eigenvalue}-e^{i\omega_{0}}&:\;\;\;\phi_{a0}^{\prime}\otimes d_{+1}\text{ and
}\phi_{b+}\otimes d_{0},\\\ \text{Group 3,
eigenvalue}-e^{-i\omega_{0}}&:\;\;\;\phi_{a0}^{\prime}\otimes d_{-1}\text{ and
}\phi_{b+}\otimes d_{+1}.\end{split}$ (168)

Therefore, any linear combination of those two eigen-states belonging to the
same group will not change the matrix representation of $C_{3}$ in Eq. (167).
By choosing specific transformations, the 6-by-6 matrix representation of
$S_{2x}$ in Eq. (167) can be reduced into a direct sum of three 2-by-2
matrices. To show that, we define the following new basis

$\displaystyle\begin{cases}\Phi_{t_{2g}}^{0,1}&=i\frac{1}{\sqrt{3}}e^{-i\omega_{0}}\phi_{a0}^{\prime}\otimes
d_{0}+\sqrt{\frac{2}{3}}e^{i\omega_{0}}\phi_{b+}\otimes d_{-1},\\\
\Phi_{t_{2g}}^{0,2}&=-i\sqrt{\frac{2}{3}}e^{i\omega_{0}}\phi_{a0}^{\prime}\otimes
d_{0}+\frac{1}{\sqrt{3}}\phi_{b+}\otimes d_{-1},\\\ \end{cases}$ (169)
$\displaystyle\begin{cases}\Phi_{t_{2g}}^{+,1}&=-i\sqrt{\frac{2}{3}}e^{i\omega_{0}}\phi_{a0}^{\prime}\otimes
d_{+1}+\frac{1}{\sqrt{3}}\phi_{b+}\otimes d_{0},\\\
\Phi_{t_{2g}}^{+,2}&=\frac{1}{\sqrt{3}}\phi_{a0}^{\prime}\otimes
d_{+1}-i\sqrt{\frac{2}{3}}e^{-i\omega_{0}}\phi_{b+}\otimes d_{0},\\\
\end{cases}$ (170)
$\displaystyle\begin{cases}\Phi_{t_{2g}}^{-,1}&=i\frac{1}{\sqrt{3}}e^{i\omega_{0}}\phi_{a0}^{\prime}\otimes
d_{-1}+\sqrt{\frac{2}{3}}\phi_{b+}\otimes d_{+1},\\\
\Phi_{t_{2g}}^{-,2}&=\sqrt{\frac{2}{3}}\phi_{a0}^{\prime}\otimes
d_{-1}+i\frac{1}{\sqrt{3}}e^{-i\omega_{0}}\phi_{b+}\otimes d_{+1},\\\
\end{cases}$ (171)

where the indices $0,+,-$ labels the eigenvalues
$1,e^{i\omega_{0}},e^{-i\omega_{0}}$ of the matrix $C_{3}$, and the indices
$1,2$ label the number of wave functions. Within the new basis

$\displaystyle\\{\Phi_{t_{2g}}^{0,1},\;\Phi_{t_{2g}}^{+,1},\;\Phi_{t_{2g}}^{0,2},\;\Phi_{t_{2g}}^{-,1},\;\Phi_{t_{2g}}^{+,2},\;\Phi_{t_{2g}}^{-,2}\\},$
(172)

the 6-by-6 matrix representation of $S_{2x}$ in Eq. (167) becomes

$\displaystyle S_{2x}$ $\displaystyle=\begin{pmatrix}S_{2x,1}&0&0\\\
0&S_{2x,2}&0\\\ 0&0&S_{2x,3}\end{pmatrix},$ (173)

where

$\displaystyle
S_{2x,1}=\begin{pmatrix}-\frac{i}{\sqrt{3}}&-\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{6}}\\\
\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{6}}&\frac{i}{\sqrt{3}}\end{pmatrix},S_{2x,2}$
$\displaystyle=\begin{pmatrix}\frac{i}{\sqrt{3}}&-\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{6}}\\\
\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{6}}&-\frac{i}{\sqrt{3}}\end{pmatrix},S_{2x,3}=\begin{pmatrix}-\frac{i}{\sqrt{3}}&-\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{6}}\\\
\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{6}}&\frac{i}{\sqrt{3}}\end{pmatrix}.$ (174)

Therefore, the three 2D Irreps generated by $C_{3}$ and $S_{2x}$ are

$\displaystyle A-\text{type:
}\\{\Phi_{t_{2g}}^{0,1},\;\Phi_{t_{2g}}^{+,1}\\}\text{ and
}\\{\Phi_{t_{2g}}^{0,2},\;\Phi_{t_{2g}}^{-,1}\\},$ (175a) $\displaystyle
B-\text{type: }\\{\Phi_{t_{2g}}^{+,2},\;\Phi_{t_{2g}}^{-,2}\\}.$ (175b)

Moreover, the time-reversal symmetry leads to the 4D Irrep,

$\displaystyle A_{2}-\text{Irrep:
}\\{\Phi_{t_{2g}}^{0,1},\;\Phi_{t_{2g}}^{+,1}\\}\,\oplus\\{\Phi_{t_{2g}}^{0,1},\;\Phi_{t_{2g}}^{+,1}\\}^{\ast},$
(176a) $\displaystyle A_{3}-\text{Irrep:
}\\{\Phi_{t_{2g}}^{0,2},\;\Phi_{t_{2g}}^{-,1}\\}\,\oplus\\{\Phi_{t_{2g}}^{0,2},\;\Phi_{t_{2g}}^{-,1}\\}^{\ast},$
(176b) $\displaystyle B_{2}-\text{Irrep:
}\\{\Phi_{t_{2g}}^{+,2},\;\Phi_{t_{2g}}^{-,2}\\}\oplus\\{\Phi_{t_{2g}}^{+,2},\;\Phi_{t_{2g}}^{-,2}\\}^{\ast}.$
(176c)

We further check the matrix representations for $C_{3}$, $S_{2x}$, $S_{2y}$
and $\mathcal{T}$ for both the A2 Irrep and A3 Irrep as

$\displaystyle C_{3}$ $\displaystyle=\left(\begin{array}[]{cccc}1&0&0&0\\\
0&\exp\left(i\omega_{0}\right)&0&0\\\ 0&0&1&0\\\
0&0&0&\exp\left(-i\omega_{0}\right)\\\
\end{array}\right),\;\;\;\;\;\;\;\;S_{2x}=\left(\begin{array}[]{cccc}-\tfrac{i}{\sqrt{3}}&-\sqrt{\tfrac{2}{3}}&0&0\\\
\sqrt{\tfrac{2}{3}}&\tfrac{i}{\sqrt{3}}&0&0\\\
0&0&\tfrac{i}{\sqrt{3}}&-\sqrt{\tfrac{2}{3}}\\\
0&0&\sqrt{\tfrac{2}{3}}&-\tfrac{i}{\sqrt{3}}\\\ \end{array}\right),$ (185)
$\displaystyle S_{2y}$
$\displaystyle=\left(\begin{array}[]{cccc}-\frac{i}{\sqrt{3}}&\frac{1-i\sqrt{3}}{\sqrt{6}}&0&0\\\
\frac{-1-i\sqrt{3}}{\sqrt{6}}&\frac{i}{\sqrt{3}}&0&0\\\
0&0&\frac{i}{\sqrt{3}}&\frac{1+i\sqrt{3}}{\sqrt{6}}\\\
0&0&\frac{-1+i\sqrt{3}}{\sqrt{6}}&-\frac{i}{\sqrt{3}}\\\
\end{array}\right),\;\;\;\;\;\;\;\;\mathcal{T}=\left(\begin{array}[]{cccc}0&0&1&0\\\
0&0&0&1\\\ 1&0&0&0\\\ 0&1&0&0\\\ \end{array}\right)\mathcal{K}.$ (194)

They are completely the same as the matrix representation of the A1 Irrep in
Eq. (156i). Please notice that both $A_{2}$-Irrep and $A_{3}$-Irrep are the
$R_{1}R_{3}$-Irrep, which is the same as the $A_{1}$-Irrep constructed in Sec.
B.2.2, and the $B_{2}$-Irrep is the $R_{2}R_{2}$-Irrep that is the same as the
$B_{1}$-Irrep in Sec. B.2.2.

##  Appendix C Approach I for the hierarchy of the quasi-symmetry

In this section, we use the perturbation theory to identify the hierarchy of
quasi-symmetry. This is Approach I mentioned in the main text. Here we explain
the necessary details. The linear-$k$-order SOC-free Hamiltonian
$\mathcal{H}_{1}({\bf k})$,

$\displaystyle\mathcal{H}_{1}(\mathbf{k})$
$\displaystyle=C_{0}\sigma_{0}\tau_{0}+2A_{1}(\mathbf{k}\cdot\mathbf{L}),$
(195)

where the angular momentum operators are
$L_{x}=\tfrac{1}{2}\sigma_{y}\tau_{0},\;L_{y}=\tfrac{1}{2}\sigma_{x}\tau_{y},\;L_{z}=-\tfrac{1}{2}\sigma_{z}\tau_{y}$
defined in Eq. (28). Notice that $\mathcal{H}_{1}(\mathbf{k})$ is invariant
under the spin SU${}_{\text{s}}$(2) symmetry group. Moreover, we notice that
an additional hidden SU${}_{\text{o}}$(2) symmetry also exists for
$\mathcal{H}_{1}(\mathbf{k})$ in the “orbital” space that can be generated by
the following operators

$\displaystyle\mathcal{M}_{1,2,3}=\tfrac{1}{2}\\{s_{0}\sigma_{y}\tau_{z},s_{0}\sigma_{y}\tau_{x},s_{0}\sigma_{0}\tau_{y}\\},$
(196)

which is shown in the main text (see Eq. (2)), and $\mathcal{M}_{1,2,3}$ all
commute with $\mathcal{H}_{1}(\mathbf{k})$ and satisfy the commutation
relations

$\displaystyle[\mathcal{M}_{i},\mathcal{M}_{j}]$
$\displaystyle=i\epsilon_{i,j,k}\mathcal{M}_{k},$ (197a)
$\displaystyle\\{\mathcal{M}_{i},\mathcal{M}_{j}\\}$
$\displaystyle=\frac{1}{2}\delta_{i,j}.$ (197b)

Here $i,j=1,2,3$, $\epsilon_{i,j,k}$ is the 3D Levi-Civita symbol, and
$\delta_{i,j}$ is the Kronecker delta function. Thus, we refer to it as the
orbital SU${}_{\text{o}}$(2) quasi-symmetry group for $\mathcal{H}_{1}({\bf
k})+\mathcal{H}_{\text{soc}}$ because of

$\displaystyle[\mathcal{M}_{i},\mathcal{H}_{1}({\bf
k})+\mathcal{H}_{\text{soc}}]=0,$ (198)

where the on-site SOC Hamiltonian is given by Eq. 51,

$\displaystyle\mathcal{H}_{\text{soc}}=4\lambda_{0}(\mathbf{S}\cdot\mathbf{L}),$
(199)

where we use ${\bf S}=\tfrac{1}{2}(s_{x},s_{y},s_{z})$ for the spin-1/2
angular momentum operators. And $[S_{i},S_{j}]=i\epsilon_{ijk}S_{k}$. In
addition, we consider the $k^{2}$-order Hamiltonian that can be represented in
a compact form,

$\displaystyle\mathcal{H}_{2}(\mathbf{k})=\mathcal{H}_{2,\mathcal{M}_{1}}(\mathbf{k})+\mathcal{H}_{2,\mathcal{M}_{2}}(\mathbf{k})+\mathcal{H}_{2,\mathcal{M}_{3}}(\mathbf{k}),$
(200)

where each part of $\mathcal{H}_{2}(\mathbf{k})$ is given by

$\displaystyle\mathcal{H}_{2,\mathcal{M}_{i}}(\mathbf{k})={\bf g}_{i}\cdot{\bf
J}_{i},$ (201)

for $i=1,2,3$. Here we define the $k$-dependent vectors as

$\displaystyle\begin{split}{\bf g}_{1}({\bf
k})&=(C_{2}k_{x}k_{y},-C_{3}k_{x}k_{z},C_{1}k_{y}k_{z}),\\\ {\bf g}_{2}({\bf
k})&=(C_{3}k_{x}k_{y},C_{1}k_{x}k_{z},-C_{2}k_{y}k_{z}),\\\ {\bf g}_{3}({\bf
k})&=(C_{1}k_{x}k_{y},C_{2}k_{x}k_{z},-C_{3}k_{y}k_{z}).\end{split}$ (202)

And the corresponding vectors of operators

$\displaystyle\begin{split}{\bf
J}_{1}&=(\sigma_{x}\tau_{x},-\sigma_{z}\tau_{x},\sigma_{0}\tau_{z}),\\\ {\bf
J}_{2}&=(\sigma_{x}\tau_{z},\sigma_{z}\tau_{z},\sigma_{0}\tau_{x}),\\\ {\bf
J}_{3}&=(\sigma_{z}\tau_{0},\sigma_{x}\tau_{0},\sigma_{y}\tau_{y}).\end{split}$
(203)

In addition, we also realize that the $k^{2}$ terms of $\mathcal{H}_{2}({\bf
k})$ break this orbital SU${}_{\text{o}}$(2) quasi-symmetry generated by
$\\{\mathcal{M}_{1,2,3}\\}$ and lead to the splitting of all bands. However,
different parts of the entire $k^{2}$-terms can lead to the reduction from
SU${}_{\text{o}}$(2) to a orbital U(1). To show that, as we discussed in the
main text, we find that

$\displaystyle[{\bf J}_{i},\mathcal{M}_{i}]=0\text{ and }\\{{\bf
J}_{i},\mathcal{M}_{j}\\}=0\text{ for }i\neq j,$ (204)

which implies

$\displaystyle[\mathcal{H}_{2,\mathcal{M}_{i}}({\bf k}),\mathcal{M}_{i}]=0.$
(205)

This can be also found in the main text and Sec. A.2. Below, we focus on how
to identity the quasi-symmetry based on the effective perturbation theory
(Approach I). Note that alternative Approach II will be presented in Sec. D.

Furthermore, we notice there is a conservation of total angular momentum at
fixed ${\bf k}$ for $\mathcal{H}_{1}({\bf
k})+\mathcal{H}_{\text{soc}}=C_{0}+2A_{1}({\bf
k}\cdot\mathbf{L})+4\lambda_{0}({\bf S}\cdot{\bf L})$. As we have discussed in
Sec. A.4, the total angular momentum ${\bf S}+{\bf L}$ is conserved at ${\bf
k}=0$. But it does not commute with the Hamiltonian $\mathcal{H}_{1}({\bf
k})+\mathcal{H}_{\text{soc}}$ at the nonzero ${\bf k}$. However, we notice
that

$\displaystyle[({\bf S}+{\bf L})\cdot\vec{n}_{\bf k},\mathcal{H}_{1}({\bf
k})+\mathcal{H}_{\text{soc}}]=0,$ (206)

where $\vec{n}_{\bf k}=\frac{{\bf k}}{k}$ is the direction of the momentum
${\bf k}$. And $({\bf S}+{\bf L})\cdot\vec{n}_{\bf k}$ is physically similar
to the helicity operator at nonzero ${\bf k}$. Also, Eq. (206) can be proved
as follows,

  * •

Spin and orbital are independent degree of freedoms, so we have $[{\bf S},{\bf
L}]=0$. Therefore, $[({\bf S}+{\bf L})\cdot\vec{n}_{\bf
k},\mathcal{H}_{1}({\bf k})]=[{\bf L}\cdot\vec{n}_{\bf k},C_{0}+2A_{1}({\bf
k}\cdot{\bf L})]=0$.

  * •

We can define total angular momentum ${\bf L}_{\text{tot}}={\bf S}+{\bf L}$,
so that $\mathcal{H}_{\text{soc}}=2\lambda_{0}({\bf L}_{\text{tot}}^{2}-{\bf
S}^{2}-{\bf L}^{2})=2\lambda_{0}({\bf
L}_{\text{tot}}^{2}-\frac{3}{4}-\frac{3}{4})$. Then one can check

$\displaystyle[{\bf L}_{\text{tot},i},{\bf
L}_{\text{tot},j}]=[S_{i},S_{j}]+[L_{i},L_{j}]=i\epsilon_{ijk}S_{k}+i\epsilon_{ijk}L_{k}=i\epsilon_{ijk}{\bf
L}_{\text{tot},k}.$ (207)

Besides, we have $[{\bf L}_{\text{tot}},{\bf L}_{\text{tot}}^{2}]=0$ (i.e.,
the square of the angular momentum commutes with any of the components), which
leads to $[{\bf L}_{\text{tot}},\mathcal{H}_{\text{soc}}]=0$. Therefore, we
have

$\displaystyle[{\bf L}_{\text{tot}}\cdot\vec{n}_{{\bf k}},{\bf
L}_{\text{tot}}^{2}]=0.$ (208)

This proves that ${\bf L}_{\text{tot}}\cdot\vec{n}_{{\bf k}}$ is a symmetry
operator of $\mathcal{H}_{1}({\bf k})+\mathcal{H}_{\text{soc}}$ at any nonzero
${\bf k}$.

###  C.1 Approach I: the U(1) quasi-symmetry protected nodal planes

We solve the eigen-problem of the Hamiltonian $\mathcal{H}_{1}(\mathbf{k})$ in
Eq. (26) or Eq. (28). Due to the full rotation symmetry, we choose the
spherical coordinate with the momentum ${\bf
k}=(k_{x},k_{y},k_{z})=k(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)$.
The eigen-energies of $\mathcal{H}_{1}(\mathbf{k})$ have two branches
$E_{\pm}=C_{0}\pm A_{1}k$ with each branch twofold degeneracy (fourfold if
spin degeneracy is involved). In this work, we assume $A_{1}>0$, and the two
degenerate eigen-wave functions $|\Psi_{A/B+}(\theta,\phi)\rangle$ of the
positive energy branch ($E_{+}$) and $|\Psi_{A/B-}(\theta,\phi)\rangle$ of the
negative energy branch ($E_{-}$) are given by

$\displaystyle\mathcal{H}_{1}(\mathbf{k})\begin{cases}|\Psi_{A/B+}(\theta,\phi)\rangle=E_{+}|\Psi_{A/B+}(\theta,\phi)\rangle,\\\
|\Psi_{A/B-}(\theta,\phi)\rangle=E_{-}|\Psi_{A/B-}(\theta,\phi)\rangle,\end{cases}$
(209)

where the index $\pm$ represent the eigenvalues of ${\bf L}\cdot\vec{n}_{{\bf
k}}$, and the eigen-states in the spherical coordinator are given by

$\displaystyle\begin{cases}|\Psi_{A+}(\theta,\phi)\rangle&=\tfrac{1}{\sqrt{2}}\left(\cos\theta\cos\phi-i\sin\phi,-\cos\theta\sin\phi-i\cos\phi,0,\sin\theta\right)^{T},\\\
|\Psi_{B+}(\theta,\phi)\rangle&=\tfrac{1}{\sqrt{2}}\left(-i\sin\theta\cos\phi,i\sin\theta\sin\phi,1,i\cos\theta\right)^{T}.\end{cases}$
(210a)
$\displaystyle\begin{cases}|\Psi_{A-}(\theta,\phi)\rangle&=\tfrac{1}{\sqrt{2}}\left(i\sin\theta\cos\phi,-i\sin\theta\sin\phi,1,-i\cos\theta\right)^{T},\\\
|\Psi_{B-}(\theta,\phi)\rangle&=\tfrac{1}{\sqrt{2}}\left(\cos\theta\cos\phi+i\sin\phi,-\cos\theta\sin\phi+i\cos\phi,0,\sin\theta\right)^{T}.\end{cases}$
(210b)

The solution is not unique, since there exists a twofold degeneracy between
$|\Psi_{A+}(\theta,\phi)\rangle$ and $|\Psi_{B+}(\theta,\phi)\rangle$ at
arbitrary ${\bf k}$, protected by the orbital SU(2) symmetry generated by
$\\{\mathcal{M}_{1},\mathcal{M}_{2},\mathcal{M}_{3}\\}$ in Eq. (196).
Moreover, the subscript $A(B)$ can represent the eigenvalues of the quasi-
symmetry operator defined as
$(\mathcal{M}_{1},\mathcal{M}_{2},\mathcal{M}_{3})\cdot\vec{n}$ with a
specific real normalized vector $\vec{n}=(n_{1},n_{2},n_{3})$. For the basis
in Eq. (210a) and Eq. (210b), the $\vec{n}$-vector reads

$\displaystyle\vec{n}=\left(-\sin(\theta)\cos(\phi),\sin(\theta)\sin(\phi),-\cos(\theta)\right).$
(211)

And one can check

$\displaystyle({\bf\mathcal{M}}\cdot\vec{n})|\Psi_{A+}(\theta,\phi)\rangle$
$\displaystyle=\;\;\,|\Psi_{A+}(\theta,\phi)\rangle,$ (212a)
$\displaystyle({\bf\mathcal{M}}\cdot\vec{n})|\Psi_{A-}(\theta,\phi)\rangle$
$\displaystyle=\;\;\,|\Psi_{A-}(\theta,\phi)\rangle,$ (212b)
$\displaystyle({\bf\mathcal{M}}\cdot\vec{n})|\Psi_{B+}(\theta,\phi)\rangle$
$\displaystyle=-|\Psi_{B+}(\theta,\phi)\rangle,$ (212c)
$\displaystyle({\bf\mathcal{M}}\cdot\vec{n})|\Psi_{B-}(\theta,\phi)\rangle$
$\displaystyle=-|\Psi_{B-}(\theta,\phi)\rangle.$ (212d)

But $\vec{n}$ can be arbitrary due to this twofold degeneracy (fourfold if
spin degeneracy is accounted). This means the choice of eigen-states are not
unique. Besides, another set of wavefunctions chosen as eigen-states of
$\mathcal{M}_{3}$ will be shown in Sec. D.2 by fixing $\vec{n}=(0,0,1)$.
Furthermore, we can project the angular momentum operator $\mathbf{L}$ into
the eigenstate subspace,

$\displaystyle\langle\Psi_{A\pm}(\theta,\phi)|\mathbf{L}|\Psi_{A\pm}(\theta,\phi)\rangle$
$\displaystyle=\langle\Psi_{B\pm}(\theta,\phi)|\mathbf{L}|\Psi_{B\pm}(\theta,\phi)\rangle=\pm\frac{\mathbf{k}}{2k},$
(213)
$\displaystyle\langle\Psi_{A\pm}(\theta,\phi)|\mathbf{L}|\Psi_{B\pm}(\theta,\phi)\rangle$
$\displaystyle=\langle\Psi_{B\pm}(\theta,\phi)|\mathbf{L}|\Psi_{A\pm}(\theta,\phi)\rangle=0.$
(214)

Here, Eqs. (213) and (214) mean the emergent angular momentum operator
$\mathbf{L}$ is along the momentum direction after the projection. Besides,
with involving the spin degree of freedom, the corresponding four-fold
degenerate wave-functions are labelled as

$\displaystyle|\Psi_{+}\rangle=\\{|\Psi_{A+\uparrow}(\theta,\phi)\rangle,|\Psi_{B+\uparrow}(\theta,\phi)\rangle,|\Psi_{A+\downarrow}(\theta,\phi)\rangle,|\Psi_{B+\downarrow}(\theta,\phi)\rangle\\},$
(215)

where

$\displaystyle\begin{split}|\Psi_{A+\uparrow}(\theta,\phi)\rangle&=(1,0)^{T}\otimes|\Psi_{A+}(\theta,\phi)\rangle,\\\
|\Psi_{B+\uparrow}(\theta,\phi)\rangle&=(1,0)^{T}\otimes|\Psi_{B+}(\theta,\phi)\rangle,\\\
|\Psi_{A+\downarrow}(\theta,\phi)\rangle&=(0,1)^{T}\otimes|\Psi_{A+}(\theta,\phi)\rangle,\\\
|\Psi_{B+\downarrow}(\theta,\phi)\rangle&=(0,1)^{T}\otimes|\Psi_{B+}(\theta,\phi)\rangle,\end{split}$
(216)

where $(1,0)^{T}$ and $(0,1)^{T}$ label the spin-up and spin-down wave
functions, respectively. The above set of wave functions serves as the basis
for the projected 4-band perturbation model via the first-order perturbation
theory, dubbed as “the $P$-model”. We treat both the on-site SOC in Eq. (51)
and the $k^{2}$-terms in Eq. (29) as perturbations,

$\displaystyle\mathcal{H}_{perb}({\bf k})$
$\displaystyle=\mathcal{H}_{\text{soc}}+s_{0}\otimes\mathcal{H}_{2}(\mathbf{k}).$
(217)

The projected Hamiltonian
$\left\langle\Psi_{+}|\mathcal{H}_{perb}|\Psi_{+}\right\rangle$ is given by

$\displaystyle\mathcal{H}_{P}^{eff(1)}({\bf k})$
$\displaystyle=(E_{+}+B_{1}k^{2})s_{0}\omega_{0}+\mathcal{H}_{\text{soc},+}^{eff(1)}({\bf
k})+\mathcal{H}_{k^{2},+}^{eff(1)}({\bf k}),$ (218)

which is marked as the $P$-model around $R$-point. And

$\displaystyle\mathcal{H}_{\text{soc},+}^{eff(1)}({\bf k})$
$\displaystyle=\lambda_{0}\left(\lambda_{x}s_{x}+\lambda_{y}s_{y}+\lambda_{z}s_{z}\right)\otimes\omega_{0},$
(219a) $\displaystyle\mathcal{H}_{k^{2},+}^{eff(1)}({\bf k})$
$\displaystyle=\tilde{C}k^{2}s_{0}\otimes\left(d_{x}\omega_{x}+d_{y}\omega_{y}+d_{z}\omega_{z}\right).$
(219b)

where $\tilde{C}=C_{1}-C_{2}+C_{3}$, and $\omega_{x,y,z}$ are Pauli matrices
for the $\\{A+,B+\\}$ band subspace. The coefficients $\lambda_{x,y,z}$ are
defined as

$\displaystyle(\lambda_{x},\lambda_{y},\lambda_{z})$
$\displaystyle=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)=\tfrac{\mathbf{k}}{|\mathbf{k}|},$
(220)

On the other hand, the coefficients $d_{x,y,z}$ are given by

$\displaystyle\begin{split}d_{x}&=\tfrac{1}{4}\sin\theta\sin(2\theta)\sin(2\phi)(\cos\phi+\sin\phi),\\\
d_{y}&=\tfrac{1}{4}\sin\theta\sin(2\theta)\sin(2\phi)(\sin\theta+\cos\theta(\cos\phi-\sin\phi)),\\\
d_{z}&=\tfrac{1}{4}\sin\theta\sin(2\theta)\sin(2\phi)(\cos\theta-\sin\theta(\cos\phi-\sin\phi)).\end{split}$
(221)

Furthermore, we use the symmetry to understand the above first-order
perturbation Hamiltonian. The basis function in Eq. (215) can be labeled by
eigen-values of symmetries,

$\displaystyle|\Psi_{A+\uparrow}(\theta,\phi)\rangle$
$\displaystyle=|\uparrow\rangle\otimes|p=+\tfrac{1}{2},q=+1\rangle,$ (222a)
$\displaystyle|\Psi_{B+\uparrow}(\theta,\phi)\rangle$
$\displaystyle=|\uparrow\rangle\otimes|p=+\tfrac{1}{2},q=-1\rangle,$ (222b)
$\displaystyle|\Psi_{A+\downarrow}(\theta,\phi)\rangle$
$\displaystyle=|\downarrow\rangle\otimes|p=+\tfrac{1}{2},q=+1\rangle,$ (222c)
$\displaystyle|\Psi_{B+\downarrow}(\theta,\phi)\rangle$
$\displaystyle=|\downarrow\rangle\otimes|p=+\tfrac{1}{2},q=-1\rangle,$ (222d)

Here we take $p=\pm 1/2$ as the eigenvalues of ${\bf L}\cdot\vec{n}_{{\bf k}}$
and $q=\pm 1$ as the eigenvalues of $\vec{\mathcal{M}}\cdot\vec{n}$. All these
four states are degenerate with eigen-energy of $\mathcal{H}_{1}$ as
$C_{0}+A_{1}k$. The linear-$k$ Hamiltonian has both orbital SU(2) symmetry and
spin SU(2) symmetry. Specifically, the orbital SU(2) symmetry generated by
$\\{\mathcal{M}_{1,2,3}\\}$ indicates that the vector $\vec{n}$ is arbitrary.
And the spin SU(2) symmetry implies that $|\uparrow/\downarrow\rangle$ can be
any direction in the spin subspace. Then, to show the origin of the hidden
quasi-symmetry of the $P$-model, we individually do the projection for the on-
site SOC and $k^{2}$-order Hamiltonian,

  * •

Only do the projection of the on-site SOC Hamiltonian. In this case, the
orbital SU(2) symmetry generated by $\\{\mathcal{M}_{1,2,3}\\}$ preserves,
because of $[\mathcal{M}_{i},\mathcal{H}_{1}({\bf
k})+\mathcal{H}_{\text{soc}}]=0$ with $i=1,2,3$. Thus, we obtain

$\displaystyle\text{SU}_{\text{s}}(2)\times\text{SU}_{\text{o}}(2)\stackrel{{\scriptstyle\left\langle\mathcal{H}_{\text{soc}}\right\rangle_{\mathcal{H}_{1}}}}{{\xhookrightarrow{\quad\quad\quad\quad\quad}}}\text{U}_{\text{s}}(1)\times\text{SU}_{\text{o}}(2),$
(223)

where the spin U(1) symmetry generator depends on ${\bf k}$. To understand
this, we recall the conservation of $({\bf S}+{\bf L})\cdot\vec{n}_{{\bf k}}$
for $\mathcal{H}_{1}({\bf k})+\mathcal{H}_{\text{soc}}=C_{0}+2A_{1}{\bf
k}\cdot{\bf L}+4\lambda_{0}{\bf S}\cdot{\bf L}$. Then, for a fixed nonzero
${\bf k}$, one can take the eigen-states in Eq. (222) as eigen-states of the
helicity operator $({\bf S}+{\bf L})\cdot\vec{n}_{{\bf k}}$ by choosing

$\displaystyle|\Psi_{A+\uparrow}(\theta,\phi)\rangle$
$\displaystyle=|s=+\tfrac{1}{2}\rangle\otimes|p=+\tfrac{1}{2},q=+1\rangle,$
(224a) $\displaystyle|\Psi_{B+\uparrow}(\theta,\phi)\rangle$
$\displaystyle=|s=+\tfrac{1}{2}\rangle\otimes|p=+\tfrac{1}{2},q=-1\rangle,$
(224b) $\displaystyle|\Psi_{A+\downarrow}(\theta,\phi)\rangle$
$\displaystyle=|s=-\tfrac{1}{2}\rangle\otimes|p=+\tfrac{1}{2},q=+1\rangle,$
(224c) $\displaystyle|\Psi_{B+\downarrow}(\theta,\phi)\rangle$
$\displaystyle=|s=-\tfrac{1}{2}\rangle\otimes|p=+\tfrac{1}{2},q=-1\rangle.$
(224d)

Here we use $s=\pm 1/2$ as eigenvalues of ${\bf S}\cdot\vec{n}_{{\bf k}}$.
Here we use $\vec{n}_{{\bf k}}$, $\vec{n}_{{\bf k}}^{\prime}$, and
$\vec{n}_{{\bf k}}^{\prime\prime}$ to be the set of 3D orthogonal coordinates
at the fixed ${\bf k}$ and $\vec{n}_{{\bf k}}$ is the direction of ${\bf k}$
(i.e., $\vec{n}_{{\bf
k}}=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)$). Namely,
$|\vec{n}_{{\bf k}}|=|\vec{n}_{{\bf k}}^{\prime}|=|\vec{n}_{{\bf
k}}^{\prime\prime}|=1$, and $\vec{n}_{{\bf k}}\cdot\vec{n}_{{\bf
k}}^{\prime}=\vec{n}_{{\bf k}}\cdot\vec{n}_{{\bf
k}}^{\prime\prime}=\vec{n}_{{\bf k}}^{\prime}\cdot\vec{n}_{{\bf
k}}^{\prime\prime}=0$. Then, the first-order projection of ${\bf S}\cdot{\bf
L}=[({\bf S}\cdot\vec{n}_{{\bf k}})({\bf L}\cdot\vec{n}_{{\bf k}})+({\bf
S}\cdot\vec{n}_{{\bf k}}^{\prime})({\bf L}\cdot\vec{n}_{{\bf
k}}^{\prime})+({\bf S}\cdot\vec{n}_{{\bf k}}^{\prime\prime})({\bf
L}\cdot\vec{n}_{{\bf k}}^{\prime\prime})]$ onto these four states in Eq. (224)
leads to

$\displaystyle\begin{split}[\left\langle\mathcal{H}_{\text{soc}}\right\rangle_{\mathcal{H}_{1}}]_{i,j}&=4\lambda_{0}\langle\Psi_{i}(\theta,\phi)|{\bf
S}\cdot{\bf L}|\Psi_{j}(\theta,\phi)\rangle,\\\
&=4\lambda_{0}\langle\Psi_{i}(\theta,\phi)|({\bf S}\cdot\vec{n}_{{\bf
k}})({\bf L}\cdot\vec{n}_{{\bf k}})+({\bf S}\cdot\vec{n}_{{\bf
k}}^{\prime})({\bf L}\cdot\vec{n}_{{\bf k}}^{\prime})+({\bf
S}\cdot\vec{n}_{{\bf k}}^{\prime\prime})({\bf L}\cdot\vec{n}_{{\bf
k}}^{\prime\prime})|\Psi_{j}(\theta,\phi)\rangle,\\\
&=4\lambda_{0}\langle\Psi_{i}(\theta,\phi)|({\bf S}\cdot\vec{n}_{{\bf
k}})({\bf L}\cdot\vec{n}_{{\bf k}})|\Psi_{j}(\theta,\phi)\rangle,\end{split}$
(225)

where $i,j=\\{A+\uparrow,B+\uparrow,A+\downarrow,B+\downarrow\\}$. Thus, the
first-order perturbation for on-site SOC Hamiltonian is

$\displaystyle\left\langle\mathcal{H}_{\text{soc}}\right\rangle_{\mathcal{H}_{1}}=\lambda_{0}\begin{pmatrix}1&0&0&0\\\
0&1&0&0\\\ 0&0&-1&0\\\ 0&0&0&-1\end{pmatrix}.$ (226)

Please notice that
$\left\langle\mathcal{H}_{\text{soc}}\right\rangle_{\mathcal{H}_{1}}$ can be
also obtained after diagonalizing $\mathcal{H}_{\text{soc},+}^{eff(1)}({\bf
k})$ in Eq. (219). At nonzero ${\bf k}$, we find that the fourfold degenerate
states [$C_{0}+A_{1}k$ for $\mathcal{H}_{1}({\bf k})$] are split by the on-
site SOC Hamiltonian $\mathcal{H}_{\text{soc}}$ into two states and each state
has tow fold degeneracy stemming from the orbital SU(2) symmetry,
$E=C_{0}+A_{1}k\pm\lambda_{0}$. And the eigen-states are just Eq. (224). And
the spin-polarization along the direction of ${\bf k}$ represents the spin
U(1) symmetry.

  * •

Only do the projection of the $k^{2}$-order Hamiltonian. In this case, the
spin SU(2) symmetry preserves. Since the $k^{2}$-order Hamiltonian is spin-
independent. Thus,

$\displaystyle\text{SU}_{\text{s}}(2)\times\text{SU}_{\text{o}}(2)\stackrel{{\scriptstyle\left\langle\mathcal{H}_{2}\right\rangle_{\mathcal{H}_{1}}}}{{\xhookrightarrow{\quad\quad\quad\quad\quad}}}\text{SU}_{\text{s}}(2)\times\text{U}_{\text{o}}(1).$
(227)

We do not have an elegant picture for the
$\left\langle\mathcal{H}_{2}\right\rangle_{\mathcal{H}_{1}}$. But we know it
does not dependent on spin, as shown in Eq. (219). The four-fold degeneracy is
split into two states, and each state has two fold degeneracy (i.e. spin
degeneracy). And, the two-by-two matrix in the $\\{A,B\\}$-subspace itself
severs as the orbital U(1) symmetry.

Therefore, up to the first order perturbation, we obtain a hierarchy of quasi-
symmetry for CoSi mentioned in the main text (see Eq. (7)),

$\displaystyle\text{SU}_{\text{s}}(2)\times\text{SU}_{\text{o}}(2)\stackrel{{\scriptstyle\left\langle\mathcal{H}_{\text{soc}}+\mathcal{H}_{2}\right\rangle_{\mathcal{H}_{1}}}}{{\xhookrightarrow{\quad\quad\quad\quad\quad\quad\quad}}}\text{U}_{\text{s}}(1)\times\text{U}_{\text{o}}(1).$
(228)

And $[\text{U}_{\text{s}}(1),\text{U}_{\text{o}}(1)]=0$, as shown below.

###  C.2 The analytical properties of the effective perturbation 4-band
$P$-model

The $P$-model has the self-commuting structure at every ${\bf k}$. Explicitly,

$\displaystyle\left[\mathcal{H}_{\text{soc},+}^{eff(1)}({\bf
k}_{1}),\mathcal{H}_{k^{2},+}^{eff(1)}({\bf k}_{2})\right]=0,\quad\forall{\bf
k}_{1}\;\&\;\forall{\bf k}_{2}\text{ in the whole momentum space}$ (229)

as discussed in the main text, which is known as the projected stabilizer code
Hamiltonian and directly leads to the U(1) quasi-symmetry group. Below, we
provide some analytical study on the properties of the effective $P$-model.
The eigen-energies of the $P$-model in Eq. (218) are

$\displaystyle\begin{split}E_{\alpha\beta}(k,\theta,\phi)=C_{0}+B_{1}k^{2}+A_{1}k+\alpha\lambda_{0}+\beta\tfrac{\sqrt{3}}{4}\tilde{C}k^{2}|\sin
2\phi\sin 2\theta\sin\theta|,\end{split}$ (230)

where $\alpha=\pm$ and $\beta=\pm$. Here $\sin 2\phi\sin
2\theta\sin\theta=4k_{x}k_{y}k_{z}/k^{3}$, indicating that there are two-fold
degeneracy on high symmetry planes ($k_{x}=0$ or $k_{y}=0$ or $k_{z}=0$).

For the $\Gamma-R-M$ plane, $\phi=\pi/4$ so $\sin 2\phi=1$. The eigen-energies
along high symmetry lines are listed as follows:

  * 1.)

Along the $R-M$ line.
$\theta=0$, so $E_{\alpha}=C_{0}+B_{1}k^{2}+A_{1}k+\alpha\lambda_{0}$. All the
bands are two-fold degenerate.

  * 2.)

Along the $R-Z$ line.
$\theta=\pi/2$, so $E_{\alpha}=C_{0}+B_{1}k^{2}+A_{1}k+\alpha\lambda_{0}$. All
the bands are two-fold degenerate.

Figure 5:  The four distinct Fermi surfaces of CoSi are numerically
calculated, as shown in (a) for the $\Gamma-R-M$ plane. These high-symmetry
points are marked in the first Brillouin zone [left panel in (a)]. And these
four Fermi surfaces are labeled by four different colors. The middle two have
a “crossing” behavior with a tiny gap. And the analytical results for the band
dispersions (see Eq. (230)) of the $P$-model along the $R-M$ line in (b) and
the $R-Z$ line in (c). The two dashes black lines represents the Fermi
surfaces with $\alpha=\pm 1$ and $\beta=0$. The $\beta$-term leads to the
quadratic band splitting around the $R-M$ line in (a) with
$\delta_{k}\sim\delta_{\theta}^{2}$, and the linear band splitting around the
$R-Z$ line in (b) with $\delta_{k}\sim\delta_{\theta}$. Here, the orange line
is for $\alpha=-1,\beta=-1$, the blue line is for $\alpha=-1,\beta=+1$, the
green line is for $\alpha=+1,\beta=-1$, and the yellow line is for
$\alpha=+1,\beta=+1$. Once the band splitting caused by the $\beta$-term is
large enough, the band crossing between the blue line ($\beta=+1$) and the
green line ($\beta=-1$) may happen at arbitrary momenta, which are exact and
protected by the quasi-symmetry.

We then analyze the perturbation along these two high symmetry lines (i.e.,
the $R-M$ line and the $R-Z$ line) to identify the band crossing types of the
Fermi surface (linear or quadratic). For a given $E_{f}$, the Fermi surface
shape of the momentum $k_{\pm}$ for the upper four bands is determined by the
quadratic equation

$\displaystyle B_{1}k^{2}+A_{1}k+C_{\alpha}=0,$ (231)

where $C_{\alpha}=\alpha\lambda_{0}+C_{0}-E_{f}$ with $\alpha=\pm$. This
equation leads to the solution

$\displaystyle\begin{split}k_{\alpha,\theta=0}=k_{\alpha,\theta=\pi/2}=\frac{1}{2B_{1}}\left(-A_{1}+\sqrt{A_{1}^{2}-4B_{1}C_{\alpha}}\right).\end{split}$
(232)

We firstly focus on the $R-M$ line by expanding Eq. (230) around $\theta=0$.
In this case, we have $\theta=0+\delta_{\theta}$ and
$k=k_{\alpha,\theta=0}+\delta_{k}$. Then, the Eq. (231) should be replaced by

$\displaystyle
0=C_{\alpha}+B_{1}k^{2}+A_{1}k+\beta\tfrac{\sqrt{3}}{4}\tilde{C}k^{2}|\sin
2\phi\sin 2\theta\sin\theta|,$ (233)

which leads to

$\displaystyle\begin{split}0=C_{\alpha}+B_{1}\left(k_{\alpha,\theta=0}+\delta_{k}\right)^{2}+A_{1}\left(k_{\alpha,\theta=0}+\delta_{k}\right)+\beta\tfrac{\sqrt{3}}{4}\tilde{C}\left(k_{\alpha,\theta=0}+\delta_{k}\right)^{2}|\sin
2\phi|(2\delta_{\theta}^{2}).\end{split}$ (234)

After neglecting the $\delta^{2}_{k}$ terms, we find

$\displaystyle\begin{split}\delta_{k,\beta}=-\beta\frac{\sqrt{3}}{2}\frac{\tilde{C}|\sin
2\phi|\delta_{\theta}^{2}(k_{\alpha,\theta=0})^{2}}{2B_{1}k_{\alpha,\theta=0}+A_{1}+\beta\sqrt{3}\tilde{C}|\sin
2\phi|\delta_{\theta}^{2}k_{\alpha,\theta=0}}.\end{split}$ (235)

To the $\delta_{\theta}^{2}$ order, we have

$\displaystyle\delta_{k,\beta}=-\beta\frac{\sqrt{3}}{2}\frac{\tilde{C}|\sin
2\phi|(k_{\alpha,\theta=0})^{2}}{2B_{1}k_{\alpha,\theta=0}+A_{1}}\times\delta_{\theta}^{2},$
(236)

which indicates that the exact crossing along the $R-M$ line is quadratic in
momentum $k=\sqrt{k_{x}^{2}+k_{y}^{2}}$, as illustrated in Fig. 5 (a) around
the $R-M$ line.

For $\theta=\pi/2$ (i.e., the $R-Z$ line), we then expand Eq. (230) around
$\theta=\pi/2+\delta_{\theta}$. Because

$\displaystyle\begin{split}\sin
2\theta\sin\theta\to\sin(\pi+2\delta_{\theta})\sin(\pi/2+\delta_{\theta})\to-2\delta_{\theta}+\mathcal{O}(\delta_{\theta}^{3}),\end{split}$
(237)

It leads to the solution for $k$ at a fixed $E_{f}$,

$\displaystyle\delta_{k,\beta}=-\beta\frac{\sqrt{3}}{2}\frac{\tilde{C}|\sin
2\phi|(k_{\alpha,\theta=\pi/2})^{2}}{2B_{1}k_{\alpha,\theta=0}+A_{1}}\times|\delta_{\theta}|,$
(238)

which indicate the exact crossing along $k_{z}$ is linear, as illustrated in
Fig. 5 (b) around the $R-Z$ line.

Figure 6: Comparison of the Fermi surfaces with the $R$-model in (a) and the
$P$-model in (b) around the quasi nodal points in the $\Gamma-R-M$ plane. (a)
shows a tiny gap in the blue circle, and (b) shows the exact quasi-symmetry
protected degeneracy.

In addition, we discuss the emergent nodal lines on Fermi surfaces for the
$P$-model. The comparison of the FSs between the 8-band ${\bf k}\cdot{\bf p}$
effective Hamiltonian, the $R$-model (e.g. see Eq. (54) in Sec. A.3) and the
first-order-perturbation 4-band Hamiltonian, the $P$-model (e.g. see Eq. (218)
in Sec. C.1) are shown in Fig. 6. The $P$-model shows exact degeneracy at non-
high-symmetry points. The crossings from two bands can be obtained from the
constraint equation

$\displaystyle E_{\alpha=+1,\beta=-1}=E_{\alpha=-1,\beta=+1},$ (239)

of which the solution

$\displaystyle\lambda_{0}=\tfrac{\sqrt{3}}{4}\tilde{C}k^{2}|\sin 2\phi\sin
2\theta\sin\theta|=\sqrt{3}\tilde{C}\frac{|k_{x}k_{y}k_{z}|}{k}$ (240)

generally leads to nodal planes. The obtained nodal planes do not intersect
with high symmetry planes, e.g. $k_{x}=0$ or $k_{y}=0$ or $k_{z}=0$. At the
Fermi energy, we require an additional constraint equation

$\displaystyle E_{f}=E_{\alpha=+1,\beta=-1}=E_{\alpha=-1,\beta=+1}.$ (241)

By solving the above equation, we find the curve equation for the emergent
nodal lines at the Fermi energy

$\displaystyle\begin{split}\sqrt{3}\tilde{C}\frac{|k_{x}k_{y}k_{z}|}{k^{3}}=\tfrac{\sqrt{3}}{4}\tilde{C}|\sin
2\phi\sin
2\theta\sin\theta|=\frac{A_{1}^{2}\lambda_{0}}{2(E_{f}-C_{0})^{2}}\Big{[}1+\frac{2B_{1}(E_{f}-C_{0})}{A_{1}^{2}}+\sqrt{1+\frac{4B_{1}(E_{f}-C_{0})}{A_{1}^{2}}}\Big{]}.\end{split}$
(242)

For a fixed Fermi energy $E_{f}$, solving Eq. (242) generally gives rise to a
line solution in the $\theta-\phi$ plane. It represents a nodal line because
of Eq. (240): solving $\theta$ and $\phi$ at fixed $E_{f}$ from Eq. (242) will
fix $k$ simultaneously. However, the existence of such a line solution for Eq.
(242) depends on the value of $E_{f}$. For a critical $E_{f}$, there is no
line solution from the curve equation (242), instead, we can only get a point
solution. To find this minimal $E_{f}$, we set $\phi=\pi/4$, and notice that
the function $|\sin 2\theta\sin\theta|$ reaches its maximum when
$\theta\to\arcsin\sqrt{\tfrac{2}{3}}$, we find the minimal $E_{f}$ as

$\displaystyle
E_{f,c}=\frac{A_{1}\sqrt{C_{4}\lambda_{0}}+B_{1}\lambda_{0}+C_{0}C_{4}}{C_{4}}.$
(243)

Thus, $E_{f,c}$ is the energy for the single nodes with twofold degeneracy on
the corresponding nodal planes. Based on this analysis, we realize that the
quasi-nodal-line will emerge into single nodes when decreasing the Fermi
energy. For $E_{f}<E_{f,c}$, no solution of Eq. (242) can be obtained anymore.
Therefore, one can conclude that each nodal line emerges into a single point
after decreasing $E_{f}$ down to $E_{f,c}$. Moreover, when the nodal plane is
split due to high-order perturbations, this point becomes a Weyl point pinned
along the (111)-axis. Thus, we conclude that, nodal lines exist only when
$E_{f}\geq E_{f,c}$.

###  C.3 The second-order perturbation: gap out the quasi-nodal planes

Next we consider the second-order perturbation corrections for the 4-band
$P$-model (e.g. see Eq. (218) in Sec. C.1), which can open a tiny gap for the
emergent nodal lines obtained from Eq. (242) at generic momenta. Recall that
the first order perturbation Hamiltonian, the $P$-model, is based on the basis
in Eq. (215)

$\displaystyle|\Psi_{+}\rangle\triangleq|\Psi_{upper}\rangle=\\{(1,0)^{T}\otimes|\Psi_{A+}(\theta,\phi)\rangle,(1,0)^{T}\otimes|\Psi_{B+}(\theta,\phi)\rangle,(0,1)^{T}\otimes|\Psi_{A+}(\theta,\phi)\rangle,(0,1)^{T}\otimes|\Psi_{B+}(\theta,\phi)\rangle\\},$
(244)

which are all eigen-states of $\mathcal{H}_{1}$ with the same eigen-energy
$E_{+}=C_{0}+A_{1}k$. And the spinless wavefunctions
$|\Psi_{A/B+}(\theta,\phi)\rangle$ are given by Eq. (210a),

$\displaystyle|\Psi_{A+}(\theta,\phi)\rangle$
$\displaystyle=\tfrac{1}{\sqrt{2}}\left(\cos\theta\cos\phi-i\sin\phi,-\cos\theta\sin\phi-i\cos\phi,0,\sin\theta\right)^{T},$
(245a) $\displaystyle|\Psi_{B+}(\theta,\phi)\rangle$
$\displaystyle=\tfrac{1}{\sqrt{2}}\left(-i\sin\theta\cos\phi,i\sin\theta\sin\phi,1,i\cos\theta\right)^{T},$
(245b)

where $A/B$ represent the eigen-values of $\vec{\mathcal{M}}\cdot\vec{n}$. The
second-order perturbation theory has been presented in the supplementary
materials in Ref. [1]. To make sure the completeness of this appendix, we
repeat the discussion of the second-order perturbation in this sub-section.
Here, we consider the inter-band correction via second-order perturbation. The
fourfold degenerate eigen-states of $\mathcal{H}_{1}$ with lower energy
($E_{-}=C_{0}-A_{1}k$) are given by

$\displaystyle|\Psi_{lower}\rangle=\left\\{(1,0)^{T}\otimes|\Psi_{B-}(\theta,\phi)\rangle,(1,0)^{T}\otimes|\Psi_{A-}(\theta,\phi)\rangle,(0,1)^{T}\otimes|\Psi_{B-}(\theta,\phi)\rangle,(0,1)^{T}\otimes|\Psi_{A-}(\theta,\phi)\rangle\right\\},$
(246)

which are all eigen-states of of $\mathcal{H}_{1}$ with the same eigen-energy
$E_{-}=C_{0}-A_{1}k$. And the spinless wavefunctions
$|\Psi_{A/B-}(\theta,\phi)\rangle$ are given by Eq. (210b),

$\displaystyle|\Psi_{A-}(\theta,\phi)\rangle$
$\displaystyle=\tfrac{1}{\sqrt{2}}\left(i\sin\theta\cos\phi,-i\sin\theta\sin\phi,1,-i\cos\theta\right),$
(247a) $\displaystyle|\Psi_{B-}(\theta,\phi)\rangle$
$\displaystyle=\tfrac{1}{\sqrt{2}}\left(\cos\theta\cos\phi+i\sin\phi,-\cos\theta\sin\phi+i\cos\phi,0,\sin\theta\right)^{T}.$
(247b)

Therefore, the second-order perturbed Hamiltonian is given by

$\displaystyle\Delta\mathcal{H}_{P}^{eff(2)}({\bf k})=\frac{1}{\Delta
E}\left(\langle\Psi_{upper}|(\mathcal{H}_{soc}+\mathcal{H}_{2}(\mathbf{k}))\hat{P}_{lower}(\mathcal{H}_{soc}+\mathcal{H}_{2}(\mathbf{k}))|\Psi_{upper}\rangle\right),$
(248)

where $\Delta E=E_{+}-E_{-}=2A_{1}k$ is the energy difference between the
upper-energy-band the lower-energy-band, and the projection operator
$\hat{P}_{lower}=|\Psi_{lower}\rangle\langle\Psi_{lower}|$ onto the lower four
bands in Eq. (246). The mixed terms of $\mathcal{H}_{soc}$ and
$\mathcal{H}_{2}(\mathbf{k})$ for the second order perturbation are given by

$\displaystyle\Delta\mathcal{H}_{P}^{eff(2)}({\bf k})=\frac{1}{\Delta
E}\left(\langle\Psi_{upper}|\mathcal{H}_{soc}|\Psi_{lower}\rangle\langle\Psi_{lower}|\mathcal{H}_{2}(\mathbf{k})|\Psi_{upper}\rangle\right)+\text{h.c.},$
(249)

with the matrix elements of $\Delta\mathcal{H}_{P}^{eff(2)}$

$\displaystyle\left[\Delta\mathcal{H}_{P}^{eff(2)}\right]_{1,1}$
$\displaystyle=2\sin^{2}(\theta)\sin(\phi)\cos(\phi)\left(\sin^{3}(\theta)\sin(\phi)+\sin(\theta)\cos^{2}(\theta)\cos(\phi)-\cos^{3}(\theta)\right),$
(250) $\displaystyle\left[\Delta\mathcal{H}_{P}^{eff(2)}\right]_{1,2}$
$\displaystyle=\sin^{2}(\theta)\cos(\phi)(\sin^{2}(\theta)\sin(2\phi)+2i\cos(\theta)\sin(\phi)(\cos(\theta)(\cos(\theta)\cos(\phi)$
$\displaystyle+\sin(\theta))+\sin(\phi)(\sin^{2}(\theta)+i\cos(\theta)))),$
(251) $\displaystyle\left[\Delta\mathcal{H}_{P}^{eff(2)}\right]_{1,3}$
$\displaystyle=\tfrac{1}{4}e^{-i\phi}\sin(\theta)\cos(\theta)(\sin(2\phi)(\sin(3\theta)\sin(\phi)+4\cos^{3}(\theta))-8\sin(\theta)\cos^{2}(\theta)$
$\displaystyle\sin(\phi)\cos^{2}(\phi)+8i\cos(\theta)\sin^{2}(\phi)+(-\tfrac{3}{2}+6i)\sin(\theta)\cos(\phi)+(\tfrac{3}{2}+2i)\sin(\theta)$
$\displaystyle\cos(3\phi)),$ (252)
$\displaystyle~{}\left[\Delta\mathcal{H}_{P}^{eff(2)}\right]_{1,4}$
$\displaystyle=e^{-i\phi}\sin(\theta)((\cos(\theta)-1)(\cos^{2}(\theta)\cos(\phi)+i\sin(\theta)\sin(\phi)\cos(\phi)(-\cos^{2}(\theta)$
$\displaystyle+\sin(\theta)(\cos(\theta)+i)\cos(\phi))+\sin(\theta)\cos(\theta)\sin^{2}(\phi)(-1-i\sin(\theta)\cos(\phi)))$
$\displaystyle-2i\cos^{2}(\tfrac{\theta}{2})(-i\cos^{2}(\theta)\cos(\phi)\cos(2\phi)+\sin(\phi)\cos(\phi)((\cos^{3}(\theta)+\cos(\theta))$
$\displaystyle\cos(\phi)-i\sin^{2}(\theta)\cos(\phi)+\sin(\theta)\cos^{2}(\theta))+\sin(\theta)\cos(\theta)\sin^{2}(\phi)(\sin(\theta)$
$\displaystyle\cos(\phi)+i)))$ (253)

and

$\displaystyle\left[\Delta\mathcal{H}_{P}^{eff(2)}\right]_{2,2}$
$\displaystyle=-\tfrac{1}{4}\sin^{5}(\theta)\sin(\phi)\sin(2\phi)$
$\displaystyle~{}~{}~{}~{}\left(\cot^{2}(\theta)(-4\cot(\theta)\csc(\phi)+4\cot(\phi)-1)+\csc^{2}(\theta)+3\right),$
(254) $\displaystyle\left[\Delta\mathcal{H}_{P}^{eff(2)}\right]_{2,3}$
$\displaystyle=\tfrac{1}{2}e^{-i\phi}\sin(\theta)\cos(\theta)(-4\sin(\phi)(\sin(\theta)\sin(\phi)+(2+2i)\sin^{2}(\frac{\theta}{2})\cos^{2}(\phi))+$
$\displaystyle\cos^{2}(\theta)\sin(2\phi)(-4i\sin^{2}(\tfrac{\theta}{2})\cos(\phi)+2i\sin(\theta)+\csc(\phi))+\cos(\theta)\csc(\phi)$
$\displaystyle(-(2+2i)\sin^{2}(\tfrac{\theta}{2})\sin^{2}(2\phi)+\sin^{2}(\tfrac{\theta}{2})\sin(4\phi)+4i\sin^{2}(\theta)\sin^{3}(\phi)\cos(\phi)$
$\displaystyle+\sin(2\phi))),$ (255)
$\displaystyle\left[\Delta\mathcal{H}_{P}^{eff(2)}\right]_{2,4}$
$\displaystyle=-\tfrac{1}{8}e^{-i\phi}\sin(\theta)\cos(\theta)(8\cos^{3}(\theta)\sin(2\phi)-16\sin(\theta)\cos^{2}(\theta)\sin(\phi)\cos^{2}(\phi)$
$\displaystyle+16i\cos(\theta)\sin^{2}(\phi)+\cos(\phi)(4\sin(3\theta)\sin^{2}(\phi)+(-3+12i)\sin(\theta))$
$\displaystyle+(3+4i)\sin(\theta)\cos(3\phi)),$ (256)

and

$\displaystyle\left[\Delta\mathcal{H}_{P}^{eff(2)}\right]_{3,3}$
$\displaystyle=-\tfrac{1}{4}\sin^{5}(\theta)\sin(\phi)\sin(2\phi)$
$\displaystyle\left(\cot^{2}(\theta)(-4\cot(\theta)\csc(\phi)+4\cot(\phi)-1)+\csc^{2}(\theta)+3\right),$
(257) $\displaystyle\left[\Delta\mathcal{H}_{P}^{eff(2)}\right]_{3,4}$
$\displaystyle=-i\sin^{2}(\theta)\sin(\phi)\cos(\phi)(2\cos^{3}(\theta)\cos(\phi)+2\cos^{2}(\theta)(\sin(\theta)+i\sin(\phi))$
$\displaystyle+\sin(\theta)(\sin(2\theta)\sin(\phi)-2i\sin(\theta)\cos(\phi))),$
(258)

and

$\displaystyle\left[\Delta\mathcal{H}_{P}^{eff(2)}\right]_{4,4}$
$\displaystyle=2\sin^{2}(\theta)\sin(\phi)\cos(\phi)(\sin^{3}(\theta)\sin(\phi)+\sin(\theta)\cos^{2}(\theta)\cos(\phi)-\cos^{3}(\theta)).$
(259)

The other parts are related by complex conjugation
$\left[\Delta\mathcal{H}_{P}^{eff(2)}\right]_{i,j}=\left[\Delta\mathcal{H}_{P}^{eff(2)}\right]_{j,i}^{\ast}$.

###  C.4 Additional terms due to the linear $k$ SOC

In the first-order perturbation for the $P$-model in Sec. C.1, we only
consider the contribution from on-site SOC Hamiltonian. Here, we further
discuss the additional first-order perturbation corrections stemming from the
linear-$k$ SOC in Eq. (53),

$\displaystyle\begin{split}\mathcal{H}_{k,\text{soc}}({\bf
k})&=\lambda_{1}(k_{x}s_{x}+k_{y}s_{y}+k_{z}s_{z})\otimes\sigma_{0}\tau_{0}+\lambda_{2}(k_{x}s_{y}\sigma_{x}\tau_{x}-k_{y}s_{z}\sigma_{0}\tau_{x}+k_{z}s_{x}\sigma_{x}\tau_{0})\\\
&+\lambda_{3}(k_{y}s_{x}\sigma_{x}\tau_{x}-k_{z}s_{y}\sigma_{0}\tau_{x}+k_{x}s_{z}\sigma_{x}\tau_{0})+\lambda_{4}(k_{x}s_{y}\sigma_{x}\tau_{z}-k_{y}s_{z}\sigma_{y}\tau_{y}-k_{z}s_{x}\sigma_{z}\tau_{x})\\\
&+\lambda_{5}(k_{y}s_{x}\sigma_{x}\tau_{z}-k_{z}s_{y}\sigma_{y}\tau_{y}-k_{x}s_{z}\sigma_{z}\tau_{x})+\lambda_{6}(k_{x}s_{y}\sigma_{z}\tau_{0}+k_{y}s_{z}\sigma_{0}\tau_{z}+k_{z}s_{x}\sigma_{z}\tau_{z})\\\
&+\lambda_{7}(k_{y}s_{x}\sigma_{z}\tau_{0}+k_{z}s_{y}\sigma_{0}\tau_{z}+k_{x}s_{z}\sigma_{z}\tau_{z}).\end{split}$
(260)

Then, we only need to compute $\langle\Psi_{+}|\mathcal{H}_{k,\text{soc}}({\bf
k})|\Psi_{+}\rangle$ based on the basis in Eq. (216),

$\displaystyle|\Psi_{+}\rangle=\\{(1,0)^{T}\otimes|\Psi_{A+}(\theta,\phi)\rangle,(1,0)^{T}\otimes|\Psi_{B+}(\theta,\phi)\rangle,(0,1)^{T}\otimes|\Psi_{A+}(\theta,\phi)\rangle,(0,1)^{T}\otimes|\Psi_{B+}(\theta,\phi)\rangle\\},$
(261)

which are all eigen-states of $\mathcal{H}_{1}({\bf k})$ with the same eigen-
energy $E_{+}=C_{0}+A_{1}k$. And the spinless wavefunctions
$|\Psi_{A/B+}(\theta,\phi)\rangle$ are given by Eq. (210a).

First, a term denoted by $\lambda_{1}$ only renormalizes $\lambda_{0}$ as
$\lambda_{0}\to\lambda_{0}+\lambda_{1}k$. Second, all the others terms denoted
by $\lambda_{2,3,4,5,6,7}$ generally gap the quasi-nodal plane obtained from
Eq. (240), i.e., $\lambda_{0}=\tfrac{\sqrt{3}}{4}\tilde{C}k^{2}|\sin 2\phi\sin
2\theta\sin\theta|$. For an illustration, we consider the $\lambda_{2}$ term
as an example, and the projected matrix is

$\displaystyle\mathcal{H}_{k,\lambda_{2},eff}(k,\theta,\phi)=\lambda_{2}k\left(\begin{array}[]{cccc}h_{11}&h_{12}&h_{13}&h_{14}\\\
h_{21}&h_{22}&h_{23}&h_{24}\\\ h_{31}&h_{32}&h_{33}&h_{34}\\\
h_{41}&h_{42}&h_{43}&h_{44}\\\ \end{array}\right),$ (266)

where the spherical coordinator is used, and

$\displaystyle\begin{split}h_{11}&=-\sin^{3}(\theta)\sin^{2}(\phi)\cos(\phi),\\\
h_{12}&=\frac{1}{2}\sin^{2}(\theta)\sin(2\phi)(-\cos(\phi)-i\cos(\theta)\sin(\phi)),\\\
h_{13}&=-\frac{1}{2}\sin(2\theta)(\cos(\theta)\sin(\phi)+i\sin(\theta)\cos^{2}(\phi)),\\\
h_{14}&=\sin(\theta)\cos(\theta)(\cos(\theta)\cos^{2}(\phi)+i\sin(\phi)(\sin(\theta)+\cos(\phi))),\\\
h_{22}&=\sin^{3}(\theta)\sin^{2}(\phi)\cos(\phi),\\\
h_{23}&=-\sin(\theta)\cos(\theta)(\cos(\theta)\cos^{2}(\phi)-i\sin(\phi)(\cos(\phi)-\sin(\theta))),\\\
h_{24}&=\frac{1}{2}\sin(2\theta)\left(\cos(\theta)\sin(\phi)+i\sin(\theta)\cos^{2}(\phi)\right),\\\
h_{33}&=\sin^{3}(\theta)\sin^{2}(\phi)\cos(\phi),\\\
h_{34}&=\frac{1}{2}\sin(2\phi)\sin^{2}(\theta)(\cos(\phi)+i\cos(\theta)\sin(\phi)),\\\
h_{44}&=-\sin^{3}(\theta)\sin^{2}(\phi)\cos(\phi).\end{split}$ (267)

The remaining parts can be obtained by complex conjugation. We find

$\displaystyle[\mathcal{M}_{eff},\mathcal{H}_{k,\lambda_{2},eff}({\bf k})]\neq
0\text{ for nonzero }{\bf k},$ (268)

where the U(1) quasi-symmetry $\mathcal{M}_{eff}$ is given by
$\mathcal{H}_{\text{soc},+}^{eff(1)}({\bf k})$ or
$\mathcal{H}_{k^{2},+}^{eff(1)}({\bf k})$. This suggests that the nodal planes
are generally gapped by involving the linear $k$ SOC terms. But this effect is
negligible.

##  Appendix D Approach II for the hierarchy of the quasi-symmetry

In this section, we discuss the approach II for the hierarchy of the quasi-
symmetry and the perturbation theory based on the solution of
$\mathcal{H}_{1}({\bf k})+\mathcal{H}_{\text{soc}}$ for the
$\mathcal{H}_{2}({\bf k})$. Please note that the algebra of the orbital SU(2)
quasi-symmetry defined in Eq. (196) in Sec. D.1,

$\displaystyle\mathcal{M}_{1,2,3}=\tfrac{1}{2}\\{s_{0}\sigma_{y}\tau_{z},s_{0}\sigma_{y}\tau_{x},s_{0}\sigma_{0}\tau_{y}\\},$
(269)

which all commute with $\mathcal{H}_{1}({\bf k})+\mathcal{H}_{\text{soc}}$,
but do not commute between themselves. And
$[\mathcal{M}_{i},\mathcal{M}_{j}]=i\epsilon_{i,j,k}\mathcal{M}_{k}$.
Moreover, we can also defined the rotation for this orbital SU(2) quasi-
symmetry group

$\displaystyle\begin{split}\mathcal{U}(\theta,\phi)&=e^{i\theta\mathcal{M}_{3}}e^{i\phi\mathcal{M}_{2}}\\\
&=\cos\left(\tfrac{\theta}{2}\right)\cos\left(\tfrac{\phi}{2}\right)s_{0}\sigma_{0}\tau_{0}+i\sin\left(\tfrac{\theta}{2}\right)\cos\left(\tfrac{\phi}{2}\right)\mathcal{M}_{3}+i\cos\left(\tfrac{\theta}{2}\right)\sin\left(\tfrac{\phi}{2}\right)\mathcal{M}_{2}+i\sin\left(\tfrac{\theta}{2}\right)\sin\left(\tfrac{\phi}{2}\right)\mathcal{M}_{1}.\end{split}$
(270)

Clearly, $\mathcal{U}(\theta,\phi)$ commutes with
$s_{0}\otimes\mathcal{H}_{1}(\mathbf{k})+\mathcal{H}_{\text{soc}}$ for any
values of $\theta$ and $\phi$. The $\mathcal{U}(0,\frac{\pi}{2})$ can rotate
$\mathcal{M}_{1}$ to $\mathcal{M}_{3}$,

$\displaystyle\begin{split}&\mathcal{U}(0,\frac{\pi}{2})=e^{i\frac{\pi}{2}\mathcal{M}_{2}}=\frac{\sqrt{2}}{2}s_{0}\otimes\left(\sigma_{0}\tau_{0}+i\sigma_{y}\tau_{x}\right)\\\
\Rightarrow\quad&\mathcal{U}(0,\frac{\pi}{2})\mathcal{M}_{1}\mathcal{U}^{\dagger}(0,\frac{\pi}{2})=\mathcal{M}_{3}.\end{split}$
(271)

For better readability of this section, here, we first repeat the linear-$k$
Hamiltonian, on-site SOC Hamiltonian and $k^{2}$-order Hamiltonian,

$\displaystyle\mathcal{H}_{1}(\mathbf{k})$
$\displaystyle=C_{0}\sigma_{0}\tau_{0}+2A_{1}(\mathbf{k}\cdot\mathbf{L}),$
(272a) $\displaystyle\mathcal{H}_{\text{soc}}$
$\displaystyle=4\lambda_{0}(\mathbf{S}\cdot\mathbf{L}),$ (272b)
$\displaystyle\mathcal{H}_{2}(\mathbf{k})$
$\displaystyle=\mathcal{H}_{2,\mathcal{M}_{1}}(\mathbf{k})+\mathcal{H}_{2,\mathcal{M}_{2}}(\mathbf{k})+\mathcal{H}_{2,\mathcal{M}_{3}}(\mathbf{k}),$
(272c)

where the spin angular momentum ${\bf S}=\tfrac{1}{2}(s_{x},s_{y},s_{z})$, and
the orbital angular momentum operators are ${\bf
L}=(\tfrac{1}{2}\sigma_{y}\tau_{0},\tfrac{1}{2}\sigma_{x}\tau_{y},-\tfrac{1}{2}\sigma_{z}\tau_{y})$.
And each part of $\mathcal{H}_{2}(\mathbf{k})$ is given by

$\displaystyle\mathcal{H}_{2,\mathcal{M}_{i}}(\mathbf{k})={\bf g}_{i}({\bf
k})\cdot{\bf J}_{i},$ (273)

for $i=1,2,3$. Here we define the $k$-dependent vectors as

$\displaystyle\begin{split}{\bf g}_{1}({\bf
k})&=(C_{2}k_{x}k_{y},-C_{3}k_{x}k_{z},C_{1}k_{y}k_{z}),\\\ {\bf g}_{2}({\bf
k})&=(C_{3}k_{x}k_{y},C_{1}k_{x}k_{z},-C_{2}k_{y}k_{z}),\\\ {\bf g}_{3}({\bf
k})&=(C_{1}k_{x}k_{y},C_{2}k_{x}k_{z},-C_{3}k_{y}k_{z}).\end{split}$ (274)

and the corresponding vectors of operators

$\displaystyle\begin{split}{\bf
J}_{1}&=(\sigma_{x}\tau_{x},-\sigma_{z}\tau_{x},\sigma_{0}\tau_{z}),\\\ {\bf
J}_{2}&=(\sigma_{x}\tau_{z},\sigma_{z}\tau_{z},\sigma_{0}\tau_{x}),\\\ {\bf
J}_{3}&=(\sigma_{z}\tau_{0},\sigma_{x}\tau_{0},\sigma_{y}\tau_{y}).\end{split}$
(275)

In addition, we also realize that the $k^{2}$ terms of $\mathcal{H}_{2}({\bf
k})$ break this orbital SU${}_{\text{o}}$(2) quasi-symmetry generated by
$\\{\mathcal{M}_{1,2,3}\\}$ and lead to the splitting of all bands. However,
different parts of the entire $k^{2}$-terms can lead to the reduction from
SU${}_{\text{o}}$(2) to a orbital U(1). To show that, as we discussed in the
main text, we find that

$\displaystyle[{\bf J}_{i},\mathcal{M}_{i}]=0\text{ and }\\{{\bf
J}_{i},\mathcal{M}_{j}\\}=0\text{ for }i\neq j,$ (276)

which implies

$\displaystyle[\mathcal{H}_{2,\mathcal{M}_{i}}({\bf k}),\mathcal{M}_{i}]=0.$
(277)

###  D.1 The algebra for the orbital SU(2) quasi-symmetry for the
$k^{2}$-order Hamiltonian

In this subsection, we discuss the algebra of the orbital SU(2) quasi-symmetry
in Eq. (196). To show the generality of the breaking of the orbital SU(2)
quasi-symmetry down to U(1) quasi-symmetry by the $k^{2}$-order Hamiltonian.
We can rewrite the $\mathcal{H}_{2}({\bf k})$ into the form

$\displaystyle\mathcal{H}_{2}(\mathbf{k})=k_{x}k_{y}\left(\vec{C}_{\mathcal{J}}\cdot\vec{\mathcal{J}}\right)+k_{x}k_{z}\left(\vec{C}_{\mathcal{P}}\cdot\vec{\mathcal{P}}\right)+k_{y}k_{z}\left(\vec{C}_{\mathcal{Q}}\cdot\vec{\mathcal{Q}}\right),$
(278)

where the three ${\bf k}$-independent parameter-vectors are

$\displaystyle\begin{split}\vec{C}_{\mathcal{J}}&=(C_{2},-C_{3},C_{1}),\\\
\vec{C}_{\mathcal{P}}&=(-C_{3},C_{1},C_{2}),\\\
\vec{C}_{\mathcal{Q}}&=(C_{1},-C_{2},-C_{3}),\\\ \end{split}$ (279)

and the corresponding operator-vectors are given by

$\displaystyle\begin{split}\vec{\mathcal{J}}&=(\sigma_{x}\tau_{x},-\sigma_{x}\tau_{z},\sigma_{z}\tau_{0}),\\\
\vec{\mathcal{P}}&=(-\sigma_{z}\tau_{x},\sigma_{z}\tau_{z},\sigma_{x}\tau_{0}),\\\
\vec{\mathcal{Q}}&=(\sigma_{0}\tau_{z},\sigma_{0}\tau_{x},\sigma_{y}\tau_{y}).\end{split}$
(280)

Moreover, we notice that

$\displaystyle\begin{cases}\;[\mathcal{J}_{a},\mathcal{M}_{b}]=i\epsilon_{abc}\mathcal{J}_{c},\\\
\;[\mathcal{P}_{a},\mathcal{M}_{b}]=i\epsilon_{abc}\mathcal{P}_{c},\\\
\;[\mathcal{Q}_{a},\mathcal{M}_{b}]=i\epsilon_{abc}\mathcal{Q}_{c},\end{cases}$
(281)

where $\epsilon_{abc}$ is the three-dimensional Levi-Civita symbol with
$a,b,c=1,2,3$. Therefore, for arbitrary real normalized vector
$\vec{n}=(n_{1},n_{2},n_{3})$, we have the following commutation relations

$\displaystyle[\vec{\mathcal{J}}\cdot\vec{n},\vec{\mathcal{M}}\cdot\vec{n}]=[\vec{\mathcal{P}}\cdot\vec{n},\vec{\mathcal{M}}\cdot\vec{n}]=[\vec{\mathcal{Q}}\cdot\vec{n},\vec{\mathcal{M}}\cdot\vec{n}]=0.$
(282)

These can be easily shown, for example,

$\displaystyle\begin{split}[\vec{\mathcal{J}}\cdot\vec{n},\vec{\mathcal{M}}\cdot\vec{n}]&=\sum_{a=1}^{3}\sum_{b=1}^{3}n_{a}n_{b}[\mathcal{J}_{a},\mathcal{M}_{b}]=\sum_{a=1}^{3}\sum_{b=1}^{3}n_{a}n_{b}\left(i\epsilon_{abc}\mathcal{J}_{c}\right)\\\
&=\sum_{a=1}^{3}\sum_{b=a+1}^{3}n_{a}n_{b}\left(i\epsilon_{abc}\mathcal{J}_{c}+i\epsilon_{bac}\mathcal{J}_{c}\right)=0.\end{split}$
(283)

Here we have used $\epsilon_{abc}+\epsilon_{bac}=0$. Therefore, we find a
general symmetry-breaking case with the orbital SU(2) quasi-symmetry down to
the U(1) quasi-symmetry. For any three normalized and orthogonal vector,
$\vec{n}$, $\vec{n}^{\prime}$ and $\vec{n}^{\prime\prime}$, satisfy
$|\vec{n}|=|\vec{n}^{\prime}|=|\vec{n}^{\prime\prime}|=1$ and
$\vec{n}\perp\vec{n}^{\prime}$, $\vec{n}\perp\vec{n}^{\prime\prime}$ and
$\vec{n}^{\prime}\perp\vec{n}^{\prime\prime}$. Then we can do the projection
for the $k^{2}$-order Hamiltonian,

$\displaystyle\mathcal{H}_{2}({\bf k})=\mathcal{H}_{2,\vec{n}}({\bf
k})+\mathcal{H}_{2,\vec{n}^{\prime}}({\bf
k})+\mathcal{H}_{2,\vec{n}^{\prime\prime}}({\bf k}),$ (284)

where we project the parameter-vectors
($\vec{C}_{\mathcal{J}},\vec{C}_{\mathcal{P}},\vec{C}_{\mathcal{Q}}$) into the
$\\{\vec{n},\vec{n}^{\prime},\vec{n}^{\prime\prime}\\}$ space,

$\displaystyle\vec{C}_{\mathcal{J}}=\vec{n}(\vec{C}_{\mathcal{J}}\cdot\vec{n})+\vec{n}^{\prime}(\vec{C}_{\mathcal{J}}\cdot\vec{n}^{\prime})+\vec{n}^{\prime\prime}(\vec{C}_{\mathcal{J}}\cdot\vec{n}^{\prime\prime}),$
(285a)
$\displaystyle\vec{C}_{\mathcal{P}}=\vec{n}(\vec{C}_{\mathcal{P}}\cdot\vec{n})+\vec{n}^{\prime}(\vec{C}_{\mathcal{P}}\cdot\vec{n}^{\prime})+\vec{n}^{\prime\prime}(\vec{C}_{\mathcal{P}}\cdot\vec{n}^{\prime\prime}),$
(285b)
$\displaystyle\vec{C}_{\mathcal{Q}}=\vec{n}(\vec{C}_{\mathcal{Q}}\cdot\vec{n})+\vec{n}^{\prime}(\vec{C}_{\mathcal{Q}}\cdot\vec{n}^{\prime})+\vec{n}^{\prime\prime}(\vec{C}_{\mathcal{Q}}\cdot\vec{n}^{\prime\prime}),$
(285c)

Therefore, the first term in Eq. (284) is given by

$\displaystyle\mathcal{H}_{2,\vec{n}}({\bf
k})=k_{x}k_{y}(\vec{C}_{\mathcal{J}}\cdot\vec{n})\times\left(\vec{n}\cdot\vec{\mathcal{J}}\right)+k_{x}k_{z}(\vec{C}_{\mathcal{P}}\cdot\vec{n})\times\left(\vec{n}\cdot\vec{\mathcal{P}}\right)+k_{y}k_{z}(\vec{C}_{\mathcal{Q}}\cdot\vec{n})\left(\vec{n}\cdot\vec{\mathcal{Q}}\right),$
(286)

which commutes with $\mathcal{M}\cdot\vec{n}$,

$\displaystyle[\mathcal{M}\cdot\vec{n},\mathcal{H}_{2,\vec{n}}({\bf k})]=0.$
(287)

Especially, in the main text, we have mentioned three cases,

  * •

$\vec{n}=(1,0,0)$. The $\mathcal{H}_{2,\vec{n}}({\bf k})$ Hamiltonian is given
by $\mathcal{H}_{2,\mathcal{M}_{1}}({\bf k})$ in Eq. (200).

  * •

$\vec{n}=(0,1,0)$. The $\mathcal{H}_{2,\vec{n}}({\bf k})$ Hamiltonian is given
by $\mathcal{H}_{2,\mathcal{M}_{2}}({\bf k})$ in Eq. (200).

  * •

$\vec{n}=(0,0,1)$. The $\mathcal{H}_{2,\vec{n}}({\bf k})$ Hamiltonian is given
by $\mathcal{H}_{2,\mathcal{M}_{3}}({\bf k})$ in Eq. (200).

Moreover, the U(1) quasi-symmetry protected nodal-plane for
$\mathcal{H}_{2,\mathcal{M}_{1}}$ is discussed in the main text. More details
will be discussed in the following Sec. D.2, Sec. D.3, and Sec. D.4.

###  D.2 Analytical solutions by using the U(1) quasi-symmetry

Furthermore, we show the important role of the orbital SU(2) quasi-symmetry
operators (i.e. $\mathcal{M}_{1,2,3}$) in analytically solving the eigen-state
problem. Here, we take $\mathcal{H}_{1}(\mathbf{k})$ in Eq. (26) or Eq. (28)
as an example,

$\displaystyle\mathcal{H}_{1}(\mathbf{k})=C_{0}+A_{1}(k_{x}\sigma_{y}\tau_{0}+k_{y}\sigma_{x}\tau_{y}-k_{z}\sigma_{z}\tau_{y}),$
(288)

which commutes with the orbital SU(2) quasi-symmetry. Here we choose the
eigen-states of $\mathcal{H}_{1}(\mathbf{k})$ to be the common eigen-state of
$\mathcal{M}_{3}=\sigma_{0}\tau_{y}$ in Eq. (196). It is equivalently to apply
a unitary transformation

$\displaystyle\mathcal{U}=\frac{1}{\sqrt{2}}s_{0}\otimes\sigma_{0}\otimes\begin{pmatrix}1&-i\\\
1&i\end{pmatrix},$ (289)

to the $R$-model, which only leads to a rotation in the
$(\tau_{x},\tau_{y},\tau_{z})$ subspace

$\displaystyle\mathcal{U}\tau_{0}\mathcal{U}^{\dagger}=\tau_{0},\mathcal{U}\tau_{x}\mathcal{U}^{\dagger}=\tau_{y},\mathcal{U}\tau_{y}\mathcal{U}^{\dagger}=\tau_{z},\mathcal{U}\tau_{z}\mathcal{U}^{\dagger}=\tau_{x}.$
(290)

Therefore, the quasi-symmetry ${\cal M}_{3}$ becomes ${\cal U}{\cal
M}_{3}{\cal U}^{\dagger}=\sigma_{0}\tau_{z}$. Note that the spin Pauli matrix
is dropped here. As a result, the linear-$k$ Hamiltonian becomes

$\displaystyle{\cal U}[\mathcal{H}_{1}(\mathbf{k})-C_{0}]{\cal U}^{\dagger}$
$\displaystyle=\left[\mathcal{H}_{1,\mathcal{M}_{3}=+1}(\mathbf{k})\right]_{2\times
2}\oplus\left[\mathcal{H}_{1,\mathcal{M}_{3}=-1}(\mathbf{k})\right]_{2\times
2}$
$\displaystyle=\begin{pmatrix}\mathcal{H}_{1,\mathcal{M}_{3}=+1}(\mathbf{k})&0\\\
0&\mathcal{H}_{1,\mathcal{M}_{3}=-1}(\mathbf{k})\end{pmatrix},$ (291a)
$\displaystyle\mathcal{H}_{1,\mathcal{M}_{3}=+1}(\mathbf{k})$
$\displaystyle=A_{1}(k_{x}\sigma_{y}+k_{y}\sigma_{x}-k_{z}\sigma_{z}),$ (291b)
$\displaystyle\mathcal{H}_{1,\mathcal{M}_{3}=-1}(\mathbf{k})$
$\displaystyle=A_{1}(k_{x}\sigma_{y}-k_{y}\sigma_{x}+k_{z}\sigma_{z}).$ (291c)

It is easy to analytically find the eigen-states of the 2-by-2 Hamiltonian
$\mathcal{H}_{1,\mathcal{M}_{3}=\pm 1}(\mathbf{k})$, in the spherical
coordinate with the momentum ${\bf
k}=(k_{x},k_{y},k_{z})=k(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)$.
The eigen-states are given by

$\displaystyle\mathcal{H}_{1,\mathcal{M}_{3}=+1}(\mathbf{k})|A-\rangle=-A_{1}k|A-\rangle,\;\text{
with
}|A-\rangle=\frac{1}{\sqrt{2}\sqrt{1+\cos\theta}}(i(\cos\theta+1),e^{-i\phi}\sin\theta)^{T},$
(292a)
$\displaystyle\mathcal{H}_{1,\mathcal{M}_{3}=+1}(\mathbf{k})|A+\rangle=\;\;\;A_{1}k|A+\rangle,\;\text{
with
}|A+\rangle=\frac{1}{\sqrt{2}\sqrt{1-\cos\theta}}(i(\cos\theta-1),e^{-i\phi}\sin\theta)^{T},$
(292b)

and by substituting the replacement
$(k_{x},k_{y},k_{z})\to(k_{x},-k_{y},-k_{z})$ (i.e.,
$(\theta,\phi)\to(\theta+\pi,\pi-\phi)$), then
$\cos\theta\to-\cos\theta,\sin\theta\to-\sin\theta,e^{-i\phi}\to-e^{i\phi}$
into the above solution, it results in

$\displaystyle\mathcal{H}_{1,\mathcal{M}_{3}=-1}(\mathbf{k})|B-\rangle=-A_{1}k|B-\rangle,\;\text{
with
}|B-\rangle=\frac{1}{\sqrt{2}\sqrt{1-\cos\theta}}(i(-\cos\theta+1),e^{i\phi}\sin\theta)^{T},$
(293a)
$\displaystyle\mathcal{H}_{1,\mathcal{M}_{3}=-1}(\mathbf{k})|B+\rangle=\;\;\;A_{1}k|B+\rangle,\;\text{
with
}|B+\rangle=\frac{1}{\sqrt{2}\sqrt{1+\cos\theta}}(i(-\cos\theta-1),e^{i\phi}\sin\theta)^{T},$
(293b)

where the subscripts $A(B)$ represent the eigenvalues $+1(-1)$ of the quasi-
symmetry $\mathcal{M}_{3}$. Therefore, the eigen-state of
$\mathcal{H}_{1}(\mathbf{k})$ are given by

$\displaystyle\mathcal{H}_{1}(\mathbf{k})|\Psi_{A/B+}(\theta,\phi)\rangle=E_{+}|\Psi_{A/B+}(\theta,\phi)\rangle,$
(294a)
$\displaystyle\mathcal{H}_{1}(\mathbf{k})|\Psi_{A/B-}(\theta,\phi)\rangle=E_{-}|\Psi_{A/B-}(\theta,\phi)\rangle,$
(294b)

where $E_{\pm}=\pm A_{1}k$ and the corresponding eigen-states are given by

$\displaystyle|\Psi_{A-}(\theta,\phi)\rangle$
$\displaystyle=|A-\rangle\otimes(-\frac{i}{\sqrt{2}},\frac{1}{\sqrt{2}})^{T},$
(295a) $\displaystyle|\Psi_{A+}(\theta,\phi)\rangle$
$\displaystyle=|A+\rangle\otimes(-\frac{i}{\sqrt{2}},\frac{1}{\sqrt{2}})^{T},$
(295b) $\displaystyle|\Psi_{B-}(\theta,\phi)\rangle$
$\displaystyle=|B-\rangle\otimes(\frac{i}{\sqrt{2}},\frac{1}{\sqrt{2}})^{T},$
(295c) $\displaystyle|\Psi_{B+}(\theta,\phi)\rangle$
$\displaystyle=|B+\rangle\otimes(\frac{i}{\sqrt{2}},\frac{1}{\sqrt{2}})^{T}.$
(295d)

Due to the presence of the orbital SU(2) symmetry, one can also find the
common eigen-states of ${\cal H}_{1}({\bf k})$ and ${\cal M}\cdot\vec{n}$ for
arbitrary real vector $\vec{n}$, which is a rotation acting on the solution in
Eq. 295. Thus, the eigen-state solution is not unique due to this twofold
degeneracy, protected by the orbital SU(2) symmetry. Note that the degeneracy
will be doubled if spin degeneracy is taken into account.

Similarly, we can further analytically solve $\mathcal{H}_{1}({\bf
k})+\mathcal{H}_{\text{soc}}$ by using the U(1) quasi-symmetry generator
$\mathcal{M}_{3}$. The $R$-model in Eq. (54) becomes

$\displaystyle\mathcal{U}\mathcal{H}_{R}\mathcal{U}^{\dagger}=\mathcal{U}\left[s_{0}\otimes\mathcal{H}_{1}(\mathbf{k})+\mathcal{H}_{\text{soc}}+s_{0}\otimes\mathcal{H}_{2}(\mathbf{k})\right]\mathcal{U}^{\dagger}.$
(296)

The first two parts, $s_{0}\otimes\mathcal{H}_{1}({\bf
k})+\mathcal{H}_{\text{soc}}$, preserves the orbital SU(2) symmetry, while
some specific terms of the $k^{2}$-order Hamiltonian can break the SU(2)
quasi-symmetry down to U(1). This shows the hierarchy structure of the quasi-
symmetry, which will be discussed in details in Sec. D. Here we focus on
$s_{0}\otimes\mathcal{H}_{1}({\bf k})+\mathcal{H}_{\text{soc}}$, which becomes
block-diagonal after this unitary transformation defined in Eq. (289),

$\displaystyle\begin{split}\mathcal{U}\left[s_{0}\otimes\mathcal{H}_{1}({\bf
k})+\mathcal{H}_{\text{soc}}\right]\mathcal{U}^{\dagger}=C_{0}+\begin{pmatrix}\mathcal{H}_{A}(\mathbf{k})&0\\\
0&\mathcal{H}_{B}(\mathbf{k})\end{pmatrix},\end{split}$ (297)

where the index $A(B)$ represent the eigenvalues $+1(-1)$ of quasi-symmetry
${\cal M}_{3}$, and $\mathcal{H}_{A}(\mathbf{k})$ and
$\mathcal{H}_{B}(\mathbf{k})$ are given by

$\displaystyle\mathcal{H}_{A}(\mathbf{k})$
$\displaystyle=A_{1}s_{0}\otimes\left(k_{x}\sigma_{y}+k_{y}\sigma_{x}-k_{z}\sigma_{z}\right)+\lambda_{0}\left(s_{x}\sigma_{y}+s_{y}\sigma_{x}-s_{z}\sigma_{z}\right),$
(298) $\displaystyle\mathcal{H}_{B}(\mathbf{k})$
$\displaystyle=A_{1}s_{0}\otimes\left(k_{x}\sigma_{y}-k_{y}\sigma_{x}+k_{z}\sigma_{z}\right)+\lambda_{0}\left(s_{x}\sigma_{y}-s_{y}\sigma_{x}+s_{z}\sigma_{z}\right).$
(299)

First, one can check that the Hamiltonian can be reduced back to that in Eq.
(291) by setting $\lambda_{0}=0$. In addition, $\mathcal{H}_{A}(\mathbf{k})$
and $\mathcal{H}_{B}(\mathbf{k})$ are related to each other by time-reversal
symmetry

$\displaystyle\mathcal{H}_{B}(\mathbf{k})=\mathcal{T}[\mathcal{H}_{A}(-\mathbf{k})]\mathcal{T}^{\dagger}.$
(300)

Here $\mathcal{T}=is_{y}\mathcal{K}$ with $\mathcal{K}$ the complex conjugate.
Therefore, we only need to solve the four eigen-states for
$\mathcal{H}_{A}(\mathbf{k})$. After straightforward calculation, the four
eigen-energies of $\mathcal{H}_{A}(\mathbf{k})$ are given by

$\displaystyle E({\bf k})=C_{0}+\left\\{\pm
A_{1}k+\lambda_{0},\quad\pm\sqrt{A_{1}^{2}k^{2}+4\lambda_{0}^{2}}-\lambda_{0}\right\\}.$
(301)

Drop the constant $C_{0}$ for short, the two positive upper bands are

$\displaystyle E_{A,+}(\mathbf{k})$ $\displaystyle=A_{1}k+\lambda_{0},$ (302a)
$\displaystyle E_{A,-}(\mathbf{k})$
$\displaystyle=\sqrt{A_{1}^{2}k^{2}+4\lambda_{0}^{2}}-\lambda_{0},$ (302b)

which both have increasing energy as $k$ increases, and $\pm$ are the band
index. Note we have assumed $A_{1}>0$. And we notice that
$E_{A,+}(\mathbf{k}=0)=E_{A,-}(\mathbf{k}=0)$ and
$E_{A,+}(\mathbf{k})>E_{A,-}(\mathbf{k})$ for any nonzero ${\bf k}$. It
indicates that the SOC-induced gap between them,
$E_{A,+}(\mathbf{k})-E_{A,-}(\mathbf{k})$, approaches to $2\lambda_{0}$ as
$k\to\infty$. By time-reversal symmetry, the four eigen-energies of
$\mathcal{H}_{B}(\mathbf{k})$ are the same. Therefore, each state has twofold
degeneracy at any nonzero ${\bf k}$. This is due to the presence of the
orbital SU(2) symmetry.

Moreover, we can also solve the eigen wavefunctions of
$\mathcal{U}\left[s_{0}\otimes\mathcal{H}_{1}({\bf
k})+\mathcal{H}_{\text{soc}}\right]\mathcal{U}^{\dagger}$. For instance, for
the two positive upper bands of $\mathcal{H}_{A}(\mathbf{k})$, the
corresponding wavefunctions are given by,

$\displaystyle\mathcal{H}_{A}(\mathbf{k})\begin{cases}|E_{A,+}(\mathbf{k})\rangle=E_{A,+}(\mathbf{k})|E_{A,+}(\mathbf{k})\rangle,\\\
|E_{A,-}(\mathbf{k})\rangle=E_{A,-}(\mathbf{k})|E_{A,-}(\mathbf{k})\rangle.\end{cases}$
(303)

And the corresponding eigen-wavefunctions are given by

$\displaystyle|E_{A,+}(\mathbf{k})\rangle=\frac{1}{\mathcal{N}_{A,+}(\mathbf{k})}\left(-i,\frac{k_{x}-ik_{y}}{k-k_{z}},\frac{-ik_{x}+k_{y}}{k+k_{z}},1\right)^{T},$
(304a)
$\displaystyle|E_{A,-}(\mathbf{k})\rangle=\frac{1}{\mathcal{N}_{A,-}(\mathbf{k})}\Big{(}i(-A_{1}k_{z}-\lambda_{0}+E_{A,-}(\mathbf{k})),A_{1}(-k_{x}+ik_{y}),A_{1}(-ik_{x}+k_{y}),A_{1}k_{z}-\lambda_{0}+E_{A,-}(\mathbf{k})\Big{)}^{T},$
(304b)

where the normalization factors are

$\displaystyle\mathcal{N}_{A,+}(\mathbf{k})$
$\displaystyle=2k/\sqrt{k_{x}^{2}+k_{y}^{2}},$ (305a)
$\displaystyle\mathcal{N}_{A,-}(\mathbf{k})$
$\displaystyle=2\sqrt{(E_{A,-}(\mathbf{k}))^{2}-\lambda_{0}^{2}}.$ (305b)

In addition, we discuss the band index $\pm$ that are actually eigen-values of
symmetry. To show that, we emphasize that one can also obtain the common
eigen-states for $\mathcal{H}_{1}({\bf k})+\mathcal{H}_{\text{soc}}$ and
${\cal M}\cdot\vec{n}$. As we discussed in Eq. (206), there is a helicity-type
symmetry operator that commutes with $\mathcal{H}_{1}({\bf
k})+\mathcal{H}_{\text{soc}}$, which indicates the index $\pm$ in the eigen-
state solution $|E_{A,\pm}(\mathbf{k})\rangle$ are eigen-values of $({\bf
S}+{\bf L})\cdot\vec{n}_{{\bf k}}$. Explicitly, one can check that

$\displaystyle[({\bf S}+{\bf L})\cdot\vec{n}_{{\bf
k}}]|E_{A,+}(\mathbf{k})\rangle$ $\displaystyle=|E_{A,+}(\mathbf{k})\rangle,$
(306a) $\displaystyle[({\bf S}+{\bf L})\cdot\vec{n}_{{\bf
k}}]|E_{A,-}(\mathbf{k})\rangle$ $\displaystyle=0.$ (306b)

It represents the eigen-values of the z-component angular moment of the total
angular momentum at nonzero ${\bf k}$.

###  D.3 Approach II: The U(1) quasi-symmetry protected nodal planes

As discussed in Sec. D.1, parts of the entire $k^{2}$-order Hamiltonian break
the SU(2) quasi-symmetry down to the U(1) quasi-symmetry. For example, we
consider

$\displaystyle\mathcal{H}_{2,\vec{n}}({\bf
k})=k_{x}k_{y}(\vec{C}_{\mathcal{J}}\cdot\vec{n})\times\left(\vec{n}\cdot\vec{\mathcal{J}}\right)+k_{x}k_{z}(\vec{C}_{\mathcal{P}}\cdot\vec{n})\times\left(\vec{n}\cdot\vec{\mathcal{P}}\right)+k_{y}k_{z}(\vec{C}_{\mathcal{Q}}\cdot\vec{n})\left(\vec{n}\cdot\vec{\mathcal{Q}}\right),$
(307)

which commutes with $\mathcal{M}\cdot\vec{n}$,

$\displaystyle[\mathcal{M}\cdot\vec{n},\mathcal{H}_{2,\vec{n}}({\bf k})]=0.$
(308)

For an illustration, we can choose $\vec{n}=(0,0,1)$ without loss of
generality, so that $\mathcal{M}_{3}$ is the symmetry generator for the
remaining U(1) quasi-symmetry. Therefore, we consider the Hamiltonian
$\mathcal{H}_{qsR}({\bf k})$ that commute with $\mathcal{M}_{3}$ as,

$\displaystyle\mathcal{H}_{qsR}({\bf k})$ $\displaystyle=\mathcal{H}_{1}({\bf
k})+\mathcal{H}_{\text{soc}}+\mathcal{H}_{2,\mathcal{M}_{3}}(\mathbf{k}),$
(309a) $\displaystyle\mathcal{H}_{1}({\bf k})$
$\displaystyle=C_{0}\sigma_{0}\tau_{0}+2A_{1}(\mathbf{k}\cdot\mathbf{L}),$
(309b) $\displaystyle\mathcal{H}_{\text{soc}}$
$\displaystyle=4\lambda_{0}(\mathbf{S}\cdot\mathbf{L}),$ (309c)
$\displaystyle\mathcal{H}_{2,\mathcal{M}_{3}}(\mathbf{k})$ $\displaystyle={\bf
g}_{3}\cdot{\bf
J}_{3}=(C_{1}k_{x}k_{y},C_{2}k_{x}k_{z},-C_{3}k_{y}k_{z})\cdot(\sigma_{z}\tau_{0},\sigma_{x}\tau_{0},\sigma_{y}\tau_{y}),$
(309d)

where
$L_{x}=\tfrac{1}{2}\sigma_{y}\tau_{0},\;L_{y}=\tfrac{1}{2}\sigma_{x}\tau_{y},\;L_{z}=-\tfrac{1}{2}\sigma_{z}\tau_{y}$.
Notice that $\mathcal{M}_{3}=\sigma_{0}\tau_{y}$ is the quasi-symmetry
operator. It is easy to check $[\mathcal{M}_{3},\mathcal{H}_{qsR}({\bf
k})]=0$. We next solve the common eigen-states of $\mathcal{M}_{3}$ and
$\mathcal{H}_{qsR}({\bf k})$. To do that, we only need to diagonalize the
$\tau_{y}$-term for $\mathcal{H}_{qsR}({\bf k})$. In other words, we apply a
unitary transformation

$\displaystyle\mathcal{U}_{\mathcal{M}_{3}}=\frac{1}{\sqrt{2}}s_{0}\otimes\sigma_{0}\otimes\begin{pmatrix}1&-i\\\
1&i\end{pmatrix},$ (310)

which leads to the rotating in the $(\tau_{x},\tau_{y},\tau_{z})$ subspace

$\displaystyle\mathcal{U}_{\mathcal{M}_{3}}\tau_{0}\mathcal{U}_{\mathcal{M}_{3}}^{\dagger}=\tau_{0},\mathcal{U}_{\mathcal{M}_{3}}\tau_{x}\mathcal{U}_{\mathcal{M}_{3}}^{\dagger}=\tau_{y},\mathcal{U}_{\mathcal{M}_{3}}\tau_{y}\mathcal{U}_{\mathcal{M}_{3}}^{\dagger}=\tau_{z},\mathcal{U}_{\mathcal{M}_{3}}\tau_{z}\mathcal{U}_{\mathcal{M}_{3}}^{\dagger}=\tau_{x}.$
(311)

Therefore, after this unitary transformation, the U(1) quasi-symmetry
generator $\mathcal{M}_{3}$ becomes

$\displaystyle\mathcal{U}_{\mathcal{M}_{3}}\mathcal{M}_{3}\mathcal{U}_{\mathcal{M}_{3}}^{\dagger}\triangleq\mathcal{M}_{qs}=\begin{pmatrix}\sigma_{0}&0\\\
0&-\sigma_{0}\end{pmatrix}.$ (312)

And we take this unitary transformation on the $\mathcal{H}_{qsR}(\mathbf{k})$
Hamiltonian and obtain

$\displaystyle\mathcal{U}_{\mathcal{M}_{3}}\mathcal{H}_{qsR}(\mathbf{k})\mathcal{U}_{\mathcal{M}_{3}}^{\dagger}\triangleq\mathcal{H}_{qsR}(\mathbf{k})=C_{0}+B_{1}k^{2}+\begin{pmatrix}\mathcal{H}_{+}^{\prime}(\mathbf{k})&0\\\
0&\mathcal{H}_{-}^{\prime}(\mathbf{k})\end{pmatrix},$ (313a)

where

$\displaystyle\mathcal{H}_{+}^{\prime}(\mathbf{k})$
$\displaystyle=s_{0}\otimes\left[(A_{1}k_{x}-C_{3}k_{y}k_{z})\sigma_{y}+(A_{1}k_{y}+C_{2}k_{x}k_{z})\sigma_{x}-(A_{1}k_{z}-C_{1}k_{x}k_{y})\sigma_{z}\right]+\lambda_{0}\left(s_{x}\sigma_{y}+s_{y}\sigma_{x}-s_{z}\sigma_{z}\right),$
(314) $\displaystyle\mathcal{H}_{-}^{\prime}(\mathbf{k})$
$\displaystyle=s_{0}\otimes\left[(A_{1}k_{x}+C_{3}k_{y}k_{z})\sigma_{y}-(A_{1}k_{y}-C_{2}k_{x}k_{z})\sigma_{x}+(A_{1}k_{z}+C_{1}k_{x}k_{y})\sigma_{z}\right]+\lambda_{0}\left(s_{x}\sigma_{y}-s_{y}\sigma_{x}+s_{z}\sigma_{z}\right),$
(315)

This means that we have chosen the eigen-states of
$\mathcal{H}_{qsR}(\mathbf{k})$ to be the common eigen-state of
$\mathcal{M}_{3}$ or $\mathcal{M}_{qs}$. Due to the presence of the U(1)
quasi-symmetry, we dubbed $\mathcal{H}_{qsR}$ as the quasi-symmetric R-model
(“qsR”). Therefore, the subscript $\pm$ for
$\mathcal{H}_{\pm}^{\prime}(\mathbf{k})$ represent the different eigenvalues
of $\mathcal{M}_{3}$ or $\mathcal{M}_{qs}$. Besides,
$\mathcal{H}_{+}^{\prime}(\mathbf{k})$ and
$\mathcal{H}_{-}^{\prime}(\mathbf{k})$ are related to each other by TR
$\mathcal{T}=is_{y}\mathcal{K}$ with $\mathcal{K}$ complex conjugate,

$\displaystyle\mathcal{H}_{-}^{\prime}(\mathbf{k})=\mathcal{T}\mathcal{H}_{+}^{\prime}(-\mathbf{k})\mathcal{T}^{\dagger}.$
(316)

Next, we compute the eigen-energy of the $qsR$-model. The energy of the upper
four bands are given by

$\displaystyle E_{+,1}(\mathbf{k})$
$\displaystyle=C_{0}+B_{1}k^{2}+\sqrt{A_{1}^{2}k^{2}+C_{1}^{2}k_{x}^{2}k_{y}^{2}+C_{2}^{2}k_{x}^{2}k_{z}^{2}+C_{3}^{2}k_{y}^{2}k_{z}^{2}-2A_{1}\tilde{C}k_{x}k_{y}k_{z}}+\lambda_{0},$
(317) $\displaystyle E_{+,2}(\mathbf{k})$
$\displaystyle=C_{0}+B_{1}k^{2}+\sqrt{A_{1}^{2}k^{2}+C_{1}^{2}k_{x}^{2}k_{y}^{2}+C_{2}^{2}k_{x}^{2}k_{z}^{2}+C_{3}^{2}k_{y}^{2}k_{z}^{2}-2A_{1}\tilde{C}k_{x}k_{y}k_{z}+4\lambda_{0}^{2}}-\lambda_{0},$
(318) $\displaystyle E_{-,1}(\mathbf{k})$
$\displaystyle=C_{0}+B_{1}k^{2}+\sqrt{A_{1}^{2}k^{2}+C_{1}^{2}k_{x}^{2}k_{y}^{2}+C_{2}^{2}k_{x}^{2}k_{z}^{2}+C_{3}^{2}k_{y}^{2}k_{z}^{2}+2A_{1}\tilde{C}k_{x}k_{y}k_{z}}+\lambda_{0},$
(319) $\displaystyle E_{-,2}(\mathbf{k})$
$\displaystyle=C_{0}+B_{1}k^{2}+\sqrt{A_{1}^{2}k^{2}+C_{1}^{2}k_{x}^{2}k_{y}^{2}+C_{2}^{2}k_{x}^{2}k_{z}^{2}+C_{3}^{2}k_{y}^{2}k_{z}^{2}+2A_{1}\tilde{C}k_{x}k_{y}k_{z}+4\lambda_{0}^{2}}-\lambda_{0},$
(320)

where $\tilde{C}=C_{1}-C_{2}+C_{3}$. The $qsR$-model breaks the $C_{3,(111)}$
rotation symmetry because of $C_{1}\neq C_{2}\neq C_{3}$. Moreover, the $\pm$
index for $E_{\pm,i}$ means the eigenvalues of $\mathcal{M}_{qs}$,

$\displaystyle\begin{split}&\mathcal{H}_{qsR}(\mathbf{k})|E_{\pm,i}(\mathbf{k})\rangle=E_{\pm,i}(\mathbf{k})|E_{\pm,i}(\mathbf{k})\rangle,\\\
&\mathcal{M}_{qs}|E_{\pm,i}(\mathbf{k})\rangle=\pm|E_{\pm,i}(\mathbf{k})\rangle,\end{split}$
(321)

where $i=1,2$ is the band index. $E_{+,i}(\mathbf{k})=E_{-,i}(-\mathbf{k})$ is
required by TR symmetry. The quasi-symmetry protected nodal planes are given
by

$\displaystyle\begin{cases}k_{x}k_{y}k_{z}>0,E_{+,1}(\mathbf{k})=E_{-,2}(\mathbf{k}),\\\
k_{x}k_{y}k_{z}<0,E_{+,2}(\mathbf{k})=E_{-,1}(\mathbf{k}).\end{cases}$ (322)

The crossings are between the bands with different eigenvalue of quasi-
symmetry $\mathcal{M}_{3}$ or $\mathcal{M}_{qs}$, and thus we find nodal
planes at generic momenta with the protection from the quasi-symmetry
$\mathcal{M}_{qs}$. By further imposing the Fermi energy constraint, there
will be nodal lines on the Fermi surface, shown in Fig. 7. In Fig. 7(a), the
four Fermi surfaces are plotted in the $\Gamma-R-M$ plane, where two inner FSs
intersect with each other and generate the quasi-symmetry protected exact
crossings (marked as purple circles). In Fig. 7 (b) and (c), the solution of
Eq. (322) are explicitly depicted, showing the exact nodal lines on the FSs.

Figure 7: The exact crossings of the 8-band $qsR$-model. In (a), the Fermi
surfaces with exact crossing are shown in the $\Gamma-R-M$ plane. In (b) and
(c), we show the exact nodal lines on the FSs. These nodal lines are protected
by the U(1) quasi-symmetry.

Next, let us discuss the terms that break the quasi-symmetry. By including the
remaining $k^{2}$-order Hamiltonian
$\mathcal{H}_{2,\mathcal{M}_{1}}(\mathbf{k})+\mathcal{H}_{2,\mathcal{M}_{2}}(\mathbf{k})$,
the full Hamiltonian is given by

$\displaystyle\mathcal{H}_{R}({\bf k})=\mathcal{H}_{qsR}({\bf
k})+\mathcal{U}_{\mathcal{M}_{3}}[\mathcal{H}_{2,\mathcal{M}_{1}}(\mathbf{k})+\mathcal{H}_{2,\mathcal{M}_{2}}(\mathbf{k})]\mathcal{U}_{\mathcal{M}_{3}}^{\dagger},$
(323)

where the U(1) quasi-symmetry breaking terms are given by

$\displaystyle\mathcal{H}_{2,qsb}(\mathbf{k})$
$\displaystyle=\mathcal{U}_{\mathcal{M}_{3}}[\mathcal{H}_{2,\mathcal{M}_{1}}(\mathbf{k})+\mathcal{H}_{2,\mathcal{M}_{2}}(\mathbf{k})]\mathcal{U}_{\mathcal{M}_{3}}^{\dagger}=\begin{pmatrix}0&s_{0}\otimes
H_{(2)}\\\ s_{0}\otimes H_{(2)}^{\dagger}&0\end{pmatrix},$ (324a)
$\displaystyle H_{(2)}$
$\displaystyle=C_{1}(k_{y}k_{z}\sigma_{0}+k_{x}k_{z}\sigma_{z})+C_{2}(-ik_{x}k_{y}\sigma_{x}+ik_{y}k_{z}\sigma_{0})+C_{3}(k_{x}k_{y}\sigma_{x}+ik_{x}k_{z}\sigma_{z})$
$\displaystyle+(C_{1}+iC_{2})k_{y}k_{z}\sigma_{0}+(C_{1}+iC_{3})k_{x}k_{z}\sigma_{z}+(-iC_{2}+C_{3})k_{x}k_{y}\sigma_{x}.$
(324b)

Therefore, we have

$\displaystyle[\mathcal{M}_{qs},\mathcal{H}_{2,qsb}(\mathbf{k})]=2\begin{pmatrix}0&h_{(2)}(\mathbf{k})\\\
-h_{(2)}^{\dagger}(\mathbf{k})&0\end{pmatrix}\neq 0.$ (325)

where

$\displaystyle\begin{split}h_{(2)}=(C_{1}+iC_{2})k_{y}k_{z}\sigma_{0}+(C_{1}+iC_{3})k_{x}k_{z}\sigma_{z}+(-iC_{2}+C_{3})k_{x}k_{y}\sigma_{x}.\end{split}$
(326)

For this Hamiltonian, $[\mathcal{M}_{qs},\mathcal{H}_{2,qsb}(\mathbf{k})]=0$
only occurs for $\mathbf{k}=(0,0,0)$ (the $R$-point). Therefore, for any Fermi
surface that does not cross the $R$-point, the nodal planes will be gapped.
This leads to the quasi-symmetry hierarchy mentioned in the main text (see Eq.
(10)),

$\displaystyle\text{SU}_{\text{s}}(2)\times\text{SU}_{\text{o}}(2)\stackrel{{\scriptstyle\mathcal{H}_{\text{soc}}}}{{\xhookrightarrow{\quad\quad\quad\;}}}\text{SU}_{\text{o}}(2)\stackrel{{\scriptstyle\mathcal{H}_{2,\mathcal{M}_{i}}}}{{\xhookrightarrow{\quad\quad\quad\quad\;}}}\text{U}_{\text{o}}(1).$
(327)

More interestingly, as we discussed in Sec. D.1, we have shown that the choice
of $\mathcal{H}_{2,\mathcal{M}_{3}}$ is just one specific case, instead, for
any real vector $\vec{n}$ of $\mathcal{H}_{2,\vec{n}}$ that commute with
$\mathcal{M}\cdot\vec{n}$ (see Sec. D.1 for details) can generally lead to the
same quasi-symmetry hierarchy results. Because we have

$\displaystyle[\mathcal{M}\cdot\vec{n},s_{0}\otimes\mathcal{H}_{1}({\bf
k})+\mathcal{H}_{\text{soc}}+s_{0}\otimes\mathcal{H}_{2,\vec{n}}({\bf k})]=0.$
(328)

Thus, one can conclude that this analysis for the reduction from SU(2) quasi-
symmetry down to U(1) quasi-symmetry is general, which can help to protect
quasi-nodal-plane at generic momenta.

To show this nodal plane ude to the remaining U(1) quasi-symmetry in our
approach II more explicitly, we next consider the perturbation to
$\mathcal{H}_{2,\vec{n}}({\bf k})$. Here we emphasize that one can also obtain
the common eigen-states for $\mathcal{H}_{1}({\bf
k})+\mathcal{H}_{\text{soc}}$ and ${\cal M}\cdot\vec{n}$. Different from
Approach I in Sec. C, here we take the eigen-state solution of
$s_{0}\otimes\mathcal{H}_{1}({\bf k})+\mathcal{H}_{\text{soc}}$, and do the
perturbation for the $k^{2}$-order Hamiltonian. Below we choose
$\vec{n}=(0,0,1)$ and the corresponding U(1) quasi-symmetry opeator is ${\cal
M}\cdot\vec{n}={\cal M}_{3}$. Recall that $\mathcal{H}_{1}({\bf
k})+\mathcal{H}_{\text{soc}}$ becomes block-diagonal after this unitary
transformation defined in Eq. (289),

$\displaystyle\begin{split}\mathcal{U}\left[s_{0}\otimes\mathcal{H}_{1}({\bf
k})+\mathcal{H}_{\text{soc}}\right]\mathcal{U}^{\dagger}=C_{0}+\begin{pmatrix}\mathcal{H}_{A}(\mathbf{k})&0\\\
0&\mathcal{H}_{B}(\mathbf{k})\end{pmatrix},\end{split}$ (329)

where $\mathcal{H}_{A/B}(\mathbf{k})$ are given by

$\displaystyle\mathcal{H}_{A}(\mathbf{k})$
$\displaystyle=A_{1}s_{0}\otimes\left(k_{x}\sigma_{y}+k_{y}\sigma_{x}-k_{z}\sigma_{z}\right)+\lambda_{0}\left(s_{x}\sigma_{y}+s_{y}\sigma_{x}-s_{z}\sigma_{z}\right),$
(330) $\displaystyle\mathcal{H}_{B}(\mathbf{k})$
$\displaystyle=A_{1}s_{0}\otimes\left(k_{x}\sigma_{y}-k_{y}\sigma_{x}+k_{z}\sigma_{z}\right)+\lambda_{0}\left(s_{x}\sigma_{y}-s_{y}\sigma_{x}+s_{z}\sigma_{z}\right).$
(331)

Again, note that the index $A(B)$ represent the eigenvalues $+1(-1)$ of the
quasi-symmetry ${\cal M}_{3}$. Therefore, after this unitary transformation,
the U(1) quasi-symmetry generator $\mathcal{M}_{3}$ becomes

$\displaystyle\mathcal{U}_{\mathcal{M}_{3}}\mathcal{M}_{3}\mathcal{U}_{\mathcal{M}_{3}}^{\dagger}\triangleq\mathcal{M}_{qs}=\begin{pmatrix}\sigma_{0}&0\\\
0&-\sigma_{0}\end{pmatrix}.$ (332)

The analytical solution for the two upper bands (i.e. Eq. (304a)) of
$\mathcal{H}_{1}({\bf k})+\mathcal{H}_{\text{soc}}$ in Sec. D.2 are given by

$\displaystyle|E_{A,+}(\mathbf{k})\rangle=\frac{1}{\mathcal{N}_{A,+}(\mathbf{k})}\left(-i,\frac{k_{x}-ik_{y}}{k-k_{z}},\frac{-ik_{x}+k_{y}}{k+k_{z}},1\right)^{T},$
(333a)
$\displaystyle|E_{A,-}(\mathbf{k})\rangle=\frac{1}{\mathcal{N}_{A,-}(\mathbf{k})}\Big{(}i(-A_{1}k_{z}-\lambda_{0}+E_{A,-}(\mathbf{k})),A_{1}(-k_{x}+ik_{y}),A_{1}(-ik_{x}+k_{y}),A_{1}k_{z}-\lambda_{0}+E_{A,-}(\mathbf{k})\Big{)}^{T}.$
(333b)

where the index $\pm$ in the eigen-state solution
$|E_{A,\pm}(\mathbf{k})\rangle$ are eigen-values of $({\bf S}+{\bf
L})\cdot\vec{n}_{{\bf k}}$ for nonzero ${\bf k}$. And the normalization
factors are given by
$\mathcal{N}_{A,+}(\mathbf{k})=2k/\sqrt{k_{x}^{2}+k_{y}^{2}}$, and
$\mathcal{N}_{A,-}(\mathbf{k})=2\sqrt{(E_{A,-}(\mathbf{k}))^{2}-\lambda_{0}^{2}}$.
Moreover, note that they are the two positive upper bands with eigen-energies,

$\displaystyle E_{A,+}(\mathbf{k})$ $\displaystyle=A_{1}k+\lambda_{0},$ (334a)
$\displaystyle E_{A,-}(\mathbf{k})$
$\displaystyle=\sqrt{A_{1}^{2}k^{2}+4\lambda_{0}^{2}}-\lambda_{0},$ (334b)

which have increasing energy as $k$ increases. And, the eigen-states of the
$\mathcal{H}_{B}(\mathbf{k})$ block can be related to those of
$\mathcal{H}_{A}(\mathbf{k})$ by time-reversal symmetry,

$\displaystyle|E_{B,+}(\mathbf{k})\rangle$
$\displaystyle=\mathcal{T}|E_{A,+}(-\mathbf{k})\rangle,$ (335a)
$\displaystyle|E_{B,-}(\mathbf{k})\rangle$
$\displaystyle=\mathcal{T}|E_{A,-}(-\mathbf{k})\rangle,$ (335b)

and

$\displaystyle\mathcal{H}_{B}(\mathbf{k})|E_{B,\pm}(\mathbf{k})\rangle=E_{A,\pm}(\mathbf{k})|E_{B,\pm}(\mathbf{k})\rangle,$
(336)

where $E_{A,\pm}(-\mathbf{k})=E_{A,\pm}(\mathbf{k})$ has been used. Therefore,
we have the four bands as a basis,

$\displaystyle\\{|\Psi_{+}(\mathbf{k})\rangle\\}$
$\displaystyle=\Big{\\{}(1,0)^{T}\otimes|E_{A,+}(\mathbf{k})\rangle,(1,0)^{T}\otimes|E_{A,-}(\mathbf{k})\rangle,(0,1)^{T}\otimes|E_{B,+}(\mathbf{k})\rangle,(0,1)^{T}\otimes|E_{B,-}(\mathbf{k})\rangle\Big{\\}}.$
(337)

In this basis, the $\mathcal{H}_{1}({\bf k})+\mathcal{H}_{\text{soc}}$ is
diagonal,

$\displaystyle\begin{split}\mathcal{H}_{1}({\bf
k})+\mathcal{H}_{\text{soc}}=C_{0}+\text{Diag}[&E_{A,+}(\mathbf{k}),E_{A,-}(\mathbf{k}),E_{A,+}(\mathbf{k}),E_{A,-}(\mathbf{k})].\end{split}$
(338)

After straightforward calculation, we next project the $k^{2}$-term
$\mathcal{H}_{2,\mathcal{M}_{3}}({\bf k})$ onto the basis in Eq. (337), and we
arrive at the quasi-symmetry $P$-model ($qsP$-model) as,

$\displaystyle\mathcal{H}_{qsP}(\mathbf{k})=\begin{pmatrix}E_{A,+}^{\prime}(\mathbf{k})+f_{1}(\mathbf{k})&d_{1}(\mathbf{k})-id_{2}(\mathbf{k})&0&0\\\
d_{1}(\mathbf{k})+id_{2}(\mathbf{k})&E_{A,-}^{\prime}(\mathbf{k})+f_{2}(\mathbf{k})&0&0\\\
0&0&E_{A,+}^{\prime}(\mathbf{k})-f_{1}(\mathbf{k})&d_{1}(\mathbf{k})-id_{2}(\mathbf{k})\\\
0&0&d_{1}(\mathbf{k})+id_{2}(\mathbf{k})&E_{A,-}^{\prime}(\mathbf{k})-f_{2}(\mathbf{k})\end{pmatrix},$
(339)

where $E_{A,\pm}^{\prime}(\mathbf{k})=C_{0}+B_{1}k^{2}+E_{A,\pm}(\mathbf{k})$.
For the projected four-band $qsP$-model, we check the protection from quasi-
symmetry as

$\displaystyle[\mathcal{M}_{sq},\mathcal{H}_{qsP}({\bf k})]=0.$ (340)

And all the other components are given by

$\displaystyle f_{1}(\mathbf{k})$
$\displaystyle=-\tilde{C}\frac{k_{x}k_{y}k_{z}}{k},$ (341a) $\displaystyle
f_{2}(\mathbf{k})$
$\displaystyle=-A_{1}\tilde{C}\frac{k_{x}k_{y}k_{z}}{\sqrt{A_{1}^{2}k^{2}+4\lambda_{0}^{2}}},$
(341b) $\displaystyle d_{1}(\mathbf{k})$
$\displaystyle=\frac{2(E_{A,+}(\mathbf{k})-E_{A,-}(\mathbf{k}))k_{x}k_{y}}{\mathcal{N}_{A,+}\mathcal{N}_{A,-}}\left(C_{1}+(C_{2}-C_{3})\frac{k_{z}^{2}}{k_{x}^{2}+k_{y}^{2}}\right),$
(341c) $\displaystyle d_{2}(\mathbf{k})$
$\displaystyle=\frac{2(E_{A,+}(\mathbf{k})-E_{A,-}(\mathbf{k}))k_{z}k}{\mathcal{N}_{A,+}\mathcal{N}_{A,-}}\frac{C_{2}k_{x}^{2}+C_{3}k_{y}^{2}}{k_{x}^{2}+k_{y}^{2}}.$
(341d)

The eigen-energies of the $qsP$-model are given by

$\displaystyle E_{\alpha,\beta}(\mathbf{k})$
$\displaystyle=\frac{1}{2}\left[\Delta
E_{+}(\alpha\mathbf{k})+\beta\sqrt{(\Delta
E_{-}(\alpha\mathbf{k}))^{2}+E_{d_{12}}}\right],$ (342)

with $\alpha,\beta=\pm$. And $\alpha=\pm$ are eigenvalues of
$\mathcal{M}_{qs}$ and $\beta=\pm$ are for the band index. Here we have
defined

$\displaystyle E_{d_{12}}(\mathbf{k})$
$\displaystyle=4\left[(d_{1}(\mathbf{k}))^{2}+(d_{2}(\mathbf{k}))^{2}\right],$
(343a) $\displaystyle\Delta E_{\pm}(\mathbf{k})$
$\displaystyle=E_{A,+}^{\prime}(\mathbf{k})+f_{1}(\mathbf{k})\pm(E_{A,-}^{\prime}(\mathbf{k})+f_{2}(\mathbf{k})).$
(343b)

Similar to the discussion for the nodal plane of the $qsR$-model (see Eq.
(322)), the quasi-symmetry protected nodal planes of the $qsP$-model are give
by

$\displaystyle\begin{cases}k_{x}k_{y}k_{z}<0,E_{+,-}(\mathbf{k})=E_{-,+}(\mathbf{k}),\\\
k_{x}k_{y}k_{z}>0,E_{+,+}(\mathbf{k})=E_{-,-}(\mathbf{k}).\end{cases}$ (344)

The crossings happen for bands with different eigenvalue of quasi-symmetry
$\mathcal{M}_{qs}$. We numerically check these exact crossings on the $\Gamma-
R-M$ planes, which is consistent with the results in Fig. 7 (a).

###  D.4 Perturbation theory for tiny gap of the nodal plane in Approach II

The above discussion on the nodal planes in the approach II requires a choice
of specific $k^{2}$-order terms, but in real materials, all the coefficients
before the $k^{2}$-order terms can generally be non-zero and at the same
order. Thus, our current approach does not directly explain the near nodal
plane seen in real materials. We notice that we treat the SOC terms accurately
in our approach II without any approximation, while the existence of near
nodal planes in real materials actually require the SOC strength to be much
smaller than the Fermi energy. Therefore, we below consider the limit of the
SOC strength $\lambda_{0}\ll A_{1}k_{F}$ ($k_{F}$ is the Fermi momentum) in
our Approach II, taking into account all non-zero $k^{2}$-order terms.

Following the same procedure of the perturbation projection as in the last
section, we project all the $k^{2}$-order terms
$\mathcal{H}_{2,\mathcal{M}_{1}}(\mathbf{k})+\mathcal{H}_{2,\mathcal{M}_{2}}(\mathbf{k})+\mathcal{H}_{2,\mathcal{M}_{3}}(\mathbf{k})$
onto the basis in Eq. (337), and get the entire perturbation Hamiltonian,

$\displaystyle\mathcal{H}_{qsP+qsB}({\bf
k})=\begin{pmatrix}E_{A,+}^{\prime}(\mathbf{k})+f_{1}(\mathbf{k})&d_{1}(\mathbf{k})-id_{2}(\mathbf{k})&g_{1}(\mathbf{k})&g_{2}(\mathbf{k})\\\
d_{1}(\mathbf{k})+id_{2}(\mathbf{k})&E_{A,-}^{\prime}(\mathbf{k})+f_{2}(\mathbf{k})&g_{3}(\mathbf{k})&g_{4}(\mathbf{k})\\\
g_{1}^{\ast}(\mathbf{k})&g_{3}^{\ast}(\mathbf{k})&E_{A,+}^{\prime}(\mathbf{k})-f_{1}(\mathbf{k})&d_{1}(\mathbf{k})-id_{2}(\mathbf{k})\\\
g_{2}^{\ast}(\mathbf{k})&g_{4}^{\ast}(\mathbf{k})&d_{1}(\mathbf{k})+id_{2}(\mathbf{k})&E_{A,-}^{\prime}(\mathbf{k})-f_{2}(\mathbf{k})\end{pmatrix}.$
(345)

Here the terms $d_{1},d_{2},f_{1},f_{2}$ have been given by Eq. (341). In
addition, the off-diagonal terms $g_{1,2,3,4}$ are generally to open a gap for
the quasi-nodal plane. Because these off-diagonal terms break the quasi-
symmetry $\mathcal{M}_{qs}$. And, they are given by

$\displaystyle g_{1}(\mathbf{k})$
$\displaystyle=(1-i)\tilde{C}\frac{k_{x}k_{y}k_{z}}{k},$ (346a) $\displaystyle
g_{4}(\mathbf{k})$
$\displaystyle=(i-1)A_{1}\tilde{C}\frac{k_{x}k_{y}k_{z}}{E_{A,-}(\mathbf{k})+\lambda_{0}},$
(346b) $\displaystyle g_{2}(\mathbf{k})$
$\displaystyle=\frac{2}{(k_{x}^{2}+k_{y}^{2})\mathcal{N}_{A,+}(\mathbf{k})\mathcal{N}_{A,-}(\mathbf{k})}\Big{\\{}(C_{2}+iC_{3})k_{x}k_{y}(k_{x}^{2}+k_{y}^{2})(E_{A,-}(\mathbf{k})-E_{A,+}(\mathbf{k}))$
$\displaystyle+(C_{1}+iC_{3})k_{x}k_{z}(A_{1}k_{x}k^{2}-ik_{y}k_{z}(E_{A,-}(\mathbf{k})-E_{A,+}(\mathbf{k}))-k_{x}k(E_{A,-}(\mathbf{k})-\lambda_{0}))$
$\displaystyle+(C_{2}-iC_{1})k_{y}k_{z}(A_{1}k_{y}k^{2}+ik_{x}k_{z}(E_{A,-}(\mathbf{k})-E_{A,+}(\mathbf{k}))-k_{y}k(E_{A,-}(\mathbf{k})-\lambda_{0})),$
(346c) $\displaystyle g_{3}(\mathbf{k})$
$\displaystyle=\frac{2}{(k_{x}^{2}+k_{y}^{2})\mathcal{N}_{A,+}(\mathbf{k})\mathcal{N}_{A,-}(\mathbf{k})}\Big{\\{}-(C_{2}+iC_{3})k_{x}k_{y}(k_{x}^{2}+k_{y}^{2})(E_{A,-}(\mathbf{k})-E_{A,+}(\mathbf{k}))$
$\displaystyle+(C_{1}+iC_{3})k_{x}k_{z}(A_{1}k_{x}k^{2}+ik_{y}k_{z}(E_{A,-}(\mathbf{k})-E_{A,+}(\mathbf{k}))-k_{x}k(E_{A,-}(\mathbf{k})-\lambda_{0}))$
$\displaystyle+(C_{2}-iC_{1})k_{y}k_{z}(A_{1}k_{y}k^{2}-ik_{x}k_{z}(E_{A,-}(\mathbf{k})-E_{A,+}(\mathbf{k}))-k_{y}k(E_{A,-}(\mathbf{k})-\lambda_{0}))\Big{\\}}.$
(346d)

Here $g_{3}(\mathbf{k})=-g_{2}(-\mathbf{k})$. The nodal planes are completely
gapped out. Now we use perturbation to explain why the gap is tiny by
realizing that the SOC in CoSi is weak enough for doing a perturbation
expansion for the coefficients $f_{1,2}({\bf k})$, $d_{1,2}({\bf k})$, and
$g_{1,2,3,4}({\bf k})$. By setting $\lambda_{0}/A_{1}k_{F}\to 0$, we obtain

$\displaystyle f_{1}({\bf k})$ $\displaystyle=f_{2}({\bf
k})=-\tilde{C}\frac{k_{x}k_{y}k_{z}}{k},$ (347a) $\displaystyle
d_{1}(\mathbf{k})$ $\displaystyle=d_{2}(\mathbf{k})=0,$ (347b) $\displaystyle
g_{1}(\mathbf{k})$
$\displaystyle=-g_{2}(\mathbf{k})=(1-i)\tilde{C}\frac{k_{x}k_{y}k_{z}}{k},$
(347c) $\displaystyle g_{2}(\mathbf{k})$ $\displaystyle=g_{3}(\mathbf{k})=0.$
(347d)

for the zeroth-order terms. Please notice that only the diagonal energies
$E_{A,\pm}^{\prime}$ that are eigen-energies of
$s_{0}\otimes\mathcal{H}_{1}+\mathcal{H}_{\text{SOC}}$ involve the SOC
$\lambda_{0}$. As a result, the perturbation Hamiltonian in Eq. (345) to the
zeroth-order in $\lambda_{0}$ becomes

$\displaystyle\mathcal{H}^{(0)}_{qsP+qsB}({\bf k})$
$\displaystyle=\begin{pmatrix}E_{A,+}^{\prime}(\mathbf{k})+f_{1}(\mathbf{k})&0&g_{1}(\mathbf{k})&0\\\
0&E_{A,-}^{\prime}(\mathbf{k})+f_{1}(\mathbf{k})&0&-g_{1}(\mathbf{k})\\\
g_{1}^{\ast}(\mathbf{k})&0&E_{A,+}^{\prime}(\mathbf{k})-f_{1}(\mathbf{k})&0\\\
0&-g_{1}^{\ast}(\mathbf{k})&0&E_{A,-}^{\prime}(\mathbf{k})-f_{1}(\mathbf{k})\end{pmatrix},$
(348a)
$\displaystyle=\begin{pmatrix}E_{A,+}^{\prime}(\mathbf{k})+f_{1}(\mathbf{k})&g_{1}(\mathbf{k})\\\
g_{1}^{\ast}(\mathbf{k})&E_{A,+}^{\prime}(\mathbf{k})-f_{1}(\mathbf{k})\end{pmatrix}\oplus\begin{pmatrix}E_{A,-}^{\prime}(\mathbf{k})+f_{1}(\mathbf{k})&-g_{1}(\mathbf{k})\\\
-g_{1}^{\ast}(\mathbf{k})&E_{A,-}^{\prime}(\mathbf{k})-f_{1}(\mathbf{k})\end{pmatrix}$
(348b)

where the diagonal terms are
$E_{A,+}^{\prime}(\mathbf{k})=C_{0}+B_{1}k^{2}+A_{1}k+\lambda_{0}$ and
$E_{A,-}^{\prime}(\mathbf{k})=C_{0}+B_{1}k^{2}+\sqrt{A_{1}^{2}k^{2}+4\lambda_{0}^{2}}-\lambda_{0}$.
Thus, the eigen-energies are given by

$\displaystyle E_{1,\pm}(\mathbf{k})$
$\displaystyle=C_{0}+B_{1}k^{2}+A_{1}k+\lambda_{0}\pm\sqrt{|f_{1}({\bf
k})|^{2}+|g_{1}({\bf k})|^{2}},$ (349a) $\displaystyle E_{2,\pm}(\mathbf{k})$
$\displaystyle=C_{0}+B_{1}k^{2}+\sqrt{A_{1}^{2}k^{2}+4\lambda_{0}^{2}}-\lambda_{0}\pm\sqrt{|f_{1}({\bf
k})|^{2}+|g_{1}({\bf k})|^{2}},$ (349b)

which leads to the equation for the nodal-plane solution

$\displaystyle E_{1,-}(\mathbf{k})=E_{2,+}(\mathbf{k})\quad\Rightarrow\quad
A_{1}k-\sqrt{A_{1}^{2}k^{2}+4\lambda_{0}^{2}}+2\lambda_{0}=2\sqrt{|f_{1}({\bf
k})|^{2}+|g_{1}({\bf k})|^{2}}.$ (350)

In the $k\to\infty$ limit, this equation becomes

$\displaystyle\lambda_{0}=\sqrt{|f_{1}({\bf k})|^{2}+|g_{1}({\bf
k})|^{2}}=\sqrt{3}\tilde{C}\left|\frac{k_{x}k_{y}k_{z}}{k}\right|.$ (351)

Note that $\tilde{C}>0$ in this work. And it is exactly the same Eq. (240)
obtained from the Approach I in Sec. C. At small ${\bf k}$, they differs from
each other. Moreover, the first-order correction from $\lambda_{0}$ will open
a tiny gap for the nodal planes, which is actually the second-order
perturbation theory in Approach I. Based on this analysis, we conclude that
the results of Approach II is equivalent from those of Approach I.

##  Appendix E The projected two-band model

In this section, we further project the first-order-perturbation 4-band
Hamiltonian, the $P$-model (e.g. see Eq. (218) in Sec. C.1) onto an effective
model that only consists of the two bands forming the nodal plane. To do that,
we consider the two eigen-states of the $P$-model Hamiltonian in Eq. (218),

$\displaystyle\mathcal{H}_{P}^{eff(1)}({\bf k})$
$\displaystyle=(E_{+}+B_{1}k^{2})s_{0}\omega_{0}+\mathcal{H}_{\text{soc},+}^{eff(1)}({\bf
k})+\mathcal{H}_{k^{2},+}^{eff(1)}({\bf k}),$ (352)

which is marked as the $P$-model around $R$-point. And

$\displaystyle\begin{split}\mathcal{H}_{\text{soc},+}^{eff(1)}({\bf
k})&=\lambda_{0}\left(\lambda_{x}s_{x}+\lambda_{y}s_{y}+\lambda_{z}s_{z}\right)\omega_{0},\\\
\mathcal{H}_{k^{2},+}^{eff(1)}({\bf
k})&=\tilde{C}k^{2}s_{0}\left(d_{x}\omega_{x}+d_{y}\omega_{y}+d_{z}\omega_{z}\right).\end{split}$
(353)

where $\tilde{C}=C_{1}-C_{2}+C_{3}$, and $\omega_{x,y,z}$ are Pauli matrices
for the $\\{A+,B+\\}$ band subspace. The eigen-energies are give by Eq. (230)

$\displaystyle\begin{split}E_{\alpha\beta}(k,\theta,\phi)=C_{0}+B_{1}k^{2}+A_{1}k+\alpha\lambda_{0}+\beta\tfrac{\sqrt{3}}{4}\tilde{C}k^{2}|\sin
2\phi\sin 2\theta\sin\theta|,\end{split}$ (354)

where $\alpha=\pm$, $\beta=\pm$, and $\sin 2\phi\sin
2\theta\sin\theta=4k_{x}k_{y}k_{z}/k^{3}$. Therefore, these two states
$\\{|\alpha=+,\beta=-\rangle,|\alpha=-,\beta=+\rangle\\}$ that give the eigen-
energies $E_{\alpha=+,\beta=-}$ and $E_{\alpha=-,\beta=+}$. The basis eigen-
wave function for the two-band model can be derived by solving the four-band
$P$ model, $\mathcal{H}_{P}^{eff(1)}$. After straightforward calculation, we
find

$\displaystyle|E_{+,-}\rangle$
$\displaystyle=\frac{1}{N_{+,-}}(\lambda_{z}+E_{\lambda},\lambda_{x}+i\lambda_{y})^{T}\otimes(d_{z}-E_{d},d_{x}+id_{y})^{T},$
(355a) $\displaystyle|E_{-,+}\rangle$
$\displaystyle=\frac{1}{N_{-,+}}(\lambda_{z}-E_{\lambda},\lambda_{x}+i\lambda_{y})^{T}\otimes(d_{z}+E_{d},d_{x}+id_{y})^{T},$
(355b)

where we have

$\displaystyle N_{s,\omega}^{2}$
$\displaystyle=((\lambda_{z}+sE_{\lambda})^{2}+\lambda_{x}^{2}+\lambda_{y}^{2})((d_{z}+\omega
E_{d})^{2}+d_{x}^{2}+d_{y}^{2}),$ (356a) $\displaystyle E_{\lambda}$
$\displaystyle=\sqrt{\lambda_{x}^{2}+\lambda_{y}^{2}+\lambda_{z}^{2}},$ (356b)
$\displaystyle E_{d}$ $\displaystyle=\sqrt{d_{x}^{2}+d_{y}^{2}+d_{z}^{2}},$
(356c)

where we define $\lambda_{x}=\sin\theta\cos\phi$,
$\lambda_{y}=\sin\theta\sin\phi$, $\lambda_{z}=\cos\theta$,
$d_{x}=\cos\theta\sin^{2}\theta\sin\phi\cos\phi(\cos\phi+\sin\phi)$,
$d_{y}=\cos\theta\sin^{2}\theta\sin\phi\cos\phi(\sin\theta+\cos\theta(\cos\phi-\sin\phi))$,
and
$d_{z}=\sin^{2}\theta\sin\phi\cos\phi(\cos^{2}\theta+\sin\theta\cos\theta(-\cos\phi+\sin\phi))$.

Within the sub-space formed by these two eigen-states, the effective
Hamiltonian $\mathcal{H}_{P}^{eff(1)}$ can be reduced to a two-band model

$\displaystyle\mathcal{H}_{eff}=\epsilon_{0}+d_{z}(\mathbf{k})\sigma_{z},$
(357)

where $\epsilon_{0}=C_{0}+B_{1}k^{2}+A_{1}k$ and
$d_{z}(\mathbf{k})=\lambda_{0}-\tfrac{\sqrt{3}}{4}\tilde{C}k^{2}|\sin
2\phi\sin
2\theta\sin\theta|=\lambda_{0}-\sqrt{3}\tilde{C}\frac{|k_{x}k_{y}k_{z}|}{k}$.
This Hamiltonian commutes with $\sigma_{z}$, which in turn generates the U(1)
quasi-symmetry group for the two-band model. The physical viewpoint of this
quasi-symmetry is related to the spin texture of the Fermi surfaces in the
momentum space. Due to the SOC term with $\mathbf{k}\cdot\mathbf{s}$ in the
$P$-model, the spin texture is in a hedgehog form on each Fermi surface
($\langle\mathbf{s}\rangle\sim\alpha\mathbf{k}$ with $\alpha=\pm$). It
indicates that the crossings between these two Fermi surfaces have opposite
spin-polarization. Therefore, according to the low-energy effective
Hamiltonian, the quasi-symmetry that protects the approximate nodal planes can
be viewed as the spin U(1) symmetry along the momentum direction.

Now we discuss the breaking of the quasi-symmetry by second-order perturbation
corrections. To see that, we can further project $\mathcal{H}_{P}^{eff(2)}$
into the subspace and the effective Hamiltonian includes two more terms

$\displaystyle\mathcal{H}_{eff(2)}=\delta d_{x}(\mathbf{k})\sigma_{x}+\delta
d_{y}(\mathbf{k})\sigma_{y},$ (358)

which are due to the spin-flipping terms. As a result, the co-dimension of the
nodal plane becomes 3 instead of 1, explaining the gap opening. While the
detailed expressions for $\delta d_{x}$ and $\delta d_{y}$ are complex and not
orthogonality, we give an estimate of the typical magnitude of the gap
$\Delta_{eff(2)}=2\sqrt{(\delta d_{x})^{2}+(\delta d_{y})^{2}}\sim 0.8$ meV
for $k_{F}=0.13$ Å-1, being the same order with the DFT estimations [1].

Therefore, perturbation theory leads to a general two-band model

$\displaystyle\mathcal{H}_{2\times
2}=\delta_{d_{x}}\sigma_{x}+\delta_{d_{y}}\sigma_{y}+\delta_{d_{z}}\sigma_{z}.$
(359)

The gap of this model is given by

$\displaystyle\Delta_{k,\text{soc}}=2\sqrt{(\delta_{d_{x}})^{2}+(\delta_{d_{y}})^{2}+(\delta_{d_{z}})^{2}}.$
(360)

If we impose the constraint of the quasi-symmetry $\sigma_{z}$ that commutes
with $\mathcal{H}_{2\times 2}$, we can always obtain the constraint equation
for the nodal planes by solving $\delta_{d_{z}}=0$.

Next, let us consider the other terms that break the quasi-symmetry, by taking
Eq. (266) as an example. The projected two-band model is then numerically
calculated directly. The three components are given by

$\displaystyle\begin{split}\delta_{d_{x}}&=\text{Re}[\langle
E_{+,-}|\mathcal{H}_{k,\lambda_{2},eff}|E_{-,+}\rangle],\\\
\delta_{d_{y}}&=-\text{Im}[\langle
E_{+,-}|\mathcal{H}_{k,\lambda_{2},eff}|E_{-,+}\rangle],\\\
\delta_{d_{z}}&=\frac{1}{2}[\langle
E_{+,-}|\mathcal{H}_{k,\lambda_{2},eff}|E_{+,-}\rangle-\langle
E_{-,+}|\mathcal{H}_{k,\lambda_{2},eff}|E_{-,+}\rangle],\end{split}$ (361)

The numerical results are shown in Fig. 8. The results only depend on $\theta$
and $\phi$, then, on the $k_{r}=0.15$ Å-1 sphere with $\lambda_{2}=0.005$ eV,
the numerical calculated $\delta_{d_{x}}$ and $\delta_{d_{y}}$ are shown in
(a) and (b), respectively. In the whole space, $\delta_{d_{x}}\neq 0$ and
$\delta_{d_{y}}$ expect the Weyl point along the $\Gamma-R$ line, which is
shown in Fig. 8 (c) by log-plotting the inverse of the gap
$\log[\frac{1}{\sqrt{\delta_{d_{x}}^{2}+\delta_{d_{y}}^{2}}}]$ (i.e., the gap
vanishes only at the Weyl point). The entire quasi-nodal plane are gapped out
by including the general liner-$k$ SOC Hamiltonian.

Figure 8: The projected two-band model with $\delta d_{x}$ in (a), $\delta
d_{y}$ in (b) and the corresponding gap in (c). We find that the gap only
closes at the Weyl point. It indicates the quasi-symmetry protected nodal
plane is completely gapped out by the linear-k SOC Hamiltonian. Here
$k_{r}=0.15$ Å-1 and $\lambda_{2}=0.005$ eV.