Does anyone know whether $(x,y)\rightarrow (-x,-y)$ is an inversion transformation or not?
I know that the standard inversion (parity) transformation in two dimensions should be something like $(x,y)\rightarrow (x,-y)$ which only flips an odd number of spatial coordinates. However, if we have a plane embedded in three-dimensional space and it is located at $z=0$, we may have such an inversion transformation for it: $(x,y,0)\rightarrow (-x,-y,0)$. Does it imply $(x,y)\rightarrow (-x,-y)$ is also a kind of inversion symmetry in two dimensions?
In fact, I found this question when I study the Bernevig-Hughes-Zhang model in physics. Anyone who is familiar with this model may have a better understanding of my question. The Bernevig-Hughes-Zhang model is a famous two-dimensional model for topological insulators with inversion symmetry. It's Bloch Hamiltonian is $$H(\mathbf{k})=\mathbf{\Gamma}\cdot \mathbf{d}$$ where the components of $\mathbf{\Gamma}$ are just some Dirac matrices: $$ \begin{array}{l} \Gamma^{0}=\mathbb{1} \otimes \mathbb{1} \\ \Gamma^{1}=\tau_{z} \otimes \mathbb{1} \\ \Gamma^{2}=\tau_{y} \otimes \mathbb{1} \\ \Gamma^{3}=\tau_{x} \otimes s_{x} \\ \Gamma^{4}=\tau_{x} \otimes s_{y} \\ \Gamma^{5}=\tau_{x} \otimes s_{z} \end{array} $$ and $\mathbf{d}$ reads $$ \begin{array}{lc} \hline \hline d_{0} & \left(\varepsilon_{s}+\varepsilon_{p}\right) / 2-\left(t_{s s}-t_{p p}\right)\left(\cos x_{1}+\cos x_{2}\right) \\ d_{1} & \left(\varepsilon_{s}-\varepsilon_{p}\right) / 2-\left(t_{s s}+t_{p p}\right)\left(\cos x_{1}+\cos x_{2}\right) \\ d_{2} & 2 t_{s p} \sin x_{1} \\ d_{3} & 0 \\ d_{4} & 0 \\ d_{5} & 2 t_{s p} \sin x_{2} \\ \hline \hline \end{array} $$ with $x_k=\mathbf{k}\cdot\mathbf{e}_k$.
In this system, the inversion operator is $P=\tau_z\otimes \mathbb{1}$. We can easily verify $PH(k_x,k_y)P^{-1}=H(-k_x,-k_y)$. However, if we assume the inversion transformation in two dimensions only flips one spatial coordinate (we take $(x,y)\rightarrow (x,-y)$ as the example), then the real inversion symmetry restriction for the Bloch Hamiltonian should be $PH(k_x,k_y)P^{-1}=H(k_x,-k_y)$, which obviously contradicts what we have gotten. Only if $(x,y)\rightarrow (-x,-y)$ is viewed as an inversion transformation, then inversion symmetry restriction becomes $PH(k_x,k_y)P^{-1}=H(-k_x,-k_y)$ and everything meets. However, I doubt about this point because as far as I know, a matrix representation of $P$ (in any number of dimensions) should have determinant equal to $-1$.
You can find this model in Fu and Kane's paper Topological insulators with inversion symmetry.