11
$\begingroup$

$\exp_nX$ is the space whose underlying set is the set of nonempty subsets $S\subseteq X$ with $|S|\le n$. Its topology is the quotient one inherited from the map $X^{\oplus n}\rightarrow\exp_nX$ given by $(x_1,\ldots,x_n)\mapsto\lbrace x_1\rbrace\cup\cdots\lbrace x_n\rbrace$. And $\exp_{m\le n}X$ is canonically embedded in it.

Interestingly, for the case $X=S^1$, we have $\exp_2S^1\approx M\ddot{o}$ (homeomorphic mobius band). Etienne Ghys saw this by considering the mobius band as $\mathbb{R}P^2$ minus the open disk (with $S^1$ as the disk's boundary) and mapping $p\in M\ddot{o}$ to the set of tangency points of the lines tangent to $S^1$ and intersecting $p$. And from this we see that $\exp_1S^1$ is the boundary of the band and not the meridian circle. Now Raul Bott showed that $\exp_3S^1\approx S^3$ (On the Third Symmetric Potency of $S^1$), and someone else showed that $\exp_1S^1\subset S^3$ is the trefoil knot. Furthermore, $\exp_2S^1$ is a Seifert surface of $\exp_1S^1\subset S^3$.

My two curious questions: What happens for $n\ge 3$ and the corresponding embeddings? Are there interesting results for other $X$?

Improve this question
$\endgroup$
6
  • 2
    $\begingroup$ If you consider the subsets of each finite cardinality together, an important result is the Dold-Thom theorem: en.wikipedia.org/wiki/Dold%E2%80%93Thom_theorem $\endgroup$
    – Qiaochu Yuan
    Commented Dec 24, 2011 at 22:51
  • 1
    $\begingroup$ Though if you don't use basepoints (as in Dold-Thom), the limiting result is (weakly) contractible. $\endgroup$
    – Moosbrugger
    Commented Dec 25, 2011 at 0:33
  • $\begingroup$ btw, the symmetric square was also studied by Marston Morse via Morse theory, of course, in connection with the extremal chord problem. $\endgroup$
    – Pietro Majer
    Commented Dec 25, 2011 at 8:46
  • 3
    $\begingroup$ @Qiaochu: But isn't our scenario different? Although $\exp_2X$ is precisely the second symmetric product (so that Pietro's comment relates here), $\exp_nX$ is a proper quotient of $SP^n(X)$ for $n\ge 3$. $\endgroup$
    – Chris Gerig
    Commented Dec 25, 2011 at 16:52
  • $\begingroup$ The description by Ghys that you mention is quite nice! I was wondering if you have a reference for it? I thought perhaps "A Singular Mathematical Promenade," since there is a whole chapter on the Mobius band, but I've had no luck finding it there. $\endgroup$
    – Nikhil Sahoo
    Commented Feb 12, 2021 at 21:16

1 Answer 1

Reset to default
16
$\begingroup$

The spaces $\exp_n(S^1)$, as well as the embeddings $\exp_n(S^1) \subset \exp_{n+2}(S^1)$ were studied by Christopher Tuffley in Finite subset spaces of $S^1$, Algebr. Geom. Topol. 2 (2002), 1119–1145, http://dx.doi.org/10.2140/agt.2002.2.1119; MR1998017 (2004f:54008), and, more recently, by Sadok Kallel and Denis Sjerve in Remarks on finite subset spaces, Homology, Homotopy Appl. 11 (2009), no. 2, 229–-250, http://www.intlpress.com/hha/v11/n2/a12/; MR2591920 (2011a:55019).

In particular, based on an argument from Clifford H. Wagner's thesis (Symmetric, cyclic, and permutation products of manifolds, Dissertationes Math. (Rozprawy Mat.) 182 (1980); MR0605369 (82h:55021)), Kallel and Sjerve show that $\exp_n(S^1)$ is a closed manifold if and only if $n=1$ or $n=3$. Furthermore, Tuffley shows that $$ \pi_1(\exp_{n+2}(S^1) \setminus \exp_{n}(S^1)) = \langle x, y \mid x^{n+2} = y^{n+1} \rangle. $$

Improve this answer
$\endgroup$
3
  • 2
    $\begingroup$ If I am remembering correctly, the pattern is this: ${exp}_d(S^1)$ for $ d =1,2,\dots$ is $S^1,S^1,S^2,S^2,S^3,S^3,\dots$ up to homotopy. $\endgroup$
    – John Klein
    Commented Dec 26, 2011 at 12:19
  • 1
    $\begingroup$ See also: Handel, David: Some homotopy properties of spaces of finite subsets of topological spaces. Houston J. Math. 26 (2000), 747–764. $\endgroup$
    – John Klein
    Commented Dec 26, 2011 at 12:23
  • 2
    $\begingroup$ John: It's kind of like you say, up to some shift. The homotopy type of $\exp_n(S^1)$ for $n=1,2,\dots$ is $$ S^1, S^1, S^3, S^3, S^5, S^5, \dots $$ This is proved by Tuffley in his Ph.D. thesis (Berkeley, 2003). $\endgroup$
    – Alex Suciu
    Commented Dec 27, 2011 at 4:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .