# Permutation symmetry - a continuous symmetry?

From quantum mechanics it is known that permutation between identical particles does not change the Hamiltonian. Assuming that the quantum system consists of a very high number of particles such that the action of the permutation group can be regarded as continuous (similar to other well-known symmetry groups).

Is it possible to define a symmetry group operator $P (x)$ (this is an operator in the infinite set symmetric group $S_\infty$) acting on a state on the spacetime point $x$ in the way that a gauge connection like the photon field can be defined?

Moreover Cayley's theorem states that every general group is isomorphic to a set of permutations.

Can such a concept be defined under above conditions?

## 1 Answer

[I somewhat haphazardly pieced this answer together, so I'm not absolutely certain the conclusion is correct.]

1. Cayley's theorem is useless here, because the group isomorphism it produces is not required to preserve any kind of topology on the groups, in particular not notions of continuity or differentiability.

2. On the infinite symmetric group $S_\infty$ on a countable set, according to this MO answer, there is the unique non-discrete topology of pointwise convergence, and Polish groups are homeomorphic to closed subgroups of it if and only if they admit a compatible left-invariant ultrametric.

3. Lie groups, which are the kinds of smooth symmetries one requires for the notion of a gauge field, are Polish groups, but they are manifolds over $\mathbb{R}$, which is not an ultrametric field. In particular, they are locally diffeomorphic to $\mathbb{R}^n$, which is not ultrametric in its standard topology.

Therefore, the infinite symmetric group, which is a closed subgroup of itself, is ultrametric, and hence not a Lie group. It carries a notion of continuity, but it is not a Lie group, so it lacks the differentiable structure that is necessary for us to define a gauge theory with it.

Another way to see that $S_\infty$ is not a Lie group is that it is totally disconnected, i.e. it has no non-trivial connected subsets, so its Lie algebra, which exponentiates to the connected component of the identity, would be zero-dimenisonal.

Thus, we are forced to conclude there is no gauge theory in the usual sense for the infinite symmetric group, because it does not seem to be a Lie group, and hence carries no differentiable structure.

(Weak supporting evidence is that I found various papers on continuous functions on $S_\infty$, but none about differentiable structures)

• Thanks for the answer! Even if $S_\infty$ is not a Lie group can there be assumed that there exists a covariant derivative $\partial_\mu+\Sigma_\mu$ where the operator $\Sigma_\mu$ ensures that the action is invariant under the action of $P (x)$ on the fermion or boson fields? Is it possible also for general continuous symmetry groups? – kryomaxim Aug 15 '15 at 8:41
• @kryomaxim: The gauge field that is added to the covariant derivative is Lie algebra-valued, so you can't added such a $\Sigma$. More worryingly, you can't even define what $\partial_\mu(P(x)\phi(x))$ is because the group carries no differentiable structure, so $(\partial_\mu P(x))\phi(x)$, one of the two terms from the product rule, is not defined. We really need a Lie group to do the constructions of gauge theory in the sense of the gauge covariant derivative. – ACuriousMind Aug 15 '15 at 13:07
• I understand that non-differentiable groups have no gauge connection. But is there an alternative concept for introducing functions for preserving local symmetry? Maybe some tools of nonstandard analysis? Nonstandard analysis deals with non-finite numbers (also called hypercomplex numbers). – kryomaxim Aug 15 '15 at 14:08
• @kryomaxim: I've never seen something like that (which doesn't mean it doesn't exist, but it's at least not standard lore). – ACuriousMind Aug 15 '15 at 14:35