Superspaces for heterotic pure spinor string compactifications

Osvaldo Chandia♣ and Brenno Carlini Vallilo♠

♣ Departamento de Ciencias, Facultad de Artes Liberales
Universidad Adolfo Ibáñez, Chile

♠ Departamento de Ciencias Físicas, Universidad Andrés Bello
Sazié 2212, Santiago, Chile

Abstract

We study supersymmetry conditions for the heterotic pure superstring
preserving and $N=1$ supersymmetry in four dimensions directly from the curved
superspace defined by the Berkovits-Howe constraints.

†† ♣<EMAIL_ADDRESS>♠<EMAIL_ADDRESS>
###### Contents

  1. 1 Introduction
  2. 2 Heterotic supergravity and the Berkovits-Howe constraints
  3. 3 Killing supervectors
    1. 3.1 Flat superspace
  4. 4 Invariance conditions
    1. 4.1 Four dimensional Poincaré symmetry
    2. 4.2 Global four dimensional $N=1$ supersymmetry

##  1 Introduction

Understanding string dynamics in curved backgrounds is still one of the most
important topics of research in string theory. It is particularly crucial to
study dualities and compactifications. Most of the knowledge in string
compactifications comes from the Ramond-Neveu-Schwarz (RNS) formalism or the
supergravity limit. In the case of heterotic string or Type II strings the RNS
formalism allowed significant advances such as non-renormalization theorems
[1] and exact results from topological strings (see, e.g., [2] for a review).
For compactification backgrounds involving Ramond-Ramond (RR) fields most of
the work done was restricted to the supergravity limit. The difficulty comes
from the RR vertex operators associated with the linearized RR flux.

The pure spinor formalism [3] introduced by Berkovits has manifest
supersymmetry in flat space-time and made possible to study RR backgrounds
without the subtleties of the RR vertex operators of the RNS
formalism.111Another promising way to study RR backgrounds is to use Closed
Superstring Field Theory [4] as in the work [5]. In two previous papers [6, 7]
we started the study of compactifications using the pure spinor formalism222It
was also studied in [8] using orbifolds. However, in those works we started
with an ansatz for the covariant super derivative algebra without relating it
the supergravity background defined by the Berkovits-Howe supergravity
constraints [9]. The assumption was guided by requiring a nilpotent BRST
charge in the supergravity limit. In this work we want to study supergravity
backgrounds for the heterotic pure spinor string preserving Poincaré symmetry
or supersymmetry in four dimensions directly from the curved superspace
defined by the Berkovits-Howe constraints. The idea is to construct the
Killing supervectors and derive the conditions imposed on the supergeometry
such that their lowest components satisfy some general requirements expected
for Poincaré or supersymmetry parameters. Knowledge of the $\theta$ expansion
of Killing supervectors is also useful since they define conserved currents on
the worldsheet which can be used to have an explicit form for the supergravity
vertex operators. Killing spinors and their algebra were already studied in,
e.g. [10] but not directly from superspace point of view.

The superspace approach taken here is considerably more involved that the
component approach historically used in the literature on superstring
compactifications. However if we want to apply covariant formalisms for the
superstring to this problem we must understand all steps in the description of
curved superspaces for string compactifications. Although the case studied
here is one of the oldest compactification models it will serve the base to
apply the same ideas to more general compactifications, e.g. [11, 12, 13, 14]
and also for Type II strings. Curved superspace methods were also used in [15]
to describe flux compactifications.

This paper is organized as follows. In the next section we give a brief review
for the heterotic string in the pure spinor formalism in a general curved
background and some of the main consequences of the Berkovits-Howe
constraints. In the Section 3 we study general Killing supervectors in ten
dimensional supergravity with the Berkovits-Howe constraints. In Section 4 we
study the consequences of imposing four dimensional Poincaré symmetry and
supersymmetry.

##  2 Heterotic supergravity and the Berkovits-Howe constraints

In this section we review the heterotic string in a curved background in the
description of the pure spinor formalism. The world-sheet action is given by

$\displaystyle S=\int d^{2}z~{}$
$\displaystyle\left(\frac{1}{2}\Pi_{\underline{a}}{\overline{\Pi}}^{\underline{a}}+\frac{1}{2}\Pi^{A}{\overline{\Pi}}^{B}B_{BA}+d_{\alpha}{\overline{\Pi}}^{\alpha}+\omega_{\alpha}{\overline{\nabla}}\lambda^{\alpha}+\rho_{\cal
A}{\nabla}\rho_{\cal A}\right.$ (2.1)
$\displaystyle\left.+d_{\alpha}{\overline{J}}^{I}W_{I}{}^{\alpha}+\lambda^{\alpha}\omega_{\beta}{\overline{J}}^{I}U_{I\alpha}{}^{\beta}\right)+S_{FT},$
(2.2)

where the superspace coordinates $Z^{M}=(X^{\underline{m}},\theta^{\mu})$
(with ${\underline{m}}=0,\dots,9,\mu=1,\dots,16$) defines $\Pi^{A}=\partial
Z^{M}E_{M}{}^{A}(Z)=(\Pi^{\underline{a}},\Pi^{\alpha})$, where
$E_{M}{}^{A}(Z)$ is the vielbein superfield. The world-sheet field
$d_{\alpha}$ is the generator for superspace translations, $\lambda^{\alpha}$
is the pure spinor variable and $\omega_{\alpha}$ is its momentum conjugate
variable. The world-sheet fields $\rho_{\cal A}$ are the heterotic fermions.
They are in the fundamental representation of the gauge group $SO(32)$ or
$E_{8}\times E_{8}$. These fermions define the current
${\overline{J}}^{I}=\frac{1}{2}K^{I}_{\cal{AB}}\rho_{\cal A}\rho_{\cal B}$,
where $K_{I}$ are the generators of the gauge group Lie algebra in the
fundamental representation. The covariant derivatives in (2.2) are defined by

$\displaystyle{\overline{\nabla}}\lambda^{\alpha}={\overline{\partial}}\lambda^{\alpha}+\lambda^{\beta}\left({\overline{\partial}}Z^{M}\Omega_{M\beta}{}^{\alpha}\right),\quad{\nabla}\rho_{\cal
A}=\partial\rho_{\cal A}+K^{I}_{\cal{AB}}\left(\partial
Z^{M}A_{IM}\right)\rho_{\cal B},$ (2.3)

where $\Omega_{M\beta}{}^{\alpha}$ is the background Lorentz connection and
$A_{IM}$ is the gauge group connection. The background fields are
$B,W_{I},U_{I}$ and the connections in (2.3). Finally, $S_{FT}$ is the
Fradkin-Tseytlin term and it is given by

$\displaystyle S_{FT}=\int d^{2}z~{}\alpha^{\prime}r^{(2)}\Phi,$ (2.4)

where $r^{(2)}$ is the world-sheet curvature and $\Phi$ is the dilaton
superfield. Note that this term breaks classical conformal invariance but it
helps to restore it at the quantum level [16].

The action (2.2) is the most general expression that is invariant under
background Lorentz and gauge transformations such that it is also classically
conformal invariant. There is another symmetry that constrains the background
fields to satisfy the equations of supergravity in ten dimensions and the
equations of super Yang-Mills in a curved background [9]. This symmetry is
generated by the pure spinor BRST charge

$\displaystyle Q=\oint dz~{}\lambda^{\alpha}d_{\alpha}.$ (2.5)

The nilpotency of $Q$ and the holomorphicity of $\lambda^{\alpha}d_{\alpha}$,
imply a set of constraints for the background superfields. These constraints
are functions of the torsion, curvature, field-strength and $H=dB$ components.
Let us remind how they are defined. They are given after constructing the
super one-forms
$E^{A}=dZ^{M}E_{M}{}^{A},\Omega_{\alpha}{}^{\beta}=dZ^{M}\Omega_{M\alpha}{}^{\beta},A_{I}=dZ^{M}A_{IM}$
and the super two-form $B=\frac{1}{2}dZ^{M}dZ^{N}B_{NM}$. Note that the
Lorentz connection has the form

$\displaystyle\Omega_{\alpha}{}^{\beta}=\delta_{\alpha}^{\beta}\Omega^{(s)}+\frac{1}{4}(\gamma^{\underline{ab}})_{\alpha}{}^{\beta}\Omega_{\underline{ab}},$
(2.6)

where $\Omega^{(s)}$ is the connection for scalings and
$\Omega_{\underline{ab}}$ is the usual Lorentz connection in ten-dimensional
superspace. In this way, one can define the connection $\Omega_{A}{}^{B}$ with
non-zero components $\Omega_{\underline{a}}{}^{\underline{b}}$ and
$\Omega_{\alpha}{}^{\beta}$. The matrix
$\gamma^{\underline{ab}}=\frac{1}{2}(\gamma^{\underline{a}}\gamma^{\underline{b}}-\gamma^{\underline{b}}\gamma^{\underline{a}})$,
where $(\gamma^{\underline{a}})_{\alpha\beta}$ and
$(\gamma^{\underline{a}})^{\alpha\beta}$ are the symmetric $16\times 16$
$\gamma$-matrices in ten dimensions that satisfy the Dirac algebra

$\displaystyle(\gamma^{\underline{a}})_{\alpha\gamma}(\gamma^{\underline{b}})^{\gamma\beta}+(\gamma^{\underline{b}})_{\alpha\gamma}
(\gamma^{\underline{a}})^{\gamma\beta}=2\eta^{\underline{ab}}\delta_{\alpha}^{\beta}.$
(2.7)

The map of a tensor with superspace curved indices to local target space
indices is performed with the use of the vielbein and its inverse. For
example,

$\displaystyle B_{NM}=(-1)^{N(M+A)}E_{M}{}^{A}E_{N}{}^{B}B_{BA},$ (2.8)

where $(A,M)$ assigns a sign $(+1)$ for bosonic directions and $(-1)$ for
fermionic directions (for more details see [17]). The torsion, curvature and
field-strength super two-forms are given by

$\displaystyle T^{A}={\nabla}E^{A}=dE^{A}+E^{B}\Omega_{B}{}^{A},\quad
R_{B}{}^{A}=d\Omega_{B}{}^{A}+\Omega_{B}{}^{C}\Omega_{C}{}^{A},\quad
F_{I}=dA_{I}+f^{JK}{}_{I}A_{J}A_{K},$ (2.9) $\displaystyle
T^{A}=\frac{1}{2}E^{B}E^{C}T_{CB}{}^{A},\quad
R_{B}{}^{A}=\frac{1}{2}E^{C}E^{D}R_{DCB}{}^{A},\quad
F_{I}=\frac{1}{2}E^{B}E^{A}F_{IAB},$ (2.10)

where the product between forms is the wedge product and $f^{JK}{}_{I}$ are
the structure constants of the gauge group Lie algebra. The covariant
derivative $\nabla$ is

$\displaystyle\nabla=E^{A}\nabla_{A}=dZ^{M}(\partial_{M}+\frac{1}{2}\Omega_{M}{}^{\underline{ab}}M_{\underline{ab}}+\Omega^{(s)}_{M}S),$
(2.11)

where $M_{ab}$ are the Lorentz generators and $S$ is the scaling generator.

Together with

$\displaystyle H=\frac{1}{6}E^{C}E^{B}E^{A}H_{ABC},$ (2.12)

the torsion, the curvature and the field-strength satisfy Bianchi identities.
They are given by

$\displaystyle{\nabla}_{[A}T_{BC]}{}^{D}+T_{[AB}{}^{E}T_{EC]}{}^{D}-R_{[ABC]}{}^{D}=0,$
(2.13)
$\displaystyle{\nabla}_{[A}R_{BC]D}{}^{E}+T_{[AB}{}^{F}R_{FC]D}{}^{E}=0,$
(2.14) $\displaystyle{\nabla}_{[A}F_{IBC]}+T_{[AB}{}^{D}F_{IDC]}=0,$ (2.15)
$\displaystyle{\nabla}_{[A}H_{BCD]}+\frac{3}{2}T_{[AB}{}^{E}H_{ECD]}=0,$
(2.16)

where the (anti)-symmetrization is on $(ABC)$ in the first three equations and
it is on $(ABCD)$ in the last equation.

The nilpotency of the pure spinor BRST charge and the holomorphicity of the
BRST current, together with gauge fixing gauge symmetries of the action, imply
that some of the torsion components are

$\displaystyle
T_{\alpha\beta}{}^{\underline{a}}=-(\gamma^{\underline{a}})_{\alpha\beta},\quad
T_{A\alpha}{}^{\beta}=0,$ (2.17)

some of the $H$ components are

$\displaystyle H_{\alpha\beta\gamma}=H_{\underline{ab}\alpha}=0,\quad
H_{\alpha\beta\underline{a}}=-(\gamma_{\underline{a}})_{\alpha\beta},$ (2.18)

and for the field-strength components

$\displaystyle F_{I\alpha\beta}=0.$ (2.19)

Using the Bianchi identities of (2.16) and the torsion components in (2.17),
one can prove that the constraints for the curvature are implied. Also, the
Bianchi identity involving
${\nabla}_{(\alpha}T_{\beta\gamma)}{}^{\underline{a}}$ implies that

$\displaystyle
T_{\alpha\underline{ab}}=2(\gamma_{\underline{ab}}\Omega^{(s)})_{\alpha}.$
(2.20)

It is important to note that since $\Omega^{(s)}_{\alpha}\neq 0$ one has to be
careful when acting with $\nabla_{\alpha}$ on gamma matrices. For example,

$\displaystyle\nabla_{\alpha}(\gamma^{\underline{a}})_{\beta\delta}=-2\Omega^{(s)}_{\alpha}(\gamma^{\underline{a}})_{\beta\delta}.$
(2.21)

Combining the Bianchi identities involving
${\nabla}_{[\alpha}T_{\underline{ab}]}{}^{\underline{c}}$ and
${\nabla}_{[\alpha}H_{\beta\underline{ab}]}$ one can show that

$\displaystyle
T_{\underline{abc}}=-H_{\underline{abc}},\quad\Omega^{(s)}_{\underline{a}}=0.$
(2.22)

The Bianchi identity involving ${\nabla}_{[\alpha}H_{\underline{abc}]}$
together with the equation
$(\gamma^{\underline{b}})_{\alpha\beta}T_{\underline{ab}}{}^{\beta}=8{\nabla}_{\underline{a}}\Omega_{\alpha}$
(which is derived from the Bianchi identity involving
${\nabla}_{[\underline{a}}T_{\alpha\beta]}{}^{\beta}$ [16]) implies that

$\displaystyle
T_{\underline{ab}}{}^{\alpha}=-\frac{1}{6}(\gamma^{\underline{c}}\nabla)^{\alpha}H_{\underline{abc}}-\frac{4}{3}\nabla_{[\underline{a}}(\gamma_{\underline{b}]}\Omega^{(s)})^{\alpha}.$
(2.23)

Finally, the Bianchi involving $R_{(\alpha\beta\gamma)}{}^{\rho}$ implies

$\displaystyle
T_{\underline{abc}}=-(\gamma_{\underline{abc}})^{\alpha\beta}{\nabla}_{\alpha}\Omega^{(s)}_{\beta},$
(2.24)

So we have that

$\displaystyle
H_{\underline{abc}}=(\gamma_{\underline{abc}})^{\alpha\beta}{\nabla}_{\alpha}\Omega^{(s)}_{\beta}.$
(2.25)

Note that $\Omega^{(s)}_{\alpha}=\frac{1}{4}{\nabla}_{\alpha}\Phi$ as it is
required by ghost number anomaly cancellation [9] (also necessary for
vanishing of the beta function at one-loop [16]). One can also use the Bianchi
identities to find the following expressions for the curvatures

$\displaystyle
R_{{\alpha\beta}}{}^{\underline{ab}}=\nabla_{(\alpha}T_{{\beta})\underline{c}}{}^{\underline{b}}\eta^{\underline{ca}}+T_{({\alpha\underline{d}}}{}^{\underline{c}}T_{\beta)\underline{c}}^{b}\eta^{\underline{da}}+\gamma_{{\alpha\beta}}^{\underline{d}}H_{\underline{d}}{}^{\underline{ab}},$
(2.26) $\displaystyle
R_{\alpha\underline{b}}{}^{\underline{cd}}=-\nabla_{\underline{b}}T_{\alpha}{}^{\underline{cd}}-T_{\underline{b}}{}^{[{\underline{c}\beta}}\gamma_{{\beta\alpha}}^{\underline{d}]},$
(2.27) $\displaystyle
R_{\underline{ab}}{}^{\underline{cd}}=-\frac{1}{8}(\gamma^{\underline{cd}})_{\alpha}{}^{\beta}\left(\nabla_{\beta}T_{\underline{ab}}{}^{\alpha}-T_{\beta[\underline{a}}{}^{\underline{e}}T_{\underline{b}]\underline{e}}{}^{\alpha}
\right),$ (2.28) $\displaystyle
R^{(s)}_{\alpha\beta}=\nabla_{(\alpha}\Omega_{\beta)}=\frac{1}{4}\gamma^{\underline{a}}_{\alpha\beta}\nabla_{\underline{a}}\Phi,$
(2.29) $\displaystyle
R^{(s)}_{\underline{a}\alpha}=\nabla_{\underline{a}}\Omega_{\alpha}=\frac{1}{4}\nabla_{\underline{a}}\nabla_{\alpha}\Phi,$
(2.30) $\displaystyle
R^{(s)}_{\underline{ab}}=T_{\underline{ab}}{}^{\alpha}\Omega_{\alpha}=\frac{1}{4}T_{\underline{ab}}{}^{\alpha}\nabla_{\alpha}\Phi.$
(2.31)

All the gauge covariant background fields depend on the $\Phi$ superfield.
This is not something new and it is also true with a different set of
constraints [18]. It is interesting to note that for the Type IIB pure spinor
string this is not true [19]. It can be shown that using the Berkovits-Howe
constraints the Ramond-Ramond $5$-form, as well as curvature generated from
it, does not appear as a higher component of $\Phi$ [20]. An explicit example
of this is the $AdS_{5}\times S^{5}$ background [3, 21] where all covariant
derivatives of $\Phi$ vanish.

##  3 Killing supervectors

As discussed in the previous section all Lorentz covariant tensors can be
computed from $\Phi$. Under a general super reparametrization $\xi^{M}(Z)$ it
transforms as

$\displaystyle\delta\Phi=\xi^{M}\nabla_{M}\Phi=\xi^{A}\nabla_{A}\Phi=0,$ (3.1)

where we are defining $\xi^{A}=\xi^{M}E_{M}{}^{A}$. However it is not enough
to impose that the transformation above vanishes. Generically the covariant
tensors are given by

$\displaystyle T_{A_{1}\cdots
A_{n}}\sim\nabla_{A_{1}}\cdots\nabla_{A_{n}}\Phi,$ (3.2)

therefore in order to have $\delta T_{A_{1}\cdots A_{n}}=0$ we must also
impose that $\delta\nabla_{A}=0$. Another way to see it is that particular
components of the vielbein cannot be obtained from $\Phi$ so to impose that
the full background is invariant under some particular super diffeomorphism
the covariant derivatives (2.11), which depend on $E_{A}{}^{M}$, should also
be invariant.

The textbook way, see e.g. [22], to study super diffeomorphisms is to
introduce a vector superfield containing $\xi^{A}$, a compensating Lorentz
rotation $\Lambda^{\underline{ab}}$ and scale transformation $\sigma$

$\displaystyle K=\xi^{A}\nabla_{A}+{\textstyle{1\over{\raise
0.3014pt\hbox{$\scriptstyle{2}$}}}}\Lambda^{\underline{ab}}M_{\underline{ab}}+\sigma
S,$ (3.3)

Although the pure spinor sigma model has two independent local Lorentz
symmetries, acting on vectors and spinors separately, a combination of the two
is fixed in the process of solving the Berkovits-Howe constraints. The local
scale symmetry is also used to fix the dimension zero torsions, however since
we are not eliminating the scale connection from $\nabla_{\alpha}$ we still
have to include the scale transformation in $K$. As we will see this will
effectively reduce the structure group of symmetry transformations from
Lorentz times scale to only Lorentz, as expected.

A generic tensor superfield $\mathcal{O}^{A_{1}\cdots A_{n}}$ transforms under
local reparametrizations and local Lorentz transformations as generated by $K$
as

$\displaystyle\delta{\mathcal{O}}^{A_{1}\cdots
A_{n}}=K\mathcal{O}^{A_{1}\cdots A_{n}}.$ (3.4)

Since the covariant derivatives map tensors to tensors we have that the
covariant derivatives themselves transform as

$\displaystyle\delta\nabla_{A}=[K,\nabla_{A}].$ (3.5)

If we apply this idea to $\nabla_{\alpha}$ and use the Berkovits-Howe
constraints we get

$\displaystyle\delta\nabla_{\alpha}=$
$\displaystyle\left(-\nabla_{\alpha}\xi^{\beta}{-\frac{1}{4}}\Lambda^{\underline{ab}}(\gamma_{\underline{ab}})_{\alpha}{}^{\beta}-\delta_{\alpha}^{\beta}\sigma\right)\nabla_{\beta}+\left(-\nabla_{\alpha}\xi^{\underline{b}}+\xi^{\beta}\gamma^{\underline{b}}_{{\alpha\beta}}-\xi^{a}T_{{\underline{a}\alpha}}{}^{\underline{b}}\right)\nabla_{\underline{b}}$
(3.6)
$\displaystyle+\frac{1}{2}\left(-\nabla_{\alpha}\Lambda^{\underline{ab}}+\xi^{C}R_{C\alpha}{}^{\underline{ab}}\right)M_{\underline{ab}}+\left(\xi^{C}R^{(s)}_{C\alpha}-\nabla_{\alpha}\sigma\right)S=0,$
(3.7)

where $R^{(s)}_{AB}$ is the scaling curvature and $R_{AB}{}^{{ab}}$ is the
Lorentz curvature. The first term in (3.6) defines $\Lambda^{\underline{ab}}$
in terms of $\xi^{\alpha}$ it also implies that

$\displaystyle\nabla_{\alpha}\xi^{\alpha}=-16\sigma,\quad(\nabla\gamma^{\underline{abcd}}\xi)=0.$
(3.8)

Before studying the consequences of these conditions let us first analyze the
second term of (3.6), which has the lowest mass dimension. From it we can
write an expression for $\xi^{\alpha}$ in terms of $\xi^{\underline{a}}$ as

$\displaystyle\xi^{\alpha}=\frac{1}{10}\gamma_{\underline{a}}^{{\alpha\beta}}\nabla_{\beta}\xi^{\underline{a}}+\frac{9}{5}\xi^{\underline{a}}(\gamma_{\underline{a}}\Omega)^{\alpha}=\frac{1}{10}\nabla_{\beta}\left(\gamma_{\underline{a}}^{{\alpha\beta}}\xi^{\underline{a}}\right)+\frac{8}{5}\xi^{\underline{a}}(\gamma_{\underline{a}}\Omega)^{\alpha}.$
(3.9)

We can also write the fundamental equation for $\xi^{\underline{a}}$ using the
usual ten dimensional gamma matrix identity

$\displaystyle\gamma_{\underline{b}({\beta\delta}}\left(-\nabla_{\alpha)}\xi^{\underline{b}}+\xi^{\underline{a}}T_{{\alpha}){\underline{a}}}{}^{\underline{b}}\right)=0,$
(3.10)

it depends only on $\xi^{\underline{a}}$ and the background. Using the
explicit form of $T_{{\alpha\underline{a}}}{}^{\underline{b}}$ and the fact
that the $\nabla_{\alpha}$ derivative of $\gamma_{\underline{b}{\beta\delta}}$
is not zero, simplifies this equation to just

$\displaystyle\nabla_{(\alpha}\left((\gamma_{\underline{b}})_{{\beta\delta})}\xi^{\underline{b}}\right)=0.$
(3.11)

A nice way to summarize this is that the isometries of a general supergravity
background with the Berkovits-Howe constraints [9] are generated by a
symmetric bi-spinor satisfying

$\displaystyle\lambda^{\alpha}\lambda^{\beta}\xi_{{\alpha\beta}}=0,\quad\nabla_{(\alpha}\xi_{{\beta\delta})}=0.$
(3.12)

Let us see what consequences are obtained by using the conditions (3.8) and
the expression for (3.9) for $\xi^{\alpha}$. Acting with ${\nabla}_{\alpha}$
on (3.9) one obtains

$\displaystyle{\nabla}_{\alpha}\xi^{\alpha}=\frac{1}{20}\gamma_{\underline{a}}^{\alpha\beta}\\{{\nabla}_{\alpha},{\nabla}_{\beta}\\}\xi^{\underline{a}}+\frac{1}{5}\gamma_{\underline{a}}^{\alpha\beta}\Omega_{\alpha}{\nabla}_{\beta}\xi^{\underline{a}}+\frac{9}{5}({\nabla}_{\alpha}\xi^{\underline{a}})(\gamma_{\underline{a}}\Omega)^{\alpha}+\frac{9}{5}\xi^{\underline{a}}{\nabla}_{\alpha}(\gamma_{\underline{a}}^{\alpha\beta}\Omega_{\beta})$
(3.13)

Using the equation for ${\nabla}_{\alpha}\xi^{\underline{a}}$ from (3.6), the
anticommutator for the for the first term and
$\Omega_{\alpha}=\frac{1}{4}{\nabla}_{\alpha}\Phi$ one obtains

$\displaystyle{\nabla}_{\alpha}\xi^{\alpha}=\frac{1}{20}\gamma_{\underline{a}}^{\alpha\beta}(\gamma^{\underline{b}}_{\alpha\beta}{\nabla}_{\underline{b}}\xi^{\underline{a}}+\xi^{\underline{b}}R_{\alpha\beta\underline{b}}{}^{\underline{a}})+\frac{1}{5}\gamma_{\underline{a}}^{\alpha\beta}\Omega_{\alpha}(\xi^{\gamma}\gamma^{\underline{a}}_{\gamma\beta}+\xi^{\underline{b}}T_{\beta\underline{b}}{}^{\underline{a}})$
(3.14)
$\displaystyle+\frac{9}{5}(\xi^{\beta}\gamma^{\underline{a}}_{\beta\alpha}+\xi^{\underline{b}}T_{\alpha\underline{b}}{}^{\underline{a}})(\gamma_{\underline{a}}\Omega)^{\alpha}+\frac{9}{40}\xi^{\underline{a}}\gamma_{\underline{a}}^{\alpha\beta}\\{{\nabla}_{\alpha},{\nabla}_{\beta}\\}\Phi.$
(3.15)

Now we use the equation for ${\nabla}_{\underline{a}}\xi^{\underline{b}}$ from
(3.24) below, the Bianchi identity involving
$R_{\alpha\beta\underline{b}}{}^{\underline{a}}$ and note that all the terms
with two factors of $\Omega_{\alpha}$ vanish (because they form a factor of
the form $(\Omega\gamma_{\underline{a}}\Omega)$), we obtain

$\displaystyle{\nabla}_{\alpha}\xi^{\alpha}=16~{}\xi^{\alpha}\Omega^{(s)}_{\alpha}\Rightarrow\sigma=-\xi^{\alpha}\Omega^{(s)}_{\alpha}.$
(3.16)

As we mentioned below equation (3.3) this condition on $\sigma$ is effectively
eliminating the scale connection from $\nabla_{\alpha}$, reducing the
structure group of the symmetry generators. Note that $\Omega_{\underline{a}}$
already vanishes as a consequence of the Berkovits-Howe constraints.

The second condition in (3.8) does not provide information: in fact, using
(3.9) one obtains

$\displaystyle(\gamma^{\underline{abcd}})_{\alpha}{}^{\beta}{\nabla}_{\beta}\xi^{\alpha}=\frac{1}{10}(\gamma^{\underline{abcd}}\gamma^{\underline{e}})^{\alpha\beta}{\nabla}_{\alpha}{\nabla}_{\beta}\xi_{\underline{e}}+\frac{1}{5}(\gamma^{\underline{abcd}}\gamma^{\underline{e}})^{\alpha\beta}\Omega_{\alpha}{\nabla}_{\beta}\xi_{\underline{e}}$
(3.17)
$\displaystyle+\frac{9}{5}(\gamma^{\underline{abcd}})_{\alpha}{}^{\beta}({\nabla}_{\beta}\xi_{\underline{e}})(\gamma^{\underline{e}}\Omega)^{\alpha}+\frac{9}{5}(\gamma^{\underline{abcd}})_{\alpha}{}^{\beta}\xi_{\underline{e}}{\nabla}_{\beta}(\gamma^{\underline{e}}\Omega)^{\alpha}.$
(3.18)

The first term becomes

$\displaystyle-\frac{1}{5}(\gamma^{\underline{abcd}})_{\alpha}{}^{\beta}
{\nabla}_{\beta}\xi^{\alpha}-\frac{8}{5}(\xi\gamma^{\underline{abcd}}\Omega)+\frac{9}{5}\xi^{[\underline{a}}(\gamma^{\underline{bcd}]})^{\alpha\beta}{\nabla}_{\alpha}\Omega_{\beta}+\frac{54}{5}\xi^{[\underline{a}}(\Omega\gamma^{\underline{bcd}]}\Omega)$
(3.19)

after anticommuting the fermionic derivatives, using the equation for
${\nabla}_{\alpha}\xi_{\underline{e}}$ derived from (3.6) and the Bianchi
identity involving $R_{\alpha\beta\underline{e}f}$. For the second term, after
using the equation for ${\nabla}_{\beta}\xi_{\underline{e}}$ from (3.6) one
obtains that this term is equal to

$\displaystyle-2(\xi\gamma^{\underline{abcd}}\Omega)-\frac{18}{5}\xi^{[\underline{a}}(\Omega\gamma^{\underline{bcd}]}\Omega).$
(3.20)

For the third term, we use again the equation for
${\nabla}_{\beta}\xi_{\underline{e}}$ to get

$\displaystyle\frac{18}{5}(\xi\gamma^{\underline{abcd}}\Omega)-\frac{54}{5}\xi^{[\underline{a}}(\Omega\gamma^{\underline{bcd}]}\Omega).$
(3.21)

For the fourth term we just contract the gamma matrices to obtain

$\displaystyle-\frac{9}{5}\xi^{[\underline{a}}(\gamma^{\underline{bcd}]})^{\alpha\beta}{\nabla}_{\alpha}\Omega_{\beta}-\frac{18}{5}\xi^{[\underline{a}}(\Omega\gamma^{\underline{bcd}]}\Omega).$
(3.22)

Adding (3.19), (3.20), (3.21) and (3.22) we obtain that

$\displaystyle(\gamma^{\underline{abcd}})_{\alpha}{}^{\beta}{\nabla}_{\beta}\xi^{\alpha}=-\frac{1}{5}(\gamma^{\underline{abcd}})_{\alpha}{}^{\beta}
{\nabla}_{\beta}\xi^{\alpha}\Rightarrow(\gamma^{\underline{abcd}})_{\alpha}{}^{\beta}{\nabla}_{\beta}\xi^{\alpha}=0.$
(3.23)

Therefore the second equation in (3.8) is identically satisfied.

The vanishing of $\delta\nabla_{\underline{a}}$ should be implied by
$\delta\nabla_{\alpha}=0$. Nevertheless, it is still useful to have its
explicit form

$\displaystyle\delta\nabla_{\underline{a}}=$
$\displaystyle\left(-\nabla_{\underline{a}}\xi^{\alpha}{-}\xi^{\underline{c}}T_{\underline{ca}}{}^{\alpha}\right)\nabla_{\alpha}+\left(-\nabla_{\underline{a}}\xi^{\underline{b}}{-}\xi^{\alpha}T_{\alpha\underline{a}}{}^{\underline{b}}{-}\xi^{\underline{c}}T_{\underline{ca}}{}^{\underline{b}}{-}\Lambda_{\underline{a}}{}^{\underline{b}}\right)\nabla_{\underline{b}}$
(3.24)
$\displaystyle+\frac{1}{2}\left(-\nabla_{\underline{a}}\Lambda^{\underline{bc}}+\xi^{D}R_{D\underline{a}}{}^{\underline{bc}}\right)M_{\underline{bc}}+\left(\xi^{C}R_{C\underline{a}}^{(s)}-\nabla_{\underline{a}}\sigma\right)S=0.$
(3.25)

One can use the equations in (3.6) to derive each term above. For example,
consider first $\\{\nabla_{\alpha},\nabla_{\beta}\\}\xi^{A}$, which is equal
to

$\displaystyle\\{\nabla_{\alpha},\nabla_{\beta}\\}\xi^{A}=\gamma^{\underline{a}}_{\alpha\beta}\nabla_{\underline{a}}\xi^{A}+\xi^{B}R_{\alpha\beta
B}{}^{A},$ (3.26)

and use the equations for $\xi$ and $\Lambda$ derived from
$\delta\nabla_{A}=0$ in (3.6) and (3.24). For $\xi^{\underline{a}}$, (3.26)
implies

$\displaystyle-\frac{1}{4}\Lambda_{\underline{bc}}[\gamma^{\underline{bc}},\gamma^{\underline{a}}]_{\alpha\beta}+\xi^{\underline{b}}\left(\nabla_{(\alpha}T_{\beta)\underline{b}}{}^{a}+T_{(\alpha\underline{b}}{}^{\underline{c}}T_{\beta)\underline{c}}{}^{\underline{a}}\right)+\xi^{\gamma}\left(2\gamma^{\underline{a}}_{\gamma(\alpha}\Omega_{\beta)}+\gamma^{\underline{b}}_{\gamma(\alpha}T_{\beta)\underline{b}}{}^{\underline{a}}\right)-2\sigma\gamma^{\underline{a}}_{\alpha\beta}$
(3.27)
$\displaystyle=-\gamma^{\underline{b}}_{\alpha\beta}\Lambda_{\underline{b}}{}^{\underline{a}}+\xi^{\underline{b}}\left(R_{\alpha\beta\underline{b}}{}^{\underline{a}}-\gamma^{\underline{c}}_{\alpha\beta}T_{\underline{bc}}{}^{\underline{a}}\right)-\xi^{\gamma}\gamma^{\underline{b}}_{\alpha\beta}T_{\gamma\underline{b}}{}^{\underline{a}}.$
(3.28)

The terms with $\Lambda$ cancel after computing the commutator. The term with
$\xi^{\underline{b}}$ is also zero because of the Bianchi identity involving
$R_{\alpha\beta\underline{b}}{}^{\underline{a}}$. The terms with
$\xi^{\gamma}$ are equal to

$\displaystyle\xi^{\gamma}\left(2\gamma^{\underline{a}}_{\gamma(\alpha}\Omega_{\beta)}+\gamma^{\underline{b}}_{(\alpha\beta}T_{\gamma)\underline{b}}{}^{\underline{a}}\right)=-2\gamma^{\underline{a}}_{\alpha\beta}(\xi^{\gamma}\Omega_{\gamma}),$
(3.29)

where we have used the fact that
$T_{\beta\underline{b}}{}^{\underline{a}}=2(\gamma_{\underline{b}}{}^{\underline{a}}\Omega)_{\beta}$
and the Fierz identity
$\gamma^{\underline{b}}_{(\gamma\alpha}(\gamma_{\underline{b}})_{\beta)\rho}=0$.
The result of (3.29) has to cancel the term with $\sigma$ in (3.28).
Therefore,

$\displaystyle\sigma=-\xi^{\gamma}\Omega_{\gamma},$ (3.30)

which is a condition already found before (see (3.16)).

As we already mentioned before, the scalar superfield $\Phi$ should also be
invariant under the transformations generated by $K$. Since
$\xi^{\gamma}\Omega_{\gamma}=-\sigma$ and
$\Omega_{\alpha}=\frac{1}{4}{\nabla}_{\alpha}\Phi$, invariance under $K$
implies that

$\displaystyle\xi^{\underline{a}}\nabla_{\underline{a}}\Phi=4\sigma.$ (3.31)

Finally, one can calculate the commutator of two different Killing
transformations parametrized by
$(\xi^{A}_{1},\Lambda^{\underline{ab}}_{1},\sigma_{1})$ and
$(\xi^{A}_{2},\Lambda^{\underline{ab}}_{2},\sigma_{2})$

$\displaystyle
K_{1}=\xi_{1}^{A}\nabla_{A}+\frac{1}{2}\Lambda_{1}^{\underline{ab}}M_{\underline{ab}}+\sigma_{1}S,\quad
K_{2}=\xi_{2}^{A}\nabla_{A}+\frac{1}{2}\Lambda_{2}^{\underline{ab}}M_{\underline{ab}}+\sigma_{2}S.$
(3.32)

Both $K$’s leave $\nabla_{A}$ invariant, so the parameters $\xi$ and $\Lambda$
satisfy the equations the come from $[K,\nabla_{A}]=0$. The commutator
$K_{3}=[K_{1},K_{2}]$ is then given by

$\displaystyle
K_{3}=\left(\xi_{1}^{B}\nabla_{B}\xi_{2}^{A}-\xi_{2}^{B}\nabla_{B}\xi_{1}^{A}-\xi_{2}^{B}\xi_{1}^{C}T_{CB}{}^{A}+\xi^{\underline{b}}_{[1}\Lambda_{2]\underline{b}}{}^{\underline{a}}\delta_{\underline{a}}^{A}+\frac{1}{4}(\gamma_{\underline{ab}})_{\alpha}{}^{\beta}\xi^{\alpha}_{[1}\Lambda_{2]}^{\underline{ab}}\delta_{\beta}^{A}\right)\nabla_{A}$
(3.33)
$\displaystyle+\frac{1}{2}\left(\xi_{2}^{B}\xi_{1}^{A}R_{AB}{}^{\underline{ab}}+\xi_{[1}^{A}\nabla_{A}\Lambda_{2]}^{\underline{ab}}+\Lambda_{1}^{\underline{[ac}}\Lambda_{2\underline{c}}{}^{\underline{b}]}\right)M_{\underline{ab}}+{\left(\xi^{A}_{[1}{\nabla}_{A}\sigma_{2]}\right)S}.$
(3.34)

Using the equations from (3.6) and (3.24) for the derivatives of
$\xi_{1},\xi_{2},\Lambda_{1},\Lambda_{2}$ the expression above can be
simplified to

$\displaystyle
K_{3}=\left(-\xi^{\underline{b}}_{[1}\xi^{\alpha}_{2]}T_{\alpha\underline{b}}{}^{\underline{a}}+\xi^{\alpha}_{1}\xi^{\beta}_{2}\gamma_{\alpha\beta}^{\underline{a}}-\xi^{\underline{b}}_{1}\xi^{\underline{c}}_{2}T_{\underline{cb}}{}^{\underline{a}}\right)\nabla_{\underline{a}}-\xi_{1}^{\underline{a}}\xi_{2}^{\underline{b}}T_{\underline{ab}}{}^{\alpha}\nabla_{\alpha}$
(3.35)
$\displaystyle+\frac{1}{2}\left(\xi_{1}^{A}\xi_{2}^{B}R_{BA}{}^{\underline{ab}}+\Lambda_{1}^{\underline{ac}}\Lambda_{2\underline{c}}{}^{\underline{b}}\right)M_{\underline{ab}}$
(3.36)
$\displaystyle=-\xi_{1}^{B}\xi_{2}^{C}T_{CB}{}^{A}\nabla_{A}+\frac{1}{2}\left(\xi_{1}^{A}\xi_{2}^{B}R_{BA}{}^{\underline{ab}}+\Lambda_{1}^{\underline{ac}}\Lambda_{2\underline{c}}{}^{\underline{b}}\right)M_{\underline{ab}}+\left(\xi^{A}_{[1}{\nabla}_{A}\sigma_{2]}\right)S$
(3.37)

So we finally obtain that

$\displaystyle\xi^{A}_{3}=-\xi_{1}^{B}\xi_{2}^{C}T_{CB}{}^{A},\quad\Lambda_{3}^{\underline{ab}}=\frac{1}{2}\left(\xi_{1}^{A}\xi_{2}^{B}R_{BA}{}^{\underline{ab}}-\Lambda_{1}^{\underline{c}[\underline{a}}\Lambda_{2\underline{c}}{}^{\underline{b}]}\right),\quad\sigma_{3}=\left(\xi^{A}_{[1}{\nabla}_{A}\sigma_{2]}\right).$
(3.38)

For the heterotic string it is also possible to include a gauge field
background. The superfields present in the sigma model are
$(\mathbf{A}_{B},\mathbf{W}^{\alpha},\mathbf{U},\mathbf{U}^{\underline{ab}})$.
They are valued in the Lie algebra of $E_{8}\times E_{8}$ or $SO(32)$ and
$\mathbf{A}_{B}$ is the usual gauge superfield. The Berkovits-Howe constraints
imply that they can be written in terms of $\mathbf{A}_{\alpha}$. One could
include the gauge superfield in the covariant derivatives together with a
compensating gauge transformation in $K$ and include their contributions to
(3.6) and (3.24). However, a gauge background is better described by its
dimension $\frac{3}{2}$ covariant field strength $\mathbf{W}^{\alpha}$ and its
transformation under $K$ is

$\displaystyle\delta\mathbf{W}^{\alpha}=\xi^{A}\nabla_{A}\mathbf{W}^{\alpha}+\frac{1}{4}\Lambda^{\underline{ab}}\left(\gamma_{\underline{ab}}\mathbf{W}\right)^{\alpha},$
(3.39)

where the $\nabla_{A}$ above includes the gauge connection. In what follows we
will focus purely on the geometry and will not include a gauge background.

###  3.1 Flat superspace

In flat space $\nabla_{A}\Phi=0$, the solution to (3.11) looks like

$\displaystyle\xi^{\underline{a}}=\epsilon^{\underline{a}}+\Lambda^{\underline{ab}}x_{\underline{b}}+(\theta\gamma^{\underline{a}}\eta)-\frac{1}{4}(\theta\gamma^{\underline{abc}}\theta)\Lambda_{\underline{bc}}.$
(3.40)

where $\epsilon^{\underline{a}}$ is the translation parameter,
$\Lambda^{\underline{ab}}$ is the Lorentz rotation and $\eta^{\alpha}$ is the
supersymmetry parameter. There are no higher order $\theta$ terms in the
expansion. One way to see this is to note that the first term in the second
line of (3.6) in flat space implies that $\Lambda^{\underline{ab}}$ is
constant. It also implies that $\nabla_{\delta}\nabla_{\alpha}\xi^{\beta}=0$,
which means $\xi^{\beta}$ is constant in $x$ and that
$\nabla_{[\alpha_{1}}\cdots\nabla_{\alpha_{n}]}\xi^{\beta}\big{|}_{\theta=0}=0$
for $n\geq 2$ so $\xi^{\beta}$ is at most linear in $\theta$. This is the
complete set of isometries of flat ten dimensional superspace.

We would like to know how the existence of (3.40) implies a flat ten
dimensional superspace. First we have to define the basic properties of
$\xi^{\underline{a}}$ in an appropriate way if the superspace is curved. In
this case there is no notion of a reparametrization invariant $\theta$
expansion. The higher components of $\xi^{\underline{a}}$ should be defined as
an expansion using the Grassmann odd covariant derivatives. Using the
$\Theta$-variable notation of Wess-Bagger [17] we can represent the $\theta$
expansion of $\xi^{\underline{a}}$ as

$\displaystyle\xi^{\underline{a}}=\xi^{\underline{a}}\big{|}_{\theta=0}+\Theta^{\alpha}(\nabla_{\alpha}\xi^{\underline{a}})\big{|}_{\theta=0}+\frac{1}{2}\Theta^{\beta}\Theta^{\alpha}(\nabla_{[\alpha}\nabla_{\beta]}\xi^{\underline{a}})\big{|}_{\theta=0}+\cdots$
(3.41)

where $\cdots$ are higher order $\Theta$ terms. For flat superspace we could
say $\epsilon^{\underline{a}}$, $\Lambda^{\underline{ab}}$ and $\eta^{\alpha}$
were constants, but that is a frame dependent notion. For a general curved
superspace we will impose that the components of $\xi^{\underline{a}}$ satisfy

$\displaystyle(\xi^{\underline{b}})\big{|}_{\theta=0}=\epsilon^{\underline{b}},\quad(\nabla_{\underline{a}}\epsilon^{\underline{b}})\big{|}_{\theta=0}=-\Lambda_{\underline{a}}{}^{\underline{b}},$
(3.42)
$\displaystyle(\nabla_{\alpha}\xi^{\underline{b}})\big{|}_{\theta=0}=\gamma^{\underline{b}}_{{\alpha\beta}}\eta^{\alpha},\quad(\nabla_{\underline{a}}\eta^{\alpha})\big{|}_{\theta=0}=0,$
(3.43)
$\displaystyle(\nabla_{\underline{a}}(\nabla_{[\alpha}\nabla_{\beta]}\xi^{\underline{a}})\big{|}_{\theta=0})\big{|}_{\theta=0}=0,$
(3.44)

where $\Lambda_{\underline{ab}}=-\Lambda_{\underline{ba}}$ and also that

$\displaystyle(\nabla_{[\alpha}\nabla_{\beta]}\xi^{\underline{a}})\big{|}_{\theta=0}=-\frac{1}{4}\gamma_{{\alpha\beta}}^{\underline{abc}}\Lambda_{\underline{bc}},$
(3.45)

is the same $\Lambda$ defined by (3.42). Because of the nested $\theta=0$
projections it is difficult to work with the conditions above. However, it is
possible to simplify them in the case where the gravitino (and dilatino)
vanishes. The bosonic covariant derivative is defined as

$\displaystyle\nabla_{\underline{a}}=E_{\underline{a}}{}^{\underline{m}}(x,\theta)\nabla_{\underline{m}}+E_{\underline{a}}{}^{\mu}(x,\theta)\nabla_{\mu}=(e_{\underline{a}}{}^{\underline{m}}(x)+\cdots)\nabla_{\underline{m}}+(\psi_{\underline{a}}{}^{\mu}(x)+\cdots)\nabla_{\mu}.$
(3.46)

where $\cdots$ are higher order $\theta$ terms. Since $\nabla_{\mu}$ is the
only operator that has a $\theta$ derivative, when
$\psi_{\underline{a}}{}^{\mu}=0$ we can write

$\displaystyle(\nabla_{\underline{a}}({\mathcal{O}}\big{|}_{\theta=0}))\big{|}_{\theta=0}=(\nabla_{\underline{a}}{\mathcal{O}})\big{|}_{\theta=0}.$
(3.47)

this allows us to simplify the projections in the conditions for the
components of $\xi^{\underline{a}}$ and use them in (3.6) and (3.24). In
summary, if we want to impose Poincaré invariance in the full ten dimensional
superspace we require that the background is invariant under transformations
generated by ten linearly independent vectors defined by
$\xi^{\underline{a}}|_{\theta=0}$ satisfying
$\nabla_{\underline{a}}\xi^{\underline{b}}|_{\theta=0}=-\Lambda_{\underline{a}}{}^{\underline{b}}|_{\theta=0}$,
with the further conditions that $\Lambda_{(\underline{ab})}=0$ and
$\nabla_{\underline{c}}\Lambda_{\underline{ab}}|_{\theta=0}=0$. It is clear
that these conditions in (3.24) automatically imply the space is flat and
without bosonic torsion.

##  4 Invariance conditions

Before we start let us summarize the results of the previous section. There
are only two independent invariance conditions from which all others can be
derived, they are

$\displaystyle\xi^{\alpha}\nabla_{\alpha}\Phi+\xi^{\underline{a}}\nabla_{\underline{a}}\Phi=0,$
(4.1)
$\displaystyle\nabla_{\alpha}\xi^{\underline{b}}=\xi^{\beta}\gamma^{\underline{b}}_{\alpha\beta}+\frac{1}{2}\xi_{\underline{a}}(\gamma^{\underline{ab}})_{\alpha}{}^{\beta}\nabla_{\beta}\Phi.$
(4.2)

We already saw that the second equation can be written purely in terms of
$\xi^{\underline{a}}$ and $\Phi$, the same can be done with the first and it
is simply

$\displaystyle\gamma_{\underline{a}}^{\alpha\beta}\nabla_{\alpha}\xi^{\underline{a}}\nabla_{\beta}\Phi+10\xi^{\underline{a}}\nabla_{\underline{a}}\Phi=0.$
(4.3)

All other equations can be obtained from the ones above and the Bianchi
identities. The last equations in both (3.6) and (3.24) can be obtained from

$\displaystyle\nabla_{\alpha}\left(\xi^{\underline{a}}\nabla_{\underline{a}}\Phi-4\sigma\right)=0,\quad\nabla_{\underline{a}}\left(\xi^{\alpha}\nabla_{\alpha}\Phi+4\sigma\right)=0,$
(4.4)

using that $\Omega_{\alpha}=\frac{1}{4}\nabla_{\alpha}\Phi$, the scaling
curvatures $R^{(s)}_{AB}$ in (2.26) and the first lines of (3.6) and (3.24).

If we can find a set of superfields
$(\xi^{A},\Lambda^{\underline{ab}},\sigma)$ such that
$\delta\nabla_{\alpha}=0$ the supergravity background will remain invariant.
The idea is to find the superspace conditions for these equations to be
satisfied for some specific set of parameters
$(\xi^{A},\Lambda^{\underline{ab}},\sigma)$ corresponding to translations in
four dimensions and one global supersymmetry. From now on we will separate
vector indices into four dimensional and six dimensional parts
$\underline{a}=(a,i)$. We will decompose the parameters as

$\displaystyle\xi^{A}=(\xi^{\alpha},\xi^{a},\xi^{i}),\quad\Lambda^{\underline{ab}}=(\Lambda^{ab},\Lambda^{ai},\Lambda^{ij}).$
(4.5)

Later, when imposing four dimensional supersymmetry we will also use complex
indices $(I,\bar{I})$ for the internal space.

###  4.1 Four dimensional Poincaré symmetry

Let us first impose only four dimensional Poincaré invariance. Of course the
conditions on the background are obvious, however it is still instructive to
see how they appear from (4.1) and (4.2). It is not possible in this case to
fix $\xi^{A}=(0,\xi^{a},0)$ as superfield conditions. This would constrain the
full ten-dimensional superspace to be flat. A less restrictive way to do it is
to impose

$\displaystyle\xi^{A}\big{|}_{\theta=0}=(0,\epsilon^{a},0),\quad\Lambda^{{ab}}|_{\theta=0}=(l^{ab},0,0),$
(4.6)

where $\epsilon^{a}$ is a four dimensional vector and $l^{ab}$ a four
dimensional Lorentz rotation. All higher components of the superfields
$(\xi^{A},\Lambda^{{ab}})$ should depend only on $(\epsilon^{a},l^{ab})$ and
the tensors of the geometry. Since we are dealing with a purely bosonic
background we will use that

$\displaystyle\left(\nabla_{\alpha_{1}}\cdots\nabla_{\alpha_{2n+1}}\Phi\right)\Big{|}_{\theta=0}=0.$
(4.7)

Staring from (4.1) we see that the first component of $\Phi$ should
independent of the four dimensional coordinates but it is unconstrained on the
internal coordinates. From using (3.6), (4.6) and (4.7) we fix the next order
in $\theta$ expansion

$\displaystyle\nabla_{\alpha}\xi^{\beta}|_{\theta=0}=-\frac{1}{4}l^{ab}(\gamma_{ab})_{\alpha}{}^{\beta},$
(4.8) $\displaystyle\nabla_{\alpha}\xi^{a}|_{\theta=0}=0,$ (4.9)
$\displaystyle\nabla_{\alpha}\xi^{i}|_{\theta=0}=0,$ (4.10)
$\displaystyle\nabla_{\alpha}\Lambda^{ab}|_{\theta=0}=0,$ (4.11)
$\displaystyle\nabla_{\alpha}\Lambda^{ai}|_{\theta=0}=0,$ (4.12)
$\displaystyle\nabla_{\alpha}\Lambda^{ij}|_{\theta=0}=0.$ (4.13)

To get the second term in the expansion we first use that

$\displaystyle\nabla_{\beta}\nabla_{\alpha}\xi^{A}$
$\displaystyle=\frac{1}{2}[\nabla_{\beta},\nabla_{\alpha}]\xi^{A}+\frac{1}{2}\\{\nabla_{\beta},\nabla_{\alpha}\\}\xi^{A}$
(4.14)
$\displaystyle=\frac{1}{2}[\nabla_{\alpha},\nabla_{\beta}]\xi^{A}+\frac{1}{2}\gamma^{\underline{a}}_{\alpha\beta}\nabla_{\underline{a}}\xi^{A}+\frac{1}{2}\xi^{B}R_{\alpha\beta
B}{}^{A}$ (4.15)

From the second term of (3.6) with $A=\underline{a}$ we get

$\displaystyle\frac{1}{2}[\nabla_{\beta},\nabla_{\alpha}]\xi^{\underline{a}}=$
$\displaystyle-\frac{1}{2}\gamma^{\underline{b}}_{\alpha\beta}\nabla_{\underline{b}}\xi^{\underline{a}}-\frac{1}{2}\xi^{B}R_{\alpha\beta
B}{}^{\underline{a}}+\nabla_{\beta}\xi^{\delta}\gamma_{\alpha\delta}^{\underline{a}}$
(4.16)
$\displaystyle+2\xi^{\delta}\Omega_{\beta}\gamma_{\alpha\delta}^{\underline{a}}-\nabla_{\beta}\xi^{\underline{b}}T_{\underline{b}\alpha}{}^{\underline{a}}-\xi^{\underline{b}}\nabla_{\beta}T_{\underline{b}\alpha}{}^{\underline{a}}.$
(4.17)

Before projecting to $\theta=0$ we must find the consequences of Poincaré
symmetry conditions $\nabla_{a}\xi^{b}|_{\theta=0}=-l_{a}{}^{b}$,
$\nabla_{a}\Lambda^{bc}|_{\theta=0}=0$ from (3.24). The first term in (3.24)
gives no information at leading order in $\theta$. Choosing the indices in the
second term to be $(a,b)$ we find that $T_{abc}|_{\theta=0}=0$. From the
choice $(a,i)$ and using that $(\nabla_{a}\xi^{i})|_{\theta=0}=0$ we have that
$T_{cai}|_{\theta=0}=0$. On the other hand, this will imply that
$(\nabla_{i}\xi^{b})|_{\theta=0}=0$, as expected. Finally, from the choice
$(i,j)$ we have that $T_{cij}|_{\theta=0}=0$. From the third term in (3.24) we
get a vanishing four dimensional curvature $R_{ab}{}^{cd}|_{\theta=0}=0$ and
also that $R_{ab}{}^{cj}|_{\theta=0}=R_{ab}{}^{ij}|_{\theta=0}=0$. Using
Bianchi identities we have that $\nabla_{i}\Lambda^{cd}|_{\theta=0}$ and
$R_{di}{}^{bc}|_{\theta=0}=0$. Using all this information we get the third
term in the covariant theta expansion of $\xi^{a}$ and $\xi^{i}$

$\displaystyle\frac{1}{2}[\nabla_{\beta},\nabla_{\alpha}]\xi^{a}|_{\theta=0}$
$\displaystyle=\left(\frac{1}{2}\gamma^{b}_{\alpha\beta}l_{b}{}^{a}-\frac{1}{2}\epsilon^{b}R_{\alpha\beta
b}{}^{a}-\frac{1}{4}(\gamma^{cd})_{\beta}{}^{\delta}\gamma_{\alpha\delta}^{a}l_{cd}-\epsilon^{b}\nabla_{\beta}T_{{b}\alpha}{}^{a}\right)\Big{|}_{\theta=0}$
(4.18)
$\displaystyle=\frac{1}{4}\gamma^{acd}_{\alpha\beta}l_{cd}-\frac{1}{2}\epsilon^{c}\left(\nabla_{[\alpha}T_{\beta]c}{}^{a}\right)|_{\theta=0},$
(4.19)
$\displaystyle\frac{1}{2}[\nabla_{\beta},\nabla_{\alpha}]\xi^{i}|_{\theta=0}$
$\displaystyle=\frac{1}{4}\gamma^{icd}_{\alpha\beta}l_{cd}-\frac{1}{2}\epsilon^{c}\left(\nabla_{[\alpha}T_{\beta]c}{}^{i}\right)|_{\theta=0}.$
(4.20)

The next steps would be to calculate
$[\nabla_{\alpha},\nabla_{\beta}]\Lambda^{ab}$ and
$\nabla_{[\alpha}\nabla_{\beta}\nabla_{\delta]}\xi^{\sigma}$. For the latter
we first calculate $[\nabla_{\beta},\nabla_{\alpha}]\xi^{\sigma}$ using the
same method as above and then applying another covariant derivative and anti
symmetrizing in all indices. This will give terms depending on the curvature.
Additional constraints on the background from imposing Poincaré symmetry in
four dimensions will come from (4.1). At lowest order in $\theta$ we only have
that

$\displaystyle(\xi^{\alpha}\nabla_{\alpha}\Phi+\xi^{\underline{a}}\nabla_{\underline{a}}\Phi)|_{\theta=0}=\epsilon^{a}(\nabla_{a}\Phi)|_{\theta=0}=0,$
(4.21)

so the first component of $\Phi$, the dilaton, is constant in the four
dimensional variables. For the next order we have to compute

$\displaystyle\nabla_{[\alpha}\nabla_{\beta]}\left(\xi^{\gamma}\nabla_{\gamma}\Phi+\xi^{\underline{a}}\nabla_{\underline{a}}\Phi\right)\Big{|}_{\theta=0}=0.$
(4.22)

Using the conditions on the symmetry parameters and evaluating at $\theta=0$
this expression becomes

$\displaystyle\left(2({\nabla}_{\beta}\xi^{\gamma}){\nabla}_{\alpha}{\nabla}_{\gamma}\Phi-2({\nabla}_{\alpha}\xi^{\gamma}){\nabla}_{\beta}{\nabla}_{\gamma}\Phi+([{\nabla}_{\alpha},{\nabla}_{\beta}]\xi^{\underline{a}}){\nabla}_{\underline{a}}\Phi+\xi^{\underline{a}}[{\nabla}_{\alpha},{\nabla}_{\beta}]{\nabla}_{\underline{a}}\Phi\right)\Big{|}_{\theta=0}=0.$
(4.23)

Using (4.8), (4.14) and (4.16) we simplify to

$\displaystyle\left(\frac{1}{2}l_{ab}(\gamma^{ab})_{[\alpha}{}^{\gamma}({\nabla}_{\beta]}{\nabla}_{\gamma}\Phi)-\frac{1}{2}l_{ab}\gamma^{ab\underline{c}}_{\alpha\beta}{\nabla}_{\underline{c}}\Phi+\epsilon^{b}\left({\nabla}_{[\alpha}T_{\beta]b}{}^{\underline{a}}\right){\nabla}_{\underline{a}}\Phi+\epsilon^{a}[{\nabla}_{\alpha},{\nabla}_{\beta}]{\nabla}_{a}\Phi\right)\Big{|}_{\theta=0}=0.$
(4.24)

Commuting the derivatives in the last term we obtain

$\displaystyle[{\nabla}_{\alpha},{\nabla}_{\beta}]{\nabla}_{a}\Phi|_{\theta=0}=-\left({\nabla}_{[\alpha}T_{\beta]a}{}^{\underline{b}}\right){\nabla}_{\underline{b}}\Phi|_{\theta=0}+{\nabla}_{a}[{\nabla}_{\alpha},{\nabla}_{\beta}]\Phi|_{\theta=0},$
(4.25)

plugging this into (4.24), it becomes

$\displaystyle\frac{1}{2}l_{ab}(\gamma^{ab})_{[\alpha}{}^{\gamma}({\nabla}_{\beta]}{\nabla}_{\gamma}\Phi)|_{\theta=0}-\frac{1}{2}l_{ab}\gamma^{ab\underline{c}}_{\alpha\beta}{\nabla}_{\underline{c}}\Phi|_{\theta=0}+{\nabla}_{a}[{\nabla}_{\alpha},{\nabla}_{\beta}]\Phi|_{\theta=0}=0.$
(4.26)

Now we use the relation

$\displaystyle{\nabla}_{\alpha}{\nabla}_{\beta}\Phi=\frac{1}{2}\gamma^{\underline{a}}_{\alpha\beta}{\nabla}_{\underline{a}}\Phi-\frac{1}{24}\gamma^{\underline{abc}}_{\alpha\beta}H_{\underline{abc}}=\frac{1}{2}\gamma^{\underline{a}}_{\alpha\beta}{\nabla}_{\underline{a}}\Phi-\frac{1}{24}\gamma^{ijk}_{\alpha\beta}H_{ijk},$
(4.27)

to finally obtain

$\displaystyle\left(\frac{1}{4}l_{ab}\\{\gamma^{ab},\gamma^{\underline{c}}\\}_{\alpha\beta}{\nabla}_{\underline{c}}\Phi+\frac{1}{48}l_{ab}H_{ijk}[\gamma^{ab},\gamma^{ijk}]_{\alpha\beta}-\frac{1}{2}(\gamma^{ab\underline{c}})_{\alpha\beta}{\nabla}_{\underline{c}}\Phi-\frac{1}{12}\gamma^{ijk}{\nabla}_{a}H_{ijk}\right)\Big{|}_{\theta=0}=0.$
(4.28)

The first term cancels the third term. The second term is vanishes because
$\gamma^{ab}$ commutes with $\gamma^{ijk}$. Finally we obtain the equation

$\displaystyle({\nabla}_{a}H_{ijk})|_{\theta=0}=0,$ (4.29)

which is the expected condition at mass dimension one from (4.1). The
calculation at fourth order in $\theta$’s is significantly more involved and
will not be presented here

###  4.2 Global four dimensional $N=1$ supersymmetry

We now turn to the conditions imposed by four dimensional global supersymmetry
and the calculation of the corresponding Killing supervector. First we want to
explain the notation we will use. In the breaking of $SO(1,9)$ to
$SO(1,3)\times SO(6)$ the sixteen component spinors will factorize into
$SL(2,{\mathbb{C}})\times SU(4)$ spinors

$\displaystyle\mathbf{16}\to({\mathbf{2}},{\mathbf{4}})+(\bar{\mathbf{2}},\bar{\mathbf{4}}).$
(4.30)

The global supersymmetry parameters are the spinors of $SL(2,{\mathbb{C}})$.
All the spinors considered in this section will be assumed to be a Grassman
odd $SL(2,{\mathbb{C}})$ spinor times a Grassmann even $SU(4)$ spinor

$\displaystyle\tilde{\eta}=\varepsilon\otimes\chi+\bar{\varepsilon}\otimes\bar{\chi}.$
(4.31)

Will also assume that $\chi$ is normalizable such that $\chi\bar{\chi}=1$.
Furthermore, when considering a different spinor $\eta^{\prime}$ we will
assume that only the $SL(2,{\mathbb{C}})$ to be different

$\displaystyle\tilde{\eta}^{\prime}=\varepsilon^{\prime}\otimes\chi+\bar{\varepsilon}^{\prime}\otimes\bar{\chi}.$
(4.32)

This will, for example, imply that

$\displaystyle\eta^{\alpha}\gamma^{a}_{\alpha\beta}\eta^{\prime\beta}=(\varepsilon\gamma^{a}\varepsilon^{\prime}),\quad\eta^{\alpha}\gamma^{i}_{\alpha\beta}\eta^{\prime\beta}=0.$
(4.33)

When we assumed that $\chi$ is normalizable it was already implicit that
$\chi$ is a nowhere vanishing spinor of the internal manifold which means the
internal manifold has $SU(3)$ structure. Locally we can choose a tangent space
basis such that an $\mathfrak{su}(3)$ subalgebra of $so(1,9)$ annihilates
$\eta$. To make that explicitly we will choose complex tangent space indices
for the internal manifold $(I,\bar{I})$. The antisymmetric product of two
gamma matrices can be decomposed as

$\displaystyle\gamma^{\underline{ab}}=(\gamma^{ab},\gamma^{aI},\gamma^{a\bar{I}},\gamma^{IJ},\gamma^{\bar{I}\bar{J}},\gamma^{I\bar{J}},\gamma_{\mathbf{1}}).$
(4.34)

The spinor $\mathfrak{su}(3)$ generators are $\gamma^{I\bar{J}}$ and any
spinor of the type (4.31) will satisfy

$\displaystyle(\gamma^{I\bar{J}})_{\alpha}{}^{\beta}\tilde{\eta}^{\alpha}=0.$
(4.35)

It is clear that we can decompose the real spinor $\tilde{\eta}$ as

$\displaystyle\tilde{\eta}^{\alpha}=\eta^{\alpha}+\bar{\eta}^{\alpha}.$ (4.36)

We will choose the normalization of $\gamma_{\mathbf{1}}$ such that

$\displaystyle\gamma_{\mathbf{1}}\eta=\eta,\quad\gamma_{\mathbf{1}}\bar{\eta}=-\bar{\eta}.$
(4.37)

Finally, the complex pair of spinors $(\eta,\bar{\eta})$ also satisfy

$\displaystyle\gamma^{\bar{I}}\eta=0,\quad\gamma^{I}\bar{\eta}=0.$ (4.38)

With all these set up the Killing supervector and local Lorentz transformation
corresponding to $N=1$ supersymmetry transformations will satisfy

$\displaystyle\xi^{A}|_{\theta=0}=(\tilde{\eta}^{\alpha},0,0),\quad\Lambda^{{ab}}|_{\theta=0}=(0,0,0),\quad\sigma|_{\theta=0}=0.$
(4.39)

Again we can go through the equations imposed by (3.6) and (3.24) finding the
higher components of $(\xi^{A},\Lambda^{{ab}})$ and the conditions imposed in
the geometry. In this case the only consequences from the $\theta=0$
projection coming from (3.24) are

$\displaystyle(\nabla_{\underline{a}}\tilde{\eta}^{\alpha})|_{\theta=0}=0$
(4.40)

It is well known (see, e.g., [23, 24]) that this means it is always possible
to choose a connection with $SU(3)$ holonomy. However it does not yet imply
Ricci flatness since the torsion is not constrained. Furthermore, it is
possible to have curvature terms like $R_{ab}{}^{I\bar{J}}|_{\theta=0}$ with
$\delta_{I\bar{J}}R_{ab}{}^{I\bar{J}}|_{\theta=0}=0$. The next order in the
$\theta$ expansion is given by (3.6)

$\displaystyle\nabla_{\alpha}\xi^{\beta}|_{\theta=0}=0,$ (4.41)
$\displaystyle\nabla_{\alpha}\xi^{\underline{a}}|_{\theta=0}=\gamma^{\underline{a}}_{\alpha\beta}\tilde{\eta}^{\beta},$
(4.42)
$\displaystyle\nabla_{\alpha}\Lambda^{\underline{ab}}|_{\theta=0}=\tilde{\eta}^{\beta}\left(\nabla_{(\alpha}T_{\beta)\underline{c}}{}^{\underline{b}}\Big{|}_{\theta=0}\eta^{\underline{ca}}+\gamma^{\underline{c}}_{\alpha\beta}H_{\underline{c}}{}^{\underline{ab}}\Big{|}_{\theta=0}\right),$
(4.43)
$\displaystyle\nabla_{\alpha}\sigma|_{\theta=0}=\frac{1}{4}\tilde{\eta}^{\beta}\gamma^{\underline{a}}_{\alpha\beta}\nabla_{\underline{a}}\Phi|_{\theta=0}.$
(4.44)

With this information we can compute the first non-trivial supersymmetry
condition from (4.1), which is the supersymmetry transformation for the
dilatino

$\displaystyle\left(-\xi^{\alpha}\nabla_{\beta}\nabla_{\alpha}\Phi+(\nabla_{\beta}\xi^{\underline{a}})\nabla_{\underline{a}}\Phi\right)\Big{|}_{\theta=0}=0.$
(4.45)

Using (4.27) we get

$\displaystyle\tilde{\eta}^{\alpha}\left(\gamma_{\beta\alpha}^{\underline{a}}\nabla_{\underline{a}}\Phi+\frac{1}{12}\gamma_{\beta\alpha}^{\underline{abc}}H_{\underline{abc}}\right)\Big{|}_{\theta=0}=0.$
(4.46)

The different sign from what is usually obtained from the supersymmetry
transformation for the dilatino comes from (2.22). At the next order we have
to compute
$\nabla_{[\beta}\nabla_{\gamma}\nabla_{\delta]}\left(\xi^{\alpha}{\nabla}_{\alpha}\Phi+\xi^{\underline{a}}{\nabla}_{\underline{a}}\Phi\right)|_{\theta=0}$.
This a long and tedious computation. To simplify it we will restrict to the
cases where $\nabla_{\underline{a}}\Phi|_{\theta=0}$ vanishes. First, acting
with ${\nabla}$’s we get

$\displaystyle{\nabla}_{\beta}{\nabla}_{\gamma}{\nabla}_{\delta}\left(\xi^{\alpha}{\nabla}_{\alpha}\Phi+\xi^{\underline{a}}{\nabla}_{\underline{a}}\Phi\right)|_{\theta=0}=\left[-\xi^{\alpha}{\nabla}_{\beta}{\nabla}_{\gamma}{\nabla}_{\delta}{\nabla}_{\alpha}\Phi\right.$
(4.47)
$\displaystyle-\left({\nabla}_{\gamma}{\nabla}_{\delta}\xi^{\alpha}\right){\nabla}_{\beta}{\nabla}_{\alpha}\Phi+\left({\nabla}_{\beta}{\nabla}_{\delta}\xi^{\alpha}\right){\nabla}_{\gamma}{\nabla}_{\alpha}\Phi-\left({\nabla}_{\beta}{\nabla}_{\gamma}\xi^{\alpha}\right){\nabla}_{\delta}{\nabla}_{\alpha}\Phi$
(4.48)
$\displaystyle\left.+\left({\nabla}_{\delta}\xi^{\underline{a}}\right){\nabla}_{\beta}{\nabla}_{\gamma}{\nabla}_{\underline{a}}\Phi-\left({\nabla}_{\gamma}\xi^{\underline{a}}\right){\nabla}_{\beta}{\nabla}_{\delta}{\nabla}_{\underline{a}}\Phi+\left({\nabla}_{\beta}\xi^{\underline{a}}\right){\nabla}_{\gamma}{\nabla}_{\delta}{\nabla}_{\underline{a}}\Phi\right]_{\theta=0},$
(4.49)

here we used (4.7). Note that the terms with an expression like
${\nabla}_{\beta}{\nabla}_{\underline{a}}\Phi$ go away because
${\nabla}_{\beta}{\nabla}_{\underline{a}}\Phi=[{\nabla}_{\beta},{\nabla}_{\underline{a}}]\Phi+{\nabla}_{\underline{a}}{\nabla}_{\beta}\Phi$
vanishes at $\theta=0$. Use something similar with the terms with
${\nabla}_{\beta}{\nabla}_{\gamma}{\nabla}_{\underline{a}}\Phi$,

$\displaystyle{\nabla}_{\beta}{\nabla}_{\gamma}{\nabla}_{\underline{a}}\Phi|_{\theta=0}={\nabla}_{\beta}\left([{\nabla}_{\gamma},{\nabla}_{\underline{a}}]\Phi+{\nabla}_{\underline{a}}{\nabla}_{\gamma}\Phi\right)|_{\theta=0}={\nabla}_{\beta}\left(-T_{\gamma\underline{a}}{}^{\underline{b}}{\nabla}_{\underline{b}}\Phi+{\nabla}_{\underline{a}}{\nabla}_{\gamma}\Phi\right)|_{\theta=0}$
(4.50)
$\displaystyle=\left([{\nabla}_{\beta},{\nabla}_{\underline{a}}]{\nabla}_{\gamma}\Phi+{\nabla}_{\underline{a}}{\nabla}_{\beta}{\nabla}_{\gamma}\Phi\right)|_{\theta=0}={\nabla}_{\underline{a}}{\nabla}_{\beta}{\nabla}_{\gamma}\Phi|_{\theta=0},$
(4.51)

and (4.49) becomes,

$\displaystyle{\nabla}_{\beta}{\nabla}_{\gamma}{\nabla}_{\delta}\left(\xi^{\alpha}{\nabla}_{\alpha}\Phi+\xi^{\underline{a}}{\nabla}_{\underline{a}}\Phi\right)|_{\theta=0}=\left[-\xi^{\alpha}{\nabla}_{\beta}{\nabla}_{\gamma}{\nabla}_{\delta}{\nabla}_{\alpha}\Phi\right.$
(4.52)
$\displaystyle-\left({\nabla}_{\gamma}{\nabla}_{\delta}\xi^{\alpha}\right){\nabla}_{\beta}{\nabla}_{\alpha}\Phi+\left({\nabla}_{\beta}{\nabla}_{\delta}\xi^{\alpha}\right){\nabla}_{\gamma}{\nabla}_{\alpha}\Phi-\left({\nabla}_{\beta}{\nabla}_{\gamma}\xi^{\alpha}\right){\nabla}_{\delta}{\nabla}_{\alpha}\Phi$
(4.53)
$\displaystyle\left.+\left({\nabla}_{\delta}\xi^{\underline{a}}\right){\nabla}_{\underline{a}}{\nabla}_{\beta}{\nabla}_{\gamma}\Phi-\left({\nabla}_{\gamma}\xi^{\underline{a}}\right){\nabla}_{\underline{a}}{\nabla}_{\beta}{\nabla}_{\delta}\Phi+\left({\nabla}_{\beta}\xi^{\underline{a}}\right){\nabla}_{\underline{a}}{\nabla}_{\gamma}{\nabla}_{\delta}\Phi\right]_{\theta=0},$
(4.54)

The first term in the rhs of (4.54) is

$\displaystyle-\frac{1}{32}(\tilde{\eta}\gamma^{\underline{abc}})_{\delta}(\gamma^{\underline{de}}\gamma_{\underline{a}})_{\beta\gamma}R_{\underline{bcde}}-\frac{1}{24\times
48}(\tilde{\eta}\gamma^{\underline{abc}})_{\delta}\gamma^{\underline{def}}_{\beta\gamma}H_{\underline{abc}}H_{\underline{def}}$
(4.55) $\displaystyle+\frac{1}{24\times
16}(\tilde{\eta}\gamma^{\underline{abc}})_{\delta}(\gamma^{\underline{efg}}\gamma_{\underline{ad}})_{\beta\gamma}H_{\underline{bc}}{}^{\underline{d}}H_{\underline{efg}}.$
(4.56)

The last line in (4.54) is

$\displaystyle-\frac{1}{24}(\tilde{\eta}\gamma^{\underline{a}})_{\delta}\gamma^{\underline{bcd}}_{\beta\gamma}{\nabla}_{\underline{a}}H_{\underline{bcd}}+\frac{1}{24}(\tilde{\eta}\gamma^{\underline{a}})_{\gamma}\gamma^{\underline{bcd}}_{\beta\delta}{\nabla}_{\underline{a}}H_{\underline{bcd}}-\frac{1}{24}(\tilde{\eta}\gamma^{\underline{a}})_{\beta}\gamma^{\underline{bcd}}_{\gamma\delta}{\nabla}_{\underline{a}}H_{\underline{bcd}}.$
(4.57)

Using (4.27) we obtain

$\displaystyle{\nabla}_{\alpha}{\nabla}_{\beta}\xi^{\gamma}|_{\theta=0}=-\frac{3}{16}(\tilde{\eta}\gamma^{c})_{\alpha}(\gamma^{\underline{ab}})_{\beta}{}^{\gamma}H_{\underline{abc}}+\frac{1}{96}(\tilde{\eta}\gamma^{\underline{abc}})_{\alpha}\delta_{\beta}^{\gamma}H_{\underline{abc}}-\frac{1}{96}(\tilde{\eta}\gamma_{\underline{abcde}})_{\alpha}(\gamma^{\underline{ab}})_{\beta}{}^{\gamma}H^{\underline{cde}}.$
(4.58)

Using this, the second line in (4.54 )is equal to

$\displaystyle\frac{1}{8\times
16}(\tilde{\eta}\gamma^{\underline{a}})_{\gamma}(\gamma^{\underline{bcd}}\gamma^{\underline{ef}})_{\beta\delta}H_{\underline{aef}}H_{\underline{bcd}}+\frac{3}{32}(\tilde{\eta}\gamma^{\underline{a}})_{\beta}(\gamma^{\underline{bcd}})_{\gamma\delta}H_{\underline{bc}}{}^{\underline{e}}H_{\underline{eda}}$
(4.59) $\displaystyle+\frac{1}{96\times
24}(\tilde{\eta}\gamma^{\underline{abc}})_{\gamma}(\gamma^{\underline{def}})_{\beta\delta}H_{\underline{abc}}H_{\underline{def}}-\frac{1}{96\times
12}(\tilde{\eta}\gamma^{\underline{abc}})_{\beta}(\gamma^{\underline{def}})_{\gamma\delta}H_{\underline{abc}}H_{\underline{def}}$
(4.60) $\displaystyle+\frac{1}{96\times
24}(\tilde{\eta}\gamma_{\underline{abcde}})_{\gamma}(\gamma^{\underline{fgh}}\gamma^{\underline{ab}})_{\beta\delta}H^{\underline{cde}}H_{\underline{fgh}}-\frac{1}{24\times
8}(\tilde{\eta}\gamma_{\underline{abcde}})_{\beta}(\gamma^{\underline{agh}})_{\gamma\delta}H^{\underline{cde}}H_{\underline{gh}}{}^{\underline{b}}.$
(4.61)

Here we have used the identity
$[\gamma_{\underline{ab}},\gamma_{\underline{cde}}]=2\left(\eta_{\underline{a}[\underline{c}}\gamma_{\underline{de}]\underline{b}}-\eta_{\underline{b}[\underline{c}}\gamma_{\underline{de}]\underline{a}}\right)$.
Using all this we obtain

$\displaystyle
6{\nabla}_{[\beta}{\nabla}_{\gamma}{\nabla}_{\delta]}\left(\xi^{\alpha}{\nabla}_{\alpha}\Phi+\xi^{a}{\nabla}_{\underline{a}}\Phi\right)|_{\theta=0}=-\frac{1}{16}(\tilde{\eta}\gamma^{\underline{abc}})_{\\{\beta}(\gamma_{\underline{ade}})_{\gamma\delta\\}}R_{\underline{bc}}{}^{\underline{de}}-\frac{1}{24\times
24}(\tilde{\eta}\gamma^{\underline{abc}})_{\\{\beta}(\gamma^{\underline{def}})_{\gamma\delta\\}}H_{\underline{abc}}H_{\underline{def}}$
(4.62)
$\displaystyle-\frac{1}{64}(\tilde{\eta}\gamma^{\underline{abc}})_{\\{\beta}(\gamma_{\underline{fgd}})_{\gamma\delta\\}}H_{\underline{bc}}{}^{\underline{d}}H_{\underline{a}}{}^{\underline{fg}}+\frac{1}{64}(\tilde{\eta}\gamma^{\underline{abc}})_{\\{\beta}(\gamma_{\underline{fga}})_{\gamma\delta\\}}H_{\underline{bce}}H^{\underline{efg}}$
(4.63)
$\displaystyle-\frac{1}{4}(\tilde{\eta}\gamma^{\underline{a}})_{\\{\beta}(\gamma^{\underline{bcd}})_{\gamma\delta\\}}{\nabla}_{\underline{a}}H_{\underline{bcd}}-\frac{3}{32}(\tilde{\eta}\gamma^{\underline{a}})_{\\{\beta}(\gamma^{\underline{bcd}})_{\gamma\delta\\}}H_{\underline{ad}}{}^{\underline{e}}H_{\underline{ebc}}+\frac{3}{16}(\tilde{\eta}\gamma^{\underline{a}})_{\\{\beta}(\gamma^{\underline{bcd}})_{\gamma\delta\\}}H_{\underline{bc}}{}^{\underline{e}}H_{\underline{eda}}$
(4.64) $\displaystyle-\frac{1}{96\times
4}(\tilde{\eta}\gamma^{\underline{abc}})_{\\{\beta}(\gamma^{\underline{def}})_{\gamma\delta\\}}H_{\underline{aef}}H_{\underline{bcd}}-\frac{1}{96\times
2}(\tilde{\eta}\gamma_{\underline{abcde}})_{\\{\beta}(\gamma^{\underline{agh}})_{\gamma\delta\\}}H^{\underline{cde}}H_{\underline{gh}}{}^{\underline{b}}$
(4.65) $\displaystyle-\frac{1}{24\times
4}(\tilde{\eta}\gamma_{\underline{abcde}})_{\\{\beta}(\gamma^{\underline{agh}})_{\gamma\delta\\}}H^{\underline{cde}}H_{\underline{gh}}{}^{\underline{b}}=0.$
(4.66)

where $\\{\alpha\beta\gamma\\}$ means cyclic permutation. The equation (4.66)
can be simplified to

$\displaystyle(\tilde{\eta}\gamma^{\underline{a}})_{\\{\beta}(\gamma^{\underline{bcd}})_{\gamma\delta\\}}\left(4{\nabla}_{\underline{a}}H_{\underline{bcd}}+\frac{9}{2}H_{\underline{ad}}{}^{\underline{e}}H_{\underline{ebc}}\right)$
(4.67)
$\displaystyle+(\tilde{\eta}\gamma^{\underline{abc}})_{\\{\beta}(\gamma^{\underline{def}})_{\gamma\delta\\}}\left(\eta_{\underline{ad}}R_{\underline{bc}}{}_{\underline{ef}}+\frac{1}{36}H_{\underline{abc}}H_{\underline{def}}+\frac{7}{24}H_{\underline{ade}}H_{\underline{bcf}}-\frac{1}{4}\eta_{\underline{ad}}H_{\underline{bc}}{}^{\underline{g}}H_{\underline{gef}}\right)$
(4.68)
$\displaystyle+\frac{1}{4}(\tilde{\eta}\gamma_{\underline{abcde}})_{\\{\beta}(\gamma^{\underline{agh}})_{\gamma\delta\\}}H^{\underline{cde}}H_{\underline{gh}}{}^{\underline{b}}=0.$
(4.69)

So far the conditions found only imply there exists a global Grassmann odd
symmetry generated by (4.39). In order to have $N=1$ supersymmetry we must
impose that at zero order in $\theta$ the commutator of two Killing vectors
with different parameters in (4.39) generate only a four dimensional
translation. From (3.37) we find

$\displaystyle\tilde{\eta}_{1}^{\alpha}\tilde{\eta}_{2}^{\beta}\left(\nabla_{(\alpha}T_{\beta)\underline{a}}{}^{\underline{b}}+\gamma_{\alpha\beta}^{\underline{c}}H_{\underline{ca}}{}^{\underline{b}}\right)|_{\theta=0}=0,$
(4.70)
$\displaystyle\tilde{\eta}_{1}^{\alpha}\tilde{\eta}_{2}^{\beta}\gamma_{\alpha\beta}^{\underline{a}}\nabla_{\underline{a}}\Phi|_{\theta=0}=0.$
(4.71)

The second equation implies that the dilaton is constant in the four
dimensional variables. The first can be written as

$\displaystyle\tilde{\eta}_{1}^{\alpha}\tilde{\eta}_{2}^{\beta}\left(2\gamma_{\alpha\beta}^{[\underline{a}}\eta^{\underline{b}]i}\nabla_{i}\Phi+\frac{1}{6}\gamma_{\alpha\beta}^{\underline{abcde}}H_{\underline{cde}}\right)\Big{|}_{\theta=0}=0.$
(4.72)

This equation can be combined with one obtained from (4.46) with a spinor
$\tilde{\eta}_{1}$ multiplying it by
$\tilde{\eta}_{2}^{\delta}(\gamma^{\underline{de}})_{\delta}{}^{\beta}$ and
then subtracting it from the same equation but with the order of spinors
reversed we get

$\displaystyle\tilde{\eta}_{1}^{\gamma}\tilde{\eta}_{2}^{\alpha}\left(2[\gamma^{\underline{de}},\gamma^{\underline{a}}]_{\alpha\gamma}\nabla_{\underline{a}}\Phi+\frac{1}{6}\\{\gamma^{\underline{de}},\gamma^{\underline{abc}}\\}_{\alpha\gamma}H_{\underline{abc}}\right)\Big{|}_{\theta=0}=0$
(4.73)

Computing the (anti-)commutators, the above expression together with (4.72)
implies that

$\displaystyle H_{a\underline{bc}}|_{\theta=0}=0.$ (4.74)

It would also be possible to require an $AdS$ super algebra from (4.72) by
requiring it to be proportional to a four dimensional Lorentz rotation,
however it is not possible to have four dimensional $AdS$ solutions without
including a gaugino condensate [25, 14]. Furthermore, without considering a
warp factor in the four dimensional metric, all $H$ flux has to vanish. In
this particular case, for spinors satisfying (4.38) the equation (4.69) will
imply that

$\displaystyle
R_{a\underline{bcd}}=R_{IJ\underline{cd}}=R_{\bar{I}\bar{J}\underline{cd}}=\delta^{I\bar{J}}R_{I\bar{J}\underline{cd}}=0.$
(4.75)

Finally, these conditions together with 4.39 and Bianchi identities imply the
four dimensional space is flat and the internal space is Ricci flat and
Kähler.

Acknowledgements Oc and Bcv would like to thank William D. Linch III for
discussions and commments on the draft and Fondecyt grants 1200342 and 1201550
for partial financial support.

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