Conditions of general $Z_{2}$ symmetry and TM$_{1,2}$ mixing for the minimal type-I seesaw mechanism in an arbitrary basis

In this paper, using a formula for the minimal type-I seesaw mechanism by $LDL^{T}$ (or generalized Cholesky) decomposition, conditions of general $Z_{2}$-invariance for the neutrino mass matrix $m$ is obtained in an arbitrary basis. The conditions are found to be $(M_{22} a_{i}^{+} - M_{12} b_{i}^{+}) \, ( M_{22} a_{j}^{-} - M_{12} b_{j}^{-}) = - \det M \, b_{i}^{+} \, b_{j}^{-}$ for the $Z_{2}$-symmetric and -antisymmetric part of a Yukawa matrix $Y_{ij}^{\pm} \equiv (Y \pm T Y )_{ij} /2 \equiv (a_{j}^{\pm}, b_{j}^{\pm})$ and the right-handed neutrino mass matrix $M_{ij}$. In other words, the symmetric and antisymmetric part of $b_{i}$ must be proportional to those of the quantity $\tilde a_{i} \equiv a_{i} - {M_{12} \over M_{22}} b_{i}$. They are equivalent to the condition that $m$ is block diagonalized by eigenvectors of the generator $T$. These results are applied to three $Z_{2}$ symmetries, the $\mu-\tau$ symmetry, the TM$_{1}$ mixing, and the magic symmetry which predicts the TM$_{2}$ mixing. For the case of TM$_{1,2}$, the symmetry conditions become $ M_{22}^{2} \, \tilde {a}_{1}^{\rm TBM} \tilde a_{2}^{\rm TBM} = - \det M \, b_{1}^{\rm TBM} b_{2}^{\rm TBM}$ and $ M_{22}^{2} \, \tilde {a}_{1,2}^{\rm TBM} \tilde a_{3}^{\rm TBM} = - \det M \, b_{1,2}^{\rm TBM} b_{3}^{\rm TBM}$ with components $\tilde a_{i}^{\rm TBM}$ and $b_{i}^{\rm TBM}$ in the TBM basis $\mathbf{v}_{1,2,3}$. In particular, for the TM$_{2}$ mixing, the magic (anti-)symmetric Yukawa matrix with $S_{2} Y = \pm Y$ is phenomenologically excluded because it predicts $m_{2}=0$ or $m_{1}, m_{3} = 0$. In the case where Yukawa is not (anti-)symmetric, the mass singular values are displayed without a root sign.

mass matrix m is obtained in an arbitrary basis. The conditions are found to be (M 22 j for the Z 2 -symmetric and -antisymmetric part of a Yukawa matrix Y ± ij ≡ (Y ± T Y ) ij /2 ≡ (a ± j , b ± j ) and the right-handed neutrino mass matrix M ij . In other words, the symmetric and antisymmetric part of b i must be proportional to those of the quantityã i ≡ a i − M12 M22 b i . They are equivalent to the condition that m is block diagonalized by eigenvectors of the generator T .
These results are applied to three Z 2 symmetries, the µ − τ symmetry, the TM 1 mixing, and the magic symmetry which predicts the TM 2 mixing. For the case of TM 1 Yukawa is not (anti-)symmetric, the mass singular values are displayed without a root sign.
In this paper, using a recently discovered seesaw formula by LDL T decomposition, we investigate conditions of a Z 2 symmetry in the mass of light neutrinos m for the minimal type-I seesaw mechanism  in an arbitrary basis. Such formulation can be applied to other generalized CP symmetries (GCP)  and seesaw mechanisms.
This paper is organized as follows. The next section gives a formula by LDL T decomposition for the minimal type-I seesaw mechanism and conditions of general Z 2 symmetry in an arbitrary basis. In Sec. 3, we analyze eigensystems of the Z 2 -symmetric m and its applications. The final section is devoted to a summary.

CONDITIONS OF GENERAL Z 2 SYMMETRY IN AN ARBITRARY BASIS
Here we review a formula by LDL T decomposition [95,96] in the minimal seesaw models [45][46][47]. The Yukawa matrix of neutrinos Y and the mass matrix of right-handed neutrinos M are defined as follows where a i , b i , and M ij are general complex parameters. By setting the vacuum expectation value of the Higgs field to unity, the mass dimension of Y ij becomes one. This M can be diagonalized by LDL T (or generalized Cholesky) decomposition; where L is a lower unitriangular matrix that has all the diagonal entries equal to one By redefining the phase of the second right-handed neutrino ν R2 , we can choose a basis such that M 22 is real-positive. A further phase transformation of the first right-handed neutrino . Thus, the phase of det M can be absorbed Although it does not correspond to a physical basis for not hierarchical M , this formula is valid in any basis.
If the two three-dimensional vectorsã ≡ (ã i ) and b ≡ (b i ) are linearly independent, the rank of m becomes two and a massless mode appears. The eigenvector belonging to the zero mode of m m † is proportional toã * × b * (the complex conjugation of the cross product).
In this paper, we investigate general conditions under which m has a Z 2 symmetry, using the formula (6) and similar arguments as in the previous paper [96]. We obtain more general results than the previous analysis of CP -symmetry.
When a Lagrangian has a Z 2 symmetry, m satisfies T m T T = m by a matrix T . Such Z 2 symmetries include µ-τ symmetry [2][3][4] and magic symmetry [40], which has been studied extensively. T is unitary because the symmetry does not change the kinetic term, and T is also Hermitian because T 2 = 1 leads to T = T −1 = T † . Then the three eigenvectors of T belonging to the eigenvalues ±1 form an orthonormal basis. The vectorsã and b can be expanded by the eigenvectors asã =ã + +ã − and b = b + + b − . This means that the vectors are divided into symmetric and antisymmetric parts under the transformation; Note that either eigenspace is one-dimensional because the eigenvalues are {+1, +1, −1} or {+1, −1, −1} for a nontrivial T .
Since det T can be changed by redefinitions of fields, it does not lose generality by choosing det T = −1. In this case, the normalized eigenvectors of T consists of e − belonging to the eigenvalue −1 and e +1 , e +2 belonging to +1. Using the vector e − , the generator can be written If e +1 † e +2 = 0 holds, T is diagonalized by a unitary matrix U = (e − , e +1 , e +2 ); In this basis, a Z 2 -symmetric m must have the following form, The conditions for m to be such a block diagonal matrix are where (u, v) ≡ u † v is the Hermitian inner product. By multiplying e − and (e +1,2 ) T from left and right of Eqs. (10) and (11) respectively, and adding two equations, conditions for m to have a Z 2 symmetry by T are summarized as Thus,ã + andã − are proportional to b + and b − respectively and their coefficients are determined by Eq. (12). Otherwise, two of the four components will be zero vectors. This kind of alignment also seems to be necessary for naturalness in the seesaw mechanism [97].
Ifã ± i and b ± i = 0 i holds, we can find solutions toã ± for given b ± ; Here, r is a complex constant defined by the non-zeroã ± i and b ± j , At first glance, the solution (13) seems to have a degree of freedom of sign ±. Indeed, it is correct that someã i and −ã i are solutions to each other. However, they are treated as independent solutions because all of b ± i and one ofã ± i are required as input parameters to determine the solution uniquely. This fact is also manifest in representations (B6) and (B7) by orthogonal matrices, as seen in the Appendix.
There are four trivial cases with each ofã ± and b ± are 0, in which the denominator of r or 1/r can not be defined andã i ± need not be proportional to b ± i . In such cases,Ỹ itself has definite symmetry, where we obtain expressions for Y and a from Eq. (4); That is, T Y = Y holds and the original Y also has the T symmetry. Although the same is true for antisymmetricỸ , the other two situations are somewhat different. For example, the case of Then, while the antisymmetric components of a i and b i must be proportional, a i can have independent symmetric components. These different behavior influences representations of mass values in the analysis of eigensystems.

III. ANALYSIS OF EIGENSYSTEM AND ITS APPLICATION
Due to the Z 2 symmetry, the mass values and eigenvectors can be formally determined for each solution (13) and (15). If Eq.
Since the projective componentsã ± and b ∓ are orthogonal to each other, the sum of respective projections must be rank one unlessã ± and b ± (or b ∓ ) are zero vectors. Thus, each of projections must be proportional to the others, asã ± ∝ b ± . The behavior of the solutions can be classified into the following three cases.
a. TỸ = ±Ỹ σ 3 . First, if each ofã ∓ and b ± is 0 in the trivial solutions (15), the mass matrix is The Hermitian matrix m m † becomes where In this situation, a matrix S = 1−2v + ⊗v + T defined by the remaining v + =ã + or b + generates another Z 2 symmetry of m [42].
b. TỸ = ±Ỹ . WhenỸ and Y are T -symmetric or antisymmetric in Eq. (15), the mass matrix is Since the eigenvalues of nontrivial T are (+1, +1, −1) or (−1, −1, +1), the solution of T Y = (det T )Y is phenomenologically excluded because the rank of m is one. In the other solution, there is no guarantee that the two vectorsã ± and b ± are orthogonal. Thus the general representation of mass singular values becomes complicated expressions as displayed in Ref. [98]. This is because thatã and b belong to the same eigenvalue of T . If we can specify another Z 2 symmetry by S,ã and b can be decomposed into projections with different eigenvalues of S, so that mass singular values can be determined as in the previous case.
c. Nontrivial solutions. In other general situations, by substituting the solution (13) into Eq. (18), the mass matrix m is The eigenvectors of m m † are There is another Z 2 symmetry generated by either S ± = 1 − 2b ± ⊗ b ± T . In the following subsections, the results obtained above will be applied to three specific symmetries, the µ − τ symmetry, the TM 1 mixing, and the TM 2 mixing predicted from the magic symmetry.
A. Bi-maximal mixing and µ − τ symmetry Although the exact µ − τ symmetry [2] that predicts θ 13 = 0 has now been excluded, let us consider this as a simple example for practice. First, the basis of the TBM mixing is given by Expanding the vectorsã and b in the TBM basis yields whereã TBM i and b TBM i are components in this basis.

The eigensystem of T is
Since the third eigenvector of m m † is fixed to v 3 , m 3 can be formally determined as This expression generally includes complex phases. However, the absolute value |m 3 | must coincide with the singular value because v 3 belongs to one-dimensional eigenspace of T (28).
The symmetry condition (12) becomes Since v 1,2 are orthogonal, the equal sign holds for each component; This is equivalent to the block diagonalization conditions (10) and (11) in the TBM basis.
Second, for the other trivial solution (22) in whichã and b have the same eigenvalues, Y and The latter is excluded because Y is also rank one. Since the eigenvector v 3 = 1 √ 2 (0, 1, −1) belongs to the zero eigenvalue, this solution is inverted hierarchy (IH) like. In this case, representations of two mass values would be cumbersome. However, it can be simpler by combining it with another Z 2 symmetry that accompanies TM 1,2 mixing described in the next subsections.
and the antisymmetric part is Since there are two independent conditions,ã 2,3 are determined for givenã 1 and b 1,2,3 ; The original Yukawa matrix Y is obtained by the inverse "rotation" L −1 . Since Y andỸ have both symmetric and antisymmetric components, they do not have definite symmetry.
The mass matrix (23) is Since the v 3 direction has a nonzero singular value, this is also an NH-like solution.

B. TM 1 mixing
Recent observations of the non-zero θ 13 stimulate studies of mixing matrices called TM 1,2 [99][100][101][102][103][104][105]; where with c θ ≡ cos θ, s θ ≡ sin θ. The absolute values of sin θ 13 are The following formalism is similar to Ref. [62] that gives a detailed analysis of TM 1 and TM 2 mixing in the minimal seesaw model. New points in this paper are that it is presented in an arbitrary basis and the existence of non-trivial solutions.
The matrix m that predicts TM 1 has a Z 2 symmetry by the following S 1 ; The eigensystem of S 1 are From this, the symmetry conditions (10) and (11) are By transformation to the TBM basis with U TBM = (v 1 , v 2 , v 3 ), the mass matrix m with the symmetry conditions is This m is indeed symmetric under S TBM 1 = diag(−1 , 1 , 1) in this basis. From this, θ 13 and m 1 (with a complex phase) are formally obtained as Hereafter mass matrices will be omitted because of their complexity, and we will only focus on forms of Yukawa matrices and the mass singular values. There are four possibilities for the trivial solutions of TM 1 ; where x 1,2,3 and y 1,2,3 are complex coefficients. Although S 1Ỹ =Ỹ leads to m 1 = 0 and a NH solution, mass singular values cannot be displayed without solving a complicated quadratic equation. A solution with S 1Ỹ = −Ỹ is excluded because it is rank one.
Solutions S 1Ỹ = ±Ỹ σ 3 lead to finite m 1 and IH. The mass singular values are Next, nontrivial solutions are investigated. In order for the two singlular values (24) to be non-zero,ã and b must have v 1 components. Thus we can set (v 1 ,ã) = 0 and (v 1 , b) = 0. The parameter r (14) can be determined as By Eq. (49),ã 2,3 can be determined from the other components. Explicitly, The mass singular values are obtained as C. TM 2 mixing and magic symmetry Similarly, a mass matrix predicting TM 2 has a Z 2 symmetry generated by the following S 2 ; This symmetry is called the magic symmetry and the matrix m is called magic in which the row sums and the column sums are all equal to a number α [40]. The eigenvalues of S 2 are From this, the symmetry conditions are In the TBM basis, the mass matrix m with these conditions is This is indeed symmetric under S TBM 2 = diag(1 , −1 , 1) in this basis. From this, θ 13 and m 2 are obtained as In the case of the magic symmetry, there are three types of solutions as well. The first four trivial solutions are The solutions with S 2Ỹ = ±Ỹ are phenomenologically excluded because they predict m 2 = 0 or m 1,3 = 0. In other words, Y cannot have magic (anti-)symmetry in this meaning. In the case of S 2Ỹ = ±Ỹ σ 3 , x 1 , y 1 = 0 (x 3 , y 3 = 0) leads to a NH (IH) solution.
The mass singular values of the trivial solutions are Furthermore, if we impose the conditionã 2 = ±ã 3 (or b 2 = ±b 3 ), m has the µ − τ symmetry and the respective signs correspond to IH and NH.
Finally, the non-trivial solution (13) is analysed. Since m 2 cannot be zero, we can set (26) and (27), the parameter r (14) is determined as From the symmetry conditions (67),ã 1,3 are determined. Explicitly, and the mass values are If we further impose the condition b 2 + b 3 = 2b 1 (b 2 = b 3 ), m has the µ − τ symmetry, corresponding to a NH (IH) solution.

IV. SUMMARY
In this paper, using a formula for the minimal type-I seesaw mechanism by LDL T (or generalized Cholesky) decomposition, conditions of general Z 2 -invariance of the neutrino mass matrix m is obtained in an arbitrary basis. The conditions are found to be ( for the Z 2 -symmetric and -antisymmetric part of a Yukawa matrix Y ± ij ≡ (Y ± T Y ) ij /2 ≡ (a ± j , b ± j ) and the right-handed neutrino mass matrix M ij . In other words, the symmetric and antisymmetric part of b i must be proportional to those of the quan- They are equivalent to the condition that m is block diagonalized by eigenvectors e i of the generator T .
Since e i are orthogonal, we can analyze the eigensystem of m m † for a Z 2 -symmetric m.
Two eigenvectors u 1,2 of m m † coincide with any of those of T , and the remaining one is a vector (u 1 × u 2 ) * orthogonal to them. Furthermore, if the Yukawa matrix does not have the where Since the asymmetric parts of the two matrices with rank one must cancel for T -invariant m, Eq. (A2) is obtained as the condition. Also, by considering these conditions from symmetries, the solution can be displayed by a complex orthogonal matrix O. This point is discussed in the next section.
Appendix B: Understanding from complex orthogonal matrices The solutions (13) and (15) can also be understood from orthgonal matrices. By defining The symmetry condition (12) satisfied by these solutions, i.e. M 2 22ã + iã − j = −|M | b + i b − j is obviously equivalent to X + X − T = 0 for X ± ≡ (X ± T X)/2. This condition can be rewritten as and it is a solution because O T = O holds.
Furthermore, a matrix representation of the solution T X = ±XO is explored. In the case of det O = +1, T X ± = ±X ± holds and X is symmetric or antisymmetric under T , as discussed above. In the case of det O = −1, we consider the following solution by a complex orthogonal matrix Q and σ 3 ≡ diag(1, −1).
This X 1 = (u + , v − ) has T -symmetric and -antisymmetric vectors 2 . The T transformation for Thus, if we define Q T σ 3 Q ≡ O 1 , this is a symmetric orthogonal matrix with det O 1 = −1 and X ′ 1 satisfies the symmetry (B2).
Indeed O 1 is symmetric, and it agrees with Eq. (B3) by a suitable redefinition.
Since we can write X = X ± or X 1 Q, the general Yukawa matrix for a T -invariant m can be written by these matrices. In the case of X = X ± , Yukawa matrices Y andỸ arẽ In other words, Y ± is symmetric or antisymmetric under T , T Y ± = ±Y , and it can be expanded by eigenvectors of T with the same eigenvalues.
In the other case, Since X 1 has the form X 1 = (u − , v + ), it represents the solution (B6) and (B7).
From the diagonalization L −1 M −1 (L T ) −1 =M −1 (2), we finally obtain and m is T -invariant. Therefore, in order to predict T -invariant m, the Yukawa matrix Y must be T -symmetric or antisymmetric, or has degrees of freedom complex orthogonal matrix Q multiplied to T -covariant X 1 that satisfies T X 1 = X 1 σ 3 .