# Spin-statistics theorem on spaces with non-integer dimensions

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What would be the spin-statistics relation for particles (including quasiparticles cf. anyons in 2+1 dimensions) in a space with non-integer dimension, $2 \lt D \lt 3$? In other words (cf. stackexchange questions here and here), what is the first homotopy group for a space with non-integer dimension, $2 \lt D \lt 3$?

Although it is not useful for my immediate work, I would also like to know for fractional dimensions $0 \lt D \lt 2$ and $3 \lt D \lt \infty$

References/reviews, written for physicists, will also be greatly appreciated!

This post imported from StackExchange Physics at 2014-08-11 14:13 (UCT), posted by SE-user crackjack

edited May 4, 2015
Here is something on the first homotopy group (fundamental group) of graphs: homepage.math.uiowa.edu/~jsimon/COURSES/M201Fall08/… Since fractional spaces may be represented through fractal network structures, may be the above link has something to do with spin of particles on fractional space.

This post imported from StackExchange Physics at 2014-08-11 14:13 (UCT), posted by SE-user crackjack

Update following the answer and discussion by @RonMaimon :

Fractal spaces are generally non-homogeneous. So, any symmetry group of such a fractal space will not be continuous except, as @RonMaimon pointed out, for trivial ones like SO(2) symmetry of $R^2 \times \mathcal{C}$ where $\mathcal{C}$ is a fractal set). So for spins, the first Homotopy group of such symmetry groups with finite- (and possibly even discrete-) topology, has to be defined accordingly. Look here for more on homotopy theory of such finite/discrete topologies. Another way, as suggested by RonMaimon is to find a clever way to homogenize a fractal space.

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There is no such thing as a homogenous space of noninteger dimension, at least nobody has defined such a thing so far, and any construction would not be a point set. The notion of a fractal space requires that space is non-homogenous, at least in standard constructions. For example, to construct a space of dimension between 0 and 1, you can make a Cantor set. The points of a cantor set are not homogenous, because the gaps are of different sizes, and the size of the nearest gap tells you where you are. There is no translation symmetry on the Cantor set considered as a metric space.

This makes it impossible to speak meaningfully of a unique space of dimension between 2 and 3. You can take the product of a bunch of copies of a Cantor set and get a space with no translation or rotational symmetry, and no notion of spin-statistics.

answered Apr 30, 2015 by (7,720 points)

Thanks for the comment. I dont quite understand the connection though. Are you implying that there is no notion of a homotopy group (since, first homotopy group ~ spins) for non-homogeneous spaces?

@crackjack; If by "first homotopy group", you mean the fundamental group of curves related by homotopy, this has nothing to do with spin. Spin is about symmetry under rotation, and these fractal spaces have no rotations, because they have no rotation or translation symmetry, aside from possibly some discrete groups. This means you can't formulate the question. The fundamental group/first-homotopy group doesn't make sense on the Cantor set, it is disconnected, it's just the trivial group "1". There is no known notion of "a homogenous fractal space" or a distinguished space of non-integer dimension, and any such space would be very tricky to construct, and I think it is impossible. But go ahead and try, there might be some clever way of defining such a thing, but it would require a clever idea.

@Ron Maimon: I am sorry. I should have clarified (which I now have) that, I was referring to generalized spin-statistics theorem, that also includes quasiparticles of non-local fields. So, in 2+1 dimensions, I am also referring to the spin-statistics of anyons, which is explained by the fundamental group of SO(2,1) being $\mathbb{Z}$, in addition to localized fields of fermions and bosons.

@crackerjack: This is the fundamental group of the symmetry group, not the fundamental group of the space. The fractal spaces can have all sorts of symmetry groups, for example, R^2 x C where C is a Cantor set has fractal dimension between 2 and 3 and has symmetry group SO(2). The question is fundamentally ill posed for the reason I gave--- there is no general notion of the symmetry group of a fractal space, they are generally asymmetric. If you find a highly symmetric fractal space, it would be interesting, but I can't imagine how to construct such a thing. It would be some weirdness of non point geometry at best. A better question would be "Is there any mathematical construction of a homogenous space of noninteger fractal dimension?". This would be a reasonable question.

@Ron Maimon: True. Or, one can also ask "Is there an analogous homotopy theory for non-homogenous space?" , and hence my earlier comment following my question. And sorry - my usage of first homotopy group of 'space' (instead of 'symmetry group of space') is really sloppy.

The homotopy is not for the space, it is for the symmetry group. There is no symmetry group on these non-homogenous space. The proper question is whether it is possible to define a homogenous space interpolating between R^n of different dimensions, you didn't ask this, but you should have. Your question has no answer, because it doesn't make sense.

@RonMaimon: That is one way (homogenizing fractal dimensions). Like I said, the other way is to extend homotopy to (symmetry groups of) fractal spaces. In the case of non-homogenous fractal spacetime, my understanding (which may be wrong) is that the SO(1,n) symmetry group of a homogenous spacetime may break down to a discrete version. Which is why I linked that MO Q&A, in my earlier comment, which talks of homotopy of discrete topologies (and another comment referring to homotopy of graphs). Also, I am sure that was a typo in your comment, but just in case: One can indeed talk of homotopy for a space as well as the symmetry group of that space. (Sorry for my late responses. I think my notification preferences are set to 'no email').

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