Non-Abelian Discrete Groups from the Breaking of Continuous Flavor Symmetries
aa r X i v : . [ h e p - ph ] A ug Non-Abelian Discrete Groups from the Breaking ofContinuous Flavor Symmetries
A. Adulpravitchai , A. Blum and M. Lindner Max-Planck-Institut f¨ur KernphysikPostfach 10 39 80, 69029 Heidelberg, Germany
Abstract
We discuss the possibility of obtaining a non-abelian discrete flavor symmetry froman underlying continuous, possibly gauged, flavor symmetry SU (2) or SU (3) throughspontaneous symmetry breaking. We consider all possible cases, where the continuoussymmetry is broken by small representations. “Small” representations are thesewhich couple at leading order to the Standard Model fermions transforming as two-or three-dimensional representations of the flavor group. We find that, given thislimited representation content, the only non-abelian discrete group which can ariseas a residual symmetry is the quaternion group D ′ . Ever since the discovery of large neutrino mixing, non-abelian discrete flavor symmetrieshave been a popular and quite successful approach towards describing the mixing patternsof the Standard Model of Particle Physics (SM). Based on the success of symmetries, it iswell-motivated to invent a new global flavor symmetry as explanation for the existence ofgenerations. The breaking of such symmetry could then justify phenomenological successfulnon-abelian discrete symmetries as unbroken subgroups. We want to take a closer lookat the possible origin of discrete symmetries governing the structure of the SM Yukawacouplings. A natural scenario is to assume that the discrete flavor symmetry is connected tosome underlying space-time or internal gauge symmetries. Connecting a flavor symmetryto the symmetries of space-time necessitates an extension of space-time itself. Thus flavorsymmetries have been connected with discrete symmetries arising in compactified extradimensions, with [1] or without [2, 3] string theory.In this paper, we will consider the other possibility, i.e. that a discrete flavor symmetry isa conserved residual subgroup of a spontaneously broken gauged flavor symmetry. The ideaof embedding a discrete flavor symmetry in a larger continuous group has been discussedin the literature, for example in [4, 5]. However no complete models exist, in the sense E-mail: [email protected] E-mail: [email protected] E-mail: [email protected] D ′ , which has been used as a flavorsymmetry [14], but does not have a rich enough structure to predict by itself very specificmixing patterns, such as tri-bimaximal mixing for neutrinos [13].We want to break a hypothetical continuous flavor symmetry (gauged or not is irrelevantfor this discussion) at a high energy scale. This flavor symmetry should commute with theSM and transform the three generations of fermions into each other. If we limit ourselvesto three generations, we only need to consider the groups SU (2) and SU (3) as all othersemi-simple Lie groups do not have two- or three-dimensional representations. We do notneed to discuss an SO (3) separately, since the SO (3) gauge theory can simply be consideredas an SU (2) theory with a limited representation content. For SU (2) the fermions willtransform as + or . The relevant Kronecker products are thus × = + , × = + , (1) × = + + . For a flavor symmetry SU (3) the possible representations for three fermion generations are The dihedral group D being an abelian group has four one-dimensional irreducible representations.The order of D is 4. The double valued counterpart of the dihedral group D is D ′ . D ′ has four one-and one two-dimensional irreducible representations. The order of D ′ is 8. D ′ is the simplest non-abeliandouble valued dihedral group and is also called the quaternion group. For more information see [13]. and , with Kronecker products × = + , × = + , (2) × = + . We will discuss the breaking of these flavor symmetries by small representations, wheresmallness means dimension not larger than five for SU (2) and eight for SU (3). This choiceof representations is motivated by the fact that only these representations can couple tofermions at leading order, as can be read off from the Kronecker products above. The VEVsof these representations can easily be discussed using linear algebra. We discuss how thecontinuous symmetry is broken by VEVs of scalars transforming under these representa-tions and show that no non-abelian discrete symmetries, apart from D ′ , can be conservedwith these representations alone, and thus one generically needs larger representations andgroup theory beyond simple linear algebra to model such a scenario. In fact, D ′ itself onlyarises as the double-valued group of the abelian D if we break SU (2) with the unfaithfulfive-dimensional representation, which is also a representation of SO (3).To determine whether a certain VEV structure conserves a subgroup of the flavorsymmetry SU (2) or SU (3), we test which elements of the flavor symmetry leave the VEVinvariant. We will assume the minimal scalar content for any representation, i.e. real scalarsfor real representations, complex scalars for pseudo-real and complex representations. Wethen check for each representation, which subgroups the VEV of a scalar field transformingunder this representation can conserve. We also consider combinations of VEVs, but onlywhere such a combination could lead to a non-abelian subgroup. We begin by discussingthose representations which can be written with one index, i.e. which can be written asvectors in our linear algebra treatment. We continue with those representations with twoindices, i.e. matrix representations. First we take the most familiar (from the SM gluons),the adjoint representation of SU (3), then we continue with the very similar of SO (3), of SU (3) and of SU (2) at the end of the paper. SU (2) with a doublet In the two-dimensional representation of SU(2) the group elements are mapped onto the2 × λ and λ . They must obey the constraint that λ λ = 1, as the product of theeigenvalues is just the determinant. Hence if one of the eigenvalues is 1, then so is theother one. The only 2 × SU (2) always breaks the entire group. This of course does not change if we add furtherscalars of any sort. 3 .2 Breaking SU (2) with a triplet The triplet is the fundamental representation of SO (3) and an unfaithful representationof SU (2). The group elements are mapped onto the 3 × SU (2) down to Spin(2), the double covering of SO (2), which is in fact isomorphicto SO (2) and U (1).Note that there is an SO (2) for each possible axis, i.e. infinitely many SO (2)’s thatare all mutually disjoint (up to the identity element). If we introduce two triplets, theirVEVs will either be linearly dependent or not. If they are linearly dependent they breakto the same subgroup. If they are linearly independent they break to disjoint subgroups,hence fully breaking SO (3). As the triplet is an unfaithful representation of SU (2), wewill always conserve a subgroup Z under which all components of the triplet transformtrivially, while both components of the doublet transform non-trivially.We conclude from this that if we use three-dimensional representations to break SU (2),we either leave invariant a U (1) ∼ = SO (2) symmetry or a Z . In particular no non-abeliansubgroups can be conserved. We therefore do not need to consider combining a tripletVEV with a VEV of a different representation. SU (3) with a triplet The argumentation for SU (3) is in fact very similar to the one for SU (2) broken by atriplet. As the intuitive geometric derivation used above is not so readily applied in thecomplex three-dimensional Euclidean space, we give a more elaborate proof using linearalgebra. This derivation (with the obvious modifications) can also be applied to SU (2).In the three-dimensional representation of SU (3) the group elements are mapped ontothe 3 × λ , λ and λ . If one of these eigenvalues, say λ , is 1 (i.e. if the element isable to be part of some conserved subgroup), then the other two eigenvalues have to fulfill λ λ = 1, since the matrix has a unit determinant. This means that if λ is also equal to1, then λ = 1 as well. That is, the only element with more than one eigenvalue equal to1 is the identity element, the only element with three 1 eigenvalues.This means that each element which is not the identity will have at most one eigenvectorcorresponding to an eigenvalue of 1. For a simple example, the matrix e iφ e − iφ
00 0 1 (3)4ill have the eigenvector (4)corresponding to an eigenvalue of 1. As no direction in three-dimensional complex space isfavored, there will exist for each complex 3-vector non-trivial group elements having thisvector as an eigenvector with eigenvalue 1. These elements form the subgroup conservedby a VEV proportional to that eigenvector. As each non-trivial element has at most onesuch eigenvector, these subgroups will all be disjoint.What is the subgroup conserved by such a VEV? We can already guess that it will be SU (2), but this can be motivated by considering the group of elements that leave invarianta vector ~v . We then make a unitary similarity transformation U → U ′ = (cid:0) ~x ~y ~v (cid:1) † U (cid:0) ~x ~y ~v (cid:1) , (5)where U is an element of the group and ~x and ~y are arbitrary mutually orthogonal vectorsthat are also orthogonal to ~v . We obtain U ′ = (cid:18) U ′ ×
00 1 (cid:19) . (6)As U ′ is unitary by itself and also has unit determinant, we see that the three-dimensionalrepresentation reduces to the two-dimensional plus the one-dimensional representation of SU (2). Since all the SU (2) subgroups are disjoint, introducing two or more triplet scalarseither breaks to an SU (2) (in case their VEVs are linearly dependent) or breaks the entire SU (3) group (if they are not).What about anti-triplets? The arguments are the same as for the triplets, if we considerthem separately, as the two representations can only be distinguished if they show uptogether. But even if we introduce scalars transforming as triplets and scalars transformingas anti-triplets, we do not find any new subgroups: The reasoning is the same as above,each scalar VEV breaks to a specific SU (2) and they are all disjoint. The only thing weobserve is that if we introduce a scalar triplet and a scalar anti-triplet they break to thesame SU (2) if the VEV of the triplet is proportional to the complex conjugated VEV ofthe anti-triplet. If this is not the case, they break to disjoint SU (2)’s, i.e. they fully break SU (3).We conclude that an arbitrary collection of scalar triplets and anti-triplets either con-serves an SU (2) subgroup of our original SU (3) symmetry or fully breaks that symmetry. SU (3) with the adjoint representation For discussing the breaking of a continuous group with matrix representations, we beginwith the eight-dimensional adjoint representations of SU (3), as it is probably the bestknown. We can write the VEV of a scalar transforming under the adjoint representation5f SU (3) as a Hermitian 3 × V . It then transforms under SU (3) in thefollowing way: V → V ′ = U V U † , (7)where U is a special, unitary matrix. As V is traceless, we need to consider two distinctcases: Either V has three distinct eigenvalues, or it has two degenerate eigenvalues λ , thethird eigenvalue being − λ . The only possible VEV with three degenerate eigenvalues isthe zero matrix, i.e. a vanishing VEV, which naturally does not break SU (3).We first consider the case of a V with three distinct eigenvalues. We are looking forthe subgroup of SU (3) formed by those elements U which leave V invariant, i.e. for which V = V ′ . This set is just the set of all matrices U that commute with V . What does itmean if U commutes with V ? Let ~v i be the eigenvector V associated with the eigenvalue λ i , which we have assumed to be nondegenerate. Then V ( U ~v i ) = U ( V ~v i ) = λ i ( U ~v i ) . (8)Hence U ~v i is also an eigenvector of V with eigenvalue λ i . As this eigenvalue is non-degenerate U ~v i must linearly depend on ~v i . Therefore ~v i is also an eigenvector of U . Thisholds for all three eigenvectors of V . We can thereby specify the subgroup conserved bythis VEV: It is the set of all U having the same set of eigenvectors as V . The most generalform for an element of this group is then U = (cid:0) ~v ~v ~v (cid:1) e iα e iβ
00 0 e − i ( α + β ) (cid:0) ~v ~v ~v (cid:1) † . (9)This representation is clearly unitarily equivalent to a diagonal representation, i.e. it re-duces to three representations of U (1). As α and β are however independent, there areactually two distinct U (1) groups. Therefore an adjoint VEV with three distinct eigen-values breaks SU (3) down to U (1) × U (1). Note that such a VEV can never conserve anon-abelian subgroup of SU (3) and we do not need to consider it any further.We now proceed to VEVs V having two degenerate eigenvalues. The eigenvectors of V are now no longer uniquely defined. If ~v and ~v are two orthonormal eigenvectors of V corresponding to the same eigenvalue, we can find an arbitrary orthonormal basis of thecorresponding eigenspace as a ~v + b ~v and − b ~v + a ~v , with a and b two complex numbersobeying | a | + | b | = 1. Defining X ≡ (cid:0) ( a ~v + b ~v ) ( − b ~v + a ~v ) ~v (cid:1) , we can thereforewrite any matrix in SU(3) that commutes with V in the following form: X e iα e iβ
00 0 e − i ( α + β ) X † (10)To reduce this representation, we do a unitary equivalence transformation by multiplyingon the right by (cid:0) ~v ~v ~v (cid:1) (11)6nd on the left with the Hermitian conjugate. The resulting matrix is | a | e iα + | b | e iβ ab ( e iα − e iβ ) 0 ab ( e iα − e iβ ) | a | e iβ + | b | e iα
00 0 e − i ( α + β ) . (12)We now show that this is a representation of SU (2) × U (1). To do this we factorize thematrix (12): e i ( α + β )2 e i ( α + β )2
00 0 e − i ( α + β ) x y − y ∗ x ∗
00 0 1 , (13)where x = | a | e i ( α − β )2 + | b | e − i α − β , (14) y = 2 iab ∗ sin (cid:18) α − β (cid:19) . (15)These two matrices commute. The first matrix is the representation of U(1), with the firsttwo generations transforming in the same way, and the third with double and oppositecharge. If we observe that | x | + | y | = 1, we see that the second matrix furnishes arepresentation of SU(2), under which the first two generations form a doublet and the thirdgeneration is a singlet. We also note that we have the correct number of free parameters:The absolute value of a (or b ), the phase of ab ∗ and the phase difference ( α − β ).We consider the case of two adjoint VEVs, where both VEVs have degenerate eigenval-ues. First of all, their non-degenerate eigenvalues could correspond to the same eigenvector.In this case, they will naturally break to the same subgroup. Then we could have the case,where the non-degenerate eigenvalue of the second VEV corresponds to an eigenvectorlying in the eigenspace of the degenerate eigenvalue of the first VEV. This, in a way, sin-gles out a basis of that eigenspace and thereby coincides with the VEV of an octet withthree distinct eigenvalues, i.e. conserves a subgroup U (1) × U (1). Therefore, if there isno relation between the eigenvectors of the two VEVs, we only conserve the subgroup Z , corresponding to the three third roots of unity, which can never be broken by adjointscalars.Finally, combining a degenerate adjoint VEV with a triplet VEV, we find three possi-bilities: First, the triplet VEV can coincide with the non-degenerate eigenvector. In thiscase e − i ( α + β ) must be equal to 1 and we break down to the same SU (2) conserved by thetriplet VEV alone. If the triplet VEV lies in the degenerate eigenspace, we break the SU (2)conserved by the octet VEV and are left with only a residual U (1). If the triplet VEV isnot an eigenvector of the adjoint VEV we again break the entire group.Thus, the only new non-abelian subgroup of SU (3) we can conserve with the VEV of ascalar transforming under the adjoint representation is the subgroup SU (2) × U (1) if theVEV has two degenerate eigenvalues. 7 .5 Breaking SU (2) with the five-dimensional representation The calculations here are very similar to those of the last section, so we will be brief. TheVEV V of a scalar transforming under the five-dimensional representation can be writtenas a 3 × SU (2) as V → V ′ = OV O T , (16)with O a special orthogonal matrix. Again the question of invariance can be reduced to aquestion of commutation and hence coincident eigenspaces. For a VEV with nondegenerateeigenvalues the general form of elements in the conserved subgroup is O = (cid:0) ~v ~v ~v (cid:1) ( − n − m
00 0 ( − n + m (cid:0) ~v ~v ~v (cid:1) T , (17)with n , m integers (as V is symmetric it can be diagonalized by a real orthogonal matrix,hence O will have only real eigenvectors and therefore only real eigenvalues). After asimilarity transformation this is a representation of Z × Z ∼ = D . However, since weactually break SU (2) with an unfaithful representation, we actually conserve the double-valued group D ′ . The SU (2) doublet will transform as a doublet in this group as well,while the triplet, as can be seen from the matrix above, decomposes into the three non-trivial one-dimensional representations. D ′ has no non-abelian subgroups, so we do notneed to consider a combination of this VEV with others.Next we consider VEVs V with two degenerate eigenvalues. The elements of the con-served subgroup must still have ~v as an eigenvector with a real eigenvalue. There are,however, two possibilities to do this. One is to assign the eigenvalue 1 to ~v . These are allelements having ~v as axis of rotation. They form SO (2) subgroup. We also have thoseelements, where the eigenvalue − ~v . These are of the form Y ( − n − n +1
00 0 − Y T , (18)using Y ≡ (cid:0) ( c ~v + s ~v ) ( − s ~v + c ~v ) ~v (cid:1) , where s and c are the sine and cosine, re-spectively, of some undefined angle. By multiplying (18) on the right by (cid:0) ~v ~v ~v (cid:1) , (19)and on the left by its transpose, we perform a unitary transformation and end up with ( − n ( c − s ) 2 cs ( − n cs ( − n ( − n ( s − c ) 00 0 − . (20)Since the above matrix must still have a unit determinant, we know that the upper left2 × − O (2) and the one-dimensional representation, where each element ismapped onto its determinant. As our original group was SU (2), we are actually breakingto the double covering of O (2), which is the group Pin(2). Combining several such VEVs,they can coincide in the non-degenerate eigenvector, in which case Pin(2) is conserved,the non-degenerate eigenvector of one can lie in the degenerate eigenspace of the other, inwhich case the conserved subgroup is D ′ , or their eigenbases could be unrelated, in whichcase only Z is conserved.There are thus only two non-abelian groups which can be the residual subgroup of SU (2) after breaking with the VEV of a five-dimensional representation: The group D ′ for non-degenerate eigenvalues and the group Pin(2) for degenerate eigenvalues. Some ofthese results can also be found in [4]. SU (3) with the six-dimensional representation Writing the VEV of the six-dimensional representation as a complex, symmetric 3 × V , it transforms under SU (3) in the following way: V → V ′ = U V U T . (21)Demanding invariance can then be rewritten as the condition U V = V U ∗ . (22)We now note that V need not necessarily be diagonalizable. However, since V is complexand symmetric can be written in the form W T V W = V diag , (23)with W unitary [15]. We can write W ≡ ( ~w , ~w , ~w ). The ~w i are then singular vectors of V obeying the relation V ~w i = σ i ~w i ∗ , (24)with σ i the diagonal elements of V diag , i.e. the singular values of V . The condition ofequation (22) then leads to V ( U ∗ ~w i ) = U V ~w i = σ i U ~w i ∗ = σ i ( U ∗ ~w i ) ∗ . (25)If V has three distinct singular values, this means that all ~w i need to be eigenvectors of U ∗ . Also, the corresponding eigenvalue of U ∗ needs to be real. Therefore the discussionis the same as for the quintuplet of SO (3): The conserved subgroup is D . If V hastwo degenerate eigenvalues, then U ∗ should act on the corresponding singular space withonly real coefficients, that is it should be block-diagonalizable to give an orthogonal 2 × O (2). As V need not to be traceless, weencounter the additional case of three degenerate singular values. Here U ∗ needs to act9n all singular vectors with real coefficients, so the conserved subgroup in this case is SO (3). Of these subgroups only the last two are non-abelian and need to be considered incombination with other VEVs.We demanded above that the eigenvalues of U need to be real. This condition stemsfrom equation (24): If ~w i obeys that relation, then α ~w i only obeys the same relation if α is real or alternatively σ i must be zero. Thereby VEVs with zero eigenvalues are alge-braically special: The group elements preserving this VEV can have complex eigenvaluescorresponding to the singular vectors of V with singular value 0. A special unitary matrixcannot have only one non-real eigenvalue. Hence the case of interest is a VEV with twozero eigenvalues. The group elements preserving this VEV are of the same form as thoseof equation (10), with the additional condition that e − i ( α + β ) , the eigenvalue correspondingto the non-zero singular value, must be real. We can therefore substitute the parameter α + β by the integer parameter m defined by α + β = mπ and using definitions (14, 15)write the group elements conserving V in the form i m i m
00 0 ( − m x y − y ∗ x ∗
00 0 1 . (26)The conserved subgroup is therefore SU (2) × Z , where the first two generations form adoublet of SU (2) and a faithful representation of Z , while the third generation is a singletof SU (2) and an unfaithful, non-trivial representation of Z .What if we combine two six-dimensional VEVs? If they coincide in all three singularvectors, the subgroup is determined by the VEV with less degenerate eigenvalues. If theyhave only one singular vector in common, we break to the subgroup of elements havingtwo degenerate real eigenvalues, that is Z . If they have no singular vectors in common,we break SU (3) fully. Zero eigenvalues are only relevant if the VEVs coincide in all threesingular vectors anyway and the zero eigenvalues correspond to the same eigenspace. Inthis case the full subgroup SU (2) × Z is conserved.We thus have three non-abelian groups that can be conserved by a sextet VEV, O (2) fortwo degenerate singular values, SO (3) for three degenerate singular values, and SU (2) × Z for two zero eigenvalues. We now need to consider combinations of these three cases withthe other VEVs we have discussed so far, triplets and octets.What if both a and a triplet acquire a VEV? If the triplet VEV is not a singularvector of the , then SU (3) is fully broken. What if it is a singular vector? If V has twodegenerate singular values, the triplet can correspond to the non-degenerate singular value.In this case, the determinant of the 2 × SO (2) or U (1). If the triplet VEV corresponds to a degenerate singular value,the degeneracy becomes irrelevant and the subgroup is Z . If V has three degeneratesingular values, the triplet, which is in the defining representation of SO (3), breaks thatsubgroup in the usual way down to U (1) or fully breaks it, if the real and imaginary partsof the triplet VEV are not parallel. If we combine a V with two zero eigenvalues with atriplet VEV, we again have two possibilities: The triplet VEV can correspond to the non-zero singular value. In this case m is fixed to be 0 or 2, and we break down to SU (2) (the10ormer Z element can just be multiplied into the SU (2) element, without changing thedeterminant). If the triplet VEV is an eigenvector of V corresponding to a zero eigenvalue,we need to take a closer look. The SU (2) element in equation (26) has eigenvalues e ± i ( α − β )2 .So, without loss of generality, we must now demand ( i ) m e i ( α − β )2 = 1. The resulting elementthen has in addition two eigenvalues of −
1, corresponding to fixed vectors. The conservedsubgroup is then Z .We proceed by combining a VEV of a with an adjoint VEV. The adjoint VEV musthave two degenerate eigenvalues, as only then do we have the possibility of conserving a non-abelian subgroup. If there does not exist a basis of singular vectors for the sextet VEV, thatis also a basis of eigenvectors for the octet VEV, SU (3) will be fully broken. In particular,the eigenvector of the octet VEV corresponding to the non-degenerate eigenvalue, ~v , mustalways be a singular vector of the sextet VEV V . The discussion is then similar to that forthe triplet, with the triplet VEV replaced by ~v . If V has two degenerate singular values, ~v can correspond to the non-degenerate singular value. In this case nothing changesand the conserved subgroup is O (2). If ~v corresponds to a degenerate singular value,the degeneracy becomes irrelevant and the subgroup is D . If V has three degeneratesingular values, one of the degeneracies becomes irrelevant and we break down to O (2).Finally, considering the case of a sextet VEV V with two zero eigenvalues, we again havetwo possibilities: ~v can correspond to the non-zero singular value. In this case nothingchanges, and SU (2) × Z is still the conserved subgroup. If ~v is an eigenvector of V corresponding to a zero eigenvalue, a specific basis is singled out for the elements of theconserved subgroup. It is thus only determined by the possible eigenvalues, and cannot benon-abelian. In this case it will be U (1) × Z . Thus, no new non-abelian subgroups canbe attained by combining the VEVs of these different SU (3) representations. SU (2) with the four-dimensional representation We finally deal with the most complicated of the matrix representations, the of SU (2).As it arises from the product of a vector and a spinor, it can be written as a 3 × as a matrix, it acts on a spinor and transforms it into a vector. As theClebsch Gordan coefficients are normally given in spherical coordinates we start with these,later switching back to Cartesian coordinates, where the scalar product of two vectors issimply matrix multiplication. In spherical coordinates, we can give the four degrees offreedom of the as φ (m= ), φ (m= ), φ (m= − ) and φ (m= − ). Correspondinglywe write the two components of the spinor we want to transform into a vector as ψ (m= )and ψ (m= − ). Using the Clebsch Gordan coefficients for SU (2) [16] we find that they11ombine in the following way to form a vector:12 ( √ φ ψ − φ ψ ) ( m = 1) , (27)1 √ φ ψ − φ ψ ) ( m = 0) , (28)12 ( φ ψ − √ φ ψ ) ( m = − . (29)Switching to Cartesian coordinates, this corresponds to a vector √ [( φ − √ φ ) ψ + ( φ − √ φ ) ψ ] i √ [ − ( φ + √ φ ) ψ + ( φ + √ φ ) ψ )] √ ( φ ψ − φ ψ ) . (30)This vector arises from multiplying a spinor by the following matrix:1 √ ( φ − √ φ ) ( φ − √ φ ) − i ( φ + √ φ ) i ( φ + √ φ ) − φ φ , (31)or, in another simpler parameterization V = a bc d − b + id a + ic , (32)where a , b , c and d are complex. This is then the most general form for the VEV of a .It transforms in the following way: V → V ′ = OV U † , (33)as it has one vector and one spinor index. O and U are of course not independent butdescribe a rotation of the same magnitude around the same axis. It can be checked byexplicit calculation that V ′ can be parameterized in the same way as V for an arbitraryrotation.Again we reformulate the condition of invariance as a condition on the eigensystems.We first observe that we can deduce from equation (33) the following two equations: V V † = OV V † O T (34) V † V = U V † V U † , (35)from which we immediately deduce that the eigenvectors of V V † (i.e. the left singularvectors of V , denoted by ~u i ) must also be eigenvectors of O (with the usual ambiguities fordegenerate singular and eigenvalues) and the right singular vectors of V , denoted by ~w i ,12ust be eigenvectors of U . Using this knowledge, we go back to equation (33). We findthat V U ~w i = V µ i ~w i = σ i µ i ~u i , (36)for i = 1 , µ i the eigenvalues of U and σ i the singular values of V . We also have V U ~w i = OV ~w i = Oσ i ~u i = λ i σ i ~u i , (37)with λ i the eigenvalues of O . From the last two equations we can deduce that λ i = µ i . As O and U are rotations by the same angle θ , their eigenvalues are e ± iθ and 1 for O and e ± i θ for U . How can they be made to coincide? Apart from the trivial case of both being theunit matrix, we are only left with the possibility of identifying the exponential eigenvalues,which is only possible for θ = ± π . The final left singular vector ~u is then the eigenvectorof O corresponding to the eigenvalue 1, i.e. it defines the axis of rotation. If it is real, wethen break to all rotations around that axis, with the angles given above. This is a Z subgroup of SU (2). If the axis is complex (and real and imaginary parts are not linearlydependent), no such elements exist and SU (2) is fully broken.Is the subgroup enlarged if V has degenerate singular values? We first take the case σ = σ = σ = 0. ~u is still an eigenvector of O , the ~u i and ~w i however need not beeigenvectors of O and U , respectively. Rather we have V U ~w i = V ( α i ~w + β i ~w ) = σ ( α i ~u + β i ~u ) (38) V U ~w i = OV ~w i = σO ~u i = σ ( α ′ i ~u + β ′ i ~u ) , (39)from which we can immediately infer that α i = α ′ i and β i = β ′ i . This means that againtheir eigenvalues need to coincide, and we again break to Z or nothing. Finally, we needto consider the case, where one of the non-trivial singular values is zero, say σ = 0. Inthis case O and U need only coincide in one eigenvalue, but this condition is already strongenough to constrain the elements in the same way, i.e. giving Z as the conserved subgroup.We thus find that a VEV for the four-dimensional representation of SU (2) can never leadto the conservation of a non-abelian subgroup, and we thus do not need to combine it withany other VEVs. The results of this paper can be summarized in one sentence: The only non-abelian discretesubgroup that can be conserved by VEVs of the small representations (dimension equal orless than five and eight, respectively) of SU (2) and SU (3) is the group D ′ .To arrive of this result, we have assumed that the continuous flavor symmetry is very“cleanly” broken - what we mean by this, is that there are no low-mass remnants of thesymmetry breaking floating around in the resulting low energy model, such as (Pseudo-)Nambu-Goldstone bosons (which can result either from the continuous symmetry itself,if it is not gauged, or from accidental symmetries of the symmetry-breaking potential)or low mass scalar degrees of freedom related with (approximately) flat directions of the13ymmetry-breaking potential, such as often appear in supersymmetric frameworks. Thisclean breaking is what is needed to reproduce models using a discrete non-abelian symmetryas starting point, as they only relate to a possible larger symmetry through embeddingand consistency constraints.It may, however, be possible to define intermediate non-abelian symmetries in a some-what less strict sense. Here it would not be possible to describe an intermediate modelwith a discrete symmetry independently of the underlying continuous symmetry, but itcan still be interesting to examine the role of the discrete symmetry in such models moreclosely. An example is the A symmetry appearing in [17].What further implications do the results presented in this paper have for flavoredmodel-building with non-abelian discrete flavor symmetries? Our results naturally stillallow for more complicated schemes for breaking a continuous flavor symmetry to a discretesubgroup. The symmetry breaking fields would then not couple to the fermions at leadingorder and a dedicated study would be necessary to check whether a lower energy modelwith a discrete flavor symmetry could be reproduced.Our results may be taken as an indication that one may have to think differently aboutintermediate discrete flavor symmetries. If one favors discrete flavor symmetries in theirown right at some scale, one should look towards other possible origins, such as from extradimensions, which would also protect such a global symmetry from quantum gravity effectsthat might otherwise eradicate it [18]. The main lesson is that the embedding of a discreteflavor symmetry in a continuous one is not nearly as simple as is often tacitly assumed. Acknowledgments
This work has been supported by the DFG-Sonderforschungsbereich Transregio 27 “Neu-trinos and beyond – Weakly interacting particles in Physics, Astrophysics and Cosmology”.14 eferences [1] T. Kobayashi, H. P. Nilles, F. Ploger, S. Raby and M. Ratz, Nucl. Phys. B , 135(2007) [arXiv:hep-ph/0611020].[2] G. Altarelli, F. Feruglio and Y. Lin, Nucl. Phys. B , 31 (2007)[arXiv:hep-ph/0610165].[3] A. Adulpravitchai, A. Blum and M. Lindner, [arXiv:0906.0468 [hep-ph]].[4] P. H. Frampton and A. Rasin, Phys. Lett. B , 424 (2000) [arXiv:hep-ph/9910522].[5] C. Hagedorn, M. Lindner and R. N. Mohapatra, JHEP , 042 (2006)[arXiv:hep-ph/0602244].[6] M. Tanimoto, T. Watari and T. Yanagida, Phys. Lett. B , 345 (1999)[arXiv:hep-ph/9904338].[7] T. Araki, T. Kobayashi, J. Kubo, S. Ramos-Sanchez, M. Ratz and P. K. S. Vau-drevange, Nucl. Phys. B , 124 (2008) [arXiv:0805.0207 [hep-th]].[8] C. Luhn and P. Ramond, JHEP , 085 (2008) [arXiv:0805.1736 [hep-ph]].[9] Ling-Fong Li, Phys. Rev. D , 1723 (1974).[10] G. Etesi, J. Math. Phys. , 1596 (1996) [arXiv:hep-th/9706029].[11] M. Koca, M. Al-Barwani and R. Koc, J. Phys. A , 2109 (1997).[12] M. Koca, R. Koc and H. Tutunculer, Int. J. Mod. Phys. A , 4817 (2003)[arXiv:hep-ph/0410270].[13] A. Blum, C. Hagedorn and M. Lindner, Phys. Rev. D , 076004 (2008)[arXiv:0709.3450 [hep-ph]].[14] M. Frigerio, S. Kaneko, E. Ma and M. Tanimoto, Phys. Rev. D , 011901 (2005)[arXiv:hep-ph/0409187].[15] W. Grimus and G. Ecker, J. Phys. A , 2825 (1988).[16] C. Amsler et al. [Particle Data Group], Phys. Lett. B , 1 (2008).[17] B. Stech and Z. Tavartkiladze, Phys. Rev. D , 076009 (2008) [arXiv:0802.0894 [hep-ph]].[18] L. M. Krauss and F. Wilczek, Phys. Rev. Lett.62