# Quotient of Three Dimensional Torus by Permutation on Coordinates

The Mobius Strip can be realized as a quotient of $$T = (S^1)^2$$ via the identifications $$(x,y) \sim (y,x)$$.

I tried to generalized this concept to a higher dimension, and consider the quotient of $$(S^1)^3$$ by the action of the symmetric group $$S_3$$ on the coordinates.

I was able to compute the homology of this space: $$H_n = \mathbb{Z}$$ for $$n = 0,1$$, and 0 otherwise (with reduced homology being 0 at $$n=0$$ as well).

Even with this information I wasn't able to identify said space in any other way. Is it well known, or, can it be described in any other fashion? What can be said about higher dimensions?

• You may be interested to know that, unlike the case of $S^1$, the $n^{\text{th}}$ symmetric product of $S^2$ is actually a manifold, namely $\mathbb{CP}^n$. May 17, 2019 at 1:50

What you are describing are the symmetric powers of $$S^1$$. The symmetric power $$SP^n(S^1)$$ is a fibre bundle over $$S^1$$ with fibre an $$(n-1)$$-simplex. (In particular your calculation of the homology for $$n=3$$ checks out.) This bundle is orientable if $$n$$ is odd, and non-orientable if $$n$$ is even.
• I also conjecture that the homology is the same for any power $SP^k(S^1)$, is that also true? May 16, 2019 at 16:31
• Hi Adi, yes, that follows from the description as a fibre bundle over $S^1$ with contractible fibre. (The long exact sequence in homotopy of the bundle gives that $SP^n(S^1)\to S^1$ is a weak homotopy equivalence, hence induces an isomorphism on homology. With a little more work you can show it's a homotopy equivalence.) May 16, 2019 at 16:54