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I need to show some properties of the topological expression involving a map $\vec{n}(x): S^2 \rightarrow S^2$

$$W=\frac{1}{4\pi}\int \vec{n} \cdot (d\vec{n} \wedge d\vec{n}),$$

but I am not very familiar with differential forms. Namely, show that $$\partial_1 \vec{n} \times \partial_2 \vec{n} \ \parallel \vec{n},$$ that $W$ is an integer, and that it is invariant under the $O(3)$ transformations of $\vec{n}$.

I know the integrand is somehow the area on the $S^2$ sphere (spanned by $x$) and intuitively I can see how this should be integer times $4\pi$, and also that the wedge product is an oriented surface area on the unit sphere, thus must be parallel to $\vec{n}$ but I'm having trouble showing these formally. Can someone give me a clue how to go about it?

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The vector $\vec{n} = (n^1, n^2, n^3)$ is the radius vector of the sphere. One can either parametrize it by spherical coordinates: $$ n^1 = \cos\phi \sin\theta $$ $$ n^2 = \sin\phi \sin\theta $$ $$ n^3 = \cos\theta $$ In order to evaluate the wedge in the formula you need just to perform the cross product with care and put $d\phi \wedge d\theta = - d \theta \wedge d\phi$ . By an evaluation of the cross and dot products you should obtain the sphere surface element in the integrand. $$\vec{n} \cdot (d\vec{n} \wedge d\vec{n}) = 2 \sin\theta d\theta \wedge d\phi$$ Another possibility is to parameterize the sphere by Cartesian coordinates: $${B^1}^2+{B^2}^2+{B^3}^2 = |\mathbf{B}|^2$$ And take $ n^i = \frac{ B^i}{|\mathbf{B}|}$ And perform the integral in Cartesian coordinates. I have performed a detailed evaluation of the integral in Cartesian coordinates in the following Questions about Berry Phase answer.

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  • $\begingroup$ Thank you, the Cartesian parametrization was easy to follow. However I'm still not getting what $\vec{n} \wedge \vec{n}$ means. Wouldn't that be a matrix, since I have a 1-form ($\vec{n}$) and I take the exterior derivative, arriving to a 2-form, then somehow taking an antisymmetric product by wedging them? $\endgroup$ Commented Sep 8, 2018 at 16:00
  • $\begingroup$ $dn \wedge dn$ means the following: $(dn \wedge dn)^i = \epsilon^{ijk} dn^i dn^j$ (with summation convention). In order to compute each element $i = 1,2,3$ you need to sum up the six terms for them the completely antisymmetric tensor $\epsilon^{ijk}$ is non-vanishing. Please observe that this wedge product is non-vanishing only because the components are forms, thus anti-symmetric upon order change. A wedge product of a function valued vector with itself is vanishing: $\epsilon^{ijk} n^i n^j = 0$ due to symmetry. $\endgroup$ Commented Sep 9, 2018 at 6:37

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