# Building a manifold from a CW complex inductively

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.

• What does $M$ have to do with the CW complex?
– Bma
Dec 13, 2021 at 3:44
• 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. Dec 13, 2021 at 5:12
• @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? Dec 13, 2021 at 5:58
• 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? Dec 13, 2021 at 6:06
• 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 $T$. Dec 13, 2021 at 6:10

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 $$\varphi$$.

• @JensReinhold by tubular neighbourhood in the above answer you meant the closed tubular neighbourhood right? Dec 14, 2021 at 2:33