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's consider the space of long knots in $\mathbb R^n, n>3$. I know that there are many results (Vassiliev, Turchin, Sinha, Kontsevich) about different expressions of cohomology of this space. I think the last result is about convergency in $E^1$ term (

But my question is about what these cohomologies are precise. So, is it true that $H^1(long\ knots\ in\ \mathbb R^4)=0$ ?

Does there exist some table with $H^i(space\ of\ long\ knots\ in\ \mathbb R^j)$ at least for small $i,j$ ?

share|cite|improve this question
Is $H^∗$ the usual cohomology? Can't you compute everything applying Alexander duality (or something similar) to the pair $(S^j,S^1)$? Or maybe I'm just misinterpreting everything... – Marco Golla Nov 14 '11 at 17:47
@Marco: he's talking about the space of all embeddings, as a function space. I think you might be interpreting the space as the knot complement. – Ryan Budney Nov 14 '11 at 17:51
@Marco: I've modified question for clarity (added word "space"). – Nikita Kalinin Nov 14 '11 at 18:24
up vote 9 down vote accepted

Long knots in $\mathbb R^4$ form a simply-connected space. I pointed this out in my survey paper A Family of Embedding Spaces. The primary tool used to prove it is what's called the embedding calculus due to Goodwillie, Klein and Weiss.

Let $\mathcal K_{n,j}$ denote the space of long embeddings of $\mathbb R^j$ into $\mathbb R^n$. The same survey paper above shows that $\mathcal K_{n,j}$ is $(2n-3j-4)$-connected, and the $(2n-3j-3)$-rd homotopy group is computed. It turns out to be either $\mathbb Z$ or $\mathbb Z_2$ depending on a parity issue, assuming $2n-3j-3 \geq 0$.

When $2n-3j-3<0$ there are a few cases where some of these homotopy groups are computed, due to Haefliger and Kervaire. When $2n-3j-3>0$ many of the rational groups have been computed by Victor Turchin in this reference. But very few integral homotopy or homology groups have been computed as of yet.

share|cite|improve this answer
It seems that it is possible to prove simply-connectedness for knots in simply-connected four-dimensional manifold just by geometric observations, so I've asked for to be convinced that it is true. Thank you! – Nikita Kalinin Nov 14 '11 at 18:22
I wonder, perhaps $Emb(S^1,\mathbb CP^2)$ is not simply connected? A loop in this embedding space could sweep-out the generator of $H_2 \mathbb CP^2 \simeq \mathbb Z$. Perhaps that's non-trivial? – Ryan Budney May 19 '14 at 21:03

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.