Intersecting D6-branes on the Z_{12}-II orientifold

Much work has been done by a number of authors with the aim of constructing the supersymmetric Standard Model in type IIA intersecting-brane theories compactified on an orientifold with various Z_N or Z_M x Z_N point groups. Here we consider the Z_{12} point group which has previously received comparatively little attention. We consider intersecting D6-branes that wrap 3-cycles consisting of a 2-cycle of the 4-dimensional lattice upon which the Z_{12} is realised times a 1-cycle of the remaining 2-torus. Our discussion is restricted to the case when these 2-cycles are"factorisable"in the sense discussed in section 3. Although it is possible to find models with the correct supersymmetric Standard Model quark-doublet content, we have not found it possible to obtain the correct quark-singlet content.


Introduction
The use of intersecting D6-branes in Type IIA string theory offers an attractive route to constructing the Standard Model in string theory [1,2], and indeed an attractive model having just the spectrum of the (non-supersymmetric) Standard Model has been obtained by Ibañéz et al. [3]. In this approach one starts with two stacks a, with N a = 3 D6-branes, and b with N b = 2 D6-branes, each wrapping the three large spatial dimensions plus 3-cycles of the six-demensional compactified space Y . Open strings beginning and ending on the stack a generate the gauge group U (3) = SU (3) colour ×U (1) a , while those that begin and end on the stack b generate the gauge group U (2) = SU (2) L ×U (1) b . Thus the non-Abelian component of the Standard Model gauge group is immediately assured. Further, (four-dimensional) chiral fermions in the bi-fundamental (N a ,N b ) = (3,2) representation of U (3) × U (2) appear at the multiple intersections of the two stacks. (Here the 3 representation of U (3) has charge Q a = +1 with respect to U (1) a , and the2 representation of U (2) has charge Q b = −1 with respect to U (1) b .) This is just the representation needed for the Standard Model quark doublet Q L . However, non-supersymmetric intersecting-brane models lead to flavour-changing neutral-current (FCNC) processes that can only be suppressed to levels consistent with the current bounds by making the string scale rather high, of order 10 4 TeV, which in turn leads to fine-tuning problems [4]. Further, in non-supersymmetric theories, the complex structure moduli are generally unstable [5]. Both of these problems are avoided if instead we seek intersecting-brane models that yield the supersymmetric Standard Model. This is the strategy that we shall pursue in this paper.
To ensure that we obtain N = 1 supersymmetry in the four space-time dimensions, it is necessary that the compactified space Y should be a Calabi-Yau 3-fold or a toroidal orbifold Ω = T 6 /P , where the (discrete) point group P must be a subgroup of SU (3) [6]. (We shall only consider the latter possibility.) The requirement that the point-group generator θ acts crystallographically on the lattice Γ that defines the torus T 6 then restricts P to be either Z N , with N = 3, 4, 6, 7, 8, 12, or Z M × Z N , with N a multiple of M and N = 2, 3, 4, 6 [7,8]. The first question is whether one can find stacks a and b, as above, whose intersections yield just the three Standard Model quark doublets. However, before proceeding further it should be noted that both of these stacks are positively charged with respect to the Ramond-Ramond (RR) 7-form gauge field to which they are "electrically" coupled. Since Y is a compact space, the electrical flux lines associated with the RR charges must close, which can only happen if the RR charges sum to zero. This in turn requires the introduction of negative RR charge. Anti D-branes,D6-branes, annihilate D6-branes, and the only feasible alternative is to use the O6-planes. These are topological defects that arise when Y is an orientifold, i.e. Y = Ω/R, where R is the embedding of the world-sheet parity operator in the compactified space. This means that every stack κ = a, b, ... has an orientifold image κ = Rκ, and that the stack a will in general intersect with both b and its orientifold image b . As with the intersections of a with b, the intersections of a with b also yield chiral fermions but they are now in the representation (N a , N b ) = (3,2) representation of U (3) × U (2), where the 2 of U (2) has charge Q b = +1 with respect to U (1) b . Then in order to get just the 3Q L quark doublets, we require that the numbers of intersections, a • b of a with b, and a • b of a with b , satisfy Of course, we must also ensure that these states have weak hypercharge Y (Q L ) = 1/6. In general, Y is a linear combination of all of the U (1) κ charges Q κ . A quark doublet arising as a (3,2) representation of U (3) × U (2) has Y (3,2) = y a − y b , whereas the alternative has Y (3, 2) = y a + y b . If quark doublets of both types occur, then y a = 1/6 and y b = 0. However, if there is only one type then, depending upon which, all we know is that y a ∓ y b = 1/6. There have been many attempts to construct the supersymmetric Standard Model, or something like it, using a variety of orientifolds [9]- [23]. None has been completely successful, but the closest approach has probably come using the Z 6 orientifold. The question then arises as to whether one can do better with a different orientifold. In this paper, we address that question using the Z 12 -II orientifold. This orbifold (and the Z 12 -I orbifold) is not completely factorisable; that is, it cannot be realised on T 2 × T 2 × T 2 . Some of the technical problems associated with such orbifolds have been discussed in [24]. In that paper the authors determine the non-chiral solutions of the RR tadpole cancellation conditions when the D6-branes lie on top of the orientifold O6-planes, the whole system satisfying (twisted) sector-by-sector RR tadpole cancellation; this is more stringent than necessary, as the vanishing of RR flux just requires overall tadpole cancellation. In what follows we consider more general configurations of intersecting (fractional) D6-branes, and attempt to construct the chiral quark, lepton and Higgs spectrum of the supersymmetric Standard Model, with the strategy of imposing overall tadpole cancellation at the end to constrain any such configurations that generate the required spectrum.
In the θ 6 -twisted sector there are 16 fixed tori T 2 3 at the Z 2 fixed points f σ 1 ,σ 2 ,σ 3 ,σ 4 on the SO(8) lattice, where with σ a = 0, 1. For ease of reference, we use the same notation as in the Z 6 case [15], denoting the fixed points by f i,j with the pairs (σ 1 , σ 2 ) and (σ 3 , σ 4 ) given the labels i, j = 1, 4, 5, 6 respectively for the values (0, 0), (1, 0), (0, 1), (1,1). Under the action of the pointgroup the 16 fixed points split into four sets, each set transforming into itself as follows: There are then four non-zero invariant exceptional 3-cycles: which is consistent with the cohomology of the θ 6 -twisted sector. The self-intersection number of a (Z 2 ) collapsed 2-cycle is, as before, given by Then, (The corresponding results for the Z 12 -I case are given in the Appendix.) The general exceptional brane Π ex κ is then given by where the coefficients e κ i are determined by the fixed points wrapped by the 2-cycle used to construct Π bulk κ , as we shall see in the following section. For two general exceptional branes of this form Exceptional cycles also arise in other twisted sectors. For example, in the θ 4 -sector there are 9 fixed tori at the Z 3 fixed points with m, p = 0, 1, 2, and, as above, collapsed 2-cycles at these fixed points may be combined with 1-cycles in T 2 3 to construct further twisted 3-cycles. However, only bulk cycles and exceptional cycles at Z 2 fixed points have a known interpretation in terms of partition functions [25] . In what follows we have therefore only considered the exceptional 3-cycles defined in eqns (2.41) ... (2.44).

Factorisable 2-cycles
The general 2-cycle on the SO(8) lattice that appears in eq. (2.21) has the form with a, b = 1, 2, ..., 4 and n a,b six arbitrary integers. Now suppose that Π 2 is the product of two 1-cycles a n a π a and b m b π b , where n a and m b are integers. In this case the six integers n a,b are expressible in terms of the eight integers n a and m b as They then satisfy the constraint A general set of six wrapping numbers n a,b will generally not satisfy this constraint, and even if they do it is not sufficient to ensure that Π 2 is "factorisable" in this way. If it is, it is straightforward to identify the four fixed points f i,j that are wrapped by Π 2 . For example, if such a factorisable 2-cycle has (n 1,2 , n 1,3 , n 1,4 , n 2,3 , n 2,4 , n 3,4 ) = (1, 0, 0, 0, 0, 0) mod 2, then (n 3 , n 4 ) = (0, 0) mod 2 = (m 3 , m 4 ) and either (n 1 , n 2 ) = (1, 0) mod 2 and (m 1 , m 2 ) = (0, 1) or (1, 1) mod 2, or vice versa. Evidently Π 2 , like π 1,2 , wraps the four fixed points f 1,j , f 4,j , f 5,j , f 6,j with j = 1, 4, 5, 6 arbitrary. Henceforth we shall only consider such factorisable 2-cycles. A priori, there are 2 6 cases to consider for the set (n 1,2 , n 1,3 , n 1,4 , n 2,3 , n 2,4 , n 3,4 ) mod 2. However, the case in which all n i,j are even is of no physical interest, since we require the wrapping numbers to have no common factor. The action of θ splits the remaining 63 cases into sets as follows: and we only need to keep one representative of each of the 19 sets. In fact, only 9 of these can satisfy the factorisation constraint given in eq. (3.3). They are listed in Table 1 together with the associated values of a 1,2 mod 2; these are defined in eqs (2.28) and (2.29). Each of these classes is associated with four sets of four fixed points, as illustrated above. The bulk part Π bulk κ of a fractional brane κ, where is determined by the 3-cycle given in eq. (2.21). Supersymmetry requires that it wraps the four fixed points that determimine the exceptional part Π ex κ as follows. The four fixed points (n 1,2 , n 1,3 , n 1,4 , n 2,3 , n 2,4 , n 3,4 ) mod 2 (a 1 , a 2 ) mod 2 contribute with a sign determined by the Wilson lines t κ 0 , t κ 1 , t κ 2 = ±1. In the example given above, the four fixed points f 1,1 , f 4,1 , f 5,1 , f 6,1 are associated with the invariant exceptional 3-cycle generated by t κ (3.8) and in this example The fixed points for all 9 classes, together with the corresponding values for e κ 1 and e κ 2 , are listed in Table 2.

Supersymmetric bulk 3-cycles
The action of the point group generator given in eq. (2.3) ensures that the closed-string sector is supersymmetric, but to avoid supersymmetry breaking in the open-string sector the D6-branes must wrap special Lagrange cycles. That is to say, we require that where Table 2. The fixed points and coefficients e κ i of the exceptional cycles associated with the 9 classes of factorisable 2-cycles; an overall factor of t 0 is omitted. is the holomorphic 3-form. The complex coordinates z 1 and z 2 are those which diagonalise the action of θ as in eq. (2.1) with v 1 and v 2 as given in eq. (2.3). The 2-cycle π a,b may be parametrised as π a,b = λπ a + µπ b with 0 ≤ λ, µ < 1 (4.4) so to evaluate dz 1 ∧ dz 2 on π a,b we need a representation of the four simple roots α a in this complex basis: 2 ) (4.5) Defining the central root by the general form so that α 2 .α 2 = 2, it is easy to verify that the remaining roots are given by where β := e iπ/6 and cos 2θ = −1/ √ 3. We parametrise the 1-cycle in T 2 3 by where e 5 and e 6 define the SU (2) × SU (2) lattice. Then, with π κ as defined in eq. (2.21), we find where τ 3 := e 6 /e 5 is the complex structure of T 2 3 . The phases of e 5 and e 6 as well as φ 1 and φ 2 are constrained by the requirement that the orientifold embedding of the world-sheet parity operator also acts as an automorphism of the lattice.

The Z 12 -II orientifold
The embedding R of the world-sheet parity operator acts on the three complex coordinates z k as complex conjugation Rz k =z k (k = 1, 2, 3) (5.1) In particular, since we require that R acts crystallographically on the root lattice, this requires that where N ab ∈ Z. This leads to six independent solutions which are displayed in Table 3. For Lattice Table 3. The action of R and the phases φ 1 and φ 2 for crystallographic action of R on α a (a = 1, 2, 3, 4); an overall sign of = ±1 is undisplayed.
the bulk 3-cycles ρ p (p = 1, 2, ..., 4) defined in eqs (2.17)-(2.20), only two combinations σ 1,2 of 2-cycles enter the invariant bulk 3-cycles: It is easy to verify that the six different lattices reduce to just two classes when acting on these combinations: (a, e, f ) : Note too that, independently of the overall sign , the product of the phases given in Table  3 restricts the hitherto unknown phase in eq. As in the Z 6 case, the action of R on the basis 1-cycles π 5,6 in T 2 3 is given by A : Rπ 5 = π 5 , Rπ 6 = −π 6 (5.9) B : Rπ 5 = π 5 , Rπ 6 = π 5 − π 6 (5.10) Thus, in both cases e 5 is real and chosen to be positive, and the complex structure of T 2 3 is given by  Table 4. Choosing the lower signs in eqs (5.7) and (5.8), the functions X κ and Y κ defined in eqs (4.1) and (4.2) are then given in Table 5.
As already noted, the orientifold action leads to the formation of O6-planes. To determine these we must first identify the two R-and two θR-invariant 1-cycles on each configuration of the SO(8) lattice. These are displayed in Table 6, as is the single R-and Table 4. The action of R on the invariant 3-cycles.
which gives: (a, e, f )A : π O6 = 2(ρ 2 + sρ 3 ) where s is the sign of Im τ 3 . We also need the action of R on the exceptional cycles j and˜ j , which in turn depends upon the action of R on the sixteen Z 2 fixed points f i,j (i, j = 1, 4, 5, 6) in the θ 6 -twisted sector. This may be determined using the action of R on the simple roots α a of the SO(8) lattice, which is displayed in Table 3 Table 8. Action of R on the θ 6 -sector fixed points f i,j (i, j = 1, 4, 5, 6).
Lattice Table 9. Action of R on the invariant exceptional 3-cycles j and˜ j .  [25]; this is most easily seen by considering the action of R on the Kähler form J := idz k ∧ dz k . The results are displayed in Table 9.

Fractional branes
As noted earlier, in order to obtain stacks which intersect at an odd number of points it is necessary to use fractional branes of the form given in eq. (3.5), where the bulk part Π bulk κ is of the form given in eq. (2.23), and determined by the 2-cycle wrapping numbers n κ a,b and the 1-cycle wrapping numbers (n κ 3 , n κ 3 ) on T 2 3 . The exceptional part Π ex κ is of the form given in eq. (2.47), in which, to ensure supersymmetry, the coefficients e κ i are determined in the manner described in §3 by the fixed points f κ i,j on the SO(8) lattice that are wrapped by the bulk 2-cycle. It follows from eqs (2.35) and (2.48) that Similarly, using the results given in Tables 4 and 9, on the (a,e,f )A lattice we find that Now, by inspection of Table 2 we see that in all cases e κ 1 = a κ 2 mod 2 and e κ 2 = a κ 1 mod 2 (6.4) Thus, on the (a,e,f )A lattice we cannot satisfy eq. (1.1). It is apparent from Tables 4 and 9 that on the (b,c,d)A lattice the orientifold image b differs only by an overall sign from that on the (a,e,f )A lattice. Thus the expression on the right-hand side of eq. (6.3) applies to a • b + a • b on the (b,c,d)A lattice. Hence we cannot satisfy eq. (1.1) on this lattice either. Proceeding similarly, on the (a,e,f )B lattice we find instead that It follows from eq. (6.4) that For the reasons given above, the same conclusions apply in the case of the (b,c,d)B lattice. The general solution of eq. (6.9) is given by (a a 1 a b 1 , a a 2 a b 2 , e a 1 e b 1 , e a 2 e b 2 ) = (x, y, y, x + 2) or (x, y, y + 2, x) mod 4 (6.10) with x, y = 0, 1, 2, 3 mod 4.
Besides the requirements of supersymmetry and factorisability discussed earlier, there are two further constraints that must be imposed upon the non-abelian stacks a and b. The first derives from the fact that on an orientifold chiral matter in the symmetric S κ and antisymmetric A κ representations of the gauge group may arise at the interesections of any stack κ with its orientifold image κ . The dimensionality of these is given by Thus, on the U (3) stack a, this gives unobserved symmetric 6-dimensional representations. Likewise, on the U (2) stack b unobserved 3-dimensional chiral representations may arise. Clearly, we must demand the absence of such symmetric representations on both of these stacks. The antisymmetric representation on the a stack is the3 representation. In principle such states are acceptable as quark singlets q c L states, provided that the hypercharge Y (q c L ) = 2y a is right. Evidently, this require that y a = 1/6 or −1/3, corresponding respectively to d c L and u c L states. On the b stack the antisymmetric representation is the singlet representation. Again, such states are acceptable as charged lepton singlets c L , provided that y b = 1/2, or as neutrino singlets ν c L , if y b = 0. It follows from the considerations at the end of §1 that only (y a , y b ) = (1/6, 0) or (−1/3, 1/2) are consistent with getting the correct weak hypercharge for the quark doublets. The numbers of such chiral representations are given by Since we must demand the absence of the symmetric S a and S b representations, the numbers of surviving anti-symmetric representations are So the first additional constraint is that since there are only 3 quark singlets and 3 lepton singlets of each flavour in the Standard Model. It follows from eqs (5.13) and (5.15), using the supersymmetry constraint Y κ = 0, with the forms of Y κ as displayed in Table 5, that Since the bulk wrapping numbers A κ p are all integers, it is evident from the middle equations that #(A κ ) = 0 mod 6. Thus, we cannot satisfy eq. (6.16) unless #(A a ) = 0 = #(A b ). On both lattices and both stacks this requires that A κ 3 = A κ 4 mod 2. It follows from eq. (6.8) that this in turn requires that on both lattices. If |Im τ 3 | = 1/2, then on both stacks and on both lattices (a κ 1 , a κ 2 ) = (0, 0) mod 2, and all terms on the left-hand side of eq. (6.10) are 0 mod 4 so cannot satisfy eq. (6.9). The alternative is to require that The solutions given in eq. (6.10) are now restricted to the form and eq. (6.21) requires that only an odd number of a a 1 , a a 2 , e a 1 , e a 2 can be 2 mod 4. However, in this case it is easy to verify that a • a = 0, and hence #(S a ) = 0. The conclusion is that only if (a κ 1 , a κ 2 ) = (1, 1) mod 2 for both stacks κ = a and b can this constraint be satisfied if we allow only the Standard Model spectrum.
Should we succeed in finding supersymmetric (factorisable) stacks a and b satisfying the constraints detailed above, it is desirable that the the (four-dimensional) SU (3) and SU (2) gauge couplings strengths unify, i.e. α a = α b (6.23) although we do not impose this as a constraint. For the gauge group U (N κ ), the fourdimensional fine structure constant α κ of a stack κ of N κ D6-branes wrapping a 3-cycle π κ is given by [26,27] 1 where m P is the Planck mass, and Y = T 6 /R × Z 12 -II is the compactified space in this case. For fractional branes κ as defined in eq. (3.5) since the consistency of the supergravity approximation requires that the contribution of the bulk part is large compared to the contribution from the exceptional part. Then, as shown in [21], for supersymmetric stacks where X κ is defined in eq. (4.1) and for the various lattices takes the values displayed in Table 5.

Computations
We have shown in §6 that the only way that we might satisfy all of the constraints is if a κ 1 and a κ 2 are both odd for both stacks, i.e. if they are of type II or III in Table 2; then x in eq. (6.21) is odd. The numerical search produced no solutions satisfying the constraints in which (a • b, a • b ) = (1, 2) or (2, 1). The only solutions that satisfy eq. (1.1) (with (a • b, a • b ) = (0, 3) or (3, 0)) and the constraints have the wrapping numbers (n κ 3 , m κ 3 ) of T 2 3 equal to (0, ±3) for one of the stacks, i.e. the wrapping numbers are not coprime; such solutions are unacceptable. The conclusion is that the Z 12 -II orientifold cannot yield just the spectrum of the supersymmetric Standard Model.
Since there are no solutions with just the supersymmetric Standard Model spectrum, it is of interest to study models that approximate to it. Instead of demanding that #(A κ ) = 0 for both stacks, suppose that we allow just one, a say, to have |#(A a )| = |a • π O6 | = 6, the minimal non-zero number. On the (a,e,f )B lattice, it then follows from eq. (6.17) that where = ±1. Further, since A a 3 −A a 4 = 1 mod 2, it follows that (a 1 , a 2 ) = (1, 0) or (0, 1) mod 2. Thus a is of type I/VIII or of type IV/VI in Table 2. For the other stack, it follows that So if there are no antisymmetric representations on this stack, we require that 3 . This means that X b = 0, which gives an infinite value for the gauge coupling strength α b . We are therefore compelled to have antisymmetric matter on both stacks. If we also require the minimal amount on b too, then the stack b must be of the same type as a with Again, if we demand that #(A b ) = 0, then A b p = 0 (p = 1, 2, 3, 4), and α b is infinite. Likewise, if instead we require the minimal amount on b too, then it must be of the same type as a with Solutions for a and b satisfying even these weaker constraints are fairly limited. For example, on the (a,e,f )B lattice, when both a and b are of type I, we find solutions of the required type with (a a 1 , a a 2 ) = (2x a , y a ), (n a 3 , m a 3 ) = (0, y a ), (e a 1 , e a 2 ) = (z a , 2t a ) (7.8) where x κ , y κ , z κ , t κ = ±1. Then A a p = (0, 0, 2x a y a , 1), (α a i ,α a i ) = (0, 0, y a z a , 2y a t a ) (7.10) and x a y a = Im τ 3 = −x b y b (7.12) Then from eq. (6.17), it follows that #(A a ) = 6 = −#(A b ) (7.14) and the required intersection numbers (a • b, a • b ) = (3, 0) arise provided that Similarly, on the (a,e,f )B lattice, when both a and b are of type IV, there are solutions of the form A a p = (x a y a , 2, −x a y a , −2), (α a i ,α a i ) = (2y a z a , y a t a , −2y a z a , −y a t a ) (7.16) These too satisfy eqs (7.14) and have the required intersection numbers when Without loss of generality, we identify a as the SU (3) stack, and b as the SU (2) stack. To avoid further non-abelian gauge symmetries, all remaining stacks λ must consist of a single D6-brane with N λ = 1. Given the fairly limited number of solutions for a and b, the intersection numbers (a • λ, a • λ ) and (b • λ, b • λ ) with an arbitrary (supersymmetric) stack λ are also limited in number and highly correlated. As already noted, unavoidably we have 6q c L states arising in the antisymmetric3 representation of SU (3) on the stack a; if y a = 1/6 these are 6d c L , whereas if y a = −1/3 they are 6u c L . Thus in these models the minimal quark-singlet spectrum arising from the intersections of a with other stacks λ, and their orientifold images λ , is 3d c L + 3u c L when y a = 1/6, and 3ū c L + 3d c L when y a = −1/3. In both cases we must therefore impose the constraint |a•λ|+|a•λ | ≤ 6 on any one of the other stacks. The intersections of the b with other stacks λ yield doublets that must be identified either as lepton L and Higgs H d doublets, if Y = −1/2, or H u doublets if Y = 1/2. The supersymmetric Standard Model has 3L + H u + H d , so we should also impose the constraint |b • λ| + |b • λ | ≤ 5 on any single stack. With a and b both of the same type, I or IV, and on both the (a,e,f )B and (b,c,d)B lattices, the allowed intersection numbers, subject to the constraints described above, are displayed in Table 10.
In both cases, since the only negative intersection numbers for a • λ are invariably accompanied by negative intersection numbers a • λ , and vice versa, it is clear that we can Table 10. Correlations between intersection numbers of the SU (3) stack a and those of the SU (2) stack b when (a • b, a • b ) = (3, 0). never get just the required 3(3) + 3(3) quark-singlet states. When a and b are both of type IV, this conclusion is true even if we do not impose the latter constraint |b • λ| + |b • λ | ≤ 5. However, if they are both of type I, then it can be satisfied, but only at the expense of having at least 12 doublets at the intersections of b with λ and λ . The conclusion is that, at least within the range of parameters searched, we cannot get the quark-singlet spectrum even of this Standard-like model.

Discussion
We have investigated whether there is scope to construct supersymmetric Standard Models in type IIA intersecting-brane theories compactified on an orientifold with a Z 12 point group. We focussed on the Z 12 -II case because, as discussed in §2, the Z 12 -I case does not have enough independent 3-cycles to make a viable model likely. The SO(8) × SU (2) × SU (2) lattice has been used; the F 4 × SU (2) × SU (2) case is equivalent. A bulk 3-cycle then consists of a 2-cycle on the SO(8) lattice times a 1-cycle on the SU (2) × SU (2) torus T 2 3 , and we have restricted attention to the case when the 2-cycle is factorisable in the sense discussed in §3. It is possible to find models with the correct supersymmetric Standard Model quark-doublet content. All examples have (a • b, a • b ) = (3, 0) or (0, 3) and possess 6 copies of either d c L or u c L quark singlets, depending on the values of y a . Thus, some vector-like matter is inevitable. All examples have non-abelian gauge coupling constant unification in the sense that α a = α b at the string scale, but we have not found it possible to obtain the minimal quark-singlet structure described in the previous section.. With R acting as complex conjugation, as in eq. (5.1), it acts crystallographically on this lattice in the 6 orientations displayed in Table 11. R acts crystallographically on the basis 1-cycles π 5,6 of the SU (3) lattice in T 2 3 in 2 orientations: A : Rπ 5 = π 5 , Rπ 6 = π 5 − π 6 (A.13) B : Rπ 5 = π 6 , Rπ 6 = π 5 (A.14) Then the action of R on the invariant bulk 3-cycles defined in eqs (2.15) and (2.16) is given in Table 12. In this case, instead of eq. (4.10), we parametrise the 1-cycle on T 2 3 by dz 3 = e 5 (n κ 3 + m κ 3 β 2 )dν (A. 15)