Winding number in Su-Schrieffer-Heeger (SSH) model

Question

I'm studying the SSH model from this review and on page 14, equation (1.38), they give a formula from evaluating the winding number saying it's easy to check it. Now, I've done the math and came up with: \begin{equation} \left(\boldsymbol{\tilde{d}}(k)\times\frac{d}{dk}\boldsymbol{\tilde{d}}(k)\right)_z=\frac{\omega^2+v\omega\cos k}{v^2+\omega^2+2v\omega\cos k} \end{equation} and to me this is not a function I'd say is easy to integrate to check if the definition of the winding number is correct.

Is there a simpler way to check that $$\frac{1}{2\pi}\int \left(\boldsymbol{\tilde{d}}(k)\times\frac{d}{dk}\boldsymbol{\tilde{d}}(k)\right)_z dk = \nu, \,\, \nu\in \mathbb{Z}$$

is true (eq 1.38) ?

The definition of $\boldsymbol{d}$ is given by (1.18)

$$d_x(k) = \nu + w \cos k; \qquad \qquad d_y(k) = w \sin k; \qquad \qquad d_z(k)=0. $$

Answer 1

We can write this a little differently to make things more clear:

\begin{align} w &= \frac{1}{2\pi}\int \frac{\vec{d}}{d}\times \partial_k \frac{\vec{d}}{d} dk\\ &= \frac{1}{2\pi}\int \frac{\vec{d}}{d}\times \left( \frac{\partial_k\vec{d}}{d}-\frac{\vec{d}\partial_k d}{d^2}\right) dk &\qquad\text{(product rule)}\\ &= \frac{1}{2\pi}\int \frac{\vec{d}\times \partial_k\vec{d}}{d^2} dk&\qquad\text{($\vec d\times \vec d=0$)}\\ \end{align}

Now, draw a picture to convince yourself of the following: $$ \frac{\vec{d}\times \partial_k\vec{d}dk}{d^2}=\frac{\vec{d}\times\delta \vec{d}}{d^2}$$ is the angle, in radians, between $\vec{d}$ and $\vec{d}+\delta\vec d$. Then by integrating this, you find the total number of radians that the vector $\vec{d}$ rotates through. This (divided by $2\pi$) is exactly the winding number.