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If $F$ is a suitably nice functor from manifolds to spaces, it has a degree $k$ "polynomial" approximation $T_k F$ in the sense of embedding calculus. We set $T_\infty F := \mathrm{holim} T_k F$.

The functor of most interest is the functor of embeddings $\mathrm{Emb}(-,N)$ into a fixed target $N$. By work of Goodwillie, Klein, and Weiss, $T_\infty\mathrm{Emb}(M,N)\simeq\mathrm{Emb}(M,N)$ as long as $\dim N-\dim M\geq3$. One says that "the Taylor tower converges."

I am interested in examples below codimension 3. One such example is knot theory, where it is known that $\mathrm{Emb}(S^1,\mathbb{R}^3)\not\simeq T_\infty\mathrm{Emb}(S^1,\mathbb{R}^3).$ This is the only example I know, and I would like to change this. Specifically, I am looking for examples of manifolds $M$ and $N$ such that either

1) $\dim N-\dim M<3$ and the Taylor tower is known not to converge, or

2) $\dim N-\dim M<3$ and the Taylor tower converges anyway.

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Are you asking for examples of specific manifolds $M$ and $N$? –  Fernando Muro Jun 8 '13 at 22:26
    
Yes. I've edited the question in order, I hope, to make it clearer. –  Ben Knudsen Jun 9 '13 at 1:12
    
@Ben: May I ask you how you come to the conclusion that the Taylor tower for knots does not converge? The only proof I know is fairly indirect. (1) The limit of the embedding tower for long knots is a (2-fold) loop space, e.g. by the work of Sinha. (2) However, long knots do not admit inverses (since usual knots do not admit inverses). (3) Thus, the embedding tower for long knots does not converge. (4) It follows that the embedding tower for knots does not converge. Do you happen to know of a more direct or simpler proof? –  Ricardo Andrade Jun 9 '13 at 1:16
    
@Ricardo: That was the argument I had in mind, as a matter of fact. It's the only one I've heard, and I don't find it very satisfactory. –  Ben Knudsen Jun 9 '13 at 1:27
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I think you can argue that the limit of the tower for knots has uncountably many components--the map from $T_k$ to $T_{k-1}$ is surjective but not injective on $\pi_0$. –  Tom Goodwillie Jun 9 '13 at 2:42
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