Nuclear Physics B301 (1988) 381-425 North-Holland, Amsterdam
T W I S T E D S U P E R S T R I N G S IN F O U R D I M E N S I O N S A. H. CHAMSEDDINE L2 and J.-P. DERENDINGER Institut ffir Theoretisehe Physik, ETH-HiSnggerberg, CH-8093 Ziirich, Switzerland
Received 24 August 1987
We study the formulation of string theories in four dimensions with explicit space-time supersymmetry, allowing the string coordinates to have both shifted and twisted, orbifold-like, boundary conditions. The theories we construct are automatically tachyon-free and can have gauge groups of small ranks. As an example, we recover the Z 3 orbifold model. We describe the construction of superstring theories with rank eight gauge groups, illustrated by an SO(10)® SU(3) ® U(1) model.
1. Introduction O n e of the most interesting recent developments [1] in string theories has been to realize that one can in fact easily escape large space-time dimensions and construct string theories directly in four dimensions. M a n y such models have already been constructed [1-7]. They generalize the concept of compactification of string theories, which gives some phenomenologically attractive models by considering, for instance, heterotic strings [8] on orbifolds [9-11]. T h e crucial observation is the fact that string fields characterizing the coordinates of the right- and left-moving sectors for the dimensions higher than four can live on different manifolds, and have different b o u n d a r y conditions. This suggests [1] constructing these theories in a manner similar to the ten-dimensional heterotic string. Because of the vast n u m b e r of possible models that can be constructed, one needs some guidelines to help sort out the phenomenologically appealing candidates. In most of the presently k n o w n theories, the rank of the gauge group is twenty-two, which is much larger than the group of the standard model, S U ( 3 ) ® SU(2) ® U(1). This may definitely be a source of problem, when, for instance, the question of s y m m e t r y breaking is considered. A " h i d d e n sector" is in general useful for p h e n o m e n o l o g y , but there is no need for such a large extension of the standard model. Also, rank twenty-two gauge groups are certainly not esthetically attractive. Thus, one of the requirements we will impose ourselves will be to obtain four1 On leave of absence from the American University of Beirut, Beirut, Lebanon. 2 Supported by the Schweizerische Nationalfonds. 0550-3213/88/$03.50©Elsevier Science Pubfishers B.V. (North-Holland Physics Publishing Division)
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dimensional string theories with lower rank gauge groups. This can be achieved by allowing the (complexified) bosonic fields to have twisted boundary conditions. We shall also demand to obtain N = 1 supersymmetry in four dimensions. This requirement is motivated by the many attractive features of supersymmetry when discussing lower energy phenomenology. To make space-time supersymmetry explicit in all our formalism, we use the Green-Schwarz construction [12] of the right-moving sector, and in the light-cone gauge. Our fields in four dimensions will be identical to those of the heterotic string of Gross et al. [8], but now with the freedom of not identifying the boundary conditions of the six "internal" right-moving coordinates with six of the left-moving ones. The requirement of supersymmetry is not fundamental in this approach. It can be easily removed to get non supersymmetric models. Our strategy will be as follows. We shall first specify the field variables necessary for the cancellation of the two-dimensional conformal anomaly, as well as for explicit space-time supersymmetry. We then investigate the structure of the N = 1 supersymmetry algebra and the related conditions on the twisted boundary conditions applied to string coordinates. This is sect. 2. In sect. 3, we derive the one-loop partition functions for bosons and fermions with twisted boundary conditions and their transformations under the modular group. Having obtained the partition functions, we use them as building blocks for the construction of the total modular invariant partition function associated to a set of (commuting) boundary conditions. This is the main issue discussed in sect. 4. The solution to the constraints imposed by modular invariance is obtained by generalizing the concept of GSO projection, in close analogy with the work of Kawai, Lewellen and Tye [2]. Our formalism is naturally close to this work. Although we use quite general boundary conditions, we have some reasons to believe that our solution is not the most general one can obtain in this formalism. In sect. 5, we illustrate our method by first deriving some models corresponding to Narain compactifications [1] of the ten-dimensional heterotic string. Introducing the twisted boundary conditions, we then obtain as an example the full spectrum of the Z 3 orbifold compactification [9] in both twisted and untwisted sectors. We use this well known case as an example of the many features of our formalism. We then construct some new models, with lower rank gauge groups. As a first example, we show how to modify the conditions for the orbifold case in order to obtain a rank eight gauge group. Sect. 6 contains our conclusions and some remarks. An appendix contains the Lorentz and supersymmetry generators in four dimensions relevant to the formalism of twisted boundary conditions.
2. The Green-Schwarz formalism in four dimensions Having set the requirement of explicit N = 1 space-time supersymmetry, we will choose the right-moving sector to carry this symmetry. Right-moving degrees of
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freedom will then be those of the Green-Schwarz superstring [12]. The left-moving sector will be purely bosonic. In this respect the construction we use is completely analogous to the ten-dimensional heterotic string [8]. The conformal anomaly is cancelled in the same way as for the heterotic string: we are using the same set of fields. Manifest space-time supersymmetry can only be easily realized in the light-cone gauge where the world-sheet time-like parameter r is identified with X += ~ - ( X ° + X3). In this light-cone gauge, the physical degrees of freedom of the two dimensional field theory are as follows. For the right-moving sector we have two transverse coordinates X"(r-o), ( a = 1 , 2 ) and their two supersymmetric partners S " ( r - o) (S ~ is a four-component spinor subject to the Majorana and light-cone gauge conditions). In addition, we take as "internal" coordinates the complex fields Zk(T--O), (k = 1,2,3) and their supersymmetric partners S ~ k ( ~ - o) (also subject to the Majorana and light-cone gauge conditions). The left-moving sector consists of two transverse coordinates X"(r+ o) and 22 internal coordinate XA(~r + o). We begin by discussing space-time supersymmetry, Lorentz invariance and their implications for the boundary conditions applicable to the string coordinates. It is convenient to take full advantage of Green and Schwarz formulation of the ten-dimensional superstring in the right-moving sector [12]. We can do so since our fields are linearly related to those of Green and Schwarz. The only difference will be in the boundary conditions. To find an explicit relation between both sets of fields we first consider the ten-dimensional light-cone action
s= -
1
f aod ( a . x % x " + o x'a.x' + i
gr-a+s),
(2.1)
where X M are the light-cone transverse coordinates ( M = 1. . . . . 8), S ~ is a 32-component spinor subject to the Weyl, Majorana and light-cone gauge conditions. The F-matrices are the 32 × 32 Dirac matrices in ten dimensions. The X I ( I = 1 . . . . . 16) are coordinates constrained to be left-moving only. The action (2.1) is invariant under the space-time supersymmetry transformations
6xM= ~
1
6SS=~F
~FMS'
F~( O X;')e,
(2.2)
VP where X ~ = ( X °, X M, X9). Considering the above fields as four-dimensional fields is similar to the reduction of ten-dimensional supergravity into four dimensions [13]. The case here is however even simpler since dimensional reduction corresponds only
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to field redefinitions. It is convenient to use the following F-matrix representation: F ~:
T ~'® I8,
/-'i=y5 ®
FJ = T5 ®
~t= 0,1,2, 3,
(0) '
0
_Oti
0 flJ
,
,
i=4,5
'
6,
j=7,8,9,
(2.3)
where (oti)k,l , and (flJ)k'r are 4 X 4 antisymmetric representations of SU(2) ® SU(2). We have exchanged indices 3 and 9: the light-cone gauge condition will be applied in four dimensions. In the above representation, the spinor S s is decomposed according to
S e, = - - ~
_ iTss,,k'
,
where a is an SO(l, 3) index and k' an SO(6) - SU(4) index. The four-dimensional spinors S '~k' are subject to the Majorana as well as the light-cone gauge conditions: (y+)~'#S #k" = 0. The three complex fields Z k are defined by the equation
Z k= --~((oLi)klXi-F (~J)klXJ).
(2.5)
To make N = 1 supersymmetry evident, we decompose the SU(4) spinor index k' into (1, k), where k is now an SU(3) index. The transformations (2.2) for the transverse four-dimensional space-time coordinates and their superpartners now read
8 X '~ = i
8S o =
(8"lag q- Ek'~ask), 2
~((,~
,~e)a
x,-i~r_(~+a
where
~+__k _ ½(1 _+y5)~ k.
z * + e ~ _ O _ z * * ) ) °,
(2.6)
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For the remaining fields Z k and S k, we have ~Zk----- p2~ ( ~ S [ _ ) _ ~ k S ( _ ) + ~ k , , , ~ Sl , "(+)l' 2 3sk=~(y_y,
ekO_X"+ivt2y_(e+O_Zk+e
0 Z *k)
+i~/2ektmy (et+O Z *m + e' 0 Z m)),
(2.7)
where S~+ ) = ½(1 ___,/s)S k. From eqs. (2.2), it is evident that to preserve supersymmetry in ten dimensions, both X M and S a must obey the periodic boundary conditions xM(¢ + o - - ~ ) = S~(~- o-,)
XM(¢ + o ) , = S~(~- o).
(2.8)
This implies that if we employ periodic boundary conditions for the four-dimensional fields X", Z k, S ~ and S ~k, we will obtain N = 4 right-moving superstrings with parameters e and e k. Since we are mainly interested in N = 1 supersymmetric string models, we shall only keep the supersymmetry generated by e and eliminate supersymmetries generated by e k with the help of boundary conditions. The transverse coordinates X a must obey periodic boundary conditions, and supersymmetry transformations (2.6) will then imply that S ~ also obeys periodic boundary conditions. Also, from eq. (2.7), O Z k and Sk have the same boundary conditions, but in order to break the supersymmetries generated by e k, these must be taken to be twisted (orbifold-like). For simplicity we shall only consider abelian twists of the form g Z k ( r -- o) = zk(~ " -- o -- ~r) = exp(--2¢riOk)zk('r -- o) + v k , gSk ( ~ -- o ) = s k ( $ -- a -- Tr) = exp(-- 27riOk ) S k ('r -- o ) ,
(2.9)
where g is a group element defined in ref. [9], made out of rotations 0 and translations v, and represented by g = (0, v). The remaining N = 1 supersymmetry transformations are i 6X" ~ ~3'~S' 2 ~S~= ~(~, Cp
~)"0
x~,
2 =
~Sg_ = 2
,
i3' e + O Z k.
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In terms of the four-dimensional fields, the action in the light-cone gauge is sl~ =
- --
1
4 ~0l t
f d r d o ( O~XOO.XO - i ~ , -
O+S
+20,~zko'~z*k--iJ~ Sky-O+sk + O~,XAO"Xa), (2.11) where the coordinates X ~ contain both left- and right-movers, while the X a (A = 1 . . . . . 22) obey a constraint restricting them to be left-movers. Having fixed periodic boundary conditions for both X a and S% their mode expansions are given by 1 a
g a ( T -- O)
o)+
--o~ne n ¢ O F/
1
=1
i x a + ½ P a ( r - o) + ~i E --a~ e-2i'~(¢ "), n~O
S~(r- o) = ~
/7
Sffe -2i"~'-°),
(2.12)
with the quantization conditions
[ x ~ , p b ] = i 8 ab,
[<,=~]=[<,s~l=nS~+m,08°~,
=0,
(2.13)
In the case of coordinates Z k (which can always be thought of as the right-moving part of the coordinates of a six-dimensional torus), we can have untwisted as well as twisted sectors. In the untwisted sector, the boundary condition is ZkOr) = Zk(O) + vk, where v k is a complex translation vector. We then have winding sectors with quantized momenta. The coordinate expansion in this case is
Zk(r--o)=
½zk + ½ P k ( r - - o ) + li E nI otke " -2in(r
o)
n~O Z**(,_
o ) = i, _t , k
+ 71P , k _( . , - o ) + 7 , , . ~
1
--Rke-2'"(" °)
(2.14)
n=~O r/ Ivn
where p k = _ 2vk/rr and rink= a*_~. The corresponding quantization conditions are
[,+,<1=o, [ a~, fl~m'] = nS,+m,o 8k'k'.
(2.15)
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In the twisted sectors with the boundary conditions (2.9) and Ok ~ 0 (modulo 1), all winding states are forbidden and the expansion for Z k is given by
1
i ;~ =k ~-- 2i(n--Ok)('r--o) z k ( q" - O) = Z k-l- ~, ~.,"~ .~------~lXn_OkC nil--J7 z * k ( ' r - O) = z ' k +
1
li Z __Rk ke--2i(n+Ok)(r--o) n + 0 kt'n+O n
(2.16)
where zk~uk(1
--e--2i~'0k) -1,
]~k+0 k ~-0/ . kn Ok.
The quantization conditions are k
k'
k
k'
[OLn_Ok, Olm_Ok] = 0 ,
;+0d : ( . -0k ,/8 n + m , O 8 k'k'
[BL0 , k,
(2.17)
] :0.
Similarly, the expansion of the space-time spinors S k is
S,,k=
~ n=
S2ke-2i,¢, o)
(2.18)
OO
in the untwisted sector. The quantization condition is
{ S: k, @~ } = (7 +) ~flS,+m,o 6k'k' .
(2.19)
In the twisted sectors, remembering that S ~ ) and S[+) have the same boundary conditions as 3 Z k and O Z *k respectively, the expansions are: ~'~sak
S(a--k)= z.., ( - ) n n
0
ke--2i(n 0k)(r-- o)
ak __ ~ (,otk .~-2i(n+Ok)(r-a) S ~ + ) - Z..,~,(+)n+O~,
with the quantization condition
{ S~k(-).-0k, ~ +)ilk, m+ 0~' } = (hy +) "138,+ m,o6~, where h = ½(1 - 3'5); all other anticommutators vanish.
(2.20)
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Working in a light-cone gauge has a major drawback, namely that we have to check the closure of the supersymmetry algebra. In the right-moving sector, and in the case of untwisted boundary conditions, the fields and their quantization conditions are identical to the Green-Schwarz superstrings, where the closure of the algebra has been explicitly shown [12] (it corresponds to the N = 4 algebra in four dimensions). In the case of twisted boundary conditions, we are only interested in the subalgebra corresponding to N = 1 supersymmetry. By repeating the analysis of Green and Schwarz, it is straightforward to obtain the group generators. For instance, the N = 1 supersymmetry generator is 2 t.~ d
3~ o
(2.21)
which, when evaluated, reads 2i
(2.22) All other Lorentz generators can be explicitly evaluated. They are given in the appendix. The algebra of N = 4 supersymmetry is broken to N = 1 simply because there is no zero mode S~ k in the twisted sectors: all quantum numbers n are shifted by + Ok. Otherwise all terms present in the relevant Green-Schwarz subalgebra are also present here, with the oscillators with opposite phases coupling to each other. The only subtlety arises when considering the Lorentz generators J ~ - , since they involve trilinear couplings of oscillators with the three phases summing to zero. In particular, the generator J~- contains the term
~ dt~ ieklrn~k)ya- s[_ ) O_Z m ,
(2.23)
which must survive the a-integration to provide the trilinear couplings. This gives the constraint ekl~ (0 k + 0, + 0m) -- 0,
mod 1
(2.24)
or, in a diagonal basis like the one we are using, 01 + 02+ 03 = 0,
modl.
(2.25)
This condition is analogous to the "triplet" constraint found in refs. [2] and [3].
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It is also trivial to construct the Lorentz generators in the left-moving sector. One just combines oscillators with shifted quantum numbers for the twisted bosons such that there is no net phase shift. All steps that were used in proving the closure of the supersymmetry algebra in ten dimensions will now hold for the subset of generators in four dimensions. All terms appearing in these generators are also present in the corresponding components of the ten-dimensional generators. We have thus determined the string coordinates which must participate in a theory using the formalism of Green and Schwarz superstrings for its right-movers, and a bosonic left-moving part. Lorentz invariance will be preserved as long as eq. (2.25) is satisfied. Although we have emphasized here N = 1 supersymmetry, one could as well obtain N = 4, N = 2 or no supersymmetry at all. The N = 4 case will correspond to periodic boundary conditions for all right-moving fields. To obtain N = 2, one must twist only two complex fields, say Z 2 and Z 3, with 01=0,
02=03~---12
(2.26)
modl
so that the constraint (2.25) is satisfied. To break all supersymmetries, since we must keep periodic boundary conditions for the space-time coordinates X a, the S ~ must have twisted boundary conditions. In this case, we can see from the generator J~ that the phase 0 ° of S ~ must satisfy oOm - ~1
rood 1
to preserve Lorentz invariance. 3. One-loop partition functions
The world-sheet of a one-loop graph has the topology of a torus. Its points can be labelled by the complex variable z = % + ~-o2. The real variables % and o 2 are periodic, with period 2~r. After the identification
z - z + 2~r- z + 2~r~,
(3.1)
the torus can be considered as the complex z-plane. The modular parameter r is restricted to the fundamental region -½
I m p = T2> 0,
[r{ > 1 .
(3.2)
Periodic boundary conditions are required for the coordinates X ~ and their superpartners S~:
xa(o1 -[- 297, 02) = x a ( o l , o"2 -b 2~r) = X~(%, o2), S ~ ( % + 2~r, %) = S~(o1, % + 2Tr) = S~(Ol, o2) ,
(3.3)
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while orbifold-like conditions are satisfied by the complex coordinates Z k (and similarly b y Z~):
gZk( 01, O2) = zk( o, + 27r, %) = e- Zi'Okzk( o~, 02) + tryk, hZk(01, a2) = Zk(a,, 02 + 2rr) = e-2i~'~*Zk(ol, 02) + ~u k.
(3.4)
T h e relations
ghZk(Ol, 02) = hgZk(al, o2) = zk(Ol + 2~r, o 2 + 2~r)
(3.5)
show that the group elements g and h must commute and also that uk(1 -- e - 2'~0k) = vk(1 -- e - 2'~+~).
(3.6)
In the special case where both 0 k and epk are zero, there is no constraint on u k and v k. T h e string fields are then in an untwisted sector. As required by supersymmetry, the boundary conditions on S k_ must be identical to those of o_zk:
gSk(ol, 02) = s k ( o l + 21r, o2) = e-2i~°ksk_(ol, o2), hsk_(ol, 02) = Sk_(ox, a2 + 2~r) = e--2i~Oksk_(al, a2).
(3.7)
We now have to discuss the contributions of the different fields with their different b o u n d a r y conditions to the one-loop partition function. Firstly, for the transverse coordinates X a, the partition function arising from the path integral over the coordinate fluctuations is well known. It has been shown [10] to be obtained by evaluating Tr[
q~r'(x°)~r( x°) l
(3.8)
in the operator formalism, )fit and ~ r being the hamiltonians for the left- and right-moving parts of X ~. In this expression, q
=
q~ e -2i~r~r"
e 2i~T '
The result is
.~(xa)
1 =
_ (~(q)~/(?/))-i, ~'2
(3.9)
in terms of the Dedekind eta function oo
~ ( q ) = ql/24 1--I (1 - q " ) . n~l
(3.10)
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391
The first factor in eq. (3.9) arises from the integration of the zero modes, while the second factor is due to the summation over non-zero oscillator modes, taking into account the ground state energy due to normal ordering. We must next evaluate the partition function for the space-time spinors S ~k. Notice that the partition function for the spinor S ~ will then follow by choosing a zero twist Ok. The hamiltonian for a pair of fermions S ~k (with fixed k) obeying eq. (3.7) is of the form
f doS*y- 0
S*,
(3.11)
which gives, after insertion of the mode expansions of S*+, ~'e'(S"~)=½ ~
[(n+Ok--1)ff,+ok-,~ '
S ~ + 0 k _ x + ( n - e k ) £ _ e k v - S ~ 0k]
n=l
+ ½(02- 0k+ ~).
(3.12)
The last term is the ground state vacuum energy due to normal ordering. The corresponding number operator is defined by N(S~k) = ½ ~
(ff,+ok_lT-S,+okl- S,_okY-S, o~).
(3.13)
n=l
The partition function is most easily found by evaluating the trace Tr[ q~(S"% e x p ( - 2 iTrq5N( S"k))] ,
(3.14)
where the last operator induces a twist by an angle ~ in the o2 direction. We use the Green-Schwarz definition [12] for calculating the trace in case of space-time spinors: Trs,s ~¢= (0 [~10) - (1 [s¢[ 1).
(3.15)
It immediately follows that
L~(S"k)=q 12(°~-°k+'6)fi
(1-qn+°k-le
n~l
2i,,q,k)(l_q,-Oke2i,,ok)
---L~;°~ .
(3.16)
The expression given in eq. (3.16) holds for a left-moving fermion. For a right-moving fermion, one would get the complex conjugate. We note that eq. (3.16) is related to the Jacobi theta function by [14] 1
1
1
[1
0
o.~°=e2i~(°-2)(q~-2)--~O I ~)-- ½
(3.17)
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By setting 0 = ~ = 0, one obtains the partition function for the (periodic) spinor S ". It is then given by ~o°, which vanishes (it has a zero mode) by the usual properties of the 0 function. This is not unexpected since the periodic boundary condition for S ~ is a consequence of N = 1 supersymmetry, which also implies that the zero point loop amplitude (the cosmological constant) vanishes. Since ~0° will be a factor in the total partition function, and since it transforms up to a phase into itself under modular transformations, one will then require that the rest of the partition function also transforms into itself, up to an opposite phase. One can also be more careful and, in analogy with the study of chiral anomalies [15], keep an infinitesimally small twist angle ~ in the o2 direction and thus keep the zero mode under control. We note that the order of the zero mode (i.e. the power of ~eo entering the partition function) gives also the order of supersymmetry. We now study the complex bosons, with the boundary conditions (3.4). In a twisted sector, no winding states are allowed. To find the hamiltonian, one substitutes the mode expansion for Z k into the expression
fo"doO Z*kO_Zk
(3.18)
to get Je(z
((n +0k-
k) =
1 +(,-
° '
'*
- - ~1 ( 0 k2 - - 0 k + ~).
'
(3.19)
Compared with the expansion (2.16), we have split Otn_0% k gl ~ -- 0 0 , . . . , 00, i n t o t w o sets a and a' with positive indices n. The corresponding number operator is N(zk)
=
~ (et ~n + O k - l ~~n + O k - 1 n=l
,t , - a,_o~a~_o~ )
(3.20)
The last term appearing in eq. (3.19) is due to normal ordering and, as expected from supersymmetry, is equal and opposite to the contribution of the twisted space-time fermions. The partition function in this case is found by evaluating the trace
Tr( q~r( Z~)e- 2~*N( Z~)
(3.21)
Using the identity Tr~,,A = E
(n[A In ,
(3.22)
i/
the partition function reads
q-~:(o~-o~+~) I-I
( 1 - q "+°~
le-Zi"~'~)(1-q n °~e2i=q'k)
Ln=l = ~
- 10~
th k "
(3.23)
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The analysis presented above will cease to hold in the case where both Ok and 0 k vanish: by definition, this case corresponds to an untwisted string state for which winding states must be considered. For a left-moving real boson with the boundary condition
X(o 1 + 2~r, 02) = X(ol, 02) + ~rP,
(3.24)
where the m o m e n t u m P belongs to a shift lattice: P = k + 0,
k integer,
the hamiltonian is ,~
1 2 gP + ~_, nasa,,
24"1
(3.25)
n>O
When substituted into the trace formula
Tr[ qS%- 2i~q'(P-°)] ,
(3.26)
this hamiltonian leads to the partition function
q-a/24fi(1-q")
1 ~
n=l
q~(k+O)2e-2i'~'k.
(3.27)
k= -~
In eq. (3.26), we have assumed that the twist operator in the o2 direction is e x p ( - 2i~rq~(P - 0)) and that the number operator with integer eigenvalues is P - 0. Inserting the infinite product representation of the theta functions:
~., q~(k t +O)e-21~kq, 2 ~ 2-~,=1/(~.) , q ~(o f i (1 + qn+O k=
oo
~e-2i~r,h)
n=l
×(l+q'
0 ~e2i~,)
into the trace (3.21), one obtains
q )(e2 ~ ) f i (1 +
qn+°-~e-2i"~')(1 + q"-°-'2e2i~')
n=l _-- ~ 0 + -,+z/2
1 / 2 ~ "~"vu -
,
[-1 oQ~ ~-,.--oj
where u = 0 + ~ and v = 0 + ½. Recall however that our bosons are complex. We can also write the previous result o, ~u'xOx. 5e~:,
where vx and vy are the real and imaginary parts of v, and similarly for u.
(3.29)
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A bosonic string field can be either in a twisted or untwisted sector. To include both possibilities in a single expression, it proves convenient to introduce the projection operator (3.30)
bZo, = 8o08,0.
Then, the general partition function for a complex boson satisfying the boundary conditions (3.4) is V
or,
,
(3.31)
for g = (0, v) and h = (if, u). In the special case of a single real boson, the only allowed non trivial twist corresponds to an angle 7r (i.e. 0 = ½), and the partition function is (1 - 2 ~ e , ) ~ - ) ~ + 2 ~ 0 0 ~
;
0 , 0 , ~ {0,}}.
(3.32)
We are now ready to write the total partition function for a fixed set of boundary conditions. We shall always factor out the contribution ~e(xa) of the transverse coordinates which are always periodic, and specify only the contributions of internal degrees of freedom. We denote the set of boundary conditions by a vector W similar to the one used in ref. [2], with two blocks of coordinates corresponding to rightand left-movers respectively. The first entry gives the boundary condition for the spinor S% As we mentioned already, we will always assume a vanishing twist for this spinor, in order to preserve N = 1 supersymmetry. (The possibility of nonsupersymmetric cases can be easily studied by removing this restriction.) The next three entries correspond to the spinors S '~k and contain their twist angles Ok. The boundary conditions for the three complex bosons Z k are then denoted by the three triplets (Ok, vxx,, vY). This completes the boundary conditions for the right-movers. The entries for the eleven complex left-moving bosons Z z are denoted by (01, v], v]). To summarize, the full set of boundary conditions in the o 1 and o2 directions is given by
w = (0, 0,, (0,, w , = (o,
v;)j(0,, v;, v;)), .;,
u;,
(3.33)
The corresponding partition function is 3 k=l
~k \ \
g'k
11
(3.34)
A,H. Chamseddine, J.-P. Derendinger / Twistedsuperstrings
395
where ~0%~ has been abbreviated by ~ . The full partition function is simply obtained by multiplying the expression (3.34) by the contribution o~e(X~) of transverse bosonic coordinates. This factor can be temporarily ignored since it is by itself modular invariant.
4. M o d u l a r i n v a r i a n c e
The string one-loop amplitudes must be invariant under the modular transformations ar+b
r ~ - -
cr+d
(4.1)
'
where a, b, c and d are integers satisfying a d - b c = 1. These transformations generate the modular group. When r is restricted to the fundamental domain defined in eq. (3.2), the modular group can be generated by the two transformations 1 'r
~ r + 1.
(4.2)
It is then sufficient to check the invariance of amplitudes under these two basic transformations. Since the basic building block of the partition function (3.34) is the function ~eo defined in eq. (3.16), we must first find its behaviour under modular transformations. This behaviour can be obtained from the transformation properties of Jacobi theta functions and of the Dedekind eta function. Since eq. (3.17) defines ~eo in terms of these two functions, one then finds: _
1
e2i~-(8
I 1 ~)(~OL~e% ,
,/-
~'~r+l:
,,~z'f ~ e i~(°2
(4.3)
°+ ~)~z'° o .
The modular transformations of the complex boson partition function M~ defined in eq. (3.31) can then be deduced from these basic transformations: 1
'r --* - - - - : .~ff---) e 2 i ' [ ( u T
r~'+l:
~'~e
vo)(,,
Vo)-~-(O-~)(q~-~2)(1--@)]~hg
~<(v-~+vo)(. . . . .
>~-(e-'2~+)(e
,
~a ) ( 1 - . ~ ' ) ] ~ g
g,
(4.4)
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A . H . Chamseddine, J.-P. Derendinger /
T w i s t e d superstrings
where
~=~00~.0, -=(~,u~),
~=(Ox,O,
To simplify the subsequent analysis, we introduce the following notation• For every set of triplets
g, = (0,, v;, vr),
h, = (~,, u;, .,Y),
we define the diagonal matrix ~s by ~ i = diag(3~i- 1, ~ i , ~g) ;
(4.6)
~g = 3o,o8o,o.
Eq. (4.4) takes then the simple form
1
q- --_~ __ n
•
~,~ag; ~
e2m(g,-go)-~,(h,-go)q~_a,g
,
T
(4.7) where go = (½, ~, ~)- With eqs. (4.3) and (4.7) at hand, it is straightforward to find the transformation of ~ew,, defined by eq. (3.34):
1
~
,
T ---> -- -- : ~w, __0e2i.(w - wo).~.(w - w o ) ~ w w , 'r
¢--*T+I: Y~W-*ei~(w-~+wo)~(w-~
Wo)~W
W ,
(4.8)
where W0 is a vector with all components equal to ~, and ~ is a diagonal matrix defined by ~ w w , = diag(1,1,1,1, ~'~k]~7)l)
(4.9)
in terms of the twist angles contained in the two vectors W and W'. The scalar product between two vectors W and W' is given by
W. W ' = ~, W ' W " - ~_, W r W 'r l
(4.10)
r
with summations taken over the left (index l) and right (index r) components respectively. Notice that we have written eq. (4.8) intentionally in a manner similar to that of ref. [2] in order to use a similar strategy to construct modular invariant partition
A.H. Chamseddine, J.-P. Derendinger / Twisted superstrings
397
functions. We however warn the reader that although some expressions may look similar to those of ref. [2], the fields we are using and their boundary conditions are different. Inspecting eqs. (4.3) and (4.7), it is evident that one needs a full set of vectors { W } such that if W and W' in the partition function ~ew"; belong to this set, then so would be the vectors in ~ew~ and ~ w , _ w. To determine such a set, we first note that ~d"° is periodic in the lower argument ~, with period 1. This is apparent in eq. (3.16): (4.11)
~ey+t=~Z'f .
This implies that ~ef is not periodic in 0. If we choose the initial arguments 0 and q5 in ~ f to belong to the interval [0,1], then the arguments in ~2~% and ~ ° 0 remain in the same interval, after invoking the periodicity (4.11) when necessary. The ability to restrict our set to take values in the interval [0,1] is the main justification of our choice to work with Y.f rather than with the 0 functions. As a simple example, when both 0 and q5 are of the form 1 / n , n being an integer, then all functions ~ef with 0, q~~ {0,1/n . . . . . (n - 1 ) / n } must be present in the modular invariant partition function. At this point, it is convenient to decompose the vectors W into two orthogonal pieces, one corresponding to rotations and the other to translations: W = R ~3 V,
R- V = 0,
(4.12)
where R=
(0, 0k, (0k,0)l(0,,0)),
v= (o,o, (o, v,)l(o,
v,)).
(4.13)
The vector R has in general many entries which are fractions. The procedure is then to find the common denominator, say N, and then include all the vectors {0, R , 2 R . . . . . ( N -
1)R}.
If one starts with a set of M independent vectors { R,, }, then all vectors of the form M
aR-
Z
amRm,
a , , = 0 , 1 , . . . , Nm - 1
(4.14)
m=l
must be included. N m is now the common denominator of all vectors R m. Similarly, for the translation vectors V, if we take a set of P vectors { Vp}, then all vectors P
c V = ~ , cpVp, p=l
cp = O, 1 . . . . . Np - 1
(4.15)
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A.H. Chamseddine, J.-P. Derendinger / Twisted superstrings
are included. We thus denote the general set of vectors W by (4.16)
s/W = aR • cV.
By taking all possible (and necessary) values for a m and ce we generate all the elements of the set. Although the values a m and Cp can be independent, in order to be more general we must also allow for the important case where the action of a twist is accompanied by a shift. In this case the set of shift vectors is (E,,,a,,,Wm + Y'.pcpVp ). Obviously, we recover the previous case by setting all V,, to zero. For convenience, we shall write A W = a R + c ' V where c' = E a mVm + EcpVp - c'?,V~,. We have seen that we can limit all the entries of each vector to belong to [0,1]. This is not in general the case for all vectors of the form (4.16). We then define new vectors by
__
aR = aR
mod 1,
c V = c-V
mod 1,
(4.17)
q
with all entries in a R and c V belonging to the interval [0,1]. One subtlety arises due to the appearance of the projection matrix ~ , defined only for two fixed vectors, in the ~ew, transformations. For a set of vectors (~¢W}, there will exist matrices ~ , w . e w . We denote this set of matrices by {~ah}, with the same definition as in eq. (4.9), for a pair of vectors aR and bR, since the matrices ~ only depend on the rotation vectors. The components of ~ab a r e ~ I a b and ~kab, with ~@tab= ( $a -0-, 0' 8 - -b '0 , 0
"
We write for the general partition function .~f= ~
c~Aw ~AW ~gW ~c~W,
(4.18)
A,B
where C AW ~ are constants whose dependence on A W and B W mined from the requirement of the modular invariance of ~ . The transformations in eq. (4.8) imply 1 "r
is to be deter-
I~AW o),,BW A,B X e2i~(AW--Wo).~"b.(~--Wo)
~ ' - - ) ¢ + 1 : ~f--)~0~= Y'. r A w ~ d w
~BW~(B-A)W
A,B
× ei,~(AW-,~+Wo).~°~.(~-- Wot.
(4.19)
A,H. Chamseddine,J.-P. Derendinger / Twistedsuperstrings
399
From these transformations, we easily deduce that CB~-~-w= exp(ZiTr(A-W - W o ) . ~ b . ( B W
- Wo))C~--w~,
C~(,-A)w = exp(i~r(AW - a+ Wo) • ~ ' h . ( A - - W - a
_ Wo) ) C~--ew'4.~
(4.20)
Unfortunately, the above equations cannot determine C~w, Aw but only constrain it. To proceed further, we adopt the approach of ref. [2] to generalize the concept of the Gliozzi-Scherk-Olive (GSO) projection operator. We note that in another approach [7], these constants were determined by demanding two-loop modular invariance, but only when the rotation angles in the purely fermionic formulation are restricted to be 0 or ~r. In general, the exact connection between the two-loop modular invariance and the generalized projection is not known. Going back to eqs. (3.21) and (3.26), we can express ~ ww, as a trace:
Tr(q'~'q"~re-2i'(W'-V°)S~Nw),
~w=
(4.21)
where •~ r
=~(Sa)
,,'a'=
71- Z ~ ( S k
ak) -1- E l ( 1
-
+
k
E[(1 -,~,)~(z')+,~',.¢~(z')].
(4.22)
I
The hamiltonians in eq. (4.22) have been defined in eqs. (3.12), (3.19) and (3.25). The number operator N w is a vector with the same structure as W, and with components
Nw=(N(S~),N(S'k),(-No(Zk),N~(Zk))I(-No(ZZ),Nv(Z'))
) , (4.23)
with N ( S '~k) and No(Z k) given in eqs. (3.13) and (3.20), while No is a doublet with eigenvalues (k x, ky) related to the winding numbers. The appearance of the minus sign with No(Z ) is due to the definition of the projection matrix ~ , where g i - 1 appears instead of 1 - ~ i . Substituting eq. (4.21) into eq. (4.18), we obtain
~= y' ¢~T..[ ,.,,~'(~)Z,,~'~AW),..-2i,,(~Wvo)..~".~v~]
(4.24)
A,B
We now require that the sum on B (i.e. in the o2-direction ) must form a projection operator E I'~AW ~ - 2 i ~ - ( ~ B
=e
VO)'~ab'NAw
2'~'a)7/(A--W, ~, ( a ) , ~pp,(c', ~ a°)),
(4.25)
A.H. Chamseddine, J.-P. Derendinger / Twisted superstrings
400
where e(a) is a phase to be determined shortly (and whose value will be zero or 1) while ~/is a function that takes values zero or one, and its arguments are defined by
dp,(a) = R,. ~ab. N~w = R,. ~ b . Nh~,
+p,(C', ~ab)= Vp,.~ab. N ~ = Vp, .~abv "~" ~ .
(4.26)
In the above equation, we have decomposed the vectors N ~ and the matrix ~ into orthogonal rotation and translation directions. For future reference, we write #~b and ~#~b explicitly
=
b- 1,o,o), (0,0,0,0,(0,
~koh , ~;b)l(O ' ~ b , ~ f h ) )
(4.27)
We can now invert eq. (4.25), after multiplying it by e2iTr(B'w -
Vo).~ah.N~w
and summing over q~,(a) and @,(c', ~ b ) , to obtain AW~ C~w
m=l
Nm 1--[ Ne p=l
e-2i~*(")
X Y'~ rl(AW,~n,~bp)e 2i~'(b¢(a)+d'~(c''~'~b) %(c',~,h)).
(4.28)
¢,' q'e' The appearance of the vector V0 in eq. (4.24) forces us to assume that this vector is always present among the set of boundary vectors and in all sectors. Using eq. (4.28), we can evaluate both CA~ v0w. In this special case, the first equation vo and C~(4.20) reads C~vo = cA-Wvvo= e2i~r(AW- Wo)-~#"°,(v°- Wo)O~AwVO.
(4.29)
With the help of eq. (4.28) inserted into (4.29), one gets the relation
E (AW, +o,+,,)
ee,, +p
= exp(2iTr ( e ( a ) - ~(0) - aW. 2z~.o. Ro + Ro. ~ o . Ro)) ×
exp(2i~r (a+(0) + c'q~(Vo, ~ v ) _+o(Vo, ~ v ) ) ) . ~l(Vo, ,., @,). ¢,, ~p (4.30)
A.H. Chamseddine, J.-P. Derendinger / Twisted superstrings
401
The left-hand side of eq. (4.30) is, by definition, positive definite since 7/ can only take values 0 or 1. The obvious solution to render the right-hand side positive definite is to set all independent terms in the phases to zero, modl,
e(a) -e(O)=aW.~°°.Ro
mod 1,
~)n(O) "~-@n(O, c p = ~ p o , ~ a O ) = o
@(0, Vo, ~aa°) = 0
modl,
Ro. ~ o . Ro = 0
rood 1,
(4.31)
and we can also take ~/(V0, q~,(0), @,(V0, ¢~a0)) to be one and zero for the other values of its arguments. The last of these equations could have been absorbed in e(a), but must hold since it can be proven independently from the second equation in (4.20), by setting A W to zero. It is not easy to find a different set of solutions, although we suspect that some special solutions exist. We can now conclude that ~/(A W, q~,, @) is non-vanishing for a unique value of the phases q~n and @, and that eq. (4.28) simplifies to
C w=
I =IN
=
e 2i~e(a)
× e 2i~r(b~(a)+ d'+(c', .~ab)_ ~o(C,"~ah)) .
(4.32)
The upshot of the projection requirement, eq. (4.25), is that the constants C are given, apart from a multiplication factor, by a unique phase. This phase must further satisfy eq. (4.20) which will then insure modular invariance. Also, substituting eq. (4.32) into the first equality (4.20) and using the first identity (4.31) leads to - (aeo(b) + c'~b(d', ~ a b ) + ~Po(d, ' ~ o o ) + b W . ~ a o . Ro ) = bq~(a) + d'g,(c', ..@~b) _ O/o(C, ' ..@oh) _ a W . ga~.b. Ro +-a-K.~. b . bR-aR . - c . v. ...~.
~ . b . R.
v b • Vo _ Vo • ~ b
o
R. o . ~ . b . b R
+c'V
. d V + W o • ~ a b . W o.
~v~b'dV~
modl. (4.33)
This identity must hold for all possible values of a n, bn, Cp and dp, and will split into two equations, one of them involving a and b only, while the others will involve all variables because of the mixing between a, c and b, d. The dependence
A.H. Chamseddine, J.-P. Derendinger /
402
Twisted superstrings
on ~ , b of the above equation makes finding its general solution a very difficult task: ~ depends non-linearly on a and b. We will only solve it under the assumptions of equating aW. ~ab. Wo with aW. ~aO. Wo' and of the vanishing of Wo. ~ a b . W0' since its vanishing when b = 0 is a consequence of the last identity in (4.31): W o - ~ "b. Wo= 3 R o . ~ b .
Ro
(11
=-34 E ~ ' ; b -
~l~;b)
=0
modl.
(4.34)
I=1
The assumption we have just made can be proved when the Vm are zero, but not otherwise. As we shall see later on, this will generally constrain our ability to reduce the rank of the group. Two further equations can be derived using eq. (4.33), when considering the special cases where c and d or a and c or d and b are zero:
bq~(a)+aq~(b)= - a W . : ~ a b . b W + 2SW.~ab. Ro
modl,
(4.35)
where + ( a , ~ a b ) = + ( a ) + + ( a , c = 0, ~ a b ) and
d[¢(a,c,~ab)--+(a,O,~ab)] +c[~(b,d,~ab)-~(b,O,~ab)]
+b [+(a, c, ~,~a) _ +(a,O, ,.~aa)] + a[¢(b, d, ~ab) __ ~(b,O, = __ c t V
"~vab " d V + a V ' ~
b'~-~
modl.
~ab)] (4.36)
These equations hold for the different choices of the coefficients. From the symmetry between a and b in eq. (4.33), we obtain 0m(0) = 2 R 0 . ~ a°. Win=0
modl,
(4.37)
which justifies our assertion that the phase e(a) in eq. (4.31) takes only the values zero or one. We can also verify that the general solution of eq. (4.25) is given by Om(a,~ab)
ab = ETn~na
n-
W,,.~"a.aw,
(4.38)
n
where the matrices Tin, are defined by T ~ = - *re(b, = 3,,)
(4.39)
403
A.H. Chamseddine, J.-P. Derendinger / Twisted superstrings
and must satisfy Tm~ 6 = Win"~ b .
W,
(4.40)
rood 1.
Similarly, eq. (4.36) can be solved to give t~p(a, C, .~) = ~p(O,O, ~ ) q- E k p q c q - k E k p m am q m
-- Vp . ~ b " a V + c V
modl,
(4.41a)
t~m ( a , C, ..q~) = @m( a , O, .~ ) -+- E k m q c q - V,. . sg ~b . a-F--4 c V q
+ Vm. ~ b .
~
(4.41b)
rood 1,
and 2k~ob = 0
(4.42)
rood 1.
The matrices k~ b are defined by kpb, = l~q,(O,O, ~i~ab) - lpq,(O, dr= 6rp, ~ab )
rood 1,
(4.43a)
k mabq -__ ~bq(O,O, ~
mod 1,
(4.43b)
ab
) - t~q(O, bt-- 6t,~,
~ab)
and must satisfy
k~bq + kq~ = Vp. ~ b .
Vq,
(4.44a)
k ~ + k ~ = Vp. ~ b .
Vm"
(4.44b)
Further constraints on the above solutions will result from the second equality in eq. (4.20), following from the ~"~ ~"+ 1 invariance. After substituting in eq. (4.31), we obtain: - (a~(a)
+ c'~ ( c ' , ~ ° b ) ) = 1-"~a~• ~ ° b • aR - ~R - ~ "
+ T177"~1Z c _ . ~ ~ a vb • "d'V - c v .,.r~ ~ b v . V o
Ro
(4.45)
modl.
Again we can split the above equation into two parts by first setting c to zero and then substituting back. After some algebra, one obtains the constraints W,, k~-
~;o~ = ~' G ~ "
~pp,(0,0, ~ob) --_
k o~b p,
W, + Wm Vo
Wo
mod 1,
(4.46)
rood 1,
(4.47)
rood 1.
(4.48)
A,H. Chamseddine, J.-P. Derendinger / Twisted superstrings
404
Collecting all results, we can now write for the constants: AW
CBW =
e - 2iw~ (HNmHN~) m p - '
bm
Xexp 2i,,'r
. ~ . o . wo
T~,a,-
Wm'~ab'aW
q-
EkmqCq q
kpqc + Ekpma
Vp'~v" aV+ cV+ k~ph]) . -
-
m
(4.49) The total partition function, including the contributions ~ ( x ") of space-time coordinates, is then of the form: 1
-1
T2"q2 ~ 2
,
× z e'"~;~°°'~°Tr(,~""w~'°'q ~'''AW'~'o,) a,b,c,d
2i~ b,,,
×exp
T,~.a~-
Wn •
aWq-
2kmqCq q
2iw~(dp-
×exp \
po
kpqcq+
p
d,.,Kpm a
--
gp
aV+cV
m
+k~hp- Vp.~V. N,4w]),
(4.50)
where ~egtl=~l( xa) q-~I( AW, l~la),
yc,,r = ) f , r ( x a )
+ovfr(AW, ~ , ) .
(4.51)
A.H. Chamseddine, J.-P. Derendinger /
405
Twisted superstrings
By construction all states will be projected out except those satisfying the equations Wm • ~"b" NAw= ~ T,,,,.a.ab W , , , . ~ " b . a W o + Ek~qCq n
q
- Vm • ~ b ( a V +
cV - a l / )
mod 1,
(4.52)
and ab ab Vp.~"t'.NAw= Ekpqcqq~kpm a mq
Vp . ~ b
.
a V + c V + k o pab
modl.
(4.53)
m
We note at this point that the second and third equations in (4.31) are special cases of these equations. When evaluating the one-loop amplitude, we must perform the integral f d "/'1
d T2
ff
~ total
(4.54)
where the integration measure is modular invariant by itself. The ~'1 variable appears only through q, and the integral over ~'1 is given by f d ' r e 2i~rrl( ' ~ ' ' - Jr''') .
(4.56)
This integral produces then the constraint Jt°'z -- J~ 'r .
(4.57)
On the other hand, the integration over ~'2 has the form f d ~-2e - 2~'2(~"'+a~"r)
q
and the masses of on-shell states can be read off the exponent mass formula is then m 2 = ½(yg,,, + jf,,~) _ p2.
(4.58)
e -4~t'c2(Pz+m2).
The
(4.59)
When associated with eq. (4.47), the mass formula becomes m 2=
m 2(right) = m 2(left),
where m 2(right) = jf~,~
p2,
m2(left) = y { , , / p 2 .
(4.60)
406
A.H. Chamseddine, J.-P. Derendinger / Twisted superstrings
The computation of the eigenvalues of the above operators immediately gives 2 a=l
n=l 3
+
n=l
--k k ~ ((n +~k--1)SP~k+~_1+(n-a0 )Sa,,_~)
k=l
n=l
+ k=l
n=l
k~a I ~ ka
_~-f"y-k
gx
2
,
(4.61)
where the quantities (written in cursive letters) are, in the order they appear in the mass formula (4.61), the number operators contained in the harniltonians for X ~, S ~, S ~k and Z k. Similarly
m 2 (left, a-W) =
- 1 '~
,~,;~a + E ( 1 - ~ , o )
E a=l
n=l
( - ~ ((a-0') 2
~'
)
I=1
+ .~= l ((n + ~ - ~ t 1)a~:+~_ 1 +
+ Y'~,, ½(K x'+CVx -1) + I=1
(n_a--Oi)~._~,))
n~x+(x~y)
•
(4.62)
n=l
The relations between the number operators defined in eq. (4.32) and the eigenvalues appearing in (4.61) and (4.62) are as follows:
n=l
N~(Zk) = ~ (d:+~--1-Ja¢~--~), No(z~) = (K~, ~#),
(4.63)
and similarly for Z I. We note that the ground state energy in the fight-handed
A.H. Chamseddine, J.-P. Derendinger / Twisted superstrings
407
sector is zero because the Fermi and Bose contributions always cancel as required by supersymmetry. The constraint in eq. (4.60) then implies that all tachyonic states are projected out. We can obtain some restrictions on the allowed rotation and translation vectors that will depend only on these vectors. By using the fact that all components in NAW take integer values and defining N, and Np as the smallest integers satisfying the equations
N~W~ = 0
rood 1,
NpVp = 0
mod 1,
(4.64)
we obtain from eq. (4.26): Nm~,~( a, ~ ) = 0
Np@(c,~)=O
rood 1, modl.
(4.65)
Using eq. (4.38) in these last equations, we deduce that
~_,r~ma. +
kp..Cp
=0
mod 1,
n
Npa...~'-qp Y'k~bc q = 0
modl.
(4.66)
q
Inserting these results into eqs. (4.40) and (4.44) leads to
NmT/,ab__ m- 0
mod 1,
ab__ N p k qp - 0
mod 1,
ab __ Nmkpm - 0
mod 1.
(4.67)
Stronger restrictions are also obtained from eqs. (4.46) and (4.47) after using the above equations
Um(½Wm "~ab" Win+ W m ' ~ ab" 14/o) = 0 Np(½Vp'~
b" J/p q- V p ' ~
b" 1 7 o ) = 0
modl, modl.
(4.68)
For convenience, we will close this section by collecting all the equations that will
408
A.H, Chamseddine, J.-P. Derendinger / Twisted superstrings
be frequently used in the next section. All the following equations are satisfied modulo 1: ab
ab __
T,~n+ T,~en- Wen
~ab.
Wn ,
(4.69)
ab __ T,~en -- ~ W m .dj~ab. Wen + W o • ~ab • W m ,
(4.70)
Wen = O,
(4.71)
NmT, m = 0,
(4.72)
2Wo. ~ab.
N m ( ~ W , , . ~ ab. W, + W o . ~ ab. Win) = 0 ,
(4.73)
kpq,"b + kq,pab= Vp" ~ , b " Vq,,
(4.74)
k pabe - k p 0,b=
!2,p v . ~ a--v h " Vp,
(4.75)
2k~ob = 0,
(4.76)
Np,kqb,=o,
(4.77)
Np(Vp • ~ b ' V o +
11/" . o ~ a b .
2"p ~ v
Vp)=0.
(4.78)
All states will be projected out except those satisfying Wm • ~ a b " N ~ w = Z., ~ " T a.,~a~ b __ W m . ~ a b . a W - k n
- Ven. ~ h . V p . ~ ab • N ~
E k m qabc q q
( a V + cV - aV) + k~bm,
(4.79)
~ ~L.KpqCq tab ~_ ~kpmaen ab q m
- V p - ~ b . (aV--VG--~) + kop~b.
(4.80)
We finally recall the important constraint 1
E ~Dlab- £ ~kab 1=1 k=l
= 0
(4.81)
5. Applications We now proceed to use the formalism developed in the previous sections to construct some four-dimensional superstring theories. To familiarize the reader with
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our method, we will first reproduce some models that have been already obtained using different approaches. We begin by recovering some of the four-dimensional heterotic strings constructed by Narain [1], possessing rank 22 gauge groups. We consider them as first simple examples since they contain only untwisted sectors. The rotation vectors R are all set to zero, and thus all #k~b and ~i~b are equal to one. We have seen that the vector V0 (with all components v k and v I equal to 1) must always be present. If we take coVo as our set of vectors, c o has only the two values zero or one, and we get two subsectors corresponding to 0 and V0. For the right-moving modes, it will be clear by examining eq. (4.61) that the zero subsector corresponds to massive excitations only. The mass formula of the right-movers contains only non-negative terms. It is then enough to observe that the last term is strictly positive in the zero subsector: 3 1
E (K~-~)
1 2
+(xoy)>0.
(5.1)
k=l
In the v0 subsector, all terms in m 2 (right) are non-negative, the vacuum energy is zero and the massless ground state corresponds to the N = 4 supersymmetric states la)R, Is)R, I k + ~ : ) R and lak)R. The constraint (4.60), equating left and right masses, tells us that we only need to consider the V0 subsector to enumerate the massless modes of the left-movers. These massless states will be solutions of the equation
l =l ~
K~2 + Ky
I=l
+ ~ a=l
n~ n=l
11
+ 2
(5.2)
1=1 n=l
At this point, and to simplify our notation for the subsequent analysis, it is convenient to double the domain of the indices of the vector components with x and y coordinates. Thus (KxJ, Ky/) --* (K'*), where I* = 1 . . . . . 22, and (K~, K ) ) --, (Kk*), with k * = 1 . . . . . 6. Also, since we have decomposed the vectors W into orthogonal R and V parts, we will only write the components which are allowed to be non-vanishing. For instance
R = ( O , Ok,Ok[01) , V=(Vk,[VI,),
k=1,2,3,
I=1,...,11,
k * = a . . . . . 6, I * = 1 . . . . . 22.
(5.3)
We now list the solutions of eq. (5.2). First, we can set either s ~ or ~ ( to one while taking all other numbers to be zero. This is the vector ground state la)L"
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Next, we can set one of the 2 2 ~ * to one and keep all other numbers to zero. We get in that way the 22 states lI*)L" These states correspond to the Cartan generators of the gauge group (which will then have rank 22). Finally, if we set two of the K i* to + 1 and all the other numbers to zero, we find the states characterized by K 1 . = (0 . . . . . +_1,0 . . . . . + 1 , 0 , . . . ) ,
I = 1 . . . . . 22.
(5.4)
These states span a 22-dimensional lattice. Adding these 924 states to the 22 Cartan generators, one finally finds the states IG)L, completing the adjoint representation of the gauge group SO(44). The full massless spectrum is obtained by taking the cross product of the left and right massless modes: ( ] a ) R + lOt)R+ [k + k ) R + ]°tk)R) ® (]b)L + [SO(44))L)This product leads to the spectrum of N = 4 supergravity coupled to N = 4 supersymmetric SO(44) Yang-Mills multiplet. We can obtain different rank-22 groups by enlarging our set of shift vectors V, but without introducing any rotation vector R. For example, the group E 8 ® E 8 ® SO(12) corresponds to the choice coVo + clV 1 + c2V2, with
::I ),
1 6 1 22
V: : (061014, (1)8) , g 2 = (06106,(12)8,08),
(5.5)
and thus Co, c: and c z can have values zero or one. We must then consider the subsectors 0, V0, V1, V2, Vo + V1, V0 + 1/2, V1 + V2 and Vo + V1 + V2. To possess massless states, a subsector must have all entries of the shift vector corresponding to the right-moving modes with value 5-: This is clear from eq. (5.1). Then, only Vo, Vo + V1, Vo + V2 and Vo+ V1 + V2 can possibly generate massless states. The massless states from V0 are the same as in the preceding example. For Vo + V~, they are given by the solutions to the equation 14
22
Z
Z
I*=1
I*=15
One finds: K I* = 0
for I* = 1 . . . . . 14,
K/*=0orl
f o r I * = 15 . . . . . 22.
(5.6)
We obtain then a total of 28-- 256 massless states belonging to the 128 + 128' spinorial representation of S0(16). Similarly, and by permutation symmetry of I1:
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and V2, we get the following massless states in the subsector Vo + V2: K I* = 0
for I* = 1 , . . . , 6 and for I* = 15 . . . . . 22,
K I* = 0 or I
f o r / * = 7 , . . . , 14.
(5.7)
Finally, the Vo + Vx + V2 subsector does not contain any massless state since the minimal ground state energy for the left-moving modes is + 1: 22
-1+~
E
( K / * - ½ ) 2>" + 1 -
I*~7
Some of the states mentioned above will be projected out by the constraint of eq. (4.53), as a consequence of the requirement of modular invariance. In the case under consideration, ~ b will act as the identity operator on the shift vectors (here a = b = 0). Before proceeding, we determine first the matrix elements kp°° from eqs. (4.44) and (4.47). We collect them in the 3 × 3 matrix
k°° kol
k°2 i°°
kl0
kn
k12
k2o
k21
k22 J
/
(s.8)
where all the k ' s are either zero or one half. From eq. (4.53), we find Vo • Nvo = 0
mod 1,
V 1 • Nvo = 0
mod 1,
V2 • Nvo = 0
mod 1,
(5.9)
where, in the modified notation we now use, Nvo reads ( K k* rK I* )v0. Eq. (5.9) now implies 22 6 1 ~, K I * - ~ 1 ~ K K*= I*=1 K*=I
0
modl,
22
12 ~
KP=0
modl
K/*=0
rood1,
I*=15 14 1
~
(5.10)
I*=7
for the solutions shown in eq. (5.4). The first of the above equations is automatically
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A.H. Chamseddine, J.-P. Derendinger / Twisted superstrings
satisfied, while the others imply that
K ' * = ( o . . . . . _+1,o ..... + 1 , o , . . . )
for 1" = 1 . . . . . 6 and zero otherwise, or for I* = 7 , . . . , 14 and zero otherwise, or for I* = 15 . . . . . 22 and zero otherwise. (5.11)
Thus, after adding the corresponding Cartan generators, one finds that the gauge g r o u p SO(44) is broken into SO(12) ® SO(16) ® SO(16). However, as we have seen, the role of V1 and V2 is not only to break the original gauge group, but also to introduce new massless states. We now determine which of these massless states survive the constraint equations. T h e massless states arising from Vo + I11 are given in eq. (5.6) and must satisfy the constraint: Vo • N ~
-= k00 + koa - V0 • ( ~ )
+ k0o = k01
m o d 1.
Similarly: V1 • N v ~ - ~ = kol
m o d 1,
V 2 • Nv~-o-ojvl = k21
mod 1.
In terms of components these constraints read
22
E K"I*=1
6 ) E K" = k*=l 22 21 E
KI* = kol
mod I
I*=15
14 ½ E K'*=kla. I*=7
(5.12)
T h e actions of the first two constraints on the states given in (5.6) are identical and we can either set kol to zero or to one-half, keeping only the spinor 128 or only the spinor 128' out of the 256 states. W h e n k01---0, the remaining massless states are defined by the following values of the components KZ*: a m o n g the eight c o m p o nents K t*, I* = 1 5 , . . . , 2 2 , we have an even n u m b e r of zero components and an
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even n u m b e r of one components. In other words 22
28"
1--I ( K ' * - ½ ) =
+1.
(5.13)
I * = 15
All other components are zero. Choosing ko~ = ~ would lead to states for which the product (5.13) is - 1. The third constraint in (5.12) is satisfied provided k12 = k21 = 0. By symmetry between the V~ and V2 vectors, we can similarly show that only one 128 representation of the other SO(16) group remains from the Vo + V2 subsector. Observing that the adjoint representation 248 of E 8 contains precisely the adjoint representation 120 and one spinor 128 of the subgroup SO(16), we conclude that the gauge group corresponding to our choice of vectors is E s ® E~ ® SO(12), with a spectrum which is still N = 4 supersymmetric. Before leaving this example, it is interesting to obtain in our formalism the model corresponding to the trivial dimensional reduction of D = 10 supergravity in four dimensions. This means that we must identify the six left-moving coordinates with the six right-moving coordinates. By studying the partition function for this case as given in ref. [12], we find that we must add five new shift vectors of the form:
(1,0 .... •
O[ 1~-,O, • . .,0) ,
•
(0, ~, O1 ..... 0]0, ~, O1 .... ,0), ....
).
(5.14)
Introducing these five additional vectors will add many subsectors. However, none of them will generate new massless modes: we have already mentioned that only vectors with ½ entries for all right-moving components produce massless states. On the contrary, they will eliminate some of the massless states obtained with V0, V1 and V2 only. They impose new constraints on the solutions given in (5.11) implying that K I* = 0
for I* = 1 . . . . . 5.
This eliminates the states forming the adjoint representation of SO(12), except for the six Cartan algebra generators. The gauge group is then E 8 ® E 8 ® U(1) 6, where the six abelian groups correspond to the six-torus of the dimensional reduction. Having clarified our formalism for the case of purely periodic boundary conditions, we next consider the more interesting case where rotation vectors are also present. A well known example [9] is the orbifold compactification of ten-dimensional heterotic strings leading to a four-dimensional model with gauge group E 6 ® SU(3) ® E 8 and a surviving N = 1 supersymmetry. The rank of the gauge group is sixteen, as in the original E 8 ® E 8 heterotic string theory. The three complex internal coordinates Z k have twisted boundary conditions, and a discrete Z 3 symmetry. Our
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formalism is in fact tailored to reproduce such a model, and to give with little variations new models. We take only one rotation vector 1 3
and identify with the action of
R x a
1 3
1 3
,o 8 ),
(5.15)
shift vector V3 reading
1 2 V 3 ~ \ (061019 , i 3~ 3 ' 3 ) '
(5.16)
so that we always have the combination
a l W l = a l ( R l + V3),
with a~ = 0 , 1 , 2 .
At this point we have two possibilities: either we include the shift vector V3 in the untwisted sector by adding a term 6aoC3V3,with c 3 -= 0,1, 2, or we do not. In the first case which will correspond to the Z 3 orbifold example, introducing V3 in the untwisted sector will break the gauge group E 8 into E 6 ® SU(3), but will also introduce new massless matter states (characterized with the eigenvalues of Z3) from the new subsectors such as V0 + V3. In the second case, V3 will only break the gauge group without introducing any new subsector in the untwisted sector. For illustration, we shall explicitly exhibit the first possibility. The spectrum in the second case can easily be deduced from the first. We stress that our choice of R 1 and V3 is constrained to satisfy eqs. (4.72) and (4.73). The constraint associated to W1 is given by eq. (4.79), and reads in this case
R 1 • N a ~ + 17 , 3 , ~J va l b l . Nc~T~lv3 =
~_~k3qcq_ab
V3 . ~ b . ( c V + a l V 3 _ a l V 3 ) + k ~ + T n a l _ W l . ~ a b . a t W 1 ,
q
(5.17) where c V = c o V 0 .-~ c 1 V 1 + ~alO(C2V2 q- c3V3) ,
and c2V2 is included in the last term because in the twisted sector V0 + V1 = V2. This relation is valid because of the projection operator ~ v - V2 is then not independent. In the untwisted sector, the projection matrix ~ v acts like the identity operator while in the twisted sector one has ~b=
(06106116).
The constraints in eq. (4.7) coming from V2 and V3 hold only when a = b = 0
A.H. Chamseddine, J.-P. Derendinger / Twisted superstrings
415
(untwisted sector) and in this case the constraint corresponding to I/3 is, as it should be, consistent with eq. (5.17). F r o m the mass formulae, it is obvious that when a I = 0, the ground state energy will not be shifted by the action of R 1, and will correspond to the untwisted sector. We now give the spectrum in this sector. The subsectors are coVo + q V 1 + c2V2 + c3V3, with c 0, q , c 2 = 0,1 and c 3 = 0, 1,2. The massless states come from V0, Vo + V1, Vo + V 2, V0 + V 3 , V0 + 2 V 3, Vo + V I + V 3 a n d V o + V 1 + 2 V 3. From the mass formula for the right-movers, eq. (4.61), it is easy to see that the ground states are l a ) R + la)R which are neutral under Z 3 symmetry, as well as IZk)R + ]s"k}R with eigenvalues X = e-2i~/3 and their complex conjugates providing the opposite helicities. For the left-moving modes, we have seen that the massless states coming from Vo, Vo + V1 and Vo + V2 form the adjoint representation of SO(12) ® E 8 ® E 8. Here however, the action of the discrete group on the coordinates Z r, I = 1, 2, 3 implies that the massless states SCx1-~[x + i ~ ( y and their complex conjugates take eigenvalues X and ~ - i respectively. Again the group SO(12) will be broken when the five shift vectors (5.14) are added, leaving only the Cartan algebra U ( I ) 6. Adding V3 will impose constraints on these states, eliminating some of them and reducing the gauge group. But new states will also appear through the additional subsectors. The V3 constraints are V3 • ~ . Nv0 = 0, V3 "~ . Nv0~V]_~ = 2 + k31
I/3 • ~ " N ~
m o d 1,
(5.18)
= k32,
where we have made use of eq. (5.17) and the definitions in eq. (4.74). The first of these constraints implies that ~1 ( K 20 + K21) + . ~K22 = 0
modl,
(5.19)
This constraint will affect the states appearing in eq. (5.11) for which the three last components of K I* are non zero. There are six of such states which satisfy (5.19): K"=
_+(1,0,1) or + ( 0 , 1 , 1 ) or + ( 1 , - 1 , 0 )
for I* = 20, 21, 22, all other components being zero. Of course, all solutions given in (5.11) for which K 2° = K 21 = K 22 = 0 are not affected. The states of (5.11) having components in the last eight directions were forming the adjoint representation of SO(16), when supplemented by the relevant Cartan generators. In our case, SO(16) is broken into SO(10) ® SU(3) ® U(1), the six states we have just listed and three Cartan generators belonging to the adjoint representation of SU(3) × U(1).
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The second constraint (5.18) eliminates some of the massless states given in eq. (5.7). If we set ,~31v°)= 0 (i.e. k ~ ) = ~), the states K I* = (05) or (03,12) or (0,14) for I* = 1 5 . . . . . 19 and K l * = ( 1 , 1 , 0 ) or (0,0,1) for I* = 2 0 , 2 1 , 2 2 remain massless. These states form a 16 + 16" representation of SO(10). They have the right U(1) charge to complete the adjoint representation of the group E6, containing SO(10) ® U(1). We then obtain an E 6 ® SU(3) gauge group contained in one of the E 8 group present in our previous example without the shift vector V3. The last constraint in eq. (5.18) is eluded by setting k~2) -- k ~ = 0. The massless states from the V o + V 3 sector are solutions of the equation 19
1=½
Z
21
(K'*) 2+½
I* = 1
( K I * + ½ ) 2 + ½ ( K 2 2 * - 1 3 ) 2.
Z
(5.20)
I* = 2 0
These are either the three states K I*= (1,0,0), (0,1,0), (0,0, - 1 ) for 1" = 20, 21, 22 and zero otherwise, or the states K I*= (+_1,0 . . . . . 0) for I * = 1 .... ,19 and K I * = ( - 1, - 1, 0), (0, - 1,1), ( - 1, 0,1) for I* = 20, 21, 22. The various constraints on these solutions are ~.#.N~+~
m
2,
1
V~ " sg" N>o + V,
_
71 ,
Vz.~.N~
= 0,
V3"~'N~o+v3
_
mod
1
13 ,
(5.21)
and are all satisfied with the change I* = 1 . . . . . 19 to I* = 15 . . . . . 19. These states form the representation (1, 3) + (10, 3) of SO(10) ® SU(3). Since they come from the Vo + V 3 sector, they will have an eigenvalue X 1 under the Z 3 symmetry. One then expects and indeed can check that V0 + 2V~ will give the states in the conjugate representations. Finally, in the sector V0 + V1 + V3, the equation determining the massless states is 14
1=~
E 1" = 1
19
(KI*)2+ ½ E I* =15
21
(Kt'-
½)2+ ½ E
( K/*- 2~) 1 +
-½(K22. + g)12,
I* =20
(5.22) with the solutions K I* = (0,..., +_1 , . . . , 0) for I* = 1 , . . . , 14 and K I* -- (0 5), ( 0 4, 1), (03, 12), (02, 13), (0, 14), (15) for I* = 15 . . . . . 19 and K t* = (1,1, - 1 ) , (1,0,0), (0,1,0)
A.H. Chamseddine, J.-P. Derendinger / Twistedsuperstrings
417
for I* = 20, 21, 22. These states must obey the constraints Vo'~(1)'Nvo+
v , + v~ - - O ,
VI " :~(1) " N v o + v~ + v~ - k[~ ) ,
rood i V2. ~ o ) .
N v o + v~ + v3 - k(2]) = 0
V3 • ~ ( 1 )
o NU0+
VX+ 173
3 '
(5.23)
and after setting k ~ ) = 0, the only surviving solutions are K t* = 0 for I* = 1 . . . . . 14, K I * = (05), (03,1z), (0,14) for I* = 15 . . . . . 19 and K t * = (1,1, - 1 ) , (1,0,0), (0,1,0)
for I * = 20,21, 22. These states form the (16,3) representation of SO(10)® SU(3) and their eigenvalue under the Z 3 symmetry is X- 1. F r o m the sector V0 + V1 + 2V3, we will get the complex conjugate representations. By forming the states invariant under the discrete symmetry, one gets exactly the spectrum obtained in ref. [9] in the untwisted sector, containing the supergravity supermultiplet, the N = 1 Yang-Mills multiplet for the gauge group E 6 @ S U ( 3 ) ® Es, ten gauge singlet matter multiplets and also 3(27, 3, 1) matter multiplets. In the twisted sector, a t = 1 or 2 and the mass formula will get new contributions, as is clear from eqs. (4.61) and (4.62). Also, we note that since the six right-moving coordinates and the first six left-moving coordinates will be shielded by the projection operator, V z will be equivalent to V0 + Vx and thus is not an independent vector. It must be dropped from the set of vectors to avoid over-counting. For the right modes, the only massless states will arise from X ~ and S ". The ground state will be the scalar supermultiplet (½, 0). In the case of left-movers, the contributions ~ ( 0 - 0 2) shift the vacuum energy which becomes - ~ instead of - 1 . The subsectors we need to consider are c o V o + c l V 1 + c 3 V3, but the sectors contributing to the massless spectrum are V0, Vo + 1/3, V0 + 2V3, Vo + V1 + V3 and Vo + V1 + 2 ~ . In the case of V o, the massless states are determined by the equation 3 2 __
I=1 n=l
+
22
22
E
+ Z
I*=7
(5.24)
1"=7 n = l
This equation is solved either if two of the ~ (3 are equal to one, or if one of the ~ {3 is one, for I = 1, 2, 3. The corresponding number operator
2N'= 2 I
I
n
A.H. Chamseddine, J.-P. Derendinger / Twistedsuperstrings
418
is then either 2 or - 1 . The constraint given by eq. (5.17) is now
RI'NR, + V3' ~ ) - N v o
= ~
modl,
(5.25)
where we have substituted Tll = ~- from eq. (4.70) and ko3 + k3o = 2 from eq. (4.74). This implies that the states solving eq. (5.24) should also satisfy 3
2
1=1
3
-~ ~ N ' = -
modl.
Since it is not the case, all these states are projected out. F o r the sector Vo + V3, the equation for the massless states is 3
~
2_
2
19 E •*=7
-,
1=1 n = l 21
+½ Z (Kt*+~)2+½(K22-½) 2.
(5.26)
I * = 20
It is easy to check that the solutions are either K t * = (0 . . . . . + 1 , 0 . . . . ) for I * = 7 , . . . , 1 9 , all other numbers being zero, or K t * = 0 for I* = 7 . . . . . 19 and K ~*= (-1,-1,1) for I * = 2 0 , 2 1 , 2 2 , all the ~ i being zero, or, finally, K I * = 0 for I* = 7 . . . . . 19 and K t * = (0,0,1), ( 0 , - 1 , 0 ) , ( - 1 , 0 , 0 ) for I * = 20,21,22, while one of the z ~ _ ~ is one and all . ~ are vanishing. The constraint in eq. (5.17) now reads 1(N1 + N 2 + N 3) + ~ ( K 20+K2~ + 2K22) = 0
modl.
(5.27)
This constraint is indeed satisfied by all the solutions to eq. (5.26). The constraints from Vo and V~ will restrict the range in K I* = (0,..., _+1, 0 . . . . ) to be I* = 15 . . . . ,19 only. The surviving states are then in the representation (10,1) + (1, 1) + 3(1, 3*) of SO(10) ® SU(3). In the sector Vo + 21/3, one gets the same solutions as previously, changing only the signs of K I*, I* = 20,21,22. However, all these states will be projected out by the constraint in eq. (5.18). Finally, from Vo + V1 + V3, we find the equation 3 1=1 n = l 14
+½
E
21
(gl*)2+½
t. = 7
E
(g/'-~)
2
i* = 20 19
+½(K22+~)2+½
E I*=15
( K ' * - ½ ) 2"
(5.28)
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The solutions are K I* = 0 for I* = 7 . . . . ,14, and K t* = (05), (04,1), (03,12), (02,13), (0, 14), (15) for I* = 15 . . . . ,19, with all the other numbers vanishing. The constraint in eq. (5.17) now gives
- ~ ( N l + N2 + N3) + ~(K2° + K21+ K22)=O
rood1.
This constraint is automatically satisfied. However, the conditions from V0 and V1 reduce K I* to only (05), (03,12) and (0,14) in the coordinates where I* = 15 . . . . ,19. These states form the (16, 1) representation of SO(10) ® SU(3). Again, all solutions of the mass formula in the sector V0 + V1 + 2V3 will be projected out by eq. (5.18). Collecting all the states, we find the massless matter multiplets 3 ( 1 , 3 " ) + (1 + 10 + 16, 1) in the twisted sector, the last of which obviously falling in a (27,1) of E 6 ® SU(3). Repeating the same analysis with the rotation vector 2R 1 (i.e. a I = 2) will provide us with the states of opposite helicity. The determination of the number of fixed points is in general complicated. However, in this case, since the left and right coordinates are identified, the result is clearly identical to the Z 3 orbifold and we have 27 fixed points [9]. We can change the resulting spectrum in the above example if we choose not to a c c o m p a n y the action of R with the shift V3, but instead take independent coefficients. Enumerating the massless states becomes more complicated in the untwisted sector since there will be additional subsectors because of the presence of the v e c t o r V3. These are Vo, Vo + V1, V0 q- V2, V0 + c3V3 and Vo + V l + ¢3V3 with c 3 = 1,2. One can show that the massless states surviving the constraints are the same as before. However, in the twisted sector, all massless states will be projected out because the constraints (4.79) and (4.80) will not be satisfied. Alternatively, we can eliminate the massless states from the untwisted sector without changing those of the twisted sector by dropping the term 6a,oC3V3 in cV. This will have the effect of removing all the 1/3 contributions to the matter sector which was responsible for the 3(27, 3) states, while still breaking the E 8 gauge group into E 6 ® SU(3) through the constraints in eq. (5.18). We now illustrate our method by constructing a new example of a string theory with a rank eight gauge group. This example will be a variation of the orbifold case we have discussed above, with new rotation vectors to reduce the rank and also a new shift vector to break further the gauge group. In order to satisfy the constraint in eq. (4.81), all the a R and b R tensor products for all values of a m and bn must have net number (i.e. left minus right) of zero entries corresponding to the complex bosons to be a multiple of four. This imposes severe restrictions on the rank of the gauge group giving in particular for the case of N = 1 supersymmetry ranks 16 and 8. We recall that only the weaker form of eq. (4.81) when b = 0 was imposed to us in general, however we were only able to solve eqs. (4.33) and (4.36) after assuming eq. (4.81). The weaker constraint involves only the vectors a R and can be easily satisfied with vectors R leading to all possible ranks. The difficulty arises when one
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wants to choose the vectors R to satisfy eq. (4.81) for all values of a , and b,. Because of the consequences of this constraint, it will be worthwhile to solve eqs. (4.33) and (4.36) without assuming it, or to try replacing the generalized GSO projection requirement with that of two loop modular invariance along the lines of ref. [7]. We now add the vector
W2= R2 + V4, where 8 2
=(0 , 5!,.5., 5 ,
_5,1 5,5 ...
5,
4,45, ( j , 0 4 ) (5.29)
V4 = (06]014, (1)4, 4 , 0 3 ) ,
so that the full set of vectors is a l W 1 + a z W 2 and c V = ¢oVo q- (ClV I -l--c 2V2 ) S a2,O -J-(c3V 3 q- c 4V4 ) S a,,Oaa2.0 .
(5.30)
Notice that R 2 introduces twists for the six right-mover coordinates which are always different from those of the left-movers. The theories constructed with R 2 cannot be considered as compactification of ten-dimensional string theories in the usual sense. The effect of introducing R 2 will be to break the first E 8 completely, and V4 will break E 6 into SO(10) ® U(1). We also need to add seven vectors of the form in eq. (5.14), with the ½ components placed alternatively between I* = 6 to 12, in order to avoid a large number of gauge singlets, This, as we have seen earlier, will not introduce new massless states. We now analyse the untwisted sector. Compared with the orbifold case, the only difference is due to the presence of the new shift vector V4, which will create new subsectors and new constraints of the form (4.80). To obtain these constraints, we first need to compute the numbers kpq, using eqs. (4.74)-(4.77). These equations determine kpq in terms of k0o, k m, ko2 and k12 which can be chosen to be either 0 or ½. These choices correspond to different conjugations of the resulting gauge groups embedded in E 8 ® E 8. The new constraints on the massless states that came from Vo, V0 + V 1, V0 + V2, V0 + V3 and V0 + V1 + V3 are respectively V4.~. Nvo=O, .
=
4,
V4 • ~ - N ~
= 0,
V4 " ~ " N ~
= 0,
V4"5~'Nvo+Va+V3 _ 45"
rood 1
(5.31)
The subsectors with 2V3 need not be explicitly considered since their massless states are simply obtained by conjugation. The subsectors omitted in eqs. (5.31) do not
A.H. Chamseddine, J.-P. Derendinger / Twisted superstrings
421
generate massless states. These new constraints due to modular invariance will r e m o v e m a n y of the massless states obtained previously. The remaining states are the following. The subsectors V0 and Vo + V2 generate the adjoint representation of SO(12) ® E s ® SU(5) ® SU(3) ® U ( I ) 2. F r o m V0 + V1, only two massless states remain, defined by K ~* = (014; 1 , 1 , 1 , 1 , 0 ; 1 , 1 , 0 ) , K 1.
= (014; 0, 0, 0, 0, 1; 0, 0, 1).
(5.32)
These states are gauge singlets, but have an eigenvalue h -1 under the discrete s y m m e t r y Z 3, which is still relevant with the addition of V4. The subsectors Vo + V3 and Vo + V1 + V3 each generate a massless SU(3) triplet. A 3 of SU(3) corresponds to K 1., I* = 20, 21, 22 given by (1,0,0),
(0,1,0),
(0,0,-1),
(5.33)
or also by (-1,-
1,0),
( - 1,0, 1),
( 0 , - 1, 1),
(5.34)
while a 3* corresponds to the opposite signs. Then Vo + V3 generates a massless triplet with K I* = 0 for I* = 1 . . . . . 19, and Vo + V1 + V3 has a triplet with K r* = 0 for I* = 1 . . . . ,18 and K 19= 1. The conjugate states are obtained in subsectors V o + 2 V 3 a n d V0+ V I + 2 V 3. M a n y new massless states come from the new subsectors containing c4V4, c 4 = 1 . . . . . 4. They fall naturally in S U ( 5 ) ® SU(3) multiplets. The relevant SU(5) representations correspond to the following values of K**, I* = 15 . . . . . 19: 5:(1,0,0,0,0), (0,1,0,0,0), (0,0,1,0,0), (0,0,0,1,0),
(o,0,0,0, -1), or 5: ( - 1, - 1, - 1 , 0 , 1 ) + 3 permutations with K 19 = 1, (1, 1 , 1 , 1 , 0 ) . 1 0 : ( 1 , 1 , 0 , 0 , 0 ) + 5 permutations with K 19 = 0 , ( 1 , 0 , 0 , 0 , - 1 ) + 3 permutations with K 1 9 = - 1 , or 10:( - 1, - 1, - 1 , 0 , 0 ) + 3 permutations with K 19 = 0 , ( - 1, - 1 , 0 , 0,1) + 5 permutations with K 19--- 1.
(5.35)
and the conjugate representations 5" and 10" have opposite K t*. The only massless states which have )to eigenvalue under Z 3 arise from Vo + Va and Vo + 4V4. They
422
A.H. Chamseddine, J.-P. Derendinger / Twistedsuperstrings
form a 10 + 10" multiplet. These states are associated to those generated by V0 and V0 + V2 and enlarge the gauge group (in the untwisted sector) to SO(12)® E 8 ® SO(10) ® U(1). There is also a variety of states with non-trivial Z 3 eigenvalues, which will give rise to chiral supersymmetry multiplets. They are classified in the SU(5)® SU(3) representations
(10" + 5 + 5 + 5",3) + (10+ 5* + 5* + 5,3*) + 2(10 + 10",1). Associated with the massless states with non zero ~ arising from subsectors without V4, they can be collected in SO(10) ® SU(3) representations:
(16+ 10+ 1,3") + (16" + 10+ 1,3) + (16+ 16", 1). The SU(3) triplets and anti-triplets fall into E 6 representations 27, but E 6 is broken into SO(10) ® U(1) as is clear from the massless SU(3) singlets. Notice also that the number of left-moving massless states in the untwisted sector is the same as in the orbifold case. All these states are invariant under SO(12) ® E s. This part of the gauge group is only an artifact of the untwisted sector. Rotation vectors will remove these gauge groups to leave only SO(10) ® SU(3) ® U(1). All in all, the untwisted sector gives the super-Yang-Mills multiplet of SO(10) ® U(1) ® SU(3) with 3(16 + 10 + 1,3") + 3(16,1) matter multiplets. At this point we can go on and analyse the twisted sector. As before we will have different variations depending on whether we identify the R and V actions. We will not analyse in details here the twisted sectors for this theory. The number of generations is anyway already too large in the untwisted sector. Having already illustrated our method in full details with the previous examples we shall close at this point. Such a full analysis is really of primary interest in the case of a theory which is phenomenologicaUy promising, and hopefully with gauge groups of ranks smaller than eight, if such solutions exist.
6. Conclusions In this paper, we have fully developed string theories in four dimensions with explicit N = 1 supersymmetry in the Green-Schwarz formalism. We imposed modular invariance as a necessity to have consistently interacting strings. This we achieved by studying the one-loop partition functions and by generalizing the GSO projection operator (which under some circumstances is known to be equivalent to two-loop modular invariance) to obtain restrictions on the theories one can construct. We have allowed for twisted boundary conditions for both fermions and bosons and we have shown that the twisting can result in gauge groups of lower ranks. We were only able to solve our general equations in a special case, which unfortunately forced on us the ranks 16 and 8 (or even 0) in the case of N = 1
A.H. Chamseddine, J.-P. Derendinger / Twisted superstrings
423
supersymmetry. Work is in progress to avoid such an obstruction. Finally we have illustrated our formalism in great details and have recovered known string theories. We have shown that with little variations on the boundary conditions one can obtain new possibilities. We believe that this formalism has the potential of producing new superstring theories with phenomenologically interesting gauge groups and massless spectra. Interesting gauge groups can be easily obtained by natural breakings of E 6. A potential difficulty is the number of generations which can be very large. The role of discrete symmetries induced by the stfift and rotation vectors deserves further investigation in that respect. One of us (A.C.) would like to thank B.E. Nilsson for discussions. We also thank Daniel Wyler for contributions to the first parts of this work.
Appendix The Lorentz generators in four dimensions are
jab
=
j+-=
l~e + j~b + j~b, 1+- ,
ja+= l~+, Ja-=la-+J~
+J~
,
where and/~ is a four-dimensional space-time index, /~ = 0, a, 3. First, the right generators are
n=l
/7
+½i k n ~
(S-nTab-Sn
' 51/o f+ 1 q- 7
+ Uo~-)n-OkY k . a b - ~ ko(-)n-0* + o(+)n+0k ~k .Y ab
oo(+)n+0*), k
o~
. °
~_~
~[
(~k+
m= - o o n= -oo
ieklp
a-,~, )re+n-01
OPT
--k a+Si-,m+°+O'+O"T q ~'k
_ ab-c,k
o(-)~+,,,-ek r "--
a--
p
d*(-)m-OlOLn-OP l ~'k
P
+
.ab-~k
o(-)m-0* + O(+)m+n+0k ~' k
k
(L+
mY
,,b-~ -m ] b
o(+)m+okla~
S(_)rn_Okfl ~ 0,']]]
424
A.H. Chamseddine, J.-P. Derendinger / Twisted superstrings
where
[o~a_mota + l ( m - l n ) L _ n Y - S n m ~ --¢uQ
+fl,_,~+Okam_Ok+½(m --k - Sik )m Ok k k -- Ok - ln)g(-)m-Ok-n'Y +½(m+O k
1 --k in)Si+)m+ok_,y
_
-
k S(+)m+0k] •
F o r the left generators, we have:
j~b=_
i £ n=l
J~-=
p-+
l(~a ~b__~b ~a] l'l ~ - n n -n hi,
--(sa-.sn---S:.sa), n=l /'/
where ~-
~
Oln =
-a
~a
-1
~1
( Otn-mOtm + f l n - m + O t f l m - O ' )
"
I n the last equation, we have complexified the 22 bosons X A into 11 complex fields Z t and assumed twisted b o u n d a r y conditions for them. For the Z I with periodic b o u n d a r y conditions, we set the corresponding 01 to zero. In all the above formula, the s u m m a t i o n over repeated indices k and I is assumed. Finally, the supersymmetry generator Q~ is 2i
2~_
k
k
k
k
a
n
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A.H. Chamseddine, J.-P. Derendinger / Twisted superstrings
425
[7] I. Antoniadis, C.P. Bachas and K. Kounnas, Nucl. Phys. B289 (1987) 87 [8] D.J. Gross, J.A. Harvey, E. Martinec and R. Rohm, Phys. Rev. Lett. 54 (1985) 502; Nucl. Phys. B256 (1985) 253; B267 (1986) 75 [9] L. Dixon, J.A. Harvey, C. Vafa and E. Witten, Nucl. Phys. B261 (1985) 678; B274 (1986) 286 [10] C. Vafa, Nucl. Phys. B273 (1986) 592 [11] L.E. Ibfifiez, H.P. Nilles and F. Quevedo, Phys. Lett. 187B (1987) 25; L.E. Ibfifiez, J.E. Kim, H.P. Nilles and F. Quevedo, Phys. Lett. 191B (1987) 282 [12] M.B. Green and J.H. Schwarz, Nucl. Phys. B181 (1981) 502; B198 (1982) 252; B198 (1982) 441; J.H. Schwarz, Phys. Reports 89 (1982) 223 [13] A.H. Chamseddine, Nucl. Phys. B185 (1981) 403 [14] L. Alvarez-Gaum~, G. Moore and C. Vafa, Comm. Math. Phys. 106 (1986) 1 [15] A.N. Schellekens and N.P. Warner, Nucl. Phys. B287 (1987) 317
