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

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?

## 1 Answer

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.