alpha_{2}-3\alpha_{3}\ .$ (5.75)

Thus, the coefficients $\alpha_{I}$ are given in terms of the charges $q_{I}$
as $\alpha_{1}=q_{2}=\sigma q_{3},\alpha_{2}=3q_{1}+q_{3}=(3\rho+1)q_{3}$ and
$\alpha_{3}=q_{1}=\rho q_{3}$. The recombination factor (5.7) can be expressed
as

$R=\frac{\sqrt{V_{\rm
cr}}}{z|q_{3}|\big{(}|\sigma|\tau+|(3\rho+1)|t+|\rho|\big{)}}\ .$ (5.76)

Using the constraint (5.39) in (5.35), the recombination factor takes the form

$R=\frac{1}{2\tau\big{(}|\sigma|\tau+|(3\rho+1)|t+|\rho|\big{)}}\sqrt{\frac{\widetilde{V}}{2(\tau-9)}}\
,$ (5.77)

where we have introduced

$\displaystyle\widetilde{V}$ $\displaystyle=$
$\displaystyle\sigma^{2}\Big{(}38\tau^{5}-311\tau^{4}-258\tau^{3}+24t^{2}(\tau-9)\tau^{2}+648\tau^{2}$
(5.78) $\displaystyle+$ $\displaystyle
2t\left(25\tau^{4}-216\tau^{3}-54\tau^{2}+108\tau-243\right)-1431\tau-162\Big{)}$
(5.79) $\displaystyle+$ $\displaystyle
4\tau^{2}\left(3\rho^{2}(\tau-9)\tau^{2}+6\rho(\tau-9)\tau+2\tau^{3}+(2t+1)\tau^{2}-18\right)$
(5.80) $\displaystyle-$ $\displaystyle
4\sigma\tau\Big{(}3\rho\left(2\tau^{4}-17\tau^{3}-12\tau^{2}+36\tau-81\right)+2\tau^{4}+(2t-5)\tau^{3}$
(5.81) $\displaystyle-$ $\displaystyle 6(2t+1)\tau^{2}+18(3t+1)\tau-27\Big{)}\
.$ (5.82)

The black hole becomes stable for $R<1$. By directly substituting specific
values of charges and the corresponding solutions $t$ and $\tau$ we find
stable black holes for a wide range of charges. In the following we consider a
class of configurations for stable as well as unstable black holes.

We choose a specific value of the charge ratio $\sigma=3/2$. In this case, the
equations of motion admit a unique solution given by $(\tau_{-},t_{-})$ in
(5.58). Substituting $\sigma=3/2$ in $(\tau_{-},t_{-})$ we find the solution
to have the simple form

$\tau_{-}=9\left(\sqrt{2}-1\right)\;\;{\rm
and}\;\;t_{-}=\frac{1}{36}\left(141-161\sqrt{2}\right)-3\left(\sqrt{2}-1\right)\rho.$
(5.83)

For $t_{-}>0$ we need $\rho$ to satisfy the bound
$\rho<-(181+20\sqrt{2})/108\simeq-1.938$. We substitute these values for
$\tau_{-}$ and $t_{-}$ (and $\sigma=3/2)$) in (5.77) to obtain the following
simple expression for the recombination factor:

$R=\frac{9\left(1+\sqrt{2}\right)\sqrt{11664\left(17-12\sqrt{2}\right)\rho^{2}+216\left(1952\sqrt{2}-2765\right)\rho-317524\sqrt{2}+449721}}{2\left(324\left(\sqrt{2}-1\right)\rho^{2}+\left(591\sqrt{2}-567\right)\rho+647\sqrt{2}-627\right)}\
.$

It can be easily verified that $R<1$ provided

$\rho<-\frac{1}{216}\left(367+8\sqrt{2}+\sqrt{5\left(28101+80\sqrt{2}\right)}\right)\simeq-3.49\
.$ (5.84)

Thus, for $\sigma=3/2$ we obtain non-BPS attractors as given by (5.83). The
Kähler cone condition requires that $\rho\lesssim-1.938$. These non-BPS black
holes remain unstable in the range $-3.49\lesssim\rho\lesssim-1.938$ and
become stable when $\rho\lesssim-3.49$.

We will now take the value $\sigma=5/2$. In this case there are two solutions
$(\tau_{+},t_{+})$ and $(\tau_{-},t_{-})$ with

$\tau_{\pm}=3\left(5\pm\sqrt{10}\right),\
t_{\pm}=-\frac{1}{180}\left(108\left(5\pm\sqrt{10}\right)\rho\pm
271\sqrt{10}+1423\right)\ .$ (5.85)

The solution $(\tau_{-},t_{-})$ exists when $\rho<-(4405+68\sqrt{10})/1620$
and the solution $(\tau_{+},t_{+})$ exists when
$\rho<-(4405-68\sqrt{10})/1620$. Thus, we have multiple solution for
$\rho<-(4405+68\sqrt{10})/1620$. We can substitute these solutions in (5.77)
to find the respective recombination factors

$R_{\pm}=\frac{15\sqrt{5}\sqrt{6480\left(89\pm
28\sqrt{10}\right)\rho^{2}-216\left(13375\pm 4208\sqrt{10}\right)\rho\pm
1138324\sqrt{10}+3618013}}{2\left(540\left(7\pm
2\sqrt{10}\right)\rho^{2}+3\left(3595\pm 1026\sqrt{10}\right)\rho\pm
5426\sqrt{10}+19025\right)}$

Introducing $\rho_{s\pm}$ as

$\rho_{s\pm}=-\frac{8695\mp 8\sqrt{10}+\sqrt{87877865\pm
72560\sqrt{10}}}{3240}\ $

we find that $R_{\pm}<1$ provided $\rho<\rho_{s\pm}$. Hence, the solution
$(\tau_{-},t_{-})$ is unstable for $\rho_{s-}<\rho<-(4405+68\sqrt{10})/1620$
and becomes stable for $\rho<\rho_{s-}$ whereas the solution
$(\tau_{+},t_{+})$ is unstable for $\rho_{s+}<\rho<-(4405-68\sqrt{10})/1620$
and becomes stable for $\rho<\rho_{s+}$. Since $\rho_{s-}<\rho_{s+}$, both the
solutions in (5.85) become stable when $\rho<\rho_{s-}$.

We will now analyze the stability of solutions for (5.56). In this case it is
not possible to obtain any analytical results. We can numerically solve the
equations of motion for various choices of $\rho$ and $\sigma$ and obtain the
recombination factor for each of the solutions. As we have noted previously,
in this case there exist two singularity curves which divide the moduli space
into three regions. We first consider $\sigma>3$ and $\rho>\rho_{m}$ for which
the solutions lie in region 3. We took the value $\rho=45$ and varied $\sigma$
from $4$ to $14$. In each of these cases, there are two distinct solutions.
Upon computing the recombination factor we observe that for a fixed $\rho$ and
for small values of $\sigma$ the black holes become stable. The recombination
factor increases monotonically and the black holes become unstable for larger
values of $\sigma$. We also observe that, for fixed values of $(\rho,\sigma)$
with a smaller value $\sigma$, the recombination factors corresponding to the
two distinct solutions almost coincide with each other. In the following, we
plot the solutions for $\tau$ and $t$ in Fig.1. We can clearly see two
branches of solutions in both the cases. Since the recombination factors
corresponding to the two branches of solutions are close to each other, we
plot it for the lower branch of solutions. Further, we plot the difference of
the two recombination factors separately (see Fig.2).

Figure 1: Two branches of solutions for $\tau$ and $t$ with $\rho=45$ and
$4\leq\sigma\leq 14$.

Figure 2: Recombination factor for the lower branch of solutions and the
difference of two recombination factors. The recombination factors take
slightly higher values in the upper branch of solutions.

We will now consider the solutions in regions 1 and 2. In this case, we have a
unique non-BPS attractor in region 1 when $\sigma$ and $\rho$ lies in the
range $-26/23<\sigma<-79/257$ and $-82/23<\rho<-575/514$ and multiple non-BPS
attractors when $\sigma<-26/23,\rho<-82/23$. We will first focus on the unique
solutions in region 1. We chose $\sigma$ to be valued in the range
$-21/20\leq\sigma\leq-11/20$ and $\rho=-7/2$. The resulting solution for
$\tau$ and $t$ are plotted in Fig.3. We notice that $t$ monotonically
increases with $\sigma$, while $\tau$ decreases as we increase it. We have
numerically computed the corresponding recombination factors (see Fig.6). We
find stable non-BPS black holes for smaller values of $\sigma$.

Further, we choose the value $\rho=-15$ for various $\sigma$ in the range
$-23/5\leq\sigma\leq-13/5$. Multiple solutions occur for these values of the
charge ratios. We find one solution in region 1 and the other in region 2. The
solutions for $(\tau,t)$ in region 1 are plotted in Fig.4 and solutions for
$(\tau,t)$ in the region 2 are plotted in Fig.5. The corresponding
recombination factors are plotted in Fig.6. The lower curve in Fig.6
corresponds to recombination factors in region 1 and upper curve in it
corresponds to recombination factors in region 2. The black hole solutions
with a larger value of $\sigma$ remain stable. The value of $R$ increases
gradually as we decrease $\sigma$ with a fixed value of $\rho$ and the black
holes become unstable after a critical value.

Figure 3: Solutions for the inhomogeneous coordinates $\tau$ and $t$ in region
1 for $\rho=-7/2$ and $-21/20\leq\sigma\leq-11/20$.

Figure 4: Solutions for the inhomogeneous coordinates $\tau$ and $t$ in region
1 for $\rho=-15$ and $-23/5\leq\sigma\leq-13/5$.

Figure 5: Solutions for the inhomogeneous coordinates $\tau$ and $t$ in region
2 for $\rho=-15$ and $-23\leq\sigma\leq-13/5$.

Figure 6: Recombination factors for the above solutions. The figure in the
left corresponds to recombination factors for solutions in region 1 with
$\rho=-7/2$ and $-21/20\leq\sigma\leq-11/20$. The figure in right corresponds
to recombination factors for the multiple solutions with $\rho=-15$ and
$-23\leq\sigma\leq-13/5$.

##  6 Non-BPS black strings

We will now consider black strings in our supergravity theory. These are
magnetically charged objects with charges $p^{I}$. They are obtained upon
wrapping $M5$ branes on a four cycle of the Calabi-Yau manifold. The central
charge associated with the black string is given by

$Z=C_{IJK}p^{I}t^{J}t^{K}\ .$ (6.1)

The effective potential corresponding to the black string is

$V=3Z^{2}-2C_{IJ}p^{I}p^{J}\ .$ (6.2)

To obtain the equation of motion, we need to extremize this effective black
string potential. Note that the moduli $t^{I}$ are constrained to satisfy
(2.2). We can use the method of Lagrange multipliers to extremize the
potential. The resulting equation of motion is given by

$6Z(p^{J}-Zt^{J})-C^{JK}C_{KLM}p^{L}p^{M}+C_{LM}p^{L}p^{M}t^{J}=0\ .$ (6.3)

Upon introducing $X^{I}\equiv p^{I}-Zt^{I}$, the above equation can be
rewritten as [14]

$4ZX^{I}+X^{J}X^{K}C_{JKL}\big{(}t^{L}t^{I}-C^{LI}\big{)}=0.$ (6.4)

Setting $X^{I}=0$ we obtain BPS solutions. Upon simplification, they take the
form

$t^{I}=\frac{p^{I}}{\big{(}C_{JKL}p^{J}p^{K}p^{L}\big{)}^{1/3}}\ .$ (6.5)

Thus, for any five dimensional supergravity the resulting BPS black string
solutions are unique. Solutions of the equation of motion (6.4) with
$X^{I}\neq 0$ correspond to non-BPS black strings.

For non-BPS black strings, using (2.2) we can show that $X^{I}$ satisfies the
constraint

$C_{IJK}t^{I}t^{J}X^{K}=0\ .$ (6.6)

Multiplying $X^{I}$ on (6.4) we can show that

$4ZC_{IJ}X^{I}X^{J}-C_{IJK}X^{I}X^{J}X^{K}=0.$ (6.7)

Setting $X^{I}=\check{X}\tilde{X}^{I}$ in the above we find

$\check{X}=\frac{4ZC_{MN}\tilde{X}^{M}\tilde{X}^{N}}{C_{IJK}\tilde{X}^{I}\tilde{X}^{J}\tilde{X}^{K}}.$
(6.8)

and $\tilde{X}^{I}$ satisfies the constraint
$C_{IJK}t^{I}t^{J}\tilde{X}^{K}=0$

In the present work we will focus on three parameter models. Using the
expression for $C_{IJK}t^{J}t^{K}$ given in (3.5) we find the central charge
(6.1) to have the form

$Z=\big{(}A_{1}x+A_{2}y+A_{3}z\big{)}p^{1}+\big{(}A_{2}x+A_{4}y+A_{5}z\big{)}p^{2}+\big{(}A_{3}x+A_{5}y+A_{6}z\big{)}p^{3}\
.$ (6.9)

Similarly, using (3.2) we find

$C_{IJ}p^{I}p^{J}=A_{1}(p^{1})^{2}+A_{4}(p^{2})^{2}+A_{6}(p^{3})^{2}+2\big{(}A_{2}p^{1}p^{2}+A_{3}p^{1}p^{3}+A_{5}p^{2}p^{3}\big{)}\
.$ (6.10)

Substituting these values in (6.2) we can obtain the expression for the black
string effective potential. We will once again focus on the explicit examples
we have considered in the previous sections to study BPS and non-BPS black
holes. Let us consider the model 1 first. The effective black string potential
for this model is given by

$\displaystyle V$ $\displaystyle=$
$\displaystyle\frac{4}{3}\left(p^{1}y(y+z)+p^{2}x(2y+z)+6p^{2}y^{2}-2p^{2}z^{2}+p_{3}y(x-4z)\right)^{2}$
(6.11) $\displaystyle-$
$\displaystyle\frac{4}{3}\left(p^{1}p^{2}(2y+z)+p^{3}y(p^{1}-2p^{3})+(p^{2})^{2}(x+6y)+p^{2}p^{3}(x-4z)\right)$
(6.12)

Upon extremization with respect to the scalar fields $x,y,z$ subject to the
constraint (5.15) we find the equations of motion to be of the form

$\displaystyle(y+z)\left(p^{2}(p^{1}-4p^{3})-2(y(p^{1}-4p^{3})+p^{2}(x-4z))\left(p^{1}y(y+z)+p^{2}x(2y+z)\right.\right.$
(6.13) $\displaystyle+$
$\displaystyle\left.\left.6p^{2}y^{2}-2p^{2}z^{2}+p^{3}y(x-4z)\right)\right)+(x-4z)\left(2(p^{2}(2y+z)+p^{3}y)\left(p^{1}y(y+z)\right.\right.$
(6.14) $\displaystyle+$
$\displaystyle\left.\left.p^{2}x(2y+z)+6p^{2}y^{2}-2p^{2}z^{2}+p^{3}y(x-4z)\right)-p^{2}(p^{2}+p^{3})\right)=0,$
(6.16) $\displaystyle
y(x-4z)\left(2(p^{1}(2y+z)+2p^{2}(x+6y)+p^{3}(x-4z))\left(p^{1}y(y+z)+p^{2}x(2y+z)\right.\right.$
$\displaystyle+$
$\displaystyle\left.\left.6p^{2}y^{2}-2p^{2}z^{2}+p^{3}y(x-4z)\right)-2p^{1}p^{2}-p^{3}(p^{1}-2p^{3})-6(p^{2})^{2}\right)+\left(x(2y+z)\right.$
(6.17) $\displaystyle+$
$\displaystyle\left.6y^{2}-2z^{2}\right)\left(p^{2}(p^{1}-4p^{3})-2(y(p^{1}-4p^{3})+p^{2}(x-4z))\left(p^{1}y(y+z)\right.\right.$
(6.18) $\displaystyle+$
$\displaystyle\left.\left.p^{2}x(2y+z)+6p^{2}y^{2}-2p^{2}z^{2}+p^{3}y(x-4z)\right)\right)=0.$
(6.19)

As before we will introduce the inhomogeneous coordinates $\tau=x/z,t=y/z$ and
eliminate $z$ using the constraint (5.15). In addition, we will rescale the
magnetic charges $p^{I}$ as $p^{1}=rp^{3}$ and $p^{2}=sp^{3}$. The equations
of motion (6.19) simplify a lot upon expressing in terms of the inhomogeneous
coordinates $\tau,t$ and charge ratios $r,s$. We find

$\displaystyle(rt+r-s\tau+4s-4t-\tau)(rt+r+4st+s\tau+\tau-4)=0,$ (6.20)
$\displaystyle
t(\tau-4)\left((2rt+r+2s(6t+\tau)+\tau-4)\left(t(rt+r+\tau-4)+s\left(6t^{2}+2\tau
t+\tau-2\right)\right)\right.$ (6.21)
$\displaystyle\left.-t(t+1)\left(2rs+r+6s^{2}-2\right)(2t+\tau-2)\right)+\left(6t^{2}+2\tau
t+\tau-2\right)\left((r-4)st(t+1)\right.$ (6.22)
$\displaystyle\left.(2t+\tau-2)-((r-4)t+s(\tau-4))\left(t(rt+r+\tau-4)+s\left(6t^{2}+2\tau
t+\tau-2\right)\right)\right)=0\ .\ $ (6.23)

It is possible to solve the above equations exactly to find the attractor
values of the moduli in terms of the magnetic charges. Apart from the BPS
solution $\tau=r,t=s$, there are three additional solutions to the above
equations. These solutions correspond to non-BPS black strings. They have the
simple form

$\displaystyle\tau$ $\displaystyle=$ $\displaystyle 4-r-4s\ ,\ t=s\ ,$ (6.24)
$\displaystyle\tau$ $\displaystyle=$ $\displaystyle\frac{r+8s}{1+2s}\ ,\
t=-\frac{s}{1+2s}\ ,$ (6.25) $\displaystyle\tau$ $\displaystyle=$
$\displaystyle\frac{4-r+4s}{1+2s}\ ,\ t=-\frac{s}{1+2s}\ .$ (6.26)

Requiring the solution to satisfy the Kähler cone condition $\tau>2,t>0$ we
find the supersymmetric solutions for $r>2,s>0$. For the first branch of non-
BPS solutions (6.24), we have $s>0,r<-2(2s-1)$. For the second branch (6.25)
we have $-1/2<s<0$ and $r>2(1-2s)$. For the third branch $-1/2<s<0$ and $r<2$.
All these branches of solutions are mutually exclusive from each other.
Moreover, the resulting solution for a given set of charges is always unique.

Now we will analyze non-BPS black strings in model 2. For this case, the
effective black string potential takes the form

$V=\frac{1}{3}\Big{(}\tilde{f}_{1}(p^{1})^{2}+4x^{4}(p^{2})^{2}+\tilde{f}_{2}(p^{3})^{2}+\tilde{f}_{3}p^{1}p^{2}+\tilde{f}_{4}p^{1}p^{3}+\tilde{f}_{5}p^{2}p^{3}\Big{)}\
,$ (6.27)

where we have introduced the functions

$\displaystyle\tilde{f}_{1}$ $\displaystyle=$ $\displaystyle
36x^{4}+24x^{3}(2y+z)+16x^{2}\left(y^{2}+yz-2z^{2}\right)-12x\left(2yz^{2}+z^{3}+1\right)-4y+9z^{4}-2z,$
(6.28) $\displaystyle\tilde{f}_{2}$ $\displaystyle=$ $\displaystyle
x^{4}-12x^{3}z+90x^{2}z^{2}-324xz^{3}+6x+729z^{4}-54z,$ (6.29)
$\displaystyle\tilde{f}_{3}$ $\displaystyle=$ $\displaystyle
4x\left(6x^{3}+2x^{2}(2y+z)-3xz^{2}-2\right),$ (6.30)
$\displaystyle\tilde{f}_{4}$ $\displaystyle=$ $\displaystyle
2\left(6x^{4}+x^{3}(4y-34z)-3x^{2}z(8y-49z)+2x\left(54yz^{2}+36z^{3}-1\right)-81z^{4}+6z\right),$
(6.31) $\displaystyle\tilde{f}_{5}$ $\displaystyle=$ $\displaystyle
4x^{2}\left(x^{2}-6xz+27z^{2}\right).$ (6.32)

We will extremize this potential to obtain the equations of motion. We need to
use the method of Lagrange multipliers to incorporate the constraint (5.39).
We find

$\displaystyle(p^{1})^{2}\left(-12x^{4}(2y+z)-16x^{3}(y-z)(y+2z)+18x^{2}z^{2}(2y+z)+2x\left(2y-9z^{4}+z\right)-3z^{2}\right)$
(6.33) $\displaystyle+$ $\displaystyle
2p^{1}x\left(x^{3}z(2p^{2}-17p^{3})+x(2p^{2}+p^{3})\left(2x^{2}(3x+y)-1\right)-54p^{3}xyz^{2}-36p^{3}xz^{3}+81p^{3}z^{4}\right)$
(6.34) $\displaystyle+$ $\displaystyle
8(p^{2})^{2}x^{5}+p^{3}x^{2}\left(8p^{2}x^{3}-36p^{2}xz(x-3z)+p^{3}\left(2(x-3z)\left(x^{2}-6xz+27z^{2}\right)+3\right)\right)=0,$
(6.36)
$\displaystyle(p^{3}x-p^{1}z)\Big{(}p^{1}\left(2x\left((x-9z)\left(6x^{2}+2x(2y+z)-3z^{2}\right)-1\right)+9z\right)+4p^{2}x^{3}(x-9z)$
$\displaystyle+$ $\displaystyle
p^{3}x\left(2(x-9z)\left(x^{2}-6xz+27z^{2}\right)+9\right)\Big{)}=0\ .$ (6.37)

To simplify these equations, we will use charge ratios defined as
$r=p^{1}/p^{3}$ and $s=p^{2}/p^{3}$ and the inhomogeneous coordinates
$\tau=x/z$ and $t=y/z$. In terms of these variables the above equations read
as

$\displaystyle
r^{2}\left(4\left(2t^{2}+2t-7\right)\tau^{3}+8(2t+1)\tau^{4}-9(2t+1)\tau^{2}-9(4t+1)\tau+27\right)$
(6.38) $\displaystyle-$ $\displaystyle
2r\tau\left((8s+4)\tau^{4}+(6s+3)\tau^{2}-9\tau(2s+6t+5)-18\tau^{3}+81\right)$
(6.39) $\displaystyle-$
$\displaystyle\tau^{2}\left(8\left(s^{2}+s+1\right)\tau^{3}-3\tau^{2}(12s-2t+5)+27(4s+3)\tau-135\right)=0\
,$ (6.41)
$\displaystyle(r-\tau)\Big{(}r\left(8\tau^{4}+4(t-22)\tau^{3}-27(2t+1)\tau^{2}+9\tau+81\right)+\tau\big{(}4(s+5)\tau^{3}$
$\displaystyle-$ $\displaystyle
3\tau^{2}(12s-6t+7)+135\tau-405\big{)}\Big{)}=0$ (6.42)

We will now solve these equations to obtain black string configurations. It
can easily be seen that $\tau=r,t=s$ solves these equations. This set of
solutions corresponds to the BPS string. To obtain the non-BPS black string
solutions we will first consider the inverse problem where we express the
charge rations $r$ and $s$ in terms of the inhomogeneous coordinates $\tau$
and $t$. This can be easily obtained from the equations of motion. We find two
independent solutions for non-BPS black strings:

$\displaystyle
r=\tau,s=-\frac{1}{\tau^{2}}\big{(}2\tau^{3}+\tau^{2}(t+1)-3\tau+9\big{)}\ ,$
(6.43)

and

$\displaystyle
r=-\frac{\tau\left(56\tau^{3}-(27-54t)\tau^{2}+405\tau-1215\right)}{24\tau^{4}+4(6t-65)\tau^{3}-9(30t+13)\tau^{2}+27\tau+243}\
,$ (6.44) $\displaystyle
s=-\frac{2\tau^{2}(4\tau^{2}-34\tau^{3}+27t^{2}-24)+t\left(64\tau^{3}-99\tau^{2}+405\tau-1215\right)+144\tau-324}{24\tau^{4}+4(6t-65)\tau^{3}-9(30t+13)\tau^{2}+27\tau+243}.$
(6.45)

The first set of equations (6.43) can be easily solved for the scalar moduli
to obtain

$\tau=r,t=-\frac{1}{r^{2}}\left(2r^{3}+r^{2}(s+1)-3r+9\right)\ .$ (6.46)

Thus, for a given set of charges, we have a unique solution in this branch.
The resulting solution will lie inside the Kähler cone if $r>3$ and
$s<-(2r^{3}+r^{2}-3r+9)/r^{2}$.

We will now analyze the second branch of non-BPS solutions given in (6.44) and
(6.45). First solve (6.44) for $t$ to obtain

$t=-\frac{r\left(24\tau^{4}-260\tau^{3}-117\tau^{2}+27\tau+243\right)+\tau\left(56\tau^{3}-27\tau^{2}+405\tau-1215\right)}{6\tau^{2}(r(4\tau-45)+9\tau)}$
(6.47)

Substituting this expression for $t$ in (6.45) and simplifying we obtain

$\displaystyle
2\tau^{4}\left(12r^{2}+4r(3s+8)+27s+9\right)-r\tau^{3}(268r+270s+135)-27(r-15)r\tau^{2}$
(6.48) $\displaystyle+$ $\displaystyle 27(r-45)r\tau+243r^{2}=0\ .$ (6.49)

This is a quartic equation in $\tau$. Thus we can express the exact analytic
solutions for $\tau$ in terms of the charge ratios $r$ and $s$. However,
before we consider the exact solutions, we will qualitatively analyze the
relevant equations for this branch of solutions.

From (6.44) and (6.45) we find that the charge ratios $r$ and $s$ are
expressed as rational functions of $\tau$ and $t$. The denominators of these
rational functions vanish when

$24\tau^{4}+4(6t-65)\tau^{3}-9(30t+13)\tau^{2}+27\tau+243=0\ .$ (6.50)

This defines a singularity curve in the moduli space. To see this, solve the
above equation for $t$ to obtain

$t=-\frac{24\tau^{4}-260\tau^{3}-117\tau^{2}+27\tau+243}{6\tau^{2}(4\tau-45)}$
(6.51)

The numerator has a single zero at $\tau=\tau_{0}\simeq 11.2506$ in the
$\tau>3$ region. The denominator vanishes at $\tau=45/4$. The value of $t$ in
the above equation remains positive for $45/4<\tau<\tau_{0}$. This gives rise
to a line in the moduli space which starts at $\tau=\tau_{0}$ when $t=0$ and
asymptotes to $\tau=45/4$ line as $t\to\infty$. Points on this curve do not
correspond to any extremal black string configuration. This singularity curve
divides the moduli space into two regions, say region 1 and region 2. As we
approach it from either of the regions the charge ratios diverge. Any point
interior to either of the regions corresponds to a possible non-BPS black
string configuration. The equations of motion (6.44) and (6.45) guarantee the
existence of such attractor configurations. In region 1, at the boundary point
$(\tau=3,t=0)$ the charge ratios take values $r=141/215$ and $s=-56/215$.
These are the minimum possible values for $r$ and $s$ in region 1. These
values change from point to point and they diverge as we approach the
singularity curve from the left. Similarly we can notice that, in region 2 the
charge ratios take the maximum values $r=-7/3$ and $s=-1/3$ as
$\tau\to\infty$. For any finite $t$ these values decrease monotonically with
$\tau$ in this region and diverge as we go closer to the singularity curve
from the right.

The above analysis indicates that these solutions are mutually exclusive to
the first branch of solutions given in (6.46). We would like to investigate
whether there exist multiple black string configurations for the second branch
of solution. We first consider extremizing $s$ in (6.45) as a function of
$\tau$ and $t$. Numerically solving the equations $\partial
s/\partial\tau=\partial s/\partial t=0$ we find that they do not admit any
solution inside the Kähler cone. We can further check if there exists an
extremum in $\tau$ for a fixed value of $t$. Setting $\partial
s/\partial\tau=0$ for fixed $t$, we find

$\displaystyle
108\tau^{4}t^{2}+36\tau\left(6\tau^{4}+\tau^{3}+45\tau^{2}-459\tau+1458\right)t+\big{(}112\tau^{6}-108\tau^{5}+3561\tau^{4}-27522\tau^{3}$
(6.52) $\displaystyle+$ $\displaystyle 71685\tau^{2}+24786\tau-10935\big{)}=0\
.$ (6.53)

This is a quadratic equation in $t$ with positive coefficients in the $\tau>3$
region. Thus, it does not admit any solution with $t>0$. As a result, $s$ does
not admit an extremum in $\tau$ for a fixed $t$ inside the Kähler cone. Hence
there can’t be any multiple solution for a given set of charges in this model.

We will now consider the exact solution of the quartic (6.48) for a given
value of $r$ and $s$. Introducing the notation

$\displaystyle d_{0}=243r^{2}\ ,\;d_{1}=27r(r-45)\ ,\;d_{2}=-27r(r-15)\ ,$
(6.54) $\displaystyle d_{3}=-r(268r+270s+135)\
,\;d_{4}=2\left(12r^{2}+4r(3s+8)+27s+9\right)\ ,$ (6.55)

the equation (6.48) can be rewritten as

$\sum_{k=0}^{4}d_{k}\tau^{k}=0\ .$

Depending on the sign of the coefficients $d_{k}$ this can admit zero, two or
four real roots. As we have argued above, for a given choice of the charge
ratios at most one of these real roots will lie inside the Kähler cone.
Denoting the solutions as $\tau_{1\pm}$ and $\tau_{2\pm}$ we have

$\displaystyle\tau_{1\pm}$ $\displaystyle=$
$\displaystyle-\frac{1}{12d_{4}}\left(3d_{3}+\sqrt{3(9d_{3}^{2}-24d_{2}d_{4}-2{\mathcal{X}})}\pm\sqrt{6\mathcal{X}+\frac{18\sqrt{3}(d_{3}^{3}-4d_{2}d_{3}d_{4}+8d_{1}d_{4}^{2})}{\sqrt{3(9d_{3}^{2}-24d_{2}d_{4}-2{\mathcal{X}})}}}\right),$
$\displaystyle\tau_{2\pm}$ $\displaystyle=$
$\displaystyle-\frac{1}{12d_{4}}\left(3d_{3}-\sqrt{3(9d_{3}^{2}-24d_{2}d_{4}-2{\mathcal{X}})}\pm\sqrt{6\mathcal{X}+\frac{18\sqrt{3}(d_{3}^{3}-4d_{2}d_{3}d_{4}+8d_{1}d_{4}^{2})}{\sqrt{3(9d_{3}^{2}-24d_{2}d_{4}-2{\mathcal{X}})}}}\right).$

Here, for easy reading we have introduced the notations

$e_{1}=d_{2}^{2}-3d_{1}d_{3}+12d_{0}d_{4}\ ,\
e_{2}=2d_{2}^{3}-9d_{2}(d_{1}d_{3}+8d_{0}d_{4})+27(d_{0}d_{3}^{2}+d_{1}^{2}d_{4})\
,$

in terms of which we define

$e_{3}=e_{2}+\sqrt{e_{2}^{2}-4e_{1}^{3}}\ {\rm and}\
\mathcal{X}=3d_{3}^{2}-8d_{2}d_{4}-2^{2/3}e_{3}^{1/3}d_{4}-2^{4/3}e_{3}^{-1/3}e_{1}\
.$

Though we obtained multiple black hole solutions in these three parameters
THCY models, the black string solutions for a given set of magnetic charges
are unique. While this might be true for the examples studied here this is not
the case in general. To see this we will consider an explicit example of three
parameter model admitting multiple black string solutions. The polytope ID
associated with this specific THCY is 146. It has Hodge numbers
$h_{1,1}=3,h_{2,1}=81$ and Euler number $\chi=-156$. The resolved weight
matrix for this manifold is

$Q=\left(\begin{array}[]{ccccccc}0&0&1&1&1&0&1\\\ 0&1&0&1&0&1&1\\\
1&0&0&1&0&0&1\\\ \end{array}\right)\ .$ (6.56)

The non-vanishing intersection numbers associated with the THCY $\mathcal{M}$
are $f=1/2,g=4/3i=4/3,j=8$. Thus, the volume of $\mathcal{M}$ is given by

$\mathcal{V}=3xyz+4xz^{2}+4yz^{2}+8z^{3}\ .$ (6.57)

The corresponding Mori cone matrix is

$\mathcal{M}^{i}_{~{}j}=\left(\begin{array}[]{ccccccc}-1&0&1&0&1&0&0\\\
-1&1&0&0&0&1&0\\\ 1&0&0&1&0&0&1\\\ \end{array}\right)\ .$ (6.58)

The divisor class $\\{D_{i}\\}$ is obtained from the $(1,1)$ homology basis of
$\mathcal{M}$ as
$D_{1}=J_{3}-J_{1}-J_{2},D_{2}=J_{2},D_{3}=J_{1}D_{4}=J_{3},D_{5}=J_{1},D_{6}=J_{2},D_{7}=J_{3}$.
From the expression for the Mori cone matrix, we find the Kähler cone matrix
to be $3\times 3$ identity matrix. Thus the Kähler cone conditions are
$x>0,y>0,z>0$ or, in terms of the inhomogeneous coordinates $\tau>0,t>0$ and
$z>0$. Using $\mathcal{V}=1$ constraint we can show that the condition $z>0$
can be derived from $\tau>0,t>0$ and need not be imposed separately.

We will study black string solutions in this model. The effective black string
potential is given by

$\displaystyle
V=\frac{1}{3}\big{(}(z(3p^{1}y+4p^{1}z+3p^{2}x+4p^{2}z)+p^{3}x(3y+8z)+8p^{3}z(y+3z))^{2}$
(6.59)
$\displaystyle-2(p^{2}(3p^{1}z+3p^{3}x+8p^{3}z)+p^{3}(p^{1}(3y+8z)+4p^{3}(x+y+6z)))\big{)}\
.$ (6.60)

The equations of motion are given by

$\displaystyle(3y+4z)(-p^{3}(3p^{1}+4p^{3})+(3p^{3}x+3p^{1}z+8p^{3}z)(8p^{3}z(y+3z)+p^{3}x(3y+8z)$
(6.61)
$\displaystyle+z(3p^{2}x+3p^{1}y+4p^{1}z+4p^{2}z)))+(3x+4z)(p^{3}(3p^{2}+4p^{3})-(3p^{3}y+3p^{2}z+8p^{3}z)$
(6.62)
$\displaystyle(8p^{3}z(y+3z)+p^{3}x(3y+8z)+z(3p^{2}x+3p^{1}y+4p^{1}z+4p^{2}z)))=0,$
(6.63)
$\displaystyle(8z(y+3z)+x(3y+8z))(p^{3}(3p^{2}+4p^{3})-(3p^{3}y+3p^{2}z+8p^{3}z)(8p^{3}z(y+3z)$
(6.64)
$\displaystyle+p^{3}x(3y+8z)+z(3p^{2}x+3p^{1}y+4p^{1}z+4p^{2}z)))+z(3y+4z)(-8p^{3}(p^{2}+3p^{3})$
(6.65)
$\displaystyle-p^{1}(3p^{2}+8p^{3})+(8p^{3}(x+y+6z)+p^{2}(3x+8z)+p^{1}(3y+8z))(8p^{3}z(y+3z)$
(6.66) $\displaystyle+p^{3}x(3y+8z)+z(3p^{2}x+3p^{1}y+4p^{1}z+4p^{2}z)))=0\ .$
(6.67)

We now express these equations in terms of the rescaled quantities to find

$\displaystyle(r(3t+4)-3s\tau-4s+4t-4\tau)(r(3t+4)+s(3\tau+4)+4(t+\tau+4))$
$\displaystyle=$ $\displaystyle 0,$ (6.68) $\displaystyle
r^{2}(3t+4)^{2}(3t+8)+8r(3t+4)\left(-4s+3t^{2}+18t+16\right)$ (6.69)
$\displaystyle-4s^{2}\left(9\tau^{2}+42\tau+40\right)-32s\left(3\tau^{2}-3(t-4)\tau+20\right)-27\tau^{2}t^{3}-72\tau
t^{3}$ (6.70) $\displaystyle-108\tau^{2}t^{2}-360\tau
t^{2}+160t^{2}-128\tau^{2}-144\tau^{2}t-448\tau t+640t-512\tau$
$\displaystyle=$ $\displaystyle 0\ .$ (6.71)

It is not instructive to express the analytical expressions for $\tau,t$ in
terms of $r,s$. Instead, we will focus on the inverse problem and express the
charge ratios as rational functions of the inhomogeneous coordinates. We find
the BPS solution $r=\tau,s=t$ and three branches of non-BPS attractors. They
are given by

$\displaystyle r=\tau\ ,\ s=-t-\frac{8(\tau+2)}{3\tau+4}\ ,$ (6.72)
$\displaystyle r=-3\tau-\frac{8(t+2)}{3t+4}\ ,\ s=t\ ,$ (6.73) $\displaystyle
r=-\frac{3\tau^{2}(3t+4)^{2}+32\left(3t^{2}+4t-4\right)+36\tau\left(3t^{2}+10t+8\right)}{(3t+4)(9\tau
t+12(\tau+t)-40)}\ ,$ (6.74) $\displaystyle
s=-\frac{3t^{2}(3\tau+4)^{2}+32\left(3\tau^{2}+4\tau-4\right)+36t\left(3\tau^{2}+10\tau+8\right)}{(3\tau+4)(9\tau
t+12(\tau+t)-40)}\ .$ (6.75)

We can easily invert (6.72) and (6.73) to express the inhomogeneous
coordinates in terms of $r,s$. We find

$\displaystyle\tau=r\ ,\ t=-s-\frac{8(r+2)}{3r+4}\ ,$ (6.76)
$\displaystyle\tau=-r-\frac{8(s+2)}{3s+4}\ ,\ t=s\ .$ (6.77)

The first solution (6.76) lies within the Kähler cone for $r>0$ and
$s<-8(r+2)/(3r+4)$ where as the second solution (6.77) lies within the Kähler
cone for $s>0$ and $r<-8(s+2)/(3s+4)$.

To invert (6.75), we will first do a shift and rescale the quantities as
$\tau=4(\tilde{\tau}-1)/3,t=4(\tilde{t}-1)/3,r=4(\tilde{r}-1)/3$ and
$s=4(\tilde{s}-1)/3$. The equations take a simpler form in terms of these
variables. We find

$\tilde{r}\tilde{t}(2\tilde{t}\tilde{\tau}-7)+2\tilde{t}^{2}\tilde{\tau}^{2}+3\tilde{t}\tilde{\tau}-3=0\
,\
\tilde{s}\tilde{\tau}(2\tilde{t}\tilde{\tau}-7)+2\tilde{t}^{2}\tilde{\tau}^{2}+3\tilde{t}\tilde{\tau}-3=0\
.$ (6.78)

Solving the above equations we find
$\tilde{\tau}=\tilde{t}\tilde{r}/\tilde{s}$ with

$\displaystyle\tilde{t}_{1\pm}$ $\displaystyle=$
$\displaystyle-\frac{1}{4}\left(\tilde{s}+\sqrt{\tilde{s}^{2}+4\tilde{g}}\pm
2\sqrt{\frac{\tilde{s}^{2}(\tilde{r}\tilde{s}-34)}{2\tilde{r}\sqrt{\tilde{s}^{2}+4\tilde{g}}}+\frac{\tilde{r}\tilde{s}^{2}-2\tilde{g}\tilde{r}-6\tilde{s}}{2\tilde{r}}}\right)\
,$ (6.79) $\displaystyle\tilde{t}_{2\pm}$ $\displaystyle=$
$\displaystyle-\frac{1}{4}\left(\tilde{s}-\sqrt{\tilde{s}^{2}+4\tilde{g}}\pm
2\sqrt{\frac{\tilde{s}^{2}(\tilde{r}\tilde{s}-34)}{2\tilde{r}\sqrt{\tilde{s}^{2}+4\tilde{g}}}+\frac{\tilde{r}\tilde{s}^{2}-2\tilde{g}\tilde{r}-6\tilde{s}}{2\tilde{r}}}\right)\
.$ (6.80)

where $\tilde{g}$ is given by

$\tilde{g}=\frac{\tilde{s}}{2\
3^{2/3}\tilde{r}}\left(\big{(}225(2\tilde{r}\tilde{s}+1)+\tilde{f}\big{)}^{1/3}+\big{(}225(2\tilde{r}\tilde{s}+1)-\tilde{f}\big{)}^{1/3}-2\
3^{2/3}\right)\ .$ (6.81)

In the above we have used the notation

$\tilde{f}=\sqrt{6(13068+24489\tilde{r}\tilde{s}+39924\tilde{r}^{2}\tilde{s}^{2}-1372\tilde{r}^{3}\tilde{s}^{3})}\
.$ (6.82)

The solution will remain inside the Kähler cone for
$\tilde{\tau}>1,\tilde{t}>1$. Though there are four solutions in (6.80) for a
given value of $\tilde{r},\tilde{s}$, all of those do not satisfy the Kähler
cone condition simultaneously. We need to find how many of those solutions lie
inside the Kähler cone. This of course will depend on the values of
$\tilde{r},\tilde{s}$. Since $\tilde{\tau}=\tilde{t}\tilde{r}/\tilde{s}$ we
can’t have solutions satisfying the Kähler cone condition if $\tilde{r}$ and
$\tilde{s}$ have opposite signs. In our notation, the Kähler cone condition
for (6.76) is $\tilde{r}>1,\tilde{s}<-1-1/\tilde{r}$, and for (6.77) it is
$\tilde{s}>1,\tilde{r}<-1-1/\tilde{s}$. Thus the three solutions (6.76),(6.77)
and (6.80) are mutually exclusive. Let us now consider the case when
$\tilde{r}$ and $\tilde{s}$ have the same sign. To understand this in detail
let us eliminate $\tau$ from the equations of motion described by (6.78) to
obtain the following quartic in the variable $\tilde{t}$

$2\tilde{r}^{2}\tilde{t}^{4}+2\tilde{r}^{2}\tilde{s}\tilde{t}^{3}+3\tilde{r}\tilde{s}\tilde{t}^{2}-7\tilde{r}\tilde{s}^{2}\tilde{t}-3\tilde{s}^{2}=0\
.$ (6.83)

First consider the case when both $\tilde{r}$ and $\tilde{s}$ are positive. In
the quartic (6.83) the coefficients change sign only once. Thus, according to
Descartes rule of signs there is only one positive root for $\tilde{t}$. By
suitable choice of the charge ratios we can make $\tilde{t}>1$ to obtain a
unique non-BPS black string solution inside the Kähler cone. However, since
the BPS solutions also exist for $\tilde{r}>1,\tilde{s}>1$ region, these two
solutions are not mutually exclusive from each other. Finally consider the
case where both $\tilde{r}$ and $\tilde{s}$ are negative. For charge ratios
valued in this range the coefficients in the quartic (6.83) change three
times. Thus the equation can admit three positive roots for $\tilde{t}$. By
suitable choice of the charge ratios it might be possible to have more than
one non-BPS attractors existing inside the Kähler cone. This is indeed the
case as a numerical analysis shows. Here we numerically solve the equation
(6.83) for $\tilde{t}$ upon setting $\tilde{r}=-6n,\tilde{s}=-5n$ with
$n=2,3,\ldots,10$. In each case there are three positive roots for $\tilde{t}$
with two of them having values greater than $1$. Since
$\tilde{\tau}=\tilde{t}\tilde{r}/\tilde{s}>\tilde{t}$ for these choices of
charge ratios, we have multiple non-BPS black strings. In Fig.7 we have
presented the two black string solutions obtained for different values of $n$.

Figure 7: Multiple non-BPS black strings in model 3 for
$\tilde{r}=-6n,\tilde{s}=-5n$ with $n=2,3,\ldots,10$. In the first case the
value of $\tilde{t}$ decreases with increasing $n$ whereas in the second case
it increases with $n$.

###  6.1 Stability

We will now analyze stability of these non-BPS black strings. For this we need
to compute the recombination factor for the black string. The recombination
factor is defined as the ratio of the volume of the non-holomorphic divisor on
which the $M5$ brane wraps to give rise to the non-BPS black string, to the
volume of minimal piece-wise calibrated cycle in the same homology class as
the non-holomorphic divisor under consideration. For double extremal black
strings, the asymptotic volume of the non-holomorphic divisor is given by the
square root of the effective black string potential at the attractor value.
For a black string of charge $p^{I}$, the volume of the minimal piece-wise
calibrated cycle is $\sum_{I}|p^{I}|C_{I}$, where $C_{I}$ is the volume of the
divisor $J_{I}$:

$C_{I}=\int_{J_{I}}J\wedge J=C_{IJK}t^{J}t^{K}\ .$ (6.84)

Denoting $R$ as the recombination factor, we have

$R=\frac{\sqrt{V_{\rm cr}}}{\sum_{I}|p^{I}|C_{I}}\ .$ (6.85)

For a three parameter model, the recombination factor is given as

$R=\frac{\sqrt{V_{\rm cr}}}{|p^{1}|C_{1}+|p^{2}|C_{2}+|p^{3}|C_{3}}\ ,$ (6.86)

where the volumes $C_{I}$ are given in (3.5). We can substitute the expression
for the black string solution in it to obtain the value of $R$. We will first
compute the recombination factor for non-BPS black strings in model 1 before
turning our discussion into the stability issues solutions in model 2. In the
case of model 1, there are three branches of non-BPS solutions. Let us first
consider the recombination factor for the first branch of solutions given in
(6.24). In this case $s>0$ and $r<2-4s$. We will consider the cases $0<r<2-4s$
and $r<0$ separately. Substituting the explicit expressions for the solution
(6.24) in (6.86) and simplifying, we find the recombination factor in the case
$s>0$ and $0<r<2-4s$ to have the form

$R=3\left(2s(s+1)(2-r-2s)\right)^{1/3}\ .$ (6.87)

In this case, $R$ always remains greater than one if $s>s_{0}$, where
$s_{0}\simeq 0.09$ is a root of the equation $108(s^{3}+s^{2})=1$ and the
resulting black strings are all unstable. For $0<s<s_{0}$ the black strings
are stable if ${\rm max}\\{0,-(108s^{3}-108s+1)/54s(s+1)\\}<r<2-4s$ and
unstable otherwise.

For $s>0$ and $r<{\rm min}\\{0,2-4s\\}$ the recombination factor becomes

$R=\frac{3\left(2s(s+1)\right)^{1/3}(2-r-2s)^{4/3}}{(2-3r-2s)}\ .$ (6.88)

In this case $R>1$ for in the entire range of allowed values for $r$ and $s$.
Thus, in this case the black strings are all unstable.

For the second branch (6.25), $r$ and $s$ are valued in the range
$-1/2<s<0,r>2-4s$. The recombination factor for this branch is given by

$R=\frac{3(-2s)^{1/3}\left((s+1)(2s+r-2)\right)^{4/3}}{(2s+1)\big{(}r(s+3)-2(s^{2}+3)\big{)}}$
(6.89)

From the above expression we find that there is no stable attractor for
$-1/2<s<-1/3$ irrespective of the values taken by $r$. For $-1/3<s<0$, the
attractors become stable if $2-4s<r<r_{0}$, where the value of $r_{0}$ is
obtained by solving $R=1$ in (6.89) for $r$ for a fixed value of $s$.

Finally we consider the third branch of solutions (6.26). In this case
$-1/2<s<0$ and $r<2$. We will first consider the range $-1/2<s<0,0<r<2$. In
this case the recombination factor is

$R=\frac{3(-2s)^{1/3}\left((s+1)(r+2s-2)\right)^{4/3}}{(2s+1)(2s^{2}-rs+r+8s-2)}\
.$ (6.90)

In this case the black strings are unstable for $-1/2<s<s_{0}$, where
$s_{0}\simeq-0.26$ is a root of $108s^{2}(s+1)^{4}-(s+3)^{3}(2s+1)^{3}=0$. For
$s_{0}<s<0$ the black strings become unstable in the range $0<r<r_{0}$ and
stable for $r_{0}<r<2$, where the value of $r_{0}$ for a fixed $s$ is
determined by $54s(s+1)^{4}(r+2s-2)^{4}-(2s+1)^{3}(2s^{2}-rs+r+8s-2)^{3}=0$.

Consider now the third branch of solutions (6.26) with $-1/2<s<0$ and $r<0$.
In this case the recombination factor becomes

$R=\frac{3(-2s)^{1/3}\left((s+1)(r+2s-2)\right)^{4/3}}{(2s+1)\left(r(s+3)+2(s^{2}+4s-1)\right)}\
.$ (6.91)

In this case the attractors are unstable for $-1/2<s<s_{0}$ where
$s_{0}\simeq=-0.036$ is a root of the equation
$2048s(s+1)^{5}+(s+3)^{4}(2s+1)^{3}=0$. For $s_{0}<s<0$ the attractors are
stable for $r_{2}<r<r_{1}$ where $r_{1,2}$ are obtained by solving $R=1$ in
(6.91) for $r$ for a fixed value of $s$.

Let us focus on the recombination factor for non-BPS black strings in model 2.
We will first consider the solution (6.46). In this case, the recombination
factor (6.86) takes the simple form

$R=\frac{3r\big{(}2r^{2}+2rs+r-3\big{)}+27}{r\big{(}2r^{2}+6rs+r-3\big{)}+9}\
.$ (6.92)

Recall that the solution (6.46) exists for $r>3$ and
$s<-(2r^{3}+r^{2}-3r+9)/r^{2}$. For any given $r$, the recombination factor
takes the minimum value $R=3/5$ when $s$ takes the maximum value
$s=-(2r^{3}+r^{2}-3r+9)/r^{2}$. The value of $R$ increases as $s$ becomes more
negative and approaches the value $R=1$ as $s\to-\infty$. Thus $R<1$ for all
the black strings in this branch of solutions and hence they are all stable.

We will now consider the recombination factor for the second branch of
solutions described by (6.47) and (6.48). The resulting expression is
complicated and it is not possible to extract any insight from it. In what
follows, we will numerically solve the equations of motion for various choices
of the charge ratios and substitute the resulting solutions to obtain the
values of the respective recombination factors. We first consider a fixed
value for $s,(s=3/2)$ and vary $r$ in the range $1\leq r\leq 5/2$. This leads
solutions in region 1. We notice the the value of $\tau$ increases
monotonically with the increase of $r$ while the value of $t$ decreases. The
plot for the inhomogeneous coordinates $\tau$ and $t$ are given in Fig.8.
Further we consider the value $s=-3/2$ and vary $r$ in the range $-24\leq
r\leq-3$ to obtain solutions in region 2. In this case, both the inhomogeneous
coordinates increase with $r$. The results are given in the Fig.9.

Figure 8: Non-BPS black string solutions in region 1 with $s=3/2$ and $1\leq
r\leq 5/2$.

Figure 9: Non-BPS black string solutions in region 2 with $s=-3/2$ and
$-24\leq r\leq-3$.

The recombination factors for solutions in both the regions are depicted in
Fig.10. In region 1 the value of recombination factor increases with $r$
whereas in region 2 it decreases. We observe that the value of $R$ remains
less than one in both the regions and it tends to saturate to $1$ as we
approach the singularity curve. Thus the corresponding black strings in the
interior of regions 1 and 2 are all stable. We have considered a wide range of
charges and numerically evaluated the recombination factor for all these
choices. We could not find any case with $R>1$. This suggests that the black
strings corresponding to the second branch of solutions are also stable.

Figure 10: Recombination factors corresponding to the solutions in both the
regions. The figure in the left corresponds to recombination factors in region
1 and the one in the right corresponds to recombination factors in region 2.

Finally we turn our attention to the stability of non-BPS black strings in
model 3. We will first consider the solutions given in (6.76) and (6.77). The
recombination factors for these two cases are given respectively as

$\displaystyle R=\frac{3(3rs+4r+4s+8)}{9rs+4r+12s+8}\ {\rm and}\
R=\frac{3(3rs+4r+4s+8)}{9rs+12r+4s+8}\ .$ (6.93)

In both these cases the value of $R$ remains less than $1$ in the allowed
range of the charge ratios. We can check this easily by expressing $R$ in
terms of the inhomogeneous coordinates. In both the cases, $R$ has a minimum
value of $3/5$ and it increases monotonically with $\tau$ and $t$ and
approaches the maximum value $R=1$ when either of these coordinates becomes
infinite. Thus the corresponding black strings are all stable. For the
solutions (6.83) we have obtained the recombination factors numerically. As we
can see from Fig.11, depending the value of $n$ the recombination factor
becomes larger or smaller than one. Thus, in this case we have both stable as
well as unstable non-BPS black string solutions.

Figure 11: Recombination factors for the multiple solutions corresponding to
Fig.7.

##  7 Conclusion

In this paper we have studied BPS as well as non-BPS black branes in five
dimensional supergravity theories arising from the compactification of
M-theory on three parameter Calabi-Yau manifolds. We considered two explicit
examples of toric Calabi-Yau manifolds with $h_{1,1}=3$ and obtained all BPS
and non-BPS black hole configurations in them. For the first model, the
resulting BPS solutions were unique for a given set of charges. However, for
model 2, we found multiple BPS black hole solutions when their charges are
valued in a particular range. The non-BPS black holes for model 1 admit three
independent solutions. However, they are mutually exclusive from each other
and for a given set of black hole charges, there is a unique solution. For the
non-BPS black holes in model 2, the moduli space admits two singularity
curves. Points on the singularity curves do not correspond to any black hole
solution. They divide the moduli space into three regions. Any point in a
given region corresponds to a possible non-BPS attractor for some suitable
choice of black hole charges. As we change the charges the points move within
a given region. But they never cross the singularity curves for finite values
of black hole charges. This model also admits multiple non-BPS black holes. We
considered the stability of doubly extremal non-BPS black hole configurations
in these models and found that the black holes become stable for a range of
charges. This is in contrast to the examples studied in [10] and the two
parameter models in [14] where the non-BPS black hole configurations were all
unstable. Moreover none of the models considered in [10, 14] admitted multiple
black hole solutions.

Subsequently we have studied non-BPS black string solutions in these models.
Once again we found three independent non-BPS solutions in model 1. They were
all mutually exclusive and hence we have unique non-BPS black string solution
for a given set of charges. Model 2 admitted two branches of solutions. The
second branch of solutions admitted a singularity curve which divided the
moduli space into two regions. Points on the singularity curve did not
correspond to non-BPS attractors for any choice of the black string charges.
Further, we found that for a given set of charges the resulting black string
solution was always unique. To demonstrate the existence of multiple non-BPS
black string solutions we considered one more three parameter Calabi-Yau
model. We obtained all black string solutions in this model and observed that
it admits multiple non-BPS solutions in a given range of black string charges.
We have also analyzed the stability of the doubly extremal black string
configurations corresponding to all the attractor values we have obtained. It
is interesting to observe so much rich structure for attractor configurations
in three parameter Calabi-Yau models. It would be worth investigating the
behaviour of both BPS as well as non-BPS multiple attractor configurations in
more detail. In four dimensions, incorporating $D6$ branes gives a rich
structure both in the BPS as well as non-BPS sector [27, 28, 29, 30].
Extending the analysis of [15] to incorporate the $D6$ branes and relating
them with the solutions studied here using the 4D-5D correspondence [31] is
another interesting aspect that needs to be explored. We hope to analyze these
issues in future.

## Acknowledgments

We would like to thank Alessio Marrani for collaboration in [14] as well as
for useful correspondences.

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