# A Hamiltonian Approach for Obtaining Irreducible Projective Representations
and the $k\cdot p$ Perturbation for Anti-unitary Symmetry Groups

Zhen-Yuan Yang  Department of physics, Renmin University, Beijing 100876,
China.  Jian Yang  Beijing National Research Center for Condensed Matter
Physics, and Institute of Physics, Chinese Academy of Sciences, Beijing
100190, China  Chen Fang  Beijing National Research Center for Condensed
Matter Physics, and Institute of Physics, Chinese Academy of Sciences, Beijing
100190, China  Zheng-Xin Liu<EMAIL_ADDRESS>Department of physics,
Renmin University, Beijing 100876, China.

###### Abstract

As is known, the irreducible projective representations (Reps) of anti-unitary
groups contain three different situations, namely, the real, the complex and
quaternion types with torsion number 1,2,4 respectively. This subtlety
increases the complexity in obtaining irreducible projective Reps of anti-
unitary groups. In the present work, a physical approach is introduced to
derive the condition of irreducibility for projective Reps of anti-unitary
groups. Then a practical procedure is provided to reduce an arbitrary
projective Rep into direct sum of irreducible ones. The central idea is to
construct a hermitian Hamiltonian matrix which commutes with the
representation of every group element $g\in G$, such that each of its
eigenspaces forms an irreducible representation space of the group $G$. Thus
the Rep is completely reduced in the eigenspaces of the Hamiltonian. This
approach is applied in the $k\cdot p$ effective theory at the high symmetry
points (HSPs) of the Brillouin zone for quasi-particle excitations in magnetic
materials. After giving the criterion to judge the power of single-particle
dispersion around a HSP, we then provide a systematic procedure to construct
the $k\cdot p$ effective model.

##  I introduction

Irreducible projective representations (IPReps) of groups, including the
irreducible linear Reps as the trivial class of IPReps, play important roles
in physicsSchur [01 Jan. 1904], Pollmann et al. [2010], Chen et al. [2011a, b,
2013], Slager et al. [2013], Barkeshli et al. [2019]. In condensed matter
physics, IPReps for discrete groups are widely used in obtaining selection
rules or analyzing spectrum degeneracyHamermesh [1989]. For instance, in the
band theory of itinerant electrons hopping in a crystal, the symmetry group is
a space group whence the degeneracy of the energy spectrum at a momentum point
is determined by the dimensions of IPReps of the little co-groupChen et al.
[1985].

Owing to the importance of IPReps, it is urgent to judge if a Rep is reducible
or not. For a finite unitary group $H$, a (projective) Rep $D(H)$ is
irreducible if it satisfies the following condition, ${1\over|H|}\sum_{h\in
H}|\chi_{h}|^{2}=1,$ where $\chi^{(\nu)}(h)={\rm Tr}D^{(\nu)}(h)$ is the
character of the element $h\in H$. When $D(H)$ is reducible, then
${1\over|H|}\sum_{h\in H}|\chi_{h}|^{2}=\sum_{\nu}a_{\nu}^{2}$, where
$a_{\nu}$ is the multiplicity of the irreducible Rep $(\nu)$ contained in
$D(H)$. In this case, we need to transform it into a direct sum of irreducible
Reps. The eigenfunction methodChen et al. [1985] is an efficient way of
performing this reduction.

On the other hand, anti-unitary groups attract more and more interests. The
well known Kramers degeneracy is a consequence of time-reversal symmetry for
fermions with half-odd-integer spin. Time reversal also protects the helical
gapless edge modes in topological insulatorsHasan and Kane [2010], Qi and
Zhang [2011] or topological superconductorsRead and Green [2000], Qi et al.
[2009]. Especially, a large amount of materials in nature exhibit magnetic
long-range order, the symmetries for some of these materials are described by
anti-unitary groups called the magnetic space groupsChristopher Bradley
[2010], where the anti-unitary operations are generally combination of time
reversal operation $T$ and certain unitary space-group element. The
irreducible Reps (also called co-Reps) of the magnetic space groups are
helpful to understand the properties of these materials. Especially, the low-
energy physics of the quasi-particles at high symmetry points (HSPs) of the
Brillouin zone (BZ) are characterized by the irreducible projective Reps of
the little co-groups.

For anti-unitary groups, there are three types of irreducible Reps which are
characterized by the torsion number. Supposing that $M(G)$ is an irreducible
Rep of an anti-unitary group $G$, and $H$ is the halving unitary subgroup
$H\subset G$ with $G=H+T_{0}H$ ($T_{0}$ is anti-unitary). Then the torsion
number is given by $R={1\over|H|}\sum_{h\in H}|\chi(h)|^{2},$ where
$\chi(h)={\rm Tr}\ M(h)$ is the character of $h$. If $R=1$, the irreducible
Rep $M(G)$ belongs to the real type; if $R=2$, then $M(G)$ belongs to the
complex type; if $R=4$ then $M(G)$ belongs to the quaternion typeShaw and
Lever [1974]. This subtlety of anti-unitary groups increases the complexity in
reducing an arbitrary projective Rep into the direct sum of irreducible ones,
especially if some IPReps appea multipole times in the reucible Rep.

In the present paper, from a physical approach we derive the criterion to
judge the irreducibilityKim [1984] of a projective Rep $M(G)$ for a finite
anti-unitary group $G$,

$\displaystyle{1\over|H|}\sum_{h\in H}{1\over 2}\left[\chi(h)\chi^{*}(h)+{\rm
Tr}[M(T_{0}h)M^{*}(T_{0}h)]\right]=1,$

or equivalently

$\displaystyle{1\over|H|}\sum_{h\in H}{1\over
2}\left[\chi(h)\chi^{*}(h)+\omega_{2}(T_{0}h,T_{0}h)\chi((T_{0}h)^{2})\right]=1,$
(1)

where $\omega_{2}(T_{0}h,T_{0}h)$ is the factor system of the projective Rep.
In this approach, we consider Hermitian Hamiltonians in terms of single-
particle bilinear operators which are commuting with all of the symmetry
operations in $G$. If the only existing Hamiltonian is proportional to the
identity matrix, then the Rep $M(G)$ is irreducible. Otherwise, if there exist
other linearly independent Hamiltonian, then $M(G)$ is reducible and the
energies of the Hamiltonian can be used to distinguished each of the
irreducible subspace. This provides an efficient method to reduce an arbitrary
reducible Rep into a direct sum of irreducible ones. The advantage of the
method is that no information of the irreducible Reps of the groups need to be
known beforehand. We further generalize this approach to judge the power of
the quasi-particle dispersions in magnetic semimetals, and then to obtain the
$k\cdot p$ effective models Bardeen [1938], F. [1940] at the HSPs in the BZ.

The rest of the paper is organized as follows. In section II, we worm up by
reviewing the IPReps of unitary groups, and then derive the formula (1) for
anti-unitary groups and interpret it in a physical Hamiltonian approach. In
section III, applying the Hamiltonian approach we provide the procedure to
reduce an arbitrary Rep of finite groups (either unitary or anti-unitary) into
a direct sum of IPReps. In section IV, we provide the criterion to judge if
the degeneracy protected by IPReps of anti-unitary groups can be lift by
certain perturbations or not, and then give the method to construct $k\cdot p$
effective Hamiltonian for magnetic materials. Section V is devoted to the
conclusions and discussions.

Since any Rep of a finite group (no matter unitary or anti-unitary) can be
transformed into a unitary one, in the present work we only discuss unitary
Reps.

##  II A Hamiltonian approach: Condition for irreducible projective Reps

###  II.1 Unitary Groups

Since the character of the identity Rep $(I)$ is $\chi^{(I)}(h)=1$ for any
$h\in H$, the following quantity

$\displaystyle a_{(I)}^{(\nu\times\nu^{*})}$ $\displaystyle=$
$\displaystyle{1\over|H|}\sum_{h\in
H}|\chi^{(\nu)}_{h}|^{2}\Big{(}\chi^{(I)}(h)\Big{)}^{*}$ $\displaystyle=$
$\displaystyle{1\over|H|}\sum_{h\in H}{\rm Tr}[D^{(\nu)}(h)\otimes
D^{(\nu)*}(h)],$

stands for the multiplicity of the identity Rep appearing in the reduced Rep
of the direct product $(\nu\times\nu^{*})$, where ${(\nu)^{*}}$ is the complex
conjugate of $(\nu)$. Then the condition of irreducibility of $(\nu)$ can be
interpreted as the following: the direct product ${(\nu)\times(\nu)^{*}}$
contains only one identity Rep, namely $a_{(I)}^{(\nu\times\nu^{*})}=1$.

The expression $a_{(I)}^{(\nu\times\nu^{*})}=1$ has a physical interpretation.
Suppose the identical particle $\psi^{\dagger}$ has $d$ internal components
$\psi^{\dagger}=(\psi_{1}^{\dagger},\psi_{2}^{\dagger},...,\psi_{d}^{\dagger})$,
which carries an Rep ${(\nu)}$ of the symmetry group $H$. This means that
$\hat{h}\psi_{i}^{\dagger}\hat{h}^{-1}=\sum_{j}D^{(\nu)}_{ji}(h)\psi_{j}^{\dagger}$
or equivalently

$\hat{h}\psi^{\dagger}\hat{h}^{-1}=\psi^{\dagger}D^{(\nu)}(h).$

The hermitian conjugation gives
$\hat{h}\psi\hat{h}^{-1}=[D^{(\nu)}(h)]^{\dagger}\psi.$ The energy spectrum is
described by the single-particle Hamiltonian

$\displaystyle\hat{\mathscr{H}}=\sum_{i}\psi_{i}^{\dagger}\Gamma_{ij}\psi_{j}=\psi^{\dagger}\Gamma\psi,$
(2)

where $\Gamma$ is an $d\times d$ matrix. The symmetry group $H$ means that the
Hamiltonian is invariant under all the symmetry operations in the group $H$.
In other words, for any $h\in H$, we have
$\hat{h}\hat{\mathscr{H}}\hat{h}^{-1}=\hat{\mathscr{H}},$ which is equivalent
to

$\displaystyle D^{(\nu)}(h)\Gamma[D^{(\nu)}(h)]^{\dagger}=\Gamma.$ (3)

Schur’s lemma indicates that when $(\nu)$ is irreducible, then $\Gamma$ must
be proportional to the identity matrix $\Gamma_{0}\propto I$. If there exist
another linearly independent matrix $\Gamma_{1}$ satisfying (3), then it must
have at least two eigenvalues. The eigenspace of each eigenvalue is closed
under action of $H$ and hence form a Rep space of $H$. This means that the Rep
$(\nu)$ is reducible. Therefore, if $I$ is the only one linearly independent
matrix satisfying (3), then the $d$-fold degenerate energy level of
$\mathscr{H}$ cannot be lift and consequently $(\nu)$ is irreducible.

The equation (3) can be expanded in the following form

$\displaystyle\sum_{j,k}\\!D^{(\nu)}_{ij}\\!(h)\Gamma_{jk}D^{(\nu)*}_{lk}\\!(h)\\!\\!$
$\displaystyle=$
$\displaystyle\\!\\!\sum_{j,k}\\!\left(\\!D^{(\nu)}(h)\\!\otimes\\!D^{(\nu)*}(h)\\!\right)_{il,jk}\\!\\!\Gamma_{jk}$
$\displaystyle=$ $\displaystyle\\!\\!\Gamma_{il}$

for all $h\in H$. If we reshape the matrix $\Gamma$ into an $d^{2}$-component
column vector (if the matrix $\Gamma$ is reshaped into the $d^{2}$-component
vector column by column, then it should be transposed into $\Gamma^{T}$ before
the reshaping), then this vector is the eigenvector of $D^{(\nu)}(h)\otimes
D^{(\nu)*}(h)$ with eigenvalue 1 for all $h\in H$, $i.e.$ it carries the
identity Rep of $H$. In other words, the vector $\Gamma$ is the CG coefficient
Sakata [1974], Dirl [1979] that combines the bases of $(\nu)$ and $(\nu)^{*}$
to a irreducible basis that belongs to the identity Rep $\chi^{(I)}(h)=1$. If
$(\nu)$ is irreducible, then the CG coefficient is unique.

Above discussion is valid no matter the Rep $(\nu)$ is linear or projective.

###  II.2 Anti-unitary groups

In the following we generalize above approach to anti-unitary groups. Consider
an anti-unitary group $G$ with $G=H+T_{0}H$, where $H\in G$ is the halving
unitary subgroup and $T_{0}$ is an anti-unitary element of the lowest order.

If $G$ is of type-IYang and Liu [2017], namely, $T_{0}^{2}=E$, then $G$ is
either a direct product group $G=H\times Z_{2}^{T}$ or a semi-direct product
$G=H\rtimes Z_{2}^{T}$, where $Z_{2}^{T}=\\{E,T_{0}\\}$. If $T\in G$ (here $T$
is the time-reversal operation which commutes with all the other elements),
then we choose $T_{0}=T$; otherwise, $T_{0}=uT$, where $u\notin G$ is a
unitary operation satisfying $T_{0}^{2}=u^{2}=E$.

On the other hand, if $G$ is of type-II, then $T_{0}^{2^{n}}=E$ with $n\geq
2$, hence $G$ cannot be written in forms of direct product or semi-direct
product of a unitary group with $Z_{2}^{T}$. Obviously, the order of $T_{0}$
is at least 4 and $T_{0}^{2}\equiv\sigma$ is a unitary element in $H$,
$\sigma\in H$.

We consider an $d$-dimensional unitary projective Rep of $G$. Any element
$g\in G$ is represented as $\hat{g}=M(g)K_{s(g)}$, which satisfies the
relations $M^{\dagger}(g)M(g)=I$ and

$\displaystyle
M(g_{1})K_{s(g_{1})}M(g_{2})K_{s(g_{2})}=\omega(g_{1},g_{2})M(g_{1}g_{2})K_{s(g_{1}g_{2})},$

where $s(g)=1$, $K_{s(g)}=K$ if $g$ is anti-unitary and $s(g)=0$, $K_{s(g)}=I$
otherwise. The factor system $\omega_{2}(g_{1},g_{2})$ satisfies the cocycle
equation

$\omega^{s(g_{1})}(g_{2},g_{3})\omega^{-1}(g_{1}g_{2},g_{3})\omega(g_{1},g_{2}g_{3})\omega^{-1}(g_{1},g_{2})=1.$

Now we derive the condition for the irreducibility of $M(g)K_{s(g)}$.

####  II.2.1 General Discussion

Since unitary group elements are easier to handle, we expect that the
irreducibility can be judged from the restrict Rep of the subgroup $H$.
Noticing that $M(H)$ is possibly reducible even if $M(G)$ is irreducible, we
have

$\displaystyle{1\over|H|}\sum_{h\in H}{\rm Tr}[M(h)\otimes M^{*}(h)]\geq 1.$

Actually $P^{(I)}={1\over|H|}\sum_{h\in H}M(h)\otimes M^{*}(h)$ is the
projector onto the subspace of identity Reps contained in the direct product
Rep $M(h)\otimes M^{*}(h)$. The eigenvalues of $P^{(I)}$ are either 1 (which
occurs at least once) or 0, hence ${\rm Tr}P^{(I)}\geq 1$.

We need to find a way to include the restrictions from the anti-unitary group
elements. Adopting the physical argument as discussed in Sec. II.1, we
consider a $d$-component particle $\psi^{\dagger}$ which carries the (co-)Rep
of $g\in G$,

$\hat{g}\psi^{\dagger}\hat{g}^{-1}=\psi^{\dagger}M(g)K_{s(g)}.$

The Hamiltonian takes the same form of (2), which is invariant under the
action of all the group elements,
$\hat{g}\hat{\mathscr{H}}\hat{g}^{-1}=\hat{\mathscr{H}}$, namely,

$\displaystyle M(h)\Gamma M(h)^{\dagger}=\Gamma,\ \ \ \ \ \ \ h\in H$ (4)
$\displaystyle M(T_{0})\Gamma^{*}M(T_{0})^{\dagger}=\Gamma.$ (5)

Similar to the discussion for unitary groups, the $\Gamma$ matrix can be
considered the CG coefficient that combines the product Rep $M(g)\otimes
M^{*}(g)K_{s(g)},g\in G$ (a linear Rep) into the identity Rep. Since the
identity matrix obviously satisfies the above two equations, the product Rep
contains at least one identity Rep. We expect that the identity matrix is the
unique linearly independent matrix satisfying (4) and (5) if the Rep
$M(g)K_{s(g)},g\in G$ is irreducible.

However, above statement is too strong for anti-unitary groups. We need one
more constraint for $\Gamma$. Notice that if a matrix commutes with an
irreducible (projective) Rep of an anti-unitary group, then this matrix may
have two eigenvalues which are mutually complex conjugate to each other Yang
and Liu [2017]. To generalize the Schur’s lemma to anti-unitary groups, the
matrix $\Gamma$ needs to be Hermitian. Namely, if an Hermitian matrix commutes
with the irreducible projective Reps of all the group elements of an anti-
unitary group, then this matrix must be proportional to the identity matrix.

Hence, in addition to (4) and (5), we should further require that

$\displaystyle\Gamma^{\dagger}=\Gamma.$ (6)

If a non-hermitian matrix $\Gamma$ satisfies (4) and (5), then obviously its
hermitian conjugate $\Gamma^{\dagger}$ also does. Therefore, the linear
combination $(\Gamma+\Gamma^{\dagger})$ is the required hermitian matrix 111
The linear combinations $\Gamma_{\pm}=(\Gamma\pm\Gamma^{\dagger})/2$ still
satisfy the relations (4) and (5). Now
$\Gamma_{\pm}^{\dagger}=\pm\Gamma_{\pm}$, meaning that $\Gamma_{+}$ is
Hermitian and $\Gamma_{-}$ is anti-Hermitian. So $\Gamma_{-}$ violates (6). On
the other hand, if we transform $\Gamma_{-}$ into an hermitian matrix
$i\Gamma_{-}$, then
$M(T_{0})(i\Gamma_{-})^{*}M^{\dagger}(T_{0})=-(i\Gamma_{-})$, namely, the
hermitian matrix $i\Gamma_{-}$ forms an eigenstate of $M(T_{0})\otimes
M^{*}(T_{0})K$ with eigenvalue $-1$, which violates (5)..

Therefore, when making using of the characters of the unitary subgroup $H$ to
judge the irreducibility of $M(g)K_{s(g)},g\in G$, we need a projection
operator $P_{HT_{0}}=P_{H}P_{T_{0}}$ to project onto the subspace formed by
hermitian and $T_{0}$ symmetric matrices. $P_{HT_{0}}$ is equivalent to
project onto the eigenvectors of $M(T_{0})\otimes M^{*}(T_{0})K$ with
eigenvalue 1 with the condition that the matrix form of these eigenvectors are
hermitian.

Therefore, considering (4), (5) and (6), the irreducibility requires that

$\displaystyle{\rm Tr}(P^{(I)}P_{HT_{0}})\\!=\\!{1\over|H|}\sum_{h\in H}{\rm
Tr}[M(h)\\!\otimes\\!M^{*}(h)P_{HT_{0}}]=1,$ (7)

namely, when projecting onto the hermitian and $T_{0}$ symmetric subspace, the
identity Rep only appears once in the direct product Rep $M(H)\otimes
M^{*}(H)$.

Eq. (7) is a general expression of the criterion that a Rep of anti-unitary
groups should meet if it is irreducible. However, the construction of the
projection operator $P_{HT_{0}}$ is not straightforward. In the following we
first consider a relatively simple case, $i.e.$ the type-I anti-unitary
groups, and then generalize the conclusion to arbitrary anti-unitary groups.

####  II.2.2 type-I anti-unitary groups

For type-I anti-unitary groups with $T_{0}^{2}\equiv\sigma=E$, situations are
much simpler. For a unitary Rep, we have $M^{\dagger}(T_{0})M(T_{0})=I$. On
the other hand, $T_{0}^{2}=E$ indicates

$[M(T_{0})K]^{2}=M(T_{0})M^{*}(T_{0})=\eta_{0}\equiv\omega_{2}(T_{0},T_{0}),$

where $\eta_{0}=\pm 1$ is an invariant of the projective Rep of type-I anti-
unitary groups.

Under the hermitian condition (6), the transpose of (5) yields
$\Gamma^{T}=M^{*}(T_{0})\Gamma M^{T}(T_{0})$, namely

$M(T)\Gamma^{T}=M(T_{0})M^{*}(T_{0})\Gamma M^{T}(T_{0})=\eta_{0}\Gamma
M^{T}(T_{0}).$

Defining $\tilde{\Gamma}=\Gamma M^{T}(T_{0}),$ then we have

$\displaystyle\tilde{\Gamma}^{T}=\eta_{0}\tilde{\Gamma}.$ (8)

This means that $\tilde{\Gamma}$ is either symmetric (if $\eta_{0}=1$) or
anti-symmetric (if $\eta_{0}=-1$). This symmetry condition is a direct
consequence of the anti-unitary symmetry condition (5) and hermiticity
condition (6). In the following, we will say $\tilde{\Gamma}$ to be
$\eta_{0}$-symmetric if it satisfies (8).

Since $\Gamma=\tilde{\Gamma}M^{*}(T_{0})$, we rewrite the Hamiltonian as

$\hat{\mathscr{H}}=\psi^{\dagger}\tilde{\Gamma}M^{*}(T_{0})\psi=\psi^{\dagger}\tilde{\Gamma}\tilde{\psi},$

then the basis $\psi$ undergoes a unitary transformation
$\psi\to\tilde{\psi}=M^{*}(T_{0})\psi$. Under the action of $h\in H$,
$\tilde{\psi}$ vary as
$\hat{h}\tilde{\psi}\hat{h}^{-1}=M^{*}(T_{0})M^{\dagger}(h)\psi=M^{*}(T_{0})M^{\dagger}(h)M^{T}(T_{0})\tilde{\psi}$.
For convenience, we define the following Rep for $h\in H$,

$\displaystyle F(h)$ $\displaystyle=$ $\displaystyle
M(T_{0})M^{*}(h)M^{\dagger}(T_{0}),$ (9)

which is equivalent to $M^{*}(h)$ with ${\rm Tr\ }F(h)={\rm Tr\
}M^{*}(h)=\chi^{*}(h)$. Accordingly, $\tilde{\psi}$ vary as
$\hat{h}\tilde{\psi}\hat{h}^{-1}=F^{T}(h)\tilde{\psi}$. Hence, the condition
$\hat{h}\hat{\mathscr{H}}\hat{h}^{-1}=\hat{\mathscr{H}}$ requires that

$\displaystyle M(h)\tilde{\Gamma}F^{T}(h)$ $\displaystyle=$
$\displaystyle\tilde{\Gamma},$ (10)

which is the deformation of (4). Similarly, (5) is transformed into

$\displaystyle M(T_{0})\tilde{\Gamma}^{*}M^{T}(T_{0})=\tilde{\Gamma}.$ (11)

As before, $\tilde{\Gamma}$ can be considered as the CG coefficient that
couples the direct product Rep $V(g)=M(g)\otimes F(g)K_{s(g)},g\in G$ (a
linear Rep) to the identity Rep, namely

$\displaystyle\sum_{ij}V_{kl,ij}(g)K_{s(g)}\tilde{\Gamma}_{ij}$
$\displaystyle=$
$\displaystyle\sum_{ij}M_{ki}(g)F_{lj}(g)K_{s(g)}\tilde{\Gamma}_{ij}$ (12)
$\displaystyle=$ $\displaystyle\tilde{\Gamma}_{kl}K_{s(g)},$

with $V(T_{0})=M(T_{0})\otimes M(T_{0})$. If the CG coefficient matrix
$\tilde{\Gamma}$ satisfies the $\eta_{0}$-symmetry condition (8), then we only
need to consider the unitary elements $h\in H$.

Obviously, $\tilde{\Gamma}=M^{T}(T)$ (i.e. $\Gamma=I$) satisfies the relations
(8) and (10). The irreducibility of $M(g)K_{s(g)},g\in G$ indicates that there
is a unique linearly independent solution. In other words, when projected onto
the $\eta_{0}$-symmetric subspace by the projection operator $P_{\eta_{0}}$,
the product Rep $V(h)=M(h)\otimes F(h)$ only contains a single identity Rep,

$\displaystyle a_{(I)}^{[M\otimes F]_{\eta_{0}}}={1\over|H|}\sum_{h\in H}{\rm
Tr}\left[M(h)\otimes F(h)P_{\eta_{0}}\right]=1.\ \ \ $ (13)

To obtain the matrix form of $[M(h)\otimes F(h)P_{\eta_{0}}]$, we devide
$M(h)\otimes F(h)$ into two parts,

$\displaystyle\sum_{ij}\big{(}M(h)\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!$
$\displaystyle\otimes F(h)\big{)}_{kl,ij}\tilde{\Gamma}_{ij}$ $\displaystyle=$
$\displaystyle\sum_{ij}{1\over
2}\big{[}M_{ki}(h)F_{lj}(h)+\eta_{0}M_{kj}(h)F_{li}(h)\big{]}\tilde{\Gamma}_{ij}+$
$\displaystyle\sum_{ij}{1\over
2}\big{[}M_{ki}(h)F_{lj}(h)-\eta_{0}M_{kj}(h)F_{li}(h)\big{]}\tilde{\Gamma}_{ij}.$

Noticing that the second summation on the righthand side vanishes owing to
$\tilde{\Gamma}_{ij}=\eta_{0}\tilde{\Gamma}_{ji}$, so we have

$\displaystyle[V_{\eta_{0}}(h)]_{kl,ij}=\frac{1}{2}\big{(}M_{ki}(h)F_{lj}(h)+\eta_{0}M_{kj}(h)F_{li}(h)\big{)},$
(14)

where we have used the notation $V_{\eta_{0}}(h)=[M(h)\otimes
F(h)P_{\eta_{0}}]$.

Introducing the unit twist matrix

$(\mathscr{T})_{kl,ij}=\delta_{kj}\delta_{li}$

with $(X\mathscr{T})_{kl,ij}=X_{kl,ji}$ for an arbitrary matrix $X$, then the
projection operator $P_{\eta_{0}}$ can be expressed as

$P_{\eta_{0}}={1\over 2}(I+\mathscr{T}_{\eta_{0}}),$

where $I$ is the identity matrix and
$\mathscr{T}_{\eta_{0}}=\eta_{0}\mathscr{T}$. Hence $V_{\eta_{0}}(h)$ in (14)
can be written as

$V_{\eta_{0}}(h)=\big{(}M(h)\otimes F(h)\big{)}P_{\eta_{0}}={1\over
2}V(h)(I+\mathscr{T}_{\eta_{0}}).$

Although $V_{\eta_{0}}(h)$ does not form a Rep of $H$, the common eigenvector
of $V_{\eta_{0}}(h)$ with eigenvalue 1 does carries the identity Rep of $H$
(see Theorem 1 for the special case in which $\sigma=E$). Defining
$P^{(I)}={1\over|H|}\sum_{h\in H}V(h)$, above statement indicates that if
$\tilde{\Gamma}$ satisfies

$\displaystyle P^{(I)}P_{\eta_{0}}\tilde{\Gamma}_{ij}=\tilde{\Gamma}_{kl},$
(15)

then it simultaneously satisfies the relations (8) and (10). Furthermore, by
choosing proper bases in the supporting space of $P^{(I)}P_{\eta_{0}}$, the
$T_{0}$-symmetry condition (11) and finally the hemiticity condition (6) can
be ensured (see Appendix A for details). Thus the criterion (13) for the
irreducibility is valid.

From the matrix form in (14), the criterion (13) can be expressed in terms of
the characters $\chi(h)={\rm Tr\ }M(h)$ of the unitary elements $h\in H$,
namely

$\displaystyle{1\over|H|}\sum_{h\in H}{1\over
2}\Big{(}\chi(h)\chi^{*}(h)+\eta_{0}{\rm Tr\ }[F(h)M(h)]\Big{)}=1,$

where ${\rm Tr\ }F(h)=\chi^{*}(h)$ has been used. Furthermore, by denoting
$\bar{h}=T_{0}^{-1}hT_{0}=T_{0}hT_{0}$, we have

$\displaystyle F(h)$ $\displaystyle=$ $\displaystyle
M(T_{0})M^{*}(h)M^{\dagger}(T_{0})$ $\displaystyle=$
$\displaystyle\omega_{2}(T_{0},h)\omega_{2}(T_{0}h,T_{0})\omega_{2}(T_{0},T_{0})M(\bar{h}).$

Above can be further simplified using the cocycle relation
$\omega_{2}^{-1}(T_{0},h)\omega_{2}^{-1}(\bar{h},h)\omega_{2}(T_{0}h,T_{0}h)\omega_{2}^{-1}(T_{0}h,T_{0})=1$,
which yields
$\omega_{2}(T_{0},h)\omega_{2}(T_{0}h,T_{0})={\omega_{2}(T_{0}h,T_{0}h)\over\omega_{2}(\bar{h},h)}$.
Noticing that $\eta_{0}^{2}=1$, therefore we have

$\displaystyle\eta_{0}{\rm Tr\ }[F(h)M(h)]$ $\displaystyle=$
$\displaystyle{\omega_{2}(T_{0}h,T_{0}h)\over\omega_{2}(\bar{h},h)}{\rm Tr\
}[M(\bar{h})M(h)]$ $\displaystyle=$
$\displaystyle\omega_{2}(T_{0}h,T_{0}h){\rm Tr\ }[M(\bar{h}h)]$
$\displaystyle=$ $\displaystyle\omega_{2}(T_{0}h,T_{0}h)\chi((T_{0}h)^{2})$
$\displaystyle=$ $\displaystyle{\rm Tr\ }[M(T_{0}h)M^{*}(T_{0}h)].$

Finally, we reach the simplified irreducible condition

$\displaystyle{1\over 2|H|}\sum_{h\in
H}\\!\Big{(}\chi(h)\chi^{*}(h)+\omega_{2}(T_{0}h,T_{0}h)\chi\big{(}(T_{0}h)^{2}\big{)}\Big{)}=1.$
(16)

Above expression is independent on the gauge choice of the projective Rep. The
factor system $\omega_{2}(T_{0}h,T_{0}h)$ can be avoided by the replacement
$\omega_{2}(T_{0}h,T_{0}h)\chi\big{(}(T_{0}h)^{2}\big{)}={\rm Tr\
}[M(T_{0}h)M^{*}(T_{0}h)]$.

In the following we show that above condition of irreducibility also works for
type-II anti-unitary groups.

####  II.2.3 type-II anti-unitary groups

For type-II anti-unitary groups, we denote $T_{0}^{2}\equiv\sigma$. Similar to
previous discussion, if we define $\tilde{\Gamma}=\Gamma M^{T}(T_{0})$, then
(10) and (11) are the conditions $\tilde{\Gamma}$ should satisfy. Furthermore,
recalling $\Gamma$ is hermitian, we have

$\displaystyle\tilde{\Gamma}^{T}=M(T_{0})\Gamma^{T}$
$\displaystyle=M(T_{0})M^{*}(T_{0})\Gamma M^{T}(T_{0})$
$\displaystyle=\eta_{0}M(\sigma)\tilde{\Gamma}.$ (17)

The self-consistency condition $(\tilde{\Gamma}^{T})^{T}=\tilde{\Gamma}$
requires that $\tilde{\Gamma}=M(\sigma)\tilde{\Gamma}F^{T}(\sigma)$ with
$F(\sigma)=\eta_{0}^{2}M(\sigma)$, or equivalently

$V(\sigma)\tilde{\Gamma}=M(\sigma)\otimes
F(\sigma)\tilde{\Gamma}=\tilde{\Gamma}.$

This requirement is guaranteed if (10) is satisfied.

Therefore, the $T_{0}$ symmetry condition (5) and the hermiticity condition
(6) combine to a single restriction (II.2.3), i.e.
$\tilde{\Gamma}=[\eta_{0}M(\sigma)\tilde{\Gamma}]^{T}$. Now we define a
generalized twist operator $\mathscr{T}_{\eta_{0}}$ which transforms
$\tilde{\Gamma}$ into $[\eta_{0}M(\sigma)\tilde{\Gamma}]^{T}$,

$\displaystyle\mathscr{T}_{\eta_{0}}=\eta_{0}\mathscr{T}[M(\sigma)\otimes I].$
(18)

It holds that $(\mathscr{T}_{\eta_{0}})^{2}=M(\sigma)\otimes
F(\sigma)=V(\sigma)$. Thus $\mathscr{T}_{\eta_{0}}$ defines a generalized
‘transpose’ of $\tilde{\Gamma}$ given that
$V(\sigma)\tilde{\Gamma}=\tilde{\Gamma}$ is satisfied.

When projected to the eigenspace of $V(\sigma)$ with eigenvalue 1, the
operator $\mathscr{T}_{\eta_{0}}$ has eigenvalues $\pm 1$. Therefore the
projector onto the generalized $\eta_{0}$-symmetric subspace is given by

$\displaystyle P_{\eta_{0}}={1\over 2}P_{\sigma}(I+\mathscr{T}_{\eta_{0}}),$

where $P_{\sigma}$ is a projection onto the eigenspace of $V(\sigma)$ with
eigenvalue 1. Defining the projector onto the subspace of identity Reps as
$P^{(I)}={1\over|H|}\sum_{h\in H}V(h)$, then it is obvious that
$P^{(I)}P_{\sigma}=P^{(I)}$. Hence the irreducibility condition (13) can be
written as

$\displaystyle a_{(I)}^{[M\otimes F]_{\eta_{0}}}={\rm
Tr}(P^{(I)}P_{\eta_{0}})={1\over 2}{\rm
Tr}\big{(}P^{(I)}(I+\mathscr{T}_{\eta_{0}})\big{)}=1.$ (19)

After some calculations, above criterion of irreducibility can be simplified
to the same form as (16) (see Appendix B).

From the definition of torsion number of irreducible Reps and the equation
(16), one can easily verify the following relation for any anti-unitary group
$G=H+T_{0}H$,

$\displaystyle{1\over|H|}\\!\\!\sum_{h\in
H}\\!\\!\omega_{2}(T_{0}h,T_{0}h)\chi\big{(}(T_{0}h)^{2}\big{)}\\!\\!$
$\displaystyle=$ $\displaystyle\\!\\!{1\over|H|}\\!\\!\sum_{u\in
T_{0}\\!H}\\!\\!\\!{\rm Tr}[M(u)M^{*}u)]$ $\displaystyle=$
$\displaystyle\\!\\!\left\\{\begin{aligned} &1,&{\ \ \ \ \rm if}\ R=1\\\
&0,&{\ \ \ \ \rm if}\ R=2\\\ &-1,&{\ \ \ \ \rm if}\ R=4\end{aligned}\right.,$

which provides another way to obtain the torsion number.

##  III Hamiltonian approach for the Reduction of Projective Reps

The criterion of judging the irreducibility actually provides a practical
procedure to reduce reducible projective Reps of finite groups. In the
following, we discuss unitary groups and anti-unitary groups separately.

###  III.1 Reduction of Reps for Unitary groups

For a general hermitian Hamiltonian matrix $\Gamma$ satisfying (3), each of
its eigenspace is an irreducible subspace of the unitary group $H$. Namely,
the eigenvalues of $\Gamma$ can be used to label the irreducible projective
Reps of $H$. In order to simultaneously block diagonalize the restrict Rep of
$H$ and its subgroups, we can make use of the class operators of $H$ and those
of its subgroupsChen et al. [1985] to lift the degeneracy of $\Gamma$.

Therefore, the central step is to construct the hermitian Hamiltonian matrix
$\Gamma$. Here we summarize the reduction procedure in the following three
steps:
(1) Obtain the subspace $\mathcal{L}^{(I)}$ which carries the identity Reps of
$M(H)\otimes M^{*}(H)$, namely, find all the bases
$v^{(I)i}\in\mathcal{L}^{(I)}$ such that for any group element ${h\in H}$,

$\displaystyle M(h)\otimes M^{*}(h)v^{(I)i}=v^{(I)i};$

(2) Chose an arbitrary basis $v=\sum_{i}r_{i}v^{(I)i}\in\mathcal{L}^{(I)}$,
where $r_{i}\in\mathbb{R}$ are arbitrary real numbers, reshape $v$ into a
matrix $\Gamma_{0}$, namely

$\displaystyle(\Gamma_{0})_{ab}=v_{a(N-1)+b};$

and then construct an hermitian matrix
$\Gamma=(\Gamma_{0}+\Gamma_{0}^{\dagger})+i(\Gamma_{0}-\Gamma_{0}^{\dagger})$;
(3) Diagonalize the class operators $C$ of $M(H)$, and the class operators
$C(s)$ of its subgroup chain $H_{1}\subset H_{2}\subset...\subset H$, and the
matrix $\Gamma$ simultaneously,

$\displaystyle\left(\begin{matrix}C\\\ C(s)\\\
\Gamma\end{matrix}\right)\phi^{(\nu)_{\varepsilon}}_{m}=\left(\begin{matrix}\nu\\\
m\\\ \varepsilon\end{matrix}\right)\phi^{(\nu)_{\varepsilon}}_{m},$

then the eigenvectors $\phi^{(\nu)_{\varepsilon}}_{m}$ are the irreducible
bases. The eigenspace of ‘energy’ $\varepsilon$ is an irreducible Rep space,
thus we can use the energy $\varepsilon$ to label the multiplicity
$(\nu)_{\varepsilon}$ if the IPRep $(\nu)$ occurs more than once. The class
operators are defined as the followingChen et al. [1985], Yang and Liu [2017]

$\displaystyle C_{i}=\sum_{h_{a}\in H}M(h_{a})M(h_{i})M^{\dagger}(h_{a}),$
(20)

and $C$ is a linear combination of $C_{i}$ with $C=\sum_{i}r_{i}C_{i}$ where
$r_{i}\in\mathbb{R}$ are arbitrary real numbers. The operators $C(s)$ are
defined in a similar way, which are used to lift the degeneracy of the
eigenvalues and to reduce the restricted Reps of the subgroups on the chain
$H_{1}\subset H_{2}\subset...\subset H$.

In the first step, the eigenvectors of $M(h)\otimes M^{*}(h),h\in H$ with
eigenvalue 1 are required. When the dimension $N$ of $M(h)$ is large, it seems
that one need to solve the eigenstates of matrices with dimension $N^{2}$.
Actually, this complexity can be avoided in two ways.

One way is to obtain the eigenvectors of $M(h)\otimes M^{*}(h)$ from the
eigenstates of $M(h)$ and $M^{*}(h)$. Since the eigenvalues of $M(h)\otimes
M^{*}(h)$ are the product of the eigenvalues of $M(h)$ and $M^{*}(h)$, the
eigenvectors of the product matrix with eigenvalue 1 is the direct product of
the eigenstates of $M(h)$ and $M^{*}(h)$ whose eigenvalues are mutually
complex conjugate. For all the elements $h\in H$ we can construct the
eigenspace of $M(h)\otimes M^{*}(h)$ with eigenvalue 1 in the same way, then
any state in the intersection of such eigenspaces satisfies the condition (1).

The other way is to construct the matrix $\Gamma_{0}$ directly,

$\Gamma_{0}=\sum_{h\in H}M(h)AM^{{\dagger}}(h),$

where $A$ is an arbitrary square matrix222 If we regard $A$ as a vector, then
the above equation can be written as $(\Gamma_{0})_{ij}=\sum_{h\in
H}M_{ik}(h)M_{jl}^{*}(h)A_{kl}=|H|P_{ij,kl}^{(I)}A_{kl}$, where
$P^{(I)}={1\over|H|}\sum_{h\in H}M(h)\otimes M^{*}(h)$ is the projection
operator projecting onto the subspace of identity Reps contained in the direct
product Rep $M(H)\otimes M^{*}(H)$.. Obviously above $\Gamma_{0}$ satisfies
the commutation relation $M(h)\Gamma_{0}M^{\dagger}(h)=\Gamma_{0}$, which is
equivalent to the eigen problem $M(h)\otimes M^{*}(h)v=v$ with the vector $v$
reshaped from $\Gamma_{0}$. Therefore, thus constructed matrix satisfies the
conditions in step (1) and step (2). Practically this method is more
straightforward.

###  III.2 Reduction of Reps for anti-unitary groups

The same idea can be generalized to reduce general Reps of anti-unitary groups
$G=H+T_{0}H$. For a general matrix $\Gamma$ satisfying (4), (5) and (6), each
of its eigenspaces is an irreducible projective Rep space of $G$. To lift the
degeneracy of the eigenvalues of $\Gamma$ , we can make use of the class
operators of $H$ and those of its subgroups.

The central step is to construct the hermitian Hamiltonian matrix $\Gamma$
satisfying the restrictions (4), (5) and (6). We summarize the reduction
procedure as the following:
(1) Following the method in section III.1, obtain a matrix $\Lambda_{0}$ which
is commuting with $M(h),h\in H$, and then construct a hermitian matrix
$\Lambda=(\Lambda_{0}+\Lambda_{0}^{\dagger})+i(\Lambda_{0}-\Lambda_{0}^{\dagger})$;
(2) Construct a matrix $\Gamma$ from $\Lambda$

$\displaystyle\Gamma$ $\displaystyle=$ $\displaystyle\Lambda+M(T_{0})K\Lambda
KM^{\dagger}(T_{0})$ $\displaystyle=$
$\displaystyle\Lambda+M(T_{0})\Lambda^{*}M^{\dagger}(T_{0}).$

It is easily verified that $\Gamma M(T_{0})K=M(T_{0})K\Gamma$ because
$T_{0}^{2}\in H$, $M(T_{0}^{2})\Lambda=\Lambda M(T_{0}^{2})$, and that
$[M(T_{0})K]^{2}=\omega_{2}(T_{0},T_{0})M(T_{0}^{2})$. Furthermore, noticing
that $hT_{0}=T_{0}(T_{0}^{-1}hT_{0})$ and that $(T_{0}^{-1}hT_{0})\in H$, it
can be shown that $\Gamma$ commutes with $M(h)$ for all $h\in H$;
(3) Simultaneously diagonalize the class operators $C$ of $M(H)$ [see (20) for
definition], the class operators $C(s)$ of the subgroup chain $H_{1}\subset
H_{2}\subset...\subset H$, and the Hamiltonian matrix $\Gamma$,

$\displaystyle\left(\begin{matrix}C\\\ C(s)\\\
\Gamma\end{matrix}\right)\phi^{(\nu)_{\varepsilon}}_{m}=\left(\begin{matrix}\nu\\\
m\\\ \varepsilon\end{matrix}\right)\phi^{(\nu)_{\varepsilon}}_{m},$

then the eigenvectors $\phi^{(\nu)_{\varepsilon}}_{m}$ are the irreducible
bases, where the bases with the same ‘energy’ $\varepsilon$ belong to the same
irreducible Rep-space $(\nu)_{\varepsilon}$.

If IPReps with torsion number $R=4$ are contained in $M(G)$ after the
reduction, then the restricted Rep of $H$ in each of the $R=4$ IPRep is a
direct sum of two identical copies of irreducible Reps of $H$. However, both
$\Gamma$ and $C$ can only provide a single eigenvalue in the IRRep of $G$.
Therefore the quantum number $m$ in step (3) are doubly degenerate. In this
case, we can use the hermitian matrix $\Lambda$ to distinguish the two
identical irreducible Reps of $H$. It is obvious that $\Lambda$ commutes with
$\Gamma,C$ and $C(s)$, so we can add it to the commuting operators in step
(3),

$\displaystyle\left(\begin{matrix}C\\\ \Lambda,C(s)\\\
\Gamma\end{matrix}\right)\phi^{(\nu)_{\varepsilon}}_{\varepsilon_{H},m}=\left(\begin{matrix}\nu\\\
\varepsilon_{H},m\\\
\varepsilon\end{matrix}\right)\phi^{(\nu)_{\varepsilon}}_{\varepsilon_{H},m},$

then all the degeneracies are lifted.

Notice that we have used the class operators of $H$ to define the class
operator $C$ for simplicity. The eigenvalues $\nu$ are not necessarily real
(it is not real if $R=2$). In this case the eigenspaces of $\nu$ and $\nu^{*}$
belong to the same IPRep of $G$. One can also adopt the class operators of the
total group $G$,
$C_{i+}=C_{h_{i}}+C_{T_{0}h_{i}T_{0}^{-1}}+C_{h_{i}^{-1}}+C_{T_{0}h_{i}T_{0}^{-1}},C_{i-}=i(C_{h_{i}}+C_{T_{0}h_{i}T_{0}^{-1}}-C_{h_{i}^{-1}}-C_{T_{0}h_{i}T_{0}^{-1}})$
to construct $\mathcal{C}=\sum_{i}(r_{i+}C_{i+}+r_{i-}C_{i-})$ Yang and Liu
[2017], where $C_{h_{i}}$ is the class operator of $H$ in the restricted Rep
and $r_{i\pm}\in\mathbb{R}$ are real numbers. Then the eigenvalues of
$\mathcal{C}$ are always real numbers, but in this case the operators
$\mathcal{C}(s)$ should include the class operators of $H$ and those of its
subgroups.

##  IV Application of the Hamiltonian approach in perturbation theory

The Hamiltonian approach can be generalized to obtain the response of the
system to symmetry breaking probe fields if the low-energy physics is
dominated by particle-like excitations, such as the electron-like
quasiparticles in metals, Bogoliubov quasi-particles in superconductors or the
magnon excitations in the spin sector. We restrict our discussion to
irreducible projective Reps of anti-unitary groups.

###  IV.1 $k\cdot p$ perturbation around high symmetry points

In this section, we discuss the nodal-point and nodal-line structures in
magnetic materials whose symmetry group are either type-III or type-IV
Shubnikov magnetic space groups. The symmetry operations which keep a momentum
$\boldsymbol{k}$ invariant (up to a reciprocal lattice vector) form a magnetic
point group $G_{0}(\boldsymbol{k})$ which is called the little co-group. The
degeneracy of the energy bands at $\boldsymbol{k}$ is determined by the
irreducible (projective) Reps of the little co-group. The dispersion around
$\boldsymbol{k}$ can be obtained using the $k\cdot p$ perturbation theory.

Suppose that the little co-group $G_{0}(\boldsymbol{k})$ has a $d$-dimensional
irreducible (projective) Rep, which is carried by the quasi-particle bases
$\psi_{\boldsymbol{k}}^{\alpha},\alpha=1,2,...,d$, with

$\displaystyle\hat{g}\psi_{\boldsymbol{k}}^{\dagger}\hat{g}^{-1}$
$\displaystyle=$ $\displaystyle\psi_{\boldsymbol{k}}^{\dagger}M(g)K_{s(g)},$
(21) $\displaystyle\hat{g}\psi_{\boldsymbol{k}}\hat{g}^{-1}$ $\displaystyle=$
$\displaystyle K_{s(g)}M(g)^{\dagger}\psi_{\boldsymbol{k}},$ (22)

for $g\in G$. The degeneracy is generally lifted at the vicinity of
$\boldsymbol{k}$. When $\delta\boldsymbol{k}$ is small enough, it is expected
that $\psi_{\boldsymbol{k}+\delta\boldsymbol{k}}^{\dagger}$ and
$\psi_{\boldsymbol{k}+\delta\boldsymbol{k}}$ vary in the way similar to (21)
and (22) under the group action,

$\displaystyle\hat{g}\psi_{\boldsymbol{k}+\delta\boldsymbol{k}}^{\dagger}\hat{g}^{-1}$
$\displaystyle=$
$\displaystyle\psi_{\boldsymbol{k}+\hat{g}\delta\boldsymbol{k}}^{\dagger}M(g)K_{s(g)},$
(23)
$\displaystyle\hat{g}\psi_{\boldsymbol{k}+\delta\boldsymbol{k}}\hat{g}^{-1}$
$\displaystyle=$ $\displaystyle
K_{s(g)}M(g)^{\dagger}\psi_{\boldsymbol{k}+\hat{g}\delta\boldsymbol{k}}.$ (24)

Suppose the Hamiltonian at $\boldsymbol{k}+\delta\boldsymbol{k}$ is given by

$\displaystyle
H_{\boldsymbol{k}+\delta\boldsymbol{k}}=\psi_{\boldsymbol{k}+\delta\boldsymbol{k}}^{\dagger}\Gamma(\delta\boldsymbol{k})\psi_{\boldsymbol{k}+\delta\boldsymbol{k}},$
(25)

where $\Gamma(\delta\boldsymbol{k})$ is an Hermitian matrix
$\Gamma^{\dagger}(\delta\boldsymbol{k})=\Gamma(\delta\boldsymbol{k})$. When
summing over all the momentum variation, the total Hamiltonian should preserve
the $G$ symmetry, $i.e.$,

$\displaystyle\hat{g}\left(\sum_{\delta
k}H_{\boldsymbol{k}+\delta\boldsymbol{k}}\right)\hat{g}^{-1}=\left(\sum_{\delta
k}H_{\boldsymbol{k}+\delta\boldsymbol{k}}\right),$ (26)

for all $g\in G$. Substituting the equations (25), (23) and (24) into (26), we
obtain,

$\displaystyle
M(g)K_{s(g)}\Gamma(g^{-1}\delta\boldsymbol{k})K_{s(g)}M^{\dagger}(g)=\Gamma(\delta\boldsymbol{k}).$
(27)

which is the most general symmetry requirement.

If the leading order of $\Gamma(\delta\boldsymbol{k})$ is linear in
$\delta\boldsymbol{k}$, namely,
$\Gamma(\delta\boldsymbol{k})\sim\delta\boldsymbol{k}\cdot\boldsymbol{\Gamma}$,
then the dispersion around this high-degeneracy point froms a cone. For
fermionic systems, a conic dispersion is called a Dirac coneYoung et al.
[2012], Tang et al. [2016], Hua et al. [2018a], Armitage et al. [2018], Hua et
al. [2018b], Watanabe et al. [2018], Cano et al. [2019], Xu et al. [2020],
Elcoro et al. [2020], Bouhon et al. [2020] if $d=4$ and if
$\tilde{T}=\mathcal{I}T$ ($\mathcal{I}$ is the spacial inversion operation) is
an element of $G_{0}(\boldsymbol{k})$ such that the energy bands are doubly
degenerate away from $\boldsymbol{k}$. On the other hand, if the degeneracy
remains unchanged along a special line crossing the point $\boldsymbol{k}$,
then this line is called a nodal lineBurkov et al. [2011], Burkov and Balents
[2011], Xu et al. [2011], Chen et al. [2015], Fang et al. [2015], Weng et al.
[2015], Bzdušek et al. [2016], Geilhufe et al. [2019], Guo et al. , Cui et al.
, Yang et al. [2021].

Following the idea of the previous sections, here we provide a criteria to
judge whether the dispersion around the point $\boldsymbol{k}$ is linear or of
higher order, and whether the degeneracy is stable in a high symmetry line.

####  IV.1.1 Nodal points with linear dispersion

Firstly, we consider linear dispersion around $\boldsymbol{k}$, namely,

$\displaystyle\Gamma(\delta\boldsymbol{k})=\sum_{m=1}^{3}\delta
k_{m}\Gamma^{m}+O(\delta k^{2}).$ (28)

Here $\delta\boldsymbol{k}$ is a dual vector under the point group operations
in $H$, namely,

$\displaystyle\hat{h}\delta k_{m}=\sum_{n}D^{(\bar{v})}_{mn}(h)\delta k_{n},$
(29)

where $(\bar{v})$ is the dual Rep of the vector Rep $(v)$ of the unitary
subgroup $H$ with $D^{(\bar{v})}(h)=\left([D^{(v)}(h)]^{-1}\right)^{T}$. The
vector Rep is real, so $(\bar{v})$ is equivalent to $(v)$ [in orthonormal
bases, $(\bar{v})$ is identical to $(v)$, but we do not require the bases
$[\boldsymbol{b}_{1},\boldsymbol{b}_{2},\boldsymbol{b}_{3}]$ in the reciprocal
space to be orthonormal].

From (27)$\sim$(29), it can be shown (see Appendix C) that
$\boldsymbol{\Gamma}$ carries the dual vector Rep of $H$, namely,

$\displaystyle
M(h)\Gamma^{m}M(h)^{\dagger}=\sum_{n}D^{(\bar{v})}_{nm}(h)\Gamma^{n}.$ (30)

In the following we first assume that the vector Rep $(v)$ is irreducible. The
case $(v)$ is reducible will be mentioned later.

According to the action of $T_{0}$ on $\delta\boldsymbol{k}$, we first discuss
a special case where $T_{0}$ acts trivially on $\delta\boldsymbol{k}$, then go
to the general cases.

## The special case $T_{0}\delta\boldsymbol{k}=\delta\boldsymbol{k}$

Firstly we consider the case that $T_{0}$ acts trivially on
$\delta\boldsymbol{k}$,

$\displaystyle T_{0}\delta\boldsymbol{k}=\delta\boldsymbol{k}.$ (31)

From (27), (28) and above equation, we have,

$\displaystyle M(T_{0})K\Gamma^{m}KM(T_{0})^{\dagger}=\Gamma^{m}.$ (32)

The requirements (30) and (32) are similar to (4) and (5), respectively. If
there exists three $d\times d$ Hermitian matrices $\Gamma^{1,2,3}$ satisfying
these requirements, then the dispersion around $\boldsymbol{k}$ forms a cone.
From the discussion in II.2, we can judge the existence of $\Gamma^{1,2,3}$ by
checking if the projected space $M(h)\otimes M^{*}(h)P_{HT_{0}}$ (or
equivalently the projected space $M(h)\otimes F(h)P_{\eta_{0}}$) contains the
dual vector Rep $(\bar{v})$ of $H$.

When the vector Rep $(v)$ of $H$ is irreducible, then the existence of linear
dispersion can be checked by calculating the following quantity,

$\displaystyle a_{(\bar{v})}^{H}=\\!{1\over
2|H|}\\!\sum_{h}\\!\left[\chi(h)\chi^{*}(h)\\!+\\!\omega(hT_{0},hT_{0})\chi((hT_{0})^{2})\right]\chi^{(v)}(h),$
(33)

where $[\chi^{(\bar{v})}(h)]^{*}=\chi^{(\bar{v})}(h)=\chi^{(v)}(h)$ has been
used. If $a_{(\bar{v})}^{H}$ is a nonzero integer, then the dispersion is
linear along all directions.

The existence of $\Gamma^{1,2,3}$ under the conditions (30) and (32) can also
be checked straightforwardly by reducing the product Rep $M(g)\otimes
M^{*}(g)K_{s(g)},g\in G$ into direct sum of IPReps using the method introduced
in section III.2. If the resultant IPReps contain the dual vector Rep(s) whose
bases are hermitian when reshaped into matrix forms, then the leading order
dispersion around $\boldsymbol{k}$ is linear. Therefore, we need a projection
operator $P_{H}$ to project the bases carrying the dual vector Reps onto the
hermitian subspace (i.e. the union of real symmetric subspace and the
imaginary anti-symmetric subspace). Performing the projection $P_{H}$ and
reshaping the remaining bases into hermitian matrices, then we obtain the
explicit form of $\Gamma^{1,2,3}$,

$\displaystyle\Gamma^{m}=\sum_{i=1}^{p}r_{i}\gamma^{m}_{i},$

where $p$ is the mutiplicity of the dual vector Rep(s) contained in the
product Rep, $r_{i}\in\mathbb{R}$ are arbitrary real numbers, and
$\gamma_{i}^{1,2,3},i=1,...,p$ are the bases of the $i$th dual vector Rep.
Substituting into (28) and (25) we obtain the $k\cdot p$ effective model.

In next section, we will introduce an alternative method to obtain the
hermitian matrices $\Gamma^{1,2,3}$ without using the projection operator
$P_{H}$.

## The general case

Generally, $T_{0}$ acts on $\delta\boldsymbol{k}$ in the following way,

$\displaystyle T_{0}\delta k_{m}=\sum_{n}D^{(\bar{v})}_{mn}(T_{0})\delta
k_{n},$

where $D^{(\bar{v})}(T_{0})$ is a $3\times 3$ real matrix, and
$D^{(\bar{v})}(T_{0})K$ can be considered as part of the dual vector Rep of
the anti-unitary group $G_{0}(\boldsymbol{k})$. Accordingly, $\Gamma^{m}$
should vary in the following way in anology to (32),

$\displaystyle
M(T_{0})(\Gamma^{m})^{*}M^{\dagger}(T_{0})=\sum_{n}\Gamma^{n}D^{(\bar{v})}_{nm}(T_{0}).$
(34)

Similar to the discussion in Sec.II.2.2, we introduce a bases transformation
$\tilde{\psi}_{\boldsymbol{k}}=M^{*}(T_{0})\psi_{\boldsymbol{k}}$, then

$\displaystyle\hat{h}\tilde{\psi}_{\boldsymbol{k}}\hat{h}^{-1}=F^{T}(h)\tilde{\psi}_{\boldsymbol{k}},\
\
\hat{T}_{0}\tilde{\psi}_{\boldsymbol{k}}\hat{T}_{0}^{-1}=M^{T}(T_{0})K\tilde{\psi}_{\boldsymbol{k}},$

where $F(h)$ is defined in (9). We further define

$\displaystyle\tilde{\Gamma}^{m}=\Gamma^{m}M^{T}(T_{0}),$ (35)

then (30) and (34) deform into

$\displaystyle M(h)\tilde{\Gamma}^{m}F^{T}(h)$ $\displaystyle=$
$\displaystyle\sum_{n}D^{(\bar{v})}_{nm}(h)\tilde{\Gamma}^{n},$ (36)
$\displaystyle M(T_{0})\big{(}\tilde{\Gamma}^{m}\big{)}^{*}M^{T}(T_{0})$
$\displaystyle=$
$\displaystyle\sum_{n}D^{(\bar{v})}_{nm}(T_{0})\tilde{\Gamma}^{n},$ (37)

respectively.

Considering the set of matrices $\tilde{\Gamma}^{1,2,3}$ as a single column
vector $\tilde{\Gamma}$ with

$\displaystyle(\tilde{\Gamma})_{n\times d^{2}+i\times
d+j}=(\tilde{\Gamma}^{n})_{ij},$ (38)

then $\tilde{\Gamma}$ carries the identity Rep of $G$,

$\displaystyle W(h)\tilde{\Gamma}=\tilde{\Gamma},\ \
W(T_{0})K\tilde{\Gamma}=\tilde{\Gamma}K,$ (39)

where

$\displaystyle\\!\\!W(h)=D^{(v)}(h)\otimes V(h)=D^{(v)}(h)\otimes M(h)\otimes
F(h),$
$\displaystyle\\!\\!W(T_{0})\\!=\\!D^{(v)}(T_{0})\\!\otimes\\!V(T_{0})\\!=\\!D^{(v)}(T_{0})\\!\\!\otimes\\!\\!M(T_{0})\\!\otimes\\!M(T_{0}).$

Here the relation $\sum_{m}D^{(\bar{v})}_{mi}(h)D^{(v)}_{mj}(h)=\delta_{ij}$
has been used. Later we will alternately use the notation $\tilde{\Gamma}^{m}$
and $\tilde{\Gamma}$.

Taking transpose of (35), we have

$\displaystyle(\tilde{\Gamma}^{m})^{T}$ $\displaystyle=$ $\displaystyle
M(T_{0})(\Gamma^{m})^{T}$ (40) $\displaystyle=$ $\displaystyle
M(T_{0})\sum_{n}D^{(v)}_{mn}(T_{0})M^{*}(T_{0})\Gamma^{n}M^{T}(T_{0})$
$\displaystyle=$
$\displaystyle\eta_{0}M(\sigma)\sum_{n}D^{(v)}_{mn}(T_{0})\tilde{\Gamma}^{n},$

where we have used the transpose of (34) namely
$\sum_{n}(\Gamma^{n})^{T}D^{(\bar{v})}_{nm}(T_{0})=M^{*}(T_{0})\Gamma^{m}M^{T}(T_{0})$
and the hermitian condition $\Gamma^{m}=(\Gamma^{m})^{\dagger}$. Namely, the
symmetry condition in (40) is a consequence of the requirements that
$\tilde{\Gamma}^{m}$ should be hermitian and $T_{0}$-symmetric.

Taking the transpose of $\tilde{\Gamma}$ twice, we obtain
$\big{(}(\tilde{\Gamma}^{m})^{T}\big{)}^{T}=\eta_{0}^{2}\sum_{l}D_{ml}^{(v)}(\sigma)M(\sigma)\tilde{\Gamma}^{l}M^{T}(\sigma)=\sum_{l}D_{ml}^{(v)}(\sigma)M(\sigma)\tilde{\Gamma}^{l}F^{T}(\sigma)$.
Therefore, the self-consistency requires that
$\sum_{l}D_{ml}^{(v)}(\sigma)M(\sigma)\tilde{\Gamma}^{l}F^{T}(\sigma)=\tilde{\Gamma}^{m}$,
which is ensured by (36).

Noticing
$\tilde{\Gamma}^{m}=[\eta_{0}M(\sigma)\sum_{n}D^{(v)}_{mn}(T_{0})\tilde{\Gamma}^{n}]^{T}$,
we can define the generalized twist operator which transforms
$\tilde{\Gamma}^{m}$ to
$[\eta_{0}M(\sigma)\sum_{n}D^{(v)}_{mn}(T_{0})\tilde{\Gamma}^{n}]^{T}$,

$\displaystyle\mathbb{T}_{\eta_{0}}$ $\displaystyle=$
$\displaystyle\eta_{0}(I\otimes\mathscr{T})\Big{(}D^{(v)}(T_{0})\otimes
M(\sigma)\otimes I\Big{)}$ $\displaystyle=$ $\displaystyle
D^{(v)}(T_{0})\otimes\mathscr{T}_{\eta_{0}},$

where $\mathscr{T}_{\eta_{0}}$ is defined in (18). Thus
$\mathbb{T}_{\eta_{0}}$ defines a generalized ‘transpose’ of $\tilde{\Gamma}$,
and the self-consistency condition is
$(\mathbb{T}_{\eta_{0}})^{2}\tilde{\Gamma}=W(\sigma)\tilde{\Gamma}=\tilde{\Gamma}$.
In the eigenspace of $W(\sigma)$ with eigenvalue 1, the operator
$\mathbb{T}_{\eta_{0}}$ has eigenvalues $\pm 1$. If we call a vector
satisfying

$\displaystyle\mathbb{T}_{\eta_{0}}\tilde{\Gamma}=\tilde{\Gamma}$ (41)

to be $\eta_{0}$-symmetric, then the projector onto the $\eta_{0}$-symmetric
subspace is

$\mathbb{P}_{\eta_{0}}={1\over
2}\mathbb{P}_{\sigma}(I+\mathbb{T}_{\eta_{0}}),$

where $\mathbb{P}_{\sigma}$ is the projector onto the eigenspace of
$W(\sigma)$ with eigenvalue 1.

Denoting $\mathbb{P}^{(I)}$ as the projector onto the subspace of identity
Reps

$\mathbb{P}^{(I)}={1\over|H|}\sum_{h\in H}W(h),$

then obviously $\mathbb{P}^{(I)}\mathbb{P}_{\sigma}=\mathbb{P}^{(I)}$. The
condition for the linear dispersion is that the $\eta_{0}$-symmetric subspace
contains the identity Rep of $H$, namely

$a_{(I)}^{H}={\rm
Tr}\big{(}\mathbb{P}^{(I)}\mathbb{P}_{\eta_{0}}\big{)}={1\over 2}{\rm
Tr}\big{(}\mathbb{P}^{(I)}(I+\mathbb{T}_{\eta_{0}})\big{)}\geq 1.$

Defining the matrices

$W_{\eta_{0}}(h)={1\over 2}W(h)(I+\mathbb{T}_{\eta_{0}}),$

with the matrix entries

$\displaystyle[W_{\eta_{0}}(h)]_{mkl,nij}\\!\\!$ $\displaystyle=$
$\displaystyle\\!\\!{1\over 2}\Big{[}D_{mn}^{(v)}(h)M_{ki}(h)F_{lj}(h)$
$\displaystyle+$
$\displaystyle\\!\\!\eta_{0}\sum_{a}D_{mn}^{(v)}(hT_{0})M_{kj}(h)F_{la}(h)M_{ai}(\sigma)\Big{]},$

then the condition of linear dispersion reduces to $a_{(I)}^{H}={\rm Tr\
}(\mathbb{P}^{(I)}\mathbb{P}_{\eta_{0}})={1\over 2|H|}\sum_{h\in H}{\rm Tr\
}W_{\eta_{0}}(h)\geq 1$. From the above expression of $W_{\eta_{0}}(h)$, after
some calculations (see Appendix B) we obtain

$\displaystyle a_{(I)}^{H}\\!\\!=\\!\\!{1\over
2|H|}\\!\sum_{h}\\!\\!\Bigg{[}\\!|\chi(h)|^{2}\chi^{(v)}\\!(h)\\!+\\!{\chi^{(v)}\\!(hT_{0})}\omega(hT_{0},hT_{0})\chi((hT_{0})^{2})\\!\Bigg{]}\\!.$
(42)

Specially, if $D^{(v)}(T_{0})=I$, then ${\chi^{(v)}(hT_{0})}=\chi^{(v)}(h)$,
above formula reduces to the equation (33); if $D^{(v)}(T_{0})=-I$, then
${\chi^{(v)}(hT_{0})}=-\chi^{(v)}(h)$, above formula can be simplified as

$\displaystyle a_{(\bar{v})}^{H}\\!=\\!{1\over
2|H|}\\!\\!\sum_{h}\\!\left[\chi(h)\chi^{*}(h)\\!-\\!\omega(hT_{0},hT_{0})\chi((hT_{0})^{2})\right]\\!\chi^{(v)}\\!(h).$

When the vector Rep $(v)$ of $G$ is reducible, then the dispersions may be
different along different directions. In this case, we need to reduce the
vector Rep $(v)$ and check the resultant irreducible Reps one by one. For
instance, if $H=D_{4h}$, then the vector Rep is reduced to $(v)=E_{u}+A_{1u}$,
where $(k_{x},k_{y})^{T}$ vary in the rule of the Rep $E_{u}$ and $(k_{z})$
vary in the rule of $A_{1u}$. In this case, we need to replace $(v)$ in (33)
or (42) by $(E_{u})$ and $(A_{1u})$. If $a_{(E_{u})}^{H}$ is nonzero, then the
dispersion along $k_{x},k_{y}$ is linear, otherwise the dispersion is
quadratic or of higher order. Similarly, if $a_{(A_{1u})}^{H}\neq 0$, then the
dispersion along $k_{z}$ is linear.

####  IV.1.2 Procedure of obtaining the $\Gamma^{1,2,3}$ matrices

As $a_{I}^{H}\geq 1$, above procedure provides another way to obtain the
matrices $\Gamma^{1,2,3}$ besides the method of reducing the product Rep
$M(g)\otimes M^{*}(g)K_{s(g)},g\in G$ into direct sum of IPReps. We only
consider the case where the 3-dimensional vector Rep $(v)$ is irreducible. The
procedure is easily generalized to the cases in which $(v)$ is reducible.

Firstly, obtain the eigenspace of
$\mathbb{P}^{(I)}\mathbb{P}_{\eta_{0}}={1\over
2}\big{(}\mathbb{P}^{(I)}(I+\mathbb{T}_{\eta_{0}})\big{)}$ with eigenvalue 1.
Supposing the dimension of this eigenspace is $p\geq 1$, choose a set of
orthonormal bases $\tilde{\zeta}_{1},\tilde{\zeta}_{2},...,\tilde{\zeta}_{p}$.

Secondly, tune the bases in above subspace such that $T_{0}$ is represented as
$IK$. To this end, calculate the Rep of $T_{0}$

$\mathscr{M}_{ij}(T_{0})K=\tilde{\zeta}_{i}^{\dagger}W(T_{0})K\tilde{\zeta}_{j},$

and construct the new bases

$\tilde{\Delta}_{j}=\sum_{i}\tilde{\zeta}_{i}\mathscr{M}^{1\over
2}_{ij}(T_{0}).$

Then each of these new bases carries the identity Reps of the total group $G$
(see Appendix D for details).

Thirdly, from (38), we can decouple each eigenvector $\tilde{\Delta}_{i}$ as

$\tilde{\Delta}_{i}={1\over\sqrt{3}}e_{1}\otimes\tilde{\gamma}_{i}^{1}+{1\over\sqrt{3}}e_{2}\otimes\tilde{\gamma}_{i}^{2}+{1\over\sqrt{3}}e_{3}\otimes\tilde{\gamma}_{i}^{3},$

where $e_{1}=(1,0,0)^{T},e_{2}=(0,1,0)^{T},e_{3}=(0,0,1)^{T}$ stand for the
bases $\boldsymbol{a}_{1},\boldsymbol{a}_{2},\boldsymbol{a}_{3}$ (not
necessarily orthogonal) of the vector Rep $(v)$ respectively, and
$\tilde{\gamma}_{i}^{m}$ is the Schmidt partner of $e_{m}$. It can be shown
that (see Appendix D), $\tilde{\gamma}_{i}^{m}$ satisfies the relations (36)
and (37), and that $\gamma_{i}^{m}=\tilde{\gamma}_{i}^{m}M^{*}(T_{0})$ is a
hermitian matrices for any $i=1,...p$ and $m=1,2,3$.

The general form of the vector $\tilde{\Gamma}$ is a linear combination of the
bases $\tilde{\Delta}_{i}$,

$\displaystyle\tilde{\Gamma}$ $\displaystyle=$ $\displaystyle
r_{1}\tilde{\Delta}_{1}+r_{2}\tilde{\Delta}_{2}+...+r_{p}\tilde{\Delta}_{p}$
(43) $\displaystyle=$
$\displaystyle{1\over\sqrt{3}}e_{1}\otimes\tilde{\Gamma}^{1}+{1\over\sqrt{3}}e_{2}\otimes\tilde{\Gamma}^{2}+{1\over\sqrt{3}}e_{3}\otimes\tilde{\Gamma}^{3},$

where $r_{1},...,r_{p}\in\mathbb{R}$ are non-universal real constants. From
(43) and the relation $\Gamma^{m}=\tilde{\Gamma}^{m}M^{*}(T_{0})$, we obtain
the matrices $\Gamma^{m}$,

$\displaystyle\Gamma^{m}=\sum_{i=1}^{p}r_{i}\Big{(}\tilde{\gamma}_{i}^{m}M^{*}(T_{0})\Big{)}=\sum_{i=1}^{p}r_{i}\gamma_{i}^{m}.$

####  IV.1.3 Higher Order Dispersions and Nodal Lines

The discussion of linear dispersion can be straightforwardly generalized to
higher order dispersions. Suppose that a set of order-$N$ homogeneous
polynomials

$P^{(N)}_{i}(\delta\boldsymbol{k})=\sum_{a+b+c=N}f^{(N)}_{i(abc)}\delta
k_{1}^{a}\delta k_{2}^{b}\delta k_{3}^{c}$

carry a linear Rep $(\bar{\mu})$ of the group $G$, the existence of the
dispersion

$H_{\boldsymbol{k}+\delta\boldsymbol{k}}=\sum_{i}P^{(N)}_{i}(\delta\boldsymbol{k})\psi_{\boldsymbol{k}+\delta\boldsymbol{k}}^{\dagger}\Gamma^{(N)}_{i}\psi_{\boldsymbol{k}+\delta\boldsymbol{k}}$

can be judged using the formula (42) with the vector Rep $(v)$ replaced by the
linear Rep $(\mu)$ (see Appendix C for an example). The method of obtaining
the corresponding matrices $\Gamma^{(N)}_{i}$ is also similar.

If the vector Rep of $G$ is reducible, it is possible that the degeneracy is
lifted along some directions (such as the $k_{x},k_{y}$ directions) but are
preserved along certain direction (such as the $k_{z}$ direction) to form a
nodal line. The little co-group on the line is generally smaller than the one
on the conner of the BZ. If the IPRep of the little co-group at the conner of
the BZ is still irreducible along a certain line, then this line is a nodal
line. Therefore, the existence of the nodal line can be judge from the formula
(1) Yang et al. [2021]. The same method can be applied to judge the stability
of the degeneracy under external perturbations (see section IV.2).

###  IV.2 Response to External Probe Fields

The IPRep $M(g)K_{s(g)},g\in G$ of anti-unitary symmetry group $G$ results in
energy degeneracy in single-particle spectrum. Here we discuss the possible
lifting of the degeneracy under external probe fields, such as
$\boldsymbol{E}$ and $\boldsymbol{B}$, stain, or temperature gradience, etc.
We assume that the probe fields carry irreducible linear Reps of the group
$G$. For instance, electric fields $\boldsymbol{E}$ or magnetic fields
$\boldsymbol{B}$ carry vector Reps of the unitary subgroup $H$, but they vary
differently under the anti-unitary element $T_{0}$ since $\boldsymbol{E}$ is
invariant under time reversal while $\boldsymbol{B}$ reverse its sign under
time reversal.

There are two possible consequences under external probes. The first possible
result is that the degeneracy guaranteed by the IPRep $M(g)K_{s(g)}$ is
preserved. The other possibility is that the degeneracy is lifted in linear or
higher order terms of the probe fields.

To judge if the probe fields can lift the degeneracy or not (summing over all
orders of perturbation), we need to know the remaining symmetry group with the
presence of the perturbation, and then judge if the restrict Rep is reducible
or not. Suppose the probe field reduces the symmetry group from $G=H+T_{0}H$
to $G^{\prime}=H^{\prime}+T^{\prime}_{0}H^{\prime}$ where $T_{0}^{\prime}$ is
anti-unitary. If the irreducible Rep $M(g)K_{s(g)}$ of $G$ remains irreducible
for $G^{\prime}$, namely if

$\displaystyle{1\over|H^{\prime}|}\sum_{h\in H^{\prime}}{1\over
2}\left[\chi(h)\chi^{*}(h)+\omega_{2}(T^{\prime}_{0}h,T^{\prime}_{0}h)\chi((T^{\prime}_{0}h)^{2})\right]=1.$

holds for the group $G^{\prime}$, then the degeneracy is robust against this
perturbation.

If the left hand side of above equation is not equal to 1, then the restricted
Rep is reducible and the degeneracy can be lifted at certain order. In the
following we only discuss the linear splitting by external fields, such as
$\boldsymbol{E}$ and $\boldsymbol{B}$. The linear response is given by the
perturbed Hamiltonian in form of

$\displaystyle
H=\sum_{\boldsymbol{k}}\psi^{\dagger}_{\boldsymbol{k}}(\boldsymbol{E}\cdot\boldsymbol{P}+\boldsymbol{B}\cdot\boldsymbol{M})\psi_{\boldsymbol{k}}$
(44)

where $P^{m},M^{m}$ are CG matrices similar to the $\Gamma^{m}$ matrices
discussed before. The existence of linear coupling terms $\boldsymbol{P}$ (or
$\boldsymbol{M}$) can be checked using the criterion (42) with
$M^{(v)}(T_{0})$ the transformation matrix of $\boldsymbol{E}$ (or
$\boldsymbol{B}$) under the action of $T_{0}$.

##  V Conclusions and Discussions

In summary, from a physical approach, we derived the condition (1) for the
irreducible projective representations of anti-unitary groups. This approach
provides a practical method to reduce an arbitrary projective Rep into a
direct sum of irreducible ones, which is applicable for either unitary or
anti-unitary groups.

As a physical application of this approach, for single particle systems with
magnetic space group symmetry, we provide the method to construct the $k\cdot
p$ perturbation theory at the high symmetry point of the Brillouin zone. We
provide the criterion (42) to judge if the dispersion is linear or of higher
order, and then provide the method to obtain the corresponding $k\cdot p$
Hamiltonian up to a few non-universal constants.

In the present work, we assume that the quasiparticles vary under linear
representations of the magnetic space groups. However, in strongly interacting
systems, projective representations of the magnetic space groups can emerge in
the fractionalized low-energy quasiparticle excitations for systems with
intrinsic topological order. We leave the discussion of this situation for
future study.

Acknowledgements We thank L. J. Zou and Y. X. Zhao for helpful discussions.
Z.Y.Y and Z.X.L. are supported by the Ministry of Science and Technology of
China (Grant No. 2016YFA0300504), the NSF of China (Grants No.11574392 and No.
11974421), and the Fundamental Research Funds for the Central Universities and
the Research Funds of Renmin University of China (Grant No. 19XNLG11). J. Yang
and C. Fang are supported by Ministry of Science and Technology of China under
grant number 2016YFA0302400, National Science Foundation of China under grant
number 11674370 and Chinese Academy of Sciences under grant number
XXH13506-202 and XDB33000000.

##  Appendix A  Hermiticity of $\Gamma$ and validity of (13) for general anti-
unitary groups

In this appendix, we firstly prove a lemma, then introduce a theorem. Type-I
and type-II anti-unitary groups are treated on equal footing.

######  Lemma 1.

If the unitary representation $M(T_{0})K$ of an anti-unitary element $T_{0}$
satisfies the condition $[M(T_{0})K]^{2}=M(T_{0})M^{*}(T_{0})=I$, then there
exist a unitary matrix $U$ such that
$U^{\dagger}M(T_{0})KU=U^{\dagger}M(T_{0})U^{*}K=IK$.

######  Proof.

Since $[M(T_{0})]^{\dagger}=[M(T_{0})]^{-1}=[M(T_{0})]^{*}$, letting
$U=[M(T_{0})]^{1\over 2}$, we have $U^{\dagger}=U^{*}=[M(T_{0})]^{-{1\over
2}}$. Consequently, one can easily verify that

$\displaystyle U^{\dagger}M(T_{0})KU$ $\displaystyle=$ $\displaystyle
U^{\dagger}M(T_{0})U^{*}K$ $\displaystyle=$ $\displaystyle[M(T_{0})]^{-{1\over
2}}M(T_{0})[M(T_{0})]^{-{1\over 2}}K$ $\displaystyle=$ $\displaystyle IK.$

######  Corollary 1.

In a linear Rep of anti-unitary group $G$, one can chose bases in the
eigenspace of $\sigma\equiv T_{0}^{2}$ with eigenvalue 1 such that $T_{0}$ is
represented as $IK$ in this subspace.

######  Proof.

Suppose $D(g)K_{s(g)},g\in G$ is a $N$-dimensional linear Rep of $G$,
$\Delta_{1},...\Delta_{N}$ are the orthonormal bases of the eigenspace of
$\sigma=T_{0}^{2}$ with eigenvalue 1, namely.

$D(\sigma)\Delta_{i}=\Delta_{i}.$

Since $\sigma T_{0}=T_{0}\sigma$, and accordingly
$D(\sigma)D(T_{0})K=D(T_{0})KD(\sigma)$, we have

$\displaystyle
D(\sigma)D(T_{0})K\Delta_{i}=D(T_{0})K\big{(}D(\sigma)\Delta_{i}\big{)}=D(T_{0})K\Delta_{i},$

namely, the eigenspace of $\sigma$ is closed under the action of $T_{0}$. In
the eigenspace of $\sigma$ with eigen value 1, the Rep of $T_{0}$ takes the
form

$\displaystyle\mathscr{M}(T_{0})_{ij}K=\tilde{\Delta}_{i}^{\dagger}D(T_{0})K\tilde{\Delta}_{j},$

with $\mathscr{M}^{\dagger}(T_{0})=\mathscr{M}^{-1}(T_{0})$ and
$\mathscr{M}(T_{0})\mathscr{M}^{*}(T_{0})=\mathscr{M}(\sigma)=I$. From lemma
1, the Rep of $T_{0}$ can be transformed into $IK$ in the new bases,

$\Delta^{\prime}_{i}=\sum_{j=1}^{N}\Delta_{j}\mathscr{M}_{ji}^{1\over
2}(T_{0}),$

with

$\mathscr{M}(T_{0})K\Delta^{\prime}_{i}=\mathscr{M}(T_{0})\Delta^{{}^{\prime}*}_{i}K=\Delta^{{}^{\prime}*}_{i}K.$

Now it is ready to introduce the theorem.

######  Theorem 1.

If $\tilde{\Gamma}$ is a common eigenvector $\tilde{\Gamma}$ of
$V_{\eta_{0}}(h),h\in H$ with eigenvalue 1, namely,
$V_{\eta_{0}}(h)\tilde{\Gamma}=\tilde{\Gamma}$ for all $h\in H$, with
$V_{\eta_{0}}(h)=\frac{1}{2}[V(h)(I+\mathscr{T}_{\eta_{0}})]$ and
$V(h)=M(h)\otimes F(h)$, then $\tilde{\Gamma}$ has the following properties:
1) it carries the identity Rep of $H$;
2) it is $\eta_{0}$-symmetric;
3) if the basis satisfies $V(T_{0})\tilde{\Gamma}^{*}=\tilde{\Gamma}$ (i.e. if
$\tilde{\Gamma}$ carries the identity Rep of $G$), then
$\Gamma=\tilde{\Gamma}M^{*}(T_{0})$ is an hermitian matrix where
$\tilde{\Gamma}$ has been reshaped into a matrix.

######  Proof.

Firstly, since $V(h)=M(h)\otimes F(h)$ is a linear Rep of $H$, we define the
following projection operator

$P^{(I)}={1\over|H|}\sum_{h\in H}V(h),$

which projects from the product space $V(h)=M(h)\otimes F(h)$ onto the
identity Rep space. Accordingly, we have

${1\over|H|}\sum_{h\in H}V_{\eta_{0}}(h)=P^{(I)}P_{\eta_{0}}.$

Supposing $\tilde{\Gamma}$ is a common eigenvector of $V_{\eta_{0}}(h),h\in H$
with eigenvalue 1, then we have

$\displaystyle{1\over|H|}\sum_{h\in H}V_{\eta_{0}}(h)\tilde{\Gamma}$
$\displaystyle=$ $\displaystyle
P^{(I)}P_{\eta_{0}}\tilde{\Gamma}=\tilde{\Gamma}.$ (45)

Therefore,

$P^{(I)}\tilde{\Gamma}=P^{(I)}P^{(I)}P_{\eta_{0}}\tilde{\Gamma}=P^{(I)}P_{\eta_{0}}\tilde{\Gamma}=\tilde{\Gamma},$

namely, $\tilde{\Gamma}$ is the CG coefficient coupling the direct product Rep
$V(h)=M(h)\otimes F(h)$ to the identity Rep.

On the other hand, from $P^{(I)}\tilde{\Gamma}=\tilde{\Gamma}$ and
$P^{(I)}P_{\eta_{0}}\tilde{\Gamma}=\frac{1}{2}P^{(I)}(I+\mathscr{T}_{\eta_{0}})\tilde{\Gamma}=\tilde{\Gamma}$,
we have

${1\over 2}(I+\mathscr{T}_{\eta_{0}})\tilde{\Gamma}=\tilde{\Gamma}.$

Therefore $\mathscr{T}_{\eta_{0}}\tilde{\Gamma}=\tilde{\Gamma}$. By
definition,
$\mathscr{T}_{\eta_{0}}\tilde{\Gamma}=[\eta_{0}M(\sigma)\tilde{\Gamma}]^{T}$,
so we have

$\tilde{\Gamma}^{T}=\eta_{0}M(\sigma)\tilde{\Gamma}.$

Namely, $\tilde{\Gamma}$ is $\eta_{0}$-symmetric. Especially, for type-I anti-
unitary groups, $\sigma=E$, the $\eta_{0}$-symmetry reduces to
$\tilde{\Gamma}^{T}=\eta_{0}\tilde{\Gamma}$.

Thus we have verified that the eigenvector $\tilde{\Gamma}$ of
$P^{(I)}P_{\eta_{0}}\tilde{\Gamma}=\tilde{\Gamma}$ indeed has the properties
1) and 2).

Now we illustrate that the relation $V(T_{0})K\tilde{\Gamma}=\tilde{\Gamma}$
can be satisfied.

Firstly we shows that $V(T_{0})K$ preserves the eigenspace of $P^{(I)}$.
Noticing that $T_{0}(\sum_{h\in H}h)=(\sum_{h\in H}h)T_{0}$, and that
$V(g)K_{s(g)},g\in G$ is a linear Rep of $G$, so
$V(T_{0})KP^{(I)}=P^{(I)}V(T_{0})K$. Therefore, if $\tilde{\Gamma}$ is an
eigenvector of $P^{(I)}$ with $P^{(I)}\tilde{\Gamma}=\tilde{\Gamma}$, then

$P^{(I)}V(T_{0})K\tilde{\Gamma}=V(T_{0})KP^{(I)}\tilde{\Gamma}=V(T_{0})K\tilde{\Gamma}.$

Namely, $V(T_{0})K\tilde{\Gamma}$ is still an eigenvector of $P^{(I)}$.

Then we show that $V(T_{0})K$ also preserves the eigenspace of $P_{\eta_{0}}$.
From the definition of the unit twist operator
$[(\mathscr{T})_{kl,ij}=\delta_{kj}\delta_{li}]$, it is easily to verify that

$(X\mathscr{T})_{kl,ij}=X_{kl,ji}$

and

$(\mathscr{T}X)_{kl,ij}=X_{lk,ij}$

for arbitrary matrix $X$. Similarly, for a direct product matrix $(X\otimes
Y)_{kl,ij}=X_{ki}Y_{lj}$, the twist operator acts as

$\displaystyle\begin{cases}[(X\otimes Y)\mathscr{T}]_{kl,ij}=(X\otimes
Y)_{kl,ji}=X_{kj}Y_{li}\\\ [\mathscr{T}(X\otimes Y)]_{kl,ij}=(X\otimes
Y)_{lk,ij}=X_{li}Y_{kj}\end{cases},$

which gives

$\displaystyle[\mathscr{T}(X\\!\otimes\\!Y)\mathscr{T}]_{kl,ij}$
$\displaystyle=[(X\\!\otimes\\!Y)\mathscr{T}]_{lk,ij}$
$\displaystyle=(X\\!\otimes\\!Y)_{lk,ji}=X_{lj}Y_{ki}\\!=\\!(Y\\!\otimes\\!X)_{kl,ij}.$

Since $V(T_{0})=M(T_{0})\otimes M(T_{0})$, we have

$\displaystyle[\mathscr{T}_{\eta_{0}}V(T_{0})K\mathscr{T}_{\eta_{0}}]$
$\displaystyle=$
$\displaystyle\eta_{0}\eta_{0}^{*}\mathscr{T}[M(\sigma)\otimes
I][M(T_{0})\otimes M(T_{0})]\mathscr{T}[M^{*}(\sigma)\otimes I]K$
$\displaystyle=$ $\displaystyle[I\otimes M(\sigma)][M(T_{0})\otimes
M(T_{0})][M^{*}(\sigma)\otimes I]K$ $\displaystyle=$
$\displaystyle[M(T_{0})\otimes M(T_{0})][M(\sigma)\otimes F(\sigma)]^{*}K$
$\displaystyle=$ $\displaystyle V(T_{0})K(\mathscr{T}_{\eta_{0}})^{2},$ (46)

where we have used $(\mathscr{T}_{\eta_{0}})^{2}=V(\sigma)$. Therefore, if
$\tilde{\Gamma}$ is $\eta_{0}$-symmetric, $i.e.$, if $\
\mathscr{T}_{\eta_{0}}\tilde{\Gamma}=\tilde{\Gamma}$, then
$\mathscr{T}_{\eta_{0}}V(T_{0})K\tilde{\Gamma}=\mathscr{T}_{\eta_{0}}V(T_{0})K\mathscr{T}_{\eta_{0}}\tilde{\Gamma}=V(T_{0})K(\mathscr{T}_{\eta_{0}})^{2}\tilde{\Gamma}=V(T_{0})K\tilde{\Gamma}$,
which means that $\big{(}V(T_{0})K\big{)}\tilde{\Gamma}$ is still
$\eta_{0}$-symmetric.

Therefore, the eigenspace of $P^{(I)}P_{\eta_{0}}$ with eigenvalue 1 is
preserved under the action of $T_{0}$. Namely, this eigspace form a linear Rep
of the anti-unitary group $G$. Noticing that $V(\sigma)=V(T_{0})V^{*}(T_{0})$
is represented as an identity matrix in this eigenspace, from lemma 1, we can
‘diagonalize’ $V(T_{0})K$ as $IK$ in this subspace. Namely, we can choose
proper bases such that $M(T_{0})\otimes
M(T_{0})\tilde{\Gamma}^{*}=\tilde{\Gamma}$, or equivalently (11), holds.

From the transpose of (11), we have
$M(T)\tilde{\Gamma}^{\dagger}[M(T)]^{T}=\tilde{\Gamma}^{T}=\eta_{0}M(\sigma)\tilde{\Gamma}$,
where the symmetry equation has been used. Substituting
$\Gamma=\tilde{\Gamma}M^{*}(T_{0})$ into above equation and noticing
$M(T_{0})M^{*}(T_{0})=\eta_{0}M(\sigma)$, we finally obtain the hermitian
condition $\Gamma^{\dagger}=\Gamma$. Thus the property 3) has been verified.

Noticing that the eigenspace of $P^{(I)}$ is either $\eta_{0}$-symmetric or
$(-\eta_{0})$-symmetric, so $P^{(I)}P_{\eta_{0}}$ is also a projection
operator

$\big{(}P^{(I)}P_{\eta_{0}}\big{)}^{2}=P^{(I)}P_{\eta_{0}},$

therefore its eigenvalues are either 1 or 0. If $M(g)K_{s(g)},g\in G$ is
irreducible, then $P^{(I)}P_{\eta_{0}}$ has only one nonzero eigenvalue. The
trace of this projection operator ${\rm Tr}\ (P^{(I)}P_{\eta_{0}})=1$ yields
the irreduciblity condition (13).

##  Appendix B Derivation of (16) for Type-II anti-unitary groups

Following the same discussion of type-I anti-unitary groups, for type-II anti-
unitary groups we obtain the matrix form
$V_{\eta_{0}}(h)=\frac{1}{2}[M(h)\otimes F(h)(I+\mathscr{T}_{\eta_{0}})]$,
namely

$\displaystyle[V_{\eta_{0}}(h)]_{kl,ij}=\frac{1}{2}\Big{(}$
$\displaystyle\\!\\!\left[M(h)\\!\otimes\\!F(h)\right]_{kl,ij}$
$\displaystyle+\eta_{0}[M(h)\\!\otimes\\!F(h)M(\sigma)]_{kl,ji}\\!\\!\Big{)}.$
(47)

From theorem 1 in appendix A, we can start with the equation (19), which can
be expressed in terms of characters as

$\displaystyle
1\\!\\!=\\!\\!\frac{1}{2|H|}\\!\sum_{h,i,j}\\!\Big{(}\\!M_{ii}(h)F_{jj}(h)\\!+\\!\eta_{0}M_{ij}(h)[F(h)M(\sigma)]_{ji}\\!\Big{)}.$
(48)

Remembering that $M(\sigma)=\eta^{-1}_{0}M(T_{0})M^{*}(T_{0})$, the second
term in (48) can be transformed into

$\displaystyle\sum_{i,j}$
$\displaystyle\eta_{0}M_{ij}(h)[F(h)M(\sigma)]_{ji}=\eta_{0}{\rm
Tr}[M(h)F(h)M(\sigma)]$ $\displaystyle=$ $\displaystyle\eta_{0}{\rm
Tr}[M(h)F(h)\eta^{-1}_{0}M(T_{0})M^{*}(T_{0})]$ $\displaystyle=$
$\displaystyle{\rm
Tr}[M(h)M(T_{0})M^{*}(h)M^{\dagger}(T_{0})M(T_{0})M^{*}(T_{0})]$
$\displaystyle=$ $\displaystyle{\rm Tr}[M(h)M(T_{0})M^{*}(h)M^{*}(T_{0})]$
$\displaystyle=$ $\displaystyle\omega(h,T_{0})\omega^{*}(h,T_{0}){\rm
Tr}[M(hT_{0})M^{*}(hT_{0})]$ $\displaystyle=$
$\displaystyle\omega(h,T_{0})\omega^{*}(h,T_{0})\omega(hT_{0},hT_{0})\chi((hT_{0})^{2}).$
(49)

Finally, noticing ${\rm Tr\ }M(h)=\chi(h)$ and ${\rm Tr\ }F(h)=\chi^{*}(h)$,
(48) reduces to (16), namely,

$\displaystyle 1=\frac{1}{2|H|}\sum_{h\in
H}\left[\chi(h)\chi^{*}(h)+\omega(hT_{0},hT_{0})\chi((hT_{0})^{2})\right].$
(50)

##  Appendix C $k\cdot p$ theory: Derivation of (30), (32), (34) and
Discussion for General Dispersions

We starts with the equation (27), namely,

$M(g)K_{s(g)}\Gamma(g^{-1}\delta\boldsymbol{k})K_{s(g)}M^{\dagger}(g)=\Gamma(\delta\boldsymbol{k}).$

Letting $\delta\boldsymbol{k}^{\prime}=g^{-1}\delta\boldsymbol{k}$, then
$\delta\boldsymbol{k}=g\delta\boldsymbol{k}^{\prime}$ and (27) becomes

$M(g)K_{s(g)}\Gamma(\delta\boldsymbol{k}^{\prime})K_{s(g)}M^{\dagger}(g)=\Gamma(g\delta\boldsymbol{k}^{\prime}).$

Since the summation over $\delta\boldsymbol{k}^{\prime}$ is equivalent to the
summation over $\delta\boldsymbol{k}$, therefore we have

$\displaystyle
M(g)K_{s(g)}\Gamma(\delta\boldsymbol{k})K_{s(g)}M^{\dagger}(g)=\Gamma(g\delta\boldsymbol{k}).$
(51)

If there is a linear dispersion then
$\Gamma(\delta\boldsymbol{k})=\sum_{m=1}^{3}\delta k_{m}\Gamma^{m}$. Notice
that $\delta\boldsymbol{k}$ varies as dual vector under the action of the
unitary subgroup $H$, namely $\hat{h}\delta
k_{m}=\sum_{n}D^{(\bar{v})}_{mn}(h)\delta k_{n}$. Substituting these relations
into (51) and letting $g=h\in H$, then we have

$\displaystyle M(h)\left(\sum_{n}\Gamma^{n}\delta
k_{n}\right)M^{{\dagger}}(h)=\sum_{m,n}\Gamma^{m}D^{(\bar{v})}_{mn}(h)\delta
k_{n}.$ (52)

Thus the equation (30) is proved, $i.e.$,
$M(h)\Gamma^{n}M(h)^{\dagger}=\sum_{m}D^{(\bar{v})}_{mn}(h)\Gamma^{m}$.

Now consider the anti-unitary element $g=T_{0}$. From (51), we obtain

$M(T_{0})K\Gamma(\delta\boldsymbol{k})KM^{\dagger}(T_{0})=\Gamma(T_{0}\delta\boldsymbol{k}).$

If $T_{0}$ has a nontrivial action on $\delta\boldsymbol{k}$, namely
$T_{0}\delta k_{m}=\sum_{n}D^{(\bar{v})}_{mn}(T_{0})\delta k_{n}$, then linear
dispersion $\Gamma(\delta\boldsymbol{k})=\sum_{m=1}^{3}\delta k_{m}\Gamma^{m}$
indicates that

$\displaystyle M(T_{0})K\left(\sum_{n}\Gamma^{n}\delta
k_{n}\right)KM^{\dagger}(T_{0})=\sum_{mn}D^{(\bar{v})}_{mn}(T_{0})\Gamma^{m}\delta
k_{n},$

which is equivalent to (34), $i.e.$,
$M(T_{0})(\Gamma^{n})^{*}M^{\dagger}(T_{0})=\sum_{m}\Gamma^{m}D^{(\bar{v})}_{mn}(T_{0}).$
Here we have used the fact that $\delta k_{n}\in\mathbb{R}$ are real numbers.
(32) is a special case of (34) with $D^{(\bar{v})}_{mn}(T_{0})=I$.

Similar discussion can be generalized to the case when the vector Rep is
reducible, or to the cases where the dispersions are of higher order.
Generally, the object $\sum_{n}\Gamma^{n}\delta k_{n}$ can be replaced by
$\sum_{i}\Gamma^{(N)}_{i}P_{i}^{(N)}(\delta\boldsymbol{k})$, where

$P^{(N)}_{i}(\delta\boldsymbol{k})=\sum_{a+b+c=N}f^{(N)}_{i(abc)}\delta
k_{1}^{a}\delta k_{2}^{b}\delta k_{3}^{c},\ \ f_{abc}\in\mathbb{R}$

belongs to a set of order-$N$ homogeneous polynomials of $\delta k_{1},\delta
k_{2},\delta k_{3}$ which vary under the rule of irreducible linear Rep of
$G$.

For instance, in the case $H=\mathscr{C}_{6v}$, the quadratic polynomials
$(P^{(2)}_{1},P^{(2)}_{2})^{T}=(k_{x}^{2}-k_{y}^{2},2k_{x}k_{y})^{T}$ vary as
a two-component column vector under the irreducible Rep
$(E_{2})=(\bar{E}_{2})$, namely,

$\displaystyle
hP^{(2)}_{i}(\delta\boldsymbol{k})=\sum_{j}D^{(\bar{E}_{2})}_{ij}(h)P^{(2)}_{j}(\delta\boldsymbol{k}),$

for $h\in\mathscr{C}_{6v}$, and

$\displaystyle
T_{0}P^{(2)}_{i}(\delta\boldsymbol{k})=\sum_{j}D^{(\bar{E}_{2})}_{ij}(T_{0})P^{(2)}_{j}(\delta\boldsymbol{k}).$

Accordingly, similar to (30) and (32) we have

$\displaystyle
M(h)\Gamma^{(2)}_{i}M(h)^{\dagger}=\sum_{j}D^{(\bar{E}_{2})}_{ji}(h)\Gamma^{(2)}_{j},$
$\displaystyle
M(T_{0})\Gamma^{(2)*}_{i}M(T_{0})^{\dagger}=\sum_{j}D^{(\bar{E}_{2})}_{ji}(T_{0})\Gamma^{(2)}_{j}.$

The existence of quadratic dispersion terms with the form
$\sum_{i=1}^{2}\Gamma^{(2)}_{i}P_{i}^{(2)}(\delta\boldsymbol{k})$ can be
judged using the formula (42) by replacing the vector Rep $(v)$ with the
linear Rep $(E_{2})$.

Applying the method introduced in section IV.1.1, we can obtain the matrices
$\Gamma^{(2)}_{1,2}$.

##  Appendix D $k\cdot p$ theory: hermiticity of $\Gamma^{1,2,3}$

We define the projection operator

$\mathbb{P}^{(I)}={1\over|H|}\sum_{h\in H}W(h)$

which project onto the subspace of identity Reps in the product Rep
$W(h)=D^{(v)}(h)\otimes M(h)\otimes F(h)$. Similarly,

${1\over|H|}\sum_{h\in
H}W_{\eta_{0}}(h)=\mathbb{P}^{(I)}\mathbb{P}_{\eta_{0}}.$

Following the discussion in Appendix A, the eigenvector $\tilde{\Gamma}$ of
$\mathbb{P}^{(I)}\mathbb{P}_{\eta_{0}}$ with

$\mathbb{P}^{(I)}\mathbb{P}_{\eta_{0}}\tilde{\Gamma}=\tilde{\Gamma}$

is also an eigenvector of $\mathbb{P}^{(I)}$ and $\mathbb{P}_{\eta_{0}}$,
namely, it is a $\eta_{0}$-symmetric vector which carries the identity Rep of
$H$.

Furthermore, referring to Appendix A and noticing the facts
$W(T_{0})K\mathbb{P}^{(I)}=\mathbb{P}^{(I)}W(T_{0})K$ and
$\mathbb{T}_{\eta_{0}}W(T_{0})K\mathbb{T}_{\eta_{0}}=W(T_{0})KW(\sigma)=W(T_{0})K\mathbb{T}_{\eta_{0}}^{2}$,
it can be verified that $W(T_{0})K$ preserves the eigenspace of
$\mathbb{P}^{(I)}$ and $\mathbb{P}_{\eta_{0}}$. Namely, the eigenspace
$\mathcal{L}_{1}$ of $\mathbb{P}^{(I)}\mathbb{P}_{\eta_{0}}$ with eigenvalue 1
is closed under the action of any $g\in G$, hence forms a linear Rep space of
$G$. From the lemma 1 and its corollary, we can choose proper bases such that
each basis carries the identity Rep of $G$, namely, we can always find the
bases of $\mathcal{L}_{1}$ such that $T_{0}$ is represented as
$\mathscr{M}(T_{0})K=IK$.

Now it is ready to prove that the matrices $\Gamma^{1,2,3}$ constructed from
$\tilde{\Gamma}$ are hermitian matrices.

######  Corollary 2.

If $\tilde{\Gamma}$ satisfy the relations (39) and (41), then the resultant
matrices $\Gamma^{1,2,3}$ are hermitian, where $\tilde{\Gamma}$ and
$\Gamma^{m}$ are related by $(\tilde{\Gamma})_{n\times d^{2}+i\times
d+j}=(\tilde{\Gamma}^{n})_{ij}$ and
$\Gamma^{m}=\tilde{\Gamma}^{m}M^{*}(T_{0})$.

######  Proof.

Equation (41) indicates that $\tilde{\Gamma}^{m}$ satisfies the symmetry
condition

$\displaystyle(\tilde{\Gamma}^{m})^{T}=\eta_{0}\sum_{n}D^{(v)}_{mn}(T_{0})M(\sigma)\tilde{\Gamma}^{n}.$
(53)

Taking complex conjugation, above equation becomes

$\displaystyle(\tilde{\Gamma}^{m})^{\dagger}$ $\displaystyle=$
$\displaystyle\sum_{n}\eta_{0}^{*}D_{mn}^{(v)*}(T_{0})M^{*}(\sigma)(\tilde{\Gamma}^{n})^{*}$
$\displaystyle=$ $\displaystyle
M^{*}(T_{0})M(T_{0})\sum_{n}D_{mn}^{(v)*}(T_{0})(\tilde{\Gamma}^{n})^{*},$

which yields

$\displaystyle
M^{T}(T_{0})(\tilde{\Gamma}^{m})^{\dagger}=\sum_{n}D_{mn}^{(v)*}(T_{0})M(T_{0})(\tilde{\Gamma}^{n})^{*}.$

On the other hand, the second equation in (39) is equivalent to (37), which
indicates that

$\displaystyle\tilde{\Gamma}^{m}M^{*}(T_{0})=\sum_{k}D^{(v)}_{mk}(T_{0})M(T_{0})(\tilde{\Gamma}^{k})^{*}.$

Since the vector Rep is a real Rep, comparing above two equations we have

$M^{T}(T_{0})(\tilde{\Gamma}^{m})^{\dagger}=\tilde{\Gamma}^{m}M^{*}(T_{0}),$

namely, $\big{(}\Gamma^{m}\big{)}^{\dagger}=\Gamma^{m}$. This completes the
proof.

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