# Hermiticity of spin-orbit coupling in real space

In the Kane-Mele model, the spin-orbit coupling is defined in real space as

$$\sum_{\langle \langle i j \rangle \rangle \alpha \beta} i t_2 \nu_{ij} s^z_{\alpha \beta} c_{i \alpha}^\dagger c_{j \beta}$$

where the sum is over next-nearest-neighbor sites on a honeycomb lattice, and $$\nu_{ij} = - \nu_{ij} = \pm 1$$ depends on the orientation of the next-nearest-neighbor bonds (I don't believe the details of how $$\nu_{ij}$$ is calculated is relevant for Hermiticity). I am having difficulty understanding how this term is Hermitian. Taking the Hermitian conjugate seemingly gives

$$\left(\sum_{\langle \langle i j \rangle \rangle \alpha \beta} i t_2 \nu_{ij} s^z_{\alpha \beta} c_{i \alpha}^\dagger c_{j \beta}\right)^\dagger = \sum_{\langle \langle i j \rangle \rangle \alpha \beta} (-i) t_2 \nu_{ij} s^z_{\alpha \beta} c_{i \alpha} c_{j \beta}^\dagger = \sum_{\langle \langle i j \rangle \rangle \alpha \beta} i t_2 \nu_{ij} s^z_{\alpha \beta} c_{j \beta}^\dagger c_{i \alpha} \\ = -\sum_{\langle \langle i j \rangle \rangle \alpha \beta} i t_2 \nu_{ij} s^z_{\alpha \beta} c_{i \alpha}^\dagger c_{j \beta}$$ where in the final line we have relabeled indices and used that $$\nu_{ij} = - \nu_{ji}$$, $$s^z_{\alpha \beta} = s^z_{\beta \alpha}$$. I must be missing something obvious, but this seems to show that the term is anti-Hermitian, instead of Hermitian. What am I missing here?

In the first expression after the equals sign on the first line you have $$c_{i\alpha}c^\dagger_{j\beta}$$. It should be $$c^\dagger_{j\beta}c_{i\alpha}$$ since $$(AB)^\dagger = B^\dagger A^\dagger$$.
From the first to second expression on the first line, the Conjugation operation should transpose the matrix $$v_{ij}$$ to $$v_{ji}$$, giving the missing minus sign.