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


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*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.

*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.

*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)
