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I am a PhD student in Physics. This problem is motivated by representing a spin-$j$ state by Majorana's stellar representation. There are $2j$ points (Majorana stars) on the two-dimensional sphere (Bloch sphere) where the wave function vanishes. Hence, a spin state corresponds to a configuration of points on the sphere. If the spin is placed in an external magnetic field, the configuration of points undergoes rigid rotation.

For a general cyclic evolution, the stars will move around the sphere (like wrapping a plastic bag around a ball), and finally come back to their original position. Therefore, the mathematical question is: how to classify continuous maps from $S^2$ to $S^2$ with $n$ fixed points? Since the homotopy group $π_2(S^2)$ is $Z$, I guess the answer will be a subgroup of $Z$.

For a general reference, see e.g. https://physics.aps.org/articles/v5/65

Any explanation or reference will be appreciated.

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Let $x_1, \ldots, x_n$ be your points on the sphere. Choose an $n+1$-th point $z$ on the target sphere, and paths from $z$ to each one of the $x_i$s. By moving along these paths you can construct a homotopy from the space of maps that fix the $x_i$s to the space of maps that send all the points $x_i$ to $z$.

This latter space is the same as the space of pointed maps from the quotient space of $S^2$ by $n$ points to $S^2$. Easy exercise: this quotient space is homotopy equivalent to the wedge sum $$S^2\vee \bigvee_{n-1}S^1$$ It follows that the space of maps that you are asking about is homotopy equivalent to $$\Omega^2S^2\times (\Omega S^2)^{n-1}$$ Here $\Omega^k X$ denotes the space of pointed maps from $S^k$ to $X$.

The set of path components of this space is the same as $\pi_2(S^2)$, which is $\mathbb Z$. Its higher homotopy groups can be written in terms of those of $S^2$.

Perhaps the moral is this: the homotopy class of a map that fixes $n$ points is determined by the degree, as usual. But you have more choices in constructing a homotopy between two maps of the same degree.

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  • $\begingroup$ Thanks for your comment! Can you explain a little bit more why the homotopy groups of the union of spheres is $\mathbb{Z}$? $\endgroup$ – ChonFai Kam Sep 21 '17 at 20:08

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