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Aug 24, 2021 at 21:57 history edited Ryan Budney CC BY-SA 4.0
add differential-geometric sketch of proof of the non-triviality of the suggested family.
Aug 24, 2021 at 21:19 history edited Ryan Budney CC BY-SA 4.0
add a sketch of a simpler argument to answer the question.
Aug 24, 2021 at 20:35 comment added Ryan Budney For $Emb(S^1, \mathbb R^3)$, yes all components have non-trivial fundamental groups. The unknot component of $Emb(S^1, \mathbb R^3)$ has the homotopy-type of $SO_3$, this is a classical result of Hatcher's but also follows from the results above. In the above description you need to use the unknot component of $K_{3,1}$ which is contractible.
Aug 24, 2021 at 20:33 comment added YCor In your description, where does the precise component of the unknot appear? Does it follow from the description that every component has a trivial/nontrivial fundamental group?
Aug 24, 2021 at 20:24 history edited Ryan Budney CC BY-SA 4.0
added 32 characters in body
Aug 24, 2021 at 20:19 history edited Ryan Budney CC BY-SA 4.0
added 32 characters in body
Aug 24, 2021 at 20:13 history answered Ryan Budney CC BY-SA 4.0