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\begin{document}

\begin{titlepage}

{\hbox to\hsize{hep-th/9806080 }}
{\hbox to\hsize{June 1998 \hfill UCSD-PTH-98-16}}
\bigskip

\begin{center}

\vspace{.5cm}

\bigskip

\bigskip

\bigskip

{\Large \bf   Branes with GUTs and Supersymmetry Breaking}

\bigskip

\bigskip

\bigskip

{\bf Joseph Lykken}$^{\bf a}$,
{\bf Erich Poppitz}$^{\bf b,}$\footnote{AfterJanuary 1, 1999: Department
of Physics, Yale University, New Haven, CT 06520-8120, USA.},
and {\bf Sandip P. Trivedi}$^{\bf a}$ \\

\smallskip

{\tt lykken@fnal.gov, epoppitz@ucsd.edu, trivedi@fnal.gov}

\bigskip

\bigskip

$^{\bf a}${ \small \it Fermi National Accelerator Laboratory\\
  P.O.Box 500\\
 Batavia, IL 60510, USA\\}

\bigskip

$^{\bf b}${\small \it Department of Physics \\
University of California at San Diego\\
9500 Gilman Drive\\
La Jolla, CA 92093, USA}

\bigskip

\bigskip

{\bf Abstract}
\end{center}
{
We study Type I string theory compactified on a $\relax{\rm T}^6/\IZ_3$ orientifold.
The low-energy dynamics is most conveniently analyzed
in terms of D3-branes. We show that a sector of the theory, which
corresponds to placing an odd number of D3-branes at orientifold fixed points,
can give rise to an $SU(5)$ gauge theory with three generations of chiral
matter fields.  The resulting model is not fully realistic, but the relative ease with
which an adequate gauge group and matter content can be obtained is promising.
The model is also of interest from the point of view of supersymmetry breaking.
We show that, for fixed values of the closed string modes, the model breaks
supersymmetry due to a conflict between a non-perturbatively generated
superpotential and an anomalous $U(1)$ D-term potential.
}

\end{titlepage}

\renewcommand{\thepage}{\arabic{page}}

\setcounter{page}{1}

\mysection{Introduction and Summary}

The past few years have seen remarkable progress in our understanding of
the non-perturbative behavior of string theory \cite{sen}. D-branes have
played a vital role in these developments \cite{tasibranes}.
The consequences of this theoretical insight in
string phenomenology  are just beginning to be explored.
In this paper we attempt to take a few preliminary steps in this  direction.
For related recent work see \cite{wittenstrongcy}--\cite{kaku}.
Perhaps the simplest idea to explore is that we live on a three dimensional
brane or somewhat more precisely, that the $3+1$ dimensional spacetime
corresponds to the world volume of a set of D3 branes. This immediately
gives rise to a question: can a (grand unified) theory accommodating the
standard
model interactions and matter content be obtained in this manner?

D-brane model building is of interest from another point of view as well.
Most of the model building so far has been carried out in the
$E_8 \times E_8 $ heterotic string \cite{dienes}.
In this context, there is a well known problem in reconciling,
within the context of weakly coupled string theory,
the ``observed"
unification of gauge coupling constants in supersymmetric
extensions of the standard model \cite{mssmgut}  and the  value of Newton's
constant. Witten \cite{wittenstrongcy}
has recently suggested working with the strongly coupled
heterotic theory to avoid this problem.
Another possibility, also mentioned
in \cite{wittenstrongcy}, is to consider model building in the Type I theory.

We begin this paper
by considering, in Section 2,  the question of gauge and gravity
unification in the Type I string theory. We show
that both  the gauge coupling unification and the
value of  the Newton constant
can be obtained within the context of Type I  perturbation theory.
Moreover, the analysis indicates that in several cases the more
appropriate  description is a T-dual one with D3 branes.
This provides additional motivation
to enquire about the standard model arising from D3 branes.

In Section 3, we turn to this issue by considering a compactification
of the Type I theory on a $\relax{\rm T}^6/\IZ_3$ orientifold. This compactification
has been considered earlier by \cite{sagnotti96}.
We point out that, in addition to the sector
considered in \cite{sagnotti96}, the moduli space for this compactification
has additional  disconnected branches, similar to the ones found in
\cite{wittentori}. The different branches correspond to distinct ways
in which the branes can be placed at the various orientifold fixed points.
The additional branches of moduli space exhibit patterns of
gauge symmetry  breaking that are not otherwise  allowed.
In particular, we show in Section 3.1, that
an $SU(5)$ grand unified theory with three generations
of matter fields in the ${\bf 10}$ and $\bar {\bf 5}$ representations
can arise in this manner.
In Section 3.2,
we show that some
nonperturbative consistency conditions \cite{Star}, \cite{wittentori}
leading to the existence of these additional branches are met.
The model obtained in
this manner is not fully realistic: there are no Higgs fields present
and  there are Yukawa couplings violating
baryon and lepton number.
Even so,  we view the relative ease with which an adequate gauge group and
matter content can be obtained as encouraging.

Finally, in Section 4, we turn to another aspect of the $SU(5)$ theory
mentioned above.
The theory has an additional $U(1)$ gauge symmetry,
which is anomalous. We show that a conflict between the
non-perturbatively generated superpotential and the
D-term of the anomalous $U(1)$ gives rise to
supersymmetry breaking in this theory. Classically, the D-branes giving
rise to the gauge theory are stuck at the orientifold plane. In the
supersymmetry breaking vacuum, some of these branes are repelled by the
orientifold and come to rest away from it.  In this discussion of
supersymmetry breaking, we neglect the gravitational interactions and keep
the dilaton and a relevant orientifold blow-up mode fixed.
We show that  supersymmetry breaking occurs for any fixed values of the
dilaton and orientifold blow-up mode. Once these modes are taken to be
dynamical, there are, as usual,  runaway directions along which supersymmetry
is restored. What happens when the relaxation of the closed
string modes and the gravitational interactions is included is an interesting
question which we leave  for the future.

\mysection{Gauge Coupling  Unification on D3 Branes}

In this section, we discuss the constraints imposed on string model
building by the requirement of gauge coupling unification
(taking the values $\alpha_{GUT}$ and $M_{GUT}$
for supersymmetric extensions of the standard model) and the
observed value of Newton's constant.
For the $E_8 \times E_8$ heterotic string, these requirements
lead to the conclusion that string theory must be strongly
coupled \cite{wittenstrongcy}.
In contrast, as has been noted earlier in \cite{wittenstrongcy}, we
will see that in the case of the Type I string theory these
requirements can be met while still
working at weak string coupling.
Moreover,  the discussion below suggests that
in several Type I models
the six   compactified dimensions can have a  length somewhat bigger
than the inverse GUT scale. In these cases, the gauge group and charged
matter   would arise  from fields  living on D3-branes that fill the $3+1$
dimensional  flat spacetime.

The relation between the string scale $\alpha^{\prime}$,  Type I string
coupling $g_I$,
volume of compactification  $V_6$, gauge coupling at unification
$\alpha_{GUT}= g^2 / 4 \pi$, and Newton's constant $G_N$
is given by \cite{sen}:
\beq
\label{Newton}
G_N~=~{(2 \pi)^7 \over 16 \pi}~{\alpha^{\prime 4} \over V_6} ~g_I^2~,
\eeq
and
\beq
\label{alphagut}
\alpha_{GUT}~=~{(2 \pi)^7 \over 4 \pi}~ {\alpha^{\prime 3} \over V_6} ~g_I~ .
\eeq
Here we will consider the situation where the six compactified dimensions
have approximately the same size $R$.
The volume $V_6$ is then roughly given by
\beq
\label{defv}
V_6~=~( 2  \pi  R)^6
\eeq
To proceed,
we need to decide how to relate the unification scale
$M_{GUT} \simeq 10^{16}$ GeV to $\alpha^{\prime}$ and $R$.
In several string models the  gauge
couplings unify even in the absence of a grand unified group.
 In these cases, one
expects the  grand unification scale to correspond to the masses
of the lightest extra
charged states present in the string theory.
These extra states can be of two
kinds: Kaluza-Klein modes with a mass of order $1/R$, or higher string
modes with a mass of order  $1/\sqrt{\alpha^{\prime}}$.
If we assume that $R > \sqrt{\alpha^{\prime}}$, the lightest extra states
have a mass $m \sim 1 / R$, leading to the
relation $R \sim 1/ M_{GUT}$.
{}From eqs.~(\ref{Newton}), (\ref{alphagut}), and  (\ref{defv}) it then follows that:
\beq
\label{wronga}
{\alpha^{\prime} \over R^2} ~=~ {1 \over \sqrt 2} ~\alpha_{GUT}~ R~ M_{Pl},
\eeq
and
\beq
\label{gI}
g_I~=~4 \sqrt{2} ~{1 \over \alpha_{GUT}^2 ~R^3~ M_{Pl}^3 }.
\eeq
With the values $\alpha_{GUT} = 0.04$ and
$R \sim 1/ M_{GUT} = (10^{16} {\rm GeV})^{-1}$
for the supersymmetric standard model \cite{mssmgut},
and
$M_{Pl} = G_N^{-1/2} =
1.2 \times 10^{19}$ GeV, we get from eq.~(\ref{gI}) that :
\beq
\label{valgI}
g_I ~ \sim ~ 10^{-6}~ .
\eeq
Thus, the gauge coupling is small, as  mentioned above.
However from  eq.~(\ref{wronga})  we find  that:
\beq
\label{watwo}
{\alpha^{\prime} \over R^2} \sim 34~ .
\eeq
This  shows that our starting  assumption ($R > \sqrt{\alpha^\prime}$)
about the  lightest  extra charged states coming
from Kaluza-Klein modes is incorrect.
A consistent solution is obtained by assuming that
$R < \sqrt{\alpha^\prime}$.
The  lightest  extra  states  which
enter at  the GUT scale are  then
higher string modes with  mass $M \sim  1/\sqrt{\alpha^{\prime}}$.
In this case,  the more  appropriate geometrical picture is
obtained by  T-dualizing along the six compactified directions.
Doing so turns the D9 branes into
D3 branes. The T-dual radius and string coupling are given by:
\beq
\label{dualr}
{\tilde R} ~= ~{\alpha^{\prime} \over R}
\eeq
and
\beq
\label{dualg}
{\tilde g_I}~ =~ \left( {\alpha^{\prime} \over R^2} \right)^3 ~g_I~.
\eeq
Eq.~(\ref{alphagut}) then  implies  directly that
\beq
\label{valdualg}
{\tilde g_I} ~=~ 2 ~\alpha_{GUT}  \simeq  0.08~.
\eeq
Furthermore, since the lightest excitations are higher string modes we now
set $\alpha^{\prime} \sim (M_{GUT})^{-2}$.
Eq.~(\ref{Newton})  then  gives
\beq
\label{finalR}
{\tilde R} ~=~ {\sqrt{\alpha^\prime}}~ \left( {1 \over 8} ~
M_{Pl}^2 ~ \alpha^{\prime} ~{\tilde g_I}^2 \right)^{1 \over 6}
\sim 3  M_{GUT}^{-1}~.
\eeq

It is useful to describe the resulting picture in words. The gauge group
and charged matter arises from D3-branes.
The six compactified dimensions
have a length scale  somewhat bigger than the
inverse GUT scale.  In particular, we note that
since all the degrees of freedom
charged under the  gauge groups arise from open
strings that end on the three-branes,
there are no momentum modes with mass of
order $1/{\tilde R}$ charged under the gauge group.
Instead, there are
winding  modes with a mass of order ${\tilde R} /\alpha^{\prime}$ but these
are  somewhat heavier than the higher string modes with mass
$\sim 1/ \sqrt{\alpha^\prime}$.

We should emphasize that the  above  picture  is meant to be suggestive.
Whether it applies or not will depend on the details
of the compactification.
 It was noted in \cite{wittenstrongcy} that  in the $E_8 \times E_8$
theory  the large gauge coupling implies an extra dimension  at a scale
somewhat below the GUT scale (for recent work and a list of references,
see \cite{nilles}).
Here it is interesting to note  that  the
presence of extra large dimensions might
be true  in the Type I case as well, and more generally,
  in attempts to build string models involving branes.
We should also  note that the  conclusion
with regards to the smallness of the string
coupling is secure regardless of the
exact relation between $R$
and $\sqrt{\alpha^{\prime}}$.\footnote{For example, setting
$\sqrt{\alpha^{\prime}} = R$ in eq.~(\ref{alphagut}) still gives  (\ref{valdualg})
for the gauge coupling.}

We end this section with a few comments.
First, strictly speaking, the discussion above applies
to models where   gauge
coupling unification occurs in the absence of a grand unified group.
One can ask what happens  if the  low-energy  field theory  is  a grand unified
theory.  In this case the  lightest extra string states need not occur at the
GUT scale but could have  larger masses.  Eq.~(\ref{Newton}) then shows
that  in these cases the compactification scale $\tilde R$
should be comparable  to $\alpha^{\prime}$.
Even so, as  we see in the next section it might be
sometimes convenient to analyze  such a  model  in terms of  D3-branes.
Second, in our analysis we have taken all the compact dimensions to have
roughly the same size. This of course need not be true. For  recent
discussions of large extra dimensions and weak scale strings see \cite{dim}.
Finally, we have assumed that the gauge group and matter content arises
from the perturbative sector of Type I theory. This, too, need not be true.
One could have a situation where some of the degrees of freedom arise from
$9$-branes while others arise from $5$-branes; for
model building along these lines see \cite{AFIZ}, \cite{kaku}, \cite{kakushadze}.

\mysection{A Three Generation Model on D3 Branes}

In this section, we present a simple ``three generation" model with D3 branes
placed at a $\relax{\rm T}^6/\IZ_3$ orientifold.
The model is, admittedly, not a realistic one, but it will serve
the purpose of making
several generic points quite explicit.
In general, the moduli space of the gauge theory which governs the low-energy
 dynamics can be quite
complicated with several disconnected sectors. The D3-brane picture allows for
a geometric description of these different branches of moduli space
\cite{wittentori}.
The different branches correspond to the distinct ways in which the branes
can be placed at the various orientifold fixed points.

The additional branches of  moduli space
can have multiple uses. We will see below that they exhibit interesting
 patterns of gauge symmetry breaking that are not otherwise  possible.
 In addition, branes placed at
different  orbifold fixed points can serve as
``visible" and ``hidden" sectors; the latter can
be responsible for supersymmetry breaking. The lightest excitations of
strings stretching between branes at different fixed points transform as
fundamentals under both the ``hidden" and ``visible" gauge groups; these
could be instrumental in communicating supersymmetry breaking.

\smallskip

\mysubsection{The $\relax{\rm T}^6/\IZ_3$ orientifold}

We now turn to studying the Type I theory  compactified on  a $\relax{\rm T}^6/\IZ_3$
orientifold.  This theory has been analyzed  by \cite{sagnotti96}
and more recently
by \cite{kakushadze}, where the low energy dynamics
was shown to correspond to an
$SU(12) \times SO(8)$  gauge theory.
Our main purpose here will be to study  some
sectors  of moduli space which are disconnected from the  $SU(12) \times
SO(8)$ theory mentioned above.  For this purpose it will be often convenient
to  T-dualize the Type I theory along the six directions of $\relax{\rm T}^6$. Doing so turns
the 9-branes into  3-branes. The disconnected sectors  then correspond
to placing
an odd number  of 3-branes at  the orientifold  fixed planes and can be easily
visualized.   We show below how  an $SU(5)$ theory with three generations
of matter fields in the  ${\bf 10}$ and $\bar {\bf 5}$
representations can be obtained in this
manner.

Let us describe the  $\relax{\rm T}^6/\IZ_3$ orientifold  in more detail.  We work for the most
part  in the   T-dual description involving  D3 branes.  The D3 branes
stretch along
$X^{1,2,3}$.
We introduce complex coordinates,   $z_1 = X^4 + i X^5, z_2 = X^6 + i X^7,
z_3 = X^8 + i X^9$,  in the  compactified  six-dimensional space.
Consider the two-torus obtained  by identifying points under
\beq
\label{twotorus}
z~ \simeq  ~z ~+ ~ R
  ~\simeq ~z ~+~ R ~ e^{i 2  \pi \over 3}~.
\eeq
The $\relax{\rm T}^6$ is obtained by taking three copies of  this two torus, corresponding
to the three complex  coordinates  $z_1, z_2, z_3$.
The orientifold group is  given by:
\beq
\label{orgroup}
G~=~\{~1, ~\alpha, ~\alpha^2 ,~ \Omega  R  (-1)^{F_L} ,
~ \Omega R (-1)^{F_L} \alpha,~
 \Omega R (-1)^{F_L} \alpha^2 ~\}~.
\eeq
Here,  $\alpha$ is a spacetime symmetry  whose action is given by:
\beq
\label{orbifold}
( z_1, ~ z_2, ~z_3 ) ~\rightarrow ( \alpha  z_1, ~\alpha  z_2, ~\alpha  z_3 )~.
\eeq
$\Omega$ denotes world-sheet orientation reversal, and $R$ is a reflection
$z_i \rightarrow - z_i, i = 1, 2, 3$.
$F_L$ is an operator that flips the sign of the left-moving Ramond states.
The orientifold group G  has a $\IZ_2$ subgroup
\beq
\label{gz2}
G_{\IZ_2}~=~ \{1 , \Omega  R  (-1)^{F_L} \}
\eeq
and a $\IZ_3$ subgroup
\beq
\label{gz3}
G_{\IZ_3}~=~\{1, \alpha, \alpha^2 \}~.
\eeq
These will play a useful role in the subsequent discussion.

In addition to acting on the spacetime indices,
the orientifold group acts on the
Chan-Paton indices of the open string states stretching between
 D3 branes \cite{tasibranes}.
The action of the group elements,
$\Omega R (-1)^{F_L}$ and $\alpha$,  on the  Chan-Paton
factors $\lambda$ is:
\beq
\label{eomega}
\lambda \rightarrow \gamma_{\Omega R (-1)^{F_L}} ~ \lambda^T ~
\gamma_{\Omega R (-1)^{F_L}}^{-1}~,
\eeq
and
\beq
\label{egamma} \lambda \rightarrow \gamma_\alpha ~ \lambda ~
\gamma_\alpha^{-1}.
\eeq
The matrices   $\gamma_\alpha $ and  $\gamma_{\Omega R (-1)^{F_L}}$ must
furnish a
representation of the orientifold group.
The matrices  $\gamma_{\Omega R (-1)^{F_L}}$,
representing the action of the $\IZ_2$ part of the orientifold group
should obey \cite{tasibranes}:
\beq
\label{gammaomega}
\gamma_{\Omega R (-1)^{F_L}}\ =
\left( \gamma_{\Omega R (-1)^{F_L}} \right)^{T}~.
\eeq
In the absence of the $\IZ_3$ orbifold projection,
 the $\Omega  R (-1)^{F_L}$ projection  would lead to
 an $SO$ gauge group on the D3 brane world volume.

Tadpole cancellation conditions  play an important role in ensuring
the consistency of  the string compactification.  For the $\relax{\rm T}^6/\IZ_3$ orientifold
these were discussed in \cite{sagnotti96}.
For the sake of brevity we will not discuss
a detailed derivation of these conditions here.  Instead we will content ourselves
with stating them; as the reader will see  these  conditions
 give rise to anomaly free gauge theories.

As expected,   the untwisted   Ramond-Ramond 4-form
charge conservation conditions require
the presence of 32 D3 branes to cancel the orientifold charge.
In addition, there
are   charge cancellation conditions for the
twisted RR fields.
Before stating these, it is useful to consider the  action of the
$G_{\IZ_3}$ and $G_{\IZ_2}$ subgroups of the orientifold group on the
$\relax{\rm T}^6$.   Consider first a two torus shown in Fig.~1.

\begin{figure}[ht]
\vspace*{13pt}
\centerline{\psfig{file=torus.ps}}
\vspace*{13pt}
{\centerline{\small Figure 1}}
\end{figure}

The  origin, denoted by the triangle in Fig.~1, is  the only fixed point
with  respect to the full $\IZ_6$ orientifold group.
In addition the $G_{\IZ_3}$
subgroup has two additional fixed points, at
$z = {1\over \sqrt{3}} e^{i \pi\over 6}$
 and $z = {1\over \sqrt{3}} e^{i \pi\over 2}$, respectively. These two
fixed points are interchanged under the
action of the $G_{\IZ_2}$ subgroup and are  denoted by
circles in Fig.~1.
Similarly, the $G_{\IZ_2}$ subgroup  has three additional fixed points
at $z = {1\over 2}, {1\over 2} e^{i 2\pi\over 6}, {1\over 2} e^{i 2\pi\over 3}$,
which are denoted by squares.
These  are images of each other under the  $G_{\IZ_3}$ symmetry.
The
fixed points for the $\relax{\rm T}^6$ can  now be deduced in  a  straightforward manner.
There
is only one fixed point under the full $\IZ_6$ symmetry---the origin.
 In addition,  as in the
$\relax{\rm T}^2$ case there are fixed
points of the $G_{\IZ_3}$ symmetry which are transformed
into one another by  the $G_{\IZ_2}$ and vice-versa; these
will be referred to, in what follows, as
$G_{\IZ_3}$ and $G_{\IZ_2}$ fixed points, respectively.

We can now return to the tadpole conditions for the
twisted Ramond-Ramond fields.  If $\gamma_\alpha$ is the matrix
which  represents the action of $G_{\IZ_3}$ on the branes at the origin, one
finds \cite{sagnotti96} that:
 \beq
\label{gammaalpha}
{\rm Tr} ~ \gamma_{\alpha} ~=~ -4 ~.
\eeq
In contrast, at a $G_{\IZ_3}$ fixed point one finds that
\beq
\label{gammaz3}
{\rm Tr} ~ \gamma_{\alpha} ~=~ 0. ~
\eeq
Finally, at each $G_{\IZ_2}$ fixed point  (and consistently at its $\IZ_3$ image
points) one can choose to  place  an even or odd number of branes.

The simplest way to meet these conditions is to place all  the $32$ D3 branes
at the origin. This gives rise, from eq.~(\ref{gammaalpha}),
 to a gauge theory
with $SU(12) \times SO(8)$
 gauge group with three generations of
$(\overline{\Yfund}, \Yfund) + (\Yasymm, {\bf 1})$
 fields
which was discussed in \cite{sagnotti96}.

The rank of the $SU(12) \times SO(8)$  gauge symmetry can be reduced by
moving some of the branes away from the origin in a continuous manner.  To be
consistent with the  $\IZ_6$ orientifold symmetry,
however,  these branes can only be
moved  away from the  origin in sets of six. From  eq.~(\ref{gammaalpha}) it then
follows that the  rank of the $SU(N)$ factor  must
always be odd; this  precludes an $SU(5)$
gauge symmetry which is attractive from  a phenomenological point of view.

\smallskip

\mysubsection{The  $SU(5)$ theory}

We turn now  to exploring some branches of moduli space,
which are disconnected
from  the  $SU(12) \times SO(8)$ theory mentioned above.
We will see  how  some of these branches give rise
to an $SU(5)$ gauge theory with three generations of  fields in
the ${\bf 10}$ and $\bar{\bf 5}$
representations.   We discuss examples of such  disconnected  branches below,
but   before doing so it is worth summarizing the  essential
features responsible for the grand unified theory.

In some sectors of moduli
space, an odd number of branes can be removed from the origin.
In particular, a situation can arise where only  $11$  of the $32$ D3 branes  are
left at the origin.    Eq.~(\ref{gammaalpha})
then  implies that  one has   an  $SU(5)$ gauge symmetry.
 In addition there  is an anomalous $U(1)$ symmetry  (which is
broken at the string scale).
The theory  has $N=1$ supersymmetry  with  matter
content corresponding to three generations of matter fields which transform as:
\begin{equation} \label{su5}
\begin{array}{c|c|c}
        &SU(5)&U(1)\\ \hline
A_{i = 1,2,3} & \Yasymm & 2 \\ \hline
 \bar{Q}_{i = 1,2,3} & \overline{\Yfund} & - 1\\
\end{array} ~.
\end{equation}
The theory has a renormalizable tree-level superpotential given by
\beq
\label{superpots}
W_{tree} = \epsilon^{ijk} A_i \bar{Q}_j \bar{Q}_k.
\eeq
The three generations arise  because the  $\IZ_3$
action in eq.~(\ref{orbifold}) does not distinguish between the three
(complex) transverse coordinates,  thus one set of matter fields
in eq.~(\ref{su5}) arise from each of them.

We now turn to discussing how these disconnected branches arise.
As we saw in the discussion above,
if we start with all $32$ branes at the origin
and move  some away in a continuous manner,
one is always left with an even
number  of branes at the origin.
Thus to   get $11$ branes some of them must
be placed at  fixed points of the  $G_{\IZ_3}$  of $G_{\IZ_2}$ subgroups,
eqs.~(\ref{gz2}), (\ref{gz3}).  Now eq.~(\ref{gammaz3})
implies that  the number of branes at
  a $G_{\IZ_3}$ fixed point must be  a  multiple of three.
 In addition, as we saw above, each  $G_{\IZ_3}$ fixed point has a $G_{\IZ_2}$
image.
Thus,  one finds that  all the branes at a  $G_{\IZ_3}$ fixed point can be moved
continuously  away in a $\IZ_6$ symmetric manner back to the origin.
 Disconnected
branches of moduli  space  can be  obtained, however,
 by placing an odd number  of branes at a $G_{\IZ_2}$
 fixed point (and its two images under $G_{\IZ_3}$).   Since
there are a large  number of  $G_{\IZ_2}$ fixed points
 in $\relax{\rm T}^6$ this gives rise to  a
large  number of possibilities.
 We will not analyze all of them in detail here.  Rather, as an illustrative
example, we focus on  a case   that gives rise to  the
$SU(5)$ ``grand unified theory" mentioned above.

For this purpose, the simplest possibility is to consider a situation
where all the branes
are  at the origin  as far as the third  $\relax{\rm T}^2$ (corresponding to the
$z_3$ coordinate) is concerned, but not as far as the
other two tori are concerned.  Consider
placing   one D3 brane at  a $G_{\IZ_2}$ fixed point in the first $\relax{\rm T}^2$ and
at  a $G_{\IZ_2}$ fixed point in the second $\relax{\rm T}^2$---this
brane has two images under the $\IZ_3$ symmetry.
Next, place   one D3 brane at the  origin  of the first two-torus,
but at a  $G_{\IZ_2}$ fixed point of the second $\relax{\rm T}^2$---this
brane has two images as well.
Finally,  place  a  D3 brane  at a $G_{\IZ_2}$ fixed point of the first torus,
but at the origin  of the second.
This brane has  two images as well.
Altogether,  counting images, this gives us
 $9$ D3 branes---an odd number---which
are stuck to  orientifold planes  away from the origin (note,
however, that the number of D3 branes at the
$G_{\IZ_2}$ fixed points in each of the three two-tori is even; this is
required by the consistency conditions discussed
 in the following section).
The remaining $23$ branes at the origin
 give rise to an  $SU(9) \times SO(5)$ gauge symmetry.
The matter fields transform as  three generations of
 $(\Yasymm, {\bf 1})$ and
$(\overline{\Yfund}, \Yfund)$, under  the $SU(9) \times SO(5)$ gauge
group.\footnote{
We are using a notation where the vector of  $SO(N)$ is denoted by $\Yfund$.}

Finally,  one can  move $12$---two sets of six---of the remaining branes
away from the origin, leaving behind $11$ branes, obtaining thus
the $SU(5)$ theory mentioned  above, eq.~(\ref{su5}).
We also  should mention that for  a generic
position of  these $12$ branes,
the full gauge symmetry includes an additional $U(1) \times U(1)$ factor.
This  ``hidden sector"  gauge group can be further  enhanced
if the branes are placed at  $G_{\IZ_3}$ fixed points or at the $G_{\IZ_2}$
orientifold planes.
For example, placing  six of the $12$ branes at a  $G_{Z_3}$ fixed
point (and the
remaining  six at the image point) gives rise to  an $N=1$ theory with
$SU(2)^3$ symmetry and  three sets of chiral matter  transforming  as
bifundamentals under pairs of the  $SU(2)$'s.    Dividing the $12$
branes between  a $G_{\IZ_2}$  orientifold plane and its two images,
on the other hand, can give rise to  a theory with $N=4$ supersymmetry and
an $SO(4)$ or  $SO(5)$ gauge symmetry  (the $SO(5)$ symmetry can
arise if the orientifold planes chosen already contain a D3 brane stuck
to them, as mentioned above in the discussion of the  disconnected
moduli space).

So far, we have ignored the effects of open strings stretched between
branes at different fixed points. The lightest excitations of such
strings are massive states which transform as fundamental-antifundamental
under the respective world volume gauge groups. In the example we gave above,
there can be two world volume theories with $N=1$ supersymmetry.
If supersymmetry were dynamically broken in one of these theories,
supersymmetry breaking would be communicated to the other gauge
theory via the massive chiral multiplets just described (and, of course, by the
supergravity in the bulk). A more precise investigation of this
 would probably  involve  details of the
supersymmetry breaking dynamics and the stabilization of the  dilaton
\cite{dine}; we leave this for future investigation.

\smallskip

\mysubsection{Non-perturbative consistency conditions}

There is one subtlety concerning disconnected sectors of moduli space
that needs to be mentioned. Sometimes such sectors are not allowed,
even when they pass all the perturbative consistency conditions,
due to non-perturbative reasons.
Similar issues were addressed in \cite{Star}, \cite{wittentori}.
We will not be able to discuss this matter in full detail here,
but  will mention some salient points.
The basic idea behind the non-perturbative
consistency conditions is as follows.
 The Type I  theory  does not have any  perturbative states which  transform
in spinor  representations  of $SO(32)$.  However, such states
are present in the dual heterotic $SO(32)$ theory and
are nonperturbative in the Type I theory.
Allowing for such spinor representations imposes additional
consistency conditions---whose origin from the Type I viewpoint is
non-perturbative.

We can  verify that the example discussed above,
giving rise to  the $SU(5)$ model,
meets  various   non-perturbative conditions. However, we should caution
the reader  that there might be  other conditions, besides the ones  we have
checked that might not be met.\footnote{By way of comparison,  we note that
the conditions  we have tested are  the analogue of
 those discussed in Section 2.3
of \cite{Star}.  The authors also formulated a stronger set of conditions in
Section  5 of ref.~\cite{Star}.  We have not  investigated the
presence of such stronger consistency conditions in  the present case.}
It is in fact useful to carry out this discussion
in the original description of  Type-I theory in terms of $9$ branes.
The essential feature giving rise to the disconnected branch of
moduli space in the example above
was the fact that there were 9 D3 branes (counting images)
which were ``stuck" at the orientifold plane.
 In the  T-dual 9-brane language we are using now, the positions of
branes correspond to expectation values  for particular Wilson lines.
The question is whether the Wilson lines' expectation values
are consistent with the existence of states that transform as
$SO(32)$ spinors---the holonomies around
any contractable loop should be trivial in the appropriate  spinor
 representation.

The example discussed in Section 3.2  is
equivalent to turning on four Wilson lines
along the noncontractible loops of two of the two-tori.
It turns out in this case that the holonomy around any  contractible path
is trivial in all representations of $SO(32)$. One can show this by explicitly
writing down the Wilson lines that correspond to the brane configuration
with $9$ D3 branes removed from the origin, which was described in Section 3.2.
Moreover,  in this  example,  an  explicit periodic flat connection,
which is not constant on the  $z_1, z_2$  four-torus,
can be found. This can be done by a straightforward
generalization of  the construction of  ref.~\cite{rosly}
to the case of $\relax{\rm T}^4$.  Furthermore, as in \cite{Star}, one can
show that if $W$ is the  Wilson line  relevant for the
particular $\IZ_3$ fixed point  then
$(W \gamma_{\alpha})^3 =1$ in spinor representations as well.

Before moving  on, let us mention that the example of Section 3.1,  giving
rise to $11$ branes at the origin,
is  just one of several possibilities  consistent with the various conditions.
For example, one can easily
work out brane configurations with an odd total
 number of branes removed from the
origin that involve moving branes to the $G_{\IZ_2}$ fixed points in all three
two-tori.

\smallskip

\mysubsection{The GUT: shortcomings}

It is useful to describe the  construction of  the
$SU(5)$ theory in  group theoretic terms perhaps
more familiar to  some model builders.
The $SO(32)$ gauge symmetry is broken to  an
$SO(11)$ subgroup (times a hidden sector group).  The
orientifold projection then  further  breaks
the symmetry to $SU(5)$  (with an additional  anomalous U(1)).
 The $\bar {\bf 5}$ and ${\bf 10}$ matter fields arise
from  the adjoint representation of $SO(11)$ by   the orientifold  projection.
The three generations arise because there are
  three complex  (six real) transverse
dimensions and because the  orientifold group acts in an identical manner on the
three directions.

The relative ease with which  a realistic gauge group and matter content can
be obtained in the Type I theory is interesting. We should note,
 that even though we used  a D3 brane description
to simplify the  discussion, the construction
as such was purely in the context of  perturbative Type I string theory.
In particular, the  matter content was obtained,
  even  though we did not have any
spinor representations of $SO(32)$ to begin with---in fact all the matter fields
can be thought of as being obtained  by
truncating adjoint representations of  $SO(32)$.

However, it should  also be noted  that the model is  meant as an illustrative
example  and  is not realistic.
There are  several reasons for this. First,  there are no Higgs fields
either  to break the $SU(5)$ gauge symmetry or to
give rise to  the $SU(2)$ Higgs doublets of the
supersymmetric  standard model.
Second, and this is perhaps a more
important  limitation,
as was mentioned in passing in eq.~(\ref{superpots}) above there
is a Yukawa coupling in the theory that violates  baryon and lepton number.
The underlying reason for this coupling is that in the $N=4$ theory, which can
be thought of as the starting point for the above construction,
there is  a coupling
involving the three adjoint fields corresponding to the transverse directions.
In the case of, say, spinor representations of $SO(10)$,
a trilinear ${\bf 16}^3$ coupling is not allowed by gauge invariance and
an R-parity symmetry can often be imposed to prevent baryon- and
lepton-number violating terms.
However, in the present example, where all the matter arises
from adjoint representations no such R parity symmetry is present.
This limitation is likely to be  quite  general.

\mysection{Supersymmetry Breaking}

We turn now to another feature of the $SU(5)\times U(1)$
theory discussed above.
  As we will see below, in the world-volume field theory context, a conflict
 between  the
non-perturbatively  generated  superpotential and
the anomalous $U(1)$ D-term
results  in   the  breaking of supersymmetry in this theory.
  Our discussion of supersymmetry  breaking will  only involve
the open string sector
corresponding to the  world volume theory on the D3 branes.
 Gravity and other closed string  effects will be neglected.  In particular,
 the dilaton and  the orientifold  blow-up mode \cite{dm}, which acts
as the Fayet-Illiopoulos
term for the $U(1)$,  are regarded as coupling constants,
 and  will be  kept fixed in the discussion below.
We will establish that supersymmetry breaking occurs
for  any finite value of these couplings.\footnote{ More accurately,
supersymmetry breaking will
be shown when the string coupling is small enough to
argue with confidence that the low-energy dynamics is governed by
the $SU(5) \times U(1)$ theory.}
But, as is usually the case, once  they are
allowed to vary, we find that  there are runaway directions along which
supersymmetry is restored.  Perhaps, these could
 be stabilized by (yet poorly understood) nonperturbative
corrections to the K\" ahler potential (see, e.g. the recent discussion in
\cite{dine} and references therein).
The stability of the supersymmetry breaking ground state in the context
of the full theory is a complicated issue, about which we have nothing to
say here.

Towards the end of this section, we will briefly
comment on this runaway behavior and the gravitational back reaction.
One final comment before we get started: the discussion
in this section
only involves the branes  at  the  $\IZ_6$ fixed point,
 any details of compactification etc. are irrelevant  in this context. For
example,   the
analysis here applies to the world volume theory of 11 D3 branes
on the noncompact ${\relax\hbox{$\inbar\kern-.3em{\rm C}$}}^3/\IZ_3$ orientifold as well.

\smallskip

\mysubsection{Supersymmetry breaking in the $U(5)$ theory}

Our strategy to
establish supersymmetry breaking is as follows.  We first neglect
the anomalous $U(1)$ and  show that in the $SU(5)$ theory  the resulting
non-perturbative superpotential gives rise to runaway behavior.   Then on
incorporating the anomalous $U(1)$ we find that its D-term gives rise
to an energy that grows along the runaway
directions. This leads  to supersymmetry breaking.

The
$SU(5)$ ``three generation" model is an s-confining theory \cite{CSS}.
The infrared  degrees of freedom are the mesons and baryons
\beqa
\label{mesons}
C ~=~& A \cdot \bar{Q} \cdot \bar{Q} &~\sim~ ( {\bf 3},~ {\bf \bar{3}} ~,  0)~,
\nonumber \\
B ~=~&A^5 &~\sim~ ({\bf 6}, ~{\bf 1} ~, 10)~, \\
 M ~=~&A^3 \cdot \bar{Q} &~ \sim~ ({\bf 8}, ~{\bf 3}  ~,  5)~, \nonumber
\eeqa
where we have shown their transformation properties under the global
$SU(3)_A \times SU(3)_{\bar{Q}}$ symmetry and the last column in each entry
refers to the charges under the anomalous $U(1)$ which follow from
eq.~(\ref{su5}).
The confining  superpotential is \cite{CSS}:
\beq
\label{wconfining}
W ~=~ { C_a^\alpha ~ B^{\beta \gamma} ~ M^{\delta a}_{\gamma}
~\epsilon_{\alpha \beta \delta} ~+ ~ M^{\alpha a}_{\beta}~M^{\beta b}_\gamma
~M^{\gamma c}_\alpha ~\epsilon_{a b c} \over \Lambda^9} ~+
~\lambda ~ \delta^a_\alpha
~C_a^\alpha ~,
\eeq
where $a, b, ... (\alpha, \beta, ...)$
denote indices under the $SU(3)_{{\bar{Q}} (A)}$ symmetry,
respectively, and the last term is the tree-level superpotential.
The tree-level
superpotential breaks the global symmetry to the diagonal $SU(3)_{diag}$.
It lifts  all the  $C$ flat directions, but  does not lift  the $B$ and
some of the  $M$   directions.  The
superpotential coupling $\lambda$ in (\ref{wconfining}) is proportional to the
value of the gauge  coupling at the string scale (since the tree-level
superpotential is the projection of the $N=4$ superpotential).

We will show now that the F-term equations of motion following from
(\ref{wconfining}) have no solutions for finite field expectation values.
Consider the equations
of motion following from the superpotential (\ref{wconfining}) (suppressing
numerical constants):
\beq
\label{eomC}
M_\alpha^{\beta a} ~B^{\alpha \lambda}~\epsilon_{\beta \lambda \delta}~=~
\delta^a_\delta ~,
\eeq
\beq
\label{eomM}
\epsilon_{a b c}~M^{\beta b}_\alpha ~M^{\gamma c}_\beta ~+~\epsilon_{\alpha
\beta \delta} ~C_a^\beta ~B^{\delta \gamma} ~= ~0~,
\eeq
\beq
\label{eomB}
\epsilon_{\gamma \alpha \beta}~C_a^\alpha ~M_\delta^{\beta a} ~= ~0~.
\eeq
Multiplying the first equation (\ref{eomC}) by
$\epsilon^{\delta \mu \nu} M_\nu^{\gamma d} \epsilon_{bad}$,
summing over $\delta$ and $a$, and substituting for
$\epsilon_{a b c} M^{\beta b}_\alpha M^{\gamma c}_\beta $ from (\ref{eomM}),
we obtain:
\beq
\label{eq1}
-~\epsilon_{\alpha \delta \beta}
~B^{\alpha \mu}~ B^{\delta \gamma} ~C_b^\beta ~
+ ~ \epsilon_{bad} ~M^{\mu a} ~B^{\alpha \nu}~M_\nu^{\gamma d} ~ = ~
M_b^{\gamma \mu} ~-~\delta_b^\mu ~M_\nu^{\gamma \nu}~.
\eeq
Now under $SU(3)_{diag}$, the field $M_\beta^{\alpha a}$ decomposes as a
${\bf 3}$, given by the partial trace  $M^{\alpha a}_a$, a ${\bf \bar{6}}$
which is antisymmetric in the upper two indices,
and a ${\bf 15}$ which is  symmetric in the upper indices (and traceless).
Note that the l.h.s. of (\ref{eq1}) is antisymmetric in $\mu, \gamma$,
hence only
the r.h.s. contributes to the symmetric part of $M$. This gives rise to
the relation:
\beq
\label{symm}
M^{\gamma \mu}_b~+~M^{\mu \gamma}_b ~-
~\delta^{\mu}_b~ M^{\gamma \nu}_{\nu} ~
-~\delta^{\gamma}_b~ M^{\mu \nu}_{\nu}~=~0~,
\eeq
from which in turn it follows that $M^{\alpha a}_a$ vanishes as does the
symmetric part  of $M$. Thus the ${\bf 3}$ and ${\bf 15}$
components of $M$ are zero.  The remaining ${\bf \bar{6}}$ can be written as
\beq
\label{assym}
M^{\gamma \mu}_a~=~ \epsilon^{\gamma \mu \kappa} ~s_{\kappa a}~,
\eeq
where $s$ is symmetric in the two indices.
Substituting into eq.~(\ref{eomC}), and evaluating for $B\cdot s$ then leads
to the  relation:
\beq
\label{invert}
s ~= ~- {1 \over 2} ~B^{-1}~.
\eeq
Substituting eq.~(\ref{assym}) in (\ref{eomB})
 and noting that $s$ is invertible, eq.~(\ref{invert}) then leads to
$C_a^\alpha = \delta_a^\alpha {\rm Tr} ~C$. This  then implies that $C = 0$.
Finally, substituting into eq.~(\ref{eomM}), one similarly finds that
$s_{\alpha \beta}  = \delta_{\alpha \beta} ~{\rm Tr}~ s$. This leads to the
conclusion that
$s = 0$. But now we see from eq.~(\ref{invert}) that $B$ must go to infinity.
Thus we have established that there are no
solutions to the F flatness conditions at finite expectation values.

The equations (\ref{eomC})--(\ref{eomB}) do have runaway solutions.
The discussion above leads to the conclusion that along a runaway direction,
$C$ and $M \rightarrow 0$, while
$B \rightarrow \infty$ in an invertible manner---more precisely
$B^{-1} \rightarrow 0$.
For example, a  runaway vacuum solution
with $SU(3)_{diag} \rightarrow SO(3)$ global
symmetry  is $B^{\alpha \beta} \sim \delta^{\alpha \beta} b$,
$C^{\alpha a} \sim \delta^{\alpha a} b^{- 3}$,
$M^{\alpha a \beta} \sim \epsilon^{\alpha a
\beta} b^{-1}$, with $b \rightarrow \infty$.
 The physics along
the $B$, det $B \ne 0$ directions is easy to understand. Along these directions
the mesons $C$ and $M$
obtain mass. Upon integrating them out,
 the superpotential of the low-energy theory
is $W_{eff} = \lambda^3 \Lambda^{18}/{\rm det} B$,
showing explicitly the runaway behavior.

So far we have studied the non-perturbative superpotential. Now let us
include the anomalous $U(1)$ by ``turning it on" in the effective theory
of the mesons $C$, $M$, and $B$.
The last column in each row of eq.~(\ref{mesons})
gives the $U(1)$ charges of the three fields.
We see that $C$ has charge zero, while $B$
and $M$ both have positive charge.
We have argued above that along a runaway
direction $B$ must go to infinity (in an invertible manner) while $M$ and $C$
go to zero.
Since all components of $B$ have   positive charge with respect to the $U(1)$,
we see that along such a direction the $U(1)$ D-term contribution to the energy
blows up. Thus, the runaway behavior dictated by the non-perturbative
superpotential is in conflict with the $U(1)$ D-term potential, leading to the
breaking of supersymmetry (note that
this is similar to the mechanism of ref.~\cite{anomalousu1}).

\smallskip

\mysubsection{Remarks on supersymmetry breaking}

We end this discussion of supersymmetry breaking with a few remarks.
We first
remind the reader of an important feature of type I orbifolds:
the anomaly cancellation
of the ``anomalous" $U(1)$s is provided, as in the heterotic case, by a
Green-Schwarz mechanism. However, unlike the heterotic case,
the axion that shifts under the $U(1)$ to cancel the anomaly is a
model-dependent field---the twisted Ramond-Ramond field from the
closed string sector (this has been pointed out in \cite{dm} and recently
discussed in \cite{IRU}).  It is in the same
supermultiplet as the orbifold blow-up
mode (the twisted NS-NS field) and can be described in terms of a chiral
superfield, denoted hereafter by $C$, with a kinetic term
\begin{equation}
\label{RRlagr}
\int ~d^4 \theta ~\left( C ~ + ~ C^\dagger ~ + V \right)^2 ~ + ~ \ldots ~,
\end{equation}
where dots denote higher-order terms.
The leading
term (\ref{RRlagr}) can be written by demanding
$U(1)$ invariance and a smooth  kinetic term for
 $C$ in the orbifold limit $\langle C \rangle = 0$.
Here  $V$ is the anomalous $U(1)$ vector superfield; in addition to (\ref{RRlagr}),
the field $C$ also has a Wess-Zumino coupling to the gauge field strengths, of
the form $\int d^2 \theta C W^\alpha W_\alpha$ \cite{dm}.
In a superunitary gauge, the term (\ref{RRlagr}) represents a mass term
(of order the string scale)  for the anomalous $U(1)$ vector superfield. By
giving an expectation value to the real part of $C$ (blowing up the orbifold)
one can induce ``tree-level" FI terms, with
$\zeta_{FI}^2 \sim  \langle C + C^\dagger \rangle$, as follows from (\ref{RRlagr}).
That (\ref{RRlagr}) is correct follows from the computation of
ref.~\cite{dm} of  the coupling of the real part of $C$ (the twisted NS-NS field)
to the D-term of the vector superfield (and from a subsequent
supersymmetry transformation). This coupling
arises from the disk with two scalar vertex operators attached to the boundary,
and a closed string twisted NS-NS scalar vertex operator in the bulk \cite{dm},
and is of order ${\cal{O}}(g_{string}) \sim g^2$.

We note that the conclusion regarding supersymmetry breaking is
true  for any sign and finite value of the $U(1)$ Fayet-Illiopoulous term.
Depending on the sign of the FI term, the D-term potential can have a zero
at finite values of the fields.\footnote{Since the $U(1)$ is anomalous one
expects that the
FI term is renormalized at one loop. However, an explicit calculation
shows that the FI term is not generated in open string orbifolds, because
of cancellations between contributions of worldsheets of different topology
\cite{FItype1}.} However the
F-term potential vanishes only at infinity. Thus supersymmetry is
broken. It is possible
though  to have vanishing  D- and F-term potentials for
infinite (negative) value of the FI term.

One would like to find out where the resulting supersymmetry breaking
vacuum lies. Unfortunately, this is quite difficult---as the following
argument shows,
one expects the vacuum to lie in a strongly coupled region
where a semiclassical analysis is not applicable.
Assuming first that such an analysis is valid, upon
balancing the U(1) D-term energy with the
F-term energy one finds that the vacuum energy (with vanishing FI term)
 scales as  $g_1^{16/9} \lambda^{2/3} \Lambda^4$, while the typical
 expectation value of a field goes like
$v \sim \lambda^{1/6} g_1^{-1/18} \Lambda$.
Here $g_1$ is the $U(1)$ gauge coupling,
$\lambda$ is the tree-level Yukawa coupling, eq.~(\ref{wconfining}),
and $\Lambda$ is the strong coupling scale of the $SU(5)$ gauge theory.
For a semiclassical analysis to be valid, one requires $v \gg \Lambda$.
 If the gauge coupling $g_1$ and $\lambda$ were independent parameters,
 this could have been achieved by taking $g_1$ to zero
 keeping $\lambda$ fixed. However, in our case
$g_1 \sim \lambda$, thus for small $g_1$ the vacuum lies in the strongly
coupled region and the semi-classical analysis
 is not applicable. One could make $v \gg \Lambda$
by taking $g_1 \gg 1$, but then the string coupling would be
large, again making a semiclassical analysis invalid.

While we cannot determine the vacuum explicitly, we know that
some of the $B$, $M$, and $C$ fields must get vacuum expectation
values.
These expectation values should correspond to
displacing some of the D3 branes
away from the orientifold. We remind the reader that the
$SU(5)$ theory under consideration here is the
world volume theory for $11$ D3 branes placed at the
orientifold. Classically, the 11 D3 branes are all
stuck to the orientifold
plane and the configuration has no moduli.  This corresponds to the fact
that in order to
meet the tadpole conditions and respect
the $\IZ_6$ symmetry no branes
can be moved away from the orientifold point.
Quantum mechanically, due to non-perturbative
supersymmetry breaking effects we see that some of the branes are repelled
by the orientifold plane and come to rest away from
 it so as to minimize the
energy.
Since there are no moduli
the  configuration of D3 branes
cannot be described in terms of classical geometry.
The displacement of branes which are
classically stuck at the orientifold is somewhat reminiscent of the
splitting of orientifold 7-planes discussed in ref.~\cite{SSX}.

As was mentioned in the beginning of this section, the above analysis
neglected
all interactions with closed string sector modes. One might at first expect
that gravitational interactions are small at low-energies and so can be
neglected. But once supersymmetry is
broken and a (boundary) cosmological
constant is induced, this is not apriori true.
Also, interactions with some other closed
string modes, which determine the
couplings of the brane theory are important.
There are two modes of this
kind. A blow-up mode for the orientifold determines
 the FI term of
the $U(1)$ \cite{dm}; this mode (together with its partner) also enters in the
determination of the coupling constant (theta angle)
for the $SU(5)$ theory.
Similarly the dilaton determines the gauge coupling
of the $U(1)$ and together with the blow-up mode
mentioned above determines the $SU(5)$ gauge coupling.
As we have argued here supersymmetry
breaking occurs for any fixed values of
these couplings, but there are runaway directions
along which it can be restored. For example,
 as we saw above there is a direction along which the FI term can
go to infinity with appropriate sign.
Similarly, if the dilaton goes to infinity, supersymmetry is restored.
In fact, as has been argued recently in \cite{dine}, it is necessary
to include the dynamics of these closed string sector modes to get a complete
description of supersymmetry breaking.  Without this there is no goldstino,
which signals that the description of supersymmetry breaking is incomplete.
 It is interesting to ask how the  system  of D3-branes will evolve once
the dilaton and the blow-up mode are allowed to
relax and the gravitational
back reaction is put in. We leave this question for the future.

Finally, another natural  configuration to consider involves not $11$, but $8$
branes
placed at the $\IZ_6$ orientifold plane. This is the minimum number required to
meet the tadpole conditions (e.g. starting with $32$ branes and moving
$24=6 \times 4 $ away leaves us with $8$ branes). The corresponding
theory has an $SU(4)\times U(1)_A$ gauge symmetry and three generations of
fields which transform in the $\Yasymm $ representation of the group.
There is in addition an anomalous $U(1)$ under which each of the $\Yasymm$
fields has the same charge. In this case if the  FI term (i.e. the
orientifold blow-up parameter) vanishes,  supersymmetry is unbroken.
This is because, in contrast to the $SU(5)$ model, the $SU(4)$ theory has
a branch of moduli space where no dynamical superpotential is
generated---this can be inferred from
 \cite{IS} by noting that the $SU(4)$ theory
with three ${\bf 6}$'s  is
equivalent to the $SO(6)$ theory with three vectors.
The breaking of
supersymmetry is then purely due to the $U(1)$ D-term and vanishes for
vanishing FI parameter.

\mysection{Acknowledgments}

We would like to thank Ken Intriligator and  Witek Skiba for discussions.
E.P. was supported by DOE contract no. DOE-FG03-97ER40506. The research
of JL and ST is supported by the Fermi National Accelerator Laboratory,
which is operated by the Universities research Association, Inc., under
contract no. DE-AC02-76CHO3000.

\newcommand{\ib}[3]{ {\em ibid. }{\bf #1} (19#2) #3}
\newcommand{\np}[3]{ {\em Nucl.\ Phys. }{\bf #1} (19#2) #3}
\newcommand{\pl}[3]{ {\em Phys.\ Lett. }{\bf B#1} (19#2) #3}
\newcommand{\pr}[3]{ {\em Phys.\ Rev. }{\bf D#1} (19#2) #3}
\newcommand{\prep}[3]{ {\em Phys.\ Rep. }{\bf #1} (19#2) #3}
\newcommand{\prl}[3]{ {\em Phys.\ Rev.\ Lett. }{\bf #1} (19#2) #3}

redpajama_set_name: RedPajamaArXiv


text:
Courtesy of BBM Facebook
PH 'open for business'
account_balanceMetro News account_circleSarah Lim chat_bubble_outline0 Comments visibility
President Ferdinand R. Marcos Jr. successfully showed to the World Economic Forum (WEF) in Davos, Switzerland that the Philippines is "open for business" and can be a premier investment hub, according to a report by Philippine News Agency.
In a statement on Saturday, House of Representatives Speaker Martin Romualdez said the Philippine delegation "actually impressed the Davos attendees."
"(They) demonstrated how the Philippines, the President, his official family, the economic managers, alongside the top businessmen and women, showed a united front to the entire world. It really impressed, and it was not lost in Davos," Romualdez said.
"We are open for business, we are here listening and we are inviting everyone to see why the Philippines would be the best destination to invest."
Multinational companies, such as investment giants Morgan Stanley and BlackRock, are already eyeing setting up shop in the country, not only because of the ideal investment climate but also due to the competencies of Filipino workers, he said.
"The list goes on. It's just an amazing experience to have the Philippines in Davos. It is an event in itself," he added. "It goes to show you that the Philippines, being the star of this year, especially in this Davos conference, will definitely show the world that not only are we open for business but … under the leadership of President Marcos, leading the Filipinos, the best workers in the world are here."
The Leyte lawmaker also noted that the Maharlika Investment Fund (MIF) generated a lot of interest.
The House has already approved the bill while Senator Mark Villar, part of the delegation to Davos, authored a similar measure.
The President made a soft launch of the MIF during the Philippines' Country Strategy Dialogue on Tuesday.
The proposal to create the MIF for the Philippines gained support from WEF participants as Marcos noted that the fund would be used to diversify the country's financial portfolio.
"We have to design it very specifically to the Philippine condition. And that's what the legislators are trying to do now – to make sure na babagay para sa atin (that it will be suitable for us). So that's the process that we're undergoing now," Marcos said in a media interview before departing Switzerland on Friday.
In and around the Metro
Authorities arrest 3 fixers at Makati City Hall
account_balanceMakati chrome_reader_mode2 hours ago
Makati Mayor Abby Binay today said an entrapment operation initiated by the city government has led to the arrest of three suspects, including two city government employees. "The city government will not tolerate fixers. We will continue t...
4 Reasons to Consider Investing In a Duplex Home
Most people don't think much about anything other than the perfect home for them when building or buying. That's the way it should be too. However, that doesn't mean you make some changes or you have to have the perfect home right a...
LTFRB probes overcrowding in jeepneys
account_balanceMetro Manila chrome_reader_mode2 hours ago
The Land Transportation Franchising and Regulatory Board (LTFRB) is looking into a health incident on a modern jeepney that may have been caused by overcrowding, according to a report by GMA News. According to Oscar Oida's report on 24 Oras on M...
DOLE to welcome ILO mission
The Department of Labor and Employment (DOLE) assured the High-Level Tripartite Mission (HLTM) authorized by the International Labor Organization (ILO) of its full cooperation and assistance, particularly in addressing workers' concerns, acc...
Manila mall launches 'bike and bite'
DPWH set to construct new building in UP Diliman
DOST approves projects for Virology Institute
DOT wants age restrictions lifted for Intramuros visits
Muntinlupa receives Seal of Good Local Governance

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How to tell if a Windows 7 re-install disk (and subsequent installation) is OEM or not?
How do I tell if a Windows 7 re-install disk is OEM (and subsequent installation) or not?
Can I explore the disk to look for a certain file or setting. And once installed can I go into the control panel and look for an indication or run a utility to detect?
I have a HP laptop which does not come with a full true Windows 7 install disk, instead it comes with recovery disk(s) and partition (i.e. a image of the hard drive partition at factory install).
At some point I would prefer to wipe the laptop disk and re-install cleanly on same laptop Windows 7 from a true Windows 7 OS (re)install disk, available on ebay, and use the genuine COA product serial license key from the label on the laptop. Reason for re-install is to get rid of all the guff (trialware, utils, etc) provided as standard. I would download necessary drivers from online.
I would prefer to obtain a Windows 7 re-install disk that is not OEM. This would be handy for use on other machines later (which have their own genuine COA), hence the preference as OEM is locked to motherboard hardware. I will be looking for an install that matches the Windows 7 edition on the laptop, i.e. 64bit home premium.
Regular Windows 7 setup disks will accept OEM product keys. I've done this so many times. . .
It's not like under XP/Server 2003.
Just in case you need proof.
"Licensed only for distribution with a new PC".
The difference will also be clear on a retail edition of a Windows 7 DVD and install (no branding).
Not the answer you're looking for? Browse other questions tagged windows-7 reinstall oem or ask your own question.
How to remove the windows 7 bootloader from my OEM partition?
Does OEM partition stores Windows 7 serial key ?
How to clean (re)install of Windows 8 OEM to a new SSD?
OEM partition of 465 GB in a 1TB disk, too big - how to resize it?
Will a Windows 10 OEM installation disk differ from a "from Microsoft" install?
How to install windows 7 on a new HDD with just the old (dead) HDD's COA key?


Construction top; Monster truck bottoms.
And, there was no changing his mind, either....the boys wants what he wants.
Who cares, right? It's just PJ's. And it was so cute, anyway.
Thanks for commenting - isn't it shameless how I asked people to comment? Oh well, I do LOVE to check my inbox and see that people left messages!
I have no idea when I installed my counter. I'm guessing probably sometime in January, after I started reading Confessions of a CF Husband. I think I got my counter by clicking on his. I could be wrong though.
Hilarious. Boys will be boys and they are all alike at this age it seems. Too cute!!


text:
Анатолий Андреевич Конин (12 июня 1914, дер. Махонино, Владимирская губерния, Российская империя — 10 ноября 1986, Москва, СССР) — советский художник, живописец, член Московского Союза советских художников с 1943 года.

Биография
В 1927 году, после окончания средней школы, Анатолий поступил в ФЗУ при заводе № 67. Неизвестно, как сложилась бы его судьба, останься он в местах, где художественное образование было получить практически невозможно. Однако помог случай — в 1932 году, уже слесарем, он был мобилизован на строительство метро первой очереди станции «Комсомольская» в Москве.

В 1935 году Анатолий поступил в Московский государственный художественный институт им. В. И. Сурикова, учился у И. Э. Грабаря, А. А. Осмеркина и А. В. Моравова. В 1943 году он уже был принят в члены МССХ и в том же году был призван в РККА. Прошёл войну, демобилизовался в сентябре 1945 г.

С конца 1940-х годов постоянно участвовал в выставках, экспонируя свои работы вместе с произведениями ведущих мастеров изобразительного искусства Москвы.

Скончался в 1986 году в Москве. Похоронен на Ваганьковском кладбище. Его произведения находятся в музеях и частных собраниях в России, на Украине, в Франции, США и других странах.

Творчество
Писал преимущественно пейзажи, натюрморты, портреты, жанровые и тематические картины, этюды с натуры. Талантливый колорист. Живописи свойственна особая мелодическая тонкость, пластичность цвета, которым художник передавал ощущение материальности воздуха, его влажности и свежести. Значительную часть мира творчества Конина составляют выразительные образы природы, нашедшие отражение в многочисленных излюбленных произведениях художника — пейзажах. Своеобразным стилистическим экспериментом и появлением новых выразительных эффектов стали работы периода увлечения искусством постимпрессионизма.

Среди произведений, созданных Анатолием Кониным, картины «Весна, серый день», «Первая зелень» (обе 1954), «Девушки» (1958), «У Ленинских гор» (1959), «Море с чайками» (1960), «Горная река на Ажеке» (1959), «Начало осени» (1961), «Пасмурный день»(1962), «Утро на море» (1964), «Теплый день» (1965), «Портрет Наташи» (1967), «Ущелье»(1968), «Вечерний пейзаж» (1969), «Весенний пейзаж» (1968), «После дождя» (1970), «Натюрморт с тыквой» (1971), «Освещенный откос у речки» (1971), «Махмут» (1972), «Дом творчества» (1972), «Натюрморт с айвой», «Пейзаж» (обе 1971) «Осенний мотив» (1972), Стволы" (1973), «В саду» (1974) и другие

Примечания

Художники СССР
Похороненные на Ваганьковском кладбище

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Cleaning Service Long Beach | Green Cleaning Seal Beach | House Cleaning Los Alamitos - Eco Cleaning Solutions, Inc.
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Every one of us who believes in God desires to receive God's blessings, because God blessing a person represents that this person is approved by God and pleases God. Then, how can we receive God's blessings? This is what we are all concerned with. We all know the Bible records that Noah and Abraham were blessed by God. So, what exactly did they do before gaining God's blessings? Let us understand their experiences through studying the Bible and follow their examples.
Why did Noah gain God's blessings?
The Bible records: "And God looked on the earth, and, behold, it was corrupt; for all flesh had corrupted his way on the earth. And God said to Noah, The end of all flesh is come before me; for the earth is filled with violence through them; and, behold, I will destroy them with the earth. Make you an ark of gopher wood; rooms shall you make in the ark, and shall pitch it within and without with pitch. … Thus did Noah; according to all that God commanded him, so did he" (Genesis 6:12-14, 22).
From God's words, we can see: When God entrusted Noah to build an ark, although he had never seen an ark, and did not know how to build an ark, he did not ask why God let him do this. Instead, his only attitude was obedience to God. He acted based on God's requirements, treated God's commission as his responsibility and duty and completed it. Thus, he gained God's blessings.
Why did Abraham gain God's blessings?
Through reading Noah and Abraham's stories, we find they had one thing in common: They both listened to and obeyed God and they all followed God's will. No matter what God said, even if they did not understand God's will at that time, they were simple, sincerely believed in God, and acted according to God's words without any hint of a transaction and demand. However, compared to them, although we also believe in God, we have impurities within us and we always make deals with God. I recalled that when God's commission came upon me, what I considered first was my own benefits. I was unwilling to do duties that were not to my benefit, even if I did, I didn't truly obey and didn't have any consideration for God. I only did them with the intention of exchanging God's blessings with my expending, such as a peaceful family, a prosperous career and the blessings to enter the heavenly kingdom. As a result, what I had done could hardly win God's approval. Therefore, if we expect to live within God's blessings, we should imitate Noah to obey God absolutely. There was also Abraham, who, when facing God's trials and refinements, only sought to satisfy God's will. So he was willing to offer up Isaac to God without other purposes and intentions. But when we are faced with trials, we always misunderstand God and complain against God. We are unable to truly obey God's rule and arrangements, so we cannot gain God's blessings.


Hundreds of millions of dollars meant to provide a little relief to the nation's struggling homeowners is being diverted to plug state budget gaps.
In a budget proposed this week, California joined more than a dozen states that want to help close gaping shortfalls using money paid by the nation's biggest banks and earmarked for foreclosure prevention, investigations of financial fraud and blunting the ill effects of the housing crisis. California was awarded more than $400 million from the banks, and Gov. Jerry Brown has proposed using the bulk of that sum to pay the state's debts.
The money was part of a national settlement valued at $25 billion and negotiated with five big banks over abuses in their mortgage and foreclosure processes.
The settlement, reached in February after a year of talks and intervention by the Obama administration, was the second-largest in history involving the states, trailing the tobacco industry settlement, and represented the first large-scale commitment by banks to provide direct aid to borrowers.
As part of the settlement, the banks agreed to pay the states $2.5 billion, money intended to help homeowners and mitigate the effects of the foreclosure surge. But critics complained that this was the only cash the banks were required to pay — the rest comes in the form of "credits" for reducing mortgage debt and other activities. Even that relatively small amount has proved too great a temptation for lawmakers.
Only 27 states have devoted all their funds from the banks to housing programs, according to a report by Enterprise Community Partners, a national affordable housing group. So far about 15 states have said they will use all or most of the money for other purposes.
In Texas, $125 million went straight to the general fund. Missouri will use its $40 million to soften cuts to higher education. Indiana is spending more than half its allotment to pay energy bills for low-income families, while Virginia will use most of its $67 million to help revenue-starved local governments.
Like California, some other states with outsize problems from the housing bust are spending the money for something other than homeowner relief. Georgia, where home prices are still falling, will use its $99 million to lure companies to the state.
The $2.5 billion was intended to be under the control of the state attorneys general, who negotiated the settlement with the five banks — Bank of America, Wells Fargo, JPMorgan Chase, Citigroup and Ally. But there is enough wiggle room in the agreement, as well as in separate terms agreed to by each state, to give legislatures and governors wide latitude. The money can, for example, be counted as a "civil penalty" won by the state, and some leaders have argued that states are entitled to the money because the housing crash decimated tax collections.
Shaun Donovan, the federal housing secretary, has been privately urging state officials to spend the money as intended. "Other uses fail to capitalize on the opportunities presented by the settlement to bring real, concerted relief to homeowners and the communities in which they live," he said Tuesday.
Some attorneys general have complied quietly with requests to repurpose the money, while others have protested. Lisa Madigan, the Democratic attorney general of Illinois, said she would oppose any effort to divert the funds. Tom Horne, the Republican attorney general of Arizona, said he disagreed with the state's move to take about half its $97 million, which officials initially said was needed for prisons.
But Mr. Horne said he would not oppose the shift because the governor and the Legislature had authority over budgetary matters. The Arizona Center for Law in the Public Interest has said it will sue to stop Mr. Horne from transferring the money.
In California, Attorney General Kamala D. Harris had played hardball in the settlement negotiations, holding out until the very end for a deal guaranteeing that a large share of the benefits would go to California, and then trumpeting her success in a news conference and a flurry of interviews with national news outlets. So Mr. Brown's revised budget put her in an awkward position.
"While the state is undeniably facing a difficult budget gap," she said in a statement, "these funds should be used to help Californians stay in their homes." Both officials are Democrats.
When asked if Mr. Brown could legally appropriate the money, which is supposed to be held in a special fund "for the benefit of California homeowners affected by the mortgage/foreclosure crisis," a spokesman for Ms. Harris declined to comment.
Just last week, Ms. Harris announced plans to give about half the money to groups that provide housing counseling and legal assistance to homeowners — groups whose budgets have shrunk while demand for their services grows. The other half would be used primarily for investigation of mortgage-related crime.
States using some or all of their money for housing have designated it for a wide variety of programs, like a small fund for low-interest loans to build housing in low-income neighborhoods, in Virginia, and Ohio's sweeping plan to demolish abandoned property.
In New York, Attorney General Eric T. Schneiderman stepped in with $15 million in settlement money for housing counseling and legal assistance when state support ran out last month, and plans to spend the bulk of its $130 million on similar programs. North Dakota will use its tiny allotment, $1.9 million, to provide housing to police officers and emergency responders in its booming oil-field counties, where shelter is scarce.
In some states, redirecting the money could have a racially discriminatory effect, said Alan Jenkins, the executive director of the Opportunity Agenda, which supports homeownership, because in some cities black homeowners disproportionately lost their homes, Mr. Jenkins said.
"If you dump all of these funds into the general coffers, the African-American homeowners are not going to benefit in any real way because they represent such a small percentage of the larger state," Mr. Jenkins said.


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Glastonbury 2020: all you need to know about dates, line-up rumours, registering for tickets, and beating the rush for the 50th anniversary festival
On September 11, 2019 by e-Today.US
The Worthy Farm festival turns 50 next summer – here's everything you need to know about joining the big anniversary bash
Next summer Glastonbury celebrates the big 5-0, and the ensuing festivities are shaping up to be very special indeed. After humble beginnings – Michael Eavis hosted his first ever festival in 1970, for about 1,500 lucky attendees – the event has evolved into one of the biggest festivals in the entire world, with a site that occupies roughly the same space as Oxford city centre. If you thought the 2019 edition of the festival was huge, then buckle up for a massive half century party that'll take things up a thousand more notches.
What's the latest news?
The latest bookies' odds see Paul McCartney, Fleetwood Mac, Foals and Elton John named favourites to headline.
McCartney has responded to the speculation.
Wolf Alice are lined up to play at September's Pilton Party. Will they both be booked for Glastonbury 2020?
Supergrass reunited for the Pilton Party too – and remain coy about whether or not they could appear at Glasto next summer.
Green Day, Fall Out Boy and Weezer have announced their joint Hella Mega UK tour on the week of Glastonbury 2020. Could they also be making an appearance?
Fancy joining the enormous Worthy Farm festival? Here's absolutely everything you need to know about Glastonbury 2020.
When does Glastonbury 2020 take place?
In the aftermath of this year's Glastonbury, the festival's official account shared the following poster on July 1. Along with teasing ahead of Glastonbury's 50th birthday, it reveals the start date of the next edition as being June 24, 2020.
Thank you to everyone who helped to make it such a special Glastonbury. We'll see you in 2020! pic.twitter.com/ftpe0kyJQz
— Glastonbury Festival (@GlastoFest) July 1, 2019
This has also been confirmed on Glastonbury's official website, which confirms that next year's festival will take place June 24-28 2020.
When do Glastonbury 2020 tickets go on sale?
Organisers Emily and Micheal Eavis are yet to share an exact date for ticket sales, but according to Glastonbury's website, more information will be revealed in autumn this year.
Typically, Glastonbury tickets for the following year go on sale in October, with a resale likely to take place around April 2020.
How much are Glastonbury tickets?
Tickets for Glastonbury 2019 cost £248, plus a £5 booking fee. As ever, the price of the ticket included entry to the festival, five nights camping, a free programme and mini-guide, onsite newspaper and mobile charging points and free firewood. A portion of the ticket price is also donated to Oxfam, Greenpeace, WaterAid and hundreds of other worthy causes, while funds are also used to improve the site's infrastructure and environmental impact.
Prices last year increased by around £10 compared to the previous Glastonbury, which should give you a rough idea of what to expect.
Do I need to register to buy a ticket for Glastonbury 2020?
Yep. No registration, no ticket.
To put a stop to ticket touting, all potential ticket-buyers for Glastonbury need to sign up ahead of the sales. It's free to register online, and just takes a few minutes: you'll need to upload a portrait-orientation, passport standard photograph of yourself, and fill in a few basic details such as your postcode.
Registration for Glastonbury 2020 is open now – festival organisers recommend that applications are submitted before September 28, 2019. This is so that you have plenty of time to resubmit your registration if there are any issues with the information provided
If you're already registered for Glastonbury tickets since 2010, relax – there's no need to repeat the process. Can't remember if you've filled one in? You can check your registration status here. Organisers also recommend keeping your registration photo up to date, particularly if your appearance has changed drastically. It's now possible to update the photo attached to your account while keeping your existing registration number.
If you last registered for Glastonbury tickets before 2010, you will need to resubmit a new application ahead of October's sale in order to buy tickets.
Once set up on the system, it's worth retrieving your registration number and noting it down ready for when ticket-buying time comes around.
Will I get to see the line-up before Glastonbury tickets go on-sale?
No – Glastonbury famously don't announce any of their line-up ahead of tickets going on sale. It's worth embracing the surprise, though – Glastonbury is always bursting with amazing acts of all varieties. The 2019 edition was headlined by Stormzy, The Killers and The Cure, with a legends slot from Aussie pop icon Kylie.
There must be a few rumours, though – do we have any idea who might be booked for Glastonbury 2020 and who could headline?
Glastonbury's father daughter co-organising duo Emily and Michael Eavis have let slip with a few hints when it comes to the line-up for 2020 already. In April 2019, Michael told BBC Somerset that The Beatles legend Paul McCartney would be coming to Glastonbury, "hopefully for the 50th. Don't make a big thing of it though, will you?" he added, cat well and truly out the bag.
McCartney himself has now said a headline slot could be a "distinct possibility".
The 1975's Matt Healy has also confirmed that the band will be at the festival in 2020, and speaking earlier this year, added that they feel "ready" for the top spot next summer. "Whether we headline it or not, who knows?" he told Matt Wilkinson on Beats1. Emily Eavis later responded to the frontman's comments, telling Wilkinson ""Oh really? That's good. Interesting. They're great. Obviously we'd love to have The 1975."
As ever, there's also speculation about a potential headline spot for Fleetwood Mac – who are yet to play Worthy Farm. Last month Mick Fleetwood told fans at their Wembley show that they still "had a big field to play", and in an interview with The Independent the band reiterated their desire to play Glastonbury.
Appearing on-stage at the Acoustic Stage during Glastonbury 2019 on June 28, Michael Eavis addressed the band's comments, saying" "I can't afford them at the moment! They'll have to bring their price down because we can't afford it. But they said the other day that they really want to do it and if they don't do it before they die they'll go to hell."
Fleetwood Mac have pulled out as headliners of this year's Jazz Fest.
As for the big pop acts that tend to appear at Glastonbury, it could be the Spice Girls. That's according to Mel B, who says the group will reconvene with Victoria Beckham to discuss the idea. Say you'll be there?
As far back as 2017, Emily Eavis confirmed that two headliner bookings were already in the bag for 2020. "Ideally, we'd have a mix of old and new acts headlining," she told NME. "A bit of both. But, a lot can happen between now and then, and it's about seeing who comes through and who would be right for us. It kind of depends, we've just got a little while, we're not in any rush, but it's quite exciting. It's always good having something in the long term plans as well."
Not only that, but Wolf Alice recently played at Glasto's Pilton Party in September. Will they both be booked for Glastonbury 2020? Not only that, but Supergrass' first show since reuniting was as the "special guest" at the Pilton Party. Now, they're back with a huge run of comeback tour dates set for 2020. Asked if they could make an appearance at Worthy Farm next year, frontman Gaz Coombes told NME: "I don't think I can say anything about that! I think we shall leave that in the hands of the Gods."
Supergrass, 2019
One particularly interesting development also comes with Green Day, Weezer and Fall Out Boy announcing details of their joint Hella Mega UK and Ireland tour the week of Glastonbury in June 2020. There are gaps in their schedule on Friday June 26 and Sunday June 28. Could any or all three of the bands be booked? Perhaps Green Day will headline, play the legends' slot or play a secret show? Perhaps all of the stops will be pulled out for Glasto's 50th anniversary.
Green Day, 2019
What are the bookies' odds for Glastonbury 2020 headliners?
Now, high street bookies Betway have released odds for rumoured Glastonbury headliners next year.
"Planning for next year's Glastonbury Festival will already be underway and the Eavis' will put on something very special for the event's 50th anniversary," said Betway's Alan Alger. "Foals are one of the big early fancies to get a headline spot at just 7/4. Their performance on the Park Stage certainly qualified them to get the nod for the Pyramid and they are one of the leading contenders to headline in 2020.
"The most likely headliners at this stage are Paul McCartney, who's just evens to play for the first time since 2004, and Fleetwood Mac. Fleetwood Mac would be something really special for Glastonbury given they've never headlined before, and they're very popular in the betting at 5/4."
Check out the latest odds below – which see Macca and Fleetwood Mac leading the chase.
Paul McCartney – Evens
Fleetwood Mac – 5/4
Foals – 7/4
Elton John – 11/4
The 1975 – 3/1
Spice Girls – 5/1
Arctic Monkeys – 11/2
Coldplay – 11/2
Ariana Grande – 6/1
Blur – 7/1
The Kinks – 7/1
BAR – 8/1
How many tickets are there for Glastonbury 2020?
This year 135,000 punters bought tickets in the general sale, with a total of 203,000 people (including artists, volunteers, and festival workers) onsite.
Going forward into next summer, a change to Worthy Farm's licence means that a further 7,000 people will be allowed to go to Glastonbury – on the condition that they all get there by public transport. The increase will give the festival a total capacity of 210,000.
In 2019, around two million people registered to buy Glastonbury tickets in total – assuming all of them gave it a try in the sale, just 6.75% of them were lucky enough to bag one.
Are there any secret tips to beat the Glastonbury ticket rush?
Every seasoned Glastonbury-goer swears by a different method. Some swear that major cities have a harder time in the sale, and trek to the most obscure part of the UK possible. Some wire up their laptops to super-charged internet and manically turn off every other device in the house, while others take the opposite approach and juggle ten iPads, four mobiles, and two computers at the same time. But the real secret, according to organisers, is far simpler than that – be patient, be prepared, and note down all the relevant details to avoid any last minute panic.
"Make sure you have all your registration details (including postcodes) ready and checked in advance, that there are sufficient cleared funds on the card you intend to use to book, and that you know all the security information for that card," they advise. If you're trying for any of your mates, it's also worth noting their postcodes and registration numbers down, to avoid the worst case scenario – messing up and entering in the wrong details on the final page.
Once you reach the booking page, you have exactly 6 minutes to complete the transaction. You're a lot more likely to succeed if you don't make any mistakes at this stage.
Glastonbury organisers also recommend a bit of determination. "Be persistent and patient – there are likely to be more people trying to buy tickets than there are tickets available, so inevitably not everyone will be able to book tickets, but the longer you try the more likely you are to get a ticket."
Other than that, all you can really do is keep your fingers firmly crossed.
And with that, bring on Glastonbury 2020!
The post Glastonbury 2020: all you need to know about dates, line-up rumours, registering for tickets, and beating the rush for the 50th anniversary festival appeared first on NME.
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Status: offline Thanks for the scenarios Hailstone and wish you a speedy recovery as well!
Status: offline It's all good. Had cancerous prostate removed so now I sit around and make TotH scenarios!
stone houses of the picturesque village were quickly destroyed. Sinagoga's defenders were not.
For the Germans to win they must hold 4 VPs.
For the Irish to win they must capture 4 VPs.
Offensive was only days away.
For the New Zealanders to win they must exit 9 squads.
For the Germans to win they must eliminate 4 squads.
following day. His hopes were soon to be fulfilled.
Status: offline "It's all good. Had cancerous prostate removed so now I sit around and make TotH scenarios!"
Sam, hope you have a speedy recovery as I did. I had mine removed in 2016 just a week before Tigers on the Hunt came out.
Do what your doctors say, drink plenty of water and you'll pull through!
Good Luck, my thoughts and prayers are with you!
Status: offline Thank you, everyone, for your sentiments. I don't even know why I even mentioned it other than it's on my mind a lot.
here's another scenario just for its recuperative powers!
the defense at Carroceto, the Allied command ordered a counterattack by the U.S.
For the Germans to win they must hold 5 VP's.
For the Americans to win they must capture 10 VP's.
the tank support to manuever very slowly for fear of bogging down in the soft ground.
the Allies grudgingly cancelled all further attempts to retake the factory.
I've found this easy (maybe even _too_ easy) to get a Major Victory as the Germans, but haven't had any success as the Americans yet.
Any idea for a good American strategy?
German squad on that side that was easily dispatched with all your fire power and best leader. Then I sneak across the road and enter the woods.
was already won and my playtesting completed - it works!
I'm glad you felt those scenarios were worth downloading. Thanks!
Status: offline Hi Hailstone, I'd like to report a possible bug in your "Silence that gun" scenario.
I was finally able to get an American major victory, by employing a "right hook" strategy.
I basically left some red herrings in the woods in front of the fortified buildings to attract the MG fire and, under cover of the woods on the right, I assaulted with the rest of my force first into the stone building containing the PaK gun and then the fortified buildings.
However, as you can see from the Victory status window, I got no credit for killing the German gun crew.
I believe this is because 10VPs are attributed to a hypothetical crew on the American side (which the Americans don't have), while no VPs are attributed to the German crew.
Again, thanks for the scenarios.
need to replace the file already uploaded with the corrected version. I'll get with tech support for that. Anyway, I welcome all suggestions and/or criticisms to improve. Thanks man!
a breakthrough at the crossroads.
For the Americans to win they must prevent the Germans from exiting 24 points.
elements of the 1st Battalion stood to meet the Germans once again.
Thanks for revitalizing some of these old Squad Leader classics!
and take my mind off stuff. I have a couple of projects and I'm trying to work out the kinks which are usually getting the AI to perform. There are more scenarios coming.
Again, thanks for the compliment, Big Ivan!
Status: offline Hey Sam, know full well what you mean about getting rid of the "freaking catheter"!
I thought I was reborn on that day in 2016!!
Good luck with the rest of the recovery!!
Status: offline Hi Hailstone, sorry to be a pain, but I believe you've re-uploaded the same old version of "Silence that Gun".
The one that's up is still dated 11/30/2018 and the VP swap for the crews is still there.
Status: offline Piotre, thanks for the catch.
even greater encirclements, many potentially dangerous Russian armored groups were bypassed.
and ordered to take the city.
For the Russians to win they must capture 16 VP stone buildings.
For the Germans to win they must keep the Russians from capturing 16 VP stone buildings.
have gone the other way.
Status: offline All is good now, thanks.
overrun the next defensive line located in the small village of Kommerscheidt.
For the Germans to win they must capture the town of Kommerscheidt.
For the Americans to win they must hold on to the town.
Schmidt under heavy artillery fire.
been eliminated. Daybreak also brought the same fate to the remainder of the XXV Tank Corps.
up for the push toward the town of Glebovka, their first inland objective.
For the Russians to win they need to capture the two artillery guns on the hill and exit 5 squads.
For the Germans to win they must prevent the capture of the artillery pieces and 5 Russian squads from exiting.
by the Western Allies five months earlier at Dieppe had been disastrous.
raw Polish levies conscripted into the 1st Polish Army by the Soviets were no match for veterans.
rapidly turned the withdrawal into a rout.
For the Germans to win they must reach the Vistula River with at least 5 squads or equivalent.
For the Poles to win they must prevent 5 squads to reach the river.
forced to commit whatever units they had available in an effort to form a new defensive line.
the units of Army Group Center would have with the superior T-34 and 52-ton monster KVII tanks.
attack near Lipki like others collapsed.
out from Tiergarten after dark on May 2nd.
For the Germans to win they must exit 32 VP's through any of the three roads leaving Spandau.
For the Russians to win they must prevent the Germans from exiting out of town and meet their fate.
units, had ceased to exist.
Always look to the TOTH website each day for the latest one.
So far, Blocking Action At Lipki has thwarted 4 attemps at a major Russian victory. Did get a Minor Victory on last try. Previous attempts were thwarted by lack of coms for Russian tanks and breakdowns.
scattered about that should not be there in the final draft. Okay now. Thanks for feedback!
As far as the Super Hero Rico, I know he's watching - I could feel his presence in the force!
Status: offline I wondered what the leaders were there for.
Status: online Die For Wypolsowa.
are engaged with the infantry Regiment 295 in the area of Wypolsowa (little Hamlet with three bridges).
Best played on German SIDE first / with LOS: 6 Hexes / with RULE:---/ with MOD: Yours / etc.