#### 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
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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:
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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].
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(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
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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)
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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)
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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.
This article is distributed under the terms of the Creative Commons
Attribution License (CC-BY 4.0), which permits any use, distribution and reproduction in
any medium, provided the original author(s) and source are credited.
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