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It is said that if a lattice has inversion symmetry, then the Berry curvature, $\vec{\Omega}(\vec{k})$ is even in $\vec{k}$, i.e. $$\vec{\Omega}(\vec{k})=\vec{\Omega}(-\vec{k})$$ I have also derived this when the inversion transformation changes the sign of every coordinate, which is what happens when we have an odd number of dimensions. In $2D$, though, inversion transformation is equivalent to one reflection, i.e. it just changes the sign of one coordinate. For example $$P(x,y)^T=(x,-y)$$ where $P$ is the inversion transformation.
In this case, I find that under inversion symmetry, $$\vec{\Omega}(k_1,k_2)=-\vec{\Omega}(k_1,-k_2)$$ If this is true, then they each of the two cases (odd number of dimensions VS even number of dimensions) give very different restrictions to the Berry curvature(and thus the Chern number). All the sources that I have read about symmetries and their relation to the band structure and the Berry phase related quantities do not comment on this. Some just comment that if we have inversion symmetry (in the form of the first equation given above) the Berry curvature is even in $\vec{k}$.

So, I am wondering why don't they also comment on the case of even number of dimensions since so many models refer to the $2D$ case (such as the Qi-Wu-Zhang model, or the Haldane model for graphene)? Does the $2D$ case not give different restrictions to the Chern number?

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In 2D, inversion is not completely equivalent to reflection.

Loosely speaking, if you imagine you have a 2D plane in 3D and do this operation to the plane the result $$ (x,y,z) \to (x, -y,z) \hspace{1cm} (1) $$ would be "up-side-down" compared to the result of an in-plane parity transformation of the form $$ (x,y,z) \to (-x, -y,z) \hspace{1cm} (2) $$ as can be seen from a $\pi$ rotation around the $y$-axis. So you would have to account for this effect as well. We can illustrate this by the following example:

enter image description here

This distribution of arrows is invariant under transformation law (2). Is is invariant under inversion in the usual sense. However, when we apply law (1) the result is

enter image description here

Clearly (1) is not symmetry in this situation, while (2) is. Inversion is therefore not the same as flipping one coordinate.

As this answer receives some mixed response I want to elaborate from a different angle. It has to do with the rotational sense and is the root of the problem. If you do your parity transformation according to prescription (2) the rotational sense of the plane is preserved, while it is not for prescription (1), where $x$ and $y$ essentially change their roles.

To close the circle to the discussion of the Berry phase, there is a simple physical argument based on a semiclassical analysis. It provides an expression for the velocity of the Bloch state $n\mathbf{k}$ (in crystal momentum representation) in the presence of an external electric field $\mathbf{E}$ as

$$ \mathbf{v}_n = \frac{1}{\hbar} \frac{\partial \epsilon_n(\mathbf{k})}{\partial \mathbf{k} } - \frac{e}{\hbar} \mathbf{E} \times \boldsymbol{\Omega}_n (\mathbf{k}) . $$

Based on this result one can immediately deduce the symmetry properties. A 3D space inversion of the form $\mathbf{x}\to -\mathbf{x}$ will change the sign of $\mathbf{v}_n$, $\mathbf{E}$ and $\mathbf{k}$. This fixes $\boldsymbol{\Omega}_n (\mathbf{k}) = \boldsymbol{\Omega}_n (-\mathbf{k})$.

This analysis applies to in-plane inversion according to Eq. (2). We can see this by taking a closer look at the second term. In 2D, only the $\Omega_z$ term is finite and therefore

$$ \mathbf{E} \times \boldsymbol{\Omega}_n = \begin{pmatrix} E_x \\ E_y \\E_z \end{pmatrix} \times \begin{pmatrix} 0\\ 0\\ \Omega^z_n \end{pmatrix} = \begin{pmatrix} E_y \Omega^z_n(k_x, k_y) \\ -E_x \Omega^z_n(k_x, k_y) \\ 0 \end{pmatrix} \overset{(2)}{\longrightarrow} \begin{pmatrix} -E_y \Omega^z_n(-k_x, -k_y) \\ +E_x \Omega^z_n(-k_x, -k_y) \\ 0 \end{pmatrix}. $$ In order to find $$ \begin{pmatrix} v_x \\ v_y \\ 0 \end{pmatrix} \overset{(2)}{\longrightarrow} \begin{pmatrix} -v_x \\ -v_y \\ 0 \end{pmatrix}, $$ we require that $\boldsymbol{\Omega}_n (\mathbf{k}) = \boldsymbol{\Omega}_n (-\mathbf{k})$ still holds.

How about your proposal, i.e., Eq. (1)? Lets assume our system has this symmetry. One component of $\mathbf{v}_n$, $\mathbf{E}$ and $\mathbf{k}$ change sign while the other one does not. We can calculate this again explicitly

$$ \begin{pmatrix} E_y \Omega^z_n(k_x, k_y) \\ -E_x \Omega^z_n(k_x, k_y) \\ 0 \end{pmatrix} \overset{(1)}{\longrightarrow} \begin{pmatrix} -E_y \Omega^z_n(k_x, -k_y) \\ -E_x \Omega^z_n(k_x, -k_y) \\ 0 \end{pmatrix}. $$ Here, we want to see that $$ \begin{pmatrix} v_x \\ v_y \\ 0 \end{pmatrix} \overset{(1)}{\longrightarrow} \begin{pmatrix} v_x \\ -v_y \\ 0 \end{pmatrix} . $$ This means under symmetry Eq. (1), we require that $\Omega^z_n(k_x ,k_y) = -\Omega^z_n(k_x, -k_y)$. However, the crucial assumption is that Eq. (1) is a symmtetry of your crystal. Otherwise this argument breaks down. And as I have shown above, symmetry Eq. (1) and Eq. (2) are in fact different in general. The transformed system appears "upside-down" w.r.t to the original. Correct for this effect and the Berry curvature transforms in the expected way again.

More on this can be found in section III. B of Xiao, Di, Ming-Che Chang, and Qian Niu. "Berry phase effects on electronic properties." Reviews of modern physics 82.3 (2010): 1959.

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  • $\begingroup$ This answer appears to receive a mixed reaction in terms of up and down votes. It would be interesting to hear some arguments (in favor or not) $\endgroup$ – sagittarius_a Jun 20 '18 at 11:45
  • $\begingroup$ While I have not upvoted or downvoted your answer, I feel that it lacks with respect to my question. In particular, you do not address the restrictions that are imposed on the Berry curvature and its concequences, which lie at the heart of my question. While what you answered are correct (and would have helped me a few months back), I took these for granted and focused on the Berry curvature in my qiestion. Thank you. $\endgroup$ – TheQuantumMan Jun 21 '18 at 7:37
  • $\begingroup$ I have a couple of questions. First, equations (1) and (2) give two different transformations. Which one is the parity (or space inversion) transformation? I am a bit confused about this point. Also, when giving the semi-classical equation of motion, you say that under space inversion, $\mathbf{v}_n$, $\mathbf{k}$ and $\mathbf{E}$ change sign (all their components). This certainly does not agree with transformations (1) and (2) though. $\endgroup$ – TheQuantumMan Jun 23 '18 at 15:35
  • $\begingroup$ I rewrote some sections in order to clarify this $\endgroup$ – sagittarius_a Jun 26 '18 at 11:24
  • $\begingroup$ It is much clearer now. I have one more question if it's OK with you: everybody is saying that space inversion symmetry in 2D means that $H(k_1,k_2)=H(k_1,-k_2)$. It seems that this is compatible only with (1) rather than (2). But, then everybody says that under space inversion symmetry, $\vec{\Omega}_n(k_1,k_2)=\vec{\Omega}_n(-k_1,-k_2)$. How is the one condition (for the Hamiltonian) is compatible with the second condition (for the Berry curvature)? [Note that a reference for this is the 2018 review paper by Shankar on topological insulators] $\endgroup$ – TheQuantumMan Jun 27 '18 at 12:54

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