How does the Mermin-Ho theorem handle singularities arising from the hairy ball theorem?

Question

I am reading about the derivation of the Mermin-Ho relation in chapter III. B of (Kamien, 2002) and trying to understand how it fits with the hairy ball theorem.

Assuming you have a unit vector $\hat{n}(\vec{r})$ which covers the entire unit sphere and can be presented in the form $\hat{n}(\vec{r}) = \hat{e}_1(\vec{r}) \times \hat{e}_2(\vec{r})$, this gives rise to a spin connection $\Omega(\vec{r}) = \hat{e}_1(\vec{r}) \cdot \nabla \hat{e}_2(\vec{r})$. The Mermin-Ho relation now establishes, that it's curl is independent of the choice of $\hat{e}_1(\vec{r})$, $\hat{e}_2(\vec{r})$ via $$ [\nabla \times \Omega(\vec{r})]_i = \frac{1}{2} \varepsilon_{\alpha\beta\gamma} n^\alpha(\vec{r}) \varepsilon_{ijk} \partial_j n^\beta(\vec{r}) \partial_k n^\gamma(\vec{r}) $$

However, considering $\hat{n}(\vec{r})$ as the position on a unit sphere, then $\hat{e}_1(\vec{r})$ (as well as $\hat{e}_2(\vec{r})$) forms a tangent vector field to the unit sphere and – following the hairy ball theorem/Poicaré-Hopf theorem – has to have a singularity of index 2. This leads to a singularity in $\Omega(\vec{r})$ and taking the curl should give an additional $\delta$-function of strength $4 \pi$.

In other words, why doesn't the Mermin-Ho relation read something like $$ [\nabla \times \Omega(\vec{r})]_i = \frac{1}{2} \varepsilon_{\alpha\beta\gamma} n^\alpha(\vec{r}) \varepsilon_{ijk} \partial_j n^\beta(\vec{r}) \partial_k n^\gamma(\vec{r}) + 4 \pi \delta(\vec{r}-\vec{r}^\prime) $$ where $\vec{r}^\prime$ captures where the index-2 singularity of $\hat{e}_1(\vec{r})$, and therefore of $\Omega(\vec{r})$, is placed?

This is also a conceptual question with respect to the physics:

As a $\Omega(\vec{r})$ can be interpreted as a vector potential, its curl constitutes a magnetic field $\vec{B} = \nabla \times \Omega$. If $\hat{n}(\vec{r})$ winds exactly once around the unit sphere, as it is the case for a magnetic skyrmion, the integrated magnetic flux $\Phi = \int d^2r B(\vec{r}) = -4 \pi$ (for $e = \hbar = 1$). However, this flux would be cancelled exactly by the $\delta$-function of strength $4 \pi$. So, is there a total magnetic flux or does it vanish?

Answer 1

The whole derivation of the Mermin-Ho relation required a choice of $\hat{e}_1$ and $\hat{e}_2$ so that $\hat{n} = \hat{e}_1 \times \hat{e}_2$ everywhere. The construction of $\Omega$ requires them to be differentiable.

However, from the hairy ball theorem we know that it is impossible for $\hat{e}_1$ and $\hat{e}_2$ to be smooth and differentiable. Thus, there are places, where $\Omega$ is singular – but right there, the Mermin-Ho relation simply does not apply because our initial assumption, of $\hat{e}_1$ and $\hat{e}_2$ being smooth, does not hold.

To define the Mermin-Ho relation, poke a hole everywhere on the unit sphere where singularities occur and define $\hat{e}_1$ and $\hat{e}_2$ everywhere else apart from these holes. This way, one has excluded the singularities.