4
$\begingroup$

Given a finite dimensional finite $CW$ complex $X$ of dimension $d$, I want to build a compact manifold $M$ (with least dimension possible) with boundary with the property that,

  1. $M$ has the same homotopy type as $X$.
  2. $M$ is inductively built and each stage is a compact manifold with boundary.
  3. When we go to the next stage from the previous one, there is exactly one critical point of index $\leq d$.

Point (2) can be done using a theorem related to elementary cobordism which I found in "Lectures on the h-Cobordism theorem" (theorem 3.12) by Milnor. But (1), I don't know how to start the induction process and wish to understand the mechanism.

$\endgroup$
7
  • 1
    $\begingroup$ What does $M$ have to do with the CW complex? $\endgroup$
    – Bma
    Dec 13, 2021 at 3:44
  • 1
    $\begingroup$ Assuming you want the manifold to have the homotopy-type of the CW-complex, you would do it by performing handle attachments corresponding to your cellular attachments, and building the corresponding Morse function that has that handle attachment. $\endgroup$ Dec 13, 2021 at 5:12
  • $\begingroup$ @RyanBudney while constructing the handle decomposition, corresponding to each skeleton $X_0, X_1, X_2$ (suppose the CW has only three skeleton) how to start the induction? (For an example take $D^2$ attached to $S^1$ via degree 2 map) If I call the corresponding manifolds $M_0, M_1, M_2$ so that $M=\cup_{i=0}^2 M_i$. can you elaborate for this example? $\endgroup$
    – piper1967
    Dec 13, 2021 at 5:58
  • 1
    $\begingroup$ I'm a little confused by what you mean by "start the induction". Isn't every handle attachment an inductive step? What do you want to induct on? $\endgroup$ Dec 13, 2021 at 6:06
  • 1
    $\begingroup$ Not sure but for that example, if you embed $D^2$ into $\mathbb R^4$ in such a way that $S^1$ is its intersection with a hyperplane $\mathbb R^3\subset\mathbb R^4$, then you can take $M_1$ a tubular neighborhood of $S^1$ in $R^3$ which is a solid torus $T$, and $M^2$ a tubular neighborhood of $D^2$ in $\mathbb R^4$ whose intersection with $\mathbb R^3$ is $T$. $\endgroup$ Dec 13, 2021 at 6:10

1 Answer 1

4
$\begingroup$

There is a more 'geometric' construction that does not use induction and gives you the correct dimension bound. A rough sketch of how this works follows.

If $\text{dim}(X) = d$, then there exists a local embedding $X \rightarrow \mathbb R^{2d}$ with only finitely many self-intersection points all of which lie in top cells of $X$. Taking a 'tubular neighbourhood' (one has to make sense of this appropriately) of this embedding gives you a $2d$-manifold with boundary $M$ that is homotopy equivalent to $X$. Now pick any self-indexing Morse $\phi$ function on $M$ and consider the stages $\varphi^{-1}(i)$, $0 \leq i \leq 2d-1$ of building $M$ using $\varphi$.

$\endgroup$
1
  • 1
    $\begingroup$ @JensReinhold by tubular neighbourhood in the above answer you meant the closed tubular neighbourhood right? $\endgroup$
    – piper1967
    Dec 14, 2021 at 2:33

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.