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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 (http://palmer.wellesley.edu/~ivolic/pdf/Papers/VassilievCollapseFinal-G%26T.pdf).

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 $i,j$ ?

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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
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@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

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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.

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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 at 21:03

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