14
$\begingroup$

Let $\mathrm{Emb}(M, N)$ be the space of smooth embeddings of a closed manifold $M$ into a manifold $N$ equipped with smooth compact-open topology.

Question 1. Are there conditions ensuring that the $k$th homotopy group of $\mathrm{Emb}(M, N)$ is infinite? Nontrivial?

I care about the situation when $k>0$ and $1<\dim(M)<\dim(N)$, and most importantly, the homotopy group is based at a homotopy equivalence $M\to N$; let's assume all this is true.

The only result I know are in the metastable range (due to Dax who built on Haefliger's work): If $k\le 2\dim(N)-3\dim(M)-3$ and $k\le\dim(N)-\dim(M)-2$, then the inclusion of $\mathrm{Emb}(M, N)$ into $\mathrm{Map}(M,N)$ is $k$-connected. Here $\mathrm{Map}(M,N)$ is space of continuous maps from $M$ to $N$, which is extensively studied in homotopy theory, especially rationally.

Question 2. Is anything else known about $\pi_k(\mathrm{Emb}(M, N))$ when $N$ is a the total space of a vector bundle over $M=S^n$, and the homotopy group is based at the zero section? What happens if the vector bundle is trivial, i.e. $N=S^n\times \mathbb R^l$?

I have spent some time reading works of Goodwillie, Klein, and Weiss, see here , who build a framework for analyzing $\mathrm{Emb}(M, N)$. Unfortunately, I was unable to extract any computations that would shed light onto the above questions. It seems the questions are open and hard, is this true? Any references, hints, or heuristics would be greatly appreciated.

$\endgroup$
  • $\begingroup$ I think the map $\operatorname{Emb}(M,N) \to \operatorname{Map}(M,N)$ is actually just $(\dim N - 2 \dim M - 1)$-connected, which should follow from some parametrized version of approximation of continuous maps by embeddings. This most likely predates Dax. By the way, I think the first bound you give is for something else (which does appear to be due to Dax) related to Haefliger's refined theory: see theorem 1.2.1 and the following remark in the survey you link to. $\endgroup$ – Ricardo Andrade May 27 '14 at 4:24
  • $\begingroup$ @RicardoAndrade: I added a link to Dax's paper. The result is mentioned on the second page of the introduction. There were earlier results, but Dax's is the best available, I think. $\endgroup$ – Igor Belegradek May 27 '14 at 11:43
  • 1
    $\begingroup$ There is some literature on the rational homotopy of spaces of long knots (embeddings $\mathbb{R}^m$ to $\mathbb{R}^n$ fixed outside a compact set $C\subset \mathbb{R}^m$). In that case $M$ is not closed, of course, but at least $M\to N$ is a homotopy equivalence. I gather people are trying to extend these results to $\operatorname{Emb}(M,N)$. You might glean some information on the current (4 years ago) state-of-the-art by looking at the list of open problems on Ismar Volic's web page, palmer.wellesley.edu/~ivolic/pdf/Papers/… $\endgroup$ – Mark Grant May 27 '14 at 13:26
  • $\begingroup$ Dear @Igor Belegradek: If you meant the result stated at the top of the page numbered 305 in Dax's article, then I think you have left out one necessary condition: the connectivity of the map $M\to N$ which you take as the basepoint must be at least $2\dim M-\dim N +k+1$. Since you are taking arbitrary maps, you must then impose the condition $2\dim M-\dim N +k+1\leq 0$, i.e. $k\leq\dim N-2\dim M -1$, which is exactly the estimate I wrote in my previous comment. By the way, you will find precisely this connectivity estimate on the first page of the introduction of Dax's article. $\endgroup$ – Ricardo Andrade May 27 '14 at 14:20
  • 1
    $\begingroup$ @RicardoAndrade: my base point is a homotopy equivalence (as I stressed above), so its connectivity is not an issue. $\endgroup$ – Igor Belegradek May 27 '14 at 14:27
8
$\begingroup$

Suppose that $M$ is compact, $N$ is simply connected and has finitely generated homology, and the codimension $n-m$ is at least $3$. Then the space $Emb(M,N)$ is such that

(1) for every basepoint $\pi_1$ is solvable and $\pi_k$ is finitely generated for all $k\ge 1$.

This can be proved using the Weiss tower and the "analyticity" or "multiple disjunction" result of John Klein and myself. Or it is possible to give an argument that does not use the tower. Either way, you repeatedly use the fact that when (1) holds for the base of a fibration and for every fiber then it also holds for the total space, and the fact that if (1) holds for the base and the total space of a fibration then it holds for every fiber.

$\endgroup$
  • $\begingroup$ You're welcome. By the way, I'm working hard on final (?) revisions of our paper. $\endgroup$ – Tom Goodwillie May 28 '14 at 11:34
  • $\begingroup$ Yes, I noticed the updates. $\endgroup$ – John Klein May 28 '14 at 12:06
  • $\begingroup$ Thank you! I wish this pretty statement would appear in the literature. I gather the base of the tower equal to the space of monomorphisms $TM\to TN$, which is a nilpotent space. It seems your are saying that the homotopy fiber of ${\mathcal T}_k\to {\mathcal T}_{k-1}$ is also nilpotent. Why? $\endgroup$ – Igor Belegradek May 28 '14 at 12:11
  • 1
    $\begingroup$ Each homotopy fiber of $T_k\to T_{k-1}$ has the homotopy type of the space of sections of a fibration over a finite complex with fixed behavior on a subcomplex. The fiber of the fibration will be simply connected under these hypotheses, and will have finitely generated homology groups if $N$ does. $\endgroup$ – Tom Goodwillie May 28 '14 at 12:46
  • 1
    $\begingroup$ The description of the fiber of the tower is in Weiss's paper "Embeddings from the point of view of immersion theory". The "finite complex" that is the base of the fibration is the space of unoriented configurations of $k$ points in $M$, and the "subcomplex" is "infinity". Each fiber of the fibration is the "total" or "iterated" homotopy fiber of a $k$-dimensional cubical diagram consisting of the spaces $Emb(S,N)$ for all subsets $S\subset\lbrace 1,\dots ,k\rbrace$. $\endgroup$ – Tom Goodwillie May 28 '14 at 12:50
7
$\begingroup$

Here are some comments:

1) Concerning finiteness results for spaces of embeddings, here is what I remember. The layers of the Goodwillie-Weiss tower when $M^m$ is closed and $N= \Bbb R^n$ have finitely generated homotopy groups when $2m+2 \le n$ (roughly the Whitney range). Furthermore, these fibers are simply connected. It follows that the embedding space $E(M,\Bbb R^n)$ is of finite type (meaning that it is weak equivalent to a CW complex with finitely many cells in each degree). This will imply finite generation of the homotopy groups.

The reason the layers have finitely generated homotopy groups is that they are section spaces over a finite complex where the fibers are built out of configuration spaces by a homotopy inverse limit procedure.

A more basic related result is this: Let $F(X,Y)$ be the function space of maps $X\to Y$, where $X$ is a finite complex and $Y$ is 1-connected with finitely generated homotopy groups. Then $F(X,Y)$ has finitely generated homotopy groups in each degree. One can see this by induction on the cells of $X$.

I think that the same result above holds when $N$ is compact, possibly with boundary and $1$-connected.

I conjectured that the each component of the embedding space should be of finite type even without the hypothesis $2m+2\le n$ (but with $m \le n-3$). However, I do not know how to prove this more general statement.

2) Concerning your first question: if $M \to N$ is a homotopy equivalence, it seems to me that the homotopy fiber $E^\text{pd}(M,N)\to F(M,N)$ taken at your given homotopy equivalence is contractible. Here, $E^\text{pd}(M,N)$ is the space of Poincare embeddings.

Next, one can analyse the difference between the Poincare embedding and the block embedding spaces as a space of lifts of the Spivak bundle. More precisely, the difference is given by factorizations of $$ M \to BG(n{-}m) \times_{BG} BO $$ through $BO(n{-}m)$. Here $BG$ classifies stable spherical fibrations, $BG(k)$ classifies $(k-1)$-spherical fibrations, and the displayed target is meant to be a homotopy pullback. Rationally, the space of such lifts can probably be computed in terms of the cohomology of $M$ (note: $BG$ is rationally trivial).

Finally, the difference between block embeddings and smooth embeddings is given by concordance embedding spaces. In the concordance stable range, one usually analyses these via relative algebraic $K$-theory (a la Waldhausen).

3) A very different approach to computations of $\pi_0(E(M,N))$ appears in my paper On embeddings in the sphere, Proc. Amer. Math. Soc. 133 (2005), 2783-2793.

$\endgroup$
  • $\begingroup$ Thank you! Doesn't contractibility of the homotopy fiber of $E^{\mathrm{pd}}(M,N)\to F(M,N)$ require codimension $\ge 3$? $\endgroup$ – Igor Belegradek May 28 '14 at 1:38
  • $\begingroup$ Yes, it does require codimension $\ge 3$. $\endgroup$ – John Klein May 28 '14 at 1:59
  • $\begingroup$ I really like the strategy in 2) even though it will take me some time to digest it and see if I can make it work. One concern that I have is about the concordance stable range: does it mean that $k$ must be small in computing $\pi_k(\mathrm{Emb}(M,N))$? If so, would not then the usual (Dax-Haefliger) metastable results kick in? What happens to the strategy in the metastable range? $\endgroup$ – Igor Belegradek May 28 '14 at 17:06
  • $\begingroup$ I searched for the strategy 2) in the literature and the only source I found was "Multiple disjunction for spaces of Poincaré embeddings", and it seems to only cover the first step; in fact, I hardly ever see any other use of "the space of Poincare embeddings". Is this where things stand, or am I not looking hard enough? $\endgroup$ – Igor Belegradek May 29 '14 at 0:04
  • $\begingroup$ Igor, the joint paper that Tom alludes to below is almost done ("Multiple disjunction for spaces of smooth embeddings"). This paper is the second and last step of the program. I can send you a preliminary version if you wish to see it. $\endgroup$ – John Klein May 29 '14 at 0:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.