6
$\begingroup$

Let $X$ and $X'$ be smooth closed manifolds. Take closed subpolyhedra $D\subset X$ and $D'\subset X'$ (with respect to some triangulations) and let $f:X\to X'$ be a homotopy equivalence such that $f(D)=D'$ and the restriction of $f$ to $D$ is also a homotopy equivalence. Is it possible for the complements $X-D$ and $X'-D'$ to have different rational homotopy types, assuming all spaces ($X,X',D,D'X-D,X'-D'$) simply-connected?

Here is some motivation behind the question: if we replace the rational homotopy type with the integral one and do not require the spaces to be simply connected, then the answer is yes, as shown in a paper by R. Longoni and P. Salvatore http://arxiv.org/abs/math/0401075; a much simpler example is in Ryan's comment below. On the other hand, additively the cohomology of $X-D$ and $X'-D'$ is obviously the same.

upd: in the first version of the question the simple connectedness condition was missing. Apologies for the mix-up.

Improve this question
$\endgroup$
10
  • 1
    $\begingroup$ There is a homotopy equivalence $(S^3,K_1) \to (S^3,K_2)$ with $K_1$ the unknot, and $K_2$ the trefoil. $\endgroup$
    – Ryan Budney
    Apr 16, 2010 at 18:20
  • $\begingroup$ Thanks, indeed! What if we assume everything simply connected? $\endgroup$
    – algori
    Apr 16, 2010 at 18:57
  • 3
    $\begingroup$ I think Salvatore ?might? have a result analogous to his paper with Longoni using simply-connected manifolds instead of lens spaces. The argument proceeds much the same -- the configuration spaces aren't homotopy-equivalent even though the underlying manifolds are. I'll ask him about it in person next week. I'll visit Longoni as well. I don't think Salvatore has written up that paper yet. Longoni is a banker in Milano. $\endgroup$
    – Ryan Budney
    Apr 16, 2010 at 19:47
  • 1
    $\begingroup$ @Ryan - +1 for the "Longoni is a banker in Milano" statement! $\endgroup$
    – Somnath Basu
    Apr 16, 2010 at 20:19
  • 1
    $\begingroup$ @Paul: The map f does not necessarily induce a map between the complements. $\endgroup$
    – Oscar Randal-Williams
    Apr 17, 2010 at 2:14

1 Answer 1

Reset to default
2
$\begingroup$

Some of the best results known along these lines are in the paper

which used to be called "Algebraic models of complements...". Briefly, it seems that even if the complement is simply connected, there are examples of what you seek when the codimension is too small, and when the codimension is large enough the rational homotopy type of a complement is determined.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thanks, Dev! I am aware of that paper, but I do not know whether in the examples given in section 9 there one can construct a homotopy equivalence of couples. $\endgroup$
    – algori
    Apr 17, 2010 at 19:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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