# Supplementary material for “Emergent chiral symmetry in non-bipartite kagome
and pyrochlore lattices with spin-orbit coupling”

Hiroki Nakai  Department of Basic Science, University of Tokyo, Meguro-ku,
Tokyo 153-8902, Japan  Masataka Kawano  Department of Basic Science,
University of Tokyo, Meguro-ku, Tokyo 153-8902, Japan  Department of Physics,
Technical University of Munich, 85748 Garching, Germany  Chisa Hotta
Department of Basic Science, University of Tokyo, Meguro-ku, Tokyo 153-8902,
Japan

Figure S 1:  (a) Kagome lattice and (b) pyrochlore lattice. The corresponding
Brillouin zone and the representative $k$-points are shown together.

##  .1 Details of the SOC Hamiltonian

We consider the SOC Hamiltonian in Eq.(3) in the main text, and show their
explicit forms.

###  .1.1 kagome lattice

For the kagome lattice, we take the lattice vectors as ${\bf\it a}_{1}=(1,0)$,
${\bf\it a}_{2}=\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$, and coordinates
of three sublattices in the unit cell as ${\bf\it r}_{1}=(0,0)$, ${\bf\it
r}_{2}=\left(\frac{1}{4},\frac{\sqrt{3}}{4}\right)$ and ${\bf\it
r}_{3}=\left(\frac{1}{2},0\right)$ (Fig.S1 (a)). In the corresponding first
Brillouin zone in reciprocal space, the highly symmetric points are defined as
$\mathrm{\Gamma}=(0,0)$, $\mathrm{K}=\left(\frac{4\pi}{3},0\right)$, and
$\mathrm{M}=\left(\pi,\frac{\pi}{\sqrt{3}}\right)$.

The rotational axes of SU(2) gauge field in the Hamiltonian Eq.(3) for the
kagome lattice using the above coordinates are given as,
$\hat{{\bf\it\nu}}_{12}=(0,0,-1)$, $\hat{{\bf\it\nu}}_{13}=(0,0,1)$, and
$\hat{{\bf\it\nu}}_{23}=(0,0,-1)$. By constructing the basis in the order
$(c^{\dagger}_{{\bf\it k}1\uparrow},c^{\dagger}_{{\bf\it
k}1\downarrow},c^{\dagger}_{{\bf\it k}2\uparrow},c^{\dagger}_{{\bf\it
k}2\downarrow},c^{\dagger}_{{\bf\it k}3\uparrow},c^{\dagger}_{{\bf\it
k}3\downarrow})$, for ${\mathcal{H}}=\sum_{{\bf\it k}}{\bf\it
c}^{\dagger}_{{\bf\it k}}\mathcal{H}({\bf\it k}){\bf\it c}_{{\bf\it k}}$, the
Bloch Hamiltonian is obtained in the $6\times 6$ form;

$\mathcal{H}({\bf\it k})=\matrixquantity(0&h_{12}({\bf\it k})&h_{13}({\bf\it
k})\\\ h_{21}({\bf\it k})&0&h_{23}({\bf\it k})\\\ h_{31}({\bf\it
k})&h_{32}({\bf\it k})&0),$ (S1)

where $h_{ij}({\bf\it k})=-2tU_{ij}\cos({\bf\it k}\cdot({\bf\it r}_{i}-{\bf\it
r}_{j}))$.

###  .1.2 pyrochlore lattice

We consider four sites in a unit cell with the lattice vectors given as
${\bf\it a}_{1}=\left(\frac{1}{2},-\frac{1}{2},0\right)$ , ${\bf\it
a}_{2}=\left(0,-\frac{1}{2},\frac{1}{2}\right)$, and ${\bf\it
a}_{3}=\left(\frac{1}{2},0,\frac{1}{2}\right)$ (Fig.S1 (b)). The corresponding
reciprocal lattice vectors are ${\bf\it b}_{1}=2\pi(1,-1,-1)$ , ${\bf\it
b}_{2}=2\pi(-1,-1,1)$, ${\bf\it b}_{3}=2\pi(1,1,1)$. The coordinates of four
sublattices in the unit cell are given as ${\bf\it
r}_{1}=\left(\frac{1}{8},\frac{3}{8},\frac{1}{8}\right)$ , ${\bf\it
r}_{2}=\left(\frac{3}{8},\frac{1}{8},\frac{1}{8}\right)$, ${\bf\it
r}_{3}=\left(\frac{1}{8},\frac{1}{8},\frac{3}{8}\right)$, and ${\bf\it
r}_{4}=\left(\frac{3}{8},\frac{3}{8},\frac{3}{8}\right)$. The highly symmetric
points in the 1st Brillouin zone are $\mathrm{X}=(0,0,2\pi)$,
$\mathrm{K}=\left(\frac{3\pi}{2},0,\frac{3\pi}{2}\right)$,
$\mathrm{W}=(\pi,0,2\pi)$, and $\mathrm{L}=(\pi,\pi,\pi)$.

The rotational axes of SU(2) gauge field are taken as
$\hat{{\bf\it\nu}}_{12}=(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0)$,
$\hat{{\bf\it\nu}}_{13}=(0,-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}})$,
$\hat{{\bf\it\nu}}_{14}=(-\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}})$,
$\hat{{\bf\it\nu}}_{23}=(\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}})$,
$\hat{{\bf\it\nu}}_{24}=(0,\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}})$,
$\hat{{\bf\it\nu}}_{34}=(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},0)$, and the
representation of the Bloch Hamiltonian for the basis taken in the order,
$(c^{\dagger}_{{\bf\it k}1\uparrow},c^{\dagger}_{{\bf\it
k}1\downarrow},\ldots,c^{\dagger}_{{\bf\it k}4\uparrow},c^{\dagger}_{{\bf\it
k}4\downarrow})$, is given as,

$\mathcal{H}({\bf\it k})=\matrixquantity(0&h_{12}({\bf\it k})&h_{13}({\bf\it
k})&h_{14}({\bf\it k})\\\ h_{21}({\bf\it k})&0&h_{23}({\bf\it
k})&h_{24}({\bf\it k})\\\ h_{31}({\bf\it k})&h_{32}({\bf\it
k})&0&h_{34}({\bf\it k})\\\ h_{41}({\bf\it k})&h_{42}({\bf\it
k})&h_{43}({\bf\it k})&0),$ (S2)

where $h_{ij}({\bf\it k})=-2tU_{ij}\cos({\bf\it k}\cdot({\bf\it r}_{i}-{\bf\it
r}_{j}))$.

##  .2 Wilson loop operator

The Wilson loop operator is the path-ordered product of SU(2) gauge fields
$U_{ij}$ given in the main text as

$P(C_{i,jk})=\exp\left[-i\frac{\Phi}{2}\hat{{\bf\it
n}}_{i,jk}\cdot{\bf\it\sigma}\right].\hskip 28.45274pt(4)$

In this section, we first derive the two gauge invariant properties from this
definition. Then, we construct a chiral operator using the properties related
to the Wilson loop operator. Finally, we clarify the definition of
“Abelian/non-Abelian” used in this paper, which is different from the non-
Abelian properties discussed in other works. We show an example that the
Wilson loop operator can be used to determine the Abelian/non-Abelian
properties.

###  .2.1 Proof of gauge invariance of $\Phi$ and $\hat{{\bf\it
n}}_{i,jk}\cdot\hat{{\bf\it n}}_{i,lm}$

We prove that the rotation angle $\Phi$ and the relative angle between
rotation axes, $\hat{{\bf\it n}}_{i,jk}\cdot\hat{{\bf\it n}}_{i,lm}$, are both
gauge invariant. Using the local gauge transformation operator,
$V_{i}=\exp\left[-i\frac{\varphi_{i}}{2}\hat{{\bf\it
m}}_{i}\cdot{\bf\it\sigma}\right]$, the Wilson loop operator is transformed as

$\begin{split}V_{i}P(C_{i,jk})V^{\dagger}_{i}&=\cos\frac{\Phi}{2}I-i(\hat{{\bf\it
n}}_{i,jk}\cdot{\bf\it\sigma})\cos\varphi_{i}\sin\frac{\Phi}{2}\\\
&\quad-i(\hat{{\bf\it m}}_{i}\crossproduct\hat{{\bf\it
n}}_{i,jk})\cdot{\bf\it\sigma}\sin\varphi_{i}\sin\frac{\Phi}{2}\\\
&\quad-i(\hat{{\bf\it m}}_{i}\cdot\hat{{\bf\it n}}_{i,jk})(\hat{{\bf\it
m}}_{i}\cdot{\bf\it\sigma})(1-\cos\varphi_{i})\sin\frac{\Phi}{2}\\\
&=\cos\frac{\Phi}{2}I-i(\hat{{\bf\it
n}}^{\prime}_{i,jk}\cdot{\bf\it\sigma})\sin\frac{\Phi}{2}\\\
&=\exp\left[-i\frac{\Phi}{2}\hat{{\bf\it
n}}^{\prime}_{i,jk}\cdot{\bf\it\sigma}\right],\end{split}$ (S3)

where

$\begin{split}\hat{{\bf\it n}}^{\prime}_{i,jk}&=\cos\varphi_{i}\hat{{\bf\it
n}}_{i,jk}+\sin\varphi_{i}(\hat{{\bf\it m}}_{i}\crossproduct\hat{{\bf\it
n}}_{i,jk})\\\ &\quad+(1-\cos\varphi_{i})(\hat{{\bf\it
m}}_{i}\cdot\hat{{\bf\it n}}_{i,jk})\hat{{\bf\it m}}_{i}.\end{split}$ (S4)

We immediately see that phase $\Phi$ remains unchanged. We now take the inner
product between two rotation axes belonging to the same site-$i$ but to
different loops, $C_{i,jk}:i\rightarrow j\rightarrow k\rightarrow i$ and
$C_{i,lm}:i\rightarrow l\rightarrow m\rightarrow i$, as,

$\begin{split}&\hat{{\bf\it n}}^{\prime}_{i,jk}\cdot\hat{{\bf\it
n}}^{\prime}_{i,lm}\\\ &=\cos^{2}\varphi_{i}\hat{{\bf\it
n}}_{i,jk}\cdot\hat{{\bf\it
n}}_{i,lm}+\sin\varphi_{i}\cos\varphi_{i}(\hat{{\bf\it
m}}_{i}\crossproduct\hat{{\bf\it n}}_{i,jk})\cdot\hat{{\bf\it n}}_{i,lm}\\\
&\quad+\cos\varphi_{i}(1-\cos\varphi_{i})(\hat{{\bf\it
m}}_{i}\cdot\hat{{\bf\it n}}_{i,jk})(\hat{{\bf\it m}}_{i}\cdot\hat{{\bf\it
n}}_{i,lm})\\\ &\quad+\sin\varphi_{i}\cos\varphi_{i}(\hat{{\bf\it
m}}_{i}\crossproduct\hat{{\bf\it n}}_{i,lm})\cdot\hat{{\bf\it n}}_{i,jk}\\\
&\quad+\sin^{2}\varphi_{i}(\hat{{\bf\it m}}_{i}\crossproduct\hat{{\bf\it
n}}_{i,jk})\cdot(\hat{{\bf\it m}}_{i}\crossproduct\hat{{\bf\it n}}_{i,lm})\\\
&\quad+\sin\varphi_{i}(1-\cos\varphi_{i})(\hat{{\bf\it
m}}_{i}\cdot\hat{{\bf\it n}}_{i,jk})\hat{{\bf\it m}}_{i}\cdot(\hat{{\bf\it
m}}_{i}\crossproduct\hat{{\bf\it n}}_{i,lm})\\\
&\quad+\cos\varphi_{i}(1-\cos\varphi_{i})(\hat{{\bf\it
m}}_{i}\cdot\hat{{\bf\it n}}_{i,jk})(\hat{{\bf\it m}}_{i}\cdot\hat{{\bf\it
n}}_{i,lm})\\\ &\quad+\sin\varphi_{i}(1-\cos\varphi_{i})(\hat{{\bf\it
m}}_{i}\cdot\hat{{\bf\it n}}_{i,lm})\hat{{\bf\it m}}_{i}\cdot(\hat{{\bf\it
m}}_{i}\crossproduct\hat{{\bf\it n}}_{i,jk})\\\
&\quad+(1-\cos\varphi_{i})^{2}(\hat{{\bf\it m}}_{i}\cdot\hat{{\bf\it
n}}_{i,jk})(\hat{{\bf\it m}}_{i}\cdot\hat{{\bf\it n}}_{i,lm}).\end{split}$
(S5)

The second and fourth terms cancel each other out, and the sixth and eighth
terms are zero. Putting together the third, seventh, and ninth terms, we get
$(1-\cos^{2}\varphi_{i})(\hat{{\bf\it m}}_{i}\cdot\hat{{\bf\it
n}}_{i,jk})(\hat{{\bf\it m}}_{i}\cdot\hat{{\bf\it n}}_{i,lm})$. Therefore,

$\begin{split}\hat{{\bf\it n}}^{\prime}_{i,jk}\cdot\hat{{\bf\it
n}}^{\prime}_{i,lm}&=\cos^{2}\varphi_{i}\hat{{\bf\it
n}}_{i,jk}\cdot\hat{{\bf\it n}}_{i,lm}\\\
&\quad+(1-\cos^{2}\varphi_{i})(\hat{{\bf\it m}}_{i}\cdot\hat{{\bf\it
n}}_{i,jk})(\hat{{\bf\it m}}_{i}\cdot\hat{{\bf\it n}}_{i,lm})\\\
&\quad+\sin^{2}\varphi_{i}(\hat{{\bf\it m}}_{i}\crossproduct\hat{{\bf\it
n}}_{i,jk})\cdot(\hat{{\bf\it m}}_{i}\crossproduct\hat{{\bf\it
n}}_{i,lm}),\end{split}$ (S6)

and

$\begin{split}&(\hat{{\bf\it m}}_{i}\crossproduct\hat{{\bf\it
n}}_{i,jk})\cdot(\hat{{\bf\it m}}_{i}\crossproduct\hat{{\bf\it n}}_{i,lm})\\\
&=(\hat{{\bf\it m}}_{i}\cdot\hat{{\bf\it m}}_{i})(\hat{{\bf\it
n}}_{i,jk}\cdot\hat{{\bf\it n}}_{i,lm})-(\hat{{\bf\it m}}_{i}\cdot\hat{{\bf\it
n}}_{i,jk})(\hat{{\bf\it m}}_{i}\cdot\hat{{\bf\it n}}_{i,lm})\\\
&=\hat{{\bf\it n}}_{i,jk}\cdot\hat{{\bf\it n}}_{i,lm}-(\hat{{\bf\it
m}}_{i}\cdot\hat{{\bf\it n}}_{i,jk})(\hat{{\bf\it m}}_{i}\cdot\hat{{\bf\it
n}}_{i,lm}),\end{split}$ (S7)

and we obtain

$\hat{{\bf\it n}}^{\prime}_{i,jk}\cdot\hat{{\bf\it
n}}^{\prime}_{i,lm}=\hat{{\bf\it n}}_{i,jk}\cdot\hat{{\bf\it n}}_{i,lm}.$ (S8)

###  .2.2 How to construct a chiral operator

Let us explain how to construct the chiral operator when (i) $\Phi=\pi$ and
(ii) $\\{\hat{{\bf\it n}}_{i,jk}\\}_{i}$ is collinear or coplanar. First of
all, for rotation axes satisfying (ii), we can take a unit vector
$\hat{{\bf\it m}}_{i}$, which is perpendicular to all $\\{\hat{{\bf\it
n}}_{i,jk}\\}_{i}$ that belong to site-$i$. Then, we find
$V_{i}=\exp\left[-i\frac{\pi}{2}\hat{{\bf\it
m}}_{i}\cdot{\bf\it\sigma}\right]=-i\hat{{\bf\it m}}_{i}\cdot{\bf\it\sigma}$.
Since $\hat{{\bf\it m}}_{i}\cdot\hat{{\bf\it n}}_{i,jk}=0$,

$\begin{split}V_{i}P(C_{i,jk})V^{\dagger}_{i}&=(-i\hat{{\bf\it
m}}_{i}\cdot{\bf\it\sigma})(-i\hat{{\bf\it
n}}_{i,jk}\cdot{\bf\it\sigma})(i\hat{{\bf\it m}}_{i}\cdot{\bf\it\sigma})\\\
&=+i\hat{{\bf\it n}}_{i,jk}\cdot{\bf\it\sigma}=-P(C_{i,jk}).\end{split}$ (S9)

Next, to consistently generate the nearby $V_{j}\,(j\neq i)$ from $V_{i}$, the
form

$V_{j}=U_{jk}U_{ki}V_{i}U_{ik}U_{kj},\quad k\neq i,j,$ (S10)

gives the natural extension using the SU(2) gauge. This form immediately gives
$V_{i}U_{ij}V^{\dagger}_{j}=-U_{ij},\;\forall i,j$ e.g.,

$\begin{split}V_{i}U_{ij}V^{\dagger}_{j}&=V_{i}U_{ij}U_{jk}U_{ki}V^{\dagger}_{i}U_{ik}U_{kj}\\\
&=V_{i}P(C_{i,kj})V^{\dagger}_{i}U_{ik}U_{kj}\\\ &=-P(C_{i,kj})U_{ik}U_{kj}\\\
&=-U_{ij}U_{jk}U_{ki}U_{ik}U_{kj}\\\
&=-U_{ij},\rule{85.35826pt}{0.0pt}\end{split}$ (S11)

and

$\begin{split}&\quad V_{l}U_{lj}V^{\dagger}_{j}\\\
&=U_{lm}U_{mi}V_{i}U_{im}U_{ml}U_{lj}U_{jk}U_{ki}V^{\dagger}_{i}U_{ik}U_{kj}\\\
&=U_{lj}U_{ji}V_{i}P(C_{i,kj})V^{\dagger}_{i}U_{ik}U_{kj}\\\
&=-U_{lj}U_{ji}P(C_{i,kj})U_{ik}U_{kj}\\\
&=-U_{lj}U_{ji}U_{ij}U_{jk}U_{ki}U_{ik}U_{kj}\\\ &=-U_{lj}.\end{split}$ (S12)

In the second example, there appears a product of SU(2) gauge fields,
$U_{im}U_{ml}U_{lj}U_{jk}U_{ki}$, along the path $i\rightarrow k\rightarrow
j\rightarrow l\rightarrow m\rightarrow i$. Since the loops defined on the
pyrochlore and triangular lattice cannot return to the original site by five
independent bonds, two of the five bonds should be the same ones, which we
chose as $j$-$l$ and $l$-$m$, i.e. $U_{ml}U_{lj}=1$.

We show another construction of the chiral operator, applied to a more general
case including the parameter region off the chiral symmetric points. We take
$\hat{{\bf\it r}}_{Ci}$ as a unit vector that points from site-$i$ to the
center of triangle or tetrahedron $C$. When
$V_{i}=\exp\left[-i\frac{\pi}{2}\hat{{\bf\it
r}}_{Ci}\cdot{\bf\it\sigma}\right]$,

$V_{i}U_{ij}(\theta)V^{\dagger}_{j}=-U_{ij}(\alpha-\theta\pm 2\pi),$ (S13)

where $\alpha$ satisfy $\hat{{\bf\it r}}_{Ci}\cdot\hat{{\bf\it
r}}_{Cj}=\cos(\alpha/2)$, $\hat{{\bf\it r}}_{Ci}\crossproduct\hat{{\bf\it
r}}_{Cj}=-\sin(\alpha/2)\hat{{\bf\it\nu}}_{ij}$, and $0<\alpha<2\pi$. Because
the unitary operator $\tilde{\Gamma}=\oplus_{j=1}^{n}=iV_{j}$ satisfy

$\tilde{\Gamma}\mathcal{H}({\bf\it
k},\theta)\tilde{\Gamma}^{\dagger}=-\mathcal{H}({\bf\it k},\alpha-\theta\pm
2\pi),$ (S14)

the energy eigenvalues of $\mathcal{H}({\bf\it k},\theta)$ have opposite sign
and the same amplitude as those of $\mathcal{H}({\bf\it k},\alpha-\theta\pm
2\pi)$. Since the top bands are flat at $\theta=0$, flat bands appear at the
bottom when $\theta=\alpha\pm 2\pi$. If there exists $\theta_{0}$ that
fulfills $\theta_{0}=\alpha-\theta_{0}\pm 2\pi$, $\mathcal{H}({\bf\it
k},\theta_{0})$ has a chiral symmetry; it corresponds to
$\theta_{0}=\arccos(-1/2)-\pi$ for kagome lattice, and
$\theta_{0}=\arccos(-1/3)-\pi$ for pyrochlore lattice. In addition to that,
since $\theta_{0}$ satisfies

$\Phi(\theta_{0}+\varphi)=2\pi-\Phi(\theta_{0}-\varphi)$ (S15)

for arbitrary $\varphi$, then $\Phi(\theta)$ is symmetric about
$(\theta,\Phi)=(\theta_{0},\pi)$.

###  .2.3 Abelian and non-Abelian gauge fields

We use the terminology, “Abelian/non-Abelian” according to whether or not
there exists a gauge that makes the SU(2) gauge fields $U_{ij}$ defined on
different bonds commutative. This definition is more rigorous than the ones
used in the previous articles; in general, when the SU(2) gauge field appears
in the Hamiltonian, they simply call it “non-Abelian”Gao and Lan (2020), but
it can become Abelian with the aid of spatial symmetry, e.g. the site-centered
inversion symmetry of a kagome lattice gives rise to the U(1) symmetryKim and
Zang (2015). However, even if $U_{ij}$ is “non-Abelian”, namely noncommutative
in some certain choices of the gauge, if there is at least one way of taking
the gauge that makes $U_{ij}$ commutative, the features linked to the
noncommutativity do not appear in gauge-invariant physical quantities such as
band structure or topological numbers. Since we are interested in the
properties related to the non-commutativity of the gauge field, these
properties can in turn enable us to judge whether the gauge is non-Abelian or
not.

Let us show here that $\Phi$ is such property; If the gauge field is Abelian
in our sense, one can simultaneously invert the rotation axis
$\hat{{\bf\it\nu}}_{ij}$ of all $U_{ij}$ by a local gauge transformation,
which is equivalent to taking $\theta\rightarrow-\theta$. Then, since $\Phi$
is a gauge invariant quantity, one can say that if the gauge field is Abelian,
$\Phi(\theta)=\Phi(-\theta)$ should hold. Conversely, if
$\Phi(\theta)\neq\Phi(-\theta)$, the gauge field is non-Abelian. This
condition works in Fig. 1(e) in the main text to find immediately that the
SU(2) gauge field of the pyrochlore lattice is non-Abelian since
$\Phi(\theta)\neq\Phi(-\theta)$.

##  .3 Symmetry analysis

In this section, we clarify several aspects of underlying symmetries of the
energy eigenstates at chiral symmetric points. First, we show that the U(1)
symmetry in spin space allows us to block-diagonalize the Bloch Hamiltonian
when it is collinear ($\theta=\theta_{1}$ in the kagome lattice and
$\theta_{2}$ in the pyrochlore lattice), and that the diagonal block is
described by a sublattice pseudospin due to chiral symmetry. Partially making
use of the sublattice pseudospin picture, we next show that the chiral zero
modes in the pyrochlore lattice are protected by the chiral symmetry and the
spatial symmetry and appear at a specific point or line in momentum space,
where the bands become degenerate.

###  .3.1 sublattice pseudospin

Let us consider the collinear case, i.e., when the Wilson loop operators
starting from site-$i$ represent the rotation around the same axis independent
of their paths. Note that the rotation axis $\hat{{\bf\it n}}_{i,jk}$ is
different at the different base site $i$. Considering the local gauge
transformation $V_{i}$ represented by the rotation around its axis, we find
that $V=\oplus_{i=1}^{n}V_{i}$ is commutative with the Bloch Hamiltonian. It
is confirmed in the same way as when we construct the chiral operator. In the
basis that diagonalizes $V$, the sublattices $i=1,2,\ldots,n$ remain unchanged
from the original basis. The Bloch Hamiltonian $\mathcal{H}({\bf\it k})$ is
therefore block-diagonalized and expressed as $\mathrm{diag}(D({\bf\it
k}),D^{*}({\bf\it k}))$ due to time reversal symmetry and site-centered
inversion symmetry; $D({\bf\it k})$ is a $n\times n$ Hermitian matrix, and
serves as a Hamiltonian about the sublattice degrees of freedom. We
investigate the constraints on $D({\bf\it k})$ imposed by the chiral symmetry
for the kagome lattice ($n=3$) and for the pyrochlore lattice ($n=4$).

We first consider the kagome lattice, whose $D({\bf\it k})$ is expanded as

$D({\bf\it k})=\sum_{i=1}^{8}R_{i}({\bf\it k})\tau_{i},$ (S16)

where $(\tau_{1},\tau_{2},\ldots,\tau_{8})$ are Gell-Mann matrices:

$\begin{split}&\tau_{1}=\matrixquantity(0\hfil&1&0\\\ 1&0&0\\\
0&0&0),\quad\tau_{2}=\matrixquantity(0&-i&0\\\ i&0&0\\\
0&0&0),\quad\tau_{3}=\matrixquantity(1&0&0\\\ 0&-1&0\\\ 0&0&0),\\\
&\tau_{4}=\matrixquantity(0\hfil&0&1\\\ 0&0&0\\\
1&0&0),\quad\tau_{5}=\matrixquantity(0&0&-i\\\ 0&0&0\\\
i&0&0),\quad\tau_{6}=\matrixquantity(0&0&0\\\ 0&0&1\\\ 0&1&0),\\\
&\tau_{7}=\matrixquantity(0\hfil&0&0\\\ 0&0&-i\\\
0&i&0),\quad\tau_{8}=\frac{1}{\sqrt{3}}\matrixquantity(1&0&0\\\ 0&1&0\\\
0&0&-2).\end{split}$ (S17)

Since the electron hops only between different sublattices, diagonal element
of $\tilde{\mathcal{H}}({\bf\it k})$ is zero and $R_{3}({\bf\it
k})=R_{8}({\bf\it k})=0$.

In the above given basis, the chiral operator $\Gamma$ is expressed as

$\Gamma=\matrixquantity(&\gamma\\\ \gamma^{*}&),\quad\gamma={\rm
diag}(z^{*},i,-z),\quad z=e^{i\frac{\pi}{6}}.$ (S18)

Since

$\Gamma\mathcal{H}({\bf\it k})\Gamma^{\dagger}=\matrixquantity(\gamma
D({\bf\it k})\gamma^{*}&\\\ &\gamma^{*}D({\bf\it k})\gamma),$ (S19)

the chiral symmetry, $\\{\mathcal{H}({\bf\it k}),\Gamma\\}=0$, requires
$\gamma^{*}D({\bf\it k})\gamma=-D^{*}({\bf\it k})$. Here,

$\begin{split}\gamma^{*}\tau_{1}\gamma&=-\frac{1}{2}\tau_{1}-\frac{\sqrt{3}}{2}\tau_{2},\quad\gamma^{*}\tau_{2}\gamma=\frac{\sqrt{3}}{2}\tau_{1}-\frac{1}{2}\tau_{2},\\\
\gamma^{*}\tau_{4}\gamma&=-\frac{1}{2}\tau_{4}+\frac{\sqrt{3}}{2}\tau_{5},\quad\gamma^{*}\tau_{5}\gamma=-\frac{\sqrt{3}}{2}\tau_{4}-\frac{1}{2}\tau_{5},\\\
\gamma^{*}\tau_{6}\gamma&=-\frac{1}{2}\tau_{6}-\frac{\sqrt{3}}{2}\tau_{7},\quad\gamma^{*}\tau_{7}\gamma=\frac{\sqrt{3}}{2}\tau_{6}-\frac{1}{2}\tau_{7},\end{split}$
(S20)

and then,

$\begin{split}&\gamma^{*}D({\bf\it k})\gamma\\\
&=\big{(}-\frac{1}{2}R_{1}({\bf\it k})+\frac{\sqrt{3}}{2}R_{2}({\bf\it
k})\big{)}\tau_{1}-\big{(}\frac{\sqrt{3}}{2}R_{1}({\bf\it
k})+\frac{1}{2}R_{2}({\bf\it k})\big{)}\tau_{2}\\\
&\quad-\big{(}\frac{1}{2}R_{4}({\bf\it k})+\frac{\sqrt{3}}{2}R_{5}({\bf\it
k})\big{)}\tau_{4}+\big{(}\frac{\sqrt{3}}{2}R_{4}({\bf\it
k})-\frac{1}{2}R_{5}({\bf\it k})\big{)}\tau_{5}\\\
&\quad-\big{(}\frac{1}{2}R_{6}({\bf\it k})-\frac{\sqrt{3}}{2}R_{7}({\bf\it
k})\big{)}\tau_{6}-\big{(}\frac{\sqrt{3}}{2}R_{6}({\bf\it
k})+\frac{1}{2}R_{7}({\bf\it k})\big{)}\tau_{7}.\\\ \end{split}$ (S21)

On the other hand, since

$\begin{split}-D^{*}({\bf\it k})&=-R_{1}({\bf\it k})\tau_{1}+R_{2}({\bf\it
k})\tau_{2}-R_{4}({\bf\it k})\tau_{4}+R_{5}({\bf\it k})\tau_{5}\\\ &\quad-
R_{6}({\bf\it k})\tau_{6}+R_{7}({\bf\it k})\tau_{7},\end{split}$ (S22)

we obtain three constraints

$\begin{split}&R_{1}({\bf\it k})+\sqrt{3}R_{2}({\bf\it k})=0,\quad
R_{4}({\bf\it k})-\sqrt{3}R_{5}({\bf\it k})=0,\\\ &R_{6}({\bf\it
k})+\sqrt{3}R_{7}({\bf\it k})=0.\end{split}$ (S23)

Using these constraints, we can reduce the number of basis to three, and by
defining $\tilde{{\bf\it R}}({\bf\it k})$ and ${\bf\it S}$ as

$\begin{split}\tilde{R}_{1}({\bf\it k})&=\frac{\sqrt{3}}{2}R_{6}({\bf\it
k})-\frac{1}{2}R_{7}({\bf\it k}),\\\ \tilde{R}_{2}({\bf\it
k})&=\frac{\sqrt{3}}{2}R_{4}({\bf\it k})+\frac{1}{2}R_{5}({\bf\it k}),\\\
\tilde{R}_{3}({\bf\it k})&=\frac{\sqrt{3}}{2}R_{1}({\bf\it
k})-\frac{1}{2}R_{2}({\bf\it k}),\end{split}$ (S24)
$\begin{split}S_{x}&=\frac{\sqrt{3}}{2}\tau_{6}-\frac{1}{2}\tau_{7}=\matrixquantity(0\hfil&0&0\\\
0&0&z\\\ 0&z^{*}&0),\\\
S_{y}&=\frac{\sqrt{3}}{2}\tau_{4}+\frac{1}{2}\tau_{5}=\matrixquantity(0\hfil&0&z^{*}\\\
0&0&0\\\ z&0&0),\\\
S_{z}&=\frac{\sqrt{3}}{2}\tau_{1}-\frac{1}{2}\tau_{2}=\matrixquantity(0\hfil&z&0\\\
z^{*}&0&0\\\ 0&0&0),\end{split}$ (S25)

$D({\bf\it k})$ is expressed as

$D({\bf\it k})=\sum_{i=1}^{3}\tilde{R}_{i}({\bf\it k})S_{i}.$ (S26)

From Eq.(S25), we find that ${\bf\it S}$ fulfills the commutation relation,

$[S_{i},S_{j}]=i\varepsilon_{ijk}S_{k},\quad{\bf\it S}^{2}=2I,$ (S27)

meaning that ${\bf\it S}$ serves as a pseudospin-1 operator of the sublattice
degrees of freedom.

Next, we consider the pyrochlore lattice. In general, the Bloch Hamiltonian
for pyrochlore lattice cannot be block-diagonalized because of its non-Abelian
property. However, at $\theta=\theta_{2}=2\arctan(\sqrt{2})$, which is the
collinear case, the system recovers the Abelian property and
$\mathcal{H}({\bf\it k})$ can be block-diagonalized into
$\mathrm{diag}(D({\bf\it k}),D^{*}({\bf\it k}))$. Here, $D({\bf\it k})$ is
expanded as

$D({\bf\it k})=\sum_{i=1}^{15}R_{i}({\bf\it k})\Omega_{i},$ (S28)

where $(\Omega_{1},\Omega_{2},\ldots,\Omega_{15})$ are linearly independent
traceless matrices:

$\begin{split}&\Omega_{1}=\sigma_{0}\otimes\sigma_{x},\quad\Omega_{2}=\sigma_{0}\otimes\sigma_{y},\quad\Omega_{3}=\sigma_{0}\otimes\sigma_{z},\\\
&\Omega_{4}=\sigma_{x}\otimes\sigma_{x},\quad\Omega_{5}=\sigma_{x}\otimes\sigma_{y},\quad\Omega_{6}=\sigma_{x}\otimes\sigma_{z},\\\
&\Omega_{7}=\sigma_{y}\otimes\sigma_{x},\quad\Omega_{8}=\sigma_{y}\otimes\sigma_{y},\quad\Omega_{9}=\sigma_{y}\otimes\sigma_{z},\\\
&\Omega_{10}=\sigma_{z}\otimes\sigma_{x},\quad\Omega_{11}=\sigma_{z}\otimes\sigma_{y},\quad\Omega_{12}=\sigma_{z}\otimes\sigma_{z},\\\
&\Omega_{13}=\sigma_{x}\otimes\sigma_{0},\quad\Omega_{14}=\sigma_{y}\otimes\sigma_{0},\quad\Omega_{15}=\sigma_{z}\otimes\sigma_{0}.\end{split}$
(S29)

As in the case of the kagome lattice, we set the coefficients as
$R_{7}({\bf\it k})=R_{12}({\bf\it k})=R_{14}({\bf\it k})=0$, whose
corresponding $\Omega_{i}$ have diagonal elements. The chiral operator is
expressed as

$\Gamma=\matrixquantity(&\gamma\\\ \gamma^{*}&),\quad\gamma={\rm
diag}(z^{10},z^{10},z^{4},z^{4}),\quad z=e^{i\frac{\pi}{6}}.$ (S30)

The constraints arising from the chiral symmetry are derived in the same way
as we did for the kagome lattice, and we find that

$\gamma^{*}\Omega_{i}\gamma=\begin{cases}+\Omega_{i}&i=1,2,10,11\\\
-\Omega_{i}&i=4,5,6,7,8,9,13,14,\end{cases}$ (S31)

and from $\gamma^{*}D({\bf\it k})\gamma=-D^{*}({\bf\it k})$, we obtain

$\begin{split}&R_{1}({\bf\it k})=R_{5}({\bf\it k})=R_{7}({\bf\it
k})=R_{9}({\bf\it k})=R_{10}({\bf\it k})=R_{14}({\bf\it k})=0,\\\ &{\bf\it
S}_{1}=(S_{1,x},S_{1,y},S_{1,z})=(-\Omega_{2},-\Omega_{6},\Omega_{4})/2,\\\
&{\bf\it
S}_{2}=(S_{2,x},S_{2,y},S_{2,z})=(-\Omega_{11},-\Omega_{13},\Omega_{8})/2,\end{split}$
(S32)

where ${\bf\it S}_{1}$ and ${\bf\it S}_{2}$ are the pseudospin-$\frac{1}{2}$
operators, which satisfy,

$[S_{a,i},S_{b,j}]=i\delta_{ab}\varepsilon_{ijk}S_{a,k},\quad{\bf\it
S}_{a}^{2}=\frac{3}{4}I.$ (S33)

Figure S 2:  The symmetry operations. (a) $C_{2}$ is a two-fold rotation,
which interchange the upward and downward tetrahedra. (b)
$s_{4\mu}\,(\mu=x,y,z)$ is a rotoreflection which is a combination of a
$\pi/2$-rotation around the axis parallel to the $\mu$-axis through the center
of the tetrahedron, and a mirror reflection across the plane perpendicular to
that axis. (c) $C_{3}$ is a three-fold rotation, which fix one vertex of the
tetrahedron and replace the other three vertices. The quantization axes
$\hat{{\bf\it m}}_{j}$, which diagonalize the chiral operator, are invariant
under this rotation.

###  .3.2 chiral zero modes

We show the existence of the chiral zero modes protected by the chiral
symmetry and the spatial symmetry. First, let us discuss the point contact at
L and W points at $\theta=\theta_{2}=2\arctan(\sqrt{2})$. As discussed above,
the Bloch Hamiltonian is block-diagonalized into $\mathrm{diag}(D({\bf\it
k}),D^{*}({\bf\it k}))$ and is expanded as

$\begin{split}D({\bf\it k})&=R_{2}({\bf\it k})\Omega_{2}+R_{4}({\bf\it
k})\Omega_{4}+R_{6}({\bf\it k})\Omega_{6}\\\ &\quad+R_{8}({\bf\it
k})\Omega_{8}+R_{11}({\bf\it k})\Omega_{11}+R_{13}({\bf\it
k})\Omega_{13}.\end{split}$ (S34)

We now analyze the symmetry of the pyrochlore lattice. Let us set the axis
perpendicular to the direction connecting one of the vertices and the center
of the tetrahedron. This axis is parallel to one of the edges of the triangle
formed by the rest of the vertices. The pyrochlore lattice is invariant under
the $\pi$-rotations ($C_{2}$) about that axis (Fig.S2 (a)). The matrix
representation of this operation for the basis that block-diagonalizes
$\mathcal{H}({\bf\it k})$ is

$\begin{split}C_{2}\;\;&=\matrixquantity(\hfil&\tilde{C}_{2,1}\\\
\tilde{C}_{2,2}&),\\\ \tilde{C}_{2,1}&=\matrixquantity(e^{\pi i/3}\hfil&&&\\\
&&e^{5\pi i/6}&\\\ &e^{5\pi i/6}&&\\\ &&&e^{4\pi i/3}),\\\
\tilde{C}_{2,2}&=\matrixquantity(e^{2\pi i/3}\hfil&&&\\\ &&e^{\pi i/6}&\\\
&e^{\pi i/6}&&\\\ &&&e^{5\pi i/3}).\end{split}$ (S35)

Now, the L points in momentum space are invariant under $C_{2}$, and the Bloch
Hamiltonian $\mathcal{H}({\bf\it k})$ commutes with $C_{2}$ as,

$\begin{split}\mathcal{H}({\bf\it k})&=C_{2}\mathcal{H}({\bf\it
k})C^{\dagger}_{2}\\\ &=\matrixquantity(\hfil&\tilde{C}_{2,1}\\\
\tilde{C}_{2,2}&)\matrixquantity(D({\bf\it k})&\\\ &D^{*}({\bf\it
k}))\matrixquantity(&\tilde{C}^{\dagger}_{2,2}\\\
\tilde{C}^{\dagger}_{2,1}&)\\\ &=\matrixquantity(\tilde{C}_{2,1}D^{*}({\bf\it
k})\tilde{C}^{\dagger}_{2,1}&\\\ &\tilde{C}_{2,2}D({\bf\it
k})\tilde{C}^{\dagger}_{2,2}).\end{split}$ (S36)

Then, we find

$D({\bf\it k})=\tilde{C}_{2,1}D^{*}({\bf\it k})\tilde{C}^{\dagger}_{2,1},\quad
D^{*}({\bf\it k})=\tilde{C}_{2,2}D({\bf\it k})\tilde{C}^{\dagger}_{2,2},$
(S37)

which lead to

$\begin{split}&\tilde{C}_{2,1}\Omega^{*}_{2}\tilde{C}^{\dagger}_{2,1}=\Omega_{13},\quad\tilde{C}_{2,1}\Omega^{*}_{13}\tilde{C}^{\dagger}_{2,1}=\Omega_{2},\\\
&\tilde{C}_{2,1}\Omega^{*}_{4}\tilde{C}^{\dagger}_{2,1}=\Omega_{8},\quad\tilde{C}_{2,1}\Omega^{*}_{8}\tilde{C}^{\dagger}_{2,1}=\Omega_{4},\\\
&\tilde{C}_{2,1}\Omega^{*}_{6}\tilde{C}^{\dagger}_{2,1}=\Omega_{11},\quad\tilde{C}_{2,1}\Omega^{*}_{11}\tilde{C}^{\dagger}_{2,1}=\Omega_{6}.\end{split}$
(S38)

We thus find, $R_{2}({\bf\it k})=R_{13}({\bf\it k})$, $R_{4}({\bf\it
k})=R_{8}({\bf\it k})$ and $R_{6}({\bf\it k})=R_{11}({\bf\it k})$.

The same analysis is performed for the other $C_{2}$ axes, and we obtain
$R_{2}({\bf\it k})=-R_{4}({\bf\it k})=R_{6}({\bf\it k})=-R_{8}({\bf\it
k})=R_{11}({\bf\it k})=R_{13}({\bf\it k})(\equiv R({\bf\it k}))$. From these
relationships, the block-diagonal Hamiltonian at L points is given as

$\begin{split}D({\bf\it k})&=R({\bf\it
k})(\Omega_{2}-\Omega_{4}+\Omega_{6}-\Omega_{8}+\Omega_{11}+\Omega_{13})\\\
&=-2{\bf\it R}({\bf\it k})\cdot({\bf\it S}_{1}+{\bf\it S}_{2}),\end{split}$
(S39)

where ${\bf\it R}({\bf\it k})=R({\bf\it k})(1,1,1)$, and its eigenvalues are
$\pm 2|{\bf\it R}({\bf\it k})|$, 0 (doubly degenerate).

The next symmetry operation we consider is the rotoreflection
($s_{4\mu}\,(\mu=x,y,z)$), the $\pi/2$-rotation about the axis parallel to the
$\mu$-axis through the center of the tetrahedron combined with the mirror
operation about the plane perpendicular to that axis (Fig.S2 (b)). Let us
consider $\mu=x$ and this operation is represented by

$\begin{split}s_{4x}&=\matrixquantity(\hfil&\tilde{s}_{4x,1}\\\
\tilde{s}_{4x,2}&),\\\ \tilde{s}_{4x,1}&=\matrixquantity(\hfil&e^{23\pi
i/12}&&\\\ &&e^{17\pi i/12}&\\\ &&&e^{11\pi i/12}\\\ e^{17\pi i/12}&&&),\\\
\tilde{s}_{4x,2}&=\matrixquantity(\hfil&e^{19\pi i/12}&&\\\ &&e^{\pi i/12}&\\\
&&&e^{7\pi i/12}\\\ e^{\pi i/12}&&&).\end{split}$ (S40)

Now, the W points in momentum space are invariant under this operation and,
similar to the above discussion at L points, we find

$D({\bf\it k})=\tilde{s}_{4x,1}D^{*}({\bf\it
k})\tilde{s}^{\dagger}_{4x,1},\quad D^{*}({\bf\it
k})=\tilde{s}_{4x,2}D({\bf\it k})\tilde{s}^{\dagger}_{4x,2},$ (S41)

which lead to

$\begin{split}&\tilde{s}_{4x,1}\Omega^{*}_{2}\tilde{s}^{\dagger}_{4x,1}=-\Omega_{8},\quad\tilde{s}_{4x,1}\Omega^{*}_{8}\tilde{s}^{\dagger}_{4x,1}=-\Omega_{2},\\\
&\tilde{s}_{4x,1}\Omega^{*}_{4}\tilde{s}^{\dagger}_{4x,1}=-\Omega_{11},\quad\tilde{s}_{4x,1}\Omega^{*}_{11}\tilde{s}^{\dagger}_{4x,1}=\Omega_{4},\\\
&\tilde{s}_{4x,1}\Omega^{*}_{6}\tilde{s}^{\dagger}_{4x,1}=\Omega_{13},\quad\tilde{s}_{4x,1}\Omega^{*}_{13}\tilde{s}^{\dagger}_{4x,1}=-\Omega_{6}.\end{split}$
(S42)

We thus find, $R_{2}({\bf\it k})=-R_{8}({\bf\it k})(\equiv R({\bf\it k}))$,
$R_{4}({\bf\it k})=R_{6}({\bf\it k})=R_{11}({\bf\it k})=R_{13}({\bf\it k})=0$.
From these relationships, the block-diagonal Hamiltonian at W points is given
as

$D({\bf\it k})=R({\bf\it k})(\Omega_{2}-\Omega_{8})=-2R({\bf\it
k})(S_{1,x}+S_{2,z}).$ (S43)

Since $[S_{1,x},S_{2,z}]=0$, its eigenvalues are $\pm 2|R({\bf\it k})|$, 0
(doubly degenerate).

Finally, we consider the nodal $\mathrm{\Gamma}$-L line at
$\theta=\theta_{3}=-2\arctan(1/\sqrt{2})$. Since it is the coplanar case, the
block-diagonalization of ${\mathcal{H}}(\bm{k})$ is unavailable, but by
combining the chiral symmetry and the spatial symmetry, we can show that there
is a band touching at zero energy Koshino  _et al._ (2014). Here, we consider
the $2\pi/3$-rotation ($C_{3}$) about an axis that connects one vertex and the
center of the tetrahedron. The matrix representation of this operation for the
original basis is

$C_{3}=\tilde{C}_{3}\otimes\matrixquantity(&1&&\\\ &&1&\\\ 1&&&\\\ &&&1),$
(S44)

where $\tilde{C}_{3}=\exp\left[-i\frac{\pi}{3}\hat{{\bf\it
n}}\cdot{\bf\it\sigma}\right]$, $\hat{{\bf\it n}}=(1,1,1)/\sqrt{3}$. In the
basis that diagonalizes the chiral operator $\Gamma$, the quantization axis is
$\hat{{\bf\it m}}_{j}$ at site $j$. At $\theta=-2\arctan(1/\sqrt{2})$, it
points toward the center of the tetrahedron. Therefore, this quantization axis
is invariant under the $C_{3}$ operation, and $C_{3}$ commutes with $\Gamma$
(Fig.S2 (c)). Since the $\mathrm{\Gamma}$-L line is invariant under $C_{3}$,
$\mathcal{H}({\bf\it k})$ commutes with $C_{3}$.

Both $\Gamma$ and $\mathcal{H}({\bf\it k})$ that commute with $C_{3}$ are
block-diagonalized into the eigenspace of $C_{3}$ with eigenvalues being $-1$,
$\omega_{1}=e^{\pi i/3}$, $\omega_{2}=e^{5\pi i/3}$. They are,
$\Gamma=\mathrm{diag}(\Gamma_{-1},\Gamma_{\omega_{1}},\Gamma_{\omega_{2}})$
and $\mathcal{H}({\bf\it k})=\mathrm{diag}(\mathcal{H}_{-1}({\bf\it
k}),\mathcal{H}_{\omega_{1}}({\bf\it k}),\mathcal{H}_{\omega_{2}}({\bf\it
k}))$, which are anticommutative in their respective blocks as

$\\{\Gamma_{a},\mathcal{H}_{a}({\bf\it k})\\}=0,\quad
a=-1,\omega_{1},\omega_{2}.$ (S45)

$\nu_{a}=\Tr\Gamma_{a}$ denotes the difference in the number of zero modes,
$N_{a,\pm}$, with positive and negative chirality (eigenvalues of $\Gamma_{a}$
is $\pm 1$); $\nu_{a}=N_{a,+}-N_{a,-}$. Since $\nu_{-1}=0$,
$\nu_{\omega_{1}}=1$ and $\nu_{\omega_{2}}=-1$, there are at least two chiral
zero modes. Since this argument holds for ${\bf\it k}$-points on the
$\mathrm{\Gamma}$-L line, we find that a nodal line appears on the
$\mathrm{\Gamma}$-L line and that it is protected by the chiral symmetry and
the three-fold rotation symmetry.

##  .4 Details of perturbation Hamiltonian

We have discussed the effect of small perturbation to Eq.(3) in the main text,
and summarized the results in Fig.3 and Table I. We explain the details of
these perturbations.

###  .4.1 uniform magnetic field

We take account of the uniform magnetic field as a Zeeman effect as

$\mathcal{H}_{\rm mf}=-\sum_{j}{\bf\it h}\cdot{\bf\it s}_{j},$ (S46)

where $s_{j,\mu}={\bf\it c}^{\dagger}_{j}(\sigma_{\mu}/2){\bf\it
c}_{j}\,(\mu=x,y,z)$ is an electron spin operator at site-$j$. This magnetic
field breaks both TRS and PHS, but may or may not break CS. We express the
Zeeman term at site-$j$ as $-{\bf\it
h}\cdot{\bf\it\sigma}=-i\exp\left[-i\frac{\pi}{2}{\bf\it
h}\cdot{\bf\it\sigma}\right]$, and if ${\bf\it h}$ is perpendicular to the
quantization axis $\hat{{\bf\it m}}_{j}$ diagonalizing the chiral operator,
then $V_{j}(-{\bf\it h}\cdot{\bf\it\sigma}_{j})V^{\dagger}_{j}=+{\bf\it
h}\cdot{\bf\it\sigma}$ holds. When ${\bf\it h}\cdot\hat{{\bf\it m}}_{j}=0$ for
each site, TRS and PHS are violated, but CS is preserved.

###  .4.2 on-site potential

We set different values of on-site potentials for $n$-independent sites in the
unit cell

$\mathcal{H}_{\rm op}=\sum_{l=1}^{n}\sum_{j\in l}\omega_{l}n_{j},$ (S47)

where $n_{j}={\bf\it c}^{\dagger}_{j}{\bf\it c}_{j}$ is an electron number
operator at site $j$. This on-site potential breaks the PHS but not the TRS,
and thus the CS is lost. This can also be interpreted as follows: the term at
site-$j$ is expressed as $\omega_{j}I$, and then this term is shown to be
invariant under the local gauge transformation as
$V_{j}(\omega_{j}I)V^{\dagger}_{j}=\omega_{j}I\neq-\omega_{j}I$.

Figure S 3:  The band structure at $\theta_{2}$ for the pyrochlore lattice
when bond modulation is $\delta_{ij}=1$ for all downward tetrahedra. The band
degeneracy is not lifted even though the spatial inversion symmetry is broken.

###  .4.3 bond modulation

The effect of bond modulation influences the hopping amplitude $t_{0},\lambda$
in a nontrivial manner, because the crystal field is varied. However, for
simplicity, we consider the simplest modulation called breathing: among the
two types of tetrahedra pointing up and down, only one of them is subject to
the modulation as $t^{\rm up}_{ij}=t$, $t^{\rm down}_{ij}=\delta_{ij}t$. This
does not modify the SU(2) gauge field $U_{ij}$, which preserves the chiral
symmetry. Moreover, since the TRS is not broken, the PHS is also preserved,
and the class to which the Bloch Hamiltonian belongs remains unchanged.

This bond modulation breaks the spatial inversion symmetry of the lattice,
which lifts the band-degeneracy in general. When $\delta_{12}=\delta_{34}$,
$\delta_{13}=\delta_{24}$ and $\delta_{14}=\delta_{23}$, even though the space
inversion symmetry is broken, it is not reflected in the band structure at
$\theta_{2}$; the band degeneracy is not lifted (Fig.S3).

## References

  * Gao and Lan (2020) Z. Gao and Z. Lan, Phys. Rev. B 102, 245133 (2020).
  * Kim and Zang (2015) S. K. Kim and J. Zang, Phys. Rev. B 92, 205106 (2015).
  * Koshino  _et al._ (2014) M. Koshino, T. Morimoto, and M. Sato, Phys. Rev. B 90, 115207 (2014).