MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$\exp_nX$ 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 $X$?

share|cite|improve this question
If you consider the subsets of each finite cardinality together, an important result is the Dold-Thom theorem: – Qiaochu Yuan Dec 24 '11 at 22:51
Though if you don't use basepoints (as in Dold-Thom), the limiting result is (weakly) contractible. – Moosbrugger Dec 25 '11 at 0:33
btw, the symmetric square was also studied by Marston Morse via Morse theory, of course, in connection with the extremal chord problem. – Pietro Majer Dec 25 '11 at 8:46
@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 $n\ge 3$. – Chris Gerig Dec 25 '11 at 16:52
up vote 15 down vote accepted

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,; 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,; 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 $$ \pi_1(\exp_{n+2}(S^1) \setminus \exp_{n}(S^1)) = \langle x, y \mid x^{n+2} = y^{n+1} \rangle. $$

share|cite|improve this answer
Perfect! I will take a glance at these. – Chris Gerig Dec 25 '11 at 16:52
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. – John Klein Dec 26 '11 at 12:19
See also: Handel, David: Some homotopy properties of spaces of finite subsets of topological spaces. Houston J. Math. 26 (2000), 747–764. – John Klein Dec 26 '11 at 12:23
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). – Alex Suciu Dec 27 '11 at 4:12

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.