Conditions for the custodial symmetry in multi-Higgs-doublet models

Journal of High Energy Physics, May 2018

Abstract We derive basis-independent, necessary and sufficient conditions for the custodial symmetry in N-Higgs-doublet models (NHDM) for N ≥ 3, and apply them on some 3HDM examples.

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Conditions for the custodial symmetry in multi-Higgs-doublet models

HJE Conditions for the custodial symmetry in multi-Higgs-doublet models M.Aa. Solberg 0 1 0 7491 Trondheim , Norway 1 Department of Mechanical and Industrial Engineering , NTNU We derive basis-independent, necessary and su cient conditions for the custodial symmetry in N-Higgs-doublet models (NHDM) for N 3HDM examples. Beyond Standard Model; Global Symmetries; Higgs Physics 3, and apply them on some 3HDM examples. Keywords: Beyond Standard Model, Global Symmetries, Higgs Physics 1 Introduction 1.1 N-Higgs-doublet models 2.1 The general NHDM potential 2 Basis-independent conditions for custodial symmetry in NHDM Quadratic terms and quartic terms proportional to K0 Quartic terms not involving K0 Necessary conditions for the custodial SO( 4 ) symmetry Necessary and su cient conditions for the custodial SO( 4 ) symmetry 13 Spontaneous breaking of the custodial SO( 4 ) symmetry Applying a necessary condition The Ivanov-Silva model The necessary condition is not su cient 3 Summary A Factorizable quartic terms B A basis for su(N ) B.1 Structure constants for the 3HDM C The inclusion AdSU(N) since cos2( W ) = m2W =m2Z at tree-level. Eq. (1.2) holds at all orders of perturbation theory when the custodial symmetry is exact. The parameter is measured experimentally to be close to unity, with a minute dependence of the chosen renormalization prescription. For = m2W m2Z cos2( W ) ; = 1; { 1 { instance, if we interpret in the minimal subtraction renormalization scheme at the energy scale mZ , takes the value is the SM Higgs doublet, and the elds 1; : : : ; 4 are real scalar elds and v is the vacuum expectation value (VEV). Moreover, in (1.3) D is the covariant derivative LH = (D )y(D ) VSM( ); = +! 0 = 1 + i 2 HJEP05(218)63 the potential VSM( r), given by (1.6) with the substitution r, is evidently invariant ! under transformations with O 2 O( 4 ). The kinetic terms of (1.3) will, in the limit g0 ! 0, be invariant under SO( 4 ) transformations [3], but not under O 2 O( 4 ) with det(O) = 1 [4]. This approximate SO( 4 ) symmetry is the custodial SO( 4 ) symmetry, which we will denote SO( 4 )C . In the presence of a VEV, this symmetry is broken down to a custodial SO( 3 ) symmetry, denoted by SO( 3 )C . If we impose the custodial symmetry on the SM Lagrangian, it forces electroweak gauge bosons W and Z0, levels of perturbation theory. To see this, consider the mass-squared matrix M be zero, g0 ! 0, and then the three massive gauge bosons will transform as a triplet under SO( 4 )C , where SO( 3 )C here leaves the vacuum invariant (it leaves the VEV alone). Then mW = mZ at all orders of perturbation theory, since the massive gauge boson elds can be interchanged by a SO( 3 )C transformation. If the custodial symmetry is extended to the Yukawa sector, the mass renormalization of the gauge bosons, due to massive fermions, will yield the same result for all the massive gauge bosons, and hence the mass degeneration of W and Z0 will still be exact [5]. Moreover, W = 0 at all orders since g0 = 0 implies no electroweak mixing. Hence cos2 W = 1, which gives us orders, when the theory is custodially symmetric. 1; : : : ; N , we will construct the most general NHDM potential from the following Hermitian bilinears, linear in each eld m; n: HJEP05(218)63 To avoid double counting, we will let 1 m < n N , and we apply the following invertible encoding a = a(m; n) to label such pairs, 1 a(m; n) = (n 1) + (m 1) N (m + 2) N (N 1) k: (1.11) Abm = Bbmn = Cbmn = 1 2 i 2 ym m; ( ym n + yn m) ( ym n yn m) Bba; Cba: 1 2 The inverse of this encoding will be denoted (m(a); n(a)). The bilinears Bbmn and Cbmn will be ordered \lexicographically" in (m; n) by the encoding a(m; n), e.g. k fCbaga=1 = fCb12; Cb13; : : : ; Cb1N ; Cb23; Cb24; : : : ; CbN 1;N g: The most general NHDM potential can then be written, V ( 1; : : : ; N ) = (m1)Abm + (a2)Bba + (a3)Cba + (m1n)AbmAbn + (a2b)BbaBbb + (a3b)CbaCbb + (m4a)AbmBba + (m5a)AbmCba + (a6b)BbaCbb; where repeated indices m and n are summed from 1 to N , while repeated indices a and b are summed from 1 to k, confer (1.11). The general NHDM Lagrangian can then be written LNHDM = (D m)y(D m) V ( 1; : : : ; N ); where V was given in (1.13), and where we as usual sum over repeated indices. However, the general NHDM Lagrangian has more sources of custodial symmetry violation than just g0. The bilinears Cba all violate the custodial symmetry, and hence the parameters 1 2 (1.10) (1.12) (1.13) (1.14) try groups of bilinears Cba and quartic terms CbaCbb. The SO( 4 ) symmetry group is the custodial symmetry, U(2) = SO( 4 ) \ Sp(2; R) = SU(2)L U(1)Y is the global symmetry of the SM, Sp(2; R) is the symmetry group of bilinears Cba, while P (2; R) is the symmetry group of quartic terms CbaCbb. Finally, O( 4 ) is the symmetry group of the bilinears Bba. Taken from [4]. HJEP05(218)63 (a3); (a3b); (m5a) and (a6b) will give contributions to the real NHDM quadruplets, given by = 1. Consider transformations of m;r = Re( m) Im( m) ! ! 0m;r = S m;r; (1.15) as in (1.7) for the SM. In section 2.2 of [3] we showed that the transformations S which left the bilinears Cba invariant, were the transformations S 2 Sp(2; R), i.e. the symmetry group of the bilinears Cbmn is the real symplectic group Sp(2; R). The symmetry group of CbaCbb was the group P (2; R), with Sp(2; R) as identity component. Both symmetry groups are incompatible with the custodial symmetry, see gure 1, and hence violate the custodial symmetry. The most general custodially symmetric NHDM potential, is then given by VCS( 1; : : : ; N ) = (m1)Abm + (a2)Bba + (m1n)AbmAbn + (a2b)BbaBbb We will denote a NHDM potential of the form (1.16) manifestly SO( 4 )C -symmetric. However, since the Higgs doublets 1; : : : ; N have the same quantum numbers, we are free to rede ne the Higgs doublets through unitary Higgs basis transformations, + (m4a)AbmBba: m ! 0m = Umn n; { 4 { for an U 2 U(N ). The kinetic terms of the NHDM Lagrangian are invariant under SU(2)L U(N ) transformations (promoted to SU(2)L Sp(N ) in the limit g0 ! 0 [3]), and the basis transformations U 2 U(N ) will thus leave the kinetic terms invariant. Normally only Higgs basis transformations U 2 SU(N ) are considered, since an overall U(1) transformation do not change the parameters of the potential. However, the NHDM potential will generally not be invariant under a SU(N ) change of Higgs basis, and hence the custodial symmetry can be hidden due to a change of basis. On the other hand, a NHDM potential that is not manifestly SO( 4 )C -symmetric may be transformed into a manifestly (1.16) (1.17) SO( 4 )C -symmetric potential through a SU(N ) basis transformation. Hence we de ne a potential to be explicitly custodially symmetric, or more simple, SO( 4 )C -symmetric, if it can be transformed into a manifestly SO( 4 )C -symmetric potential by a Higgs basis transformation. Thus a SO( 4 )C -symmetric potential can be transformed to the form (1.16). On very general grounds, an implementation of the custodial symmetry in a NHDM potential can be transformed to a manifest SO( 4 )C symmetry, i.e. the potential can be transformed to the form (1.16), by a Higgs SU(N ) basis transformation [6]. We will in this article develop necessary and su cient conditions for the custodial symmetry in arbitrary NHDM potentials for N > 2. For N = 2, necessary and su cient conditions for SO( 4 )C were derived independently in [7] and [8]. Necessary and su cient conditions for SO( 4 )C in the case N = 3, and necessary conditions for the cases N = 4 and N = 5 are given in [6], in a di erent formalism than the one applied in this article. The cases N 3 are di erent from the case N = 2, since the bilinears (1.10) transform under the adjoint representation AdSU(N) of SU(N ), and AdSU(2) = SO( 3 ) while AdSU(N) ( SO(N 2 1) for N > 2: (1.18) Hence most SO(N 2 1) matrices will not be at our disposal for N > 2, and this will complicate the procedure, when we try to rotate a possibly SO( 4 )C -symmetric potential into a manifestly SO( 4 )C -symmetric potential. 2 Basis-independent conditions for custodial symmetry in NHDM For a discussion of the special case where the quartic terms can be factorized, see appendix A. 2.1 The general NHDM potential We will now nd basis-independent, su cient and necessary conditions for having a custodially symmetric potential in the general NHDM. To do this, we will adopt much of the notation applied in [9]. Now de ne K = Tr(K~ ): { 5 { and let the Hermitian N N matrix K~ be given by ~ = ( 1; 2; : : : ; N )T ; K~ = ~ ~ y = BB 0 B B y1 2 y1 1 . . . y1 N y2 1 : : : y2 2 : : : . . . y2 N : : : y N N 1 y N 1 yN 2 CC . . . C C A Moreover, let be generalized Gell-Mann matrices, which is a basis for the real vector space of Hermitian N N matrices. Then the N 2 linearly independent Hermitian bilinears in the elds 1; : : : ; N can be written (2.1) (2.2) (2.3) As in [9], we will let Greek indices like run from 0 to N 2 1, while Latin indices like a run from 1 to N 2 de ne the matrices 1. Summation of repeated indices is, as usual, also assumed. We will such that the SO( 4 )C -violating bilinears Cb are ordered rst, that is Ka = 2Cba = 2Cbm(a);n(a); for 1 a see appendix B, for a construction of such matrices. We will put the custodial symmetryviolating bilinears rst, to make the conditions for the custodial symmetry simpler to express. The very rst bilinear K0 will be de ned as in [9], with 0 = p2=N I. Now the general NHDM potential may also be written in the same manner as in [9] V = 0K0 + aKa + 0K02 + 2K0 aKa + KaEabKb; although the bilinears K are organized in a di erent order here. In (2.6) the parameters 0; a; 0; a are real, and E is a real and symmetric (N 2 1) (N 2 1) matrix. We will refer to the last term of (2.6) as VE, that is Under the basis transformation (1.17) the matrix K~ will transform as while K00 = K0; Ka = Rab(U )Kb for a 1; (2.4) (2.5) (2.6) (2.7) (2.8) (2.9) (2.10) (2.11) a = Rab(U )U bU y; matrix Rab(U ) 2 AdSU(N) is de ned by U y aU = Rab(U ) b: Here the a's may be generalized Gell-Mann matrices, or any basis for the Lie algebra su(N ), although we will stick to the generalized Gell-Mann matrices a de ned in appendix B. This means the bilinears Ka transform under the adjoint representation of SU(N ), while K0 transforms under the trivial representation of SU(N ). If the trace tr( i j ) / 1) (see appendix C), where the latter inclusion is strict for N and by multiplying (2.11) by (R 1)ca, we obtain be an orthogonal matrix with R 1 = RT , and (2.12) then yields Moreover, comparing (2.10) and (2.12) gives us R 1(U ) = R(U y). If tr( i j ) / ij , R will Rca1 a = U cU y: Rac a = U cU y; which we will apply later. If we make the substitution ~ ! U ~ , and hence Ka ! Rab(U )Kb in the potential (2.6), the potential remains invariant if we simultaneously substitute the parameters HJEP05(218)63 of the potential with and ~x ! R(U )~x and x0 ! x0 for x 2 f ; g; E ! E0 = R(U )ERT (U ): We will now nd basis-independent conditions for SO( 4 )C symmetry of the di erent parts of the potential, considered isolated. Later, we will patch these conditions together for su cient and necessary conditions for custodial symmetry in the general NHDM potential. 2.1.1 Quadratic terms and quartic terms proportional to K0 The quadratic terms given by the parameters a and the quartic terms given by the parameters a, see (2.6), transform under SU(N ) Higgs basis transformations in exactly the same manner, confer (2.14). Since the rst N (N 1)=2 bilinears Ka correspond to custodial symmetry-violating operators of the type Cb, they must be possible to transform away by some Higgs SU(N ) basis transformation, when the terms aKa and aKa are custodially symmetric. Necessary and su cient conditions for having the custodial symmetry in these terms simultaneously, is then the existence of a Higgs basis transformation given by R(U ) 2 AdSU(N), such that Rij j = 0 and Quartic terms not involving K0 If a potential is custodially symmetric, the real, symmetric matrix E has to be similar to an E0 given by VE = KaEabKb E0 = RERT ; { 7 { (2.12) (2.13) (2.14) (2.15) (2.16) (2.17) (2.18) where the N (N 1)=2 rst rows and columns of E0 consist of zeros, i.e. E0 = BBB0 : : : 0 : : : 0C ; C C . . . . . . 1 .. C . C C C A Nullity(E) to make this possible, is that where X is an arbitrary (N + 2)(N 1)=2 (N + 2)(N 1)=2 block. The matrix X will be real and symmetric, since R and E are real, and since E is symmetric. The rst 1)=2 rows and columns of E0 have to equal zero to make all terms containing custodial symmetry-violating bilinears Cb disappear. Moreover, this transformation has to be SO(N 2 1). The rst condition that has to be met (2.19) (2.20) (2.21) HJEP05(218)63 SO(N 2 which means E has an eigenvalue zero with a multiplicity of at least N (N 1)=2, or equivalently, the nullspace of E has dimension N (N 1)=2 or more. We will now prove that the matrix E can be transformed to the form given by (2.19) only by a matrix R 2 1) which has N (N 1)=2 orthonormal nullvectors of E as its rst rows. Proposition 1. Assume the matrix E has at least k = N (N 1)=2 linearly independent nullvectors, and let R 2 SO(N 2 1) be such that E0 = RERT . Then the k rst rows and columns of E0 are zero, as given in (2.19), if and only if R is of the form R = [n1; : : : ; nk; ck+1; : : : ; cN2 1]T ; where n1; : : : ; nk are nullvectors of E and n1; : : : ; nk; ck+1; : : : ; cN2 1 are orthonormal column vectors. Proof. ((): Assume R is of the form given by (2.21). Since n1; : : : ; nk are nullvectors of E, the k rst columns of the product (ERT ) are zero, and hence also the k rst columns of E0 = RERT are zero. Now E0 is symmetric since E is symmetric, and consequently the k rst rows of E0 are also zero. ()): Assume the k rst columns and rows of E0 = RERT are zero. Let RT = the product (ERT ) for a xed l Write E = [e1; : : : ; eN2 1]T . Let (hl)j = ejT cl, where l [c1; : : : ; ck; ck+1; : : : cN2 1]. Since RT R = I, the column vectors fcj g are orthonormal. k. We will now show that (hl)j = 0 for all j and l k, k. Then hl is the l'th column in which means that cl is a nullvector of E: Ei0l = (ci)j (hl)j = 0 by assumption. This means that hl?ci for all i. But the set fcigi=1 This infers that fcigi=1 N2 1 spans all RN2 1, and hence hl?ci for all i cannot be true unless hl = 0. Hence the components (hl)j = 0 for all j, and then cl is a nullvector of E for N2 1 is linearly independent, since R was invertible. all l k. { 8 { The choice, or permutations, of nullvectors n1; : : : ; nk, does not a ect E0, since the vanishing elements of E0 are the only ones to involve these nullvectors: E0 = RERT infers Ei0j = Ri E Rj = (ci) E (cj ) = 0 (2.22) when i or j 1)=2, and where fclgl=1 N2 1 = fn1; : : : ; nk; ck+1; : : : ; cN2 1g. We will now derive su cient and necessary conditions for when a matrix R, for instance of the form (2.21), is a member of the adjoint representation of SU(N ), that is R 2 1). To obtain this, we will need some results about Lie algebras. Let g be a Lie algebra. By a Lie algebra automorphism r we will mean a R-linear bijection on the vector space g, such that the Lie bracket is preserved. That is, r is a injection (1-1) from g onto g, and if X = xi i 2 g, then r(X) = xir( i), and for all X; Y 2 g. When g = su(N ), automorphisms are either similarity transformations (inner automorphisms) or combinations of similarity transformations and complex conjugation (outer automorphisms): Proposition 2. An automorphism r : su(N ) ! su(N ) for N > 2 is either an inner automorphism, i.e. a similarity transformation r(X) = U XU y; r(X) = U X U y: for an U 2 SU(N ), or a combination of complex conjugation and a similarity transformation, Proof. Similarity transformations r(X) = U XU y are Lie algebra automorphisms since they are the derivatives of Lie group automorphisms (V ) = U V U y for U; V 2 SU(N ), see e.g. [10]. These are the inner automorphisms. All non-inner automorphisms are called outer automorphisms. For su(N ) with N > 2, the outer automorphisms consist of complex conjugation, in combination with an inner automorphism: The outer isomorphism group Out(g) of the a real, simple Lie algebra g, is always given by Out(g) = Aut(g)=Inn(g), just as is the case for complex, simple Lie algebras [11]. The real, simple Lie algebra su(N ) has an outer automorphism group Out(su(N )) isomorphic to the outer automorphism group of the complexi cation of real su(N ), that is su(N; C) = sl(N; C). The outer automorphism group Out(su(N )) is hence the automorphism group of the Dynkin diagram AN 1, which is trivial for N = 2, and isomorphic to Z2 for N > 2, where the non-trivial element of Out(su(N )) in the latter case corresponds to complex conjugation. This is a well known result, but see e.g. [12] with the compact su(N ) = g as a real form of the complex, simple Lie algebra sl(N; C), where a Cartan involution of su(N ) only generate the trivial group, and hence Out(su(N )) = Aut(AN 1) = Out(sl(N; C)). Finally, since (U XU y) = U X U y, and U 2 SU(N ) when U 2 SU(N ), an outer automorphism can always be written on the form (2.24). (2.24) { 9 { See appendix D for an explanation why complex conjugation is an outer automorphism of su(N ) for N > 2, while it is an inner automorphism of su(2). We can now show the following characterization of AdSU(N) for N > 2: a matrix R is an element of AdSU(N) if and only if the linear mapping r on su(N ) associated with R preserves the commutator for all elements of su(N ), and does not involve complex conjugation. Proposition 3. Let r : su(N ) ! su(N ) be a mapping on the Lie algebra su(N ), with N > 2. Moreover, let fvigi=1 N2 1 be a basis for su(N ) with tr(vivj) / ij such that AdSU(N) SO(N 2 1), and let R be a real (N 2 1) (N 2 1) matrix such that for X = xivi 2 su(N ), we have r(X) = (Rx)ivi = Rijxjvi: Then R 2 AdSU(N) if and only if the mapping r is an inner Lie algebra automorphism, that is, r is a R-linear bijection and for all X; Y 2 su(N ), while for all U 2 SU(N ), r(X) 6= U X U y for some X 2 su(N ). Proof. ()): Assume R 2 AdSU(N). This means there is an U 2 SU(N ) such that Rijvi = U vjU y by (2.13). Then r(Z) = r(zivi) = Rijzjvi = zjU vjU y = U ZU y for any Z 2 su(N ), and hence r is an inner automorphism. It respects the commutator in the following manner: [r(X); r(Y )] = [U XU y; U Y U y] = U (XY Y X)U y = r([X; Y ]). By proposition 2 an inner automorphism means an automorphism that does not involve complex conjugation. ((): Assume r is an inner Lie algebra automorphism on su(N ). Then, by de nition of an inner automorphism, r(X) = U XU y, for an U 2 SU(N ). Then U XU y = r(X) = r(xivi) = xir(vi) = xjRijvi, and R 2 AdSU(N) by (2.13). We will now nd conditions on the matrix R equivalent with the associated linear where fvigi=1 mapping r preserving the commutator. Let X; Y 2 su(N ), with X = xii i; Y = yji j, N2 1 = fi igi=1 N2 1 is a basis for the Lie algebra su(N ) with tr(vivj) / ij, and where the matrices i are satisfying [ i; j] = 2if ijk k: [r(X); r(Y )] = 2ixiyjRaiRcjf ace e: xiRaiyjRcj( 2if ace e), and hence The constants f ijk are denoted structure constants. The matrices i may be generalized Gell-Mann matrices, or any other matrices such that fi igi=1 [X; Y ] = [xii i; yji j] = xiyj[ i; j] = xiyj(2if ijk k), hence N2 1 is a basis for su(N ). Then r([X; Y ]) = 2xiyjf ijkr(i k) = 2xiyjf ijkReki e: Note that r is a R-linear function on su(N ), with linear combinations over R of the basis vectors fi igi=1 N2 1 as domain, and not linear combinations over R of the matrices f igi=1 N2 1 On the other hand, [r(X); r(Y )] = [xiRaii a; yjRcji c] = xiRaiyjRcj[i a; i c] = (2.25) (2.26) (2.27) HJEP05(218)63 and since xi; yj are arbitrary, we get xiyj Rekf ijk = xiyj RaiRcj f ace; Rekf ijk = RaiRcj f ace: The equations (2.29) being satis ed is equivalent with the mapping r respecting the commutator. If we furthermore assume that R is a bijection, it will be invertible, and the inverse will be the transposed matrix RT , since by proposition 3 either R 2 AdSU(N) or R is a product of complex conjugation and matrices in AdSU(N). The latter means, in case the i's are generalized Gell-Mann matrices, that R 2 O(N 2 1).1 In fact, either R will be bijective or it will be zero: (2.28) (2.29) (2.30) (2.31) (2.32) Proposition 4. A Lie algebra homomorphism r : su(N ) ! su(N ) is either an automorphism (and hence bijective), or r = 0. Proof. The kernel of a Lie algebra homomorphism : g ! g0 is an ideal of the Lie algebra g. Furthermore, every Lie algebra ideal i corresponds precisely to a homomorphism with kernel i. A Lie algebra is called simple if every proper ideal iCg is trivial, that is i = 0. Now su(N ) is a simple Lie algebra, which means that only trivial, proper ideals exist. Hereby a homomorphism which is not an automorphism equals zero. Rcj , that is RiTd = Rdi and RjTg = Rgj , we get2 If we now assume R 6= 0, and multiply (2.29) by the inverses of the matrices Rai and RdiRgj Rekf ijk = f dge: l = Nullity(E) k; Unfortunately, whether the matrix R of (2.21) is an element of AdSU(N) or not, will depend on the choice of orthonormal nullvectors n1; : : : ; nk, where k = N (N 1)=2. Let and let S = fn~igli=1 be an orthonormal set of nullvectors of E. Then S spans the nullspace k of E, and any choice of orthonormal nullvectors fnigi=1 of R, can be written as a rotation of the vectors of S. This means that any linear combination of the nullvectors of S can be written as [n1; n2; : : : ; nk; nk+1; : : : ; nl]T = O [n~1; n~2; : : : ; n~l]T ; where the matrices of nullvectors are regarded as 1 l matrices with the nullvectors as elements, so that the transpose does not act on the nullvectors. The super uous vectors Tr( a b) / ab, leads to (2.30) as well. 1If the basis matrices i are either real or purely imaginary, then complex conjugation r(X) = xir( i) = xi i , which makes complex conjugation R a diagonal matrix with Rii = 1 (no sum over i), negative if i is imaginary. Hence R 2 O(N 2 1). 2An alternative characterization of AdSU(N) is that AdSU(N) is the matrices R which leave the trace Tr(XY Z XZY ) invariant for arbitrary X; Y; Z 2 su(N ). Here fX; Y; Zg are simultaneously transformed as W = wi i ! (Rw)i i = Rijwj i, for all W 2 fX; Y; Zg. Applying this characterization, and that that is, OOT = I, and we conclude that nk+1; : : : ; nl will not be applied in the construction of R, and the l l matrix O has to be an element of the orthogonal group O(l) to ensure the vectors n1; : : : ; nl are orthonormal: O 2 O(l): On the other hand, if R0 shall be an element of AdSU( 3 ), we must for instance have Matrices O that change the orders of two nullvectors are examples of elements in the orthogonal group O(l). To see that the order of the nullvectors matters, take for instance SU( 3 ) with the matrices given in appendix B as su( 3 )-basis. Let R be de ned by (2.21), i.e. with identical nullvectors as R except for the two rst rows being interchanged. Assume R satis es (2.30) for some choice of ck+1; : : : ; cN2 1 . We will show that R0 can not satisfy (2.30) for any choice of c0k+1; : : : ; c0N2 1, and hence the order of the nullvectors matters when we want to test for the custodial symmetry: all the equations given by di erent values for d; g and e in (2.30) have to be satis ed simultaneously, if the quartic terms shall be custodially symmetric. Since R satis es (2.30) for all choices of d; g and e, we must for instance have3 1 2 1 2 = f 123 = R1iR2j R3kf ijk: = f 213 = R20iR10j R30kf ijk; where R20i is the second row of R0, that is R20i = (n1)i by (2.35), while R10j = (n2)j and R30k = (n3)k (N (N in case of a custodial symmetry). Then (2.37) yields 1)=2 = 3 when N = 3, and hence we must have at least 3 nullvectors 1 2 = R20iR10j R30kf ijk = R1iR2j R3kf ijk; which contradicts (2.36), and R and R0 cannot simultaneously be elements of AdSU(N). Moreover, if the nullity of E is greater than 3, and if n1 in (2.34) is interchanged with an entirely new nullvector n1, eq. (2.36) with R1i = (n1)i may not hold anymore. And if (2.36) still holds, it will not hold anymore if n1 is interchanged with n2, as in the previous example. 3The structure constant f 123 = 1=2 with our alternatively ordered Gell-Mann matrices, de ned in appendix B, while f 123 = 1 with the standard Gell-Mann matrices. (2.33) (2.34) (2.35) (2.36) (2.37) (2.38) The equations (2.30) give us a simple test, that is, a necessary condition, for SO( 4 )C symmetry. Choosing d; g; e 1)=2, reduces all elements from the matrix R in (2.21) to elements of the chosen nullvectors, f dge = (nd)i(ng)j (ne)kf ijk: If there for any choice of the nullvectors in (2.39) is a choice of d; g; e (dependent on the choice of nullvectors), such that (2.39) does not hold, then the quartic terms VE (and the potential) are not SO( 4 )C -symmetric. The advantage with (2.39), is that it only refers to the nullvectors of E and the structure constants of su(N ), which are simple to calculate. Consider a speci c set of l = Nullity(E) orthonormal nullvectors S = fn~igii==l1, they can be rotated into any orthonormal set of nullvectors n1; n2; : : : ; nk; nk+1; : : : ; nl of E, as given by (2.32). Then the necessary condition (2.39) for the custodial symmetry expressed with speci c, orthonormal nullvectors fn~igii==l1 becomes f dge = (Odpn~p)i(Ogqn~q)j (Oern~r)kf ijk: If for all choices of O 2 O(l) there exist choices of indices d; g; e that (2.40) does not hold, then the quartic terms VE (and hence the whole potential) are not SO( 4 )C -symmetric. For the 3HDM (2.40) can be substantially simpli ed, in the case of three nullvectors, i.e. l = 3. For the 3HDM all equations (2.40) will hold automatically, perhaps except for the case where (dge) = (123): if two of the indices d; g; e are equal, the left hand side of (2.40) is zero, and the right hand side is also zero, since the expression will be odd in two of the indices. For instance, if d = g = 1, then the two rst factors of the right hand side, (O1pn~p)i and (O1qn~q)j , are the same, and since f ijk is antisymmetric in i and j, the sum over i and j will be zero for each k. Now write f ijk = ijkf (i; j; k), where ijk is the completely antisymmetric Levi-Civita symbol. Then (2.39) (2.40) (2.41) f 123 = ijkO1p(n~p)iO2q(n~q)j O3r(n~r)kf (i; j; k) = ijk pqrO1pO2qO3r(n~1)i(n~2)j (n~3)kf (i; j; k) = det(O)(n~1)i(n~2)j (n~3)kf ijk; since ijkAipAjqAkr = ijk pqrAi1Aj2Ak3 for any square matrix A, with Aab = (n~b)a in our case. We have also used that det(O) = pqrO1pO2qO3r. Hence, for the 3HDM in the case E has exactly 3 nullvectors, the necessary condition (2.40) holds for all SO( 3 )rotated nullvectors, if and only if it holds for the initial nullvectors n~1, n~2 and n~3. We will apply (2.41) to show that certain potentials are not custodially symmetric in section 2.2.1. 2.1.4 Necessary and su cient conditions for the custodial SO(4) symmetry We will now summarize our results from the previous sections as a theorem, giving necessary and su cient, basis-independent conditions for having the custodial symmetry in a NHDM. Let V ( 1; : : : ; N ) be a NHDM potential, given by (2.6). If there exists a Higgs basis transformation ~ ! ~ 0 = U ~ , such that V 0( 01; : : : ; 0N ) = V ( 1; : : : ; N ), and V 0( 01; : : : ; 0N ) is manifestly SO( 4 )C -symmetric, we called the original potential V ( 1; : : : ; N ) SO( 4 )C symmetric, cf. the discussion below (1.17). We can then show the following: Theorem 1. Let V be a NHDM potential, given by (2.6), with N 3. Then the potential V is SO( 4 )C -symmetric if and only if the following three conditions are satis ed simultaneously: i) The nullity l of the matrix E of (2.6) is equal to or greater than k = N (N 1)=2. HJEP05(218)63 ii) There exists a real (N 2 1) (N 2 1) matrix R whose N (N 1)=2 rst rows are an orthonormal set of nullvectors of E, such that f abc = RaiRbj Rckf ijk; is satis ed for all a; b and c. The constants f ijk here are the structure constants associated with the alternatively ordered, generalized Gell-Mann matrices f j gjN=21 1 of appendix B. iii) The matrix R of condition ii) also satis es Rij j = 0 and where j and j are given by (2.6). Moreover, the solutions of (2.42) will come in pairs R1 and R2 = RccR1, where Rcc = Ik k 0 0 Im m ! ; (2.42) (2.43) (2.44) Higgs elds. nullspace of E. with m = N 2 1 k, and k given above. The matrix Rcc represent complex conjugation when acting on the Lie algebra su(N ) with the basis fi j gjN=21 1 given in appendix B. Exactly one of the solutions R1 and R2 will correspond to a SU(N ) basis transformation of the Finally, condition iii) will hold if the column vectors ~ and ~ are orthogonal to the Proof. As stated in (2.20), the matrix E of the quartic terms VE must have Nullity(E) = l 1)=2, and the k rst rows of R must by (2.21) be orthonormal nullvectors of E, for E to be transformed by R to the form E0 given by (2.19). The matrix R must be a bijection, satisfy (2.42) and not involve complex conjugation on the Lie algebra su(N ) with basis fi j gjN=21 1 for R to be an element of AdSU(N), by (2.30) and proposition 3. By proposition 4, R is bijective if it does not equal zero, and R 6= 0 since the k rst rows of R are nullvectors of E. If R1 is a solution of (2.42), R2 = RccR1 will also be a solution, since Rcc represents complex conjugation which is an automorphism on su(N ), and since R2 has the same k rst rows as R1, namely the chosen nullvectors of R. Exactly one of these two solutions are elements of AdSU(N), since it can be written as an inner automorphism of su(N ), the other will be an outer automorphism: if R is an inner automorphism, we can write r(X) = (Rkj xj )(i k) = U XU y, with X = xj (i j ), and then (RlckcRkj xj )(i l) = (U XU y) for some U 2 SU(N ), which is an outer automorphism. On the other hand, if R is outer, then R = RccR will be inner since two complex conjugations will give an ordinary similarity transformation of X. By (2.16) the necessary and su cient conditions for having the custodial SO( 4 ) symmetry in the terms additional to VE in the potential is given by (2.43). Eq. (2.43) will hold if ~ and ~ are orthogonal to the nullspace of E, since the k rst rows of R are orthonormal nullvectors of E. If l = k, then (2.43) will hold if and only if ~ and ~ are orthogonal to the nullspace of E, since the k rst rows of E then spans the whole nullspace of E. The N (N 1)=2 rst columns of R had to consist of orthonormal nullvectors of E, for R to be of the form (2.21) necessary to transform the matrix E into a manifestly SO( 4 )C symmetric matrix E0. When we are searching for a matrix R as described in the theorem, l concrete, orthonormal nullvectors of E can be rotated by O(l) transformations to nd a matrix R which satis es the conditions of theorem 1. This was discussed in connection with (2.32). The system of equations given by (2.42) is overdetermined. Permutations of the indices a; b and c give equivalent equations, since the structure constants are totally antisymmetric in all indices for su(N ), cf. e.g. [14]. Moreover, if (at least) two indices among a; b and c are identical, the left hand side of (2.42) will be zero, and the right hand side will be zero too: if, for instance a = b the expression on the right hand changes sign if we interchange the indices i and j, and hence the sums over i and j equal zero for each k. Hence, there are N 2 1 3 = 1 6 N 2 3 N 2 2 N 2 1 (2.45) equations left, which equals 56 equations for the 3HDM. Moreover, there are (N 2 1)2 elements in R, and if we subtract the (N 2 nullvectors of E, we end up with (1=2)(N 1) N (N 1)=2 elements associated with 1)2(N + 1)(N + 2) variables, which equals 40 for N = 3. If we add the k2 = (1=8)(N 2)(N 1)N (N + 1), where k = N (N 1)=2, variables SO(k) rotations of the nullvectors generate (we here assume the nullity l of E is k), we get totally (1=8)(N 1)(N + 1)(N (5N + 2) 8) variables. In the 3HDM this corresponds to 43 variables. The di erence between the number of equations and variables for N 3, and the system of equations (2.42) is hence overdetermined. In condition ii) in theorem 1, we state that R should not represent complex conjugation of elements of su(N ) expressed by the basis B = fi j gj=1 1)=2 rst elements of the basis B are real, while the other elements are purely imaginary, this meant that complex conjugation is represented by a matrix R with 1's on the N (N 1)=2 rst diagonal elements, and 1 on all the other diagonal elements, while R is zero elsewhere. This matrix R will satisfy (2.42), but is not an element of AdSU(N), and N2 1 of appendix B. Since the is then 1 24 (N 1)(N + 1) N 4N 3 35N 6 + 48 > 0; (2.46) HJEP05(218)63 does hence not correspond to a Higgs basis transformation. For the 3HDM, R representing complex conjugation then reads Rcc = ii) in this case. The reason for this is that AdSU(2) = SO( 3 ), and hence any R 2 SO( 3 ) will satisfy (2.42), which will hold for all R 2 AdSU(N) also in the case N = 2. And given a normalized nullvector n1 of E, you can always construct a matrix R 2 SO( 3 ) with n1 as the rst row. Then, since R 2 SO( 3 ) = AdSU(2), (2.42) will hold. The fact that (2.42) holds for all R 2 SO( 3 ) can also be shown by an explicit calculation: for N = 2, the structure constants where abc is the Levi-Civita symbol, and hence (2.42) in this case becomes equivalent to (2.47) (2.48) (2.49) (2.50) (2.51) h0j nj0i = (0; vn)T ; the SO( 4 )C symmetry is broken or, more precisely, hidden. The complex VEVs vn here occurs in the lower elements of the doublets, due to electrical charge conservation. Given at least two non-zero VEVs in a manifestly SO( 4 )C -symmetric potential, the least amount of symmetry breaking occurs in the case of vacuum alignment, i.e. when all VEVs occur in the same direction of the real quadruplets n;r, cf. (1.15). This means that all VEVs can be written f abc / abc; abc = RaiRbj Rck ijk: But, similarly to the argument below (2.41), RaiRbj Rck ijk = R1iR2j R3k ijk abc = det(R) abc = abc, where the last equality is valid when det(R) = 1, and hence (2.49) holds for all R 2 SO( 3 ). On the other hand, condition ii) of theorem 1 implies condition i) for any N , and hence these two conditions are equivalent for N = 2. (Condition i) is in any case just a rst, simple necessary condition for SO( 4 )C symmetry.) But, for N = 2 as for all N , the matrices R which ful l condition ii) may not satisfy condition iii). Hence, if there is a matrix R which satisfy both condition ii) and iii), then the potential is SO( 4 )C -symmetric also for N = 2, although condition ii) is more trivial in this case. Finally, for N = 2 both solutions R1 and R2 = RccR1 mentioned in theorem 1, will correspond to a SU(2) Higgs basis transformation, since complex conjugation is an inner automorphism of SU(2) for N = 2. 2.1.5 Spontaneous breaking of the custodial SO(4) symmetry We have referred to a NHDM potential with a custodial SO( 4 ) symmetry as SO( 4 )C symmetric. In the presence of VEVs vn given by vn = v~nei ; HJEP05(218)63 where v~n and are real, and we can transform all VEVs to real VEVs without altering the parameters of the potential, by making an U(1) phase transformation on all scalar elds simultaneously. Normally only U 2 SU(N ) are considered as Higgs basis transformations since an overall U(1) transformation does not a ect the parameters of the potential. But when we allow for U 2 U(N ), i.e. allow overall complex phases in addition to the SU(N ) Higgs basis transformations, vacuum alignment is equivalent to all VEVs being real. Then, in case of real VEVs, SO( 4 )C symmetry is spontaneously broken down to SO( 3 )C , with three broken generators [3]. If there is a Higgs basis where the potential is manifestly SO( 4 )C -symmetric, where SO( 3 )C at the same time is intact, then the potential is simultaneously explicitly and \spontaneously" custodially symmetric, i.e. custodially symmetric. This will happen if and only if there exists a matrix R which satis es condition ii) and iii) of theorem 1, where the vacua (v1; : : : ; vN )T = h0jU(R)~ j0i (2.52) are real (i.e. aligned), for a Higgs basis transformation U(R) 2 U(N ) associated with R through (2.10). (There are N matrices U in SU(N ) associated with each R, and when U(R) 2 U(N ) an additional complex phase may be present, without a ecting the matrix R). Condition i) of theorem 1 is, as we commented in the end of section 2.1.4, a consequence of condition ii), and will hence be satis ed when condition ii) is satis ed. We will summarize the discussion in this section as a corollary of theorem 1: Corollary 1. Let V (~ ) be a NHDM potential. Then V is custodially symmetric, i.e. is SO( 4 )C -symmetric with a SO( 3 )C -symmetric vacuum in a basis where the potential is manwhere R satis es theorem 1, and where h0jU(R)~ j0i is a real vector. ifestly SO( 4 )C -symmetric, if and only if there is a Higgs basis transformation U(R) 2 U(N ) Finally, the custodial symmetry will be spontaneously broken if the potential is SO( 4 )C symmetric, but there is no Higgs basis where the potential is manifestly SO( 4 )C -symmetric and where all VEVs are real at the same time. 2.2 2.2.1 Applying a necessary condition We will now give examples of potentials that are not SO( 4 )C -symmetric, by applying the necessary condition (2.41). Consider a 3HDM potential V where the quartic terms VE = KaEabKb in (2.7) are speci ed by 0 B B B ... ... ... CC E = BB0 : : : 0 0 0CC : B0 : : : 0 0 0C 0 : : : 0 0 0 C A (2.53) Let X in (2.53) be any real, symmetric and invertible 5 5 matrix. In the 3HDM with our choice of basis for su( 3 ) (see appendix B), the column vector consisting of the bilinears Ka becomes K~ = 2 Cb12; Cb13; Cb23; Bb12; Bb13; Bb23; Ab1 2 Ab2 ; Ab1 + Ab2 p 2 3 2Ab3 !T ; (2.54) where the three rst bilinears violate SO( 4 )C (in the Higgs basis they are written). Since X is invertible, its 5 columns are linearly independent, and hence the 5 rst columns of E are linearly independent, while the 3 last columns of E are zero. Hence the dimension of the columnspace, and the rank, is 5, and the nullity (i.e. the dimension of the nullspace) is 8 5 = 3 = l. Three orthonormal nullvectors of E are then By inserting the nullvectors of (2.55) into the necessary condition for the custodial symmetry for the 3HDM (2.41) (alternatively, the more general necessary condition (2.40) can be applied), we get B B BB .. CC B0C C C . 0 i j . BB .. CC B C B B0C C B0C 1 k B0CC f ijk = B f 678; which does not hold, since f 123 = 1=2, while f 678 = 0, cf. appendix B.1. Hence a 3HDM potential containing quartic terms given by VE = KaEabKb and (2.53), is not SO( 4 )C -symmetric. de ned by We can control that the necessary condition (2.41) makes sense, by checking that evidently SO( 4 )C -symmetric terms satisfy the condition. Let the matrix E of VE now be which is of the form (2.19), and hence give manifestly SO( 4 )C -symmetric terms VE. The block X is again a real, symmetric and invertible, but otherwise arbitrary 5 5 matrix. B 00 0 0 : : : 01 B0 0 0 : : : 0C E = BB0 0 0 : : : 0CC ; BB ... ... ... 0 0 0 C C A (2.55) (2.56) (2.57) Then the nullspace of E has dimension 3, and three orthonormalized nullvectors are Now (2.41) yields f 123 = det(O)(n~1)i(n~2)j (n~3)kf ijk = n~1 = n~2 = n~3 = 0; 0; 0; 0; 0; 0; ; 1 2 0; 0; 0; p ; 0; p ; 0; 0 p ; 0; p ; 0; 0; 0; 0; 0 1 2 p !T 3 1 2 2 1 2 1 2 2 q 3 2 3 2 2 0 0 0 3 4 p 3 4 ; T T ; : which holds as long as O 2 SO( 3 ), and hence VE given by (2.57) satisfy the necessary condition (2.41) for SO( 4 )C symmetry, as it should. We will now apply theorem 1 to show that a certain 3HDM potential is SO( 4 )C -symmetric. Consider the 3HDM potential given by E = BB 0 B B B B B B B B B B B B B B B B 2 0 1 0 1 q 3 2 2 1 p 2 2 1 p 1 2 1 p 2 p p p 2 1 3 1 2 0 1 p 2 0 1 0 1 q 3 2 2 1 p p and where the other parameters dependent of Higgs basis transformations are given by ~ = ~ = 1 6 ; p ; p 2 ; 5 3 2 ; 3 p1 ; p 3 T ; ; confer (2.6). Other parameters are arbitrary. As we shall prove in this section, this potential is SO( 4 )C -symmetric. First we check that the nullity of E N (N 1)=2 = 3, which it is, since it has exactly 3 eigenvalues that equal zero. Mathematica then gives us the following orthonormal nullvectors of E, (2.58) (2.59) (2.60) (2.61) (2.62) We then check that these nullvectors satisfy the necessary condition for custodial symmetry, for the 3HDM given by (2.41), and nd that the nullvectors satisfy condition (2.41), for det O = 1. Thus the necessary condition (2.41) also holds for all SO( 3 ) rotations of the nullvectors. We now try to solve the equations (2.42) in theorem 1 for the choice (2.62) of nullvectors. As given in (2.45) and the following discussion for the 3HDM, there will be 56 distinct (but possibly dependent) equations, including one that holds according to the already satis ed necessary condition. On the other hand, there are 40 variables (we do not include SO( 3 ) rotations of the nullvectors in this rst attempt). Applying Mathemathica's Solve-command, then gives us two solutions: when we include the already xed three rst rows of R, which consist of the (transposed) nullvectors (2.62), one solution R1 that 0 0 1 p 2 2 0 0 0 0 0 0 0 0 C 0 C A C 1 2 C : R2 = RccR1; The other solution R2 equals where Rcc is given by (2.47), and represents complex conjugation on the Lie algebra su( 3 ) with our basis fi j gj8=1 given in appendix B. We now check that the parameters ~ and ~ given by (2.61) satisfy condition iii) of theorem 1, which they do for both matrices R1 and R2. Hence the potential given by (2.60) and (2.61) is SO( 4 )C -symmetric by theorem 1. Both solutions R = R1 and R2 correspond to the same, manifestly SO( 4 )C -symmetric matrix E0:4 E0 = RERT = BBBB 00 00 00 1 0 0 0 0 0 B B B 0 0 0 0 B B B 0 0 0 4 0 0 0 1 0 11 CCCC : C C 0 C 1 4The matrix R3 = R1Rcc will also transform E into a manifestly SO( 4 )C-symmetric E0, but alters the signs of the two rst rows of R1, and is hence not a solution when we have chosen the three rst rows to equal the nullvectors in (2.62). According to theorem 1, only one of the solutions R1 and R2 will correspond to a SU(N ) basis transformation of the Higgs elds. Here it turns out that R1 corresponds to a SU( 3 ) transformation of the Higgs elds, and is hence a matrix in AdSU( 3 ). 5 This also means that R1 represents an inner automorphism of su( 3 ), while R2 corresponds to an outer automorphism of su( 3 ), i.e. an inner automorphism combined with complex conjugation. A matrix U1 2 SU( 3 ) which corresponds to R1 through (2.10) is given by U1 = B 0 2 2 There will be two other SU( 3 ) matrices which correspond to R1, namely U1 and = e 23i , i.e. a third-root of unity: since the center of SU( 3 ) is Z(SU( 3 )) = f I : 3 = 1g = Z3, the kernel of the adjoint action of SU( 3 ) is Ker(Ad) = Z3, and hence AdSU( 3 ) = SU( 3 )=Z3. Then each R 2 AdSU( 3 ) will correspond to three di erent U 2 SU( 3 ) through (2.10). The same is valid in SU(N ), that is, AdSU(N) = SU(N )=ZN , and each R 2 AdSU(N) correspond to exactly N di erent U 2 SU(N ) via (2.10). 2.2.3 The Ivanov-Silva model The Ivanov-Silva model was given as a counter-example to the then widely believed, but erroneous claim that an explicitly CP -invariant NHDM necessarily has a real basis [13]. A SO( 4 )C -symmetric potential must be explicitly CP -invariant, since a SO( 4 )C -symmetric potential can be written in a basis where all coe cients are real, which in our formalism means no terms linear in Cba are present in the potential [3]. On the other hand, CP invariance does not necessarily imply SO( 4 )C symmetry. The Ivanov-Silva model is an example of the latter. Since there is no real basis for this model, there will always be terms linear in Cba present in the potential, and hence it is not SO( 4 )C -symmetric. The model's violation of the custodial symmetry is also implicitly stated in a footnote in [13], which says that the model has no other symmetries than powers of an order-4, generalized CP transformation J , de ned by where by applying theorem 1. then becomes empty. We will now con rm the result that the Ivanov-Silva model is not SO( 4 )C -symmetric, 5By applying Mathematica's NSolve-command for R2 and solving for U via (2.10). The solution space J : m ! Xmn n ; 0 1 0 0 1 X = B 0 0 i C : (2.67) (2.68) as V = V0 + V1, with and V0 = m121Ab1 m222(Ab2 + Ab3) + 1Ab12 + 2(Ab22 + Ab3) 2 with all parameters real, except . Moreover, V0, V1 and all the parameters are the same as in the original article [13]. The Ivanov-Silva potential is then given by (2.6) with the following parameters: The potential of Ivanov and Silva's 3HDM model can be written in terms of bilinears 0 = 7 = 8 = p : 1 6 1 24 7 3 1 + 2 2 + 2 3 + 03 ; 1 = 2 = 3 = 4 = 5 = 6 = 0; p p p p E = BBB 0 4 2 1 4 0 0 0 0 0 0 E = 0 14 ( 4 B B B B B B B B B B B B B B 1 2 = ( 8) =( 9) 14 p34= ( 9) 0 0 0 4 0 0 0 5 =( ) = Im( ). The other parameters of the Ivanov-Silva potential are 3 2 03) and where <( ) = Re( ), while The matrix E given by (2.71) is generally not custodially symmetric, since it is generally not singular, and hence does not satisfy condition i) of theorem 1, which requires an eigenvalue 0 of (at least) multiplicity 3. To see this, set all parameters e.g. to 1, with the result (2.69) (2.70) (2.71) (2.72) (2.73) Then det(E) = 1=65536, and E is not singular, and hence does not correspond to a SO( 4 )C symmetric potential. Now expand the scalar elds around the vacuum, i.e. let j = 1 2 + j p12 (vj + j + i j ) ! ; where v1 is real and v2 = v3 = 0 in the Ivanov-Silva model. Then the neutral mass matrix split in an -sector and a -sector, where the (tree-level) masses of the 's are and where 1 is the SM Higgs. The masses of the 's are given by and nally the charged masses are m 1 = 0; m 2; 3 = m 2; 3 ; m 1 = 0; m 2 ; 3 3v12 2 m222; where the two non-zero masses are identical. The mass degeneration between the -sector and the -sector, given by (2.76), is not the same mass degeneration that occurs in custodially symmetric potentials (i.e. potentials where the VEVs are real in some basis where SO( 4 )C symmetry is manifest, cf. corollary 1) [3]: the latter mass degeneration is between the - (CP -odd) sector and the charged sector, where the two sectors get identical masses. In contrast, the mass degeneration in the Ivanov-Silva model is only partial, and between other sectors. This mass degeneration is caused by the generalized CP symmetry J of the model, confer (2.67), which both the potential and the vacuum respect. 2.2.4 The necessary condition is not su cient We will now conclude with a 3HDM example which shows the necessary condition for the custodial symmetry, given by (2.41), is as expected not su cient: the following nullvectors satisfy the necessary condition (2.41), n~1 = (1; 0; 0; 0; 0; 0; 0; 0)T ; n~2 = (0; 0; 0; 0; 0; 0; 1; 0)T ; n~3 = 0; 0; 0; ; ; 0; 0; 0 : 1 2 p 3 2 !T Furthermore, if we apply Mathematica's NSolve-command to solve the full set of equations (2.42), with the nullvectors (2.78) as the three rst rows of R, we get a solution space which is empty. Thus the necessary condition (2.41) for SO( 4 )C symmetry, is not su cient. (2.74) (2.75) (2.76) (2.77) (2.78) We started by organizing the NHDM bilinears that vary under SU(N ) Higgs basis transformations in a vector Ka, given by (2.3), by putting the N (N 1)=2 custodial symmetryviolating bilinears rst in Ka. These custodial symmetry-violating bilinears are (proportional to) bilinears of the type Cb, see (2.4) and (1.10). We derived a Higgs basis-invariant necessary condition (2.40) for SO( 4 )C symmetry, that only involves the nullvectors of the matrix E from the quartic part of the NHDM potential (2.6). In the case of the 3HDM with Nullity(E) = 3, this necessary condition was simpli ed to (2.41), since if it holds for one choice of the nullvectors, it holds for all SO( 3 ) rotations of the nullvectors. The main result of this article, theorem 1, gave us basis-invariant necessary and su cient conditions for an explicit custodial symmetry in a general NHDM, N 3. Corollary 1 in section 2.1.5 yielded corresponding conditions for a simultaneously custodially symmetric potential and vacuum. In section 2.2 we applied both the necessary condition and the necessary and sufcient conditions of theorem 1 on some 3HDM potentials. Here we showed that a certain family of 3HDM potentials, given by (2.53) are not SO( 4 )C -symmetric, since they do not ful l the necessary condition (2.41). Moreover, we showed that another 3HDM potential, given by (2.60) and (2.61), is SO( 4 )C -symmetric since it ful ls the necessary and su cient conditions given by theorem 1. Finally, in section 2.2.3, we also applied our methods to demonstrate that the Ivanov-Silva model is not SO( 4 )C -symmetric. A Factorizable quartic terms Conditions for SO( 4 )C in the quartic terms of the NHDM potential can be formulated relatively easy if the quartic terms are factorizable. We will de ne the quartic terms to be factorizable if they can be written (A.1) (A.2) (A.3) (A.4) The general NHDM potential cannot be written this way, since A and B together contain 2N 2 free parameters, while the quartic part of the general NHDM potential contains 2 1 N 2(N 2 + 1) free parameters, which supersedes 2N 2 for N > 1. Let be the (Hermitian) mass matrix of the potential, i.e. write the quadratic terms of the potential as Then, if the quartic terms are factorizable as in (A.1), the potential is custodially symmetric if and only if there is a basis transformation where A and B are Hermitian N N Higgs doublets, N matrices, and ~ is the N 1 vector consisting of the V4 = ~ yA~ ~ yB~ ; ~ = ( 1; 2; : : : ; N )T : V2 = ~ y ~ : ~ ! ~ 0 = U ~ ; U 2 SU(N ); such that the matrices U yAU , U yBU and U y U are simultaneously real: the SO( 4 )C violating terms are terms involving one or two factors of bilinears of type Cb, and the bilinears of type Cb are generated by the imaginary parts of the matrices A, B and . We will now give criteria for when a family of Hermitian matrices can be made simulk taneously real by similarity transformations. Let fAigi=1 be a family of Hermitian N k matrices. Then there is a U 2 SU(N ) that simultaneously makes the matrices fAigi=1 N similar to real matrices, that is if and only if there is a symmetric W 2 SU(N ) such that HJEP05(218)63 We will now de ne a basis fvagj=1 N2 1 = fi agj=1 N2 1 for su(N ), appropriate for our purposes. The Lie algebra su(N ) consists of anti-Hermitian N N matrices, i.e. matrices A with the property Ay = A. Generalized Gell-Mann matrices are on the other hand Hermitian. The matrices a in our basis will be the same as the generalized Gell-Mann matrices given in e.g. [9], but their order will be di erent. We will order all the imaginary matrices rst, corresponding to the SO( 4 )C -violating bilinears Cb. Let ~e1 = (1; 0; : : : ; 0)T ; ~eN = (0; : : : ; 0; 1)T : Proof. ()): Assume U yAiU is real for all i, where U 2 SU(N ). Then U yAiU = (U yAiU )y = (U yAiU )T = U T AiT U . Hence AiU U T = U U T AiT since U U T = I. Let W = U U T , and W is unitary and symmetric. ((): If W is unitary and symmetric, there exists an unitary, symmetric matrix U , such that U 2 = W [15]. Then AiU U T = U U T AiT for all i by the assumption (A.6). Hence U yAiU = U T AiT U , which infers that (U yAiU )T = U T AiT U = U yAiU , which means U yAiU is real for all i. This leads us to the following su cient and necessary conditions for having SO( 4 )C symmetry in a potential with factorizable quartic terms: The potential is custodially symmetric if and only if there is a symmetric W 2 SU(N ) such that XW = W XT 8X 2 fA; B; g : Eq. (A.7) represents a set of 3N 2 linear equations in N 2 + N variables, if we disregard the condition W 2 SU(N ). Including the latter condition, eq. (A.7) consists of 3N 2 non-linear equations in N 2 1 (the dimension of SU(N )) variables. In any case the set of equations is overdetermined, and generally have no solution, which re ects that the potential generally is not custodially symmetric. B A basis for su(N ) U yAiU is real 8i; 1 i k; AiW = W A T i i k: (A.5) (A.6) (A.7) (B.1) Here we get where a = i~em~eny + i~en~e my for a = a(m; n); b = ~em~eny + ~en~e my for b = a(m; n) + N (N 2 1) : Kc = Tr(K~ c) = ~ y c~ ; Ka = 2Cbm(a);n(a) for 1 a Kb = 2Bbm(a);n(a) for N (N 1) 2 k; k + 1 b = k + a(m; n) 2k: The bilinears Cb and Bb are hence ordered \lexicographically", e.g. Cb12; Cb13; : : : ; Cb1N , Cb23; Cb24; : : : ; CbN 1;N , and afterwards the bilinears of type Bb in the same pattern. Finally, we de ne the diagonal (and traceless) matrices, j = s 2 m(m + 1) " m X ~el~ely l=1 # m~em+1~e my+1 ; for 1 m N 1 and j = m + N (N 1). The bilinears Kj = Tr(K~ j ) of (B.6) q 2 PN N are di erent linear combinations of the bilinears Abn, cf. (1.10), orthogonal to K0 = j=1 Abj . We denote the matrices f cgcN=21 1 constructed above, as the alternative, generalized Gell-Mann matrices. B.1 Structure constants for the 3HDM The structure constants f ijk of the alternative, generalized Gell-Mann matrices f cgcN=21 1 are given by [ i; j ] = f ijk k: The structure constants corresponding to su( 3 ), relevant for the 3HDM, with the alternative Gell-Mann matrices given by (B.2) and (B.6) will be the same as the structure constants of the ordinary Gell-Mann matrices, but the indices will be changed due to the change of the ordering of the matrices c . Our permutation of the original Gell-Mann matrices is (4; 1; 7; 5; 2; 6; 3; 8), which means that the ordinary Gell-Mann matrix 1 is 4 in our alternative order, and so on. The structure constants then become changed according to the permutation of the Gell-Mann matrices, which means that for the matrices of (B.2) and (B.6) the following hold: Then the N N matrix u = ~em~eny has elements umn = 1 for xed m and n, and ukl = 0 for all (k; l) 6= (m; n). m < n N , and a = a(m; n) be de ned as in (1.11), and let f 147 = 1; f 123 = f 156 = f 246 = f 275 = f 345 = f 367 = f 285 = f 386 = p 3 2 : (B.2) (B.3) (B.4) (B.5) (B.6) (B.7) (B.8) HJEP05(218)63 The structure constant f abc is completely antisymmetric in all its indices, i.e. interchanging two indices changes the sign of f [14]. All structure constants not derivable from (B.8) through permutations of the indices, are zero. C The inclusion AdSU(N) 1) When the basis vectors of a Lie algebra are orthogonal relative to the inner product induced by the Killing form, the matrix R(U ) will be orthogonal. The Killing form is de ned (X; Y ) = tr(ad(X)ad(Y )), where ad(X)(Y ) = [X; Y ], and is a linear transformation (a matrix) on the Lie algebra g. Let b = fvj gj=1 invariant under automorphisms of the Lie algebra g, i.e. (r(X); r(Y )) = (X; Y ) for all N2 1 be a basis for g. The Killing form is X; Y 2 g for all automorphisms r 2 Aut(g). Then, if the invariance of under r for general X; Y induce (vj ; vk) / jk; RklRkm = lm; tr(i j i k) = 2 jk; U = 1 0 : which means R is orthogonal, where R is the matrix associated with the automorphism r, given the basis b. For g = su(N ), the Killing form is If g = su(N ) and we choose b to be a basis generated by generalized Gell-Mann matrices, a \Gell-Mann basis" b = fi j gjN=21 1, we get and hence the matrices of AdSU(N) become orthogonal with our preferred basis b. Moreover, R(I) = I, R(U ) is a continuous function of U and SU(N ) is connected, hence all matrices R(U ) will be contained in the identity component of the orthogonal group O(N 2 1). Thus As indicated by the above, there are bases where AdSU(N) does not consist of orthogonal matrices. Consider a general change of Lie algebra basis from a Gell-Mann basis to another basis, given by i a ! i 0a = Mabi b, where M is a real and invertible matrix. Then the matrices of AdSU(N) relative to the primed basis will be given by Rd0a(U ) = MdcRcbMba , 1 by applying (2.10). By choosing M with determinant which di ers from 1, we easily get examples of matrices R0(U ) 2 Ad0SU(N) not being orthogonal. D Complex conjugation as an automorphism of su(N ) Complex conjugation is an inner automorphism of su(2) while it is an outer automorphism of su(N ), N > 2: for N = 2, complex conjugation will be implemented by a similarity transformation with (C.1) (C.2) (C.3) (C.4) (C.5) (D.1) The outer automorphism group of su(N ) is isomorphic to the automorphism group of the Dynkin diagram AN 1, which is Z2 for N > 2 and the trivial group Z1 for N = 2. This means Out(su(N )) for N > 2 consists of only one non-trivial element, namely complex conjugation: complex conjugation is obviously a R-linear bijection on su(N ), and respects the commutator, so complex conjugation is an automorphism on su(N ). To see that complex conjugation never equals a (unitary) similarity transformation for N > 2, consider the diagonal matrix X3 = i diag(1; 1 2) 2 su( 3 ): Assume that complex conjugation is a similarity transformation. Then there exists an U 2 SU( 3 ), such that U X3U y = X3 holds. But similar matrices have the same determinant, and the determinant of X3 and X3 di ers by a factor ( 1), and hence they can not be similar, and thus complex conjugation cannot be an inner automorphism on su( 3 ). For su(N ), N > 3, assume again that complex conjugation on su(N ) equals a similarity N matrix X = X3 0 0 0 ! ; (D.2) (D.3) where X3 is de ned above and X 2 su(N ) (it is anti-Hermitian). By assumption, there should exist an U 2 SU(N ) such that Y teristic polynomial of X is given by det(tI = U Y U y for all Y 2 su(N ). The charac X) = det(tI3 3 X3) det(tIN 3 N 3) = det(tI3 3 det(tI X3) t N 3, and by the same manner the characteristic polynomial of X is X ) = det(tI3 3 X3 ) t N 3 . We then calculate the di erence between the characteristic polynomials of the, by assumption similar, matrices X and X , det(tI X) det(tI X ) = 4itN 3 (D.4) a contradiction, since similar matrices should have the same characteristic polynomial. This again means that complex conjugation is not an inner automorphism for su(N ). Acknowledgments The author wishes to thank K. Skotheim, M. Kachelrie , S. Willenbrock, M. Zhang, C. C. Nishi, E. Straume, P. Osland and R. K. Solberg for helpful communication. Open Access. 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M. Aa. Solberg. Conditions for the custodial symmetry in multi-Higgs-doublet models, Journal of High Energy Physics, 2018, 163, DOI: 10.1007/JHEP05(2018)163