MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

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

Let $X$ be a topological space and $E_n(X)$ the space of finite sets of cardinality $\leq n$. It is a theorem of Bott that $E_3(S^1)=S^3$. What is the idea to show that the embedding $S^1\hookrightarrow E_3(S^1)$ sending $x$ to the set $\{ x\}$ is the trefoil knot?

share|cite|improve this question

closed as off-topic by Ryan Budney, Misha, Steven Sam, S. Carnahan May 20 '14 at 4:51

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – S. Carnahan
If this question can be reworded to fit the rules in the help center, please edit the question.

Is this a homework problem? Depending on what you know there's various efficient proofs. One way is to notice $E_3(S^1)$ is 3-dimensional, and it has an action of $SO_2$ since $S^1$ does. This makes it a Seifert Fibred Space, and these were classified by Seifert. So you need only identify the quotient space and classify the singular fibres, in this case there's one with isotropy of order 2, and one with isotropy of order 3. A trefoil is a (2,3)-torus knot. – Ryan Budney Jul 16 '11 at 15:56
student -- I've put back ticks around the formula with brackets. – algori Jul 16 '11 at 15:58
I think it would be best to post your question on Also, it might be a good idea to give people more background on your motivation for this question. – Ryan Budney Jul 16 '11 at 15:59
Ryan -- innocent until proven guilty, no? And I think using the classification of Seifert fibered spaces to solve this is a bit of an overkill. – algori Jul 16 '11 at 16:49
Ryan -- what I meant was: one can not be sure that this is homework until one has found a homework style solution. The solution that uses the classification of Seifert fibered spaces may or may not be such a solution, depending on what was covered in the course. So until further evidence is revealed I'm prepared to give the author the benefit of doubt. Besides, this time of year is not when people usually ask for help with homework solutions. – algori Jul 16 '11 at 20:03