# Gauge constraint in the definition of the $\mathbb{Z}_2$ invariant

In Fu and Kane's paper from 2006, the authors define the $$\mathbb{Z}_2$$ invariant for time-reversal invariant topological insulators as an obstruction to Stoke's theorem, $$$$\nu=\frac{1}{2\pi}\left[\oint_{\partial\text{HBZ}}\mathrm{d}\mathbf{k}\cdot\mathbf{A}(\mathbf{k})-\int_{\text{HBZ}}\mathrm{d}^2\mathbf{k}F(\mathbf{k})\right]\text{(mod 2)},$$$$ where $$\mathbf{A}(\mathbf{k})=\sum_{n,\sigma}\langle u_{n,\sigma}(\mathbf{k})|\mathrm{i}\nabla|u_{n,\sigma}(\mathbf{k})\rangle$$ and $$F=\nabla\times\mathbf{A}(\mathbf{k})$$ (where $$|u_{n,\sigma}(\mathbf{k})\rangle$$s are the periodic Bloch functions). The integrals are evaluated over half of the Brillouin zone. This formula is not gauge invariant and the authors impose a time reversal constraint on the periodic Bloch functions, $$$$|u_{n,\uparrow}(-\mathbf{k})\rangle=\Theta|u_{n,\downarrow}(\mathbf{k})\rangle$$$$ $$$$|u_{n,\downarrow}(-\mathbf{k})\rangle=-\Theta|u_{n,\uparrow}(\mathbf{k})\rangle$$.$$ However, I do not understand how this fixes the gauge. I can always define the new Bloch states,

$$$$|u_{n,\uparrow}(\mathbf{k})\rangle'=\mathrm{e}^{\mathrm{i}\theta_{\mathbf{k}}}|u_{n,\uparrow}(\mathbf{k})\rangle$$$$ $$$$|u_{n,\downarrow}(\mathbf{k})\rangle'=\mathrm{e}^{-\mathrm{i}\theta_{-\mathbf{k}}}|u_{n,\downarrow}(\mathbf{k})\rangle$$$$ which satisfies the gauge constraint for any arbitrary choice of $$\theta_{\mathbf{k}}$$. The Berry connection gets modified to, $$$$\mathbf{A}(\mathbf{k})\rightarrow\mathbf{A}(\mathbf{k})+\nabla(\theta_{\mathbf{k}}-\theta_{-\mathbf{k}}).$$$$ If I choose a $$\theta$$ that has a singularity in only one-half of the Brillouin zone, it can change the value of $$\nu$$ by 1. My question is, do we need any additional constraints on the Bloch functions to define $$\nu$$ unambiguously modulo 2? Thank you!

• I may be misunderstanding you, but if $\theta_\mathbf k$ is singular at some $\mathbf k_0$ then $\theta_\mathbf k -\theta_{-\mathbf k}$ is singular at both $\mathbf k_0$ and $-\mathbf k_0$, no? Commented Oct 10, 2023 at 0:23
• @J.Murray Indeed there will be a singularity at $\mathbf{k}_0$ and $-\mathbf{k}_0$, however only one of them will lie inside the HBZ, therefore when you integrate over the HBZ, the singularity at $\mathbf{k}_0$ will contribute and change the value of $\nu$ by 1. Commented Oct 10, 2023 at 1:40
• I suppose I don't really understand the question. To which term in your first expression is your gauge transformation meant to contribute? $F$ is gauge invariant, and the gradient of a scalar function vanishes under a closed loop integral. Are you imagining that your $\theta$'s have some kind of branch cut singularity that intersects the integration path? Commented Oct 10, 2023 at 3:14
• Yes, I'm trying to understand what happens when $\theta$ has a branch cut. For example, if we choose $\theta=\arctan(y/x)$, the integral of $\nabla\theta$ around a contour encircling the origin will give us 1. Commented Oct 10, 2023 at 13:11
• Sorry, the integral around a contour encircling the origin will give us $2\pi$. Commented Oct 10, 2023 at 14:11

As you have mentioned, upon a gauge transformation, we have $$$$\mathbf{A}(\mathbf{k})\rightarrow\mathbf{A}(\mathbf{k})+\nabla(\theta_{\mathbf{k}}-\theta_{-\mathbf{k}}).$$$$ so $$\nu$$ is changed by \begin{align} \Delta\nu & =\frac{1}{2\pi}\oint_{\partial\text{HBZ}}\nabla(\theta_{\mathbf{k}}-\theta_{-\mathbf{k}})\\ & = 2\cdot \frac{1}{2\pi}\oint_{\partial\text{HBZ}}\nabla\theta_{\mathbf{k}} \end{align} where the loop integral is the winding of $$\theta_{\bf k}$$ and yield integer multiple of $$2\pi$$ thus $$\Delta \nu$$ gives 0 mod 2.