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A spin structure on a Riemannian bundle of rank >2 is the lift of the structure group from $\text{SO}(n)$ to its universal cover $\text{Spin}(n).$ It may also be defined in the case $n=2$ as the lift of the structure group to a double cover of $\text{SO}(2)$, which is of course not the universal cover.

So what about lifts to other covers of $\text{SO}(2)$. Does the lift to a three-to-one cover of $\text{SO}(2)$ have a topological obstruction living in $H^2(B,\mathbb{Z}/3)$? Is the concept of a lift to the universal cover of $\text{SO}(2)$ affected by the unusual statistics of anyons?

Searching the literature for spin structures, I found only the double cover case. And searching the archive for anyons yields nothing but solid state physics articles. So I would be happy just to have a reference to any work discussing this case.

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I know nothing about the physics you have in mind, but I can tell you about the topology. The classifying space $BSO(2)$ is a $K(\mathbb{Z},2)$, so oriented 2-plane bundles are in bijection with classes in $H^2(B,\mathbb{Z})$. For any $n$, the $n$-sheeted cover $SO(2)\to SO(2)$ corresponds to multiplication by $n$ on $H^2(B,\mathbb{Z})$, so a bundle lifts to that cover iff the corresponding cohomology class divisible by $n$, which is equivalent to the vanishing of the mod $n$ reduction of the class in $H^2(B,\mathbb{Z}/n)$ (generalizing the second Stiefel-Whitney class for $n=2$). When a lift exists, the set of lifts is a torsor over $H^1(B,\mathbb{Z}/n)$, since the homotopy fiber of the map $n:BSO(2)\to BSO(2)$ is a $K(\mathbb{Z}/n,1)$.

The universal cover is contractible, so a bundle lifts to the universal cover iff the bundle is trivial. The set of such lifts is a torsor over $H^1(B,\mathbb{Z})$.

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Thank you. The input from physics is that anyons satisfy complicated statistics, due to living in a representation of the braid group rather than the symmetric group. I thought the complicated topological issues with the representation theory of anyons would necessarily imply a more complicated theory of anyon structure (analogue of spin structure). I guess I was wrong! –  Joe Hannon Apr 23 '13 at 22:42

Here is a different explanation, in terms of exact sequences. There is an exact sequences:

$0 \to \mathbb Z/n \to SO(2) \to SO(2) \to 0 $

since $SO(2)$ is the $n$-fold cover of $SO(2)$. Taking cohomology, we get:

$H^1(B, SO(2) ) \to H^1(B, SO(2) ) \to H^2(B, \mathbb Z/n)$

Since $H^1(B,G)$ is the group of principal $G$-bundles for $G$ an abelian group, the exact sequence property means that the bundle lifts if and only if the class in $H^2(B,\mathbb Z/n)$ is zero.

It is a bit more difficult to see why this method why the cohomology class is just the first Chern class of the associated complex line bundle modulo $n$.

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