Basically, your question is related to the relationship between symmetries and energy bands. In the following, I will give a summary you can refer to and give a course link to prove them.
- Time-reverasl symmetry ($\mathcal{T}$) implies:
$$E_{n,-\chi}(-\vec{k})=E_{n,\chi}(\vec{k}) \tag{1} $$
- Inversion or parity symmetry ($\mathcal{P}$) implies:
$$E_{n,\chi}(-\vec{k})=E_{n,\chi}(\vec{k}) \tag{2} $$
- Then in crystals that have inversion and time-reversal symmetry implies:
$$E_{n,-\chi}(\vec{k})=E_{n,\chi}(\vec{k}) \tag{3} $$
Here $n$ is the energy quantum number, $\vec{k}$ the crystal momentum, and $\chi$ the spin quantum number. This expression $(3)$ should answer your question "$\mathcal{P}\mathcal{T}$-symmetry leading to spin-degenerate band structures"
You may want to know how to derive these conclusions, please take a look at these slides. (All these conclusions are borrowed from that, in which all the derivations are very pedagogical if you have basic training in solid-state physics.)
The paper you cited discussed the change of energy band when considering different symmetry breaking:
- (a): $\mathcal{P}$ ($\checkmark$) and $\mathcal{T}$ ($\checkmark$)
- (b): $\mathcal{P}$ ($\times$) and $\mathcal{T}$ ($\checkmark$)
- (c): $\mathcal{P}$ ($\times$) and $\mathcal{T}$ ($\times$)
and the material realization for case (c) with bilayer CrI$_3$. And the symmetry analysis of bilayer CrI$_3$ is also stated in the main manuscript (section: Symmetry of 2D magnetic insulator CrI$_3$):
The atomic crystal of CrI$_3$ exhibits inversion symmetry ($\mathcal{P}$) with two inversion centers, one inside the monolayer and the other in between neighboring monolayers. The AFM order reduces the crystal symmetry by removing the interlayer inversion center. Therefore, an AFM bilayer breaks the inversion symmetry, although an FM bilayer does not. Another important feature is that the AFM ordered phase preserves $\mathcal{P}\mathcal{T}$ symmetry while it breaks $\mathcal{P}$ and $\mathcal{T}$ independently; hence, no polarization exists in the ordered phase.
- Now I believe you are comfortable about what the author is talking about with (1), (2), and (3).
Disclaimer: The previous answer can not be accepted. Because it doesn't solve the main point: $\mathcal{P}$ and $\mathcal{T}$ are broken but the combined $\mathcal{PT}$ is conserved.
I have sent an email to the corresponding author of CrI3 and state the problem with him. I will list the main points below although I can't prove them.
#Second email:
- TRS induces Kramers degeneracy, at k that respects TRS (like the Gamma point). This is because T is an anti-unitary operator. In other words, the Kramers theorem applies for any anti-unitary operator.
- The combined symmetry PT is also anti-unitary, because P is unitary and T is anti-U. Therefore, PT leads to Kramers degeneracy, at k that respect PT. As you know, any k-points respect PT.
#First email:
I assume that you understand the Kramers degeneracy induced by T.
Because PT is antisymmetric, PT leads to the same Kramers degeneracy at generic k point. Therefore, everywhere is doubly degenerate in the band structure.
This is irrelevant to whether P/T is the symmetry of the system.
May it helps.
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A more detailed point-to-point answer after a long time:
Is there a rigorous way to look at new material, to say if PT-symmetry would lead to a spin degeneracy? Because from what I've read, and I could totally be wrong, it looks like people do band structure calculations and then attributes such a degeneracy to PT-symmetry.
Both the time-reversal operator $(\mathcal{T})$ and $\mathcal{PT}$ operator are anti-unitary operators for spin-half particles and admit the Kramers' degeneracy theorem for the spin (One can find an exact proof for this degeneracy theorem in this book: chapter IV)
I do not hold a good theoretical physics footing, so I don't know much about PT symmetry with regard to specific systems, but I do know about the universal CPT symmetry that is preserved. Is one just a subset of the other?
$\mathcal{CPT}$ is a symmetry operator and $\mathcal{PT}$ is also a symmetry operator, therefore $\mathcal{PT}$ can not be considered as a subset of $\mathcal{CPT}$.
Finally, I am having a hard time trying to understand how the system can break inversion symmetry and time-reversal symmetry but the product of these two symmetry operators is still a 'good' quantum number i.e. one that commutes with the Hamiltonian.
This question can be explained by the following screencast:
