On insensitivity of this topological action (which is the winding number) to arbitrary variations

Question

I am studying the chapter on topology of Altland and Simons' Condensed Matter Field Theory and came across this discussion:

Consider a field theory such that $ \mathbf{n}: S^2 \mapsto S^2, \mathbf{x} \mapsto \mathbf{n}(\mathbf{x})$ and that $|\mathbf{n}|=1$. The relevant homotopy group is $\pi_2(S^2) \simeq \mathbb{Z}$. The action is given by

\begin{equation} S[\mathbf{n}] = \dfrac{i\theta}{4\pi} \int dx_1 dx_2 \{ \mathbf{n} \cdot (\partial_1 \mathbf{n} \times \partial_2 \mathbf{n} ) \} \end{equation}

Suppose the variation on $\mathbf{n}$ is such that $\mathbf{n} \rightarrow \mathbf{n} + \epsilon \mathbf{m}$, where $\epsilon$ is infinitesimal and $\mathbf{m}$ is arbitrary vector field. The variation is presented differently in the book.

Since $|\mathbf{n}|=1$, we have $\mathbf{m}\cdot \mathbf{n}=0$, i.e. they are perpendicular, and that $\epsilon^2 \simeq 0$. Moreover, $\partial_i \mathbf{n} \perp \mathbf{n}$ for $i=1,2$.

Here's my question: I managed to end up with

\begin{equation} \delta S = \dfrac{3 i\theta}{4\pi} \int dx_1 dx_2 \{ \epsilon \mathbf{m} \cdot (\partial_1 \mathbf{n} \times \partial_2 \mathbf{n} ) \} + \dfrac{ i\theta}{4\pi} \int dx_1 dx_2 \{\epsilon \partial_1 (\mathbf{n} \cdot ( \mathbf{m} \times \partial_2 \mathbf{n} )) + \epsilon \partial_2 (\mathbf{n} \cdot ( \partial_1 \mathbf{n} \times \mathbf{m})) \} \end{equation}

The first integral, which is only the term shown in the book, is obviously zero since $\mathbf{m} \perp \mathbf{n}$ while $(\partial_1 \mathbf{n} \times \partial_2 \mathbf{n} ) \parallel \mathbf{n}$.

The second integral must also be zero as it is argued that the action above is insensitive to small variations, but I cannot figure out the argument.

Answer 1

The second term "vanishes" in the sense of that it only contributes to a boundary term. If the integral is taken over a closed manifold (such as $S^2$), the boundary term should vanish because a closed manifold has no boundary.

To see this, let us first introduce the one-form Berry connection $A=A_1\mathrm{d}x_1+A_2\mathrm{d}x_2$: $$A_i=\boldsymbol{n}\cdot(\boldsymbol{m}\times\partial_i\boldsymbol{n}).$$ Upto an unimportant overall factor, the second term can be written as $$\int\mathrm{d}x_1\mathrm{d}x_2(\partial_1 A_2-\partial_2 A_1)=\int_\Omega\mathrm{d}A=\int_{\partial\Omega}A,$$ where $\Omega$ denotes the manifold to be integrated over and $\partial \Omega$ is its boundary. On closed manifold the boundary $\partial\Omega$ vanishes, so the integral is zero.