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I had asked a similar question about a calculation involving the winding number here. But i haven't got a satisfactory response. So, I am rephrasing this question in a slightly different manner. What is the winding number of a magnetic monopole solution? Why is it a topological invariant? How is it connected to the degree of a map and the vector potential? While answering please could you bear in mind the fact that I have some very little knowledge of point-set topology, and no knowledge at all of algebraic topology.

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I was intending to update my answer to your original question to give you the required additional information. It is just a little technical and takes some time to write down all the technicalities.

Actually, one just has to write down the the functional:

$\int_{S_2} \mathrm{tr}(\Phi d\Phi \wedge d\Phi)$,

where,

$\Phi = \Phi_x \sigma_x + \Phi_y \sigma_y + \Phi_z \sigma_z$, ($\sigma_x , \sigma_y, \sigma_z$ are the Pauli matrices)

explicitely in his favorite coordinate system to understand its meaning:

First observation:

Taking ito account the constraint defining the two sphere:$\Phi_x ^2+\Phi_y ^2+\Phi_z ^2=1$, (In matrix notation, this condition is equivalent to $\Phi^2 = I$)

then the integrand is just the area element of the (Higgs vacuum) two-sphere.

Second observation:

Consider the two form:

$\omega = \mathrm{tr}(\Phi d\Phi \wedge d\Phi)$.

It is easy to verify that it is closed:

$d\omega =\mathrm{tr}(d\Phi \wedge d\Phi \wedge d\Phi)= 0$ (by the antisymmetry of the wedge product)

But it cannot be exact, otherwise the area of the two sphere would have been zero by the Stokes theorem.

Third observation:

The variation of this form (with respect to any perturbation) is exact:

$\delta \omega = d \mathrm{tr}(\delta \Phi \Phi d\Phi)$

(Please notice that one needs to use the condition: $\Phi^2 = I \rightarrow \Phi \delta \Phi +\delta \Phi \Phi = 0 $)

First conclusion: This form does not change under an infinitesimal variation of the fields. Since any continuous map can be built of a series of infinitesimal maps, this form does not change for a continuous deformations of the map $\Phi$. Thus it is a topological invariant.

Second conclusion: Please look at the surface area element in spherical coordinates: $\omega = sin\theta d\theta d\phi$ and consider the map $\theta \rightarrow \theta $ , $\phi \rightarrow N \phi $ , Clearly this map winds the sphere N times and it is not difficult to verify that the integral is equal to N. This is the reason for the name winding number. In my answer to your first question another family of maps with arbitrary winding numbrers were given.

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hey..thanks a lot for the answer. But please could you tell me, why $Tr(\Phi d\Phi \wedge d\Phi)$ equals the term in my question $Tr([d\Phi,d\Phi],\Phi)$. The only thing I know about wedge products is their asymmetric nature, the fact that it is roughly the generalized the cross product, it is a 2-form, which is obtained by acting $d$ on a 1-form, and this gives the curl. How is it related to the commutator of the 1-forms? –  ramanujan_dirac Jul 4 '12 at 13:12
1  
This is because the commutator of odd valued differential forms comes with a plus sign, and I wasn't strict about normalizations . Please see Yang Zhang monograph for a clear introduction: lepp.cornell.edu/~yz98/notes/…. Please, see also the review article of Eguchi Kilkey Hansen empg.maths.ed.ac.uk/Activities/GT/EGH.pdf, where a good introduction for Lie algebra valued differential forms. –  David Bar Moshe Jul 4 '12 at 14:07
    
OK. I will have a look. +1 and answer. –  ramanujan_dirac Jul 4 '12 at 16:11
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