specific moduli points as argued before. For a concrete example, when
$N=E_{7}+\Gamma_{\\!\\!(10,2)}$, we choose moduli such that

$\Gamma_{\\!\\!(10,2)}=\Gamma_{\\!\\!(9,1)}+U_{\text{sd}},$ (5.59)

where $\Gamma_{\\!\\!(9,1)}$ is even self-dual and $U_{\text{sd}}$ is the $U$
lattice at the self-dual radius. We then take $P_{N}$ to be purely along
$U_{\text{sd}}$, concretely
$P_{N}=(0^{7},0^{9},\frac{1}{\sqrt{2}};0,\frac{1}{\sqrt{2}})$. We also choose
$P_{I}=(\sqrt{2}\ell,p_{L};p_{R})$, where $p_{L}$ and $p_{R}$ are given in
(5.47). Since $P_{N}^{2}=0$, the same values of $(\ell,j,k)$ in (5.4.3) again
satisfy the level-matching condition (5.52). For the $v_{1}$ shift we then
find

$m^{2}_{R}=m_{L}^{2}=\frac{1}{2}\Big{(}\frac{1}{2R}+\frac{1}{2}R\Big{)}^{2}-\frac{3}{4},$
(5.60)

which is negative for $(\sqrt{3}-1)<\sqrt{2}R<(\sqrt{3}+1)$. Similarly, for
the shift $v_{2}$

$m^{2}_{R}=m_{L}^{2}=\frac{1}{2}\Big{(}\frac{1}{4R}+\frac{1}{2}R\Big{)}^{2}-\frac{1}{2},$
(5.61)

which is negative in the range $(2-\sqrt{2})<2R<(2+\sqrt{2})$. Thus, for $R$
in the above regions, tachyons will appear from $g$-invariant combinations.

####  5.4.4 $s=4M$ examples

Let us first describe the invariant and normal lattices of the examples. They
are obtained by starting at a point in moduli space where

$\Gamma\simeq\Gamma_{0}=U+\widehat{\Gamma}_{(18,2)},\qquad\widehat{\Gamma}_{(18,2)}=(D_{18};D_{2})[(Sp_{18};Sp_{2}),(V_{8};V_{2})].$
(5.62)

As explained in Appendix A, $[(Sp_{18};Sp_{2}),(V_{8};V_{2})]$ are glue
vectors such that the even self-dual $\widehat{\Gamma}_{(18,2)}$ consists of
the correlated classes

$(Sp_{18};Sp_{2})+(Sc_{18};Sc_{2})+(V_{18};V_{2})+(Sp^{\prime}_{18};Sp^{\prime}_{2}).$
(5.63)

Now we define the ${\mathbb{Z}}_{2}$ involution on $\Gamma$ to act only on
$\widehat{\Gamma}_{(18,2)}$ by reflecting the 2 right-movers along $D_{2}$ as
well as $s$ directions along $D_{18}$. The structure of the resulting
invariant and normal lattices can be understood from the decomposition under
$D_{18-s}+D_{s}$ of the $D_{18}$ classes, namely from

$\begin{split}Sp_{18}&=(Sp_{18-s},Sp_{s})+(Sp^{\prime}_{18-s},Sp^{\prime}_{s}),\\\
Sc_{18}&=(Sc_{18-s},Sc_{s})+(V_{18-s},V_{s}),\\\
V_{18}&=(V_{18-s},Sc_{s})+(Sc_{18-s},V_{s}),\\\
Sp^{\prime}_{18}&=(Sp^{\prime}_{18-s},Sp_{s})+(Sp_{18-s},Sp^{\prime}_{s}).\end{split}$
(5.64)

In the invariant lattice we have to set the components along $D_{s}$ and
$D_{2}$ to zero, which is an element of the respective scalar classes. Thus,
besides the inert $U$, $I$ just descends from $(Sc_{18};Sc_{2})$ and only
contains the scalar class, i.e. the root lattice, of $D_{18-s}$. On the other
hand, in the normal lattice the components along $D_{18-s}$ must be set to
zero. From eqs. (5.63) and (5.64) we then read that $N$ is $(D_{s};D_{2})$
with correlated classes $(Sc_{s};Sc_{2})$ and $(V_{s};V_{2})$, or
equivalently, with glue vectors $[(V_{s};V_{2})]$. Summarizing, the resulting
invariant and normal lattices are

$I=U+D_{18-s},\qquad N=(D_{s};D_{2})[(V_{s};V_{2})],$ (5.65)

where $s=4M$, $M=0,1,2,3,4$. The corresponding Nikulin point is
$(r,a,\delta)=(20-4M,2,1)$.

For $s=4M$ the level-matching condition (5.11) requires
$2v^{2}\in{\mathbb{Z}}$ and there are several choices for the shift $v$, as
discussed in section 5.3.1. We will limit ourselves to the simple choice
$(m,n,\alpha)=(0,1,0)$ so that

$v=\frac{1}{4}\left(0,\frac{1}{2R};\frac{1}{2R}\right).$ (5.66)

For $w\in I^{*}/I$ we select $w=({\overbrace{1,0,\dots,0}^{18-s},0;0})$,
belonging to the vector class $V_{18-s}$. Notice that $w^{2}=1$, in agreement
with (5.44).

Gauge group

In these examples the invariant lattice is of the form $I=U+K$, with
$K=D_{18-s}$. Since the shift $v$ has no components along $K$, the gauge group
from $P\in I$ is then $U(1)\otimes SO(36-2s)$, as follows from the general
analysis in section 5.4.1.

To deduce the gauge factors arising from momenta with $P_{N}\not=0$, we need
to find the possibilities with $P_{L}^{2}=P_{IL}^{2}+P_{NL}^{2}=2$, and
$P_{R}=0$, which can be done by looking at the classes in eq. (5.64). When
$P_{I}=0$, it must be that $P_{N}\in N$ belongs either to $(Sc_{s};Sc_{2})$ or
to $(V_{s};V_{2})$, as explained before. Since $P_{R}=0$, necessarily
$P_{NL}\in Sc_{s}$ and the solutions of $P_{NL}^{2}=2$ are the $2s(s-1)$ non-
trivial roots of $D_{s}$. Forming $g$-invariant combinations then gives
$s(s-1)$ states, which correspond to the adjoint of $SO(s)\otimes SO(s)$. When
both $P_{I}$ and $P_{N}$ are not zero, but again $P_{R}=0$, besides the scalar
classes already considered, we are left with $P_{IL}\in V_{18-s}$ and
$P_{NL}\in V_{s}$. Together they give $(36-2s)\times(2s)$ additional solutions
of $P_{L}^{2}=2$, and therefore $s(36-2s)$ $g$-invariant states, which can be
thought to transform as vectors under $SO(s)\otimes SO(36-2s)$. In fact, the
total number of states, with both $P_{N}=0$ and $P_{N}\not=0$, turns out to be
$\left[\tfrac{1}{2}s(s-1)+\tfrac{1}{2}(36-s)(35-s)\right]$, which is precisely
the dimension of $SO(s)\times SO(36-s)$. The number of invariant vector
states, and their transformation properties, indicate that the full gauge
group is $U(1)\otimes SO(36-4M)\otimes SO(4M)$, adding the $U(1)$ from the $U$
lattice. As mentioned in the general discussion, when moving along moduli
space to generic moduli points we expect that the normal lattice contributions
become massive and the gauge group left will be $U(1)\otimes SO(36-8M)$.

Massless fermions

Massless fermions are absent in these examples. The candidates with $P_{IL}\in
V_{18-s}$ and $P_{NL}\in V_{s}$ are projected out because they have $e^{2i\pi
P\cdot(2v+w)}=1$.

Tachyons

In the $g$-twisted sector there are generically tachyons, arising for instance
from states with $N_{L}=N_{R}=0$ and $p=(-1,0,0,0)$, as seen from the masses
given in (5.50). Further setting
$P=(P_{18-s},\frac{k}{2R}+jR;\frac{k}{2R}-jR)$, with $P_{18-s}$ along the
$(18-s)$ invariant directions, and taking $P_{18-s}=0$, yields

$m_{L}^{2}=\tfrac{1}{2}\big{(}\\!\left(k+\tfrac{1}{4}\right)\tfrac{1}{2R}+jR\,\big{)}^{\\!2}-\tfrac{4-M}{4},$
(5.67)

which can be negative for specific values of $R$ when $M<4$. Thus, tachyons
will be present provided there is level-matching $m_{L}^{2}=m_{R}^{2}$.
Indeed, this condition has integer solutions
$j=\frac{3-M}{4k+1}\in{\mathbb{Z}}$, for each value of $M$, e.g. $k=0$,
$j=3-M$.

Let us now inspect the $g^{2}$-twisted sector. For $P=(P_{N},P_{I})$ we take
$P_{N}=(P_{s};P_{2R})$ and $P_{I}=(P_{18-s},\frac{k}{2R}+jR;\frac{k}{2R}-jR)$,
where $P_{s}$ and $P_{2R}$ are respectively along the $s$ left and the two
right reflected directions, and $P_{18-s}$ was defined before. Recall that in
the $(g^{2},1)$ and $(g^{2},g^{2})$ sectors $P\in\Gamma$, while $P\in I$ in
$(g^{2},g)$ and $(g^{2},g^{3})$. The level-matching condition (5.52) leads to

$(k+\tfrac{1}{2})j+\tfrac{1}{2}(P_{18-s}+w)^{2}-\tfrac{1}{2}=-\tfrac{1}{2}P_{s}^{2}+\tfrac{1}{2}P_{2R}^{2},$
(5.68)

which can be satisfied taking $P_{s}=P_{2R}=0$, $P_{18-s}=0$ or
$P_{18-s}=(-1,\underline{\pm 1,0\dots,0})$, and $j=0$. This data implies
$m_{L}^{2}=\frac{1}{2}\big{(}(k+\frac{1}{2})\frac{1}{2R}\big{)}^{\\!2}\\!-\\!\frac{1}{2}$,
therefore tachyons could appear for $R>\frac{1}{4}$. The projection phase
(5.53), present in $(g^{2},g)$ and $(g^{2},g^{3})$ sectors, turns out to be
$e^{-2i\pi\Phi}=-(-1)^{M}$, for $j=0$. Hence, tachyons with $P\in I$ are
invariant for $M$ even but will be projected out for $M$ odd.

Again, it is easy to see that tachyons appear when momenta have non vanishing
components along the normal lattice. For instance, since classes are
correlated we can choose momenta $P_{18-s}\in V_{18-s}$ and $P_{s}\in V_{s}$.
Now we can take $(P_{18-s}+w)=0$ but $P_{s}^{2}=1$ and the level-matching
condition is still verified with $j=0$ and $P_{2R}=0$. Thus, the same values
as above for negative masses are obtained. Since $P_{N}\not=0$, there will be
necessarily $g$-invariant tachyonic states.

####  5.4.5 $s=10$ example

In this example the invariant lattice is chosen to be

$I=U+E_{8},$ (5.69)

corresponding to $(r,a,\delta)=(10,0,0)$. It can be reached starting from a
moduli point where $\Gamma$ takes the form in eq. (5.55). The involution now
acts by reflecting the 2 right-movers as well as the $s=10$ left directions
along $\Gamma_{\\!\\!(10,2)}$. Thus, $N=\Gamma_{\\!\\!(10,2)}$. As shown in
section 5.3.1, item 3, all shifts of order 4 are connected to $v$ given in
(5.31), with $(m,n,\alpha)=(1,2,0)$. Besides, we can set $w=0$, according to
eq. (5.44) since $s=2\,{\rm{mod}}\,4$.

Gauge group

From the general analysis in section 5.4.1, we conclude that for $P\in I$,
since the shift $v$ has no components along the $K$ lattice, the gauge group
is simply the one associated to the root lattice of $K$, namely $U(1)\otimes
E_{8}$. For generic moduli in $N$ this will be the full group, but there are
extra factors for special moduli. For instance, since $\Gamma_{\\!\\!(10,2)}$
is even self-dual, there exist moduli such that
$N=U_{\text{sd}}+U_{\text{sd}}+E_{8}$, where an additional $U(1)^{2}\otimes
SO(16)$ will appear from normal directions. The $U(1)$ factors originate in
the $g$-invariant combinations of charged vectors in $U_{\text{sd}}$, whereas
the $SO(16)$ comes from symmetrized charged $E_{8}$ roots.

Tachyons

In the $g$-twisted sector tachyons arise for particular values of $R$. For
instance, from eq. (5.50) we find that states with $P=0$, $p=(-1,0,0,0)$, and
$N_{L}=N_{R}=0$, are tachyonic when $(5-2\sqrt{6})<R^{2}<(5+2\sqrt{6})$.

In the $g^{2}$-twisted sector it is easy to see that tachyons appear when
$P_{N}\not=0$. Otherwise, for $P_{N}=0$ and
$P_{I}=(\gamma,\frac{k}{2R}+jR;\frac{k}{2R}-jR)$, $\gamma\in E_{8}$, the
level-matching condition (5.52) has solutions with $\gamma=0$ and
$(k,j)=(0,0),(-2,-1)$. The resulting mass is negative for
$(3-2\sqrt{2})<R^{2}<(3+2\sqrt{2})$. Tachyons will be present in this range of
$R$, because the projection phase (5.53) can be checked to be 1.

####  5.4.6 $s=15$ example

The invariant lattice is taken to be $I=U+A_{1}^{3}$, corresponding to
$(r,a,\delta)=(5,3,1)$. The shift $w$ can be chosen as a fundamental weight in
each $A_{1}$ component, thus $w^{2}=\frac{3}{2}$, in agreement with eq. (5.44)
for $s=3\,{\rm{mod}}\,4$. From the general analysis we know that for a shift
vector with no components along $K=A_{1}^{3}$, the resulting gauge group
associated to $I$ is $U(1)\times SU(2)^{3}$. A possible such shift is
$v_{1}=\frac{1}{4}(1,1,0,0,0)$. As discussed in section 5.3.1, item 2, this
$v_{1}$ cannot be connected by boosts to other allowed shifts such as
$v_{2}=\frac{1}{4}(2,2,\sqrt{2},0,0)$. We will not go into a detailed analysis
of the spectrum, which proceeds as in the previous examples.

##  6 Moving in the moduli space of orbifold theories

A remarkable feature of the Nikulin orbifolds that we have constructed is the
existence of a web of transitions between models with different
$(r,a,\delta)$. The fundamental link is a decrease of $s$ by $1$, or
equivalently an increase of $r$ by $1$, and it is realised through shuffling
an $A_{1}$ component in the normal lattice $N$ to the invariant lattice $I$.
The question we would like to pose is: are all the points in figure 1
connected in this way? Namely, starting from one point in the figure, can one
make a sequence of transitions to reach any other point?

The answer to this question turns out to be positive. We will first prove in
subsection 6.1 that the partition functions of two models with $s$ differing
by one — through moving an $A_{1}$ lattice from $N$ to $I$ — indeed agree.
This follows our conformal field theory analysis of the $SU(2)$ flat
connections in section 4. We will see that upon turning on particular moduli
one reaches one point or another. This corresponds to a motion between the
Higgs and Coulomb branches of the moduli space of the theory and will be
analysed in detail.

In subsection 6.2 we will perform a systematic analysis of transitions between
models. We start from $\Gamma_{\\!\\!(19,3)}\simeq U^{3}+E_{8}^{2}$ and apply
the ${\mathbb{Z}}_{2}$ involution such that one or both $E_{8}$ factors are in
the normal lattice. We then shuffle the $A_{1}$ components in the $E_{8}$
lattice(s) from $N$ to $I$. This changes the value of $s$ by $-1$ and so is
referred to as an $s$-transition. The action of the orbifold also applies to
the $U^{3}$ part and will play an important role in our analysis. However, the
$U^{3}$ components come for the ride and are not affected by the shuffling of
the $A_{1}$ factors. We will end this subsection by working out explicit
examples of transitions.

It turns out that all but four points in figure 1 can be constructed using the
method in subsection 6.2. The remaining four points correspond to the triples
$(10,8,0)$, $(10,10,0)$, $(14,6,0)$ and $(18,4,0)$. The triples $(10,8,0)$ and
$(10,10,0)$ have $E_{8}(2)$ components in their normal and invariant lattices.
The $E_{8}(2)$ lattice is obtained by exchanging the two $E_{8}$ factors. The
triples $(14,6,0)$ and $(18,4,0)$ have $U(2)^{2}$ components in their normal
lattice which is obtained by starting from $\Gamma_{\\!\\!(19,3)}\simeq
U^{2}+\Gamma_{(9,1)}+E_{8}$ and defining the proper action of the
${\mathbb{Z}}_{2}$ involution on the even self-dual lattice $\Gamma_{(9,1)}$.
In subsection 6.3 we construct these four models and show how they are reached
by making $s$-transitions from other points.

We therefore conclude that all the 75 models are connected. This seems to
suggest that these theories sit in a big moduli space where each triple
$(r,a,\delta)$ and its associated moduli space form a subspace of it.
Furthermore, these subspaces are connected through $s$-transitions between the
models. There is however a caveat here: as discussed in subsection 5.3, there
are particular models for which not all the shift vectors can be connected
through automorphisms of the lattice — see case 2, part (b), in subsection
5.3.1. This therefore raises the question whether the big moduli space is
unique.

Let us consider the model with $I=U+A_{1}^{3}$ which was studied in subsection
5.3.1. We showed that there are two families of shift vectors in this theory
which cannot be connected through turning on a Wilson line. Furthermore, we
showed that for this theory the two families of shift vectors cannot be
connected by any automorphism of the lattice which may in general contain
applying T-dualities and turning on any number of Wilson lines. At this point
one might conclude that the $I=U+A_{1}^{3}$ model sits in two disconnected big
moduli spaces associated with the two families of the shift vectors. There is
however yet another interesting aspect of the $I=U+A_{1}^{3}$ model: using the
methods we develop in this section, we can make a transition from this model
to the theory with $I=U+E_{7}$ and map the two disconnected families of the
shift vectors in $I=U+A_{1}^{3}$ to the shift vectors in the $I=U+E_{7}$
theory. In the latter theory, however, all shift vectors are connected! (See
item $4.$ in subsection 5.3.1). Therefore, the disconnected families of shift
vectors of $I=U+A_{1}^{3}$ are connected through making the $s$-transition to
$I=U+E_{7}$. This means that the existence of a unique big moduli space is
still conceivable: while in a particular subspace of it there may be regions
which are disconnected in that subspace, these regions are individually
connected to another subspace within which all theories are connected. Thus in
this way the two families of the shift vectors are indeed connected in the big
moduli space.

This reasoning however applies only to one model. In order to draw a definite
conclusion on the existence of a unique big moduli space one needs to analyse
all theories with disconnected families of shift vectors. We plan to study
this further in future work.

Table LABEL:tab_IN summarises the invariant and normal lattices that we will
construct in this section for all the triples in figure 1. We stress that the
lattices in this table are found at a particular point in moduli space.

###  6.1 Changing $s$

To clarify the main idea let us first consider the $(1,g)$ sector. In the
partition function $Z_{0,1}$, given in eq. (5.7), we can identify separate
contributions from non-compact bosons, fermions, the lattice $I$ and the
lattice $N$. In the $N$ contribution, denoted $Z_{N}(1,g)$, the lattice sum is
absent and only oscillator terms enter because the $s$ left-moving and the two
right-moving normal directions are reflected. More precisely,

$Z_{N}(1,g)=\left(\frac{2\eta}{\vartheta_{2}}\right)^{\\!\\!\frac{s}{2}}\\!\\!\left(\frac{2\bar{\eta}}{\bar{\vartheta}_{2}}\right)=\left(\frac{2\eta}{\vartheta_{2}}\right)^{\\!\\!\frac{s-1}{2}}\\!\\!\left(\frac{2\bar{\eta}}{\bar{\vartheta}_{2}}\right)\,\frac{1}{\eta}\sum_{P\in
A_{1}}q^{\frac{1}{2}P^{2}}e^{2\pi iP\cdot v_{1}}\ ,$ (6.1)

where $v_{1}$ has components $(v_{1L};v_{1R})=(\frac{1}{2\sqrt{2}};0)$ and
$2v^{2}_{1}=\frac{1}{4}$. In the second equality we used the identities

$\left(\frac{2{\eta}}{{\vartheta}_{2}}\right)^{\\!\\!\frac{1}{2}}=\frac{1}{q^{\frac{1}{24}}\prod\limits_{n}(1+q^{n})}=\frac{1}{\eta}\sum_{n}q^{n^{2}}e^{i\pi
n}=\frac{1}{\eta}\sum_{P\in A_{1}}q^{\frac{1}{2}P^{2}}e^{2\pi iP\cdot v_{1}}\
.$ (6.2)

We also used the fact that $P\in A_{1}$ means $P=\sqrt{2}n$. The main take
from the second form of $Z_{N}(1,g)$ in eq. (6.1) is that a decrease of $s$ by
1 is accompanied by the emergence of a lattice sum over $A_{1}$. In turn this
lattice sum can be absorbed in the contribution of the invariant lattice
thereby increasing $r$ by 1. The analysis further shows that if the mother
theory characterized by $s$ has shift $v$, then the daughter theory with
$(s-1)$ will have shift $(v+v_{1})$. A consistency check is that
$2(v+v_{1})^{2}+\frac{s-1}{4}=2v^{2}+\frac{s}{4}$, implying that if the mother
theory satisfies the level-matching condition (5.11) so does its daughter.

So far we have argued that the partition functions of the mother and daughter
theories do match in the $(1,g)$ sector. To match the full partition functions
of the two theories, it will be enough to consider the $(1,g)$ and $(1,g^{2})$
sectors since all other sectors are obtained by modular transformations of
these two sectors. Matching in the $(1,g^{2})$ sector, however, is more
involved because we need to specify how an explicit $A_{1}$ arises in the
original lattice $N$. We will shortly elaborate on this using particular
examples in which we will also provide an interpretation of the underlying
physics.

As explained in section 5.3, the smallest value of $r$ for which a shift $v$
satisfying level matching exists is $r=2$, with two possibilities for $a$. We
will first show how models with $r\geq(a+2)$ and $\delta=1$ can be obtained
starting from $(2,0,0)$. The models with $r=a$ and/or $\delta=0$ will be
discussed later.

Let us now examine the transitions starting from $(2,0,0)$. At this point we
have $I=U$, whereas $N$ is a generic even self-dual lattice333 We denote even
self-dual lattices of signature $(8j+d,d)$ by $\Gamma_{\\!\\!(8j+d,d)}$.
$\Gamma_{\\!\\!(18,2)}$. We take $N=\Gamma_{\\!\\!(10,2)}+E_{8}$, but other
choices such as $N=U+\Gamma_{\\!\\!(17,1)}$ can also be made. The next step is
to find an $A_{1}$ inside $N$ in order to go from $r=2$ to $r=3$. The natural
candidate in this example is an $A_{1}$ inside $E_{8}$. To proceed we then
decompose $E_{8}$ in terms of $E_{7}+A_{1}$. Vectors in the $E_{8}$ lattice
split as $(R_{7},R_{1})+(F_{7},F_{1})$. Here $R_{7}$ and $R_{1}$ denote the
root lattices themselves, renamed for clarity, whereas $F_{7}$ and $F_{1}$
refer to weights in the conjugacy classes of ${\bf{56}}$ of $E_{7}$ and
${\bf{2}}$ of $A_{1}$ respectively.

In the $r=2$ theory the partition function in the $(1,g)$ sector can be
written as

$Z_{0,1}=\hat{Z}_{0,1}\left(\frac{2{\eta}}{{\vartheta}_{2}}\right)^{\\!\\!\frac{1}{2}},$
(6.3)

where $\hat{Z}_{0,1}$ represents the partition function of all the remaining
CFT, i.e. non-compact bosons, fermions, lattice $I=U$, and the remaining
pieces $\Gamma_{\\!\\!(10,2)}$ and $E_{7}$ in $N$. The $I$ part includes the
dependence on the shift $v$, which is assumed to fulfill level matching
$2v^{2}+\frac{s}{4}\in{\mathbb{Z}}$ for $s=18$. The detailed form of
$\hat{Z}_{0,1}$, which can be read from (5.7), is not needed because it does
not change in our method to go from $r=2$ to $r=3$. The remaining term
$\left(2{\eta}/{\vartheta}_{2}\right)^{\\!\\!\frac{1}{2}}$ is precisely the
partition function of the oscillator along $A_{1}$, which being part of $N$,
does not come with a lattice sum. We now use the identity (6.2) to obtain

$Z_{0,1}=\hat{Z}_{0,1}\,\times\,\frac{1}{\eta}\sum_{P\in
A_{1}}\bar{q}^{\frac{1}{2}P^{2}}e^{2\pi iP\cdot v_{1}}.$ (6.4)

Here $v_{1}$ has right component zero and left component equal to half the
fundamental weight of $A_{1}$, i.e. as before
$(v_{1L},v_{1R})=(\frac{1}{2\sqrt{2}},0)$ and $2v_{1}^{2}=\frac{1}{4}$.
Equation (6.4) is exactly the partition function for the $(3,1,1)$ model where
$I=U+A_{1}$ and $N=\Gamma_{\\!\\!(10,2)}+E_{7}$. The new shift is $v+v_{1}$.
As proven in general before (see below eq. (6.2)), since $v$ satisfies level
matching for $s=18$ so does $v+v_{1}$ for $s=17$.

We now look at the $(1,g^{2})$ sector for $(2,0,0)$. In this sector we must
take into account the decomposition of $E_{8}\subset N$ in terms of
$E_{7}+A_{1}$ because the partition function $Z_{0,2}$ effectively involves a
sum over $N$. We then write

$Z_{0,2}=\hat{Z}_{0,2}^{R_{7}}\times\frac{1}{\eta}\sum_{P\in
R_{1}}{q}^{\frac{1}{2}P^{2}}+\hat{Z}_{0,2}^{F_{7}}\times\frac{1}{\eta}\sum_{P\in
F_{1}}{q}^{\frac{1}{2}P^{2}}.$ (6.5)

Here $\hat{Z}_{0,2}^{R_{7}}$ is the partition function of all other degrees of
freedom in the CFT, including the piece from $E_{7}$ in the class $R_{7}$ that
is correlated with the $A_{1}$ class $R_{1}$. Analogously,
$\hat{Z}_{0,2}^{F_{7}}$ contains the part from $E_{7}$ in the class $F_{7}$
correlated with the $A_{1}$ class $F_{1}$. Explicit expressions for
$\hat{Z}_{0,2}^{R_{7}}$ and $\hat{Z}_{0,2}^{F_{7}}$ are not necessary as they
remain unchanged in the transition from $r=2$ to $r=3$. Note also that for the
$(2,0,0)$ model $w$ is trivial since $I^{*}_{o}$ is null for $I=U$ — see above
eq. (5.12).

In the $(3,1,1)$ model with $I=U+A_{1}$ and $N=\Gamma_{\\!\\!(10,2)}+E_{7}$,
$Z_{0,2}$ also includes sums over the two classes in the weight lattice of
$A_{1}$, which are correlated with the two classes in the weight lattice of
$E_{7}$. In fact, in $(3,1,1)$ we have

$Z_{0,2}=\hat{Z}_{0,2}^{R_{7}}\times\frac{1}{\eta}\sum_{P\in
R_{1}}{q}^{\frac{1}{2}P^{2}}e^{2\pi
iP\cdot(2v_{1}+w_{1})}+\hat{Z}_{0,2}^{F_{7}}\times\frac{1}{\eta}\sum_{P\in
F_{1}}{q}^{\frac{1}{2}P^{2}}e^{2\pi iP\cdot(2v_{1}+w_{1})}$ (6.6)

where $v_{1}$ was introduced before and $w_{1}$ is defined as $e^{2i\pi
P^{2}}=e^{2i\pi P\cdot w_{1}}$ for all $P$ in the weight lattice of $A_{1}$.
It follows that $w_{1}$ is the fundamental weight of $A_{1}$ and that
$2v_{1}+w_{1}$ is a root of $A_{1}$. Therefore, $e^{2\pi
iP\cdot(2v_{1}+w_{1})}=1$ for all $P$ in the weight lattice of $A_{1}$. We
thus conclude that equations (6.5) and (6.6) are the same. This proves that in
the $(1,g^{2})$ sector the partition functions of the $(2,0,0)$ and $(3,1,1)$
models are exactly equal.

After showing that the $(2,0,0)$ model gives the same partition function as
the $(3,1,1)$ model let us discuss the physical interpretation. In the
$(3,1,1)$ model, with $I=U+A_{1}$, the $A_{1}$ can be realized by a left-
moving boson denoted $Y$. The Kac-Moody currents are $J_{3}=\partial Y$ and
$J_{\pm}=e^{\pm i\sqrt{2}Y}$. In this case $g$ acts as $Y\to Y+2\pi v_{1}$,
where the shift $v_{1}$ has components $v_{1L}=\frac{1}{2\sqrt{2}}$ and
$v_{1R}=0$. Thus, $J_{3}\to J_{3}$ and $J_{\pm}\to-J_{\pm}$ under $g$. The
exactly marginal operators of the theory are $J_{3}\bar{\partial}\bar{X}_{3}$
and $J_{\pm}\bar{\partial}\bar{X}_{1,2}$, where $\bar{\partial}\bar{X}_{i}$
are the right-moving bosons with $i=3$ being the direction in $I$ and $i=1,2$
being the two right-moving directions in $N$ that are reflected by $g$. The
theory also has $U(1)$ massless gauge fields
$J_{3}\bar{\partial}\bar{X}^{\mu}$ where $\mu$ refers to noncompact space-time
directions.444In the $-1$ ghost picture the operators correspond to $U(1)$
massless gauge fields $J_{3}\bar{\psi}^{\mu}$ as well as massless scalars
$J_{3}\bar{\psi}^{3}$ and $J_{\pm}\bar{\psi}^{1,2}$, where $\bar{\psi}$ are
right-moving fermions.

There are two possible deformations that can be made at this point — see
subsection 4.4 for related discussion. One possibility is to give a vev to
$J_{3}\bar{\partial}\bar{X}_{3}$. This deformation is along what we call the
Coulomb branch because it leaves the $U(1)$ gauge symmetry unbroken. However,
this vev will make $J_{\pm}\bar{\partial}\bar{X}_{1,2}$ massive. This is all
part of the $(3,1,1)$ moduli space. The second possibility is to give a vev to
e.g. $(J_{+}+J_{-})\bar{\partial}\bar{X}_{1}$. This will break the $U(1)$
gauge symmetry and in particular $J_{3}\bar{\partial}\bar{X}_{3}$ will become
massive. We say that this deformation is along the Higgs branch. In order to
solve the Higgs branch at arbitrary points it is convenient to use
rebosonisation to write $(J_{+}+J_{-})=J^{\prime}_{3};=\partial Y^{\prime}$.
On $Y^{\prime}$, $g$ acts as reflection and we get to the moduli space of the
$(2,0,0)$ model, where the $A_{1}$ is part of the $N$ lattice.

Note that in all the above discussion the details of the rest of the CFT did
not play a role. All we needed in the $(2,0,0)$ model, was the appearance of
an $A_{1}$ point in the Higgs branch (the moduli space of $(18,2)$ boosts in
the $N$ direction) so that there is a massless $U(1)$ gauge field and the
corresponding scalar. Once that happens, we can give vev to the latter scalar.
Some of the Higgs branch fields become massive now and we end up with the
moduli space of the $(3,1,1)$ model.

The procedure to go from $(2,0,0$) to $(3,1,1)$ can be repeated to reach
models with larger value of $r$. For example, $(4,2,1)$ can be obtained from
$(3,1,1)$, where $I=U+A_{1}$ and $N=\Gamma_{\\!\\!(10,2)}+E_{7}$, by
decomposing $E_{7}$ in terms of $D_{6}+A^{\prime}_{1}$, where the prime is put
to distinguish from the $A_{1}$ in the previous step. The two $E_{7}$ classes
decompose as $R_{7}=(Sc_{6},R^{\prime}_{1})+(Sp_{6},F^{\prime}_{1})$ and
$F_{7}=(V_{6},F^{\prime}_{1})+(Sp^{\prime}_{6},R^{\prime}_{1})$. In the
$(1,g)$ sector we now split $\hat{Z}_{0,1}$ in eq. (6.4) into an
$A^{\prime}_{1}$ piece and the rest of the CFT and write

$\hat{Z}_{0,1}=\hat{Z^{\prime}}_{0,1}\left(\frac{2{\eta}}{{\vartheta}_{2}}\right)^{\\!\\!\frac{1}{2}}.$
(6.7)

This can be recast in the form of the $(4,2,1)$ model as

$\hat{Z}_{0,1}=\hat{Z^{\prime}}_{0,1}\times\frac{1}{\eta}\sum_{P\in
R^{\prime}_{1}}{q}^{\frac{1}{2}P^{2}}e^{2\pi iP\cdot v_{2}}$ (6.8)

where $v_{2}$ is half the fundamental weight of $A^{\prime}_{1}$. The total
shift becomes $v+v_{1}+v_{2}$, which satisfies the level matching condition
for $r=4$. Substituting back in eq. (6.4) yields

$Z_{0,1}=\hat{Z^{\prime}}_{0,1}\times\frac{1}{\eta^{2}}\sum_{P\in
D_{2}}{q}^{\frac{1}{2}P^{2}}e^{2\pi iP\cdot(v_{1}+v_{2})}$ (6.9)

where we have used the fact that the root lattice of $D_{2}$ is the scalar
class $Sc_{2}$ of $SO(4)$, which is the same as $(R_{1},R^{\prime}_{1})$. This
is exactly the $Z_{0,1}$ for the $(4,2,1)$ model in which $I=U+D_{2}$ and
$N=\Gamma_{\\!\\!(10,2)}+D_{6}$.

In the $(1,g^{2})$ sector, the $\hat{Z}_{0,2}^{R_{7}}$ and
$\hat{Z}_{0,2}^{F_{7}}$ in eq. (6.6) become

$\begin{split}\hat{Z}_{0,2}^{R_{7}}&=\hat{Z^{\prime}}_{0,2}^{Sc_{6}}\times\frac{1}{\eta}\sum_{P\in
R^{\prime}_{1}}{q}^{\frac{1}{2}P^{2}}e^{2\pi
iP\cdot(2v_{2}+w_{2})}+\hat{Z^{\prime}}_{0,2}^{Sp_{6}}\times\frac{1}{\eta}\sum_{P\in
F^{\prime}_{1}}{q}^{\frac{1}{2}P^{2}}e^{2\pi
iP\cdot(2v_{2}+w_{2})}\\\\[5.69054pt]
\hat{Z}_{0,2}^{F_{7}}&=\hat{Z^{\prime}}_{0,2}^{Sp^{\prime}_{6}}\times\frac{1}{\eta}\sum_{P\in
R^{\prime}_{1}}{q}^{\frac{1}{2}P^{2}}e^{2\pi
iP\cdot(2v_{2}+w_{2})}+\hat{Z^{\prime}}_{0,2}^{V_{6}}\times\frac{1}{\eta}\sum_{P\in
F^{\prime}_{1}}{q}^{\frac{1}{2}P^{2}}e^{2\pi
iP\cdot(2v_{2}+w_{2})}\end{split}$ (6.10)

Here $w_{2}$ is the fundamental weight in $A^{\prime}_{1}$. Substituting this
in eq. (6.6) then gives

$\begin{split}&Z_{0,2}=\frac{1}{\eta^{2}}\Big{[}\hat{Z^{\prime}}_{0,2}^{Sc_{6}}\sum_{P\in
Sc_{2}}{q}^{\frac{1}{2}P^{2}}e^{2\pi
iP\cdot(2(v_{1}+v_{2})+(w_{1}+w_{2}))}+\hat{Z^{\prime}}_{0,2}^{Sp_{6}}\sum_{P\in
Sp_{2}}{q}^{\frac{1}{2}P^{2}}e^{2\pi
iP\cdot(2(v_{1}+v_{2})+(w_{1}+w_{2}))}\\\\[5.69054pt]
&\qquad\quad\;\;+\hat{Z^{\prime}}_{0,2}^{V_{6}}\sum_{P\in
V_{2}}{q}^{\frac{1}{2}P^{2}}e^{2\pi
iP\cdot(2(v_{1}+v_{2})+(w_{1}+w_{2}))}+\hat{Z^{\prime}}_{0,2}^{Sp^{\prime}_{6}}\sum_{P\in
Sp^{\prime}_{2}}{q}^{\frac{1}{2}P^{2}}e^{2\pi
iP\cdot(2(v_{1}+v_{2})+(w_{1}+w_{2}))}\Big{]}\end{split}$ (6.11)

where we have used the decompositions $Sc_{2}=(R^{\prime}_{1},R_{1})$,
$V_{2}=(F^{\prime}_{1},F_{1})$, $Sp_{2}=(F^{\prime}_{1},R_{1})$ and
$Sp^{\prime}_{2}=(R^{\prime}_{1},F_{1})$ for the classes of $D_{2}$ in
$A^{\prime}_{1}+A_{1}$. Since $w_{1}$ and $w_{2}$ are fundamental weights of
$A_{1}$ and $A^{\prime}_{1}$, for $D_{2}$ now in $I$, it turns out that
$w=w_{1}+w_{2}$ is in the $V_{2}$ class, in agreement with our general
analysis for $w$ when $I=U+D_{4n+2}$ — see the discussion below eq. (5.12).
Furthermore, all the phases above are trivial since $2v_{1}+w_{1}$ and
$2v_{2}+w_{2}$ are roots of $A_{1}$ and $A^{\prime}_{1}$. Hence, $Z_{0,2}$ is
clearly unchanged except that eq. (6.11) is the expression corresponding to
the $(4,2,1)$ model.

Actually it is clear that at every step when we convert an $A_{1}$ which is in
the $N$ directions, to an $A_{1}$ along $I$ directions there will be
additional contributions to $v$ equal to half fundamental and to $w$ equal to
the fundamental of the latter $A_{1}$. Therefore this will automatically
ensure the level-matching condition (5.11) and the condition (5.19) on $w$.
Moreover, $2v+w$ will be trivial and the form of $Z_{0,2}$ will remain
unchanged.

So far we discussed connecting the first three points on the line $r=a+2$ in
Figure 1. We can continue the procedure to obtain various models with $I\simeq
U+K$ for other values of $r$. However, at first sight it is not clear whether
all models can be connected. As one increases $r$, the first question arises
at the transition from $r=5$ to $r=6$. Concretely, we have to check whether
$(6,4,1)$ and $(6,2,0)$, with $I\simeq U+D_{2}+D_{2}$ and $I\simeq U+D_{4}$
respectively, can both be reached starting from $(5,3,1)$. It is clear that
the first model can be obtained by taking the two $D_{2}=A_{1}+A_{1}$ in two
different $E_{8}$ lattices in $N=\Gamma_{\\!\\!(2,2)}+E_{8}^{2}$. This can be
shown as outlined above. For the second case where $D_{4}$ is embedded in one
$E_{8}$, one might suspect that the above procedure could fail for the
following reason. In the decomposition $D_{4}\rightarrow
A_{1}+A_{1}+A_{1}+A_{1}$, the scalar class $Sc_{4}$ of $D_{4}$ splits as
$(R_{1},R_{1},R_{1},R_{1})+(F_{1},F_{1},F_{1},F_{1})$ . Of course $Z_{0,2}$
will always work, as $2v+w$ in each $A_{1}$ factor is trivial. The problem
would be with $Z_{0,1}$ where from the point of view of the $(6,2,0)$ model
the lattice sum must be in $I$ which includes
$(R_{1},R_{1},R_{1},R_{1})+(F_{1},F_{1},F_{1},F_{1})$ in the $D_{4}$ part
decomposed in terms of $A_{1}+A_{1}+A_{1}+A_{1}$. However, in the above
procedure for every $A_{1}$ that we rebosonize only $R_{1}$ contributes to
$Z_{0,1}$. Of course the solution is trivial. For our procedure of obtaining
$v$, the part of $v$ in $D_{4}$, in the above decomposition, will be
$\frac{1}{2}(F_{1},F_{1},F_{1},F_{1})$, so the contribution of
$(F_{1},F_{1},F_{1},F_{1})$ to $Z_{0,1}$ will vanish because for each $A_{1}$
it involves the Jacobi function $\vartheta_{1}$ that is identically zero.

In the above we did not do anything to $U$. In the whole process of changing
$s$ step by step, $U$ was a spectator. Thus, we can proceed in the same manner
with models that contain $U(2)$ in their invariant lattice without touching
$U(2)$ — see table LABEL:tab_IN. Hence, it seems that all the models in the
$U(2)$ series are also connected, except the one involving $E_{8}(2)$ that
appears only for $r=10$ and that will be analyzed in subsection 6.3.

###  6.2 A systematic way of changing $s$ in $E_{8}$

Can all orbifold models associated with the triples $(r,a,\delta)$ in figure 1
be connected using the method of shuffling $A_{1}$ lattices? To answer this
question we start with the even self-dual lattice
$\Gamma_{\\!\\!(19,3)}\simeq\Gamma_{0}=U^{3}+E_{8}^{2}$. The idea is to focus
on the $E_{8}$ components inside $\Gamma_{0}$. Since changing $s$ involves
shuffling $A_{1}$’s, it is best to decompose $E_{8}$ in terms of $A_{1}^{8}$.
The form of $E_{8}$ vectors in the $A^{8}_{1}$ basis can be deduced in steps
starting from the decompositions of $E_{8}$ under $D_{4}+D_{4}$ with standard
correlated conjugacy classes, namely

$(Sc_{4},Sc_{4})+(V_{4},V_{4})+(Sp_{4},Sp_{4})+(Sp^{\prime}_{4},Sp^{\prime}_{4})\
.$ (6.12)

Next each $D_{4}$ lattice is decomposed into $D_{2}+D_{2}$ with correlated
classes

$\displaystyle Sc_{4}$ $\displaystyle=(Sc_{2},Sc_{2})+(V_{2},V_{2})\
,\qquad\quad\;$ $\displaystyle V_{4}$
$\displaystyle=(Sc_{2},V_{2})+(V_{2},Sc_{2})\ ,$ $\displaystyle Sp_{4}$
$\displaystyle=(Sp_{2},Sp_{2})+(Sp^{\prime}_{2},Sp^{\prime}_{2})\
,\qquad\quad$ $\displaystyle Sp^{\prime}_{4}$
$\displaystyle=(Sp_{2},Sp^{\prime}_{2})+(Sp_{2},Sp^{\prime}_{2})\ .$ (6.13)

Finally, the $D_{2}$ lattice is decomposed into $A_{1}+A_{1}$ with
correlations

$Sc_{2}=(R,R),\quad V_{2}=(F,F),\quad Sp_{2}=(R,F),\quad\quad
Sp^{\prime}_{2}=(F,R)\ ,$ (6.14)

where $R$ and $F$ are the root lattice and the fundamental class of $A_{1}$
(we are dropping a subscript 1 to simplify expressions). The end result is
that the groupings of classes are such that the number of $F$’s is multiple of
4, as required also by the fact that $E_{8}$ is even. More precisely, there is
one each of $R^{8}$ and $F^{8}$, plus 14 possible orderings of $R^{4}F^{4}$.
The latter are such that in any two groupings there is an even number of $F$
overlaps, also needed because $E_{8}$ is integral. One can also check that the
total number of vectors with length square 2 is 240, equal to the number of
$E_{8}$ roots.

Moving $A_{1}$’s from normal to invariant directions is straightforward in
terms of the decomposition under $A^{8}_{1}$. Suppose that we start from a
model where all the directions of one $E_{8}$, say the first, are reflected.
Then $I_{1}$ coming from this $E_{8}$ is null and $N_{1}=E_{8}$. By moving one
$A_{1}$ of this normal lattice to invariant directions we land at
$I_{1}=A_{1}$ and $N_{1}=E_{7}$, which can be seen by decomposing $E_{7}$ in
terms of $A_{1}^{7}$. Besides, the contribution to $a$ from this $E_{8}$,
denoted $a_{1}$, is 1. Thus, after the decrease $\Delta_{s_{1}}=-1$ we have
data $(I_{1},N_{1},a_{1})=(A_{1},E_{7},1)$. We can further move two and three
$A_{1}$’s in a similar way. If we move four $A_{1}$’s to invariant directions
there are two possiblities depending on the relative positions. Moving $s_{1}$
$A_{1}$’s is the same as moving $(8-s_{1})$ $A_{1}$’s but exchanging $I_{1}$
and $N_{1}$ with $a_{1}$ remaining unchanged. The results are shown in table
1. The second column in this table can also be understood as giving the
corresponding invariant and normal lattices of the possible ${\mathbb{Z}}_{2}$
automorphisms of $E_{8}$. Indeed, the results can be obtained applying the
same theorem [32] that produces Figure 1.

$\phantom{-}\Delta_{s_{1}}$ | $(I_{1},N_{1},a_{1})$
---|---
$\phantom{-}0$ | $(\text{null},E_{8},0)$
$-1$ | $(A_{1},E_{7},1)$
$-2$ | $(A_{1}^{2},D_{6},2)$
$-3$ | $(A_{1}^{3},A_{1}+D_{4},3)$
$-4$ | $(A_{1}^{4},A_{1}^{4},4)$
$-4$ | $(D_{4},D_{4},2)$
$-5$ | $(A_{1}+D_{4},A_{1}^{3},3)$
$-6$ | $(D_{6},A_{1}^{2},2)$
$-7$ | $(E_{7},A_{1},1)$
$-8$ | $(E_{8},\text{null},0)$
Table 1: Invariant and normal lattices obtained from moving $A_{1}$’s in
$E_{8}$.

The same analysis works for the second $E_{8}$ and yields
$(I_{2},N_{2},a_{2})$ as in table 1. Suppose we start from the $(2,0,0)$ model
with $I=U$ and $N=U+E_{8}^{2}$. Using the transitions in table 1 in the two
$E_{8}$’s we obtain models with $I=U+K=U+I_{1}+I_{2}$ up to a maximum value of
$a=8$. These models are located inside the dashed blue lines in figure 2.
There are 41 points inside this region and all but 5 of them are connected in
this way. The 5 points are: $(6,4,0)$, $(10,2,0)$, $(10,6,0)$, $(10,8,0)$ and
$(14,4,0)$ and are denoted by blue circles in the figure. Except for
$(10,8,0)$, the remaining four points can be reached from $I=U(2)$ and
$N=U(2)+U+E_{8}^{2}$. Note that the $(10,2,0)$ model has $I\simeq
U(2)+E_{8}\simeq U+D_{8}$ and can be obtained by replacing $E^{2}_{8}$ in
$\Gamma_{\\!\\!(19,3)}$ by the $Spin(32)/{\mathbb{Z}}_{2}$ lattice.

$0$$2$$4$$6$$8$$10$$12$$14$$16$$18$$20$$r:=20-s$$0$$1$$2$$3$$4$$5$$6$$7$$8$$9$$10$$11$$a$${\delta=1}$$\delta=0$
Figure 2: Starting from the $(2,0,0)$ model and shuffling $A_{1}$ lattices in
the two $E_{8}$ components of
$\Gamma_{\\!\\!(19,3)}\simeq\Gamma_{0}=U^{3}+E_{8}^{2}$ as in table 1, we
connect to models with $I=U+K$ and $a\leq 8$. These models are located inside
the dashed blue lines. Five points depicted as blue circles are not obtained
in this way. Starting from $(2,2,0)$ model and moving the $A_{1}$ lattices we
connect to models with $I=U(2)+K$ and $a\leq 10$ located inside the dotted red
lines. The red dots and circles in this region may not be obtained in this
way.

Larger values of $a$ (i.e. $a>8$) can be obtained by changing the action of
the ${\mathbb{Z}}_{2}$ involution on the $U^{3}$ block in
$\Gamma_{\\!\\!(19,3)}$. The contribution from this block to $I$ and $N$
lattices are denoted as $I_{3}$ and $N_{3}$, of signature $(3-s_{3},1)$ and
$(s_{3},2)$, respectively. In the previous paragraph we considered an
involution that reflects two of the $U$ factors and hence obtained $I_{3}=U$,
$N_{3}=U^{2}$, $s_{3}=2$ and $a_{3}=0$. Higher values of $a_{3}$ may be
obtained by exchanging two $U$’s which yields a $U(2)$ component in both
$I_{3}$ and $N_{3}$. Another possibility is to take the $U$ lattice at the
$SU(2)$ point, i.e. $(A_{1};A_{1})[(F_{1};F_{1})]$, and reflect only the left
or the right part of it. Combining these actions we find that
$(I_{3},N_{3},a_{3})$ are of the form given in table 2 (recall that
$U(2)+A_{1}\simeq A_{1}(-1)+A^{2}_{1}$).

$s_{3}$ | $(I_{3},N_{3},a_{3})$
---|---
$3$ | $(A_{1}(-1),U^{2}+A_{1},1)$
$2$ |  $(U,U^{2},0)$, $(A_{1}(-1)+A_{1},A_{1}(-1)+A_{1}+U,2)$, $(U(2),U(2)+U,2)$
$1$ |  $(U+A_{1},U+A_{1}(-1),1)$, $(U(2)+A_{1},U(2)+A_{1}(-1),3)$
$0$ | $(U+A_{1}^{2},A^{2}_{1}(-1),2)$
Table 2: Invariant and normal lattices embedded in $U^{3}$.

Suppose we take $I_{3}=U(2)$ and start from $I=U(2)$ and $N=U(2)+U+E_{8}^{2}$.
Moving the $A_{1}$ components in the two $E_{8}$’s we connect to models with a
maximum value of $a=10$. These models are located inside the region surrounded
by dotted red lines in figure 2. There are again some models inside this
domain which may not be obtained in this way and are denoted by red dots and
circles in the figure. Some of them may be reached from other choices of
$I_{3}$ in table 2. The remaining possible forms of the $I_{3}$ lattice yield
similar domains as the ones shown in figure 2 but with values of $r$ shifted.
We will not go through all the details here but only emphasise the important
results. It turns out that all but four models $(r,a,\delta)$ in figure 2 have
invariant and normal lattices that can be written as $I=I_{1}+I_{2}+I_{3}$,
$N=N_{1}+N_{2}+N_{3}$, $a=a_{1}+a_{2}+a_{3}$, where $(I_{3},N_{3},a_{3})$ are
given in table 2 and $(I_{1},N_{1},a_{1})$ and $(I_{2},N_{2},a_{2})$ are given
in table 1. We have collected the list of $I$ and $N$ lattices which are
constructed in this way in table LABEL:tab_IN.

The four exceptions are the entries 34, 36, 56 and 71 in the table. The former
two have $E_{8}(2)$ components in their $I$ and $N$ lattices and the latter
two have $U(2)^{2}$ components in their normal lattice. As such, they cannot
be constructed using the method developed in this subsection. We will
construct these four models in subsection 6.3 and show how they are connected
to the models we constructed here.

So far we have shown that the models inside each $I_{3}$ domain are connected
to each other through moving the $A_{1}$ components in $E_{8}^{2}$. 555Recall
that the asymmetric orbifold construction may not be realised for the triples
$(1,1,1)$ and $(2,2,1)$ — see the discussion at the beginning of section 5.3.
There are seven such domains corresponding to the entries in table 2. Are the
different $I_{3}$ domains connected to each other? The answer is yes. This is
because there are common triples in the overlap of any two domains. Consider
two domains $I_{3}$ and $I_{3}^{\prime}$. The models located in the overlap of
these two domains may be described in two ways: $I\simeq I_{3}+K\simeq
I_{3}^{\prime}+K^{\prime}$ where the two descriptions are related by a boost
transformation in $SO(r-1,1)$. The former description is connected to the
$I_{3}$ domain. Upon performing a boost transformation we obtain the latter
description which is connected to all points in the $I_{3}^{\prime}$ domain.
Therefore the two domains are indeed connected to each other.

As an example consider the two domains with $I_{3}=U$ and
$I_{3}^{\prime}=U(2)$ in figure 2. Start from the triple $(6,6,1)$ with
$I=U(2)+A_{1}^{3}+A_{1}$ and $N=U(2)+U+A_{1}+D_{4}+E_{7}$. In the invariant
lattice the $A_{1}^{3}$ components come from the same $D_{4}$ whereas the
remaining $A_{1}$ has been moved from a different $D_{4}$. Upon moving the
remaining $A_{1}$ in the former $D_{4}$, we make a transition to the triple
$(7,5,1)$ with $I=U(2)+D_{4}+A_{1}$ and $N=U(2)+U+D_{4}+E_{7}$. Next we make a
boost transformation in $SO(6,1)$ to obtain $I=U+D_{2}^{2}+A_{1}$. The
associated normal lattice is $N=U^{2}+D_{2}^{2}+E_{7}$. We can now use $A_{1}$
shifts to reach other points within the $I_{3}=U$ domain.

We end this subsection by providing an explicit example of making transitions
between models in oder to explain the ideas and the method developed so far.

####  6.2.1 Example: transitions from $(12,8,1)$

This example serves to illustrate transitions from the $r=(a+4)$ line in
figure 1. We begin by describing the invariant and normal lattices
corresponding to a ${\mathbb{Z}}_{2}$ involution acting on
$\Gamma_{\\!\\!(19,3)}\simeq\Gamma_{0}=U^{3}+E_{8}^{2}$, now with the second
and third $U$’s at the $SU(2)$ point. Under ${\mathbb{Z}}_{2}$ the first $U$
is untouched, while in the second and third only the right part is reflected.
Thus, from $U^{3}$ there is a contribution $I_{3}=U+A_{1}^{2}$ to $I$ and
$N_{3}=A_{1}(-1)^{2}$ to $N$, with $A_{1}^{2}$ classes in $I^{*}$ correlated
with $A^{2}_{1}(-1)$ classes in $N^{*}$. The action on $E^{2}_{8}$ is defined
below.

One of the $E_{8}$ lattices, say the first, is decomposed as $D_{4}+D_{4}$,
with standard correlated conjugacy classes, see eq. (6.12). We then reflect
one $D_{4}$ to obtain further contributions $I\supset D_{4}$ and $N\supset
D_{4}$, with their classes correlated in $I^{*}$ and $N^{*}$. The second
$E_{8}$ is also decomposed under $D_{4}+D_{4}$ but we resort to triality in
one factor to exchange $V_{4}$ and $Sp_{4}$. The classes are then correlated
as

$(Sc_{4},Sc_{4})+(V_{4},Sp_{4})+(Sp_{4},V_{4})+(Sp^{\prime}_{4},Sp^{\prime}_{4}).$
(6.15)

In this second $E_{8}$ we then split each $D_{4}$ factor as $D_{2}+D_{2}$,
each with correlations given in eq. (6.2), and then reflect one $D_{2}$ from
each of the $D_{4}$ factors. As explained before, the invariant piece from the
second $E_{8}$ must be $D_{2}+D_{2}=A_{1}^{4}$. Moreover, the normal piece
from the second $E_{8}$ will also be $D_{2}+D_{2}$. The conjugacy classes of
$I^{*}$ and $N^{*}$ will be correlated according to the decompositions made in
the process.

All in all, we find $I=U+A_{1}^{2}+D_{4}+D_{2}^{2}$,
$N=A_{1}(-1)^{2}+D_{4}+D_{2}^{2}$ with definite correlations between $I^{*}$
and $N^{*}$ classes. While the correlation of the classes of $A_{1}(-1)$ with
$A_{1}$ and $D_{4}$ with $D_{4}$ is standard, the correlation of classes of
$D_{2}^{2}$ of $N$ with $D_{2}^{2}$ of $I$, which is a bit non-standard, is
given by

$\displaystyle~{}(D_{2}^{*}/D_{2},D_{2}^{*}/D_{2};D_{2}^{*}/D_{2},D_{2}^{*}/D_{2})$
(6.16)
$\displaystyle=(Sc_{2},Sc_{2};Sc_{2},Sc_{2})+(V_{2},Sc_{2};V_{2},Sc_{2})+(Sc_{2},V_{2};Sc_{2},V_{2})+(V_{2},V_{2};V_{2},V_{2})$
$\displaystyle+(Sc_{2},Sp_{2};V_{2},Sp_{2})+(V_{2},Sp_{2};Sc_{2},Sp_{2})+(Sc_{2},Sp^{\prime}_{2};V_{2},Sp^{\prime}_{2})+(V_{2},Sp^{\prime}_{2};Sc_{2},Sp^{\prime}_{2})$
$\displaystyle+(Sp_{2},Sc_{2};Sp_{2},V_{2})+(Sp^{\prime}_{2},Sc_{2};Sp^{\prime}_{2},V_{2})+(Sp_{2},V_{2};Sp_{2},Sc_{2})+(Sp^{\prime}_{2},V_{2};Sp^{\prime}_{2},Sc_{2})$
$\displaystyle+(Sp_{2},Sp_{2};Sp^{\prime}_{2},Sp^{\prime}_{2})+(Sp^{\prime}_{2},Sp_{2};Sp_{2},Sp^{\prime}_{2})+(Sp_{2},Sp^{\prime}_{2};Sp^{\prime}_{2},Sp_{2})+(Sp^{\prime}_{2},Sp^{\prime}_{2};Sp_{2},Sp_{2}).$

where we used the ordering $(N^{*}/N;I^{*}/I)$.

Now we can take one of the $A_{1}$ in $N$ (it could be either from $D_{2}^{2}$
or an $A_{1}$ inside $D_{4}$) and rebosonize it so that it becomes part of $I$
with a shift equal to half the fundamental in that $A_{1}$. To find the new
$I$, we need to figure out how the fundamental of that $A_{1}$ that we have
picked in $N$ is correlated to the rest of the conjugacy classes.

First consider $A_{1}$ in one of the $D_{2}$ in $N$ and decompose that $D_{2}$
in terms of $A_{1}^{2}$. Under $A^{2}_{1}$ the various conjugacy classes of
$D_{2}^{*}/D_{2}$ are given in eq. (6.14). Finally we move the second $A_{1}$
to $I$. Then the right hand side of eq. (6.16) becomes

$\displaystyle\quad\
(Sc_{2},R;R,Sc_{2},Sc_{2})+(V_{2},R;R,V_{2},Sc_{2})+(Sc_{2},F;F,Sc_{2},V_{2})+(V_{2},F;F,V_{2},V_{2})$
(6.17)
$\displaystyle+(Sc_{2},R;F,V_{2},Sp_{2})+(V_{2},R;F,Sc_{2},Sp_{2})+(Sc_{2},F;R,V_{2},Sp^{\prime}_{2})+(V_{2},F;R,Sc_{2},Sp^{\prime}_{2})$
$\displaystyle+(Sp_{2},R;R,Sp_{2},V_{2})+(Sp^{\prime}_{2},R;R,Sp^{\prime}_{2},V_{2})+(Sp_{2},F;F,Sp_{2},Sc_{2})+(Sp^{\prime}_{2},F;F,Sp^{\prime}_{2},Sc_{2})$
$\displaystyle+(Sp_{2},R;F,Sp^{\prime}_{2},Sp^{\prime}_{2})+(Sp^{\prime}_{2},R;F,Sp_{2},Sp^{\prime}_{2})+(Sp_{2},F;R,Sp^{\prime}_{2},Sp_{2})+(Sp^{\prime}_{2},F;R,Sp_{2},Sp_{2})\
.$

In the above we have moved the second $A_{1}$ to the right of the semicolon
because we have rebosonized that $A_{1}$ so that it is along the new $I$
directions. The new invariant piece from the second $E_{8}$ will be obtained
by taking all the vectors that have zero entry on the left of the semicolon
which means that the left of the semicolon must be $(Sc_{2},R)$. Hence, this
new invariant piece will be $(R,Sc_{2},Sc_{2})+(F,V_{2},Sp_{2})$. Note that
the first class is just $A_{1}+D_{2}+D_{2}$, while the second one admits 16
vectors of square length 2. Thus, the two terms altogether give 31 vectors of
square length 2, which are actually the roots of $D_{4}+A_{1}$. This can be
seen by writing the last $D_{2}$ in $A_{1}+D_{2}+D_{2}$ as $A_{1}+A_{1}$, and
noting that $Sp_{2}$ is neutral with respect to the second $A_{1}$. Combining
all the remaining parts of $I$ we find that the new full invariant lattice is
$I^{\prime}=U+A_{1}^{3}+D_{4}^{2}$, which corresponds to the $(13,7,1)$ model.
The new normal lattice is $N^{\prime}=A_{1}(-1)^{2}+D_{4}+D_{2}+A_{1}$.

What if we had moved an $A_{1}$ inside $N$ from the first $E_{8}$, namely from
$D_{4}$ which we write as $D_{2}+A_{1}+A_{1}$ ? The $E_{8}$ conjugacy classes
before rebosonizing an $A_{1}$ are

$\displaystyle\quad\
(Sc_{2},R,R;Sc_{4})+(V_{2},F,F;Sc_{4})+(Sc_{2},F,F;V_{4})+(V_{2},R,R;V_{4})$
$\displaystyle+(Sp_{2},R,F;Sp_{4})+(Sp^{\prime}_{2},F,R;Sp_{4})+(Sp^{\prime}_{2},R,F;Sp^{\prime}_{4})+(Sp_{2},F,R;Sp^{\prime}_{4}).$
(6.18)

Next we move the second $A_{1}$ to the $I$ direction which means moving the
semicolon one place to the left. The new $I^{\prime}$ can come only from
states that have zero in the first two entries, i.e. they must be $(Sc_{2},R)$
in the normal direction. Thus, the invariant piece coming from this $E_{8}$
will be $A_{1}+D_{4}$. Combining with the remaining parts of $I$ we get
$I^{\prime}=U+A_{1}^{3}+D_{4}+D_{2}^{2}$ and
$N^{\prime}=A_{1}(-1)^{2}+D_{2}^{2}+A_{1}+D_{2}$, which give the $(13,9,1)$
model. Hence, the end point depends on which $A_{1}$ is moved from $N$ to $I$.

We can continue the procedure. For instance, suppose we move two $A_{1}$’s,
one from the first $E_{8}$ and another from the second $E_{8}$. The new
invariant lattice can be found combining the results in the above paragraphs
for the 2 $E_{8}$’s, together with $U+A_{1}^{2}$. In this way we obtain
$I^{\prime}=U+A_{1}^{2}+(A_{1}+D_{4})+(A_{1}+D_{4})$, which gives the
$(14,8,1)$ model with $N^{\prime}=A_{1}(-1)^{2}+A_{1}^{2}+D_{2}^{2}$.

###  6.3 The remaining triples

In the previous subsections we developed a method for making transitions
between orbifold models. We started from
$\Gamma_{\\!\\!(19,3)}\simeq\Gamma_{0}=U^{3}+E_{8}^{2}$, defined the action of
the ${\mathbb{Z}}_{2}$ involution to obtain $I$ and $N$, and moved $A_{1}$
components from the $N$ to $I$ to change the value of $s$. We observed that
there are four triples in figure 1 which cannot be constructed in this way.
These points are $(10,8,0)$, $(10,10,0)$, $(14,6,0)$ and $(18,4,0)$. Triples
$(10,8,0)$ and $(10,10,0)$ contain $E_{8}(2)$ in their $I$ and $N$ lattices
which is obtained by exchanging the two $E_{8}$’s. Triples $(14,6,0)$ and
$(18,4,0)$ have $U(2)^{2}$ in $N$ which cannot not be obtained from the
involutions of the $U^{3}$ block as can be seen in table 2. In this subsection
we construct these four models and show how to connect to them by making
$s$-transitions from other points inside the $I_{3}$ domains. We therefore
conclude that all the 75 triples in figure 1 are connected to each other
through $s$-transitions.

####  6.3.1 $(10,8,0)$

For the theory with $(r,a,\delta)=(9,9,1)$, the invariant lattice is
generically $I\simeq U(2)+A_{1}^{7}$. As shown in eq. (5.34) there is a point
in moduli space where $I=A_{1}(-1)+E_{8}(2)$, which happens to be more
convenient to start to survey possible transitions. This point can also be
obtained directly from
$\Gamma_{\\!\\!(19,3)}\simeq\Gamma_{\\!\\!(2,2)}+U+E_{8}^{2}$, taking the
${\mathbb{Z}}_{2}$ involution to act by reflecting $\Gamma_{\\!\\!(2,2)}$,
exchanging the two $E_{8}$’s, and reflecting the left part of $U$ at the
$SU(2)$ point, i.e. at the self-dual radius. In this case
$N=\Gamma_{\\!\\!(2,2)}+A_{1}+E_{8}(2)$. Besides, the conjugacy classes of
$I^{*}/I$ and $N^{*}/N$ are correlated between $A_{1}(-1)^{*}/A_{1}(-1)$ and
$A_{1}^{*}/A_{1}$, as well as between the $E_{8}(2)^{*}/E_{8}(2)$’s in $I$ and
$N$. Now, in the $A_{1}$ appearing in $N$ we can rebosonize $J_{1}\rightarrow
J^{\prime}_{3}$ as discussed in subsection 6.1. The result is that the $A_{1}$
of $N$ will move to $I$ and since the $A_{1}(-1)^{*}/A_{1}(-1)$ and
$A_{1}^{*}/A_{1}$ classes were correlated, the new $I$ will be $U+E_{8}(2)$
which is the $(10,8,0)$ model.

Instead of using this procedure on the $A_{1}$ of $N$, we could have taken
$\Gamma_{\\!\\!(2,2)}=U+U$, with one of the $U$’s at the $SU(2)$ point. We
could then perform the rebosonization of the $J_{1}$ current on the left-
moving $A_{1}$ of this $U$ so that it becomes part of $I$. However, now the
classes of this left-moving $A_{1}$ is _not_ correlated with the already
existing $A_{1}(-1)$ in $I$. Therefore, after this transition,
$I=A_{1}(-1)+A_{1}+E_{8}(2)$ and we arrive at the $(10,10,1)$ model, since
this $I$ is equivalent to $U(2)+A_{1}^{8}$. Indeed,
$U(2)+A_{1}^{7}+A_{1}\simeq A_{1}(-1)+E_{8}(2)+A_{1}$.

####  6.3.2 $(10,10,0)$

With our prescribed method of turning an $A_{1}$ lattice in the normal
direction to the invariant direction, it is not obvious how we can connect the
point $(10,10,0)$ with $I=U(2)+E_{8}(2)$, to another point with $r=9$.
However, we can try to connect it to a point with $r=11$. Let us then start
from $(11,11,1)$ and take $I=U(2)+A_{1}+D_{2}^{4}$. This can be realized
starting from the self dual lattice $\Gamma_{\\!\\!(19,3)}\simeq
U^{3}+E_{8}^{2}$, and writing each $E_{8}$ as $D_{4}^{2}$ with conjugacy
classes given in eq. (6.15) (observe that we have used triality in one $D_{4}$
to exchange $V_{4}$ and $Sp_{4}$). We then split each $D_{4}$ factor as
$D_{2}+D_{2}$, each with correlations given in eq. (6.2). The involution acts
by reflection of one $D_{2}$ factor in each $D_{4}$, exchange of the first two
$U$’s, and reflection of the right movers in the third $U$ at the $SU(2)$
point. This gives the desired $I=U(2)+A_{1}+D_{2}^{4}$, and
$N=U(2)+A_{1}(-1)+D_{2}^{4}$, with correlated classes implied by the defining
involution. This result follows because in $I$ each of the reflected $D_{2}$
lattice vectors must be set to zero, which is an element of $Sc_{2}$. Thus,
from eqs. (6.15) and (6.2) we see that the invariant piece from each $E_{8}$
can only come from $(Sc_{4},Sc_{4})$ and must be $D_{2}+D_{2}=A_{1}^{4}$.

As discussed in subsection 5.3.2, it can be shown that $I=U(2)+A_{1}^{9}$ is
in the same moduli space as $I=A_{1}(-1)+A_{1}^{2}+E_{8}(2)$ — see eq. (5.34).
In fact, $A_{1}(-1)+A_{1}^{2}\simeq U(2)+A_{1}$. Thus, we conclude that
$I=U(2)+A_{1}^{9}$ is equivalent to $I=U(2)+A_{1}+E_{8}(2)$. We now use the
method of rebosonization on the $A_{1}$ factor of the latter and convert it to
an $A_{1}$ in the normal direction. Note that we are _increasing_ $s$ by one
unit and therefore this method will work only if the shift vector $v$ of the
$(11,11,1)$ model has a part in this $A_{1}$ equal to half the fundamental.
After this transition we attain $I=U(2)+E_{8}(2)$, which is indeed the
$(10,10,0)$ model.

Note that we could have used the method of transition one step earlier, namely
taking $I=A_{1}(-1)+A_{1}^{2}+E_{8}(2)$. If there is a part of the shift equal
to half the fundamental in one of the $A_{1}$ factors, we could convert it
into an $N$ direction resulting in $I=A_{1}(-1)+A_{1}+E_{8}(2)$ which is the
$(10,10,1)$ model. Thus, even though we were not able to go from $r=9$
directly to $(10,10,0)$, we could go from $(9,9,1)$ to $(10,10,1)$ as
described in subsection 6.3.2. Then, inverting the process just described, we
could go to the $(11,11,1)$ model, and then as explained in the previous
paragraph we can transition to the $(10,10,0)$ model.

Note that while moving from $s$ to $(s-1)$ we do not have to worry about the
shift vector: the $A_{1}$ which moves from $N$ to $I$ automatically comes with
a consistent shift. However, when we move from $s$ to $s+1$, an $A_{1}$ must
move from $I$ to $N$, and that is possible only if the shift vector has a
component along this $A_{1}$ which is equal to half the fundamental.

####  6.3.3 $(18,4,0)$

Let us start with $\Gamma_{\\!\\!(19,3)}\simeq U^{2}+\Gamma_{(9,1)}+E_{8}$
where $\Gamma_{(9,1)}$ is an even self-dual lattice. We will choose a
particular point in the moduli space of $\Gamma_{(9,1)}$ such that it can be
expressed as $(D_{9};D_{1})[(V_{9};V_{1}),(Sp_{9};Sp_{1})]$ with correlated
classes $(Sc_{9};Sc_{1})$, $(V_{9};V_{1})$, $(Sp_{9};Sp_{1})$ and
$(Sp^{\prime}_{9};Sp^{\prime}_{1})$. Furthermore we decompose $D_{9}$ in terms
of $(D_{8},D_{1})$. Various conjugacy classes in this decomposition are
$Sc_{9}=(Sc_{8},Sc_{1})+(V_{8},V_{1})$, $V_{9}=(Sc_{8},V_{1})+(V_{8},Sc_{1})$,
$Sp_{9}=(Sp_{8},Sp_{1})+(Sp^{\prime}_{8},Sp^{\prime}_{1})$ and
$Sp^{\prime}_{9}=(Sp_{8},Sp^{\prime}_{1})+(Sp^{\prime}_{8},Sp_{1})$. Thus
$\Gamma_{(9,1)}$ at this point can be expressed in the decomposition
$(D_{8},D_{1};D_{1})$ as

$\displaystyle\Gamma_{(9,1)}=(Sc_{8},Sc_{1};Sc_{1})+(V_{8},V_{1};Sc_{1})+(Sc_{8},V_{1};V_{1})+(V_{8},Sc_{1};V_{1})+$
(6.19)
$\displaystyle\qquad\;\,+\;(Sp_{8},Sp_{1};Sp_{1})+(Sp^{\prime}_{8},Sp^{\prime}_{1},Sp_{1})+(Sp_{8},Sp^{\prime}_{1};Sp^{\prime}_{1})+,(Sp^{\prime}_{8},Sp_{1};Sp^{\prime}_{1})\
.$

The involution in $\Gamma_{\\!\\!(19,3)}$ we take to be exchange of two $U$’s
and reflection along the $D_{1}(-1)$ and $D_{1}$ directions in
$\Gamma_{(9,1)}$. The contribution to $I$ and $N$ coming from the exchange of
two $U$’s is $U(2)$ for each lattice. The contribution to $I$ coming from the
reflection in $\Gamma_{(9,1)}$ is obtained by setting momenta along
$D_{1}(-1)$ and $D_{1}$ directions to zero and as a result, it is just scalar
class of $D_{8}$ — see eq. (6.19). The normal lattice is obtained by setting
the momenta along the directions of $D_{8}$ to zero and hence, the
contribution to $N$ from this part is $(Sc_{1};Sc_{1})+(V_{1};V_{1})$. This is
indeed the $U(2)$ lattice as can be seen by setting $R=1/\sqrt{2}$ in (5.33)
and observing that the lattice has the from $(p_{L};p_{R})=(k+w,k-w)$ where
the two entries are either both even or both odd, i.e. it is the lattice
$(D_{1};D_{1})[(V_{1},V_{1})]$. Altogether we find $I=U(2)+D_{8}+E_{8}$ and
$N=U(2)^{2}$.

In the above construction we further decompose $E_{8}$ in terms of
$(E_{7},A_{1})$ and reflect $A_{1}$ — see eqs. (A.3) and (A.4) in appendix A
for this decomposition. The result is $I=U(2)+D_{8}+E_{7}$ and
$N=U(2)^{2}+A_{1}$, i.e. the $(17,5,1)$ model. Now the $s$-transition can be
performed by moving the $A_{1}$ from $N$ to $I$, resulting in the above
$(18,4,0)$ model. Note that the $(17,5,1)$ model has $I\simeq
U(2)+D_{8}+E_{7}\simeq U+A_{1}^{2}+E_{8}+A_{1}+D_{4}$ and $N\simeq
U(2)^{2}+A_{1}\simeq A_{1}^{2}(-1)+A_{1}^{2}+A_{1}$ where the second
descriptions may be obtained from our construction in section 6.1 with
$I_{3}=U+A_{1}^{2}$ in table 2. Therefore the $(18,4,0)$ model can be
connected to other triples in this domain.

####  6.3.4 $(14,6,0)$

We proceed as for the $(18,4,0)$ model just discussed. In addition, we
decompose $E_{8}$ in terms of $(D_{4},D_{4})$ as in (6.12) and reflect one
$D_{4}$. The result is $I=U(2)+D_{8}+D_{4}$ and $N=U(2)^{2}+D_{4}$, which
gives the $(14,6,0)$ model. Next we want to make an $s$-transition to reach
this point. To do so, consider the above construction but do one more
reflection inside the $D_{8}$ lattice appearing in $\Gamma_{(9,1)}$ as
follows. Decompose $D_{8}$ in terms of $(D_{6},D_{2})$ and furthermore
decompose $D_{2}=(A_{1},A_{1})$ — see eq. (6.14). The correlated classes are
given by $Sc_{8}=(Sc_{6},R,R)+(V_{6},F,F,)$,
$V_{8}=+(Sc_{6},F,F)+(V_{6},R,R)$,
$Sp_{8}=(Sp_{6},R,F)+(Sp^{\prime}_{6},F,R)$,
$Sp^{\prime}_{8}=(Sp^{\prime}_{6},R,F)+(Sp_{6},F,R)$. Now in the decomposition
$(D_{6},A_{1},A_{1},D_{1};D_{1})$, reflect one $A_{1}$ as well as
$(D_{1};D_{1})$. The contribution coming from $\Gamma_{(9,1)}$ to $I$ and $N$
is thus $D_{6}+A_{1}$ and $U(2)+A_{1}$. Combining everything together we find
$I=U(2)+D_{6}+A_{1}+D_{4}$ and $N=U(2)^{2}+A_{1}+D_{4}$ which is the
$(13,7,1)$model. Now we make an $s$-transition by moving the $A_{1}$ from $N$
to $I$. By construction above, the $I^{*}/I$ classes are correlated with the
$N^{*}/N$ classes in such a way that upon moving the $A_{1}$ from $N$ to $I$
we get back to the $(14,6,0)$ model.

##  7 Final remarks

In this work we constructed asymmetric orbifolds describing heterotic strings
on ${\mathbb{T}}^{3}/{\mathbb{Z}}_{2}$ at the worldsheet level. In the context
of M-theory/heterotic duality, these theories are dual to M-theory on K3
surfaces quotiented by a non-symplectic involution. Such involutions are
uniquely characterised by the three parameters $(r,a,\delta)$ of the
sublattice of the K3 lattice which is invariant under the involution. For a
given $(r,a,\delta)$, the heterotic orbifold is additionally characterised by
the shift vector $v$ constrained by level matching. We found that for the two
triples with $(r,a,\delta)=(1,1,1)$ and $(2,2,1)$, the asymmetric orbifold is
not viable due to the small rank of the invariant lattices, which makes it
impossible to find a shift vector satisfying the level matching condition. One
possible way to remedy this problem is to compactify the theory further on a
circle, i.e. move to six dimensions. The direction along the circle is chosen
to be left invariant under the involution and as such the circle is referred
to as the spectator circle. One would do the analysis in six dimensions and
then take the limit where the radius of the spectator circle goes to infinity
to recover the seven-dimensional theory. The rank of the invariant lattice is
bigger in six dimensions because of the presence of the spectator circle. One
then asks whether it is now possible to define shift vectors that satisfy the
level matching condition in the new theory. If so, one would classify the
inequivalent families of such shift vectors and analyse how they emerge in the
original theory as we take the radius of the spectator circle to infinity.

We proved that all the asymmetric orbifold models are connected to each other
through $s$-transitions. This raised the question whether there exists a
unique big moduli space where each model associated with the triple
$(r,a,\delta)$ is a subspace of it. The answer is not clear because we also
saw that some models contain several inequivalent shift vectors. However, the
analysis of one such model, namely the one associated with
$(r,a,\delta)=(5,3,1)$, showed that the inequivalent shift vectors are
connected by making an $s$-transition to the $(9,1,1)$ theory. Whether the
same conclusion holds for all models with inequivalent shifts remains to be
studied. It will be interesting to study the existence of inequivalent shift
vectors and the connection between the asymmetric orbifolds in one lower
dimension, upon compactifying the theory on an spectator circle. It could be
that all models are connected, as it occurs for the 10-dimensional
supersymmetric and non-supersymmetric heterotic theories compactified on a
circle [63, 6].

Although tachyons are not allowed in the untwisted sector of our perturbative
models, they generically appear in the twisted sectors in some regions of
moduli space. For instance, in some models tachyons occur only for values of
the circle radius such that $R_{min}<R<R_{max}$. They become massless at the
endpoints and massive outside this interval. When present, tachyons only
signal that the vacuum is unstable. An appealing scenario is that tachyon
condensation produces a stable lower dimensional vacuum, as shown to occur in
some non-supersymmetric string setups [66, 67, 17, 68, 21]. It would be
worthwhile to examine if such dynamics could take place in the non-
supersymmetric heterotic orbifolds that we have studied.

We have seen that in the heterotic on ${\mathbb{T}}^{3}/{\mathbb{Z}}_{2}$
there always exist untwisted fermions, massless or massive, which are
necessarily charged. On the M-theory or type IIA side, charged fermionic
states are non-perturbative, coming from membranes or D2-branes wrapped on
various 2-cycles. This result prompts the question whether the K3 quotient by
the given Nikulin involution supports fermions or not, i.e. whether it is a
spin manifold or not. As discussed at the end of section 2, for most
$(r,a,\delta)$ the answer is negative except for the two cases, $r=2$,
$a=2,0$, and $\delta=0$. Now, in M-theory or type IIA on a spin manifold one
would expect fermions that are not charged, for instance massive KK modes of
the gravitino. If the quotient is spin, absence of uncharged fermions in the
heterotic could still be explained arguing that KK modes of M-theory or type
IIA in the strong coupling limit (i.e. heterotic theory) may disappear or
become infinitely massive. Another point is that, even if the 4-dimensional K3
quotient is not spin, such as the Enriques case, the branes are localized on
2-cycles that do not see the whole 4-dimensional space. It could very well be
that on these 2-cycles there exists a spin structure. Perhaps that is all that
is needed to obtain 7-dimensional or 6-dimensional fermions from these brane
states. For quantum consistency, the involution must necessarily act with some
phases on the membrane or D2-brane states, since it does on the heterotic
states that have momentum along the $\Gamma_{\\!\\!(19,3)}$ lattice and feel
the effect of the shifts $v$ and $w$. What are these quantum consistency
conditions in M-theory or IIA theory ? A way to examine this problem would be
to study type IIA on an Enriques surface. The idea would be to look at
D2-branes wrapped on some 2-cycles of K3 and see how the involution modifies
the analysis. Consistency conditions might have to be imposed on non-trivial
RR backgrounds necessary for the heterotic/type IIA duality, as found in
supersymmetric compactifications to four and six dimensions where similar
issues also arise [69, 47, 40, 45]. We plan to come back to these matters in
the future.

Finally, another possible extension of this work is to construct heterotic
${\mathbb{T}}^{3}$ asymmetric orbifolds in which the orbifold generator
includes other non-symplectic automorphisms of the $\Gamma_{\\!\\!(19,3)}$
lattice such as those analized e.g. in [70, 71, 72]. It is also conceivable to
consider higher dimensional tori ${\mathbb{T}}^{d}$ and quotient by
automorphisms of the even self-dual lattice $\Gamma_{\\!\\!16+d,d}$ that break
supersymmetry.

## Acknowledgments

We are grateful to D. Allcock, E. Andrés, L. da Rold, C. Mayrhofer, H. Parra
De Freitas, S. Theisen and G. Thompson for interesting comments and valuable
insights. A. Font and G. Aldazabal acknowledge hospitality and support from
IFT UAM-CSIC via the Centro de Excelencia Severo Ochoa Program under Grant
SEV-2016-0597. The work of G. Aldazabal is partially supported by the CONICET
grant PIP-11220200100981CO. The work of B.S. Acharya is supported by a grant
from the Simons Foundation (#488569, Bobby Acharya).

##  Appendix A Construction of $I$ and $N$ lattices for all $(r,a,\delta)$

The key player in our analysis is the even self-dual lattice $\Gamma_{(19,3)}$
which is unique up to $SO(19,3)$ transformations. We study ${\mathbb{Z}}_{2}$
involutions of $\Gamma_{(19,3)}$, denoted by $\theta$, that act by reflecting
$s$ left-moving and 2 right-moving directions. Such involutions have been
classified by Nikulin [31, 32, 33]. Each $\theta$ leaves invariant a lattice,
denoted $I$, of rank $r$ and signature $(r-1,1)$, where $r=20-s$. The normal
lattice in $\Gamma_{\\!\\!(19,3)}$, denoted $N$, has rank $(2+s)$ and
signature $(s,2)$. Both $I$ and $N$ are even sub-lattices of
$\Gamma_{\\!\\!(19,3)}$ and satisfy $I^{*}/I=N^{*}/N=({\mathbb{Z}}_{2})^{a}$
where $a\in{\mathbb{Z}}_{\geq 0}$. Lattices which satisfy the latter condition
are called 2-elementary lattices. All possible involutions, as well as the
corresponding lattices $I$ and $N$, can be completely characterized by the
triple $(r,a,\delta)$, where the invariant $\delta$ is defined in eq. (2.2).
Figure 1 depicts the 75 allowed triples found by Nikulin.

In the main body of the paper we developed the asymmetric orbifold realisation
of the involutions in the context of heterotic string theory. The orbifold
construction is performed in its full generality, i.e. it is independent of
the invariants $(r,a,\delta)$ of the models. Nonetheless at various points we
have discussed examples associated with specific triples in order to
illustrate the ideas.

For each triple $(r,a,\delta)$, there exist special points in the moduli space
of the corresponding lattices where $I$ and $N$ are given by an orthogonal
direct sums of 2-elementary lattices. The purpose of this appendix is to
provide explicit actions of involutions on $\Gamma_{(19,3)}$ that yield $I$
and $N$ of this form for all the 75 triples. The involutions are not in
general realised uniquely and what we present is one possible action of
$\theta$ for each triple. In our constructions of even self dual lattices we
shall use the method of gluing the ADE lattices which defines glue vector
generators among the conjugacy classes of the lattices such that upon
addition, they generate all other conjugacy classes. We refer the reader to
[73, Chapter 4] for details.

Notations: We denote a lattice $\Lambda$ with glue vectors $G_{1},G_{2},\cdots
G_{n}$ by $\Lambda[G_{1},G_{2},\cdots,G_{n}]$. For instance, $D_{n}[V_{n}]$
has one glue vector generator being the vector conjugacy class, two conjugacy
classes in total (together with the scalar class), and is an odd self-dual
lattice. The plain lattice $\Lambda$ with no glue vectors refers to the root
lattice of $\Lambda$. When there are lattices with correlated classes we quote
them inside parenthesis, e.g. $(D_{8},D_{2})[(V_{8},V_{2})]$. For lattices
with a Lorentzian signature, the left and right-moving components are placed
on the left and right sides of a semicolon, for instance
$(D_{8};D_{2})[(V_{8};V_{2})]$ denotes a lattice with signature $(8,2)$ and
with correlated classes. The squared norm of a vector $v$ in a lattice with
purely left (right) directions is $||v||^{2}\geq 0$ ($||v||^{2}\leq 0$).

We follow two steps to construct the invariant and normal lattices $I$ and
$N$:
1\. Start at a point in the moduli space of the even self-dual lattice
$\Gamma:=\Gamma_{(19,3)}$ where it takes a convenient form.
2\. Find an involution that leads to $I$ and $N$ with the desired values of
$(r,a,\delta)$.
The invariant lattice is generically a sum of 2-elementary lattice components
given in table 3. The value of $a$ for $I$ is the sum of those of the
component lattices. $I$ has $\delta=1$ if any of the component lattices has
$\delta=1$ and otherwise it has $\delta=0$.

$\Lambda$ U U(2) $A_{1}(-1)$ A1 E7 E8 E8(2) D4m D4m+2 $(r,a,\delta)$ (2,0,0)
(2,2,0) (1,1,1) (1,1,1) (7,1,1) (8,0,0) (8,8,0) ($4m$,2,0) ($4m+2$,2,1)

Table 3: Even 2-elementary lattice building blocks for the construction of
$\Gamma_{(r-1,1)}$ sublattices of K3 which are invariant under Nikulin’s
involutions.

The leftmost points in each row in figure 1 make the diagonal line $r=a$,
namely, the value of the rank coincides with the multiplicity of the center.
We will see that in order to reproduce these points, we need to include a
contribution $U^{2}$ in the starting even self-dual lattice. Namely,
$\Gamma=U^{2}+\Gamma^{\prime}$ with $\Gamma^{\prime}$ an even self-dual
lattice of signature $(17,1)$. By permuting the two $U$ lattices, namely,

$\theta|P_{1L},P_{2L};P_{1R},P_{2R}\rangle=|P_{1R},P_{2R};P_{1L},P_{2L}\rangle$
(A.1)

we obtain the $U(2)$ lattice where the length squares of the vectors are
scaled by $2$ with respect to the original $U$. We thus have
$(r,a,\delta)=(2,2,0)$ for $U(2)$, see table 3. To obtain the desired
invariant lattice $I$, we may need to further act on $\Gamma^{\prime}$.

Before we begin the construction, let us review different involutions in the
lattice $E_{8}\equiv D_{8}[Sp_{8}]$ which lead to invariant sublattices useful
for our constructions. In particular, we consider involutions that leave
invariant $A_{1}$ or $E_{7}$, both contributing with $a=1$ since
$E_{7}^{*}/E_{7}=A_{1}^{*}/A_{1}=\mathbb{Z}_{2}$. Note that in order to obtain
an odd value of $a$, the lattice $I$ must contain one $E_{7}$ and/or an odd
number of $A_{1}$ components. The fundamental classes in $E_{7}^{*}$ (i.e. 56)
and in $A_{1}^{*}$ have only half-integer lengths. Thus, lattices with odd
values of $a$ will _always_ have $\delta=1$.

Recall that the 240 roots of $E_{8}\equiv D_{8}[Sp_{8}]$ (i.e. vectors with
square norm equal to two) are

$(\underline{\pm 1,\pm
1,0,0,0,0,0,0}),\quad(\underbrace{\pm\tfrac{1}{2},\pm\tfrac{1}{2},\pm\tfrac{1}{2},\pm\tfrac{1}{2},\pm\tfrac{1}{2},\pm\tfrac{1}{2},\pm\tfrac{1}{2},\pm\tfrac{1}{2}}_{\text{even
number of +'s}})\ ,$ (A.2)

where underlining means permutations. Under the involution

$\displaystyle\theta|P_{1},\cdots,P_{6},P_{7},P_{8}\rangle$ $\displaystyle=$
$\displaystyle|P_{1},\cdots,P_{6},P_{8},P_{7}\rangle$ (A.3)

which permutes the last two entries (and corresponds to $s=1$), the invariant
lattice corresponds to the $E_{7}$ root lattice with the 126 roots given by

$(\underline{\pm 1,\pm 1,0,0,0,0},0,0)\ ,\quad\pm(0,0,0,0,0,0,1,1)\
,\quad\pm(\underbrace{\pm\tfrac{1}{2},\pm\tfrac{1}{2},\pm\tfrac{1}{2},\pm\tfrac{1}{2},\pm\tfrac{1}{2},\pm\tfrac{1}{2}}_{\text{even
number of +'s}},\tfrac{1}{2},\tfrac{1}{2})\ .$ (A.4)

It is also straightforward to identify the involutions that lead to
$I=D_{6-m}+A_{1}$, $m=\\{0,2,4,6\\}$. For instance, permute two momentum
entries in $D_{8}[Sp_{8}]$ and reflect the rest (hence, $s=1+6=7$):

$\displaystyle\theta|P_{1},P_{2},P_{3},\cdots,P_{8}\rangle$ $\displaystyle=$
$\displaystyle|P_{2},P_{1},-P_{3},\dots,-P_{8}\rangle\ .$ (A.5)

This involution leaves the states $|1,1,0,\dots,0\rangle$ and
$|-1,-1,0,\dots,0\rangle$ invariant which correspond to the roots of the
$A_{1}$ lattice. Likewise,

$\displaystyle\theta|P_{1},P_{2},P_{3},P_{4},P_{5},\cdots,P_{8}\rangle$
$\displaystyle=$
$\displaystyle|P_{2},P_{1},P_{3},P_{4},-P_{5},\cdots,-P_{8}\rangle$ (A.6)

(with $s=1+4=5$) leads to $I=A_{1}+D_{2}$. The invariant lattices
$I=A_{1}+D_{4}$ and $I=A_{1}+D_{6}$ are obtained in a similar way. We conclude
that by permuting the first two entries and reflecting an $m$ number of the
remaining $6$ momentum entries we find the invariant lattice $I=A_{1}+D_{6-m}$
with $\delta=1$, $a=3$ for $m\neq 6$ and $a=1$ for $m=6$.

In the remainder of this appendix we will provide explicit constructions of
the invariant lattices $I$ (and correspondingly the normal lattices $N$) for
each point in figure 1. For some constructions the invariant lattices are the
same as the entries in table LABEL:tab_IN, and for some constructions they are
isomorphic to the corresponding entries.

###  A.1 Even $a$

Let us first consider lattices with even values of $a$. An even self-dual
lattices with signature $(\gamma_{+},\gamma_{-})$ and
$\gamma_{+}-\gamma_{-}\equiv 0\mod 8$ is unique up to lattice isometries [74].
Thus, $\Gamma\cong U^{3}+E_{8}^{2}$. We shall start with an even self-dual
lattice of interest $\Gamma$ and define the specific involutions which act on
$\Gamma$ to yield the desired lattices $I$ and $N$. For $a=2$ we perform
involutions to arrive at an invariant lattice of the form $I=U+D_{2m}$. Since
$D_{2m}^{*}/D_{2m}=\mathbb{Z}_{2}^{2}$, we have $a=2$ — see table 3. Thus the
associated invariant lattices correspond to the triples $(2m+2,2,0)$ for
$m=2k$ and $(2m+2,2,1)$ for $m=2k+1$, $k\in{\mathbb{Z}}$. For $a=4$, we take
involutions which yield $I=U+D_{2m_{1}}+D_{2m_{2}}$. This gives the triples
$(2+2m_{1}+2m_{2},4,0)$ for $m_{1}=2k_{1}$, $m_{2}=2k_{2}$, and
$(2+2m_{1}+2m_{2},4,1)$ for $m_{1}=2k_{1}+1$, $m_{2}=2k_{2}+1$ where
$k_{1},k_{2}\in{\mathbb{Z}}$. We shall continue with such a decomposition to
obtain all values of $a=2,4,6,8,10$.

There are also other possibilities at hand, e.g taking involutions such that
the invariant lattice contains the $E_{8}(2)$ lattice with the triple
$(8,8,0)$ and/or the $U(2)$ lattice with $(2,2,0)$. Moreover, an even value of
$a$ with $\delta=1$ can also be obtained by taking an even number of $A_{1}$
components and/or 2 $E_{7}$ components — cf. eqs. (A.3) and (A.5).

####  A.1.1 $a=0$

Start with the even self-dual lattice
$\Gamma=U+(D_{2};D_{2})[(Sp_{2};Sp_{2}),(V_{2};V_{2})]+D_{8}[Sp_{8}]+D_{8}[Sp_{8}]$.
The lattice $D_{8}[Sp_{8}]$ is even self-dual and is equal to the $E_{8}$
lattice. Consider the involution changing $s_{0}$ signs in the left $D_{2}$,
$s_{1}$ signs in the first $E_{8}$, $s_{2}$ sings in the second $E_{8}$ and
the signs of the two right components in $D_{2}$. The total value of $s$ is
$s=s_{0}+s_{1}+s_{2}$. For instance, we choose $s_{0}=2$, $s_{1}=8$, $s_{2}=8$
with the involution

$\theta|P\rangle=|-P_{1L},-P_{2L},-Q_{1L},\dots-
Q_{8L},-Q_{1L}^{\prime},\dots,-Q_{8L}^{\prime};-P_{1R},-P_{2R}\rangle$ (A.7)

where $P_{L}$, $Q_{L}$, $Q^{\prime}_{L}$ and $P_{R}$ correspond to the left
$D_{2}$, the first $E_{8}$, the second $E_{8}$ and the right $D_{2}$,
respectively. This involution leads to $I=U$, $(r,a,\delta)=(2,0,0)$.
Similarly, taking other values of $s_{0}$, $s_{1}$, and $s_{2}$ we find
lattices of different ranks. All lattices with $a=0$ in figure 1 are obtained:
$s_{0}=2$, $s_{1}=0$, $s_{2}=0$ gives $s=2$, $I=U+D_{8}[Sp]+D_{8}[Sp]$ with
$(r,a,\delta)=(18,0,0)$.
$s_{0}=2$, $s_{1}=8$, $s_{2}=0$ gives $s=2$, $I=U+D_{8}[Sp]$ with
$(r,a,\delta)=(10,0,0)$.
$s_{0}=2$, $s_{1}=8$, $s_{2}=8$ gives $s=2$, $I=U$ with
$(r,a,\delta)=(2,0,0)$.

####  A.1.2 $a=2$

Start with the even self dual lattice
$\Gamma=U+(D_{18};D_{2})[(Sp_{18};Sp_{2}),(V_{18};V_{2})]$ with conjugacy
classes $(Sc_{18},Sc_{2})$, $(V_{18},V_{2})$, $(Sp_{18},Sp_{2})$ and
$(Sp^{\prime}_{18},Sp^{\prime}_{2})$. Besides changing the signs of 2 right
components we change $s$ signs in left weights, where $s$ is even, to obtain
$I=U+D_{18-s}$. We find that all $a=2$ points are reproduced in this way
except for $(r,a,\delta)=(2,2,0)$, $(2,2,1)$, $(10,2,1)$ and $(18,2,1)$.

Now consider
$\Gamma=U+(D_{2};D_{2})[(Sp;Sp),(V;V)]+D_{8}[Sp_{8}]+D_{8}[Sp_{8}]$.
Reflecting first the two right movers leaves us with $(D_{2},0)$ and further
reflecting one left entry in $D_{2}$ gives
$(D_{2};D_{2})[(Sp,Sp),(V,V)]\rightarrow A_{1}$. Note that the corresponding
component in the normal lattice is of the form
$(A_{1};D_{2})[(R_{1},Sp^{\prime}_{2})]$. Next, performing the involution
(A.3) on one of the $E_{8}$ lattices yields $E_{7}$ and we find
$I=U+A_{1}+E_{7}+E_{8}$ with $(r,a,\delta)=(18,2,1)$. Reflecting in addition
all the directions along the $E_{8}$ lattice gives $I=U+A_{1}+E_{7}$ with
$(r,a,\delta)=(10,2,1)$.

To obtain the triple $(2,2,1)$, consider
$\Gamma=(A_{1};A_{1})[(F_{1};F_{1})]^{2}+U+E_{8}^{2}$. Reflect the left
$A_{1}$ in one $(A_{1};A_{1})[(F_{1};F_{1})]$ and the right $A_{1}$ in the
other one. Further reflect along all the 18 directions in $U+E_{8}^{2}$. This
yields $I=(A_{1};A_{1})$ with $(r,a,\delta)=(2,2,1)$.

Finally, consider $\Gamma=U^{3}+E_{8}^{2}$. Performing the involution (A.1) on
two $U$ lattices which leads to $U(2)$, inverting the left and right
directions in the remaining $U$ as well as the $16$ left directions along
$E_{8}^{2}$ we find: $I=U(2)$, $(r,a,\delta)=(2,2,0)$, $s=18$.

####  A.1.3 $a=4$

Consider $\Gamma=U+(D_{m};D_{2})[(Sp_{m};Sp_{2}),(V_{m};V_{2})]+G_{18-m}$,
$m=2+8k=\\{2,10,18\\}$. The value of $m=18$ leads us to the construction in
subsection A.1.2. Here we consider $m=\\{2,10\\}$. The lattice $G_{18-m}$,
i.e. $G_{16}$ and $G_{8}$, must be even self dual. For $m=2$ we can choose
$G_{16}=D_{16}[Sp_{16}]$ or $G_{16}=D_{8}[Sp_{8}]+D_{8}[Sp_{8}]$ and for $m=8$
we have $G_{8}=D_{8}[Sp_{8}]$. The idea is to invert $s$ left weight
components distributed among $D_{m}$ and $G_{18-m}$ where $s$ is an even
integer.

Let us first consider $m=2$ and choose the even self dual lattice
$G_{16}=\Gamma_{16}=D_{16}[Sp]$. We invert two right entries in $D_{2}$ and
$s$ signs in $G_{16}$ ($s\neq 16$). This gives $I=U+D_{2}+D_{16-s}$ with
$(r,a,\delta)=(20-s,4,1)$. 666Note that if we were to invert the signs of $s$
entries as 2 in $D_{2}$ and $s-2$ in $G_{16}$ then $I=D_{16-(s-2)}+U$,
$(r,a,\delta)=(20-s,2,0),(20-s,2,1)$ for $s=2+8k$, $s=4k$ respectively and we
recover the $a=2$ result above . We thus obtain the triples $(18,4,1)$,
$(16,4,1)$, $(14,4,1)$, $(12,4,1)$, $(10,4,1)$, $(8,4,1)$ and $(6,4,1)$.

Next we consider $G_{16}=D_{8}[Sp]+D_{8}[Sp]$. Besides the reflection on two
right directions, we invert $s_{0}$ signs in the left $D_{2}$ lattice, $s_{1}$
signs in the first $D_{8}$ and $s_{2}$ signs in the second $D_{8}$ lattices.
where $s_{0}=2$ and $s_{1}$ and $s_{2}$ are even integers. We then obtain
$I=U+D_{8-s_{1}}+D_{8-s_{2}}$ with $s=2+s_{1}+s_{2}$, $s_{1},s_{2}\neq 0$ and
$r=18-s$. This provides an alternative way of constructing some of the
lattices with $(r,4,1)$ triples with $r>4$. Notice that for $s_{1}=s_{2}=4$
the length squares of all weight vectors in $D_{8-s_{1}=4}$ and
$D_{8-s_{2}=4}$ are integers and so $\delta=0$. Thus in this case we obtain
the triple $(10,4,0)$.

The triples $(4,4,1)$, $(6,4,0)$ and $(14,4,0)$ cannot be constructed in this
way. However, it appears that we can construct the invariant lattices for
these models using the $U(2)$ lattice. The special property of the $(6,4,0)$
triple is that it is the unique point with $a<r$ that necessarily contains the
$U(2)$ lattice in its constituent components [32]. Start with
$\Gamma=U^{3}+D_{8}[Sp_{8}]+D_{8}[Sp_{8}]$, permute two $U$ lattices to get
$U(2)$, reflect along the left and right directions of the third $U$ lattice,
change the sings of $s_{1}$ momentum entries in the first $D_{8}$ and $s_{2}$
entries in the second $D_{8}$. We obtain:
$s_{1}=6$, $s_{2}=8$ gives $I=U(2)+D_{2}$ with $(r,a,\delta)=(4,4,1)$.
$s_{1}=4$, $s_{2}=8$ gives $I=U(2)+D_{4}$ with $(r,a,\delta)=(6,4,0)$.
$s_{1}=4$, $s_{2}=0$ gives $I=U(2)+D_{4}+E_{8}$ with $(r,a,\delta)=(14,4,0)$.

Finally, we consider the triple $(18,4,0)$. This case is slightly more subtle
because the normal lattice has signature $(2,2)$ with $(r,a,\delta)=(4,4,0)$
and the only possible way to realise this is to have a lattice isomorphic to
$N=U(2)^{2}$. This model is constructed in section 6.3.3.

####  A.1.4 $a=6$

Consider the even self dual lattice
$\Gamma=U+(D_{2};D_{2})[(Sp_{2};Sp_{2}),(V_{2};V_{2})]+D_{8}[Sp_{8}]+D_{8}[Sp_{8}]$.
Following our previous constructions, we consider $s=s_{1}+s_{2}$ inversions
distributed among the two $D_{8}$ components where $s_{1},s_{2}\neq 0$. This
gives $I=U+D_{2}+D_{8-s_{1}}+D_{8-s_{2}}$ with $a=6$ and we obtain the triples
$(16,6,1)$, $(14,6,1)$, $(12,6,1)$, $(10,6,1)$ and $(8,6,1)$.

To construct the triple $(6,6,1)$ consider the even self dual lattice
$\Gamma=U^{2}+(A_{1};A_{1})[(F_{1};F_{1})]+D_{8}[Sp_{8}]+D_{8}[Sp_{8}]$.
Perform an involution that leads to $U(2)$, inverts the signs on both $A_{1}$
directions, and inverts 6 directions in each $D_{8}$ lattice. This gives
$s=1+1+6+6=14$ and we obtain $I=U(2)+D_{2}+D_{2}$ with $(r,a,\delta)=(6,6,1)$.
All triples with $a=6$ points are constructed.

The two remaining points to construct are the $a=6$ triples with $\delta=0$:
$(10,6,0)$ and $(14,6,0)$. As for the former, the invariant lattice may be
constructed by starting from $\Gamma=U^{3}+D_{8}[Sp_{8}]+D_{8}[Sp_{8}]$,
exchanging two $U$ lattices, and reflecting the third $U$ as well as four
entries in each $E_{8}$. This yields $I=U(2)+D_{4}+D_{4}$.

The construction of the triple $(14,6,0)$ is similar to the $(18,4,0)$ model
and is discussed in section 6.3.4.

####  A.1.5 $a=8$

Start with
$\Gamma=U+(D_{2};D_{2})[(Sp_{2};Sp_{2}),(V_{2},V_{2})]+(D_{4},D_{4})[(Sp_{4},V_{4}),(V_{4},Sp_{4})]+D_{8}[Sp_{8}]$.
Consider the inversion of $s_{0}$ signs in $D_{2}$, $s_{1}$ in the first
$D_{4}$, $s_{2}$ signs in the second $D_{4}$ and $s_{3}$ signs in the $D_{8}$
lattice. This gives $s=s_{0}+s_{1}+s_{2}+s_{3}$. To obtain $a=8$, we set
$s_{0}=0$, $s_{1}=s_{2}=2$ and $s_{3}\neq 0$. This gives
$I=U+D_{2}+D_{2}+D_{2}+D_{8-s_{3}}$. The triples $(14,8,1)$, $(12,8,1)$ and
$(10,8,1)$ are constructed in this way.

Next consider the triples $(10,8,0)$ and $(8,8,1)$. The former can be obtained
from $I=U+E_{8}(2)$. To construct the latter, consider
$\Gamma=U^{3}+D_{8}[Sp]+(D_{4},D_{4})[Sp_{4},V_{4}]$. Obtain $U(2)$ under the
involution as before and switch $s_{1}=6$ signs in $D_{8}$ as well as
$s_{2}=s_{3}=2$ signs in the two $D_{4}$ lattices. We find
$I=U(2)+D_{2}+D_{2}+D_{2}$ with $(r,a,\delta)=(8,8,1)$.

####  A.1.6 $a=10$

Start with
$\Gamma=U+(D_{2};D_{2})[(Sp_{2};Sp_{2}),(V_{2};V_{2})]+(D_{4},D_{4})[(Sp_{4},V_{4}),(V_{4},Sp_{4})]^{2}$.
Consider as before reflections $s_{i}$ acting on each factor with $s_{0}=0$
along the left $D_{2}$. Choosing $s_{1}=s_{2}=s_{3}=s_{4}=2$ we find we find
$I=U+D_{2}+D_{2}+D_{2}+D_{2}+D_{2}$ with $(r,a,\delta)=(12,10,1)$.

There remains the two points $(10,10,0)$ and $(10,10,1)$ to be constructed.
Consider $\Gamma=U^{3}+D_{8}[Sp_{8}]+D_{8}[Sp_{8}]$. First, exchange two $U$’s
to obtain $U(2)$. Next, reflect along the directions of the remaining $U$.
Finally, permute the two $E_{8}$ lattices. We obtain $I=U(2)+E_{8}(2)$ with
$(r,a,\delta)=(10,10,0)$.

To obtain the $(10,10,1)$ triple, start from
$\Gamma=(A_{1};A_{1})[(F_{1};F_{1})]^{3}+D_{8}[Sp_{8}]+D_{8}[Sp_{8}]$. Reflect
2 right momenta in the first two $(A_{1};A_{1})$ factors. Next reflect the
left momenta in the second and third $(A_{1};A_{1})$ factors. Finally permute
the two $E_{8}$ momenta. This gives $I=(A_{1};A_{1})+E_{8}(2)$ with
$(r,a,\delta)=(10,10,1)$.

###  A.2 Odd $a$

As discussed earlier, for odd values of $a$ the invariant lattice must contain
an $E_{7}$ component and/or an odd number of $A_{1}$ components, see the
paragraph below eq. (A.1). We shall now construct the invariant lattices $I$
for odd values of $a$.

####  A.2.1 $a=1$

We start with the lattice
$\Gamma=U+(D_{2};D_{2})[(Sp_{2};Sp_{2}),(V_{2};V_{2})]+D_{8}[Sp_{8}]+D_{8}[Sp_{8}]$.
We next decompose the left $D_{2}$ lattice as $(A_{1},A^{\prime}_{1})$ with
correlated classes $Sc_{2}=(R_{1},R_{1})$, $V_{2}=(F_{1},F_{1})$,
$Sp_{2}=(F_{1},R_{1})$ and $Sp^{\prime}_{2}=(R_{1},F_{1})$. Perform an
involution that reflects the two right directions along $D_{2}$ as well as one
left direction along $A_{1}^{\prime}$. This gives $I=U+A_{1}+E_{8}+E_{8}$ with
$(r,a,\delta)=(19,1,1)$. If in addition we reflect eight directions of one of
the $E_{8}$ lattices, we obtain the triple $(11,1,1)$. Similarly, reflecting
eight directions of both $E_{8}$’s gives the triple $(3,1,1)$.

Start with the same even self dual lattice, reflect the two left and the two
right directions along $(D_{2};D_{2})$, apply the involution on $E_{8}$ which
leads to $E_{7}$ — see eq. (A.3). We obtain $I=U+E_{8}+E_{7}$ with
$(r,a,\delta)=(17,1,1)$. Reflect in addition along the directions of the
$E_{8}$ lattice to obtain $I=U+E_{7}$ with $(r,a,\delta)=(9,1,1)$.

In order to obtain the triple $(1,1,1)$ consider
$\Gamma=[(A_{1};A_{1})(F_{1};F_{1})]+U^{2}+E_{8}^{2}$. Reflect the left
$A_{1}$ direction, as well as all directions along $U^{2}$ and $E_{8}^{2}$. We
obtain $I=A_{1}(-1)$ with $(r,a,\delta)=(1,1,1)$.

####  A.2.2 $a=3$

Start with
$\Gamma=U+(D_{10};D_{2})[(Sp_{10};Sp_{2}),(V_{10};V_{2})]+D_{8}[Sp_{8}]$.
Reflect $s$ directions along $D_{10}$ where $s$ is even and $s_{1}\neq 10$.
Perform the involution (A.3) to obtain $I=U+D_{10-s_{1}}+E_{7}$. This yields
the triples $(19,3,1)$, $(17,3,1)$, $(15,3,1)$, $(13,3,1)$ and $(11,3,1)$.

Alternatively, use the $D_{8}[Sp_{8}]$ base in eq. (A.2) and apply the
involution (A.5) to obtain $I=U+D_{10-s_{1}}+A_{1}$. We obtain the triples
$(5,3,1)$, $(7,3,1)$, $(9,3,1)$, $(11,3,1)$ and $(13,3,1)$ where the last two
were already constructed in the above paragraph.

Finally, to construct the $(3,3,1)$ triple start with
$\Gamma=U^{2}+(A_{1};A_{1})(F_{1};F_{1})+D_{8}[Sp_{8}]+D_{8}[Sp_{8}]$. Permute
the two $U$’s to obtain $U(2)$, reflect the left and the right $A_{1}$
components, apply the involution (A.5) on the first $D_{8}[Sp_{8}]$ lattice
and reflect all eight entries in the second $D_{8}[Sp_{8}]$. We find
$I=U(2)+A_{1}$ with $(r,a,\delta)=(3,3,1)$.

####  A.2.3 $a=5$

Consider
$\Gamma=U+(D_{2};D_{2})[(Sp_{2};Sp_{2}),(V_{2},V_{2})]+D_{8}[Sp_{8}]+D_{8}[Sp_{8}]$.
Following the above constructions, perform involutions to obtain
$I=U+D_{2}+D_{8-s_{1}}+A_{1}$ and $I=U+D_{2}+D_{8-s_{1}}+E_{7}$, where $s$ is
even and $0<s<8$. For $I=U+D_{2}+D_{8-s_{1}}+A_{1}$ we find the triples
$(11,5,1)$, $(9,5,1)$ and $(7,5,1)$ and for $I=U+D_{2}+D_{8-s_{1}}+E_{7}$ we
obtain $(17,5,1)$, $(15,5,1)$ and $(13,5,1)$.

To obtain the only remaining triple $(5,5,1)$ start from
$\Gamma=U^{2}+(A_{1};A_{1})(F_{1};F_{1})+D_{8}[Sp_{8}]+D_{8}[Sp_{8}]$. Perform
the permutation leading to $U(2)$, reflect the right $A_{1}$ direction and
perform involutions on each $D_{8}[Sp]$ to obtain $A_{1}^{2}$. We find
$I=U(2)+A_{1}^{3}$ with $(r,a,\delta)=(5,5,1)$.

####  A.2.4 $a=7$

Consider
$\Gamma=U+(D_{2};D_{2})[(Sp_{2};Sp_{2}),(V_{2};V_{2})]+D_{8}[Sp_{8}]+D_{8}[Sp_{8}]$.
Reflect the two right directions along $D_{2}$, perform $s_{1}$ reflections in
the first $D_{8}[Sp_{8}]$ lattice where $s_{1}$ is even and $0<s_{1}<8$. Next
apply the involution of the type (A.6) to the second $D_{8}[Sp_{8}]$ to select
the invariant lattice $D_{6-s_{2}}+A_{1}$, where $s_{2}$ is even and
$0<s_{2}<6$. We obtain $I=U+D_{2}+D_{8-s_{1}}+D_{6-s_{2}}+A_{1}$ corresponding
to the triples $(15,7,1)$, $(13,7,1)$, $(11,7,1)$ and $(9,7,1)$.

To obtain the $(7,7,1)$ triple start with
$\Gamma=U^{2}+(A_{1};A_{1})(F_{1};F_{1})+D_{8}[Sp_{8}]+D_{8}[Sp_{8}]$. Permute
the two $U$ components to obtain the $U(2)$ lattice, reflect the right $A_{1}$
direction as well as six directions along each $D_{8}[Sp_{8}]$. We find
$I=U(2)+D_{2}+D_{2}+A_{1}$ with $(r,a,\delta)=(7,7,1)$.

####  A.2.5 $a=9$

Start with
$\Gamma=U+(D_{2};D_{2})[(Sp_{2};Sp_{2}),(V_{2};V_{2})]+(D_{4},D_{4})[(Sp_{4},V_{4}),(V_{4},Sp_{4})]+D_{8}[Sp_{8}]$.
Following our approach in subsection A.1.5, we reflect the two right
directions along $D_{2}$, $s_{1}$ left directions in the first the $D_{4}$,
$s_{2}$ left directions in the second $D_{4}$, and perform an involution on
the $D_{8}[Sp_{8}]$ component which leaves $D_{6-s_{3}}+A_{1}$ invariant. Here
$s_{1},s_{2},s_{3}>0$ and even. We find
$I=U+D_{2}+D_{4-s_{1}}+D_{4-s_{2}}+D_{6-s_{3}}+A_{1}$ which yields the triples
$(13,9,1)$ and $(11,9,1)$.

Next consider
$\Gamma=U^{2}+(A_{1};A_{1})[(F_{1};F_{1})]+(D_{4},D_{4})[(Sp_{4},V_{4}),(V_{4},Sp_{4})]+D_{8}[Sp_{8}]$.
Perform the involution on $U^{2}$ to obtain the $U(2)$ lattice, reflect the
right component $A_{1}$, and reflect $s_{1}=2$, $s_{2}=2$ and $s_{3}=6$
directions along $D_{4}$, $D_{4}$ and $D_{8}$. This yields
$I=U(2)+A_{1}+D_{2}+D_{2}+D_{2}$ with $(r,a,\delta)=(9,9,1)$.

####  A.2.6 $a=11$

Start with
$\Gamma=U^{2}+(A_{1};A_{1})[(F_{1};F_{1})]+(D_{4},D_{4})[(Sp_{4},V_{4}),(V_{4},Sp_{4})]+(D_{4},D_{4})[(Sp_{4},V_{4}),(V_{4},Sp_{4})]$.
Perform the involution on the $U$ lattices to obtain $U(2)$, reflect the right
direction along $A_{1}$, reflect 2 entries in each $D_{4}$ component. We find
$I=U(2)+A_{1}+D_{2}+D_{2}+D_{2}+D_{2}$ with $(r,a,\delta)=(11,11,1)$.

This concludes the construction of the $I$ and $N$ lattices for all 75 points
in figure (1).

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