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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 (Lambrechts, Turchin, and Volić - The rational homology of spaces of long knots in codimension $> 2$).

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

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

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  • $\begingroup$ 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... $\endgroup$
    – Marco Golla
    Commented Nov 14, 2011 at 17:47
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    $\begingroup$ @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. $\endgroup$
    – Ryan Budney
    Commented Nov 14, 2011 at 17:51
  • $\begingroup$ @Marco: I've modified question for clarity (added word "space"). $\endgroup$
    – Nikita Kalinin
    Commented Nov 14, 2011 at 18:24

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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 On the other side of the bialgebra of chord diagrams. But very few integral homotopy or homology groups have been computed as of yet.

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  • $\begingroup$ 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! $\endgroup$
    – Nikita Kalinin
    Commented Nov 14, 2011 at 18:22
  • $\begingroup$ 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? $\endgroup$
    – Ryan Budney
    Commented May 19, 2014 at 21:03

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