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?

  • 2
    $\begingroup$ 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 $\mathbb{CP}^n$. $\endgroup$ May 17, 2019 at 1:50

1 Answer 1


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.

This result is due to H. R. Morton, and is nicely written up at the nLab page https://ncatlab.org/nlab/show/symmetric+product+of+circles

  • $\begingroup$ I also conjecture that the homology is the same for any power $SP^k(S^1)$, is that also true? $\endgroup$
    – Adi Ostrov
    May 16, 2019 at 16:31
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    $\begingroup$ 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.) $\endgroup$
    – Mark Grant
    May 16, 2019 at 16:54

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