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?


1 Answer 1


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.